All finite initial segments of natural numbers, FISONs F(n) = {1, 2,
3, ..., n} as well as their union are less than the set ℕ of natural numbers.
Proof: Assume UF(n) = ℕ. The small FISONs are not necessary. What is the first necessary FISON? There is none! All can be dropped. But according
to Cantor's Theorem B, every non-empty set of different numbers of the
first and the second number class has a smallest number, a minimum. This proves that the set of indices n of necessary F(n), by not having a
first element, is empty.
Regards, WM
On 1/21/25 6:45 AM, WM wrote:
All finite initial segments of natural numbers, FISONs F(n) = {1, 2,
3, ..., n} as well as their union are less than the set ℕ of natural
numbers.
Proof: Assume UF(n) = ℕ. The small FISONs are not necessary. What is
the first necessary FISON? There is none! All can be dropped. But
according to Cantor's Theorem B, every non-empty set of different
numbers of the first and the second number class has a smallest
number, a minimum. This proves that the set of indices n of necessary
F(n), by not having a first element, is empty.
Which is a proof of ANY, not ALL together,
All finite initial segments of natural numbers,
FISONs F(n) = {1, 2, 3, ..., n}
as well as their union
are less than the set ℕ of natural numbers.
Proof:
Assume UF(n) = ℕ.
The small FISONs are not necessary.
What is the first necessary FISON?
There is none!
All can be dropped.
But according to Cantor's Theorem B,
every non-empty set of different numbers of
the first and the second number class
has a smallest number, a minimum.
This proves that
the set of indices n of necessary F(n),
by not having a first element,
is empty.
On 21.01.2025 13:17, Richard Damon wrote:
On 1/21/25 6:45 AM, WM wrote:
All finite initial segments of natural numbers, FISONs F(n) = {1, 2,
3, ..., n} as well as their union are less than the set ℕ of natural
numbers.
Proof: Assume UF(n) = ℕ. The small FISONs are not necessary. What is
the first necessary FISON? There is none! All can be dropped. But
according to Cantor's Theorem B, every non-empty set of different
numbers of the first and the second number class has a smallest
number, a minimum. This proves that the set of indices n of necessary
F(n), by not having a first element, is empty.
Which is a proof of ANY, not ALL together,
It is a proof of not any. The proof that not all together are necessary
is this: U{F(1), F(2), F(3), ...} = U{F(2), F(3), F(4), ...}.
Regards, WM
On 1/21/2025 6:45 AM, WM wrote:
For each finite set, there is
a (finite) FISON larger than it,
and that FISON subsets ℕ
For each finite set, there is
a (finite) FISON larger than it.
All can be dropped.
Each is not all.
Each can be dropped.
All cannot be dropped.
But according to Cantor's Theorem B,
every non-empty set of different numbers of
the first and the second number class
has a smallest number, a minimum.
This proves that
the set of indices n of necessary F(n),
by not having a first element,
is empty.
ℕ = ⋃{FISON}
On 1/21/25 7:44 AM, WM wrote:
On 21.01.2025 13:17, Richard Damon wrote:
On 1/21/25 6:45 AM, WM wrote:
All finite initial segments of natural numbers, FISONs F(n) = {1, 2,
3, ..., n} as well as their union are less than the set ℕ of natural >>>> numbers.
Proof: Assume UF(n) = ℕ. The small FISONs are not necessary. What is >>>> the first necessary FISON? There is none! All can be dropped. But
according to Cantor's Theorem B, every non-empty set of different
numbers of the first and the second number class has a smallest
number, a minimum. This proves that the set of indices n of
necessary F(n), by not having a first element, is empty.
Which is a proof of ANY, not ALL together,
It is a proof of not any. The proof that not all together are
necessary is this: U{F(1), F(2), F(3), ...} = U{F(2), F(3), F(4), ...}.
which doesn't prove your claim about the Natural Numbers.
But this doesn't say that the infinite doesn't exist, and that we can't
make the Natural Numbers from a union of an infinite set of FISONs.
And, because FISONs are finite, no less than an infinite number of them should be expected to be needed.
This doesn't mean we need ALL of them, just an infinite number of them.
On 22.01.2025 00:41, Richard Damon wrote:
On 1/21/25 7:44 AM, WM wrote:
On 21.01.2025 13:17, Richard Damon wrote:
On 1/21/25 6:45 AM, WM wrote:
All finite initial segments of natural numbers, FISONs F(n) = {1,
2, 3, ..., n} as well as their union are less than the set ℕ of
natural numbers.
Proof: Assume UF(n) = ℕ. The small FISONs are not necessary. What
is the first necessary FISON? There is none! All can be dropped.
But according to Cantor's Theorem B, every non-empty set of
different numbers of the first and the second number class has a
smallest number, a minimum. This proves that the set of indices n
of necessary F(n), by not having a first element, is empty.
Which is a proof of ANY, not ALL together,
It is a proof of not any. The proof that not all together are
necessary is this: U{F(1), F(2), F(3), ...} = U{F(2), F(3), F(4), ...}.
which doesn't prove your claim about the Natural Numbers.
It proves what I said: not all are required.
But this doesn't say that the infinite doesn't exist, and that we
can't make the Natural Numbers from a union of an infinite set of FISONs.
According to Cantor's Theorem B, every non-empty set of different
numbers of the first and the second number class has a smallest number,
a minimum. This proves that the set of indices n of necessary FISONs,
by not having a first element, is empty.
And, because FISONs are finite, no less than an infinite number of
them should be expected to be needed.
Infinitely many fail like infinitely many traiangles would fail.
This doesn't mean we need ALL of them, just an infinite number of them.
Contradicted by Cantor's theorem.
Regards, WM
On 1/22/25 5:34 AM, WM wrote:
On 22.01.2025 00:41, Richard Damon wrote:
On 1/21/25 7:44 AM, WM wrote:
On 21.01.2025 13:17, Richard Damon wrote:
On 1/21/25 6:45 AM, WM wrote:
All finite initial segments of natural numbers, FISONs F(n) = {1,
2, 3, ..., n} as well as their union are less than the set ℕ of
natural numbers.
Proof: Assume UF(n) = ℕ. The small FISONs are not necessary. What >>>>>> is the first necessary FISON? There is none! All can be dropped.
But according to Cantor's Theorem B, every non-empty set of
different numbers of the first and the second number class has a
smallest number, a minimum. This proves that the set of indices n
of necessary F(n), by not having a first element, is empty.
Which is a proof of ANY, not ALL together,
It is a proof of not any. The proof that not all together are
necessary is this: U{F(1), F(2), F(3), ...} = U{F(2), F(3), F(4), ...}.
which doesn't prove your claim about the Natural Numbers.
It proves what I said: not all are required.
And no one said you needed to take the Union of ALL the FISONs, just ALL
of an infinite set of FISONs.
On 21.01.2025 21:35, Jim Burns wrote:
On 1/21/2025 6:45 AM, WM wrote:
[...]
For each finite set, there is
a (finite) FISON larger than it,
and that FISON subsets ℕ
It is a proper subset
and therefore not necessary
but completely useless in the union.
For each finite set, there is
a (finite) FISON larger than it.
That means the set is potentially infinite.
All can be dropped.
Each is not all.
All consists of each and not more.
Each can be dropped.
All cannot be dropped.
If you were right,
there would be a first FISON required
according to Cantor's theorem.
But according to Cantor's Theorem B,
every non-empty set of different numbers of
the first and the second number class
has a smallest number, a minimum.
This proves that
the set of indices n of necessary F(n),
by not having a first element,
is empty.
ℕ = ⋃{FISON}
Contradicted by mathematics,
namely Cantor's theorem.
On 22.01.2025 13:10, Richard Damon wrote:
On 1/22/25 5:34 AM, WM wrote:
On 22.01.2025 00:41, Richard Damon wrote:
On 1/21/25 7:44 AM, WM wrote:
On 21.01.2025 13:17, Richard Damon wrote:which doesn't prove your claim about the Natural Numbers.
On 1/21/25 6:45 AM, WM wrote:
All finite initial segments of natural numbers, FISONs F(n) = {1, >>>>>>> 2, 3, ..., n} as well as their union are less than the set ℕ of >>>>>>> natural numbers.
Proof: Assume UF(n) = ℕ. The small FISONs are not necessary. What >>>>>>> is the first necessary FISON? There is none! All can be dropped. >>>>>>> But according to Cantor's Theorem B, every non-empty set of
different numbers of the first and the second number class has a >>>>>>> smallest number, a minimum. This proves that the set of indices n >>>>>>> of necessary F(n), by not having a first element, is empty.
Which is a proof of ANY, not ALL together,
It is a proof of not any. The proof that not all together are
necessary is this: U{F(1), F(2), F(3), ...} = U{F(2), F(3), F(4), ...}. >>>
It proves what I said: not all are required.
And no one said you needed to take the Union of ALL the FISONs, just ALL
of an infinite set of FISONs.
Can't you read?
Assume UF(n) = ℕ. The small FISONs are not necessary. What is the first necessary FISON? There is none! All can be dropped. But according to
Cantor's Theorem B, every non-empty set of different numbers of the
first and the second number class has a smallest number, a minimum. This proves that the set of indices n of necessary F(n), by not having a
first element, is empty.
Regards, WM
On 1/22/2025 5:29 AM, WM wrote:
For each (finite) FISON, there is
a (finite) FISON larger than it,
and
it is a proper subset of that larger FISON,
which is a proper subset of ℕ
and therefore not necessary
but completely useless in the union.
For each FISON,
each natural number in it
is also in a later, fuller.by.one, and larger FISON,
and
is not dropped by dropping that (earlier) FISON.
For the union of all (finite) FISONs,
there isn't any
(finite) FISON larger than it.
The union of all (finite) FISONs is not finite.
"All" is not more than the repeated "each". If all could not be dropped,Each can be dropped.
All cannot be dropped.
But according to Cantor's Theorem B,
every non-empty set of different numbers of
the first and the second number class
has a smallest number, a minimum.
This proves that
the set of indices n of necessary F(n),
by not having a first element,
is empty.
ℕ = ⋃{FISON}
Contradicted by mathematics,
namely Cantor's theorem.
Le 22/01/2025 à 16:04, WM a écrit :first necessary FISON? There is none! All can be dropped. But according
Assume UF(n) = ℕ. The small FISONs are not necessary. What is the
Being "necessary" in an union of sets is not a property of a givenelement. It is a property of a subset.
If you have three coins of 2 euros not a single one is "necessary" topay a 3 euros drink
On 22.01.2025 13:10, Richard Damon wrote:
On 1/22/25 5:34 AM, WM wrote:
On 22.01.2025 00:41, Richard Damon wrote:
On 1/21/25 7:44 AM, WM wrote:
On 21.01.2025 13:17, Richard Damon wrote:
On 1/21/25 6:45 AM, WM wrote:
All finite initial segments of natural numbers, FISONs F(n) = {1, >>>>>>> 2, 3, ..., n} as well as their union are less than the set ℕ of >>>>>>> natural numbers.
Proof: Assume UF(n) = ℕ. The small FISONs are not necessary. What >>>>>>> is the first necessary FISON? There is none! All can be dropped. >>>>>>> But according to Cantor's Theorem B, every non-empty set of
different numbers of the first and the second number class has a >>>>>>> smallest number, a minimum. This proves that the set of indices n >>>>>>> of necessary F(n), by not having a first element, is empty.
Which is a proof of ANY, not ALL together,
It is a proof of not any. The proof that not all together are
necessary is this: U{F(1), F(2), F(3), ...} = U{F(2), F(3),
F(4), ...}.
which doesn't prove your claim about the Natural Numbers.
It proves what I said: not all are required.
And no one said you needed to take the Union of ALL the FISONs, just
ALL of an infinite set of FISONs.
Can't you read?
Assume UF(n) = ℕ. The small FISONs are not necessary. What is the first necessary FISON? There is none! All can be dropped. But according to
Cantor's Theorem B, every non-empty set of different numbers of the
first and the second number class has a smallest number, a minimum. This proves that the set of indices n of necessary F(n), by not having a
first element, is empty.
Regards, WM
On 22.01.2025 18:17, Jim Burns wrote:
On 1/22/2025 5:29 AM, WM wrote:
[...]
For each (finite) FISON, there is
a (finite) FISON larger than it,
and
it is a proper subset of that larger FISON,
which is a proper subset of ℕ
Right.
The sequence of FISONs is potentially infinite.
For each FISON F(n) there exists F(n^n^n).
For the union of all (finite) FISONs,
there isn't any
(finite) FISON larger than it.
There is no constant "all" in potential infinity.
The union of all (finite) FISONs is not finite.
Anyhow
there is no set of FISONs
the union of which would be ℕ.
Each can be dropped.
All cannot be dropped.
"All" is not more than the repeated "each".
But There is no single UF(n) that equals N, because you can ony get
there from the union of an infinite set of FISONs.
Your question about the "first" necessary fission is an invalid
question, as it is the same as asking for the highest Natural Number,
which doesn't exist,
On 1/23/2025 3:43 AM, WM wrote:
ℕ is a superset of each FISON
Each set which is a superset of each FISON
is a superset of ℕ
That's a longer.winded way to say that
ℕ is the union of all FISONs
Each FISON is a proper subset of ℕ
Each FISON is not ℕ
"All" is not more than the repeated "each".
'All' is complete,
whatever you (WM) mean by 'complete'.
From the finite.length description of a FISON,
we know that up to a FISON
is not complete, is not all.
On 23.01.2025 13:01, Richard Damon wrote:
But There is no single UF(n) that equals N, because you can ony get
there from the union of an infinite set of FISONs.
The union of all FISONs does not cover ℕ. Otherwise Cantor's theorem
would require the existence of a first necessary FISON. That is
mathematics.
Your question about the "first" necessary fission is an invalid
question, as it is the same as asking for the highest Natural Number,
which doesn't exist,
No, it is not. But there are only invalid handwaving answers.
Regards, WM
On 23.01.2025 16:18, Jim Burns wrote:
ℕ is a superset of each FISON
Each set which is a superset of each FISON
is a superset of ℕ
Wrong.
That's a longer.winded way to say that
ℕ is the union of all FISONs
The union of all FISONs does not cover ℕ.
Otherwise Cantor's theorem would require
the existence of a first necessary FISON.
Each FISON is a proper subset of ℕ
Each FISON is not ℕ
Therefore each FISON can be dropped from
the set of candidates.
Nothing remains.
"All" is not more than the repeated "each".
'All' is complete,
whatever you (WM) mean by 'complete'.
From the finite.length description of a FISON,
we know that up to a FISON
is not complete, is not all.
Up to every FISON
|ℕ \ {1, 2, 3, ..., n}| = ℵo.
Since every FISON is
the union of all its predecessors
we get
F(n): |ℕ \ UF(n)| = ℵo.
If you don't believe in
the union of all F(n),
find the first exception.
On 1/24/2025 4:37 AM, WM wrote:
On 23.01.2025 16:18, Jim Burns wrote:
The union of all FISONs covers UF(n)
Each FISON is a proper subset of another FISON.
Each FISON is a proper subset of UF(n)
No FISON is UF(n)
Whatever contains each FISON contains UF(n)
Otherwise Cantor's theorem would require
the existence of a first necessary FISON.
Each FISON is a proper subset of ℕ
Each FISON is not ℕ
Therefore each FISON can be dropped from
the set of candidates.
Candidates for what office? For UF(n) ?
Up to every FISON
|ℕ \ {1, 2, 3, ..., n}| = ℵo.
For any two FISONs {1,2,...,j} {1,2,...,k}
their sum {1,2,...,j,j+1,j+2,...,j+k} is a FISON
⎛ Consider Bob such that,
⎜ before all FISON.end.swaps n⇄n+1
⎜ Bob is in the first FISON.end 0
⎜
⎜ If Bob is in FISON.end n
⎜ then
⎜ it is after n-1⇄n and before n⇄n+1
⎜
⎜ If it is after all FISON.end.swaps
⎜ then Bob is not.in any FISON.end,
⎜ even though
⎜ no FISON.end.swap takes Bob
⎝ anywhere else.
On 1/24/25 4:41 AM, WM wrote:
On 23.01.2025 13:01, Richard Damon wrote:
But There is no single UF(n) that equals N, because you can ony get
there from the union of an infinite set of FISONs.
The union of all FISONs does not cover ℕ. Otherwise Cantor's theorem
would require the existence of a first necessary FISON. That is
mathematics.
Sure it does,
you just need to take the union of an infinite number of
them.
On 24.01.2025 13:29, Richard Damon wrote:Cantor’s theorem does not force a necessary FISON. We already established that no finite set of FISONs suffices.
On 1/24/25 4:41 AM, WM wrote:
On 23.01.2025 13:01, Richard Damon wrote:
But There is no single UF(n) that equals N, because you can ony getThe union of all FISONs does not cover ℕ. Otherwise Cantor's theorem
there from the union of an infinite set of FISONs.
would require the existence of a first necessary FISON.
What the FUCK that makes NO sense.you just need to take the union of an infinite number of them.FISONs enumerate themselves. There is no infinite FISON and hence no
infinite number of them.
On 24.01.2025 16:44, Jim Burns wrote:[assuming UF(n) means {1, 2, ..., n}]
On 1/24/2025 4:37 AM, WM wrote:
On 23.01.2025 16:18, Jim Burns wrote:
The union of all FISONs covers UF(n)
Simply contradicted by:
∀n ∈ UF(n): |ℕ \ {1, 2, 3, ..., n}| = ℵo
No, it’s just a factual description.Each FISON is a proper subset of another FISON. Each FISON is a properThat is potential infinity.
subset of UF(n). No FISON is UF(n)
No. Can you not conceive of an infinite set?Whatever contains each FISON contains UF(n)Alas this is not a set but a (potentially in-) finite changing
collection.
There is obviously no single FISON that is equal to N.Otherwise Cantor's theorem would require the existence of a first
necessary FISON.
Do you agree that or every FISON the question whether it is necessaryYes, in the negative. That does not imply the nonexistence of a
can be answered?
N is not a FISON.Up to every FISON |ℕ \ {1, 2, 3, ..., n}| = ℵo.
The union is infinite, since each FISON adds a different number.For any two FISONs {1,2,...,j} {1,2,...,k} their sumTherefore the union cannot be larger than a FISON.
{1,2,...,j,j+1,j+2,...,j+k} is a FISON
The infinite union is the infinite FISON. But there is no infinite FISONThe limit of FISONs is N.
An infinite sequence of swaps may not correspond to a single swap.⎛ Consider Bob such that,Swaps cannot eliminate Bob. He remains but i the darkness.
⎜ before all FISON.end.swaps n⇄n+1 ⎜ Bob is in the first FISON.end 0 ⎜
⎜ If Bob is in FISON.end n ⎜ then ⎜ it is after n-1⇄n and before n⇄n+1
⎜ If it is after all FISON.end.swaps ⎜ then Bob is not.in any
FISON.end,
⎜ even though ⎜ no FISON.end.swap takes Bob ⎝ anywhere else.
On 24.01.2025 13:29, Richard Damon wrote:
On 1/24/25 4:41 AM, WM wrote:
On 23.01.2025 13:01, Richard Damon wrote:
But There is no single UF(n) that equals N, because you can ony get
there from the union of an infinite set of FISONs.
The union of all FISONs does not cover ℕ. Otherwise Cantor's theorem
would require the existence of a first necessary FISON. That is
mathematics.
Sure it does,
Give the first.
you just need to take the union of an infinite number of them.
FISONs enumerate themselves. There is no infinite FISON and hence no
infinite number of them.
Regards, WM
On 24.01.2025 16:44, Jim Burns wrote:
The union of all FISONs covers UF(n)
Simply contradicted by:
∀n ∈ UF(n): |ℕ \ {1, 2, 3, ..., n}| = ℵo
Try to find a counter example. Fail.
Each FISON is a proper subset of another FISON.
Each FISON is a proper subset of UF(n)
No FISON is UF(n)
That is potential infinity.
Whatever contains each FISON contains UF(n)
Alas this is not a set but
a (potentially in-) finite changing collection.
Otherwise Cantor's theorem would require
the existence of a first necessary FISON.
Do you agree that
or every FISON
the question whether it is necessary
can be answered?
Swaps cannot eliminate Bob.
He remains but i the darkness.
On 1/25/2025 6:02 AM, WM wrote:
Swaps cannot eliminate Bob.
He remains but i the darkness.
No swaps are into the darkness.
There is obviously no single FISON that is equal to N.
Do you agree that or every FISON the question whether it is necessaryYes, in the negative. That does not imply the nonexistence of a
can be answered?
sufficient set.
Cantor’s theorem does not force a necessary FISON. We already established that no finite set of FISONs suffices.The union of all FISONs does not cover ℕ. Otherwise Cantor's theorem >>>> would require the existence of a first necessary FISON.
I said:
Sure it does, you just need to take the union of an infinite number of
them.
FISONs enumerate themselves. There is no infinite FISON and hence no
infinite number of them.
Then, what is the highest FISON?
If there is only a finite number of them, THEN there is a maximum
If it is after all swaps,
then Bob isn't in any FISON.end room,
even though swaps only place him in FISON.end rooms.
Do you agree that
for every FISON
the question whether it is necessary can be answered?
If Bob is somewhere after all swaps,
how did he get there?
Swaps cannot eliminate Bob.
He remains but i the darkness.
No swaps are into the darkness.
On 1/25/2025 2:35 PM, Jim Burns wrote:
On 1/25/2025 6:02 AM, WM wrote:
Swaps cannot eliminate Bob.
He remains but i the darkness.
A set larger than
any set with
⎛ fuller.by.one sets larger and
⎝ emptier.by.one sets smaller
is not
any set with
⎛ fuller.by.one sets larger and
⎝ emptier.by.one sets smaller.
No swaps are into the darkness.
On 25.01.2025 14:56, joes wrote:*finite
There is obviously no single FISON that is equal to N.Obviously every* union of FISONs is finite, because there are never two consecutive actually infinite sets in ℕ, and every FISON has an infinite set as successors.
Obviously every* unbion of FISONs is a FISON.
Yes, every single FISON can be dropped.It implies that there is not a first FISON. That proves by inductionDo you agree that or every FISON the question whether it is necessaryYes, in the negative. That does not imply the nonexistence of a
can be answered?
sufficient set.
that every FISON can be dropped. Your belief is dysfunctional logic.
On 1/25/2025 5:59 AM, joes wrote:
Am Sat, 25 Jan 2025 12:09:56 +0100 schrieb WM:
FISONs enumerate themselves. There is no infinite FISON and hence noWhat the FUCK that makes NO sense.
infinite number of them.
WOW!
Am Sun, 26 Jan 2025 09:41:22 +0100 schrieb WM:
On 25.01.2025 14:56, joes wrote:*finite
There is obviously no single FISON that is equal to N.Obviously every* union of FISONs is finite, because there are never two
consecutive actually infinite sets in ℕ, and every FISON has an infinite >> set as successors.
Obviously every* union of FISONs is a FISON.
Every infinite union of FISONs is obviously infinite.
Yes, every single FISON can be dropped.It implies that there is not a first FISON. That proves by inductionDo you agree that or every FISON the question whether it is necessaryYes, in the negative. That does not imply the nonexistence of a
can be answered?
sufficient set.
that every FISON can be dropped. Your belief is dysfunctional logic.
On 25.01.2025 21:23, Jim Burns wrote:Infinite swaps can.
On 1/25/2025 2:35 PM, Jim Burns wrote:No swaps can complete actual infinity of ℕ.
On 1/25/2025 6:02 AM, WM wrote:
Swaps cannot eliminate Bob. He remains but i the darkness.
A set larger than any set with ⎛ fuller.by.one sets larger and ⎝
emptier.by.one sets smaller is not any set with ⎛ fuller.by.one sets
larger and ⎝ emptier.by.one sets smaller.
No swaps are into the darkness.
On 25.01.2025 21:12, Chris M. Thomasson wrote:No. There are infinitely many rows.
On 1/25/2025 5:59 AM, joes wrote:
Am Sat, 25 Jan 2025 12:09:56 +0100 schrieb WM:
FISONs enumerate themselves. There is no infinite FISON and hence noWhat the FUCK that makes NO sense.
infinite number of them.
{1}
{2, 1}
{3, 2, 1}
...
Can you see that the first column is not longer than all finite rows?
On 25.01.2025 15:16, Richard Damon wrote:Yes there is. There are no consecutive infinities. Every FISON is,
Sure it does, you just need to take the union of an infinite number of
them.
But that is impossible because there are not two consecutive actually infinite sets in ℕ. Since every FISON is followed by an actually
infinite set, ∀n ∈ U(F(n)): |ℕ \ {1, 2, 3, ..., n}| = ℵo, there is no actually infinite set of FISONs.
The „system” of mathematics knows inf. many.That depends on the system. All we know is that it is finite.FISONs enumerate themselves. There is no infinite FISON and hence noThen, what is the highest FISON?
infinite number of them.
Therefore, not a finite number of FISONs.If there is only a finite number of them, THEN there is a maximumA variable maximum, "something becoming, emerging, produced, i.e., as we
put it, the potential infinite." [Hilbert]
On 23.01.2025 16:18, Jim Burns wrote:For every natural, {1, …, n} exists.
On 1/23/2025 3:43 AM, WM wrote:
ℕ is a superset of each FISON Each set which is a superset of eachWrong.
FISON is a superset of ℕ
Cantor's theorem does not force the existence of a necessary set.That's a longer.winded way to say that ℕ is the union of all FISONsThe union of all FISONs does not cover ℕ. Otherwise Cantor's theorem
would require the existence of a first necessary FISON.
N is not a FISON."All" is not more than the repeated "each".
'All' is complete, whatever you (WM) mean by 'complete'.
From the finite.length description of a FISON,
we know that up to a FISON is not complete, is not all.
Up to every FISON |ℕ \ {1, 2, 3, ..., n}| = ℵo.
Since every FISON is theIs there supposed to be a universal quantifier there?
union of all its predecessors we get F(n): |ℕ \ UF(n)| = ℵo.
If you don't believe in the union of all F(n), find the first exception.WDYM believe in the union?
Am Sun, 26 Jan 2025 10:42:48 +0100 schrieb WM:
{1}No.
{2, 1}
{3, 2, 1}
...
Can you see that the first column is not longer than all finite rows?
On 25.01.2025 14:59, joes wrote:
Cantor’s theorem does not force a necessary FISON. We already established >> that no finite set of FISONs suffices.The union of all FISONs does not cover ℕ. Otherwise Cantor's theorem >>>>> would require the existence of a first necessary FISON.
Cantor's theorem concerns also infinite sets. Without a first element
the set is empty.
Regards, WM
On 25.01.2025 15:16, Richard Damon wrote:
I said:
Sure it does, you just need to take the union of an infinite number of
them.
But that is impossible because there are not two consecutive actually infinite sets in ℕ. Since every FISON is followed by an actually
infinite set, ∀n ∈ U(F(n)): |ℕ \ {1, 2, 3, ..., n}| = ℵo, there is no actually infinite set of FISONs.
FISONs enumerate themselves. There is no infinite FISON and hence no
infinite number of them.
Then, what is the highest FISON?
That depends on the system. All we know is that it is finite.
If there is only a finite number of them, THEN there is a maximum
A variable maximum, "something becoming, emerging, produced, i.e., as we
put it, the potential infinite." [Hilbert]
Regards, WM
Am Fri, 24 Jan 2025 10:37:51 +0100 schrieb WM:
The union of all FISONs does not cover ℕ. Otherwise Cantor's theoremCantor's theorem does not force the existence of a necessary set.
would require the existence of a first necessary FISON.
WM pretended :
On 25.01.2025 21:12, Chris M. Thomasson wrote:
On 1/25/2025 5:59 AM, joes wrote:
Am Sat, 25 Jan 2025 12:09:56 +0100 schrieb WM:
FISONs enumerate themselves. There is no infinite FISON and hence no >>>>> infinite number of them.What the FUCK that makes NO sense.
WOW!
{1}
{2, 1}
{3, 2, 1}
...
Can you see that the first column is not longer than all finite rows?
Those are not FISONs.
On 1/26/25 3:43 AM, WM wrote:
On 25.01.2025 14:59, joes wrote:No, without a first element, the set doesn't exist.
Cantor’s theorem does not force a necessary FISON. We alreadyThe union of all FISONs does not cover ℕ. Otherwise Cantor's theorem >>>>>> would require the existence of a first necessary FISON.
established
that no finite set of FISONs suffices.
Cantor's theorem concerns also infinite sets. Without a first element
the set is empty.
On 1/26/25 3:41 AM, WM wrote:
It implies that there is not a first FISON. That proves by induction
that every FISON can be dropped. Your belief is dysfunctional logic.
No, it implies that the "set of necessary FISONs" doesn't exist.
Only by having the FALSE premise of the existance of such a set, can you
do your induction,
On 1/26/25 3:51 AM, WM wrote:
On 25.01.2025 15:16, Richard Damon wrote:
I said:
Sure it does, you just need to take the union of an infinite number
of them.
But that is impossible because there are not two consecutive actually
infinite sets in ℕ. Since every FISON is followed by an actually
infinite set, ∀n ∈ U(F(n)): |ℕ \ {1, 2, 3, ..., n}| = ℵo, there is no
actually infinite set of FISONs.
Why do we need "consecutive" infinite sets,
FISONs enumerate themselves. There is no infinite FISON and hence no
infinite number of them.
Then, what is the highest FISON?
That depends on the system. All we know is that it is finite.
No, it doesn't exist.
If there is only a finite number of them, THEN there is a maximum
A variable maximum, "something becoming, emerging, produced, i.e., as
we put it, the potential infinite." [Hilbert]
Which is about KNOWLEDGE, not the actual existance.
On 26.01.2025 10:37, joes wrote:The length? You gave an infinite list. Every column is infinite.
Am Sun, 26 Jan 2025 09:41:22 +0100 schrieb WM:
On 25.01.2025 14:56, joes wrote:*finite Every infinite union of FISONs is obviously infinite.
There is obviously no single FISON that is equal to N.Obviously every* union of FISONs is finite, because there are never
two consecutive actually infinite sets in ℕ, and every FISON has an
infinite set as successors.
Obviously every* union of FISONs is a FISON.
But no actually infinite set is possible.
{1}
{2, 1}
{3, 2, 1}
...
The length of the first column gives the cardinality.
It is bounded by the length of all rows all of which are finite.The rows are unbounded.
No actual infinity is represented by the first column.
Am Sun, 26 Jan 2025 10:59:09 +0100 schrieb WM:
On 26.01.2025 10:37, joes wrote:The length? You gave an infinite list. Every column is infinite.
Am Sun, 26 Jan 2025 09:41:22 +0100 schrieb WM:
On 25.01.2025 14:56, joes wrote:*finite Every infinite union of FISONs is obviously infinite.
There is obviously no single FISON that is equal to N.Obviously every* union of FISONs is finite, because there are never
two consecutive actually infinite sets in ℕ, and every FISON has an
infinite set as successors.
Obviously every* union of FISONs is a FISON.
But no actually infinite set is possible.
{1}
{2, 1}
{3, 2, 1}
...
The length of the first column gives the cardinality.
It is bounded by the length of all rows all of which are finite.The rows are unbounded.
No actual infinity is represented by the first column.
On 26.01.2025 18:55, joes wrote:All columns are infinite.
Am Sun, 26 Jan 2025 10:59:09 +0100 schrieb WM:No row is infinite. No column is longer than all rows. All rows and
On 26.01.2025 10:37, joes wrote:The length? You gave an infinite list. Every column is infinite.
Am Sun, 26 Jan 2025 09:41:22 +0100 schrieb WM:
On 25.01.2025 14:56, joes wrote:*finite Every infinite union of FISONs is obviously infinite.
There is obviously no single FISON that is equal to N.Obviously every* union of FISONs is finite, because there are never
two consecutive actually infinite sets in ℕ, and every FISON has an >>>>> infinite set as successors.
Obviously every* union of FISONs is a FISON.
But no actually infinite set is possible.
{1}
{2, 1}
{3, 2, 1}
...
The length of the first column gives the cardinality.
columns belong to potentially infinite collections.
What are you getting at?Right. The columns are unbounded too. With n also n^n^n is contained. Nevertheless the fixed number ℵo of elements is neither in a row (by definition) nor in a columns (by symmetry).It is bounded by the length of all rows all of which are finite.The rows are unbounded.
No actual infinity is represented by the first column.
Now try to find yourself the solution: What is unbounded but smallerWhat do you want to derive from the answer that makes this an
than the first transfinite quantity?
On 26.01.2025 13:38, Richard Damon wrote:
On 1/26/25 3:51 AM, WM wrote:
On 25.01.2025 15:16, Richard Damon wrote:
I said:
Sure it does, you just need to take the union of an infinite number
of them.
But that is impossible because there are not two consecutive actually
infinite sets in ℕ. Since every FISON is followed by an actually
infinite set, ∀n ∈ U(F(n)): |ℕ \ {1, 2, 3, ..., n}| = ℵo, there is no
actually infinite set of FISONs.
Why do we need "consecutive" infinite sets,
They would not exist even if they were needed.
{1}FISONs enumerate themselves. There is no infinite FISON and hence
no infinite number of them.
Then, what is the highest FISON?
That depends on the system. All we know is that it is finite.
No, it doesn't exist.
{2, 1}
{3, 2, 1}
...
The first column never gets larger than a FISON.
If there is only a finite number of them, THEN there is a maximum
A variable maximum, "something becoming, emerging, produced, i.e., as
we put it, the potential infinite." [Hilbert]
Which is about KNOWLEDGE, not the actual existance.
FISONs are about knowledge.
Regards, WM
All finite initial segments of natural numbers, FISONs F(n) = {1, 2,
3, ..., n} as well as their union are less than the set ℕ of natural numbers.
Proof: Assume UF(n) = ℕ. The small FISONs are not necessary. What is the first necessary FISON? There is none! All can be dropped. But according
to Cantor's Theorem B, every non-empty set of different numbers of the
first and the second number class has a smallest number, a minimum. This proves that the set of indices n of necessary F(n), by not having a
first element, is empty.
Regards, WM
On 25.01.2025 21:23, Jim Burns wrote:
On 1/25/2025 2:35 PM, Jim Burns wrote:
On 1/25/2025 6:02 AM, WM wrote:
Swaps cannot eliminate Bob.
He remains but i the darkness.
A set larger than
any set with
⎛ fuller.by.one sets larger and
⎝ emptier.by.one sets smaller
is not
any set with
⎛ fuller.by.one sets larger and
⎝ emptier.by.one sets smaller.
No swaps are into the darkness.
No swaps can complete actual infinity of ℕ.
No actual infinity is represented by the first column.
No swaps can complete actual infinity of ℕ.
Now try to find yourself the solution: What is unbounded but smaller
than the first transfinite quantity?
Regards, WM
Am Sun, 26 Jan 2025 19:08:08 +0100 schrieb WM:
All columns are infinite.No row is infinite. No column is longer than all rows. All rows andBut no actually infinite set is possible.The length? You gave an infinite list. Every column is infinite.
{1}
{2, 1}
{3, 2, 1}
...
The length of the first column gives the cardinality.
columns belong to potentially infinite collections.
On 1/26/25 9:28 AM, WM wrote:
{1}
{2, 1}
{3, 2, 1}
...
The first column never gets larger than a FISON.
Sure it does,
FISONs are about knowledge.
So, you agree that they don't tell you about the actual existance of the number, just what you can know about them.
you can't show that the first FISON isn't a member of the
set of necessary FISONs without assuming the set of necessary FISONs
exists.
I make no claim that there is a specific set of "necessary" FISONs,
Thinking a bit about this, the "Set of Necessary FISONs" will be empty,
becuase no particular FISON is needed, you just need an infinite set of
them.
A simple proof of this is that we can build up the set at least two
different ways using two infinite sets with no members in common.
The first set it the set of all ODD FISONs, i.e. FISONs whoes highest
member is an odd number.
There is no Natural Number not covered by this union, as for every
Natural Number, either it is itself odd, and thus part of its own FISON,
or the number one greater than itself (which does exist) will be odd,
and this number will exist in that FISON, and thus in the union of them.
We can also do that with the set of even FISONs.
For a FISON to be in the set of "Necessary" it would need to be in EVERY
set that meets the requriment, but since no set does, there are no "necessary" FISONs.
Just like there are no "necessaery" Numbers to have the sum of the set
to be zero.
Sure, but the 'S' in FISON stands for segment, not set, and ordinals
have order and are constructed non-arbitrarily.
On 1/26/2025 1:42 AM, WM wrote:
{1}
{2, 1}
{3, 2, 1}
...
Can you see that the first column is not longer than all finite rows?
As you take this to infinity, your rows become infinite... ;^)
On 26.01.2025 21:26, Chris M. Thomasson wrote:
On 1/26/2025 1:42 AM, WM wrote:
{1}
{2, 1}
{3, 2, 1}
...
Can you see that the first column is not longer than all finite rows?
As you take this to infinity, your rows become infinite... ;^)
No. FISONs are finite. The infinite is completed by the succeeding
numbers: ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
Regards, WM
On 26.01.2025 20:06, joes wrote:That’s just wrong. There are OBVIOUSLY infinitely many rows.
Am Sun, 26 Jan 2025 19:08:08 +0100 schrieb WM:
No column is greater than all rows. All rows are finite. No column is actually infinite, i.e., has length |ℕ|.All columns are infinite.No row is infinite. No column is longer than all rows. All rows andBut no actually infinite set is possible.The length? You gave an infinite list. Every column is infinite.
{1}
{2, 1}
{3, 2, 1}
...
The length of the first column gives the cardinality.
columns belong to potentially infinite collections.
On 1/26/2025 4:49 AM, WM wrote:
No swaps can complete actual infinity of ℕ.
No actual infinity is represented by the first column.
No appendix 𝔻 such that ∀d ∈ 𝔻: f(d) = d
completes potential.infinity to actual.infinity.
No swaps can complete actual infinity of ℕ.
For each FISON, there is
a fuller.by.one FISON which is larger.
That is complete enough
in order to allow Bob to disappear.
The sets aren't larger and aren't smaller.
They are the same size,
which allows Bob to disappear by lossless swaps.
Am Mon, 27 Jan 2025 12:13:38 +0100 schrieb WM:
On 26.01.2025 20:06, joes wrote:That’s just wrong. There are OBVIOUSLY infinitely many rows.
Am Sun, 26 Jan 2025 19:08:08 +0100 schrieb WM:No column is greater than all rows. All rows are finite. No column is
All columns are infinite.No row is infinite. No column is longer than all rows. All rows andBut no actually infinite set is possible.The length? You gave an infinite list. Every column is infinite.
{1}
{2, 1}
{3, 2, 1}
...
The length of the first column gives the cardinality.
columns belong to potentially infinite collections.
actually infinite, i.e., has length |ℕ|.
On 26.01.2025 23:31, Richard Damon wrote:Obviously, but not for infinite sets of FISONs.
you can't show that the first FISON isn't a member of the set of
necessary FISONs without assuming the set of necessary FISONs exists.
Logic? The set ℕ exists. And I can show that every FISON is neither necessary nor sufficient to accomplish that aim.
WM has brought this to us :
On 26.01.2025 20:29, FromTheRafters wrote:
Sure, but the 'S' in FISON stands for segment, not set, and ordinals
have order and are constructed non-arbitrarily.
A segment written as a set has no order.
Then it is not a FISON.
Google's AI says: (and everyone's gonna love this one)
=====================================================
AI Overview
FISONs are a sequence of numbers that prove dark natural numbers. Each
FISON ends with a natural number, and every finite union of FISONs is
also a FISON.
Explanation
FISONs: A sequence of numbers that prove dark natural numbers.
Dark natural numbers: Proved by the sequence of FISONs.
Natural number: A number that ends a FISON.
Finite union: A union of FISONs that is also a FISON.
Example
The sequence of FISONs can be represented as {1}, {1, 2}, {1, 2, 3}, and so on.
Each FISON ends with a natural number, such as 1, 2, or 3.
Every finite union of FISONs is also a FISON, such as {1} » {1, 2} = {1, 2}.
Dark numbers
Dark natural numbers proved by the sequence of FISONs ... actual infinity exchanges quantifiers and states $ع "Fn: |Fn| < |ع| ⁄ Fn...
Technische Hochschule Augsburg
Am Mon, 27 Jan 2025 12:35:56 +0100 schrieb WM:
On 26.01.2025 23:31, Richard Damon wrote:
Obviously, but not for infinite sets of FISONs.you can't show that the first FISON isn't a member of the set of
necessary FISONs without assuming the set of necessary FISONs exists.
Logic? The set ℕ exists. And I can show that every FISON is neither
necessary nor sufficient to accomplish that aim.
On 27.01.2025 15:14, joes wrote:No problem, any infinite set of FISONs has one.
Am Mon, 27 Jan 2025 12:35:56 +0100 schrieb WM:
On 26.01.2025 23:31, Richard Damon wrote:Obviously, but not for infinite sets of FISONs.
you can't show that the first FISON isn't a member of the set of
necessary FISONs without assuming the set of necessary FISONs exists.
Logic? The set ℕ exists. And I can show that every FISON is neither
necessary nor sufficient to accomplish that aim.
Also an infinite set needs a first element.
On 26.01.2025 13:38, Richard Damon wrote:Quantifier shift: yes, every FISON, but the successor of the SET of all
On 1/26/25 3:51 AM, WM wrote:
On 25.01.2025 15:16, Richard Damon wrote:
Sure it does, you just need to take the union of an infinite number
of them.
But that is impossible because there are not two consecutive actually
infinite sets in ℕ. Since every FISON is followed by an actually
infinite set, ∀n ∈ U(F(n)): |ℕ \ {1, 2, 3, ..., n}| = ℵo, there is no
actually infinite set of FISONs.
Neither does it stop.{1}No, it doesn't exist.That depends on the system. All we know is that it is finite.FISONs enumerate themselves. There is no infinite FISON and hence no >>>>> infinite number of them.Then, what is the highest FISON?
{2, 1}
{3, 2, 1}
...
The first column never gets larger than a FISON.
On 27.01.2025 13:49, joes wrote:It is not obvious to me.
Am Mon, 27 Jan 2025 12:13:38 +0100 schrieb WM:Obviously is obviously is not correct.
On 26.01.2025 20:06, joes wrote:That’s just wrong. There are OBVIOUSLY infinitely many rows.
Am Sun, 26 Jan 2025 19:08:08 +0100 schrieb WM:No column is greater than all rows. All rows are finite. No column is
All columns are infinite.No row is infinite. No column is longer than all rows. All rows andBut no actually infinite set is possible.The length? You gave an infinite list. Every column is infinite.
{1}
{2, 1}
{3, 2, 1}
...
The length of the first column gives the cardinality.
columns belong to potentially infinite collections.
actually infinite, i.e., has length |ℕ|.
There are infinitely many rows with infinitely many numbers.No. There are infinitely many rows, with finitely many numbers each.
Nevertheless all rows are of length smaller than |ℕ| and therefore by symmetry also the first column is of length smaller than |ℕ|.What is the symmetry?
On 26.01.2025 12:11, joes wrote:No, I was talking about sufficient sets.
Am Fri, 24 Jan 2025 10:37:51 +0100 schrieb WM:
The union of all FISONs does not cover ℕ. Otherwise Cantor's theoremCantor's theorem does not force the existence of a necessary set.
would require the existence of a first necessary FISON.
But your claim that a set exists forces the existence of a first FISON
that cannot be discarded. Otherwise all can be discarded.
On 26.01.2025 23:54, Jim Burns wrote:
On 1/26/2025 4:49 AM, WM wrote:
No swaps can complete actual infinity of ℕ.
No actual infinity is represented by the first column.
No appendix 𝔻 such that ∀d ∈ 𝔻: f(d) = d
completes potential.infinity to actual.infinity.
∀n ∈ U(F(n)): |ℕ \ {1, 2, 3, ..., n}| = ℵo
ℕ \ {1, 2, 3, ...} = { }
∀n ∈ U(F(n)): |ℕ \ {1, 2, 3, ..., n}| = ℵo
ℕ \ {1, 2, 3, ...} = { }
On 27.01.2025 13:49, joes wrote:
Am Mon, 27 Jan 2025 12:13:38 +0100 schrieb WM:
On 26.01.2025 20:06, joes wrote:That’s just wrong. There are OBVIOUSLY infinitely many rows.
Am Sun, 26 Jan 2025 19:08:08 +0100 schrieb WM:No column is greater than all rows. All rows are finite. No column is
All columns are infinite.No row is infinite. No column is longer than all rows. All rows andBut no actually infinite set is possible.The length? You gave an infinite list. Every column is infinite.
{1}
{2, 1}
{3, 2, 1}
...
The length of the first column gives the cardinality.
columns belong to potentially infinite collections.
actually infinite, i.e., has length |ℕ|.
Obviously is obviously is not correct.
There are infinitely many rows with infinitely many numbers.
Nevertheless all rows are of length smaller than |ℕ| and therefore by symmetry also the first column is of length smaller than |ℕ|.
Regards, WM
On 27.01.2025 15:14, joes wrote:
Am Mon, 27 Jan 2025 12:35:56 +0100 schrieb WM:
On 26.01.2025 23:31, Richard Damon wrote:Obviously, but not for infinite sets of FISONs.
you can't show that the first FISON isn't a member of the set of
necessary FISONs without assuming the set of necessary FISONs exists.
Logic? The set ℕ exists. And I can show that every FISON is neither
necessary nor sufficient to accomplish that aim.
Also an infinite set needs a first element. But no FISON is necessary or sufficient.
Regards, WM
WM wrote :
On 27.01.2025 15:14, joes wrote:
Am Mon, 27 Jan 2025 12:35:56 +0100 schrieb WM:
On 26.01.2025 23:31, Richard Damon wrote:Obviously, but not for infinite sets of FISONs.
you can't show that the first FISON isn't a member of the set ofLogic? The set ℕ exists. And I can show that every FISON is neither
necessary FISONs without assuming the set of necessary FISONs exists. >>>>
necessary nor sufficient to accomplish that aim.
Also an infinite set needs a first element.
The real numbers are an infinite set, which one is first?
Am Mon, 27 Jan 2025 15:33:01 +0100 schrieb WM:
Also an infinite set needs a first element.No problem, any infinite set of FISONs has one.
Am Sun, 26 Jan 2025 15:28:14 +0100 schrieb WM:
{1}Neither does it stop.
{2, 1}
{3, 2, 1}
...
The first column never gets larger than a FISON.
On 1/27/2025 7:55 AM, WM wrote:
∀n ∈ U(F(n)): |ℕ \ {1, 2, 3, ..., n}| = ℵo
ℕ \ {1, 2, 3, ...} = { }
What is ⋃{F(n)} ?
What is {1,2,3,...} ?
{1,2,3,...} = (⋃{F(n)})∪𝔻 = ⋃{F(n),𝔻}
Certainly, an appendix 𝔻 is possible.
∀n ∈ ⋃{F(n)}:
|⋃{F(n)}\{1,2,3,...,n}| = |⋃{F(n)}\{1,2,3,...,n-1}|
There is no first end.segment smaller than ⋃{F(n)}
The end.segments are well.ordered.
No 𝔻 completesᵂᴹ ⋃{F(n)}
On 27.01.2025 18:30, joes wrote:My mistake. It doesn’t „get” or „evolve”, it just is - infinitely long.
Am Sun, 26 Jan 2025 15:28:14 +0100 schrieb WM:
True. It evolves between finite and ℕ. A variable maximum, "something becoming, emerging, produced, i.e., as we put it, the potential{1}Neither does it stop.
{2, 1}
{3, 2, 1}
...
The first column never gets larger than a FISON.
infinite." [Hilbert]
On 1/27/25 9:04 AM, WM wrote:
There are infinitely many rows with infinitely many numbers.
Nevertheless all rows are of length smaller than |ℕ| and therefore by
symmetry also the first column is of length smaller than |ℕ|.
Which is obviously incorrect, as it contains the indicator ... which
mean to continue infinitely.
Yes, each individual row is finite, as it isn't the "last" row, and more
are below it.
On 27.01.2025 18:10, joes wrote:Yes it does. Any infinite set of FISONs has a first element.
Am Mon, 27 Jan 2025 15:33:01 +0100 schrieb WM:
But no set of FISONs the union of which is ℕ has a first element.Also an infinite set needs a first element.No problem, any infinite set of FISONs has one.
If anNo, as you have shown, no element is necessary.
infinite set was existing, you could easily find a first not completely useless element.
On 1/27/25 9:33 AM, WM wrote:
Also an infinite set needs a first element. But no FISON is necessary
or sufficient.
Which means the set of "necessary FISONs" doesn't exist.
That doesn't mean what you want it to mean, it just shows your logic is broken.
we can build an infinite number of infinite sets of FISONs whose union
is the Natural Numbers,
Am Tue, 28 Jan 2025 10:02:59 +0100 schrieb WM:
On 27.01.2025 18:30, joes wrote:My mistake. It doesn’t „get” or „evolve”, it just is - infinitely long.
Am Sun, 26 Jan 2025 15:28:14 +0100 schrieb WM:True. It evolves between finite and ℕ. A variable maximum, "something
{1}Neither does it stop.
{2, 1}
{3, 2, 1}
...
The first column never gets larger than a FISON.
becoming, emerging, produced, i.e., as we put it, the potential
infinite." [Hilbert]
WM wrote on 1/28/2025 :
On 27.01.2025 16:23, FromTheRafters wrote:
WM wrote :
On 27.01.2025 15:14, joes wrote:
Am Mon, 27 Jan 2025 12:35:56 +0100 schrieb WM:
On 26.01.2025 23:31, Richard Damon wrote:Obviously, but not for infinite sets of FISONs.
you can't show that the first FISON isn't a member of the set of >>>>>>> necessary FISONs without assuming the set of necessary FISONs
exists.
Logic? The set ℕ exists. And I can show that every FISON is neither >>>>>> necessary nor sufficient to accomplish that aim.
Also an infinite set needs a first element.
The real numbers are an infinite set, which one is first?
Here we talk about FISONs.
You said 'an infinite set needs a first element' and FISONs are not
infinite sets.
Am Tue, 28 Jan 2025 09:58:14 +0100 schrieb WM:
On 27.01.2025 18:10, joes wrote:Yes it does. Any infinite set of FISONs has a first element.
Am Mon, 27 Jan 2025 15:33:01 +0100 schrieb WM:But no set of FISONs the union of which is ℕ has a first element.
Also an infinite set needs a first element.No problem, any infinite set of FISONs has one.
If anNo, as you have shown, no element is necessary.
infinite set was existing, you could easily find a first not completely
useless element.
On 28.01.2025 10:29, FromTheRafters wrote:
WM wrote on 1/28/2025 :
On 27.01.2025 16:23, FromTheRafters wrote:
WM wrote :
On 27.01.2025 15:14, joes wrote:
Am Mon, 27 Jan 2025 12:35:56 +0100 schrieb WM:
On 26.01.2025 23:31, Richard Damon wrote:Obviously, but not for infinite sets of FISONs.
you can't show that the first FISON isn't a member of the set of >>>>>>>> necessary FISONs without assuming the set of necessary FISONs
exists.
Logic? The set ℕ exists. And I can show that every FISON is
neither necessary nor sufficient to accomplish that aim.
Also an infinite set needs a first element.
The real numbers are an infinite set, which one is first?
Here we talk about FISONs.
You said 'an infinite set needs a first element' and FISONs are not
infinite sets.
An infinite set of FISONs that has the union ℕ needs a first element.
On 28.01.2025 00:32, Richard Damon wrote:He didn’t say they were necessary.
we can build an infinite number of infinite sets of FISONs whose unionThat is the diploma of stupidity. Name the first useful FISON!
is the Natural Numbers,
On 28.01.2025 10:17, joes wrote:
Am Tue, 28 Jan 2025 10:02:59 +0100 schrieb WM:Not as long as ℕ.
On 27.01.2025 18:30, joes wrote:My mistake. It doesn’t „get” or „evolve”, it just is - infinitely long.
Am Sun, 26 Jan 2025 15:28:14 +0100 schrieb WM:True. It evolves between finite and ℕ. A variable maximum, "something
{1}Neither does it stop.
{2, 1}
{3, 2, 1}
...
The first column never gets larger than a FISON.
becoming, emerging, produced, i.e., as we put it, the potential
infinite." [Hilbert]
Regards, WM
On 27.01.2025 23:07, Jim Burns wrote:
On 1/27/2025 7:55 AM, WM wrote:
∀n ∈ U(F(n)): |ℕ \ {1, 2, 3, ..., n}| = ℵo
ℕ \ {1, 2, 3, ...} = { }
What is ⋃{F(n)} ?
It is the union of FISONs, ℕ_def.
What is {1,2,3,...} ?
ℕ
{1,2,3,...} = (⋃{F(n)})∪𝔻 = ⋃{F(n),𝔻}
Certainly, an appendix 𝔻 is possible.
Yes, it is the dark domain.
∀n ∈ ⋃{F(n)}:
|⋃{F(n)}\{1,2,3,...,n}| = |⋃{F(n)}\{1,2,3,...,n-1}|
There is no first end.segment smaller than ⋃{F(n)}
The end.segments are well.ordered.
But potentially infinite.
No 𝔻 completesᵂᴹ ⋃{F(n)}
Then actual infinity does not exist.
Am Tue, 28 Jan 2025 10:53:04 +0100 schrieb WM:
An infinite set of FISONs that has the union ℕ needs a first element.
Where’s the problem? Every set of FISONs has a first element.
WM expressed precisely :
An infinite set of FISONs that has the union ℕ needs a first element.
It has one defined already, the initial element.
Am Tue, 28 Jan 2025 10:21:20 +0100 schrieb WM:
On 28.01.2025 00:32, Richard Damon wrote:
He didn’t say they were necessary.we can build an infinite number of infinite sets of FISONs whose unionThat is the diploma of stupidity. Name the first useful FISON!
is the Natural Numbers,
On 1/28/25 3:58 AM, WM wrote:
Many sets of FISONs whose union is N exist, and each of those sets has a first element.
On 28.01.2025 13:45, Richard Damon wrote:Nobody said it was necessary.
On 1/28/25 3:58 AM, WM wrote:
Many sets of FISONs whose union is N exist, and each of those sets hasName the first element of only one of those sets which is required in
a first element.
that set.
On 1/28/25 4:56 AM, WM wrote:
When every not necessary or not useful element has been discarded
nothing remains.
But that operation can't be done, your logic removes one by one, and
thus can't remove ALL.
On 1/28/25 4:21 AM, WM wrote:
On 28.01.2025 00:32, Richard Damon wrote:
On 1/27/25 9:33 AM, WM wrote:
Also an infinite set needs a first element. But no FISON is
necessary or sufficient.
Which means the set of "necessary FISONs" doesn't exist.
The set of useful FISONs does not exist. Otherwise name the first one.
You need to define "useful".
On 28.01.2025 11:54, joes wrote:Naturally. No finite, but every infinite does.
Am Tue, 28 Jan 2025 10:53:04 +0100 schrieb WM:
Not every set of FISONs has the union ℕ.An infinite set of FISONs that has the union ℕ needs a first element.Where’s the problem? Every set of FISONs has a first element.
Name the first FISON thatThat is a complete non sequitur.
cannot be discarded to yield the union union ℕ.
If all can be discarded, then there is no set with union ℕ.
A set such that a larger FISON does NOT exist
is sufficiently large in order for Bob
to disappear purely from swaps within the set.
I agree that
your (WM's) actualᵂᴹ infinity does not exist.
That's not our infinity.
Am Tue, 28 Jan 2025 16:15:35 +0100 schrieb WM:
On 28.01.2025 13:45, Richard Damon wrote:Nobody said it was necessary.
On 1/28/25 3:58 AM, WM wrote:Name the first element of only one of those sets which is required in
Many sets of FISONs whose union is N exist, and each of those sets has
a first element.
that set.
Am Tue, 28 Jan 2025 16:10:59 +0100 schrieb WM:
On 28.01.2025 11:57, joes wrote:Do you get that at TH Augsburg?
Am Tue, 28 Jan 2025 10:21:20 +0100 schrieb WM:He said it is possible. That is the diploma of stupidity.
On 28.01.2025 00:32, Richard Damon wrote:He didn’t say they were necessary.
we can build an infinite number of infinite sets of FISONs whoseThat is the diploma of stupidity. Name the first useful FISON!
union is the Natural Numbers,
Am Tue, 28 Jan 2025 16:07:51 +0100 schrieb WM:
On 28.01.2025 11:54, joes wrote:Naturally. No finite, but every infinite does.
Am Tue, 28 Jan 2025 10:53:04 +0100 schrieb WM:Not every set of FISONs has the union ℕ.
An infinite set of FISONs that has the union ℕ needs a first element. >>> Where’s the problem? Every set of FISONs has a first element.
Name the first FISON thatThat is a complete non sequitur.
cannot be discarded to yield the union union ℕ.
If all can be discarded, then there is no set with union ℕ.
On 28.01.2025 14:10, Jim Burns wrote:
A set such that a larger FISON does NOT exist
is sufficiently large in order for Bob
to disappear purely from swaps within the set.
He disappears from visibility.
I agree that
your (WM's) actualᵂᴹ infinity does not exist.
That's not our infinity.
I use Cantors actual infinity:
|ℕ| is a fixed quantity
larger than all natural numbers.
What is your infinity?
On 28.01.2025 13:45, Richard Damon wrote:
On 1/28/25 3:58 AM, WM wrote:
Many sets of FISONs whose union is N exist, and each of those sets has
a first element.
Name the first element of only one of those sets which is required in
that set.
Regards, WM
On 1/28/2025 10:29 AM, WM wrote:
On 28.01.2025 14:10, Jim Burns wrote:
A set such that a larger FISON does NOT exist
is sufficiently large in order for Bob
to disappear purely from swaps within the set.
He disappears from visibility.
There are no swaps into a room,
except for rooms with a later swap.out.
Darkᵂᴹ or visibleᵂᴹ,
there are no swaps into a room,
except for rooms with a later swap.out.
I use Cantors actual infinity:
Only when it pleases you to do so.
|ℕ| is a fixed quantity
larger than all natural numbers.
What is your infinity?
The union of all FISONs is infinite.
What are you (WM) most recently calling it?
WM expressed precisely :
On 28.01.2025 11:26, FromTheRafters wrote:The only needed one would be the last one but
WM expressed precisely :
An infinite set of FISONs that has the union ℕ needs a first element. >>>
there is no last one just like there is no last natural number.
YOU are the one that says there needs to be a "required" FISON.
Omega has no immediate predecessor.
on 1/29/2025, WM supposed :
On 29.01.2025 05:07, Richard Damon wrote:
YOU are the one that says there needs to be a "required" FISON.
If there is an infinite set, then it has a first element
Still wrong.
On 28.01.2025 17:29, Jim Burns wrote:Or the other way around: there is no room where she can stay,
On 1/28/2025 10:29 AM, WM wrote:No.
On 28.01.2025 14:10, Jim Burns wrote:There are no swaps into a room,
A set such that a larger FISON does NOT exist is sufficiently largeHe disappears from visibility.
in order for Bob to disappear purely from swaps within the set.
except for rooms with a later swap.out.
No. If it were finite, it had an initial segment.Darkᵂᴹ or visibleᵂᴹ,One exception exists: ω-1.
there are no swaps into a room,
except for rooms with a later swap.out.
Exactly, when it pleases you.No I use it in order to show its consequences. In my personal opinion potential infinity appears preferable.I use Cantors actual infinity:Only when it pleases you to do so.
Aha!Yes, it has no last FISON.|ℕ| is a fixed quantity larger than all natural numbers. What is yourThe union of all FISONs is infinite.
infinity?
But its union cannot be ℕ because a non-emptyWhy should a necessary set exist?
set of FISONs would be necessary.
But the set has no elements, becauseBut not all at once, only every single one. Maybe you can prove that
for every FISON of the assumed set we can prove that it is useless and
could be dropped.
Am Wed, 29 Jan 2025 09:26:35 +0100 schrieb WM:
But its union cannot be ℕ because a non-emptyWhy should a necessary set exist?
set of FISONs would be necessary.
On 28.01.2025 18:18, FromTheRafters wrote:
WM expressed precisely :
On 28.01.2025 11:26, FromTheRafters wrote:The only needed one would be the last one but there is no last one
WM expressed precisely :
An infinite set of FISONs that has the union ℕ needs a first element. >>>>
just like there is no last natural number.
The only needed one would be a FISON that does not obey
|ℕ \ {1, 2, 3, ..., n}| = ℵo
but
|ℕ \ {1, 2, 3, ..., n}| = 0
But it is not existing. Therefore you claim that an infinite set is necessary. But that is pure matheology.
Assume any set of FISONs with uninon U(F(n)) = ℕ. Then every FISON can
be dropped as completely useless. Nothing remains. Therefore:
IF U(F(n)) = ℕ THEN ℕ = { }.
FISONs can accomplish only a potentially infinite collection.
Regards, WM
On 1/29/25 6:47 AM, WM wrote:
On 29.01.2025 11:53, joes wrote:
Am Wed, 29 Jan 2025 09:26:35 +0100 schrieb WM:
But its union cannot be ℕ because a non-emptyWhy should a necessary set exist?
set of FISONs would be necessary.
The set cannot be empty. Therefore it has elements. For each element we
can show that it is not leading to the aim. Therefore no such set
U(F(n)) = ℕ exists.
Regards, WM
The set of "required" FISONs is empty.
That doesn't mean we can't build an infinite set of FISONs whose union
is the set of Natual Numbers.
We can in fact build an infinite set of infinite sets of FISONs whose
union is the set of Natural Numbers, it just is a fact that no single
FISON is in all of them, so none are "required"
The fact that you can't understand that FACT, just shows your stupidity.
Do you disagree that the union of all FISONs that end in a odd number
will cover N, if not what number doesn't get covered.
By the same token, the union of all FISONs that end in a even number
will cover N.
Thus, no individual FISON is needed, we just need to select an infinite
set of FISONs to cover N.
Your failure to process that statement and even try to refute it says
that you know in the back of your head that you are wrong, but you won't listen to that voice, and thus make yourself an idiot.
WM explained on 1/29/2025 :
On 29.01.2025 10:38, FromTheRafters wrote:
Omega has no immediate predecessor.
What is immediately before ω?
Nothing, it is the initial member of the transfinite ordinals. just as
zero is the initial member of the finite ordinals.
And how long is the distance from ω to a natnumber?
This makes no sense.
WM was thinking very hard :
It is Cantor's theorem that every set of ordinals has a first element.
FISONs are v. Neumann ordinals.
Yes, but some infinite sets don't have a first element.
We can in fact build an infinite set of infinite sets of FISONs whose
union is the set of Natural Numbers,
On 29.01.2025 15:04, FromTheRafters wrote:No ordinal.
WM explained on 1/29/2025 :There is something before zero.
On 29.01.2025 10:38, FromTheRafters wrote:Nothing, it is the initial member of the transfinite ordinals. just as
Omega has no immediate predecessor.What is immediately before ω?
zero is the initial member of the finite ordinals.
AHAHA there are stupid questions.It is research.And how long is the distance from ω to a natnumber?This makes no sense.
On 29.01.2025 15:00, FromTheRafters wrote:
WM was thinking very hard :
It is Cantor's theorem that every set of ordinals has a first
element. FISONs are v. Neumann ordinals.
Yes, but some infinite sets don't have a first element.
Every finite or infinite set of ordinals or FISONs has a first element!
Other sets are irrelevant in the present context.
Regards, WM
On 1/30/25 4:07 AM, WM wrote:
On 29.01.2025 15:04, FromTheRafters wrote:
WM explained on 1/29/2025 :
On 29.01.2025 10:38, FromTheRafters wrote:
Omega has no immediate predecessor.
What is immediately before ω?
Nothing, it is the initial member of the transfinite ordinals. just as
zero is the initial member of the finite ordinals.
There is something before zero.
Not in the set of Natural Numbers.
And how long is the distance from ω to a natnumber?
This makes no sense.
It is research.
It is stupidity, as it is "research" that starts from provably false assumptions.
Regards, WM
Sorry, you are just proving your stupidity, and that you are just
totally unqualified to speak on the topics you speak on.
Perhaps someone who relies on your statements will sue you for your
false statements.
Am Thu, 30 Jan 2025 10:07:13 +0100 schrieb WM:
On 29.01.2025 15:04, FromTheRafters wrote:No ordinal.
WM explained on 1/29/2025 :There is something before zero.
On 29.01.2025 10:38, FromTheRafters wrote:Nothing, it is the initial member of the transfinite ordinals. just as
Omega has no immediate predecessor.What is immediately before ω?
zero is the initial member of the finite ordinals.
WM formulated the question :
On 29.01.2025 15:04, FromTheRafters wrote:
WM explained on 1/29/2025 :
On 29.01.2025 10:38, FromTheRafters wrote:
Omega has no immediate predecessor.
What is immediately before ω?
Nothing, it is the initial member of the transfinite ordinals. just
as zero is the initial member of the finite ordinals.
There is something before zero.
Not in the ordinals, which is what you say the context is this time.
And how long is the distance from ω to a natnumber?
This makes no sense.
It is research.
Is there a 'distance' metric in the ordinals?
============================================
AI Overview
No, there is no inherent "distance" metric in ordinal numbers
On 28.01.2025 17:29, Jim Burns wrote:
On 1/28/2025 10:29 AM, WM wrote:
On 28.01.2025 14:10, Jim Burns wrote:
A set such that a larger FISON does NOT exist
is sufficiently large in order for Bob
to disappear purely from swaps within the set.
He disappears from visibility.
There are no swaps into a room,
except for rooms with a later swap.out.
No.
Darkᵂᴹ or visibleᵂᴹ,
there are no swaps into a room,
except for rooms with a later swap.out.
One exception exists: ω-1.
Darkᵂᴹ or visibleᵂᴹ,
there are no swaps into a room,
except for rooms with a later swap.out.
One exception exists: ω-1.
On 1/30/25 4:14 AM, WM wrote:
On 29.01.2025 13:46, Richard Damon wrote:
We can in fact build an infinite set of infinite sets of FISONs whose
union is the set of Natural Numbers,
Do it. Build such a set. I will show that it fails because all FISONs
are useless for reaching the aim.
I did.
On 1/30/25 4:05 AM, WM wrote:
On 29.01.2025 15:00, FromTheRafters wrote:Right, but the "Set of Neccessary FISONs" isn't actually a set,
WM was thinking very hard :
It is Cantor's theorem that every set of ordinals has a first
element. FISONs are v. Neumann ordinals.
Yes, but some infinite sets don't have a first element.
Every finite or infinite set of ordinals or FISONs has a first
element! Other sets are irrelevant in the present context.
A set that doesn't exist doesn't need to have a first element.
On 30.01.2025 11:41, FromTheRafters wrote:There is nothing before ω.
WM formulated the question :Not in the known ordinals. But if anything else, then name it, and how
On 29.01.2025 15:04, FromTheRafters wrote:Not in the ordinals, which is what you say the context is this time.
WM explained on 1/29/2025 :There is something before zero.
On 29.01.2025 10:38, FromTheRafters wrote:Nothing, it is the initial member of the transfinite ordinals. just
Omega has no immediate predecessor.What is immediately before ω?
as zero is the initial member of the finite ordinals.
many of these are there.
Ah, you mean the difference. That can not be defined withoutWrong. The ordinals 7 and 27 have the distance 27 - 7 = 20.Is there a 'distance' metric in the ordinals?It is research.And how long is the distance from ω to a natnumber?This makes no sense.
============================================
AI Overview
No, there is no inherent "distance" metric in ordinal numbers
Am Thu, 30 Jan 2025 20:15:16 +0100 schrieb WM:
On 30.01.2025 11:41, FromTheRafters wrote:
There is nothing before ω.Not in the known ordinals. But if anything else, then name it, and howNot in the ordinals, which is what you say the context is this time.There is something before zero.Nothing, it is the initial member of the transfinite ordinals. justOmega has no immediate predecessor.What is immediately before ω?
as zero is the initial member of the finite ordinals.
many of these are there.
WM explained on 1/30/2025 :
On 30.01.2025 11:41, FromTheRafters wrote:
WM formulated the question :
On 29.01.2025 15:04, FromTheRafters wrote:
WM explained on 1/29/2025 :
On 29.01.2025 10:38, FromTheRafters wrote:
Omega has no immediate predecessor.
What is immediately before ω?
Nothing, it is the initial member of the transfinite ordinals. just
as zero is the initial member of the finite ordinals.
There is something before zero.
Not in the ordinals, which is what you say the context is this time.
Not in the known ordinals. But if anything else, then name it, and how
many of these are there.
And how long is the distance from ω to a natnumber?
This makes no sense.
It is research.
Is there a 'distance' metric in the ordinals?
============================================
AI Overview
No, there is no inherent "distance" metric in ordinal numbers
Wrong. The ordinals 7 and 27 have the distance 27 - 7 = 20.
Finite ordinals are like finite cardinalities.
On 29.01.2025 15:04, FromTheRafters wrote:
WM explained on 1/29/2025 :
On 29.01.2025 10:38, FromTheRafters wrote:
Omega has no immediate predecessor.
What is immediately before ω?
Nothing, it is
the initial member of the transfinite ordinals.
just as zero is
the initial member of the finite ordinals.
There is something before zero.
On 1/30/2025 4:07 AM, WM wrote:
On 29.01.2025 15:04, FromTheRafters wrote:
WM explained on 1/29/2025 :
On 29.01.2025 10:38, FromTheRafters wrote:
Omega has no immediate predecessor.
What is immediately before ω?
Nothing, it is
the initial member of the transfinite ordinals.
just as zero is
the initial member of the finite ordinals.
There is something before zero.
I was right!
You (WM) have negative cardinality:
emptier than {}.
----
I wonder.
If the volume of references to negative cardinality
<waves> becomes large enough,
will Google AI start to answer questions with it,
whatever the quality of those references?
WM explained on 1/30/2025 :
On 30.01.2025 20:54, FromTheRafters wrote:
WM explained on 1/30/2025 :
On 30.01.2025 11:41, FromTheRafters wrote:
WM formulated the question :
On 29.01.2025 15:04, FromTheRafters wrote:
WM explained on 1/29/2025 :
On 29.01.2025 10:38, FromTheRafters wrote:
Omega has no immediate predecessor.
What is immediately before ω?
Nothing, it is the initial member of the transfinite ordinals.
just as zero is the initial member of the finite ordinals.
There is something before zero.
Not in the ordinals, which is what you say the context is this time.
Not in the known ordinals. But if anything else, then name it, and
how many of these are there.
And how long is the distance from ω to a natnumber?
This makes no sense.
It is research.
Is there a 'distance' metric in the ordinals?
============================================
AI Overview
No, there is no inherent "distance" metric in ordinal numbers
Wrong. The ordinals 7 and 27 have the distance 27 - 7 = 20.
Finite ordinals are like finite cardinalities.
Fine. Why do you mention this triviality?
Because you treat ordinals as if they were cardinals.
On 1/29/2025 3:26 AM, WM wrote:
ω-1, if ω-1 existed,
would need to be both growable¹ and shrinkable¹.
However,
in that case,
ω would need to be both growable¹ and shrinkable¹,
ω+1 would need to be both growable¹ and shrinkable¹,
On 1/30/25 4:14 AM, WM wrote:
On 29.01.2025 13:46, Richard Damon wrote:I did.
We can in fact build an infinite set of infinite sets of FISONs whose
union is the set of Natural Numbers,
Do it. Build such a set. I will show that it fails because all FISONs
are useless for reaching the aim.
On 30.01.2025 15:30, Richard Damon wrote:
On 1/30/25 4:14 AM, WM wrote:
On 29.01.2025 13:46, Richard Damon wrote:
We can in fact build an infinite set of infinite sets of FISONs
whose union is the set of Natural Numbers,
Do it. Build such a set. I will show that it fails because all FISONs
are useless for reaching the aim.
I did.
Show the first element that cannot be discarded without changing the union.
Regards, WM
On 30.01.2025 19:44, Jim Burns wrote:
On 1/29/2025 3:26 AM, WM wrote:
ω-1, if ω-1 existed,
would need to be both growable¹ and shrinkable¹.
It is both.
However,
in that case,
ω would need to be both growable¹ and shrinkable¹,
It is both.
ω+1 would need to be both growable¹ and shrinkable¹,
It is both.
On 30.01.2025 21:58, joes wrote:No natural number is immediately before ω, because a natural number
Am Thu, 30 Jan 2025 20:15:16 +0100 schrieb WM:
On 30.01.2025 11:41, FromTheRafters wrote:
There are all natural numbers before ω. Afterwards follows ω orThere is nothing before ω.Not in the known ordinals. But if anything else, then name it, and howNot in the ordinals, which is what you say the context is this time.There is something before zero.Nothing, it is the initial member of the transfinite ordinals. just >>>>>> as zero is the initial member of the finite ordinals.Omega has no immediate predecessor.What is immediately before ω?
many of these are there.
something else between them and ω.
Ordinals aren't guaranteed to increase.
On 1/30/25 2:30 PM, WM wrote:
Do it. Build such a set. I will show that it fails because all
FISONs are useless for reaching the aim.
I did.
Show the first element that cannot be discarded without changing the
union.
Why?
On 1/30/25 2:28 PM, WM wrote:
But a set that has union ℕ is a set.
But since the "Set of Necessary FISONs" isn't actually a "Set"
WM explained :
On 31.01.2025 11:02, FromTheRafters wrote:
Ordinals aren't guaranteed to increase.
What ordinal has not a greater successor?
Ordinals n are well-ordered. The distance from 0 is n - 0 = n.
Starting with omega, they are all countably infinite until you reach the uncountably infinite.
On 30.01.2025 15:30, Richard Damon wrote:Not necessarily. You may add the smallest segment that is not in the
On 1/30/25 4:14 AM, WM wrote:If there is a set with U(F(n)) = ℕ, then it has a first element that is
On 29.01.2025 13:46, Richard Damon wrote:
We can in fact build an infinite set of infinite sets of FISONs whose
union is the set of Natural Numbers,
not completely useless.
But all F(n) can be shown to be completelyAgain: every single one, or even an arbitrary finite number.
useless because infinitely many natnumbers are missing.
Am Thu, 30 Jan 2025 23:32:52 +0100 schrieb WM:
On 30.01.2025 15:30, Richard Damon wrote:Not necessarily.
On 1/30/25 4:14 AM, WM wrote:If there is a set with U(F(n)) = ℕ, then it has a first element that is
On 29.01.2025 13:46, Richard Damon wrote:
We can in fact build an infinite set of infinite sets of FISONs whose >>>>> union is the set of Natural Numbers,
not completely useless.
set does not imply a necessary subset;
But all F(n) can be shown to be completelyAgain: every single one, or even an arbitrary finite number.
useless because infinitely many natnumbers are missing.
If you have inf. many segments, you obviously have inf. many
numbers.
On 31.01.2025 02:02, Richard Damon wrote:It is a sufficient set. The necessary set is empty.
On 1/30/25 2:28 PM, WM wrote:
take the set O of v. Neumann ordinals A(n) that you claim satisfiesBut a set that has union ℕ is a set.But since the "Set of Necessary FISONs" isn't actually a "Set"
U(A(n)) = ℕ.
From this set O every finite subset can be subtracted without changingKey word „finite”.
the result.
Therefore, by induction, no finite A(n) remains.Please formalise.
Therefore the set O has no first ordinal.The original set O of all ordinals has a first (it is also a superset of
On 31.01.2025 11:43, FromTheRafters wrote:
WM explained :
On 31.01.2025 11:02, FromTheRafters wrote:
Ordinals aren't guaranteed to increase.
What ordinal has not a greater successor?
Ordinals n are well-ordered.
The distance from 0 is n - 0 = n.
Starting with omega,
they are all countably infinite
until you reach the uncountably infinite.
ω < ω + 1, < ω + 2 < ... < ω + ω = ω2 < ω2 + 1 < ...
Am Fri, 31 Jan 2025 11:21:58 +0100 schrieb WM:
On 31.01.2025 02:02, Richard Damon wrote:It is a sufficient set. The necessary set is empty.
On 1/30/25 2:28 PM, WM wrote:take the set O of v. Neumann ordinals A(n) that you claim satisfies
But a set that has union ℕ is a set.But since the "Set of Necessary FISONs" isn't actually a "Set"
U(A(n)) = ℕ.
From this set O every finite subset can be subtracted without changingKey word „finite”.
the result.
Therefore, by induction, no finite A(n) remains.Please formalise.
WM formulated on Friday :
On 31.01.2025 11:43, FromTheRafters wrote:
WM explained :
On 31.01.2025 11:02, FromTheRafters wrote:
Ordinals aren't guaranteed to increase.
What ordinal has not a greater successor?
Ordinals n are well-ordered. The distance from 0 is n - 0 = n.
Starting with omega, they are all countably infinite until you reach
the uncountably infinite.
ω < ω + 1, < ω + 2 < ... < ω + ω = ω2 < ω2 + 1 < ...
Each of which is countably infinite.
On 1/31/2025 6:08 AM, WM wrote:
On 31.01.2025 11:43, FromTheRafters wrote:
WM explained :
On 31.01.2025 11:02, FromTheRafters wrote:
Ordinals aren't guaranteed to increase.
What ordinal has not a greater successor?
Ordinals n are well-ordered.
The distance from 0 is n - 0 = n.
Starting with omega,
they are all countably infinite
until you reach the uncountably infinite.
ω < ω + 1, < ω + 2 < ... < ω + ω = ω2 < ω2 + 1 < ...
[0,0) ⊆ [0,1) ⊆ [0,2) ⊆ ... ⊆ [0,ω)
[0,ω) ⊆ [0,ω+1) ⊆ [0,ω+2) ⊆ ... ⊆ [0,ω+ω) =
[0,ω⋅2) ⊆ [0,ω⋅2+1) ⊆ [0,ω⋅2+2) ⊆ ... ⊆ [0,ω⋅3) [0,ω⋅3) ⊆ ... ⊆ [0,ω⋅4) ⊆ ... ⊆ [0,ω⋅ω) =
[0,ω^2) ⊆ ... ⊆ [0,ω^3) ⊆ ... ⊆ [0,ω^ω) =
[0,ω^^2) ⊆ ... ⊆ [0,ω^^3) ⊆ ... ⊆ [0,ω^^ω) =
[0,ω^^^2) ⊆ ... ⊆ [0,ω^^^^2) ⊆ ... ⊆ [0,Ω)
[0,Ω) ⊆ [0,Ω+1) ⊆ [0,Ω+2) ⊆ ... ⊆ [0,Ω+ω)
#[0,0) < #[0,1) < #[0,2) < ... < #[0,ω)
#[0,ω) = #[0,ω+1) = #[0,ω+2) = ... = #[0,ω+ω) =
#[0,ω⋅2) = #[0,ω⋅2+1) = #[0,ω⋅2+2) = ... = #[0,ω⋅3)
#[0,ω⋅3) = ... = #[0,ω⋅4) = ... = #[0,ω⋅ω) =
#[0,ω^2) = ... = #[0,ω^3) = ... = #[0,ω^ω) =
#[0,ω^^2) = ... = #[0,ω^^3) = ... = #[0,ω^^ω) =
#[0,ω^^^2) = ... = #[0,ω^^^^2) = ... = ... < #[0,Ω)
#[0,Ω) = #[0,Ω+1) = #[0,Ω+2) = ... = #[0,Ω+ω)
Requiring #[0,k) < #[0,k+1)
restricts k to the first row < ω
On 31.01.2025 02:02, Richard Damon wrote:
On 1/30/25 2:28 PM, WM wrote:
But a set that has union ℕ is a set.
But since the "Set of Necessary FISONs" isn't actually a "Set"
take the set O of v. Neumann ordinals A(n) that you claim satisfies
U(A(n)) = ℕ.
From this set O every finite subset can be subtracted without changing
the result. Therefore, by induction, no finite A(n) remains. Therefore
the set O has no first ordinal. Therefore it is not a set of ordinals. Therefore your claim is wrong.
Regards, WM
On 31.01.2025 12:31, joes wrote:???
Am Thu, 30 Jan 2025 23:32:52 +0100 schrieb WM:Necessarily, because otherwise all elements can be discarded.
On 30.01.2025 15:30, Richard Damon wrote:Not necessarily.
On 1/30/25 4:14 AM, WM wrote:If there is a set with U(F(n)) = ℕ, then it has a first element that
On 29.01.2025 13:46, Richard Damon wrote:
We can in fact build an infinite set of infinite sets of FISONs
whose union is the set of Natural Numbers,
is not completely useless.
The empty set has no first element.A sufficient set does not imply a necessary subset;If there is no first element necessary,
then all can be discarded,No, only finitely consecutive ones.
All naturals are finite. FISONs are not naturals but sets and can’t beAll finite natural numbers as well as all FISONs obey the Peano axioms.But all F(n) can be shown to be completely useless because infinitelyAgain: every single one, or even an arbitrary finite number.
many natnumbers are missing.
If you have inf. many segments, you obviously have inf. many numbers.
Removing all leaves nothing, in particular no sufficient set forIt is obvious that N is not empty.
U(F(n)) = ℕ.
On 29.01.2025 11:53, joes wrote:Oh, but you showed the opposite:
Am Wed, 29 Jan 2025 09:26:35 +0100 schrieb WM:
The set cannot be empty.But its union cannot be ℕ because a non-empty set of FISONs would beWhy should a necessary set exist?
necessary.
Therefore it has elements. For each element we--
can show that it is not leading to the aim. Therefore no such set
U(F(n)) = ℕ exists.
On 31.01.2025 02:02, Richard Damon wrote:The empty set doesn’t.
On 1/30/25 2:30 PM, WM wrote:
Because every set of ordinals has a first element.Why?Show the first element that cannot be discarded without changing theDo it. Build such a set. I will show that it fails because allI did.
FISONs are useless for reaching the aim.
union.
On 31.01.2025 14:31, Jim Burns wrote:
On 1/31/2025 6:08 AM, WM wrote:
On 31.01.2025 11:43, FromTheRafters wrote:
WM explained :
On 31.01.2025 11:02, FromTheRafters wrote:
Ordinals aren't guaranteed to increase.
What ordinal has not a greater successor?
Ordinals n are well-ordered.
The distance from 0 is n - 0 = n.
Starting with omega,
they are all countably infinite
until you reach the uncountably infinite.
ω < ω + 1, < ω + 2 < ... < ω + ω = ω2 < ω2 + 1 < ...
[0,0) ⊆ [0,1) ⊆ [0,2) ⊆ ... ⊆ [0,ω)
[0,ω) ⊆ [0,ω+1) ⊆ [0,ω+2) ⊆ ... ⊆ [0,ω+ω) =
[0,ω⋅2) ⊆ [0,ω⋅2+1) ⊆ [0,ω⋅2+2) ⊆ ... ⊆ [0,ω⋅3)
[0,ω⋅3) ⊆ ... ⊆ [0,ω⋅4) ⊆ ... ⊆ [0,ω⋅ω) =
[0,ω^2) ⊆ ... ⊆ [0,ω^3) ⊆ ... ⊆ [0,ω^ω) =
[0,ω^^2) ⊆ ... ⊆ [0,ω^^3) ⊆ ... ⊆ [0,ω^^ω) =
[0,ω^^^2) ⊆ ... ⊆ [0,ω^^^^2) ⊆ ... ⊆ [0,Ω)
[0,Ω) ⊆ [0,Ω+1) ⊆ [0,Ω+2) ⊆ ... ⊆ [0,Ω+ω)
#[0,0) < #[0,1) < #[0,2) < ... < #[0,ω)
#[0,ω) = #[0,ω+1) = #[0,ω+2) = ... = #[0,ω+ω) =
#[0,ω⋅2) = #[0,ω⋅2+1) = #[0,ω⋅2+2) = ... = #[0,ω⋅3)
#[0,ω⋅3) = ... = #[0,ω⋅4) = ... = #[0,ω⋅ω) =
#[0,ω^2) = ... = #[0,ω^3) = ... = #[0,ω^ω) =
#[0,ω^^2) = ... = #[0,ω^^3) = ... = #[0,ω^^ω) =
#[0,ω^^^2) = ... = #[0,ω^^^^2) = ... = ... < #[0,Ω)
#[0,Ω) = #[0,Ω+1) = #[0,Ω+2) = ... = #[0,Ω+ω)
Requiring #[0,k) < #[0,k+1)
restricts k to the first row < ω
Satz B.
Jeder Inbegriff von verschiedenen Zahlen
der ersten und zweiten Zahlenklasse hat
eine kleinste Zahl, ein Minimum.
[Cantor, p. 332]
Theorem B:
Every embodiment of different numbers
of the first and the second number class has
a smallest number, a minimum.
On 28.01.2025 11:26, FromTheRafters wrote:LOL. The set missing an element has changed (their union hasn’t).
WM expressed precisely :
No, when it is discarded, nothing changes.An infinite set of FISONs that has the union ℕ needs a first element.It has one defined already, the initial element.
On 29.01.2025 05:07, Richard Damon wrote:Nobody claimed there was a necessary set.
YOU are the one that says there needs to be a "required" FISON.If there is an infinite set, then it has a first element. But for every element it is clear that it is useless to get U(F(n)) = ℕ. Hence the
claim is nonsense.
On 29.01.2025 10:38, FromTheRafters wrote:This is literally senseless. You are mistaken in even assuming
Omega has no immediate predecessor.What is immediately before ω?
And how long is the distance from ω to a natnumber?ω, if you cared to define it as that number k such that n + k = ω.
On 27.01.2025 16:23, FromTheRafters wrote:Yeah, sets of FISONs have a first element.
WM wrote :Here we talk about FISONs.
On 27.01.2025 15:14, joes wrote:The real numbers are an infinite set, which one is first?
Am Mon, 27 Jan 2025 12:35:56 +0100 schrieb WM:Also an infinite set needs a first element.
On 26.01.2025 23:31, Richard Damon wrote:Obviously, but not for infinite sets of FISONs.
you can't show that the first FISON isn't a member of the set ofLogic? The set ℕ exists. And I can show that every FISON is neither >>>>> necessary nor sufficient to accomplish that aim.
necessary FISONs without assuming the set of necessary FISONs
exists.
On 30.01.2025 15:30, Richard Damon wrote:
On 1/30/25 4:14 AM, WM wrote:
On 29.01.2025 13:46, Richard Damon wrote:
We can in fact build
an infinite set of infinite sets of FISONs
whose union is the set of Natural Numbers,
Do it.
Build such a set.
I will show that it fails because
all FISONs are useless for reaching the aim.
I did.
If there is a set with U(F(n)) = ℕ,
then it has a first element that is not completely useless.
But all F(n) can be shown to be completely useless
because infinitely many natnumbers are missing.
Therefore you did not.
You cannot.
Nobody can.
On 27.01.2025 23:07, Jim Burns wrote:Conventionally called N.
On 1/27/2025 7:55 AM, WM wrote:
It is the union of FISONs, ℕ_def.∀n ∈ U(F(n)): |ℕ \ {1, 2, 3, ..., n}| = ℵo ℕ \ {1, 2, 3, ...} = { }What is ⋃{F(n)} ?
More properly called N_undef.What is {1,2,3,...} ?ℕ
On 28.01.2025 10:19, joes wrote:There are no naturals without an initial segment.
Am Tue, 28 Jan 2025 09:58:14 +0100 schrieb WM:But any set of FISONs fails.
On 27.01.2025 18:10, joes wrote:Yes it does. Any infinite set of FISONs has a first element.
Am Mon, 27 Jan 2025 15:33:01 +0100 schrieb WM:But no set of FISONs the union of which is ℕ has a first element.
Also an infinite set needs a first element.No problem, any infinite set of FISONs has one.
No element was claimed to be necessary.When every not necessary or not useful element has been discardedIf an infinite set was existing, you could easily find a first notNo, as you have shown, no element is necessary.
completely useless element.
nothing remains.
WM has brought this to us :
On 31.01.2025 14:12, FromTheRafters wrote:
WM formulated on Friday :
On 31.01.2025 11:43, FromTheRafters wrote:
WM explained :
On 31.01.2025 11:02, FromTheRafters wrote:
Ordinals aren't guaranteed to increase.
What ordinal has not a greater successor?
Ordinals n are well-ordered. The distance from 0 is n - 0 = n.
Starting with omega, they are all countably infinite until you
reach the uncountably infinite.
ω < ω + 1, < ω + 2 < ... < ω + ω = ω2 < ω2 + 1 < ...
Each of which is countably infinite.
That is not of interest. According to Cantor, every set of ordinals
has a smallest element. Each is greater than its predecessor.
What is meant by 'greater' in this context?
On 1/31/25 5:21 AM, WM wrote:
From this set O every finite subset can be subtracted without
changing the result. Therefore, by induction, no finite A(n) remains.
Therefore the set O has no first ordinal. Therefore it is not a set of
ordinals. Therefore your claim is wrong.
Induction doesn't work that way.
There is no requirement that a "minimal" set exists.
Am Fri, 31 Jan 2025 13:28:35 +0100 schrieb WM:
then all can be discarded,No, only finitely consecutive ones.
All naturals are finite. FISONs are not naturals but sets and can’t be added together.All finite natural numbers as well as all FISONs obey the Peano axioms.But all F(n) can be shown to be completely useless because infinitelyAgain: every single one, or even an arbitrary finite number.
many natnumbers are missing.
If you have inf. many segments, you obviously have inf. many numbers.
Removing all leaves nothing, in particular no sufficient set forIt is obvious that N is not empty.
U(F(n)) = ℕ.
Am Fri, 31 Jan 2025 11:23:35 +0100 schrieb WM:
Because every set of ordinals has a first element.The empty set doesn’t.
On 1/30/2025 5:32 PM, WM wrote:
ℕ holds only completely.uselessᵂᴹ numbers.
ℕ isn't what you (WM) think it is.
That's not a logic.problem.
That's a you.problem.
On 31.01.2025 19:32, joes wrote:
Am Tue, 28 Jan 2025 09:54:38 +0100 schrieb WM:
Here we talk about FISONs.Yeah, sets of FISONs have a first element.
Unfortunately no set satisfying U(A(n)) = ℕ has no first element.
Regards, WM
Am Sat, 01 Feb 2025 13:24:31 +0100 schrieb WM:
On 31.01.2025 15:46, Richard Damon wrote:
It works that way: When n belongs to ℕ, then n+1 belongs to ℕ.It works this way: If n can be removed, n AND n+1 can be removed.
And it works also thos way:
When n can be deleted, then n+1 can be deleted. Nothing remains.
But it doesn’t work this way: „If n can be left out, all n can be.”
Am Tue, 28 Jan 2025 09:54:38 +0100 schrieb WM:
Here we talk about FISONs.Yeah, sets of FISONs have a first element.
On 31.01.2025 15:46, Richard Damon wrote:It works this way: If n can be removed, n AND n+1 can be removed.
On 1/31/25 5:21 AM, WM wrote:
It works that way: When n belongs to ℕ, then n+1 belongs to ℕ.From this set O every finite subset can be subtracted withoutInduction doesn't work that way.
changing the result. Therefore, by induction, no finite A(n) remains.
Therefore the set O has no first ordinal. Therefore it is not a set of
ordinals. Therefore your claim is wrong.
And it works also thos way:
When n can be deleted, then n+1 can be deleted. Nothing remains.
The set of all segments is not empty.There is no requirement that a "minimal" set exists.There is the assumption that a set with U(A(n)) = ℕ exists. No element remains. The set does not exist.
On 31.01.2025 18:51, joes wrote:ω, corresponding to „all”, is not natural.
Am Fri, 31 Jan 2025 13:28:35 +0100 schrieb WM:
Induction concerns all natural numbers n as well as all A(n).then all can be discarded,No, only finitely consecutive ones.
There is no FISON for ω (duh).FISONs represent ordinals.All naturals are finite. FISONs are not naturals but sets and can’t beAll finite natural numbers as well as all FISONs obey the PeanoBut all F(n) can be shown to be completely useless becauseAgain: every single one, or even an arbitrary finite number.
infinitely many natnumbers are missing.
If you have inf. many segments, you obviously have inf. many numbers.
axioms.
added together.
Wrong. Finite sets of FISONs do not result in N.But the set claimed to have the union ℕ gets empty without changing its union.Removing all leaves nothing, in particular no sufficient set forIt is obvious that N is not empty.
U(F(n)) = ℕ.
On 31.01.2025 18:52, joes wrote:The set of necessary FISONs has no first element, being empty.
Am Fri, 31 Jan 2025 11:23:35 +0100 schrieb WM:
The empty set contains no ordinals. Every set of ordinals containsBecause every set of ordinals has a first element.The empty set doesn’t.
ordinals. "Jeder Inbegriff von verschiedenen Zahlen der ersten und
zweiten Zahlenklasse hat eine kleinste Zahl, ein Minimum." [Cantor, p.
332]
WM was thinking very hard :
On 31.01.2025 18:52, joes wrote:
Am Fri, 31 Jan 2025 11:23:35 +0100 schrieb WM:The empty set contains no ordinals. Every set of ordinals contains
Because every set of ordinals has a first element.The empty set doesn’t.
ordinals. "Jeder Inbegriff von verschiedenen Zahlen der ersten und
zweiten Zahlenklasse hat eine kleinste Zahl, ein Minimum." [Cantor, p.
332]
The empty set is order type zero. It comes 'before' order type one.
Is 'before' the same as 'smaller' in the infinite sense as it is in
finite sense?
On 31.01.2025 19:32, joes wrote:True.
Am Tue, 28 Jan 2025 09:54:38 +0100 schrieb WM:
Unfortunately no set satisfying U(A(n)) = ℕ has no first element.Here we talk about FISONs.Yeah, sets of FISONs have a first element.
The set of necessary FISONs has no first element, being empty.
Am Sat, 01 Feb 2025 13:46:50 +0100 schrieb WM:
On 31.01.2025 18:51, joes wrote:ω, corresponding to „all”, is not natural.
Am Fri, 31 Jan 2025 13:28:35 +0100 schrieb WM:Induction concerns all natural numbers n as well as all A(n).
then all can be discarded,No, only finitely consecutive ones.
Wrong. Finite sets of FISONs do not result in N.But the set claimed to have the union ℕ gets empty without changing itsRemoving all leaves nothing, in particular no sufficient set forIt is obvious that N is not empty.
U(F(n)) = ℕ.
union.
On 01.02.2025 14:16, joes wrote:Wrong question. You cannot remove infinitely many. „Infinitely many”
Am Sat, 01 Feb 2025 13:24:31 +0100 schrieb WM:
On 31.01.2025 15:46, Richard Damon wrote:
Which one cannot be removed?It works that way: When n belongs to ℕ, then n+1 belongs to ℕ.It works this way: If n can be removed, n AND n+1 can be removed.
And it works also thos way:
When n can be deleted, then n+1 can be deleted. Nothing remains.
But it doesn’t work this way: „If n can be left out, all n can be.”
What is the difference between adding and subtracting?Induction does not transfer to infinities (although it DOES
On 31.01.2025 19:34, Jim Burns wrote:
On 1/30/2025 5:32 PM, WM wrote:
ℕ holds only completely.uselessᵂᴹ numbers.
ℕ isn't what you (WM) think it is.
That's not a logic.problem.
That's a you.problem.
There is the assumption that
a set with U(F(n)) = ℕ exists.
There is the assumption that
a set with U(F(n)) = ℕ exists.
Without changing the union
we can remove every element by induction.
No element remains.
The set does not exist.
On 01.02.2025 15:09, joes wrote:
The set of necessary FISONs has no first element, being empty.The set of FISONs assumed to satisfy U(A(n)) = ℕ cannot be empty. But we can prove by induction that every FISON can be discarded without
changing the union. That disproves the assumption.
Regards, WM
On 01.02.2025 15:07, joes wrote:Induction proves a sentence for every number, not for the set.
Am Sat, 01 Feb 2025 13:46:50 +0100 schrieb WM:ω describes the order type but is not the set ℕ but greater than all natnumbers.
On 31.01.2025 18:51, joes wrote:ω, corresponding to „all”, is not natural.
Am Fri, 31 Jan 2025 13:28:35 +0100 schrieb WM:Induction concerns all natural numbers n as well as all A(n).
then all can be discarded,No, only finitely consecutive ones.
It covers the elements of N, not the set itself.Induction covers all natural numbers. Otherwise it would not beWrong. Finite sets of FISONs do not result in N.But the set claimed to have the union ℕ gets empty without changingRemoving all leaves nothing, in particular no sufficient set forIt is obvious that N is not empty.
U(F(n)) = ℕ.
its union.
sufficient in the Peano axioms.
Am Sat, 01 Feb 2025 14:40:12 +0100 schrieb WM:
On 01.02.2025 14:16, joes wrote:
Am Sat, 01 Feb 2025 13:24:31 +0100 schrieb WM:
Wrong question. You cannot remove infinitely many. „Infinitely many”When n can be deleted, then n+1 can be deleted. Nothing remains.It works this way: If n can be removed, n AND n+1 can be removed.
But it doesn’t work this way: „If n can be left out, all n can be.” >> Which one cannot be removed?
is not a natural number.
Am Sat, 01 Feb 2025 19:23:29 +0100 schrieb WM:
Induction proves a sentence for every number, not for the set.
You cannot extend a sentence about numbers to sets.
It covers the elements of N, not the set itself.Induction covers all natural numbers. Otherwise it would not beWrong. Finite sets of FISONs do not result in N.But the set claimed to have the union ℕ gets empty without changingRemoving all leaves nothing, in particular no sufficient set forIt is obvious that N is not empty.
U(F(n)) = ℕ.
its union.
sufficient in the Peano axioms.
On 2/1/2025 7:56 AM, WM wrote:
There is the assumption that
a set with U(F(n)) = ℕ exists.
Without changing the union
we can remove every element by induction.
No element remains.
The set does not exist.
Each finiteᵒᵘʳ initial segment F(k) of ⋃{F(n)}
can grow¹ to another initial segment F(k+1)
which is also finiteᵒᵘʳ, and is larger than F(k),
and is not larger than ⋃{F(n)}
{F(n}} holds each finiteᵒᵘʳ initial segment F(k)
⋃{F(n)} is larger than each F(k).
On 2/1/25 1:17 PM, WM wrote:
The set of FISONs assumed to satisfy U(A(n)) = ℕ cannot be empty. ButAnd that is your problem, you make assumptions that are unwarented,
we can prove by induction that every FISON can be discarded without
changing the union. That disproves the assumption.
On 02.02.2025 01:33, Richard Damon wrote:
On 2/1/25 1:17 PM, WM wrote:
The set of FISONs assumed to satisfy U(A(n)) = ℕ cannot be empty. ButAnd that is your problem, you make assumptions that are unwarented,
we can prove by induction that every FISON can be discarded without
changing the union. That disproves the assumption.
I think you've got it! Yes, U(A(n)) is not ℕ.
Regards, WM
On 01.02.2025 19:42, joes wrote:ANd it has rules that you don't seem to understand.
Am Sat, 01 Feb 2025 14:40:12 +0100 schrieb WM:
On 01.02.2025 14:16, joes wrote:
Am Sat, 01 Feb 2025 13:24:31 +0100 schrieb WM:
Wrong question. You cannot remove infinitely many. „Infinitely many”When n can be deleted, then n+1 can be deleted. Nothing remains.It works this way: If n can be removed, n AND n+1 can be removed.
But it doesn’t work this way: „If n can be left out, all n can be.” >>> Which one cannot be removed?
is not a natural number.
Mathematical induction is a method for proving that a statement
P(n) is true for every natural number n that is, that the infinitely
many cases P(0),P(1),P(2),P(3),... all hold. [Wikipedia]
Regards, WM
On 02.02.2025 05:55, joes wrote:No, to *a* number, though which is arbitrary. Not to a *set* of numbers.
Am Sat, 01 Feb 2025 19:23:29 +0100 schrieb WM:
Induction proves a sentence for every number, not for the set.But to all numbers.
You cannot extend a sentence about numbers to sets.
N is not a natural number.Peano does not describe the set ℕ?It covers the elements of N, not the set itself.Induction covers all natural numbers. Otherwise it would not beWrong. Finite sets of FISONs do not result in N.But the set claimed to have the union ℕ gets empty without changing >>>>> its union.Removing all leaves nothing, in particular no sufficient set for >>>>>>> U(F(n)) = ℕ.It is obvious that N is not empty.
sufficient in the Peano axioms.
Anyhow all natural numbers n and all A(n) are discarded.No, there are more than any finite number.
On 01.02.2025 20:21, Jim Burns wrote:It’s not an assumption. Hint: F(n+1) = F(n) u n+1 and then
On 2/1/2025 7:56 AM, WM wrote:
There is the assumption that a set with U(F(n)) = ℕ exists.
The union of a nonempty set is not empty.Without changing the union we can remove every element by induction.
No element remains. The set does not exist.
Each finiteᵒᵘʳ initial segment F(k) of ⋃{F(n)} can grow¹ to another >> initial segment F(k+1)
which is also finiteᵒᵘʳ, and is larger than F(k),
and is not larger than ⋃{F(n)}
{F(n}} holds each finiteᵒᵘʳ initial segment F(k) ⋃{F(n)} is larger than
each F(k).
But all F(n) can be discarded without changing the union.
F(1) can be discarded. If F(n) can be discarded, then F(n+1) can be discarded.But P(ω) does not hold.
Note: Mathematical induction is a method for proving that a statement
P(n) is true for every natural number n that is, that the infinitely
many cases P(0),P(1),P(2),P(3),... all hold. [Wikipedia]
Therefore if U(F(n)) = ℕ, then { } = ℕThere are no naturals with infinite segments.
Am Sun, 02 Feb 2025 12:43:49 +0100 schrieb WM:
On 02.02.2025 05:55, joes wrote:No, to *a* number, though which is arbitrary. Not to a *set* of numbers.
Am Sat, 01 Feb 2025 19:23:29 +0100 schrieb WM:But to all numbers.
Induction proves a sentence for every number, not for the set.
You cannot extend a sentence about numbers to sets.
Peano does not describe the set ℕ?N is not a natural number.
Anyhow all natural numbers n and all A(n) are discarded.No, there are more than any finite number.
On 2/2/25 6:34 AM, WM wrote:
On 02.02.2025 01:33, Richard Damon wrote:
On 2/1/25 1:17 PM, WM wrote:
The set of FISONs assumed to satisfy U(A(n)) = ℕ cannot be empty.And that is your problem, you make assumptions that are unwarented,
But we can prove by induction that every FISON can be discarded
without changing the union. That disproves the assumption.
I think you've got it! Yes, U(A(n)) is not ℕ.
But N is in the union of any inifinite set of FISONs.
What isn't, is your A(n), as none of the FISONs are individually needed.
On 01.02.2025 19:42, joes wrote:And you want the wrong P(ω) to hold, but you cannot remove infinitely
Am Sat, 01 Feb 2025 14:40:12 +0100 schrieb WM:
On 01.02.2025 14:16, joes wrote:
Am Sat, 01 Feb 2025 13:24:31 +0100 schrieb WM:
Mathematical induction is a method for proving that a statement P(n) isWrong question. You cannot remove infinitely many. „Infinitely many”When n can be deleted, then n+1 can be deleted. Nothing remains.It works this way: If n can be removed, n AND n+1 can be removed.
But it doesn’t work this way: „If n can be left out, all n can be.” >>> Which one cannot be removed?
is not a natural number.
true for every natural number n that is, that the infinitely many cases P(0),P(1),P(2),P(3),... all hold. [Wikipedia]
On 01.02.2025 20:21, Jim Burns wrote:
On 2/1/2025 7:56 AM, WM wrote:
There is the assumption that
a set with U(F(n)) = ℕ exists.
Without changing the union
we can remove every element by induction.
No element remains.
The set does not exist.
Each finiteᵒᵘʳ initial segment F(k) of ⋃{F(n)}
can grow¹ to another initial segment F(k+1)
which is also finiteᵒᵘʳ, and is larger than F(k),
and is not larger than ⋃{F(n)}
{F(n}} holds each finiteᵒᵘʳ initial segment F(k)
⋃{F(n)} is larger than each F(k).
But all F(n) can be discarded
without changing the union.
F(1) can be discarded.
If F(n) can be discarded, then F(n+1) can be discarded.
Note:
Mathematical induction is a method for proving that
a statement P(n) is true for every natural number n
that is, that
the infinitely many cases P(0),P(1),P(2),P(3),...
all hold. [Wikipedia]
Therefore if U(F(n)) = ℕ, then { } = ℕ
Therefore if U(F(n)) = ℕ, then { } = ℕ
On 02.02.2025 13:27, Richard Damon wrote:
On 2/2/25 6:34 AM, WM wrote:
On 02.02.2025 01:33, Richard Damon wrote:
On 2/1/25 1:17 PM, WM wrote:
The set of FISONs assumed to satisfy U(A(n)) = ℕ cannot be empty.And that is your problem, you make assumptions that are unwarented,
But we can prove by induction that every FISON can be discarded
without changing the union. That disproves the assumption.
I think you've got it! Yes, U(A(n)) is not ℕ.
But N is in the union of any inifinite set of FISONs.
Then the assumption U(A(n)) = ℕ would not be unwarranted.
What isn't, is your A(n), as none of the FISONs are individually needed.
But all can be dropped without changing the union.
Regards, WM
F(1) can be discarded.
If F(n) can be discarded, then F(n+1) can be discarded.
Note:
Mathematical induction is a method for proving that
a statement P(n) is true for every natural number n
that is, that
the infinitely many cases P(0),P(1),P(2),P(3),...
all hold. [Wikipedia]
For each k ∈ ⋃{FISON}:
⋃{FISON} = ⋃({FISON}\{F(k)})
P(k) :⇔ ⋃{FISON}=⋃({FISON}\{F(k)})
For each k ∈ ⋃{FISON}: P(k)
Because
none in linearly.ordered {FISON} is last.
Am Sun, 02 Feb 2025 12:06:26 +0100 schrieb WM:
On 01.02.2025 19:42, joes wrote:
Am Sat, 01 Feb 2025 14:40:12 +0100 schrieb WM:
Mathematical induction is a method for proving that a statement P(n) isAnd you want the wrong P(ω) to hold, but you cannot remove infinitely
true for every natural number n that is, that the infinitely many cases
P(0),P(1),P(2),P(3),... all hold. [Wikipedia]
many segments.
This is different from all P(n) which together say
you can remove any finite *number of* segments (not the segments
themselves).
Do you get that?
There is no infinite segment, but
there *are* infinitely many.
On 2/2/25 11:39 AM, WM wrote:describe the set ℕ?
N is not a natural number.
Not described by the Peano axioms?
You seem to not understand the differnce between a number and a set of numbers
Since I don't recall you actually DEFINING what A(n) means, any
assumption about it would be unwarranted.
German A(n), or English F(n)
is the FISON {1, 2, 3, ..., n}.
The assumption is that U(F(n)) = ℕ.
By induction we prove that every F(n) can be removed without changing
the union. Therefore the assumption leads to { } = ℕ. Therefore the assumption is wrong.
Le 03/02/2025 à 16:11, crank Wolfang Mückenheim aka WM a écrit :
German A(n), or English F(n)
I'm quite sure A(n) is not a German expression, neither is F(n) an English one.
is the FISON {1, 2, 3, ..., n}.
The assumption is that U(F(n)) = ℕ.
By induction we prove that every F(n) can be removed without changing
the union. Therefore the assumption leads to { } = ℕ. Therefore the
assumption is wrong.
Anyone able to claim such a fallacy shouldn't be allowed to be put in front of
students, in any country.
Le 03/02/2025 à 16:11, crank Wolfang Mückenheim aka WM a écrit :
German A(n), or English F(n)
is the FISON {1, 2, 3, ..., n}.
The assumption is that U(F(n)) = ℕ.
By induction we prove that every F(n) can be removed without changing
the union. Therefore the assumption leads to { } = ℕ. Therefore the
assumption is wrong.
Anyone able to claim such a fallacy
On 2/3/2025 7:41 AM, WM wrote:
How can Peano create the complete set by induction?
Peano describes a set with induction.
It is a complete set which is described.
(We don't use any other, "incomplete" sets.)
But the claims are silent about what wasn't described.
Peano describes _the elements_ of ⋃{FISON}
⋃{FISON} isn't an element of ⋃{FISON}
On 02.02.2025 19:23, Jim Burns wrote:
On 2/2/2025 6:25 AM, WM wrote:
F(1) can be discarded.
If F(n) can be discarded, then F(n+1) can be discarded.
Note:
Mathematical induction is a method for proving that
a statement P(n) is true for every natural number n
that is, that
the infinitely many cases P(0),P(1),P(2),P(3),...
all hold. [Wikipedia]
For each k ∈ ⋃{FISON}:
No, for *all* k ∈ ⋃{FISON}.
Peano creates the set ℕ by induction.
I remove the set of FISONs by the same induction:
n ==> n+1.
What us the difference?
⋃{FISON} = ⋃({FISON}\{F(k)})
P(k) :⇔ ⋃{FISON}=⋃({FISON}\{F(k)})
For each k ∈ ⋃{FISON}: P(k)
Because
none in linearly.ordered {FISON} is last.
None of the natural numbers is last.
How can Peano create the complete set by induction?
On 03.02.2025 16:15, Python wrote:
Le 03/02/2025 à 16:11, crank Wolfang Mückenheim aka WM a écrit :
German A(n), or English F(n)
is the FISON {1, 2, 3, ..., n}.
The assumption is that U(F(n)) = ℕ.
By induction we prove that every F(n) can be removed without changing
the union. Therefore the assumption leads to { } = ℕ. Therefore the
assumption is wrong.
Anyone able to claim such a fallacy
You appear far unable to understand this discussion. Every intelligent mathematician understands: All natural numbers created by Peano
induction are subject to induction and can be removed by the same
induction. Matheology needs the wselfcontradictory assumption that the
set ℕ is constructed by induction but cannot be deconstructed by induction.
By the way, have you meanwhile understood why Rennenkampff's example
failed? Hint: First you have to enumerate the euros.
Your "argument" didn't use enumeration. So my objection stands. So does Rennenkampf's.
On 03.02.2025 19:06, Jim Burns wrote:
On 2/3/2025 7:41 AM, WM wrote:
How can Peano create the complete set by induction?
Peano describes a set with induction.
Without axioms
nothing must be used oin formal mathematics.
Therefore Peano, Zermelo, or v. Neumann
create ℕ as well as the set of all FISONs
for use in set theory.
It is a complete set which is described.
(We don't use any other, "incomplete" sets.)
Therefore all FISONs can be removed from
the set of all FISONs.
All natural numbers can be added by induction to a set A.
1 is added to A, and
if n is added to A, then n+1 is added to A.
All FISONs can be subtracted from the set of all FISONs
by the same procedure.
F(1) is subtracted.
If F(n) is subtracted, then F(n+1) is subtracted.
But the claims are silent about what wasn't described.
Peano describes _the elements_ of ⋃{FISON}
⋃{FISON} isn't an element of ⋃{FISON}
Peano creates
the set of all natural numbers as well as
the set of all FISONs.
Le 03/02/2025 à 16:15, Python a écrit :
Le 03/02/2025 à 16:11, crank Wolfang Mückenheim aka WM a écrit :
German A(n), or English F(n)
I'm quite sure A(n) is not a German expression, neither is F(n) an English one.
is the FISON {1, 2, 3, ..., n}.
The assumption is that U(F(n)) = ℕ.
By induction we prove that every F(n) can be removed without changing
the union. Therefore the assumption leads to { } = ℕ. Therefore the
assumption is wrong.
Anyone able to claim such a fallacy shouldn't be allowed to be put in front of
students, in any country.
It is not easy to answer that.
On 02.02.2025 21:11, Richard Damon wrote:
On 2/2/25 11:39 AM, WM wrote:describe the set ℕ?
N is not a natural number.
Not described by the Peano axioms?
You seem to not understand the differnce between a number and a set of
numbers
Peano creates by induction the set ℕ of all natural numbers.
Why doe I not delete by the same induction the set of all FISONs?
Regards, WM
On 02/03/2025 11:48 AM, Jim Burns wrote:
On 2/3/2025 1:36 PM, WM wrote:
On 03.02.2025 19:06, Jim Burns wrote:
On 2/3/2025 7:41 AM, WM wrote:
How can Peano create
the complete set by induction?
Peano describes a set with induction.
Without axioms
nothing must be used oin formal mathematics.
Axioms describe the domain.
Describe a finite ordinal:
⎛ Sets of them are minimummed or empty.
⎜ Each has an immediate predecessor or is zero,
⎜ and each of the priors of each
⎝ has an immediate predecessor of is zero.
That description is the axioms of the finite ordinals.
(There are other ways to describe them.)
Therefore Peano, Zermelo, or v. Neumann
create ℕ as well as the set of all FISONs
for use in set theory.
Axioms describe.
Magic spells create.
Axioms that are not _false_, define a domain.
Others define a mere contingent contrivance.
On 2/3/2025 1:36 PM, WM wrote:
Therefore Peano, Zermelo, or v. Neumann
create ℕ as well as the set of all FISONs
for use in set theory.
Axioms describe.
Magic spells create.
Therefore all FISONs can be removed from
the set of all FISONs.
We can describe the removal of all of them, sic: {}
All natural numbers can be added by induction to a set A.
Either all natural numbers are in A,
or they aren't all in A.
Those are all the choices.
1 is added to A, and
if n is added to A, then n+1 is added to A.
All FISONs can be subtracted from the set of all FISONs
by the same procedure.
F(1) is subtracted.
If F(n) is subtracted, then F(n+1) is subtracted.
For each FISON,
there is a larger FISON not larger than U{FISON}
We could decide that that isn't their behavior,
but, if we decide that, everything turns to gibberish.
On 2/3/25 10:11 AM, WM wrote:
On 02.02.2025 21:11, Richard Damon wrote:
Since I don't recall you actually DEFINING what A(n) means, any
assumption about it would be unwarranted.
German A(n), or English F(n) is the FISON {1, 2, 3, ..., n}.
The assumption is that U(F(n)) = ℕ.
Then why did you change your notation.
By induction we prove that every F(n) can be removed without changing
the union. Therefore the assumption leads to { } = ℕ. Therefore the
assumption is wrong.
No, you prove that ANY FISON can be removed, not that ALL can be.
On 2/3/25 7:45 AM, WM wrote:
On 02.02.2025 21:11, Richard Damon wrote:
On 2/2/25 11:39 AM, WM wrote:describe the set ℕ?
N is not a natural number.
Not described by the Peano axioms?
You seem to not understand the differnce between a number and a set
of numbers
Peano creates by induction the set ℕ of all natural numbers.
Why doe I not delete by the same induction the set of all FISONs?
Because you don't prove the needed induction.
You prove that no individual F(n) is needed.
Note also, Peano doesn't "create" the Naturals with induction,
On 04.02.2025 01:39, Richard Damon wrote:
On 2/3/25 7:45 AM, WM wrote:
On 02.02.2025 21:11, Richard Damon wrote:
On 2/2/25 11:39 AM, WM wrote:describe the set ℕ?
N is not a natural number.
Not described by the Peano axioms?
You seem to not understand the differnce between a number and a set
of numbers
Peano creates by induction the set ℕ of all natural numbers.
Why doe I not delete by the same induction the set of all FISONs?
Because you don't prove the needed induction.
You prove that no individual F(n) is needed.
F(1) can be subtracted, and if F(n) can be subtracted, then F(n+1) can
be subtracted. That is the needed induction.
Note also, Peano doesn't "create" the Naturals with induction,
He does. 1 or 0 ∈ ℕ, and if n is there, then n' is there.
Regards, WM
On 2/4/25 5:50 AM, WM wrote:
You prove that no individual F(n) is needed.
F(1) can be subtracted, and if F(n) can be subtracted, then F(n+1) can
be subtracted. That is the needed induction.
Which just proves that all F(n) are in the set that can be individually
not needed.
Note also, Peano doesn't "create" the Naturals with induction,
He does. 1 or 0 ∈ ℕ, and if n is there, then n' is there.
Which doesn't CREATE the set N, it shows that some other set is N.
On 03.02.2025 20:48, Jim Burns wrote:
On 2/3/2025 1:36 PM, WM wrote:
Therefore Peano, Zermelo, or v. Neumann
create ℕ as well as the set of all FISONs
for use in set theory.
Axioms describe.
Magic spells create.
To describe something
it must be existing.
If ℕ is existing,
we do not need axioms.
If ℕ is existing,
we do not need axioms.
If ℕ is existing,
we do not need axioms.
On 04.02.2025 13:17, Richard Damon wrote:
On 2/4/25 5:50 AM, WM wrote:
You prove that no individual F(n) is needed.
F(1) can be subtracted, and if F(n) can be subtracted, then F(n+1)
can be subtracted. That is the needed induction.
Which just proves that all F(n) are in the set that can be
individually not needed.
Then they all F(n) together can be individually discarded. Just like all natural numbers together can be discarded by subtracting them.
Note also, Peano doesn't "create" the Naturals with induction,
He does. 1 or 0 ∈ ℕ, and if n is there, then n' is there.
Which doesn't CREATE the set N, it shows that some other set is N.
If there is another set, then we could use it without need of Peano.
Regards, WM
On 2/4/2025 5:11 AM, WM wrote:
On 03.02.2025 20:48, Jim Burns wrote:
On 2/3/2025 1:36 PM, WM wrote:
Therefore Peano, Zermelo, or v. Neumann
create ℕ as well as the set of all FISONs
for use in set theory.
Axioms describe.
Magic spells create.
To describe something
it must be existing.
To describe something existing,
it must be existing.
Describing what doesn't exist doesn't create it.
If ℕ is existing,
we do not need axioms.
Describing it did not create it.
If ℕ is existing,
we do not need axioms.
If ℕ is described,
its description can have appended to it
not.first.false claims,
If ℕ is existing,
we do not need axioms.
If the axioms are contradictory,
ℕ is not existing.
On 2/4/25 12:04 PM, WM wrote:
Note also, Peano doesn't "create" the Naturals with induction,
He does. 1 or 0 ∈ ℕ, and if n is there, then n' is there.
Which doesn't CREATE the set N, it shows that some other set is N.
If there is another set, then we could use it without need of Peano.
There are many ways to create the set of Natual Numbers.
It isn't the induction axiom that does it though.
The set of Natural numbers (in Peano) are created by the OTHER axioms,
On 05.02.2025 00:43, Richard Damon wrote:
On 2/4/25 12:04 PM, WM wrote:
Note also, Peano doesn't "create" the Naturals with induction,
He does. 1 or 0 ∈ ℕ, and if n is there, then n' is there.
Which doesn't CREATE the set N, it shows that some other set is N.
If there is another set, then we could use it without need of Peano.
There are many ways to create the set of Natual Numbers.
Peano, Dedekind, Cantor, Zermelo, Schmidt, v. Neumann, Lorenzen did it.
By induction. In all cases "if n ∈ ℕ then n+1 ∈ ℕ" is the fundamental property.
It isn't the induction axiom that does it though.
A proof by induction consists of two cases. The first, the base case,
proves the statement for n = 0 without assuming any knowledge of other
cases. The second case, the induction step, proves that if the statement holds for any given case n = k then it must also hold for the next case
n = k+1. [Wikipedia]
The set of Natural numbers (in Peano) are created by the OTHER axioms,
They are necessary only because Peano uses the clumsy notion of
successor. Nevertheless he fails, because he describes only sequences
like 1, π, π^π, π^π^π, ... Lorenzen for instance does not need any other
axiom than the induction described above.
Regards, WM
On 2/5/2025 4:14 AM, WM wrote:
Either way, the axioms do not create,
but only describe.
On 04.02.2025 18:51, Jim Burns wrote:
On 2/4/2025 5:11 AM, WM wrote:
If ℕ is existing,
we do not need axioms.
If ℕ is described,
its description can have appended to it
not.first.false claims,
Either they did exist or they did not.
If ℕ is existing,
we do not need axioms.
If the axioms are contradictory,
ℕ is not existing.
And if they are not contradictory,
then the ℕ created by them exists.
If ℕ is existing,
we do not need axioms.
Describing it did not create it.
Then nothing further than stating ℕ
would be necessary.
That is a wrong opinion.
On 05.02.2025 18:19, Jim Burns wrote:
On 2/5/2025 4:14 AM, WM wrote:
Either way, the axioms do not create,
but only describe.
The axioms of non-standard analysis or string theory
create.
But that is irrelevant for our topic.
The axioms of Peano, Dedekind, Cantor, Zermelo, Schmidt,
v. Neumann, Lorenzen concern all natural numbers
with no exception by induction.
By the same induction
I can remove all FISONs from U(F(n))
without changing the claimed union.
It might just matter what a natural number, induction,
a FISON, and a union are.
On 05.02.2025 19:43, Jim Burns wrote:
On 2/5/2025 12:32 PM, WM wrote:
On 05.02.2025 18:19, Jim Burns wrote:
Either way, the axioms do not create,
but only describe.
The axioms of
non-standard analysis or string theory
create.
The axioms of
non-standard analysis or string theory
do not create, but only describe.
The axioms of Peano, Dedekind, Cantor,
Zermelo, Schmidt, v. Neumann, Lorenzen
concern all natural numbers
with no exception by induction.
By the same induction
I can remove all FISONs from U(F(n))
without changing the claimed union.
It might just matter what a natural number,
induction, a FISON, and a union are.
The axiom of induction:
∀P( P(1) /\ ∀k(P(k) ==> P(k+1)) ==> ∀n (P(n)))
P(1): U(F(n) \ F(1)) = ℕ.
P(k): U(F(n) \ {F(1), F(2), ..., F(k)}) = ℕ
P(k+1): U(F(n) \ {F(1), F(2), ..., F(k+1)}) = ℕ.
I can remove all FISONs from U(F(n))
without changing the claimed union.
On 05.02.2025 19:43, Jim Burns wrote:
It might just matter what a natural number, induction,
a FISON, and a union are.
The axiom of induction:∀P( P(1) /\ ∀k(P(k) ==> P(k+1)) ==> ∀n (P(n)))
P(1): U(F(n) \ F(1)) = ℕ.
P(k): U(F(n) \ {F(1), F(2), ..., F(k)}) = ℕ
P(k+1): U(F(n) \ {F(1), F(2), ..., F(k+1)}) = ℕ.
Regards, WM
On 2/5/25 2:18 PM, WM wrote:
On 05.02.2025 19:43, Jim Burns wrote:And thus you can claim that no F(n) is "necessary".
It might just matter what a natural number, induction,
a FISON, and a union are.
The axiom of induction:∀P( P(1) /\ ∀k(P(k) ==> P(k+1)) ==> ∀n (P(n))) >>
P(1): U(F(n) \ F(1)) = ℕ.
P(k): U(F(n) \ {F(1), F(2), ..., F(k)}) = ℕ
P(k+1): U(F(n) \ {F(1), F(2), ..., F(k+1)}) = ℕ.
Doesn't mean you can't use a set of them to build the set of Natural
Numbers.
On 2/5/25 12:32 PM, WM wrote:
On 05.02.2025 18:19, Jim Burns wrote:
On 2/5/2025 4:14 AM, WM wrote:
Either way, the axioms do not create,
but only describe.
The axioms of non-standard analysis or string theory create.
But that is irrelevant for our topic. The axioms of Peano, Dedekind,
Cantor, Zermelo, Schmidt, v. Neumann, Lorenzen concern all natural
numbers with no exception by induction. By the same induction I can
remove all FISONs from U(F(n)) without changing the claimed union.
Indiction is only ONE of the axioms,
On 2/5/2025 2:18 PM, WM wrote:
The axiom of induction:
∀P( P(1) /\ ∀k(P(k) ==> P(k+1)) ==> ∀n (P(n)))
P(1): U(F(n) \ F(1)) = ℕ.
P(k): U(F(n) \ {F(1), F(2), ..., F(k)}) = ℕ
P(k+1): U(F(n) \ {F(1), F(2), ..., F(k+1)}) = ℕ.
A description, not a magic spell.
...which could also be written...
I can remove all FISONs from U(F(n))
without changing the claimed union.
What you (WM) mean by 'remove all...without changing..."
is that, by induction,
On 06.02.2025 01:46, Jim Burns wrote:
On 2/5/2025 2:18 PM, WM wrote:
The axiom of induction:
∀P( P(1) /\ ∀k(P(k) ==> P(k+1)) ==> ∀n (P(n)))
P(1): U(F(n) \ F(1)) = ℕ.
P(k): U(F(n) \ {F(1), F(2), ..., F(k)}) = ℕ
P(k+1): U(F(n) \ {F(1), F(2), ..., F(k+1)}) = ℕ.
A description, not a magic spell.
...which could also be written...
Why should we use two different writings?
I can remove all FISONs from U(F(n))
without changing the claimed union.
What you (WM) mean by 'remove all...without changing..."
is that, by induction,
it is proved that all F(n) can be subtracted like by induction all
natural numbers can be subtracted from ℕ:
{1} can be subtracted because it is an element of in ℕ.
If {n} has been subtracted, then {n+1} can be subtracted because it is
an element of in ℕ.
By the axiom of induction the result is the empty set.
Regards, WM
On 06.02.2025 03:29, Richard Damon wrote:
On 2/5/25 2:18 PM, WM wrote:
On 05.02.2025 19:43, Jim Burns wrote:And thus you can claim that no F(n) is "necessary".
It might just matter what a natural number, induction,
a FISON, and a union are.
The axiom of induction:∀P( P(1) /\ ∀k(P(k) ==> P(k+1)) ==> ∀n (P(n))) >>>
P(1): U(F(n) \ F(1)) = ℕ.
P(k): U(F(n) \ {F(1), F(2), ..., F(k)}) = ℕ
P(k+1): U(F(n) \ {F(1), F(2), ..., F(k+1)}) = ℕ.
My proof is this: IF U(F(n)) = ℕ, THEN U({ }) = { } = ℕ.
Doesn't mean you can't use a set of them to build the set of Natural
Numbers.
It means precisely that. The premise is wrong because { } = ℕ is wrong.
Regards, WM
On 06.02.2025 03:29, Richard Damon wrote:
On 2/5/25 12:32 PM, WM wrote:
On 05.02.2025 18:19, Jim Burns wrote:
On 2/5/2025 4:14 AM, WM wrote:
Either way, the axioms do not create,
but only describe.
The axioms of non-standard analysis or string theory create.
But that is irrelevant for our topic. The axioms of Peano, Dedekind,
Cantor, Zermelo, Schmidt, v. Neumann, Lorenzen concern all natural
numbers with no exception by induction. By the same induction I can
remove all FISONs from U(F(n)) without changing the claimed union.
Indiction is only ONE of the axioms,
Peano needs more axioms because his approach is very clumsy and dysfunctional. He creates sequences without repetitions - nothing else,
let alone numbers. He creates quack, quackquick, quackquickquack, ...
Lorenzen uses: Make a stroke and if you have made x strokes make another stroke. Nothing else than this induction is required to get the natural numbers in the unary system.
Regards, WM
On 06.02.2025 01:46, Jim Burns wrote:
On 2/5/2025 2:18 PM, WM wrote:
On 05.02.2025 19:43, Jim Burns wrote:
On 2/5/2025 12:32 PM, WM wrote:
On 05.02.2025 18:19, Jim Burns wrote:
Either way, the axioms do not create,
but only describe.
The axioms of
non-standard analysis or string theory
create.
The axioms of
non-standard analysis or string theory
do not create, but only describe.
The axiom of induction:
A description, not a magic spell.
...which could also be written...
Why should we use two different writings?
I can remove all FISONs from U(F(n))
without changing the claimed union.
What you (WM) mean by 'remove all...without changing..."
is that, by induction,
it is proved that all F(n) can be subtracted
like
by induction
all natural numbers can be subtracted from ℕ:
{1} can be subtracted
because it is an element of in ℕ.
If {n} has been subtracted,
then {n+1} can be subtracted
because it is an element of in ℕ.
By the axiom of induction
the result is the empty set.
On 2/6/2025 3:54 AM, WM wrote:
The axiom of induction:
∀P:ℕ₁→{⊤,⊥}:
The key is that ∀ᴺ¹n: ∃ᴺ¹j′: n<j′
By the axiom of induction
the result is the empty set.
You (WM) agree that ∀ᴺ¹n: ∃ᴺ¹j′: n<j′
On 06.02.2025 15:57, Jim Burns wrote:
The key is that ∀ᴺ¹n: ∃ᴺ¹j′: n<j′
The key is that
the set ℕ is created by induction.
If the set M is described as the smallest set satisfying
1 ∈ M and n ∈ M ==> n+1 ∈ M
then ℕ\M = Ø.
There is nothing remaining.
By the axiom of induction
the result is the empty set.
You (WM) agree that ∀ᴺ¹n: ∃ᴺ¹j′: n<j′
No. I agree to what I wrote.
The axiom of induction:
∀P:ℕ₁→{⊤,⊥}:
That appears like nonsense.
I prefer Wikipedia:
∀P( P(1) /\ ∀k(P(k) ==> P(k+1)) ==> ∀n (P(n)).
On 2/6/2025 11:55 AM, WM wrote:
On 06.02.2025 15:57, Jim Burns wrote:
The key is that ∀ᴺ¹n: ∃ᴺ¹j′: n<j′
The key is that
the set ℕ is created by induction.
The set ℕ₁ is described as having induction valid for it.
Sets missing natural numbers and
sets with extra, non.inducible, un.natural numbers
are not ℕ₁
If the set M is described as the smallest set satisfying
1 ∈ M and n ∈ M ==> n+1 ∈ M
then ℕ\M = Ø.
ℕ₁ = ∅ satisfies that definition.
Better:
ℕ₁ is the emptiest set M such that
1 ∈ M and n ∈ M ⇒ n+1 ∈ M
Thus:
1 ∈ ℕ₁ and n ∈ ℕ₁ ⇒ n+1 ∈ ℕ₁
∀P:(1 ∈ P and n ∈ P ⇒ n+1 ∈ P) ⇒ ℕ₁ ⊆ P
Is ℕ₁ the emptiest set M such that
1 ∈ M and n ∈ M ⇒ n+1 ∈ M ?
I prefer Wikipedia:
∀P( P(1) /\ ∀k(P(k) ==> P(k+1)) ==> ∀n (P(n)).
That's intended to be part of the definition of ℕ₁
Which is curious, when one considers that
ℕ₁ appears nowhere in it.
On 06.02.2025 19:54, Jim Burns wrote:No. N is exactly the set of all, and only the natural numbers.
On 2/6/2025 11:55 AM, WM wrote:Then it is the collection ℕ_def of definable numbers.
On 06.02.2025 15:57, Jim Burns wrote:The set ℕ₁ is described as having induction valid for it.
The key is that ∀ᴺ¹n: ∃ᴺ¹j′: n<j′The key is that the set ℕ is created by induction.
Then it is the set ℕ of all natural numbers.
Sets missing natural numbers and sets with extra, non.inducible,
un.natural numbers are not ℕ₁
You contradict yourself.
Tautologically.Which is curious, when one considers that ℕ₁ appears nowhere in it.The axiom of induction holds for all predicates P which satisfy
induction.
If the set M is described as the smallest set satisfying F(1) ∈ M andWrong. M cannot be finite.
F(n) ∈ M ==> F(n+1) ∈ M then M contains all FISONs which can be subtracted from U(Fn)) without changing the assumed result ℕ.
N is exactly the set of all, and only the natural numbers.
It does not matter that you call it N_def.
F(n+1) ∈ M . This is an infinite set containing all FISONs.
Your extension of
that set is unclear, since you have not provided any axioms.
If the set M is described as the smallest set satisfying F(1) ∈ M andWrong. M cannot be finite.
F(n) ∈ M ==> F(n+1) ∈ M then M contains all FISONs which can be
subtracted from U(Fn)) without changing the assumed result ℕ.
On 07.02.2025 09:59, joes wrote:
N is exactly the set of all, and only the natural numbers.
It does not matter that you call it N_def.
It is described by the axioms 1 ∈ ℕ and n ∈ ℕ ==> n+1 ∈ ℕ. This is an
infinite set containing all natnumbers.
The set M of all FISONs is described by the axioms F(1) ∈ M and F(n) ∈ M
F(n+1) ∈ M . This is an infinite set containing all FISONs.
Your extension of
that set is unclear, since you have not provided any axioms.
Cantor claims that *a fixed quantity greater than all natural numbers* exists.
An increasing sequence of natural numbers is never greater than all
natural numbers.
If the set M is described as the smallest set satisfying F(1) ∈ M andWrong. M cannot be finite.
F(n) ∈ M ==> F(n+1) ∈ M then M contains all FISONs which can be
subtracted from U(Fn)) without changing the assumed result ℕ.
Inductive sets are infinite sets, according to set theory.
Regards, WM
On 06.02.2025 19:54, Jim Burns wrote:
On 2/6/2025 11:55 AM, WM wrote:
If the set M is described as
the smallest set satisfying
1 ∈ M and n ∈ M ==> n+1 ∈ M
then ℕ\M = Ø.
ℕ₁ = ∅ satisfies that definition.
No. 1 is not in ∅,
Better:
ℕ₁ is the emptiest set M such that
1 ∈ M and n ∈ M ⇒ n+1 ∈ M
Thus:
1 ∈ ℕ₁ and n ∈ ℕ₁ ⇒ n+1 ∈ ℕ₁
∀P:(1 ∈ P and n ∈ P ⇒ n+1 ∈ P) ⇒ ℕ₁ ⊆ P
Is ℕ₁ the emptiest set M such that
1 ∈ M and n ∈ M ⇒ n+1 ∈ M ?
Relevant is the set of FISONs.
I prefer Wikipedia:
∀P( P(1) /\ ∀k(P(k) ==> P(k+1)) ==> ∀n (P(n)).
That's intended to be part of the definition of ℕ₁
As well it is
the definition of the collection of all FISONs.
Which is curious, when one considers that
ℕ₁ appears nowhere in it.
The axiom of induction holds for
all predicates P which satisfy induction.
If the set M is described as the smallest set satisfying
F(1) ∈ M and F(n) ∈ M ==> F(n+1) ∈ M
then M contains
all FISONs which can be subtracted from U(Fn))
without changing the assumed result ℕ.
On 2/6/2025 2:32 PM, WM wrote:
I prefer Wikipedia:
∀P (P(1) /\ ∀k(P(k) ==> P(k+1)) ==> ∀n (P(n)).
That's intended to be part of the definition of ℕ₁
As well it is
the definition of the collection of all FISONs.
Which is curious, when one considers that
the collection of all FISONs appears nowhere in it.
If the set M is described as the smallest set satisfying
F(1) ∈ M and F(n) ∈ M ==> F(n+1) ∈ M
then M contains
all FISONs which can be subtracted from U(Fn))
without changing the assumed result ℕ.
Also, any superset of (emptiest) M
contains at least what M contains, and
thus also contains all FISONs.
On 07.02.2025 17:10, Jim Burns wrote:
On 2/6/2025 2:32 PM, WM wrote:
On 06.02.2025 19:54, Jim Burns wrote:
On 2/6/2025 11:55 AM, WM wrote:
I prefer Wikipedia:
∀P (P(1) /\ ∀k(P(k) ==> P(k+1)) ==> ∀n (P(n)).
That's intended to be part of the definition of ℕ₁
As well it is
the definition of the collection of all FISONs.
Which is curious, when one considers that
the collection of all FISONs appears nowhere in it.
The axiom of induction says:
If any property or predicate P satifies
(P(1) /\ ∀k(P(k) ==> P(k+1)),
then it describes all elements of
an inductive = infinite set.
That is satisfied by the set M of all FISONs
which are useless in U(A(n)) = ℕ.
On 07.02.2025 17:10, Jim Burns wrote:
On 2/6/2025 2:32 PM, WM wrote:
I prefer Wikipedia:
∀P (P(1) /\ ∀k(P(k) ==> P(k+1)) ==> ∀n (P(n)).
That's intended to be part of the definition of ℕ₁
As well it is
the definition of the collection of all FISONs.
Which is curious, when one considers that
the collection of all FISONs appears nowhere in it.
The axiom of induction says: If any property or predicate P satifies
(P(1) /\ ∀k(P(k) ==> P(k+1)), then it describes all elements of an inductive = infinite set. That is satisfied by the set M of all FISONs
which are useless in U(A(n)) = ℕ.
If the set M is described as the smallest set satisfying
F(1) ∈ M and F(n) ∈ M ==> F(n+1) ∈ M
then M contains
all FISONs which can be subtracted from U(Fn))
without changing the assumed result ℕ.
Also, any superset of (emptiest) M
contains at least what M contains, and
thus also contains all FISONs.
But only all FISONs can be discarded.
Regards, WM
On 2/7/25 11:39 AM, WM wrote:
On 07.02.2025 17:10, Jim Burns wrote:
On 2/6/2025 2:32 PM, WM wrote:
I prefer Wikipedia:
∀P (P(1) /\ ∀k(P(k) ==> P(k+1)) ==> ∀n (P(n)).
That's intended to be part of the definition of ℕ₁
As well it is
the definition of the collection of all FISONs.
Which is curious, when one considers that
the collection of all FISONs appears nowhere in it.
The axiom of induction says: If any property or predicate P satifies
(P(1) /\ ∀k(P(k) ==> P(k+1)), then it describes all elements of an
inductive = infinite set. That is satisfied by the set M of all FISONs
which are useless in U(A(n)) = ℕ.
So all elements are individually not needed.
On 2/7/2025 11:39 AM, WM wrote:
On 07.02.2025 17:10, Jim Burns wrote:
On 2/6/2025 2:32 PM, WM wrote:
On 06.02.2025 19:54, Jim Burns wrote:
On 2/6/2025 11:55 AM, WM wrote:
I prefer Wikipedia:
∀P (P(1) /\ ∀k(P(k) ==> P(k+1)) ==> ∀n (P(n)).
That's intended to be part of the definition of ℕ₁
As well it is
the definition of the collection of all FISONs.
Which is curious, when one considers that
the collection of all FISONs appears nowhere in it.
The axiom of induction says:
If any property or predicate P satifies
(P(1) /\ ∀k(P(k) ==> P(k+1)),
then it describes all elements of
an inductive = infinite set.
All inductive sets are infinite.
Not all infinite sets are inductive.
That is satisfied by the set M of all FISONs which are useless in
U(A(n)) = ℕ.
Only not.followed FISONs are not (your term) uselessᵂᴹ.
Each FISON is followed.
On 08.02.2025 01:14, Richard Damon wrote:
On 2/7/25 11:39 AM, WM wrote:
On 07.02.2025 17:10, Jim Burns wrote:
On 2/6/2025 2:32 PM, WM wrote:
I prefer Wikipedia:
∀P (P(1) /\ ∀k(P(k) ==> P(k+1)) ==> ∀n (P(n)).
That's intended to be part of the definition of ℕ₁
As well it is
the definition of the collection of all FISONs.
Which is curious, when one considers that
the collection of all FISONs appears nowhere in it.
The axiom of induction says: If any property or predicate P satifies
(P(1) /\ ∀k(P(k) ==> P(k+1)), then it describes all elements of an
inductive = infinite set. That is satisfied by the set M of all
FISONs which are useless in U(A(n)) = ℕ.
So all elements are individually not needed.
All not needed elements belong to an inductive set. Inductive sets have
no last element. Therefore the set of FISONs which could have the union
ℕ has no first element. That means it is empty.
Therefore U(A(n)) = ℕ ==> U{ } = { } = ℕ. This is false. By contraposition we get ~{ } = ℕ ==> ~ U(A(n)) = ℕ.
Regards, WM
On 07.02.2025 22:45, Jim Burns wrote:
On 2/7/2025 11:39 AM, WM wrote:
On 07.02.2025 17:10, Jim Burns wrote:
On 2/6/2025 2:32 PM, WM wrote:
On 06.02.2025 19:54, Jim Burns wrote:
On 2/6/2025 11:55 AM, WM wrote:
I prefer Wikipedia:
∀P (P(1) /\ ∀k(P(k) ==> P(k+1)) ==> ∀n (P(n)).
That's intended to be part of the definition of ℕ₁
As well it is
the definition of the collection of all FISONs.
Which is curious, when one considers that
the collection of all FISONs appears nowhere in it.
The axiom of induction says:
If any property or predicate P satifies
(P(1) /\ ∀k(P(k) ==> P(k+1)),
then it describes all elements of
an inductive = infinite set.
All inductive sets are infinite.
Not all infinite sets are inductive.
Irrelevant.
That is satisfied by
the set M of all FISONs which
are useless in U(A(n)) = ℕ.
Only not.followed FISONs are not (your term) uselessᵂᴹ.
Each FISON is followed.
Therefore all are useless.
All not needed elements belong to an inductive set.
Inductive sets have no last element.
Therefore
the set of FISONs which
could have the union ℕ
has no first element. That means it is empty.
Therefore U(A(n)) = ℕ
Therefore U(A(n)) = ℕ ==> U{ } = { } = ℕ.
This is false.
By contraposition we get
~{ } = ℕ ==> ~ U(A(n)) = ℕ.
On 08.02.2025 01:14, Richard Damon wrote:All FISONs do.
On 2/7/25 11:39 AM, WM wrote:
On 07.02.2025 17:10, Jim Burns wrote:
On 2/6/2025 2:32 PM, WM wrote:
I prefer Wikipedia:
∀P (P(1) /\ ∀k(P(k) ==> P(k+1)) ==> ∀n (P(n)).
That's intended to be part of the definition of ℕ₁
As well it is the definition of the collection of all FISONs.
Which is curious, when one considers that the collection of all
FISONs appears nowhere in it.
The axiom of induction says: If any property or predicate P satifies
(P(1) /\ ∀k(P(k) ==> P(k+1)), then it describes all elements of an
inductive = infinite set. That is satisfied by the set M of all FISONs
which are useless in U(A(n)) = ℕ.
So all elements are individually not needed.
All not needed elements belong to an inductive set.
Therefore the set of FISONs which could have the unionWTF? The first element of set of all FISONs is A(0) = {0}.
ℕ has no first element.
On 2/8/2025 5:35 AM, WM wrote:
On 07.02.2025 22:45, Jim Burns wrote:
On 2/7/2025 11:39 AM, WM wrote:
On 07.02.2025 17:10, Jim Burns wrote:
On 2/6/2025 2:32 PM, WM wrote:
On 06.02.2025 19:54, Jim Burns wrote:
On 2/6/2025 11:55 AM, WM wrote:
I prefer Wikipedia:
∀P (P(1) /\ ∀k(P(k) ==> P(k+1)) ==> ∀n (P(n)).
That's intended to be part of the definition of ℕ₁
As well it is
the definition of the collection of all FISONs.
Which is curious, when one considers that
the collection of all FISONs appears nowhere in it.
The axiom of induction says:
If any property or predicate P satifies
(P(1) /\ ∀k(P(k) ==> P(k+1)),
then it describes all elements of
an inductive = infinite set.
All inductive sets are infinite.
Not all infinite sets are inductive.
Irrelevant.
inductive ≠ infinite
inductive ≠ valid.for.induction
(despite the name)
minimal.inductive = valid.for.induction
That is satisfied by
the set M of all FISONs which
are useless in U(A(n)) = ℕ.
Only not.followed FISONs are not (your term) uselessᵂᴹ.
Each FISON is followed.
Therefore all are useless.
All not needed elements belong to an inductive set.
...which is also
a minimal.inductive = valid.for.induction set.
Inductive sets have no last element.
Yes.
inductive 𝕎 ⇒ ∀ᵂj ∃ᵂk j<k
Therefore
the set of FISONs which
could have the union ℕ
FISON F(k) = {i:i≤k}
Set of FISONs {{i:i≤k}:k}
Set of sets 𝒜 of FISONs
{𝒜⊆{{i:i≤k}:k}}
Set of sets 𝒜 of FISONs with union ℕ
{𝒜⊆{{i:i≤k}:k}:⋃𝒜=ℕ}
Set of FISONs common to sets with union ℕ ⋂{𝒜⊆{{i:i≤k}:k}:⋃𝒜=ℕ}
Is
⋂{𝒜⊆{{i:i≤k}:k}:⋃𝒜=ℕ}
a set of FISONs with union ℕ
?
Is
⋂{𝒜⊆{{i:i≤k}:k}:⋃𝒜=ℕ} ∈ {𝒜⊆{{i:i≤k}:k}:⋃𝒜=ℕ} ?
If you say yes, how do you know?
Union of FISONs common to sets with union ℕ ⋃⋂{𝒜⊆{{i:i≤k}:k}:⋃𝒜=ℕ}
has no first element. That means it is empty.
Therefore U(A(n)) = ℕ
Is
⋃⋂{𝒜⊆{{i:i≤k}:k}:⋃𝒜=ℕ} = ℕ
?
Therefore U(A(n)) = ℕ ==> U{ } = { } = ℕ.
This is false.
By contraposition we get
~{ } = ℕ ==> ~ U(A(n)) = ℕ.
And thus you claim that 36 can not be factored, as all of its factors
are not needed.
Am Sat, 08 Feb 2025 11:27:49 +0100 schrieb WM:
On 08.02.2025 01:14, Richard Damon wrote:All FISONs do.
On 2/7/25 11:39 AM, WM wrote:
On 07.02.2025 17:10, Jim Burns wrote:
On 2/6/2025 2:32 PM, WM wrote:
I prefer Wikipedia:
∀P (P(1) /\ ∀k(P(k) ==> P(k+1)) ==> ∀n (P(n)).
That's intended to be part of the definition of ℕ₁
As well it is the definition of the collection of all FISONs.
Which is curious, when one considers that the collection of all
FISONs appears nowhere in it.
The axiom of induction says: If any property or predicate P satifies
(P(1) /\ ∀k(P(k) ==> P(k+1)), then it describes all elements of an
inductive = infinite set. That is satisfied by the set M of all FISONs >>>> which are useless in U(A(n)) = ℕ.
So all elements are individually not needed.
All not needed elements belong to an inductive set.
Therefore the set of FISONs which could have the unionWTF? The first element of set of all FISONs is A(0) = {0}.
ℕ has no first element.
Do you use ω-1 ?
On 08.02.2025 12:51, Richard Damon wrote:
And thus you claim that 36 can not be factored, as all of its factors
are not needed.
1, 2, 3, 4, 6, 9, 12, 18, 36.
According to Cantor, the set of factors has a smallest element, 1, and
the set of necessary factors has a smallest element, 6, or if double application is not allowed, 4.
Regards, WM
On 08.02.2025 12:51, Richard Damon wrote:
And thus you claim that 36 can not be factored,
as all of its factors are not needed.
1, 2, 3, 4, 6, 9, 12, 18, 36.
According to Cantor,
the set of factors has a smallest element,
the set of factors has a smallest element, 1, and
the set of necessary factors has a smallest element, 6,
or if double application is not allowed, 4.
Regards, WM
On 08.02.2025 18:43, Jim Burns wrote:
On 2/8/2025 9:21 AM, Jim Burns wrote:
On 2/8/2025 5:35 AM, WM wrote:
Inductive sets have no last element.
Do you use ω-1 ?
Not in this proof.:
The axiom of induction says:
If any property or predicate P satifies
(P(1) /\ ∀k(P(k) ==> P(k+1)),
then it describes all elements of an
inductive = infinite set.
That is satisfied by the set M of all FISONs
which are useless in U(A(n)) = ℕ.
Inductive sets have no last element.
Therefore U(A(n)) = ℕ ==>
U{ } = { } = ℕ.
On 2/8/2025 2:44 PM, WM wrote:
On 08.02.2025 12:51, Richard Damon wrote:
And thus you claim that 36 can not be factored,
as all of its factors are not needed.
1, 2, 3, 4, 6, 9, 12, 18, 36.
According to Cantor,
the set of factors has a smallest element,
Also, according to Pythagoras.
the set of factors has a smallest element, 1, and
the set of necessary factors has a smallest element, 6,
or if double application is not allowed, 4.
What is an unnecessary factor?
On 2/8/25 2:44 PM, WM wrote:
On 08.02.2025 12:51, Richard Damon wrote:
And thus you claim that 36 can not be factored, as all of its factors
are not needed.
1, 2, 3, 4, 6, 9, 12, 18, 36.
According to Cantor, the set of factors has a smallest element, 1, and
the set of necessary factors has a smallest element, 6, or if double
application is not allowed, 4.
And which of those was "necessary"?
I don't need 6, because I can factor into 4 * 9
The problem is your recursion on FISONs can't complete
On 2/8/2025 4:54 PM, WM wrote:
The axiom of induction says:
If any property or predicate P satisfies
(P(1) /\ ∀k(P(k) ==> P(k+1)),
then it describes all elements of an inductive = infinite set.
Not for all inductive sets.
For all minimal.inductive sets.
minimal.inductive ≠ inductive ≠ infinite
That is satisfied by the set M of all FISONs which are useless in
U(A(n)) = ℕ.
Without exception,
the union of FISONs.after is ℕ
Therefore U(F(n)) = ℕ ==>
U{ } = { } = ℕ.
Why that '==>' ?
My best guess at why you claim U{ } = { } = ℕ
is that you (WM) are assuming that,
for some FISON (ie, F(ω-1)) such that
there are no FISONs.after.
You (WM) haven't given any other reason.
On 08.02.2025 18:43, Jim Burns wrote:Caveat: P is *not* satisfied by the set of all k, only by its elements.
Do you use ω-1 ?
Not in this proof.:
The axiom of induction says: If any property or predicate P satifies
(P(1) /\ ∀k(P(k) ==> P(k+1)), then it describes all elements of an inductive = infinite set. That is satisfied by the set M of all FISONs
which are useless in U(A(n)) = ℕ.
On 08.02.2025 23:28, Richard Damon wrote:
On 2/8/25 2:44 PM, WM wrote:
On 08.02.2025 12:51, Richard Damon wrote:
And thus you claim that 36 can not be factored, as all of its
factors are not needed.
1, 2, 3, 4, 6, 9, 12, 18, 36.
According to Cantor, the set of factors has a smallest element, 1,
and the set of necessary factors has a smallest element, 6, or if
double application is not allowed, 4.
And which of those was "necessary"?
I don't need 6, because I can factor into 4 * 9
4 is not necessary because it is smaller than 6, and 6 is sufficient.
The problem is your recursion on FISONs can't complete
Try to understand Cantor's theorem.
Regards, WM
Am Sat, 08 Feb 2025 22:54:46 +0100 schrieb WM:
On 08.02.2025 18:43, Jim Burns wrote:Caveat: P is *not* satisfied by the set of all k, only by its elements.
Do you use ω-1 ?
Not in this proof.:
The axiom of induction says: If any property or predicate P satifies
(P(1) /\ ∀k(P(k) ==> P(k+1)), then it describes all elements of an
inductive = infinite set. That is satisfied by the set M of all FISONs
which are useless in U(A(n)) = ℕ.
On 2/9/25 5:46 AM, WM wrote:
On 08.02.2025 23:28, Richard Damon wrote:
On 2/8/25 2:44 PM, WM wrote:
On 08.02.2025 12:51, Richard Damon wrote:
And thus you claim that 36 can not be factored, as all of its
factors are not needed.
1, 2, 3, 4, 6, 9, 12, 18, 36.
According to Cantor, the set of factors has a smallest element, 1,
and the set of necessary factors has a smallest element, 6, or if
double application is not allowed, 4.
And which of those was "necessary"?
I don't need 6, because I can factor into 4 * 9
4 is not necessary because it is smaller than 6, and 6 is sufficient.
But then your "necessary" has a application order, which the word
doesn't have.
On 09.02.2025 13:34, joes wrote:
Am Sat, 08 Feb 2025 22:54:46 +0100 schrieb WM:
On 08.02.2025 18:43, Jim Burns wrote:Caveat: P is *not* satisfied by the set of all k, only by its elements.
Do you use ω-1 ?
Not in this proof.:
The axiom of induction says: If any property or predicate P satifies
(P(1) /\ ∀k(P(k) ==> P(k+1)), then it describes all elements of an
inductive = infinite set. That is satisfied by the set M of all FISONs
which are useless in U(A(n)) = ℕ.
Peano is not satisfied by the set of all k either. But no exceptioncan
be identified.
Regards, WM
On 09.02.2025 01:00, Jim Burns wrote:
On 2/8/2025 4:54 PM, WM wrote:
That is satisfied by the set M of
all FISONs which are useless in
U(A(n)) = ℕ.
Without exception,
the union of FISONs.after is ℕ
No.
Therefore U(F(n)) = ℕ ==>
U{ } = { } = ℕ.
Why that '==>' ?
From the assumption U(F(n)) = ℕ
I have derived that { } = ℕ.
My best guess at why you claim U{ } = { } = ℕ
is that you (WM) are assuming that,
for some FISON (ie, F(ω-1)) such that
there are no FISONs.after.
No.
You (WM) haven't given any other reason.
If U(F(n)) = ℕ, then
F(1) can be omitted without changing the result.
If F(k) can be omitted,
then F(k+1) can be omitted too.
The set of FISONs which can be omitted
is an inductive set, i.e., all FISONs.
On 09.02.2025 01:00, Jim Burns wrote:
On 2/8/2025 4:54 PM, WM wrote:
The axiom of induction says:
If any property or predicate P satisfies
(P(1) /\ ∀k(P(k) ==> P(k+1)),
then it describes all elements of
an inductive = infinite set.
Not for all inductive sets.
For all minimal.inductive sets.
May be called so.
describes each of its inductive subsets.(P(1) /\ ∀k(P(k) ==> P(k+1)),
If any property or predicate P satisfies
(P(1) /\ ∀k(P(k) ==> P(k+1)),
then it describes all elements of
[a minimal.inductive] set.
minimal.inductive ≠ inductive ≠ infinite
Then you are wrong.
Every inductive set is infinite.
And all that does is say that the set of all FISONs is a set where all
the members are individually
not needed to build a set whose union is
the set of Natual Numbers.
There is nothing in the theory to let you jump from a property of the
members to a property of the set as a whole,
On 2/9/25 9:50 AM, WM wrote:
On 09.02.2025 14:08, Richard Damon wrote:
On 2/9/25 5:46 AM, WM wrote:
On 08.02.2025 23:28, Richard Damon wrote:
On 2/8/25 2:44 PM, WM wrote:
On 08.02.2025 12:51, Richard Damon wrote:
And thus you claim that 36 can not be factored, as all of its
factors are not needed.
1, 2, 3, 4, 6, 9, 12, 18, 36.
According to Cantor, the set of factors has a smallest element, 1, >>>>>> and the set of necessary factors has a smallest element, 6, or if
double application is not allowed, 4.
And which of those was "necessary"?
I don't need 6, because I can factor into 4 * 9
4 is not necessary because it is smaller than 6, and 6 is sufficient.
But then your "necessary" has a application order, which the word
doesn't have.
I apply Cantor's theorem B.
Do you mean axiom 8? I see nothing in his theorems that talk about
anything like this.
On 2/9/2025 5:59 AM, WM wrote:
minimal.inductive ≠ inductive ≠ infinite
Then you are wrong.
Every inductive set is infinite.
Some infinite sets, such as E(137),
are not inductive.
Also,
some inductive sets, such as ℝ,
are not minimal.inductive.
However,
the FISONs are inductive.
No FISON ends the FISONs.
On 09.02.2025 13:34, joes wrote:Huh? The successor axiom doesn’t talk about sets.
Am Sat, 08 Feb 2025 22:54:46 +0100 schrieb WM:Peano is not satisfied by the set of all k either. But no exceptioncan
On 08.02.2025 18:43, Jim Burns wrote:Caveat: P is *not* satisfied by the set of all k, only by its elements.
Do you use ω-1 ?
Not in this proof.:
The axiom of induction says: If any property or predicate P satifies
(P(1) /\ ∀k(P(k) ==> P(k+1)), then it describes all elements of an
inductive = infinite set. That is satisfied by the set M of all FISONs
which are useless in U(A(n)) = ℕ.
be identified.
On 09.02.2025 18:04, Richard Damon wrote:
And all that does is say that the set of all FISONs is a set where all
the members are individually
and all their predecessors
not needed to build a set whose union is the set of Natual Numbers.
There is nothing in the theory to let you jump from a property of the
members to a property of the set as a whole,
All FISONs are proven useless for changing the assumption. Therefore
they can be omitted and what remains is { } = ℕ.
Regards, WM
On 2/10/25 4:56 AM, WM wrote:
The set of useless FISONs is inductive and therefore infinite. NoWhich means that each element is not needed, but doesn't prove that you
FISON can change the assumption U(A(n)) = ℕ. Therefore every FISON can
be omitted. ==> { } = ℕ.
can't get the answer from a union of an infinite set of them.
Am Sun, 09 Feb 2025 15:48:36 +0100 schrieb WM:
On 09.02.2025 13:34, joes wrote:Huh? The successor axiom doesn’t talk about sets.
Am Sat, 08 Feb 2025 22:54:46 +0100 schrieb WM:Peano is not satisfied by the set of all k either. But no exception can
On 08.02.2025 18:43, Jim Burns wrote:Caveat: P is *not* satisfied by the set of all k, only by its elements.
Do you use ω-1 ?
Not in this proof.:
The axiom of induction says: If any property or predicate P satifies
(P(1) /\ ∀k(P(k) ==> P(k+1)), then it describes all elements of an
inductive = infinite set. That is satisfied by the set M of all FISONs >>>> which are useless in U(A(n)) = ℕ.
be identified.
On 2/10/25 4:47 AM, WM wrote:
"Theorem B: Every embodiment of different numbers of the first and the
second number class has a smallest number, a minimum."
So, your "Set of Required FISONs" isn't a set of the first or second
class, it is an empty set.
On 09.02.2025 18:20, Jim Burns wrote:
On 2/9/2025 5:59 AM, WM wrote:
On 09.02.2025 01:00, Jim Burns wrote:
On 2/8/2025 4:54 PM, WM wrote:
The set of useless FISONs is inductive
and therefore infinite.
No FISON can change the assumption
U(A(n)) = ℕ.
Therefore every FISON can be omitted.
{ } = ℕ.
The axiom of induction says:
If any property or predicate P satisfies
(P(1) /\ ∀k(P(k) ==> P(k+1)),
then it describes all elements of
an inductive = infinite set.
If any property or predicate P satisfies
(P(1) /\ ∀k(P(k) ==> P(k+1)),
then it describes all elements of
[a minimal.inductive] set.
minimal.inductive ≠ inductive ≠ infinite
Then you are wrong.
Every inductive set is infinite.
Some infinite sets, such as E(137),
are not inductive.
That is irrelevant.
Also,
some inductive sets, such as ℝ,
are not minimal.inductive.
That is irrelevant.
(P(1) /\ ∀k(P(k) ==> P(k+1)),
On 10.02.2025 13:37, Richard Damon wrote:
On 2/10/25 4:56 AM, WM wrote:
The set of useless FISONs is inductive and
therefore infinite.
No FISON can change the assumption U(A(n)) = ℕ.
Therefore
every FISON can be omitted.
{ } = ℕ.
Which means that each element is not needed,
but doesn't prove that
you can't get the answer from
a union of an infinite set of them.
Does Zermelo define a set
by induction or
only its elements?
On 09.02.2025 18:20, Jim Burns wrote:
On 2/9/2025 5:59 AM, WM wrote:
On 2/8/2025 4:54 PM, WM wrote:
Therefore U(F(n)) = ℕ ==>
U{ } = { } = ℕ.
If U(F(n)) = ℕ, then
F(1) can be omitted without changing
the result.
If F(k) can be omitted,
then F(k+1) can be omitted too.
The set of FISONs which can be omitted
is an inductive set, i.e., all FISONs.
However,
the FISONs are inductive.
No FISON ends the FISONs.
Induction holds
for all natural numbers and
for all FISONs of the infinite set.
Therefore U(F(n)) = ℕ ==>
U{ } = { } = ℕ.
On 10.02.2025 13:37, Richard Damon wrote:
On 2/10/25 4:56 AM, WM wrote:
The set of useless FISONs is inductive and therefore infinite. NoWhich means that each element is not needed, but doesn't prove that
FISON can change the assumption U(A(n)) = ℕ. Therefore every FISON
can be omitted. ==> { } = ℕ.
you can't get the answer from a union of an infinite set of them.
Does Zermelo define a set by induction or only its elements?
Regareds, WM
On 10.02.2025 13:37, Richard Damon wrote:
On 2/10/25 4:47 AM, WM wrote:
"Theorem B: Every embodiment of different numbers of the first and
the second number class has a smallest number, a minimum."
So, your "Set of Required FISONs" isn't a set of the first or second
class, it is an empty set.
So it is.
Regards, WM
On 2/10/25 10:02 AM, WM wrote:
On 10.02.2025 13:37, Richard Damon wrote:
On 2/10/25 4:47 AM, WM wrote:
"Theorem B:
Every embodiment of different numbers of
the first and the second number class has
a smallest number, a minimum."
So, your "Set of Required FISONs" isn't
a set of the first or second class,
it is an empty set.
So it is.
Which just shows that
no particular FISON is needed.
Doesn't mean that you can't use
a set of FISONs to make
the set of Natural Numbers.
As pointed out,
your logic says you can't factor 36,
as none of the factors of 35 are "required",
since {4, 9} and {2, 18} are disjoint sets,
so no one factor is necessary.
On 2/10/2025 4:56 AM, WM wrote:
The set of useless FISONs is inductive
and therefore infinite.
No FISON can change the assumption
U(A(n)) = ℕ.
Therefore every FISON can be omitted.
Do you accept
∀ᴺj′:∀ᴺi′:∃ᴺk′:
k′ = max{i′,j′+1}
?
On 2/10/2025 10:00 AM, WM wrote:
On 10.02.2025 13:37, Richard Damon wrote:
On 2/10/25 4:56 AM, WM wrote:
The set of useless FISONs is inductive and
therefore infinite.
Yes.
A FISON with FISONs.after is uselessᵂᴹ.
Each FISON is uselessᵂᴹ.
The set of FISONs is minimal.inductive.
The set of uselessᵂᴹ FISONs is minimal.inductive.
For each FISON,
the union of FISONs.after is the same set,
a set we have named ℕ.
Therefore
every FISON can be omitted.
{ } = ℕ.
No.
Only ⋃{FISONs.after} = ⋃{}
for a FISON without FISONs.after.
However,
each FISON is with FISONs.after.
Which means that each element is not needed,
but doesn't prove that you can't get the answer from
a union of an infinite set of them.
Does Zermelo define a set
by induction or
only its elements?
Zermelo defines exactly (not fuller, not emptier)
which elements are in a set,
a set of which there can only be one
(extensionality).
That defined set is minimal.inductive.
Therefore, induction is valid with it.
On 2/10/25 10:00 AM, WM wrote:
Does Zermelo define a set by induction or only its elements?Zermelo BUILDS a set by creating its members.
Note, you keep on
confusing a property of the members for a property of the set.
No individual FISON is required to build a set of FISONs whose union is
the set of Natural Numbers,
but that doesn't mean the set of FISONs you
CAN use is empty.
On 2/10/25 10:02 AM, WM wrote:
On 10.02.2025 13:37, Richard Damon wrote:
On 2/10/25 4:47 AM, WM wrote:
"Theorem B: Every embodiment of different numbers of the first and
the second number class has a smallest number, a minimum."
So, your "Set of Required FISONs" isn't a set of the first or second
class, it is an empty set.
So it is.
Which just shows that no particular FISON is needed.
Doesn't mean that you can't use a set of FISONs to make the set of
Natural Numbers.
On 11.02.2025 01:48, Jim Burns wrote:No FISON is necessary.
On 2/10/2025 7:05 PM, Richard Damon wrote:Then name the first FISON that must exist according to Cantor's theorem,
On 2/10/25 10:02 AM, WM wrote:Yes.
On 10.02.2025 13:37, Richard Damon wrote:
On 2/10/25 4:47 AM, WM wrote:
Which just shows that no particular FISON is needed.So it is."Theorem B:So, your "Set of Required FISONs" isn't a set of the first or second >>>>> class, it is an empty set.
Every embodiment of different numbers of the first and the second
number class has a smallest number, a minimum."
Doesn't mean that you can't use a set of FISONs to make the set of
Natural Numbers.
see above.
Induction does not include infinity.⎛ If no FISON is required ⎜ then no FISON is possible.That is not necessarily so. But if a FISON or set of FISONs is possible
and not empty, then it has a first element. Find it. Overcome the result
of induction, i.e., infinity.
On 2/10/2025 7:05 PM, Richard Damon wrote:
On 2/10/25 10:02 AM, WM wrote:
On 10.02.2025 13:37, Richard Damon wrote:
On 2/10/25 4:47 AM, WM wrote:
"Theorem B:
Every embodiment of different numbers of
the first and the second number class has
a smallest number, a minimum."
So, your "Set of Required FISONs" isn't
a set of the first or second class,
it is an empty set.
So it is.
Which just shows that
no particular FISON is needed.
Doesn't mean that you can't use
a set of FISONs to make
the set of Natural Numbers.
Yes.
⎛ If no FISON is required
⎜ then no FISON is possible.
On 11.02.2025 01:03, Richard Damon wrote:
On 2/10/25 10:00 AM, WM wrote:
Does Zermelo define a set by induction or only its elements?Zermelo BUILDS a set by creating its members.
So do I.
Note, you keep on confusing a property of the members for a property
of the set.
There is a set {1} smaller than ℕ.
If the set {1, 2, 3, ..., n} including its predecessors is smaller than
ℕ, then the set {1, 2, 3, ..., n+1} including its predecessors is
smaller than ℕ.
No individual FISON is required to build a set of FISONs whose union
is the set of Natural Numbers,
No FISON is available. All FISONs belong to the inductive set F of insufficient FISONs. There is no FISON remaining.
but that doesn't mean the set of FISONs you CAN use is empty.
No, but it means: IF U(F) = ℕ, THEN { } = ℕ,
Regards, WM
On 11.02.2025 01:03, Richard Damon wrote:If by „predecessor” you mean {1, …, n-1} for n>1, then yes. N is not finite and has no largest element.
On 2/10/25 10:00 AM, WM wrote:
So do I.Does Zermelo define a set by induction or only its elements?Zermelo BUILDS a set by creating its members.
Note, you keep on confusing a property of the members for a property ofThere is a set {1} smaller than ℕ.
the set.
If the set {1, 2, 3, ..., n} including its predecessors is smaller than
ℕ, then the set {1, 2, 3, ..., n+1} including its predecessors is
smaller than ℕ.
„Available”? Many FISONs exist. What is F? No single FISON suffices.No individual FISON is required to build a set of FISONs whose union isNo FISON is available. All FISONs belong to the inductive set F of insufficient FISONs. There is no FISON remaining.
the set of Natural Numbers,
Then shut up.but that doesn't mean the set of FISONs you CAN use is empty.No
but it means: IF U(F) = ℕ, THEN { } = ℕ,U(F) != {}
On 10.02.2025 18:26, Jim Burns wrote:If there were a last FISON, it would be N.
On 2/10/2025 10:00 AM, WM wrote:That is not the only reason. Also a FISON with no FISONs after it would
On 10.02.2025 13:37, Richard Damon wrote:Yes.
On 2/10/25 4:56 AM, WM wrote:
The set of useless FISONs is inductive and therefore infinite.
A FISON with FISONs.after is uselessᵂᴹ.
Each FISON is uselessᵂᴹ.
be useless because ∀F(n) ∈ F: |ℕ \ F(n)| = ℵo.
Nor are there natural numbers.The set of FISONs is minimal.inductive.And there is no FISON beyond this set.
The set of uselessᵂᴹ FISONs is minimal.inductive.
Wrong. How did you even come up with this shit?For each FISON,Then Zermelo does not describe the set ℕ but only a FISON. Then ℕ does follow upon it but, since it is not defined et all, does not exist.
the union of FISONs.after is the same set,
a set we have named ℕ.
Pick an endsegment of FISONs.Name the first one. Every set of FISONs has a definable first element,No.Therefore every FISON can be omitted.
{ } = ℕ.
Only ⋃{FISONs.after} = ⋃{}
for a FISON without FISONs.after.
unless it is empty.
All finite sets of FISONs, sure.However, each FISON is with FISONs.after.Induction covers all.
No, you try to include infinity.So do I.Zermelo defines exactly (not fuller, not emptier) which elements are inWhich means that each element is not needed,Does Zermelo define a set by induction or only its elements?
but doesn't prove that you can't get the answer from a union of an
infinite set of them.
a set.
A single set doesn’t have any successor at all.a set of which there can only be one (extensionality).That is F.
This set is infinite and has no successors after.
That defined set is minimal.inductive.
Therefore, induction is valid with it.
On 10.02.2025 16:16, Jim Burns wrote:
On 2/10/2025 4:56 AM, WM wrote:
The set F of FISONs which can be removed without
changing the assumed result UF = ℕ
is the infinite set F of all FISONs.
This is proven by just the same induction
as Zermelo proves his infinite set Z.
Either you accept both proofs or none.
But without there is no set theory.
The set of useless FISONs is inductive
and therefore infinite.
No FISON can change the assumption
U(A(n)) = ℕ.
Therefore every FISON can be omitted.
Do you accept
∀ᴺj′:∀ᴺi′:∃ᴺk′:
k′ = max{i′,j′+1}
?
No, I won't try to dive into your private notation.
How induction works is well known.
If not consult Wikipedia or my book
W. Mückenheim: "Mathematik für die ersten Semester",
4th ed., De Gruyter, Berlin (2015)
On 10.02.2025 16:16, Jim Burns wrote:Nope, any *finite* set of FISONs can be removed. The set of *those*
On 2/10/2025 4:56 AM, WM wrote:
The set of useless FISONs is inductive and therefore infinite.
No FISON can change the assumption U(A(n)) = ℕ.
Therefore every FISON can be omitted.
Do you accept ∀ᴺj′:∀ᴺi′:∃ᴺk′:
k′ = max{i′,j′+1}
?
No, I won't try to dive into your private notation. How induction works
is well known. If not consult Wikipedia.
The set F of FISONs which can be removed without changing the assumed
result UF = ℕ is the infinite set F of all FISONs. This is proven by
just the same induction as Zermelo proves his infinite set Z.
On 2/11/2025 4:31 AM, WM wrote:
On 10.02.2025 16:16, Jim Burns wrote:
The set F of FISONs which can be removed without
changing the assumed result UF = ℕ
is the infinite set F of all FISONs.
Yes.
This is proven by just the same induction
as Zermelo proves his infinite set Z.
Either you accept both proofs or none.
But without there is no set theory.
That part above is fine.
Your problem is in the step after that,
which, for some reason, you skip over.
Only a hypothetical last.FISON ͚F,
a FISON with {after.͚F} = {}
supports your reasoning:
There is no last FISON.
⋃{} ≠ ℕ
It is irrelevant what details exist. Induction covers the whole infiniteNo, I won't try to dive into your private notation.
Do you accept that,
for each two FISON.numbers j′ and i′
there exists a FISON.number maximum k′ of
i′ and the successor j′+1 of j′
?
How induction works is not well known to you (WM).
On 11.02.2025 18:42, Jim Burns wrote:
On 2/11/2025 4:31 AM, WM wrote:
The set F of FISONs which can be removed
without changing the assumed result UF = ℕ
is the infinite set F of all FISONs.
Yes.
Fine.
This is proven by just the same induction
as Zermelo proves his infinite set Z.
Either you accept both proofs or none.
There is no further step.
But without there is no set theory.
That part above is fine.
Your problem is in the step after that,
which, for some reason, you skip over.
The part above defines
the set of all FISONs that can be omitted
without changing the union.
There is no further step.
Only a hypothetical last.FISON ͚F,
a FISON with {after.͚F} = {}
supports your reasoning:
The set of all FISONs is accepted.
No further restricitons allowed.
There is no last FISON.
⋃{} ≠ ℕ
That is not claimed.
Claimed is only UF = ℕ ==> ⋃{} = ℕ.
Can't you understand the meaning of an implication?
No, I won't try to dive into your private notation.
Do you accept that,
for each two FISON.numbers j′ and i′
there exists a FISON.number maximum k′ of
i′ and the successor j′+1 of j′
?
It is irrelevant what details exist.
Induction covers the whole infinite set.
How induction works is not well known to you (WM).
What is wrong in my application in your opinion?
Induction covers the whole infinite set.You mean by that
Am Tue, 11 Feb 2025 10:31:48 +0100 schrieb WM:
The set F of FISONs which can be removed without changing the assumedNope, any *finite* set of FISONs can be removed. The set of *those*
result UF = ℕ is the infinite set F of all FISONs. This is proven by
just the same induction as Zermelo proves his infinite set Z.
is infinite, but no set of removable FISONs is.
(Zermelo or whoever don’t prove the membership of infinite elements.)
On 11.02.2025 18:42, Jim Burns wrote:Actually, no: you can not remove the whole set, only single elements.
On 2/11/2025 4:31 AM, WM wrote:
On 10.02.2025 16:16, Jim Burns wrote:
Fine.The set F of FISONs which can be removed without changing the assumedYes.
result UF = ℕ is the infinite set F of all FISONs.
The erroneous step is from „every finite number” to „an infinite number”.This is proven by just the same induction as Zermelo proves his
infinite set Z.
Either you accept both proofs or none.
But without there is no set theory.
That part above is fine.
Your problem is in the step after that,
which, for some reason, you skip over.
The part above defines the set of all FISONs that can be omitted without changing the union. There is no further step.
Lol.Only a hypothetical last.FISON ͚F,The set of all FISONs is accepted. No further restricitons allowed.
a FISON with {after.͚F} = {}
supports your reasoning:
It is wrong, which is the other possibility for a nonsensical consequent.There is no last FISON.That is not claimed. Claimed is only UF = ℕ ==> ⋃{} = ℕ.
⋃{} ≠ ℕ
Can't you understand the meaning of an implication?
No, it only covers the elements.It is irrelevant what details exist. Induction covers the whole infiniteNo, I won't try to dive into your private notation.
Do you accept that,
for each two FISON.numbers j′ and i′
there exists a FISON.number maximum k′ of i′ and
the successor j′+1 of j′ ?
set.
You incorrectly include infinity.How induction works is not well known to you (WM).What is wrong in my application in your opinion?
On 2/11/2025 2:23 PM, WM wrote:
On 11.02.2025 18:42, Jim Burns wrote:
On 2/11/2025 4:31 AM, WM wrote:
The set F of FISONs which can be removed
without changing the assumed result UF = ℕ
is the infinite set F of all FISONs.
Yes.
Fine.
This is proven by just the same induction
as Zermelo proves his infinite set Z.
Either you accept both proofs or none.
What is this "both proofs"?
There is no further step.
Then there is no further conclusion.
Be so kind as to cease claiming U{} = N
Claimed is only UF = ℕ ==> ⋃{} = ℕ.
What is your new reason for claiming UF = U{} ?
⎛ Not that Unicode character U+2115 'ℕ'
⎝ is as good as any other name for UF.
What is wrong in my application in your opinion?
You (WM) currently are ignoring that,
for each two FISON.numbers j′ and i′
there exists a FISON.number maximum k′ of
i′ and the successor j′+1 of j′
which means
you ignore that
for each FISON F'
the union of FISONs.after F' are equal.
which contradicts U{F} = U{}
More broadly,
you (WM) write
Induction covers the whole infinite set.You mean by that
that P(k) such that P(0) ∧ ∀k:P(k)⇒P(k+1)
is valid for the minimal.inductive set itself
in addition to each of its elements.
Which is usually not.even.wrong.
Am Tue, 11 Feb 2025 20:23:28 +0100 schrieb WM:
On 11.02.2025 18:42, Jim Burns wrote:Actually, no: you can not remove the whole set, only single elements.
On 2/11/2025 4:31 AM, WM wrote:Fine.
On 10.02.2025 16:16, Jim Burns wrote:
The set F of FISONs which can be removed without changing the assumedYes.
result UF = ℕ is the infinite set F of all FISONs.
The erroneous step is from „every finite number” to „an infinite number”.
Induction covers the whole infiniteNo, it only covers the elements.
set.
You incorrectly include infinity.How induction works is not well known to you (WM).What is wrong in my application in your opinion?
On 11.02.2025 21:01, joes wrote:Pretty sure he doesn’t. What else is in Z?
Am Tue, 11 Feb 2025 10:31:48 +0100 schrieb WM:
Zermelo proves the existence of the set Z which contains the infiniteThe set F of FISONs which can be removed without changing the assumedNope, any *finite* set of FISONs can be removed. The set of *those*
result UF = ℕ is the infinite set F of all FISONs. This is proven by
just the same induction as Zermelo proves his infinite set Z.
is infinite, but no set of removable FISONs is.
(Zermelo or whoever don’t prove the membership of infinite elements.)
set ℕ of all finite natural numbers.
I prove the existence of the infinite set F of all finite FISONsNo, if you remove everything, you change the union. The union of
removable without changing the result of UF.
On 11.02.2025 21:01, joes wrote:
Am Tue, 11 Feb 2025 10:31:48 +0100 schrieb WM:
The set F of FISONs which can be removed without changing the assumedNope, any *finite* set of FISONs can be removed. The set of *those*
result UF = ℕ is the infinite set F of all FISONs. This is proven by
just the same induction as Zermelo proves his infinite set Z.
is infinite, but no set of removable FISONs is.
(Zermelo or whoever don’t prove the membership of infinite elements.)
Zermelo proves the existence of the set Z which contains the infinite
set ℕ of all finite natural numbers.
I prove the existence of the infinite set F of all finite FISONs
removable without changing the result of UF.
Note: Inductive sets are infinite.
Regards, WM
WM wrote :
On 12.02.2025 01:38, Jim Burns wrote:
On 2/11/2025 2:23 PM, WM wrote:
On 11.02.2025 18:42, Jim Burns wrote:
On 2/11/2025 4:31 AM, WM wrote:
The set F of FISONs which can be removed
without changing the assumed result UF = ℕ
is the infinite set F of all FISONs.
Yes.
Fine.
This is proven by just the same induction
as Zermelo proves his infinite set Z.
Either you accept both proofs or none.
What is this "both proofs"?
One is Zermelo's proof by induction that there is an infiite set Z.
It's an axiom, not a theorem.
Am Wed, 12 Feb 2025 10:25:32 +0100 schrieb WM:
Zermelo proves the existence of the set Z which contains the infinitePretty sure he doesn’t. What else is in Z?
set ℕ of all finite natural numbers.
On 2/12/25 4:25 AM, WM wrote:
Note: Inductive sets are infinite.
Which shows that all are individually removable.
WM wrote :
On 12.02.2025 01:38, Jim Burns wrote:
On 2/11/2025 2:23 PM, WM wrote:
On 11.02.2025 18:42, Jim Burns wrote:
On 2/11/2025 4:31 AM, WM wrote:
The set F of FISONs which can be removed
without changing the assumed result UF = ℕ
is the infinite set F of all FISONs.
Yes.
Fine.
This is proven by just the same induction
as Zermelo proves his infinite set Z.
Either you accept both proofs or none.
What is this "both proofs"?
One is Zermelo's proof by induction that there is an infinite set Z.
It's an axiom, not a theorem.
On 12.02.2025 10:46, joes wrote:No, N does not include itself.
Am Tue, 11 Feb 2025 20:23:28 +0100 schrieb WM:Induction proves all elements and even the set (Zermelo, Z).
On 11.02.2025 18:42, Jim Burns wrote:Actually, no: you can not remove the whole set, only single elements.
On 2/11/2025 4:31 AM, WM wrote:
On 10.02.2025 16:16, Jim Burns wrote:
The set F of FISONs which can be removed without changing the
assumed result UF = ℕ is the infinite set F of all FISONs.
No, all naturals are finite!The erroneous step is from „every finite number” to „an infiniteInduction proves an infinite number.
number”.
Yes, there is no natural number of FISONs you can not remove.All elements. None remains.Induction covers the whole infinite set.No, it only covers the elements.
Mithilfe der Definition über induktiveSays nothing about *the set* having that property.
Mengen lässt sich die Beweismethode der vollständigen Induktion rechtfertigen (daher auch der Name induktiv): Soll gezeigt werden, dass
alle natürlichen Zahlen eine bestimmte Eigenschaft haben, so betrachte
die Menge E ... [Wikipedia]
Why do you want to remove an infinite FIS of FISONs then?No, there is no infinite FISON. All are finite.You incorrectly include infinity.How induction works is not well known to you (WM).What is wrong in my application in your opinion?
All can be omitted without changing the result.No, any one can be, or even any finite number, but there are
On 12.02.2025 01:38, Jim Burns wrote:
On 2/11/2025 2:23 PM, WM wrote:
On 11.02.2025 18:42, Jim Burns wrote:
On 2/11/2025 4:31 AM, WM wrote:
The set F of FISONs which can be removed
without changing the assumed result UF = ℕ
is the infinite set F of all FISONs.
This is proven by just the same induction
as Zermelo proves his infinite set Z.
Either you accept both proofs or none.
What is this "both proofs"?
One is Zermelo's proof by induction
that there is an infiite set Z.
The other is my proof by induction:
Assume that
the set F of FISONs F(n) = {1, 2, 3, ..., n}
has the union UF = ℕ.
Notice that
F(1) can be omitted
without changing the result.
Notice that
when F(k) can be omitted,
then also F(k+1) can be omitted.
This makes
the set of FISONs which can be omitted
without changing the result
an inductive set.
It has no last element.
It is F.
The complementary
set of FISONs which cannot be omitted,
has no first element.
It is empty.
From the assumption UF = ℕ
we have obtained U{ } = { } = ℕ.
From the assumption UF = ℕ
we have obtained U{ } = { } = ℕ.
From the assumption UF = ℕ
we have obtained U{ } = { } = ℕ.
This result is false.
On 12.02.2025 13:16, Richard Damon wrote:No. There is nothing that has all as predecessors. If you remove all,
On 2/12/25 4:25 AM, WM wrote:
All are removable together with their predecessors.Note: Inductive sets are infinite.Which shows that all are individually removable.
When a FISON can beDoes not follow.
removed as useless, then all its predecessors can be removed as useless
too. Therefore all FISONs can be removed as useless.
On 12.02.2025 12:06, joes wrote:Done and done. As I expected, either Z does not contain, but is equal
Am Wed, 12 Feb 2025 10:25:32 +0100 schrieb WM:
Consult his Untersuchungen über die Grundlagen der Mengenlehre (1908)Zermelo proves the existence of the set Z which contains the infinitePretty sure he doesn’t. What else is in Z?
set ℕ of all finite natural numbers.
after learning what induction is.
On 12.02.2025 01:38, Jim Burns wrote:
On 2/11/2025 2:23 PM, WM wrote:
What is wrong in my application in your opinion?
You (WM) currently are ignoring that,
for each two FISON.numbers j′ and i′
there exists a FISON.number maximum k′ of
i′ and the successor j′+1 of j′
which means
you ignore that
for each FISON F'
the union of FISONs.after F' are equal.
which contradicts U{F} = U{}
Proofs by induction have no reason to observe your claim.
If Zermelo's induction is valid for
his set Z including the set ℕ,
then my proof is valid for the set F too.
On 12.02.2025 13:16, Richard Damon wrote:
On 2/12/25 4:25 AM, WM wrote:
Note: Inductive sets are infinite.
Which shows that all are individually removable.
All are removable together with their predecessors. When a FISON can be removed as useless, then all its predecessors can be removed as useless
too. Or do you claim an exception? Therefore all FISONs can be removed
as useless.
Regards, WM
On 2/12/2025 4:19 AM, WM wrote:
One is Zermelo's proof by induction
that there is an infiite set Z.
https://en.wikipedia.org/wiki/Zermelo_set_theory
The proof of Z you refer to is the bare assertion
that inductiveᶻ Z exists, AXIOM VII "Infinity".
Z exists such that Z∋{} ∧ ∀a∈Z∋{a}
From the assumption UF = ℕ
we have obtained U{ } = { } = ℕ.
We have obtained that,
for each FISON F′ ∈ {F} and 𝒜 ∈ 𝒫{F}
if ⋃𝒜 = ℕ then _not.necessarily_ F′ ∈ 𝒜
From the assumption UF = ℕ
we have obtained U{ } = { } = ℕ.
From the assumption ⋃{F} = ℕ
we have obtained ⋂𝒫ᵁᐧᙿᴺ{F} = {}
and also 𝒫ᵁᐧᙿᴺ{F} ≠ {}
On 2/12/25 10:43 AM, WM wrote:
Examples are the set of All FISONs with an Odd Top number, or the set of
ALL FISONs with an Even Top Number.
On 12.02.2025 20:39, Jim Burns wrote:The result has certainly changed from a nonempty set, whatever
On 2/12/2025 4:19 AM, WM wrote:
One is Zermelo's proof by induction that there is an infiite set Z.
https://en.wikipedia.org/wiki/Zermelo_set_theory
The proof of Z you refer to is the bare assertion that inductiveᶻ Z
exists, AXIOM VII "Infinity".
Z exists such that Z∋{} ∧ ∀a∈Z∋{a}
From the assumption UF = ℕ we have obtained U{ } = { } = ℕ.
We have obtained that,
for each FISON F′ ∈ {F} and 𝒜 ∈ 𝒫{F}
if ⋃𝒜 = ℕ then _not.necessarily_ F′ ∈ 𝒜
All of us who know induction know that by induction we have obtained
that all FISONs can be removed without changing the result.
With P = „You can leave out A(n) and all preceding FISONs” you get an infinite number of such sentences, where each leaves infinitely manyFrom the assumption UF = ℕ we have obtained U{ } = { } = ℕ.
From the assumption ⋃{F} = ℕ
we have obtained ⋂𝒫ᵁᐧᙿᴺ{F} = {}
and also 𝒫ᵁᐧᙿᴺ{F} ≠ {}
Learn induction. Mathematical induction is a method for proving that a statement is true for every natural number {P(0),P(1),P(2),P(3),\dots }
all hold. No FISON remains.
Am Thu, 13 Feb 2025 12:59:56 +0100 schrieb WM:
All of us who know induction know that by induction we have obtainedThe result has certainly changed from a nonempty set, whatever
that all FISONs can be removed without changing the result.
you think the union of inf. many FISONs is.
From the assumption UF = ℕ we have obtained U{ } = { } = ℕ.
With P = „You can leave out A(n) and all preceding FISONs” you get an infinite number of such sentences, where each leaves infinitely many
FISONs in the union.
On 13.02.2025 13:54, joes wrote:Yes, but it does change when you omit *all*. UF is not empty.
Am Thu, 13 Feb 2025 12:59:56 +0100 schrieb WM:
The assumption was UF = ℕ and has not changed by omitting any FISON.All of us who know induction know that by induction we have obtainedThe result has certainly changed from a nonempty set, whatever you
that all FISONs can be removed without changing the result.
think the union of inf. many FISONs is.
Yes, you can remove any natural number of FISONs, but not all,Induction covers all natural numbers. None is remaining.With P = „You can leave out A(n) and all preceding FISONs” you get anFrom the assumption UF = ℕ we have obtained U{ } = { } = ℕ.
infinite number of such sentences, where each leaves infinitely many
FISONs in the union.
On 12.02.2025 10:46, joes wrote:
Am Tue, 11 Feb 2025 20:23:28 +0100 schrieb WM:
On 11.02.2025 18:42, Jim Burns wrote:
On 2/11/2025 4:31 AM, WM wrote:
The set F of
FISONs which can be removed
without changing the assumed result UF = ℕ
is the infinite set F of all FISONs.
Yes.
Fine.
Actually, no:
you can not remove the whole set,
only single elements.
Induction proves all elements and
even the set (Zermelo, Z).
The erroneous step is from „every finite number”
to „an infinite number”.
Induction proves an infinite number.
On 12.02.2025 20:39, Jim Burns wrote:
On 2/12/2025 4:19 AM, WM wrote:
One is Zermelo's proof by induction
that there is an infiite set Z.
https://en.wikipedia.org/wiki/Zermelo_set_theory
The proof of Z you refer to is the bare assertion
that inductiveᶻ Z exists, AXIOM VII "Infinity".
Z exists such that Z∋{} ∧ ∀a∈Z∋{a}
Then follows almost a whole page of proof.
But that is irrelevant for
the present topic.
From the assumption UF = ℕ
we have obtained U{ } = { } = ℕ.
We have obtained that,
for each FISON F′ ∈ {F} and 𝒜 ∈ 𝒫{F}
if ⋃𝒜 = ℕ then _not.necessarily_ F′ ∈ 𝒜
All of us who know induction
know that
by induction we have obtained that
all FISONs
can be removed
without changing the result.
From the assumption UF = ℕ
we have obtained U{ } = { } = ℕ.
From the assumption ⋃{F} = ℕ
we have obtained ⋂𝒫ᵁᐧᙿᴺ{F} = {}
and also 𝒫ᵁᐧᙿᴺ{F} ≠ {}
Learn induction.
Mathematical induction is
a method for proving that
a statement is true for every natural number
{P(0),P(1),P(2),P(3),\dots }
all hold.
No FISON remains.
On 13.02.2025 00:49, Jim Burns wrote:
Learn that
by A(1) and if A(n) then A(n+1)
an infinite set is created
which has no successors.
Learn that if there were successors,
they would have a fixed first element.
Learn that
{1}
{2, 1}
{3, 2, 1}
...
is not longer than broad.
Learn that
{1}
{2, 1}
{3, 2, 1}
...
is not longer than broad.
On 13.02.2025 13:54, joes wrote:
Am Thu, 13 Feb 2025 12:59:56 +0100 schrieb WM:
All of us who know induction know that by induction we have obtainedThe result has certainly changed from a nonempty set, whatever
that all FISONs can be removed without changing the result.
you think the union of inf. many FISONs is.
The assumption was UF = ℕ and has not changed by omitting any FISON.
From the assumption UF = ℕ we have obtained U{ } = { } = ℕ.
With P = „You can leave out A(n) and all preceding FISONs” you get anInduction covers all natural numbers. None is remaining.
infinite number of such sentences, where each leaves infinitely many
FISONs in the union.
Regards, WM
Am Thu, 13 Feb 2025 14:51:44 +0100 schrieb WM:
On 13.02.2025 13:54, joes wrote:Yes, but it does change when you omit *all*.
Am Thu, 13 Feb 2025 12:59:56 +0100 schrieb WM:The assumption was UF = ℕ and has not changed by omitting any FISON.
All of us who know induction know that by induction we have obtainedThe result has certainly changed from a nonempty set, whatever you
that all FISONs can be removed without changing the result.
think the union of inf. many FISONs is.
UF is not empty.
On 2/13/2025 6:59 AM, WM wrote:
All of us who know induction
know that
by induction we have obtained that
all FISONs
each FISON
The present topics are "What is induction?"
On 13.02.2025 15:02, joes wrote:„All”, i.e. inf. many, is not an element of that.
Am Thu, 13 Feb 2025 14:51:44 +0100 schrieb WM:It does because induction is valid for all elements of the inductive
On 13.02.2025 13:54, joes wrote:Yes, but it does change when you omit *all*.
Am Thu, 13 Feb 2025 12:59:56 +0100 schrieb WM:The assumption was UF = ℕ and has not changed by omitting any FISON.
All of us who know induction know that by induction we have obtained >>>>> that all FISONs can be removed without changing the result.The result has certainly changed from a nonempty set, whatever you
think the union of inf. many FISONs is.
set.
UF is not empty.When will you learn that nobody claims such a nonsense?
All FISONs can be removed without changing the result.
On 2/12/2025 5:19 AM, WM wrote:
Induction proves all elements and
even the set (Zermelo, Z).
No.
Not the set.
Each member of the set.
Set intensionality means
same.membered sets are one set.
It does not mean sets are self.members.
The erroneous step is from „every finite number”
to „an infinite number”.
Induction proves an infinite number.
Induction is valid without any infinite sets.
In ST+F, there are no infinite sets,
but there is induction.
On 2/13/2025 7:04 AM, WM wrote:
On 13.02.2025 00:49, Jim Burns wrote:
Learn that
by A(1) and if A(n) then A(n+1)
an infinite set is created
described
which has no successors.
Each FISON has a successor in the set of FISONs.
The set of FISONs isn't any FISON.
Learn that if there were successors,
they would have a fixed first element.
Since induction on the set of FISONs concerns
a predicate on FISONs, not the set of FISONs,
presence or absence of successors for
the set of FISONS and anything else not.in
the set of FISONs
is irrelevant.
Learn that
{1}
{2, 1}
{3, 2, 1}
...
is not longer than broad.
For each set A which has
emptier.by.one subsets A\{a} smaller than A
there is a FISON larger than A
The set of rows and the set of columns
are both the size of ⋃{F} and each other.
On 13.02.2025 15:02, joes wrote:
Am Thu, 13 Feb 2025 14:51:44 +0100 schrieb WM:
On 13.02.2025 13:54, joes wrote:Yes, but it does change when you omit *all*.
Am Thu, 13 Feb 2025 12:59:56 +0100 schrieb WM:The assumption was UF = ℕ and has not changed by omitting any FISON.
All of us who know induction know that by induction we have obtained >>>>> that all FISONs can be removed without changing the result.The result has certainly changed from a nonempty set, whatever you
think the union of inf. many FISONs is.
It does because induction is valid for all elements of the inductive set.
UF is not empty.
When will you learn that nobody claims such a nonsense?
Regards, WM
On 2/13/25 8:51 AM, WM wrote:
Induction covers all natural numbers. None is remaining.
Induction doesn't change the set, so the set of FISONs still remains.
You show the set of REQUIRED FISONs is empty, but that doesn't empty the
set of FISONs you can use.
Am Fri, 14 Feb 2025 13:46:53 +0100 schrieb WM:
On 13.02.2025 15:02, joes wrote:„All”, i.e. inf. many, is not an element of that.
Am Thu, 13 Feb 2025 14:51:44 +0100 schrieb WM:It does because induction is valid for all elements of the inductive
On 13.02.2025 13:54, joes wrote:Yes, but it does change when you omit *all*.
Am Thu, 13 Feb 2025 12:59:56 +0100 schrieb WM:The assumption was UF = ℕ and has not changed by omitting any FISON.
All of us who know induction know that by induction we have obtained >>>>>> that all FISONs can be removed without changing the result.The result has certainly changed from a nonempty set, whatever you
think the union of inf. many FISONs is.
set.
UF is not empty.When will you learn that nobody claims such a nonsense?
Am Thu, 13 Feb 2025 12:59:56 +0100 schrieb WM:
All FISONs can be removed without changing the result.
Your are BUILDING the set of "Not individually needed" FISONs, not the
set of FISONs you can use to take a union of to make the Natural Numbers.
On 14.02.2025 00:44, Richard Damon wrote:
On 2/13/25 8:51 AM, WM wrote:
Induction covers all natural numbers. None is remaining.
Induction doesn't change the set, so the set of FISONs still remains.
Of course. But all FISONs are proven useless.
You show the set of REQUIRED FISONs is empty, but that doesn't empty
the set of FISONs you can use.
The set of useful FISONs is empty. Show the first not useless FISON.
Regards, WM
On 13.02.2025 17:38, Jim Burns wrote:
On 2/12/2025 5:19 AM, WM wrote:
On 12.02.2025 10:46, joes wrote:
Induction proves all elements and
even the set (Zermelo, Z).
No.
Not the set.
Each member of the set.
All members of the set.
The erroneous step is from „every finite number”
to „an infinite number”.
Induction proves an infinite number.
Induction is valid without any infinite sets.
Induction creates infinite sets.
If i contruct x(1) and from x(n) follows x(n+1)
then the set of all x(n) is infinite.
In ST+F, there are no infinite sets,
but there is induction.
I apply mathematics.
Sure, the set of FISONs:,
{ 1}, {1, 2}, {1, 2, 3}, ... {1, 2, 3, ..., n}, ...}
Is a useful set of FISONs, and its first element is {1}
On 2/14/25 9:24 AM, WM wrote:
The set of useful FISONs is empty. Show the first not useless FISON.
Your set of "useful FISONs" doesn't have a proper definition,
On 2/14/2025 7:52 AM, WM wrote:
Induction creates infinite sets.
Zermelo set theory [I...VII] describes
a domain of discourse with
Z an inductiveᶻ set
If i contruct x(1) and from x(n) follows x(n+1)
then the set of all x(n) is infinite.
then the set of all x(n) is
an inductive subset of the minimal.inductive set.
There is no set of all FISON.numbers in ST+F.
On 13.02.2025 19:16, Jim Burns wrote:You are mistaken. The set N contains each natural number, nothing else,
On 2/13/2025 6:59 AM, WM wrote:
No, all FISONs. The set of all natural numbers is proved by induction,All of us who know induction know that by induction we have obtainedeach FISON
that all FISONs
not the set of each natural number.
HAHAHAHAHAThe present topics are "What is induction?"No, everybody not knowing it, like you, should first learn it.
On 14.02.2025 16:42, Jim Burns wrote:
On 2/14/2025 7:52 AM, WM wrote:
Induction creates infinite sets.
Zermelo set theory [I...VII] describes
a domain of discourse with
Z an inductiveᶻ set
Where all elements are covered by induction.
If i contruct x(1) and from x(n) follows x(n+1)
then the set of all x(n) is infinite.
then the set of all x(n) is
an inductive subset of the minimal.inductive set.
And no element is outside.
There is no set of all FISON.numbers in ST+F.
But it is the result of the proof. There is not the least doubt that no
FISON remains for constructing ℕ.
Regards, WM
On 13.02.2025 17:38, Jim Burns wrote:No, it does not include a member called „all members”.
On 2/12/2025 5:19 AM, WM wrote:
All members of the set.Induction proves all elements and even the set (Zermelo, Z).No. Not the set. Each member of the set.
Wrong. There is indeed a number, albeit not natural, of FISONs youSet intensionality means same.membered sets are one set.All members are proved. This means none remains. Zermelo obtains from
It does not mean sets are self.members.
that the existence of the infinite set Z.
No, it does not prove a sentence to hold for infinite numbersThe erroneous step is from „every finite number”Induction proves an infinite number.
to „an infinite number”.
Sure. But no x is.Induction is valid without any infinite sets.Induction creates infinite sets.
If i contruct x(1) and from x(n) follows x(n+1) then the set of all x(n)
is infinite.
lolIn ST+F, there are no infinite sets, but there is induction.I apply mathematics.
On 14.02.2025 13:52, joes wrote:That is not what I said. I am not disputing that.
Am Fri, 14 Feb 2025 13:46:53 +0100 schrieb WM:All are elements of the inductive set.
On 13.02.2025 15:02, joes wrote:„All”, i.e. inf. many, is not an element of that.
Am Thu, 13 Feb 2025 14:51:44 +0100 schrieb WM:It does because induction is valid for all elements of the inductive
On 13.02.2025 13:54, joes wrote:
Am Thu, 13 Feb 2025 12:59:56 +0100 schrieb WM:The assumption was UF = ℕ and has not changed by omitting any FISON. >>>> Yes, but it does change when you omit *all*.
All of us who know induction know that by induction we haveThe result has certainly changed from a nonempty set, whatever you >>>>>> think the union of inf. many FISONs is.
obtained that all FISONs can be removed without changing the
result.
set.
UF is not empty.When will you learn that nobody claims such a nonsense?
Am Thu, 13 Feb 2025 12:59:56 +0100 schrieb WM:
All FISONs can be removed without changing the result.
That shows that UF = ℕ is wrong.
IF UF = ℕ THEN { } = ℕ.
Not Then UF = { }.
On 14.02.2025 15:57, Richard Damon wrote:
Sure, the set of FISONs:,
{ 1}, {1, 2}, {1, 2, 3}, ... {1, 2, 3, ..., n}, ...}
Is a useful set of FISONs, and its first element is {1}
By induction every FISON and its predecessors has been shown useless.
Regards, WM
On 13.02.2025 19:16, Jim Burns wrote:
On 2/13/2025 6:59 AM, WM wrote:
All of us who know induction
know that
by induction we have obtained that
all FISONs
each FISON
No, all FISONs.
The set of all natural numbers
is proved by induction,
not the set of each natural number.
The present topics are "What is induction?"
No, everybody not knowing it, like you,
should first learn it.
On 13.02.2025 20:47, Jim Burns wrote:
On 2/13/2025 7:04 AM, WM wrote:
On 13.02.2025 00:49, Jim Burns wrote:
Learn that
by A(1) and if A(n) then A(n+1)
an infinite set is created
described
which has no successors.
Each FISON has a successor in the set of FISONs.
The set of FISONs isn't any FISON.
The set of all FISONs is created.
Learn that if there were successors,
they would have a fixed first element.
Since induction on the set of FISONs concerns
a predicate on FISONs, not the set of FISONs,
presence or absence of successors for
the set of FISONS and anything else not.in
the set of FISONs
is irrelevant.
No, induction concerns
alle the natural numbers n of the FISONs A(n).
Learn that
{1}
{2, 1}
{3, 2, 1}
...
is not longer than broad.
For each set A which has
emptier.by.one subsets A\{a} smaller than A
there is a FISON larger than A
Irrelevant.
All are finite and will never become infinite.
The set of rows and the set of columns
are both the size of ⋃{F} and each other.
Yes, that is potentially infinite.
On 14.02.2025 16:42, Jim Burns wrote:
On 2/14/2025 7:52 AM, WM wrote:
Induction creates infinite sets.
Zermelo set theory [I...VII] describes
a domain of discourse with
Z an inductiveᶻ set
Where all elements are covered by induction.
If i contruct x(1) and from x(n) follows x(n+1)
then the set of all x(n) is infinite.
then the set of all x(n) is
an inductive subset of the minimal.inductive set.
And no element is outside.
There is no set of all FISON.numbers in ST+F.
But it is the result of the proof.
You claim that UF = X and also U({}) = X.
If UF=N and N={}, then UF={}.
Am Fri, 14 Feb 2025 13:52:27 +0100 schrieb WM:
On 13.02.2025 17:38, Jim Burns wrote:No, it does not include a member called „all members”.
On 2/12/2025 5:19 AM, WM wrote:All members of the set.
Induction proves all elements and even the set (Zermelo, Z).No. Not the set. Each member of the set.
Wrong.Set intensionality means same.membered sets are one set.All members are proved. This means none remains. Zermelo obtains from
It does not mean sets are self.members.
that the existence of the infinite set Z.
On 2/14/25 11:28 AM, WM wrote:
The definition is that it is a set of FISONs which has a smallest
element that is not as useless as a cup of coffee.
Which, as I said, is a definition in Naive Set theory,
On 2/14/2025 11:40 AM, WM wrote:
On 14.02.2025 16:42, Jim Burns wrote:
On 2/14/2025 7:52 AM, WM wrote:
Induction creates infinite sets.
Zermelo set theory [I...VII] describes
a domain of discourse with
Z an inductiveᶻ set
Where all elements are covered by induction.
No,
we don't know that, in inductiveᶻ Z,
each element is covered by inductionᶻ.
Being covered by induction is
being in the only.inductive.subset.
This reasoning only applies to
elements of the minimal.inductive set.
The minimal.inductive set is not
an element of the minimal.inductive set.
This reasoning does not apply to it.
Your alleged proof takes an unjustified leap
from "each FISON is omissible"
to "each set of omissible FISONs is ommissible".
Each FISON is.
Some sets are and some sets aren't.
{F} is a set of omissible FISONs.
{F} isn't an omissible set of FISONs.
Without the leap, there is no conflict.
On 2/14/2025 8:02 AM, WM wrote:
The set of rows and the set of columns
are both the size of ⋃{F} and each other.
Yes, that is potentially infinite.
The set of rows, the set of columns, and ⋃{F}
do not change.
Bob is not anywhere he has swapped to.
On 14.02.2025 18:32, joes wrote:
You claim that UF = X and also U({}) = X.
No. You and others have claimed that there is a set of FISONs containing
all natural numbers. I have accepted that a s premise.
If UF=N and N={}, then UF={}.
No, learn what an implication is! In W. Mückenheim: "Mathematik für die ersten Semester", 4th ed., De Gruyter, Berlin (2015) for instance it is explained for beginners.
There is no FISON that is in the claimed set. Therefore
If UF = ℕ then ℕ = {}.
This does *not* prove UF = {}. F contains sets and UF contains natural numbers, but not enough natural numbers, not ℕ.
Regards, WM
On 14.02.2025 19:02, Richard Damon wrote:
On 2/14/25 11:28 AM, WM wrote:
The definition is that it is a set of FISONs which has a smallest
element that is not as useless as a cup of coffee.
Which, as I said, is a definition in Naive Set theory,
Obviously you have no clue of set theory, be it naive or advanced.
Every set of ordinals has a smallest element. Look up the notion of well-order.
Regards, WM
The set of all natural numbers is not.in
the only inductive subset of the set of
all natural numbers.
WM was thinking very hard :Each FISON and all its predecessors are proven omissible by
induction. This implies that the whole set is omissible.
Only if there were such a thing as a last FISON being omitted.
And since you build your claimed "set" by Naive Set theory, your sets
are worthless.
The set you show is empty, is the set of FISONs that are individually
not necessary.
On 2/15/25 6:50 AM, WM wrote:
On 14.02.2025 19:02, Richard Damon wrote:
On 2/14/25 11:28 AM, WM wrote:
The definition is that it is a set of FISONs which has a smallest
element that is not as useless as a cup of coffee.
Which, as I said, is a definition in Naive Set theory,
Obviously you have no clue of set theory, be it naive or advanced.
Every set of ordinals has a smallest element. Look up the notion of
well-order.
Every NON-EMPTY set of Ordinals has a smallest element.
The problem is your "claimed set" can't actually defined in anything but Naive set theory,
and the fact that you admit that it might be needed
shows you have an understanding of the problem.
The set you can actually define, is the set of FISONs that are
individually required to build up a set that can union to the Natural Numbers, but that CAN be empty, and we can still make a set that reaches there.
WM was thinking very hard :
On 15.02.2025 14:13, FromTheRafters wrote:
WM was thinking very hard :Each FISON and all its predecessors are proven omissible by
induction. This implies that the whole set is omissible.
Only if there were such a thing as a last FISON being omitted.
You are clearly wrong!
Says the person who also says that there are infinite FISONs.
On 15.02.2025 14:13, FromTheRafters wrote:Haha. There is no last natural.
WM was thinking very hard :You are clearly wrong! Induction concerns the whole set.
Each FISON and all its predecessors are proven omissible byOnly if there were such a thing as a last FISON being omitted.
induction. This implies that the whole set is omissible.
Zermelo: "In order to secure the existence of infinite sets, we need the following axiom." [Zermelo: Untersuchungen über die Grundlagen der Mengenlehre I, S. 266][actual axiom missing]
This is the axiom of infinity proved by--
induction. It ascertains the existence of an infinite set. It ascertains
the set Z, Z_0 and the union of singletons ℕ.
On 14.02.2025 20:48, Jim Burns wrote:Nah. Then it would be finite.
On 2/14/2025 8:02 AM, WM wrote:
Then it would need to contain ℕ, as an infinite FISON.The set of rows, the set of columns, and ⋃{F} do not change.The set of rows and the set of columns are both the size of ⋃{F} and >>>> each other.Yes, that is potentially infinite.
Only in the limit of the permutations, which may not be a permutationBob is not anywhere he has swapped to.Then your logic allows lossless exchanges with losses. Not acceptable.
On 15.02.2025 14:40, Richard Damon wrote:Indeed, if you remove any number of FISONs, infinitely many remain.
And since you build your claimed "set" by Naive Set theory, your setsZermelo constructs his set in modern set theory. Induction concerns the
are worthless.
whole set. Compare Zermelo: "In order to secure the existence of
infinite sets, we need the following axiom." [Zermelo: Untersuchungen
über die Grundlagen der Mengenlehre I, S. 266] This is the axiom of
infinity proved by induction. It ascertains the existence of an infinite
set. It ascertains the set Z, Z_0 and the union of singletons ℕ.
The remaining set should have at least one element.
The set you show is empty, is the set of FISONs that are individually
not necessary.
On 14.02.2025 19:02, Richard Damon wrote:No, that is not the definition. No element of the set of FISONs is
On 2/14/25 11:28 AM, WM wrote:
Obviously you have no clue of set theory, be it naive or advanced. EveryThe definition is that it is a set of FISONs which has a smallestWhich, as I said, is a definition in Naive Set theory,
element that is not as useless as a cup of coffee.
set of ordinals has a smallest element. Look up the notion of
well-order.
On 14.02.2025 18:43, joes wrote:No, you can not substitute a set into a sentence about naturals.
Am Fri, 14 Feb 2025 13:52:27 +0100 schrieb WM:Induction proves the existence of the inductive set of all its members. Induction concerns the whole set.
On 13.02.2025 17:38, Jim Burns wrote:No, it does not include a member called „all members”.
On 2/12/2025 5:19 AM, WM wrote:All members of the set.
Induction proves all elements and even the set (Zermelo, Z).No. Not the set. Each member of the set.
Are you saying there are infinite naturals?Read, learn, try to understand.Wrong. There is indeed a number, albeit not natural, of FISONs you canSet intensionality means same.membered sets are one set.All members are proved. This means none remains. Zermelo obtains from
It does not mean sets are self.members.
that the existence of the infinite set Z.
NOT leave out: this number is infinite, and induction doesn’t reach it.
From „you can leave out any finite number of FISONs” it does not follow that „you can leave out an infinite number of them”. There are noNo, it does not prove a sentence to hold for infinite numbersThe erroneous step is from „every finite number”Induction proves an infinite number.
to „an infinite number”.
substituted into it, only finite ones, which there are infinitely many
of.
Even though there are infinitely many!Sure. But no x is.Induction is valid without any infinite sets.Induction creates infinite sets.
If i contruct x(1) and from x(n) follows x(n+1) then the set of all
x(n) is infinite.
On 14.02.2025 18:32, joes wrote:So what do you claim?
You claim that UF = X and also U({}) = X.No. You and others have claimed that there is a set of FISONs containing
all natural numbers. I have accepted that a s premise.
I understand it perfectly well. You apparently don’t. Ex falso…If UF=N and N={}, then UF={}.No, learn what an implication is!
There is no FISON that is in the claimed set. Therefore If UF = ℕ then ℕ = {}.On the contrary, the set of all FISONs contains *all and only* the FISONs. Nobody but you is saying that the empty set or the union thereof should
This does *not* prove UF = {}. F contains sets and UF contains natural numbers, but not enough natural numbers, not ℕ.You could’ve gone right out and said you’re extending N with infinite numbers.
On 15.02.2025 14:40, Richard Damon wrote:
And since you build your claimed "set" by Naive Set theory, your sets
are worthless.
Zermelo constructs his set in modern set theory. Induction concerns the
whole set. Compare Zermelo: "In order to secure the existence of
infinite sets, we need the following axiom." [Zermelo: Untersuchungen
über die Grundlagen der Mengenlehre I, S. 266] This is the axiom of
infinity proved by induction. It ascertains the existence of an infinite
set. It ascertains the set Z, Z_0 and the union of singletons ℕ.
The set you show is empty, is the set of FISONs that are individually
not necessary.
The remaining set should have at least one element. But it has none.
Further it would be silly to claim that replacing a smaller FISON is
helpful after a larger FISON has been removed.
Regards, WM
On 15.02.2025 14:47, Richard Damon wrote:
On 2/15/25 6:50 AM, WM wrote:
On 14.02.2025 19:02, Richard Damon wrote:
On 2/14/25 11:28 AM, WM wrote:
The definition is that it is a set of FISONs which has a smallest
element that is not as useless as a cup of coffee.
Which, as I said, is a definition in Naive Set theory,
Obviously you have no clue of set theory, be it naive or advanced.
Every set of ordinals has a smallest element. Look up the notion of
well-order.
Every NON-EMPTY set of Ordinals has a smallest element.
So it is.
The problem is your "claimed set" can't actually defined in anything
but Naive set theory,
It is defined in modern set theory.
and the fact that you admit that it might be needed shows you have an
understanding of the problem.
Of course, better than you.
The set you can actually define, is the set of FISONs that are
individually required to build up a set that can union to the Natural
Numbers, but that CAN be empty, and we can still make a set that
reaches there.
No, that would imply that it is meaningful to replace a removed FISON by smaller FISONs.
Regards, WM
On 14.02.2025 20:48, Jim Burns wrote:
On 2/14/2025 8:02 AM, WM wrote:
On 13.02.2025 20:47, Jim Burns wrote:
swaps such that, after all swaps,
Bob is not anywhere he has swapped to.
Then your logic allows
lossless exchanges with losses.
Not acceptable.
swaps such that, after all swaps,
Bob is not anywhere he has swapped to.
Then your logic allows
lossless exchanges with losses.
Not acceptable.
The set of rows and the set of columns
are both the size of ⋃{F} and each other.
Yes, that is potentially infinite.
The set of rows, the set of columns, and ⋃{F}
do not change.
Then it would need to contain ℕ,
as an infinite FISON.
On 14.02.2025 20:22, Jim Burns wrote:
On 2/14/2025 11:40 AM, WM wrote:
On 14.02.2025 16:42, Jim Burns wrote:
On 2/14/2025 7:52 AM, WM wrote:
Induction creates infinite sets.
Zermelo set theory [I...VII] describes
a domain of discourse with
Z an inductiveᶻ set
Where all elements are covered by induction.
No,
we don't know that, in inductiveᶻ Z,
each element is covered by inductionᶻ.
Induction proves the existence of
the inductive set of all its members.
Being covered by induction is
being in the only.inductive.subset.
That is here the set of all FISONs.
That is here the set of all FISONs.
Each FISON is.
Some sets are and some sets aren't.
Which set do you have in mind?
{F} is a set of omissible FISONs.
{F} isn't an omissible set of FISONs.
Without the leap, there is no conflict.
Without this leap there are no infinite sets.
Am Sat, 15 Feb 2025 13:56:17 +0100 schrieb WM:
On 14.02.2025 20:48, Jim Burns wrote:
Only in the limit of the permutations, which may not be a permutationBob is not anywhere he has swapped to.Then your logic allows lossless exchanges with losses. Not acceptable.
itself.
Am Sat, 15 Feb 2025 15:55:55 +0100 schrieb WM:
On 15.02.2025 14:40, Richard Damon wrote:Indeed, if you remove any number of FISONs, infinitely many remain.
And since you build your claimed "set" by Naive Set theory, your setsZermelo constructs his set in modern set theory. Induction concerns the
are worthless.
whole set. Compare Zermelo: "In order to secure the existence of
infinite sets, we need the following axiom." [Zermelo: Untersuchungen
über die Grundlagen der Mengenlehre I, S. 266] This is the axiom of
infinity proved by induction. It ascertains the existence of an infinite
set. It ascertains the set Z, Z_0 and the union of singletons ℕ.
The remaining set should have at least one element.
The set you show is empty, is the set of FISONs that are individually
not necessary.
On 2/15/25 9:58 AM, WM wrote:
The set you can actually define, is the set of FISONs that are
individually required to build up a set that can union to the Natural
Numbers, but that CAN be empty, and we can still make a set that
reaches there.
No, that would imply that it is meaningful to replace a removed FISON
by smaller FISONs.
WHy?
On 2/15/2025 7:56 AM, WM wrote:
swaps such that, after all swaps,
Bob is not anywhere he has swapped to.
Then your logic allows
lossless exchanges with losses.
Not acceptable.
If you accept ST+F
The set of rows, the set of columns, and ⋃{F}
do not change.
Then it would need to contain ℕ,
as an infinite FISON.
Infinite FISONs
(finite initial segments of naturals)
are not acceptable.
No element of the set of FISONs is
necessary
Yes, but when someone says 'finite FISON' it implies thast there are
some that are not finite.
There can be an infinite number of FISON's...
However, no FISON is
infinite in and of itself,
On 2/15/2025 7:49 AM, WM wrote:
Induction proves the existence of
the inductive set of all its members.
ℕ has an only.inductive subset ℕ
{F} has only one inductive subset: {F}
{F} is a set of omissible FISONs.
{F} isn't an omissible set of FISONs.
Without the leap, there is no conflict.
You (WM) imagine a last finite step,
into the infinite.
On 16.02.2025 00:52, Jim Burns wrote:That is simply false. You may remove one or another, but not at the
On 2/15/2025 7:49 AM, WM wrote:
The set of FISONs is also an inductive set. Compare v. Neumann'sInduction proves the existence of the inductive set of all itsℕ has an only.inductive subset ℕ
members.
definition of finite cardinal numbers.
{F} has only one inductive subset: {F}Correct.
But you are wrong. If all FISONs are omissible by induction, then the
{F} is a set of omissible FISONs.
{F} isn't an omissible set of FISONs. Without the leap, there is no
conflict.
set of all FISONs is omissible.
What should remain if all F(n) are omitted? Inserting smaller F(n) after larger one have been omitted as useless?Babbage: „I am not able to rightly apprehend the kind of confusion that
More precisely: Every single element can be omitted, but not moreYou (WM) imagine a last finite step, into the infinite.No, I accept Zermelo's proof of an infinite set by the axiom of
induction: { } and a ==> {a}. That creates an infinite set.
All elements can be omitted.
The set can be omitted.No, the set is not an element of itself that can be omitted.
What should remain?Indeed, why should something remain if you remove everything? If you
And why? Note that you must define a first element.
There is no doubt that this proves the whole set Z without assuming aWDYM „prove a set”? For every *element* you can prove the statement „This, and only this segment can be omitted.” Unfortunately there
last element.
On 15.02.2025 23:00, Chris M. Thomasson wrote:The number of FISONs is definitely larger than any finite
There can be an infinite number of FISON's...Not greater than all finite numbers however.
No. The „width and height” is not a finite number. If itHowever, no FISON is infinite in and of itself,Therefore the figure {1}
{2, 1}
{3, 2, 1}
...
has width and height not larger than all finite numbers.
Am Sat, 15 Feb 2025 17:54:05 +0000 schrieb joes:No, obviously you can’t remove all („which one is necessary?” my ass). That the union of the empty set is not N has no bearing on the matter.
Am Sat, 15 Feb 2025 12:50:40 +0100 schrieb WM:Therefore all can be removed, and U(F) = ℕ ==> U(F\F) = ℕ.
On 14.02.2025 19:02, Richard Damon wrote:No, that is not the definition. No element of the set of FISONs is
On 2/14/25 11:28 AM, WM wrote:Obviously you have no clue of set theory, be it naive or advanced.
The definition is that it is a set of FISONs which has a smallestWhich, as I said, is a definition in Naive Set theory,
element that is not as useless as a cup of coffee.
Every set of ordinals has a smallest element. Look up the notion of
well-order.
necessary
--(though one could be, if we wouldn’t agree that none is)
for their union to be N. *That* set has a smallest element, as does
every other infinite set of FISONs (the nonempty finite sets do as
well, but their union is not N, but the largest FISON). Now you come
along and claim that the empty set should have a first element.
Am Sat, 15 Feb 2025 17:42:51 +0000 schrieb joes:No further rows would be needed. Actually, the first column is N.
Am Sat, 15 Feb 2025 13:56:17 +0100 schrieb WM:
On 14.02.2025 20:48, Jim Burns wrote:Nah. Then it would be finite.
On 2/14/2025 8:02 AM, WM wrote:
Then it would need to contain ℕ, as an infinite FISON.The set of rows, the set of columns, and ⋃{F} do not change.The set of rows and the set of columns are both the size of ⋃{F} >>>>>> and each other.Yes, that is potentially infinite.
Bravo, you wilfully misunderstood „limit”.There is no limit in counting if countable sets are really countable.Only in the limit of the permutations, which may not be a permutationBob is not anywhere he has swapped to.Then your logic allows lossless exchanges with losses. Not acceptable.
itself.
On 15.02.2025 18:54, joes wrote:
No element of the set of FISONs is
necessary
Therefore all can be removed, and U(F) = ℕ ==> U(F\F) = ℕ.
Regards, WM
On 15.02.2025 18:38, joes wrote:No. It proves that it doesn’t matter *which single element* is removed.
Am Sat, 15 Feb 2025 15:55:55 +0100 schrieb WM:
On 15.02.2025 14:40, Richard Damon wrote:
Induction proves an inductive set F of FISONs that can be removedIndeed, if you remove any number of FISONs, infinitely many remain.The set you show is empty, is the set of FISONs that are individuallyThe remaining set should have at least one element.
not necessary.
without changing the result U(F) = ℕ ==> U(F\F) = ℕ.
on 2/16/2025, WM supposed :
On 16.02.2025 00:52, Jim Burns wrote:
On 2/15/2025 7:49 AM, WM wrote:
Induction proves the existence of
the inductive set of all its members.
ℕ has an only.inductive subset ℕ
The set of FISONs is also an inductive set.
The set of FISONs is essentially the set of natural numbers.
Am Sun, 16 Feb 2025 12:18:28 +0100 schrieb WM:Show what should remain.
If all FISONs are omissible by induction, then the
set of all FISONs is omissible.
That is simply false.
And why? Note that you must define a first element.Indeed, why should something remain if you remove everything?
There is no doubt that this proves the whole set Z without assuming aWDYM „prove a set”?
last element.
„This, and only this segment can be omitted.” Unfortunately there
is no element that encompasses all numbers.
Am Sun, 16 Feb 2025 11:39:45 +0100 schrieb WM:
Therefore all can be removed, and U(F) = ℕ ==> U(F\F) = ℕ.No, obviously you can’t remove all
Am Sun, 16 Feb 2025 11:53:25 +0100 schrieb WM:
On 15.02.2025 23:00, Chris M. Thomasson wrote:The number of FISONs is definitely larger than any finite
There can be an infinite number of FISON's...Not greater than all finite numbers however.
number. Otherwise you could count to a largest one.
No.However, no FISON is infinite in and of itself,Therefore the figure {1}
{2, 1}
{3, 2, 1}
...
has width and height not larger than all finite numbers.
The „width and height” is not a finite number. If it
were, there would be a last FISON without a successor.
Actually, the first column is N.
Am Sun, 16 Feb 2025 11:31:49 +0100 schrieb WM:
On 15.02.2025 18:38, joes wrote:
Am Sat, 15 Feb 2025 15:55:55 +0100 schrieb WM:
On 15.02.2025 14:40, Richard Damon wrote:
No. It proves that it doesn’t matter *which single element* is removed.Induction proves an inductive set F of FISONs that can be removedIndeed, if you remove any number of FISONs, infinitely many remain.The set you show is empty, is the set of FISONs that are individually >>>>> not necessary.The remaining set should have at least one element.
without changing the result U(F) = ℕ ==> U(F\F) = ℕ.
It proves nothing about removing multiple.
On 16.02.2025 12:59, joes wrote:What are you talking about? There are always inf. many if you remove
Am Sun, 16 Feb 2025 12:18:28 +0100 schrieb WM:Show what should remain.
If all FISONs are omissible by induction, then the set of all FISONsThat is simply false.
is omissible.
You DO change the result to the empty set if you remove everything; theEverything is removed that can be removed without changing the result.And why? Note that you must define a first element.Indeed, why should something remain if you remove everything?
You mean „prove the existence”."In order to secure the existence of infinite sets, we need theThere is no doubt that this proves the whole set Z without assuming aWDYM „prove a set”?
last element.
following axiom." [Zermelo] This is the axiom of infinity or induction:
{ } and if a then {a}.
That element would have to be ω or N respectively.For every *element* you can prove the statement
„This, and only this segment can be omitted.” Unfortunately there is no >> element that encompasses all numbers.
On 2/16/25 5:39 AM, WM wrote:
On 15.02.2025 18:54, joes wrote:Which just proves there is no necesary set.
No element of the set of FISONs is
necessary
Therefore all can be removed, and U(F) = ℕ ==> U(F\F) = ℕ.
Am Sun, 16 Feb 2025 16:52:00 +0100 schrieb WM:n.
What are you talking about? There are always inf. many if you remove
a natural number of FISONs, and their union is N.
You DO change the result to the empty set if you remove everything; the
union of all FISONs is not the empty set.
You can only remove a finite
number of them, doesn’t matter which.
On 16.02.2025 13:18, joes wrote:
Am Sun, 16 Feb 2025 11:31:49 +0100 schrieb WM:
On 15.02.2025 18:38, joes wrote:No. It proves that it doesn’t matter *which single element* is removed.
Am Sat, 15 Feb 2025 15:55:55 +0100 schrieb WM:
On 15.02.2025 14:40, Richard Damon wrote:
Induction proves an inductive set F of FISONs that can be removedIndeed, if you remove any number of FISONs, infinitely many remain.The set you show is empty, is the set of FISONs that are individually >>>>>> not necessary.The remaining set should have at least one element.
without changing the result U(F) = ℕ ==> U(F\F) = ℕ.
It proves nothing about removing multiple.
Learn induction.
Show the first FISON which changes the premise U(F) = ℕ.
Regards, WM
On 15.02.2025 20:41, Jim Burns wrote:
On 2/15/2025 7:56 AM, WM wrote:
On 14.02.2025 20:48, Jim Burns wrote:
swaps such that, after all swaps,
Bob is not anywhere he has swapped to.
Then your logic allows
lossless exchanges with losses.
Not acceptable.
If you accept ST+F
I don't.
If you accept ST+F
I don't.
The set of rows, the set of columns, and ⋃{F}
do not change.
Then it would need to contain ℕ,
as an infinite FISON.
Infinite FISONs
(finite initial segments of naturals)
are not acceptable.
Correct. Therefore the figure
{1}
{2, 1}
{3, 2, 1}
...
is finitely broad and finitely high.
On 16.02.2025 00:52, Jim Burns wrote:
{F} has only one inductive subset: {F}
Correct.
{F} is a set of omissible FISONs.
{F} isn't an omissible set of FISONs.
Without the leap, there is no conflict.
But you are wrong.
If all FISONs are omissible by induction,
then the set of all FISONs is omissible.
You (WM) imagine a last finite step,
into the infinite.
All elements can be omitted.
The set can be omitted.
On 14.02.2025 19:06, Jim Burns wrote:
The set of all natural numbers is not.in
the only inductive subset of the set of
all natural numbers.
The set of all natural numbers is constructed
by induction.
On 2/16/25 11:51 AM, WM wrote:
Show the first FISON which changes the premise U(F) = ℕ.
Why does there need to be?
Just because there is not individual FISON that when removed changes the results, doesn't mean that you can remove ALL.
On 2/16/25 11:33 AM, WM wrote:
On 16.02.2025 13:11, joes wrote:
Am Sun, 16 Feb 2025 11:39:45 +0100 schrieb WM:
Therefore all can be removed, and U(F) = ℕ ==> U(F\F) = ℕ.No, obviously you can’t remove all
Proofs by induction cover all elements of the set. My proof proves
that no FISON is capable of changing the premise UF = ℕ.
There is no requirement for a "necessary" elememt for the set.
On 16.02.2025 00:52, Jim Burns wrote:
{F} has only one inductive subset: {F}
Correct.
All elements can be omitted.
The set can be omitted.
On 2/15/2025 9:51 AM, WM wrote:
On 14.02.2025 19:06, Jim Burns wrote:
The set of all natural numbers is not.in
the only inductive subset of the set of
all natural numbers.
The set of all natural numbers is constructed
by induction.
By axiom "infinity",
an inductive set exists.
If the set with an only.inductive.subset
was an element, we'd prove the set has P.
But the set isn't an element.
We don't prove the set has P.
On 16.02.2025 19:39, Richard Damon wrote:
Just because there is not
individual FISON that when removed
changes the results,
doesn't mean that you can remove ALL.
Proofs by induction cover all FISONs.
On 2/16/2025 6:18 AM, WM wrote:
All elements can be omitted.
The set can be omitted.
The set is not an element.
On 16.02.2025 21:56, Jim Burns wrote:Exactly.
On 2/16/2025 6:18 AM, WM wrote:
If every human has an end, then the human race need not have an end.All elements can be omitted.The set is not an element.
The set can be omitted.
If every human has ended, then the human race has ended.If. Induction doesn’t prove that.
On 16.02.2025 23:43, Jim Burns wrote:
On 2/15/2025 9:51 AM, WM wrote:
On 14.02.2025 19:06, Jim Burns wrote:
The set of all natural numbers is not.in
the only inductive subset of the set of
all natural numbers.
The set of all natural numbers is constructed
by induction.
By axiom "infinity",
an inductive set exists.
The axiom applies induction.
If the set with an only.inductive.subset
was an element, we'd prove the set has P.
But the set isn't an element.
We don't prove the set has P.
Sometimes this is right, sometimes it is wrong.
When all elements of a set are subject to induction
then the set is an inductive set.
When all elements of a set are removed,
then the set is removed.
Example:
If every human has an end,
then the human race need not have an end.
If every human has ended,
then the human race has ended.
On 16.02.2025 19:39, Richard Damon wrote:
Just because there is not individual FISON that when removed changes
the results, doesn't mean that you can remove ALL.
Proofs by induction cover all FISONs.
Regards, WM
Am Mon, 17 Feb 2025 20:40:47 +0100 schrieb WM:
On 16.02.2025 21:56, Jim Burns wrote:Exactly.
On 2/16/2025 6:18 AM, WM wrote:If every human has an end, then the human race need not have an end.
All elements can be omitted.The set is not an element.
The set can be omitted.
If every human has ended, then the human race has ended.If. Induction doesn’t prove that.
On 2/17/2025 2:27 PM, WM wrote:
Proofs by induction cover all FISONs.
Proofs by induction prove that
some property A(k) describes each element of
some inductive
That reasoning is silent about
whether the _set_ (not its elements) has A(k).
In an inductive set with an only.inductive.subset,
there is no element which,
upon the removal of it and its priors,
all the elements -- or even _almost_ all --
have been removed.
Example:
If every human has an end,
then the human race need not have an end.
If every human has ended,
then the human race has ended.
If each human ends
but, a day after they end, another has not ended,
then the human race does not end.
If there is a baton (named 'Bob') such that
each human receiving Bob passes it to
a human who ends a day or more after they end,
then,
After all passes, Bob isn't.
Infinity isn't finite.
It isn't even almost finite.
On 2/17/25 2:25 PM, WM wrote:
A set without elements is an empty set and not capable of producing ℕ.
But an empty set of REQUIRED elements doesn't mean we can't have a set
of sufficient elements.
We can build a sufficent set from non-required FISONs, just like we can factor 36 from non-required factors.
On 17.02.2025 21:36, joes wrote:You can’t prove that humanity dies from the fact that every human dies.
Am Mon, 17 Feb 2025 20:40:47 +0100 schrieb WM:But it is obviously true.
On 16.02.2025 21:56, Jim Burns wrote:Exactly.
On 2/16/2025 6:18 AM, WM wrote:If every human has an end, then the human race need not have an end.
All elements can be omitted.The set is not an element.
The set can be omitted.
If every human has ended, then the human race has ended.If. Induction doesn’t prove that.
Induction, as applied by Zermelo and others, proves: If every element is created or described by induction, then the set of all elements isNo.
created or described by induction.
Further: If all elements of a set are subtracted, then the set isMakes no sense. There is nothing to subtract the set from.
subtracted. What should remain?
If every element of a set is countable, then the set is a countable set.What the fuck, absolutely not.
On 17.02.2025 20:59, Jim Burns wrote:But the union has changed from a nonempty set to an empty one.
On 2/17/2025 2:27 PM, WM wrote:
Proofs by induction cover all FISONs.
Proofs by induction prove that some property A(k) describes each
element of some inductive
set.
Here this property is that FISON F(n) can be removed without changing
the premise U({F(1), F(2), F(3), ...} \ {F(1), F(2), ..., F(n)}) = ℕ.
(*)
That reasoning is silent about whether the _set_ (not its elements) hasTherefore I gave you an example that you should be able to understand:
A(k).
If every human has ended, then the human race has ended.
Analogously: If every FISON has been removed without changing the union,
then the set {F(1), F(2), F(39, ...} has been removed without changing
the union.
On 18.02.2025 04:02, Richard Damon wrote:
We can build a sufficent set from non-required FISONs, just like we
can factor 36 from non-required factors.
We can go through the factors and find the first required factor 6.
We can go through the FISONs and find that a FISON F(n) changing
U({F(1), F(2), F(3), ...} \ {F(1), F(2), ..., F(n)}) = ℕ
does not exist.
The belief that some smaller FISONs would help shows missing brain.
Regards, WM
Am Tue, 18 Feb 2025 09:54:02 +0100 schrieb WM:
Therefore I gave you an example that you should be able to understand:But the union has changed from a nonempty set to an empty one.
If every human has ended, then the human race has ended.
Analogously: If every FISON has been removed without changing the union,
then the set {F(1), F(2), F(3), ...} has been removed without changing
the union.
On 2/18/25 5:02 AM, WM wrote:
On 18.02.2025 04:02, Richard Damon wrote:
We can build a sufficent set from non-required FISONs, just like we
can factor 36 from non-required factors.
We can go through the factors and find the first required factor 6.
6 isn't a required factor, as you can use 4 * 9.
On 18.02.2025 11:08, joes wrote:No, the other way around: if humanity has ended, every human „has ended”.
Am Tue, 18 Feb 2025 10:02:53 +0100 schrieb WM:Correct! But if every human has ended, then humanity has ended.
On 17.02.2025 21:36, joes wrote:You can’t prove that humanity dies from the fact that every human dies.
Am Mon, 17 Feb 2025 20:40:47 +0100 schrieb WM:But it is obviously true.
On 16.02.2025 21:56, Jim Burns wrote:Exactly.
On 2/16/2025 6:18 AM, WM wrote:If every human has an end, then the human race need not have an end.
All elements can be omitted.The set is not an element.
The set can be omitted.
If every human has ended, then the human race has ended.If. Induction doesn’t prove that.
Indeed.Here in sci.math we should adhere to mathematics.Induction, as applied by Zermelo and others, proves: If every elementNo.
is created or described by induction, then the set of all elements is
created or described by induction.
For every set M that doesn’t contain itself, M \ {M} = M.Further: If all elements of a set are subtracted, then the set isMakes no sense. There is nothing to subtract the set from.
subtracted. What should remain?
--If every element of a set is countable, then the set is a countableWhat the fuck, absolutely not.
set.
„If every natural is finite, then there are only finitely many
naturals.”?
On 18.02.2025 11:09, joes wrote:Um, that proves that you cannot remove everything without changing
Am Tue, 18 Feb 2025 09:54:02 +0100 schrieb WM:
That's the proof. Nice that you now understand it.Therefore I gave you an example that you should be able to understand:But the union has changed from a nonempty set to an empty one.
If every human has ended, then the human race has ended.
Analogously: If every FISON has been removed without changing the
union,
then the set {F(1), F(2), F(3), ...} has been removed without changing
the union.
On 2/18/25 4:53 AM, WM wrote:
On 18.02.2025 04:02, Richard Damon wrote:
On 2/17/25 2:25 PM, WM wrote:
A set without elements is an empty set and not capable of producing ℕ. >>But an empty set of REQUIRED elements doesn't mean we can't have a
set of sufficient elements.
There is no element that could be a meaningful member of any
sufficient set. Therefore there is no sufficient set.
Of course there are, its just they are not individually needed, but are collectively sufficient.
Am Tue, 18 Feb 2025 10:02:53 +0100 schrieb WM:
On 17.02.2025 21:36, joes wrote:You can’t prove that humanity dies from the fact that every human dies.
Am Mon, 17 Feb 2025 20:40:47 +0100 schrieb WM:But it is obviously true.
On 16.02.2025 21:56, Jim Burns wrote:Exactly.
On 2/16/2025 6:18 AM, WM wrote:If every human has an end, then the human race need not have an end.
All elements can be omitted.The set is not an element.
The set can be omitted.
If every human has ended, then the human race has ended.If. Induction doesn’t prove that.
Induction, as applied by Zermelo and others, proves: If every element isNo.
created or described by induction, then the set of all elements is
created or described by induction.
Am Tue, 18 Feb 2025 16:18:52 +0100 schrieb WM:
On 18.02.2025 11:09, joes wrote:Um, that proves that you cannot remove everything without changing
Am Tue, 18 Feb 2025 09:54:02 +0100 schrieb WM:That's the proof. Nice that you now understand it.
Therefore I gave you an example that you should be able to understand: >>>> If every human has ended, then the human race has ended.But the union has changed from a nonempty set to an empty one.
Analogously: If every FISON has been removed without changing the
union,
then the set {F(1), F(2), F(3), ...} has been removed without changing >>>> the union.
the union. What else could you have meant by that?
On 18.02.2025 16:28, joes wrote:It doesn’t matter what you assume the union to be. You quite sensibly
Am Tue, 18 Feb 2025 16:18:52 +0100 schrieb WM:We can remove every FISON without changing the *assumed* union ℕ.
On 18.02.2025 11:09, joes wrote:Um, that proves that you cannot remove everything without changing the
Am Tue, 18 Feb 2025 09:54:02 +0100 schrieb WM:That's the proof. Nice that you now understand it.
Therefore I gave you an example that you should be able toBut the union has changed from a nonempty set to an empty one.
understand:
If every human has ended, then the human race has ended.
Analogously: If every FISON has been removed without changing the
union,
then the set {F(1), F(2), F(3), ...} has been removed without
changing the union.
union. What else could you have meant by that?
On 17.02.2025 20:59, Jim Burns wrote:
On 2/17/2025 2:27 PM, WM wrote:
Proofs by induction cover all FISONs.
Proofs by induction prove that
some property A(k) describes each element of
some inductive
set.
Here this property is that
FISON F(n) can be removed without changing
the premise
U({F(1),F(2),F(3),...}\{F(1),F(2),...,F(n)}) = ℕ.
(*)
That reasoning is silent about
whether the _set_ (not its elements) has A(k).
Therefore I gave you an example
that you should be able to understand:
If every human has ended,
then the human race has ended.
Analogously:
If every FISON has been removed
without changing the union, then the set
{F(1), F(2), F(39, ...}
has been removed without changing the union.
On 18.02.2025 13:25, Richard Damon wrote:
On 2/18/25 4:53 AM, WM wrote:
On 18.02.2025 04:02, Richard Damon wrote:
On 2/17/25 2:25 PM, WM wrote:
A set without elements is an empty set
and not capable of producing ℕ.
But an empty set of REQUIRED elements
doesn't mean we can't have a
set of sufficient elements.
There is no element that could be
a meaningful member of any sufficient set.
Therefore there is no sufficient set.
Therefore there is no sufficient set.
Of course there are,
its just they are not individually needed,
but are collectively sufficient.
For every FISON there is the question:
Can it belong to a collectively sufficient set.
For every FISON the answer is no.
On 17.02.2025 23:03, Jim Burns wrote:
In an inductive set with an only.inductive.subset,
there is no element which,
upon the removal of it and its priors,
all the elements -- or even _almost_ all --
have been removed.
Therefore such proofs are done by induction.
They cover all elements.
Example:
If every human has an end,
then the human race need not have an end.
If every human has ended,
then the human race has ended.
If each human ends
but, a day after they end, another has not ended,
then the human race does not end.
If every human has ended, then there is no other one.
If there is a baton (named 'Bob') such that
each human receiving Bob passes it to
a human who ends a day or more after they end,
then,
Bob passes with the last one.
After all passes, Bob isn't.
Infinity isn't finite.
It isn't even almost finite.
Nevertheless:
If all elements of a set have been omitted,
then the set has been omitted.
On 18.02.2025 13:25, Richard Damon wrote:
On 2/18/25 5:02 AM, WM wrote:
On 18.02.2025 04:02, Richard Damon wrote:
We can build a sufficent set from non-required FISONs, just like we
can factor 36 from non-required factors.
We can go through the factors and find the first required factor 6.
6 isn't a required factor, as you can use 4 * 9.
We go through the ordinals, with FISONs and with factors. There is a
first required factor. There is no first required FISON. All are useless because of
∀n ∈ U(F(n)): |ℕ \ {1, 2, 3, ..., n}| = ℵo.
Regards, WM
On 18.02.2025 13:25, Richard Damon wrote:
On 2/18/25 4:53 AM, WM wrote:
On 18.02.2025 04:02, Richard Damon wrote:
On 2/17/25 2:25 PM, WM wrote:
A set without elements is an empty set and not capable of producing ℕ. >>>But an empty set of REQUIRED elements doesn't mean we can't have a
set of sufficient elements.
There is no element that could be a meaningful member of any
sufficient set. Therefore there is no sufficient set.
Of course there are, its just they are not individually needed, but
are collectively sufficient.
For every FISON there is the question: Can it belong to a collectively sufficient set. For every FISON the answer is no.
Regards, WM
On 18.02.2025 13:25, Richard Damon wrote:
On 2/18/25 4:53 AM, WM wrote:
On 18.02.2025 04:02, Richard Damon wrote:
On 2/17/25 2:25 PM, WM wrote:
A set without elements is an empty set and not capable of producing ℕ. >>>But an empty set of REQUIRED elements doesn't mean we can't have a
set of sufficient elements.
There is no element that could be a meaningful member of any
sufficient set. Therefore there is no sufficient set.
Of course there are, its just they are not individually needed, but
are collectively sufficient.
For every FISON there is the question: Can it belong to a collectively sufficient set. For every FISON the answer is no.
Regards, WM
Am Tue, 18 Feb 2025 16:15:47 +0100 schrieb WM:
On 18.02.2025 11:08, joes wrote:No, the other way around: if humanity has ended, every human „has ended”.
Am Tue, 18 Feb 2025 10:02:53 +0100 schrieb WM:
On 17.02.2025 21:36, joes wrote:You can’t prove that humanity dies from the fact that every human dies. >> Correct! But if every human has ended, then humanity has ended.
Am Mon, 17 Feb 2025 20:40:47 +0100 schrieb WM:But it is obviously true.
On 16.02.2025 21:56, Jim Burns wrote:If. Induction doesn’t prove that.
On 2/16/2025 6:18 AM, WM wrote:If every human has an end, then the human race need not have an end. >>>>> Exactly.
All elements can be omitted.The set is not an element.
The set can be omitted.
If every human has ended, then the human race has ended.
Indeed.Here in sci.math we should adhere to mathematics.Induction, as applied by Zermelo and others, proves: If every elementNo.
is created or described by induction, then the set of all elements is
created or described by induction.
Am Tue, 18 Feb 2025 16:34:20 +0100 schrieb WM:
We can remove every FISON without changing the *assumed* union ℕ.It doesn’t matter what you assume the union to be. You quite sensibly denied that it is empty, yet you claim it would (not? I’m confused)
change.
For inductive sets with multiple.inductive.subsets,
it's inadequate to prove that a subset is inductive
in order to conclude that subset is the whole set.
The subset of {F} with that property
is inductive.
Showing that a subset is inductive
is called proof by induction.
{F} is the only.inductive.subset of {F}.
In the case of {F}, a proof by induction shows
that any subset of {F} with that property is {F},
because that subset can't be anything else..
That reasoning is silent about
whether the _set_ (not its elements) has A(k).
Am 18.02.2025 um 16:32 schrieb joes:No. Humanity can survive even though every human is mortal.
Am Tue, 18 Feb 2025 16:15:47 +0100 schrieb WM:
On 18.02.2025 11:08, joes wrote:
Am Tue, 18 Feb 2025 10:02:53 +0100 schrieb WM:
On 17.02.2025 21:36, joes wrote:
Am Mon, 17 Feb 2025 20:40:47 +0100 schrieb WM:
On 16.02.2025 21:56, Jim Burns wrote:
On 2/16/2025 6:18 AM, WM wrote:
It is an equivalence.No, the other way around: if humanity has ended, every human „hasCorrect! But if every human has ended, then humanity has ended.You can’t prove that humanity dies from the fact that every humanBut it is obviously true.Exactly.If every human has an end, then the human race need not have anAll elements can be omitted. The set can be omitted.The set is not an element.
end.
If every human has ended, then the human race has ended.If. Induction doesn’t prove that.
dies.
ended”.
They didn’t say that by induction, you can infer properties of a setThen read what they said.Indeed.Here in sci.math we should adhere to mathematics.Induction, as applied by Zermelo and others, proves: If everyNo.
element is created or described by induction, then the set of all
elements is created or described by induction.
Such as for the set of all FISONs.For every set M that doesn’t contain itself, M \ {M} = M.Further: If all elements of a set are subtracted, then the set isMakes no sense. There is nothing to subtract the set from.
subtracted. What should remain?
Word on this?If every element of a set is countable, then the set is a countableWhat the fuck, absolutely not.
set.
„If every natural is finite, then there are only finitely many
naturals.”?
Am 18.02.2025 um 17:45 schrieb joes:Nobody claimed that the union of the empty set were N.
Am Tue, 18 Feb 2025 16:34:20 +0100 schrieb WM:
It is enough to show that the claimed union is mistaken.We can remove every FISON without changing the *assumed* union ℕ.It doesn’t matter what you assume the union to be. You quite sensibly
denied that it is empty, yet you claim it would (not? I’m confused)
change.
Name a FISON that can not be put into a set that is sufficient set, one
whose union is the set of Natural Numbers.
On 2/18/2025 10:22 AM, WM wrote:
You (WM) have located a problem.
You try to work around it by not.mentioning it.
What you're not.mentioning is your assumption
that none of these sets are infinite.
⋃{F} = ℕ
Am 19.02.2025 um 13:17 schrieb Richard Damon:Dude, you cannot still have the same nonempty union after removing every element, even if you do not think it is N. Do you think the union of all
Name a FISON that can not be put into a set that is sufficient set, oneEvery FISON. There is no sufficient set. If it is assumed, then F(1) can
whose union is the set of Natural Numbers.
be omitted without changing the union of the remainder. And if F(n) can
be omitted without changing this union, then also F(n+1) can be omitted without changing this union. That makes the omitted FISONs the inductive collection of all FISONs and proves the implication: If UF = ℕ, then { }
= ℕ.
Am 18.02.2025 um 19:22 schrieb Jim Burns:Go on not mentioning it.
On 2/18/2025 10:22 AM, WM wrote:
You (WM) have located a problem.Wrong. Induction has been invented for infinite sets.
You try to work around it by not.mentioning it.
What you're not.mentioning is your assumption that none of these sets
are infinite.
Um aber die Existenz "unendlicher" Mengen zu sichern, bedürfen wir noch
des folgenden ... Axioms. [Zermelo: Untersuchungen über die Grundlagen
der Mengenlehre I, S. 266]
Disproof: UF != {}⋃{F} = ℕProof: If UF = ℕ is assumed, then F(1) can be omitted without changing
the union of the remainder. And if F(n) can be omitted without changing
this union, then also F(n+1) can be omitted without changing this union.
That makes the omitted FISONs the inductive collection of all FISONs and proves the implication: If UF = ℕ, then { } = ℕ.
Am Wed, 19 Feb 2025 15:31:54 +0100 schrieb WM:
They didn’t say that by induction, you can infer properties of a set
from its elements - like being finite or not. Compare {Q, N, R},
the set of endsegments, and {0, 1, 2}.
Word on this?„If every natural is finite, then there are only finitely many
naturals.”?
Am Wed, 19 Feb 2025 15:28:51 +0100 schrieb WM:
Am 18.02.2025 um 17:45 schrieb joes:Nobody claimed that the union of the empty set were N.
Am Tue, 18 Feb 2025 16:34:20 +0100 schrieb WM:It is enough to show that the claimed union is mistaken.
We can remove every FISON without changing the *assumed* union ℕ.It doesn’t matter what you assume the union to be. You quite sensibly
denied that it is empty, yet you claim it would (not? I’m confused)
change.
Am 19.02.2025 um 15:50 schrieb joes:Still true.
Am Wed, 19 Feb 2025 15:31:54 +0100 schrieb WM:
No. Humanity can survive even though every human is mortal.It is an equivalence.No, the other way around: if humanity has ended, every human „hasCorrect! But if every human has ended, then humanity has ended.You can’t prove that humanity dies from the fact that every human >>>>>> dies.Exactly.If every human has an end, then the human race need not have an >>>>>>>>> end.All elements can be omitted. The set can be omitted.The set is not an element.
If every human has ended, then the human race has ended.
ended”.
Sure. Nobody is arguing against that.They didn’t say that by induction, you can infer properties of a setBy induction all elements can be defined. This guarantees the existence
from its elements - like being finite or not. Compare {Q, N, R},
the set of endsegments, and {0, 1, 2}.
of an infinite set. Um aber die Existenz "unendlicher" Mengen zu
sichern, bedürfen wir noch des folgenden ... Axioms. [Zermelo: Untersuchungen über die Grundlagen der Mengenlehre I, S. 266]
The set of all FISONs does not contain the set of all FISONs.Further: If all elements of a set are subtracted, then the set is >>>>>>> subtracted. What should remain?
By „every natural” I obviously mean N, not a finite subset.The set of all natural numbers which can be defined is (potentially in-) finite. infinitesimally smaller than ℕ.Word on this?„If every natural is finite, then there are only finitely many
naturals.”?
Am 19.02.2025 um 15:51 schrieb joes:No. That would mean UF = {}, which you can’t seriously believe.
Am Wed, 19 Feb 2025 15:28:51 +0100 schrieb WM:Proven however is this: UF = ℕ ⟹ Ø = ℕ.
Am 18.02.2025 um 17:45 schrieb joes:Nobody claimed that the union of the empty set were N.
Am Tue, 18 Feb 2025 16:34:20 +0100 schrieb WM:It is enough to show that the claimed union is mistaken.
We can remove every FISON without changing the *assumed* union ℕ.It doesn’t matter what you assume the union to be. You quite sensibly >>>> denied that it is empty, yet you claim it would (not? I’m confused)
change.
Am 18.02.2025 um 19:22 schrieb Jim Burns:
You (WM) have located a problem.
You try to work around it by not.mentioning it.
What you're not.mentioning is your assumption
that none of these sets are infinite.
Wrong.
There is no element that could be
a meaningful member of any sufficient set.
Therefore there is no sufficient set.
Of course there are,
its just they are not individually needed,
but are collectively sufficient.
For every FISON there is the question:
Can it belong to a collectively sufficient set.
For every FISON the answer is no.
Induction has been invented for infinite sets.
Um aber die Existenz "unendlicher" Mengen zu sichern,
bedürfen wir noch des folgenden ... Axioms.
[Zermelo: Untersuchungen über
die Grundlagen der Mengenlehre I, S. 266]
⋃{F} = ℕ
Proof:
If UF = ℕ is assumed, then
F(1) can be omitted
without changing the union of the remainder.
And if F(n) can be omitted
without changing this union,
then also F(n+1) can be omitted
without changing this union.
That makes the omitted
That makes the [omissible] FISONs
the inductive collection of all FISONs
proves the implication:
If UF = ℕ, then { } = ℕ.
Am 18.02.2025 um 17:45 schrieb joes:
Am Tue, 18 Feb 2025 16:34:20 +0100 schrieb WM:
We can remove every FISON without changing the *assumed* union ℕ.It doesn’t matter what you assume the union to be. You quite sensibly
denied that it is empty, yet you claim it would (not? I’m confused)
change.
It is enough to show that the claimed union is mistaken.
Regards, WM
Not proven for the set {F} of FISONs,
which is not a FISON.
The sum of any two natural numbers
is a natural number.
The "sum" of all the natural numbers,
for any reasonable definition of that,
will be larger than any natural number,
and not a natural number.
WM has brought this to us :
Induction has been invented for infinite sets.
Transfinite Induction has been...
On 2/19/25 11:52 AM, WM wrote:
Proof: If UF = ℕ is assumed, then F(1) can be omitted without changing
the union of the remainder. And if F(n) can be omitted without
changing this union, then also F(n+1) can be omitted without changing
this union. That makes the omitted FISONs the inductive collection of
all FISONs and proves the implication: If UF = ℕ, then { } = ℕ.
But induction doesn't subtract elements.
All you have shown is that the no element in the set of all FISON is
neeeded.
The problem is your "set" UF isn't being defined by a proper set theory,
but just by Naive Set Theory.
Am Wed, 19 Feb 2025 18:05:01 +0100 schrieb WM:
By induction all elements can be defined. This guarantees the existenceSure. Nobody is arguing against that.
of an infinite set. Um aber die Existenz "unendlicher" Mengen zu
sichern, bedürfen wir noch des folgenden ... Axioms. [Zermelo:
Untersuchungen über die Grundlagen der Mengenlehre I, S. 266]
Am Wed, 19 Feb 2025 18:07:56 +0100 schrieb WM:
Proven however is this: UF = ℕ ⟹ Ø = ℕ.No. That would mean UF = {}, which you can’t seriously believe.
If you think UF = X != N, then you need to accept U{} = X.
On 2/19/25 11:57 AM, WM wrote:
Note, your subject line uses the word you mean, "necessary", but you
ignore the fact that a set of necessary elements doesn't need to exist.
Am Wed, 19 Feb 2025 17:16:15 +0000 schrieb joes:Will the human race go extinct if we don’t achieve immortality?
Am Wed, 19 Feb 2025 18:05:01 +0100 schrieb WM:
Am 19.02.2025 um 15:50 schrieb joes:
Am Wed, 19 Feb 2025 15:31:54 +0100 schrieb WM:
Still true.No. Humanity can survive even though every human is mortal.It is an equivalence.No, the other way around: if humanity has ended, every human „has >>>>>> ended”.Correct! But if every human has ended, then humanity has ended.You can’t prove that humanity dies from the fact that every human >>>>>>>> dies.Exactly.If every human has an end, then the human race need not have >>>>>>>>>>> an end.All elements can be omitted. The set can be omitted.The set is not an element.
If every human has ended, then the human race has ended.
Substitute: UF = X != N. Does it follow that X = {}?Fine. Induction covers all elements of an infinite inductive set. ThereSure. Nobody is arguing against that.They didn’t say that by induction, you can infer properties of a set >>>> from its elements - like being finite or not. Compare {Q, N, R},By induction all elements can be defined. This guarantees the
the set of endsegments, and {0, 1, 2}.
existence of an infinite set.
is no k+1 remaining.
Subtraction of all elements leaves the empty set.
UF = ℕ ==> Ø = ℕ.
Is that because every natural is finite?The set of all FISONs does not contain the set of all FISONs.Further: If all elements of a set are subtracted, then the set >>>>>>>>> is subtracted. What should remain?
The set of all natural numbers which can be defined is„If every natural is finite, then there are only finitely many >>>>>>>> naturals.”?
finite. infinitesimally smaller than ℕ.
--By „every natural” I obviously mean N, not a finite subset.
[spam deleted]
Am 19.02.2025 um 15:50 schrieb joes:
Am Wed, 19 Feb 2025 15:31:54 +0100 schrieb WM:
They didn’t say that by induction, you can infer properties of a set
from its elements - like being finite or not. Compare {Q, N, R},
the set of endsegments, and {0, 1, 2}.
By induction all elements can be defined. This guarantees the existence
of an infinite set. Um aber die Existenz "unendlicher" Mengen zu
sichern, bedürfen wir noch des folgenden ... Axioms. [Zermelo: Untersuchungen über die Grundlagen der Mengenlehre I, S. 266]
Word on this?„If every natural is finite, then there are only finitely many
naturals.”?
The set of all natural numbers which can be defined is (potentially in-) finite. infinitesimally smaller than ℕ.
Proof: If UF = ℕ is assumed, then F(1) can be omitted without changing
the union of the remainder.
And if F(n) can be omitted without changing this union, then also
F(n+1) can be omitted without changing this union.
That makes the omitted FISONs the inductive collection of all FISONs
and proves the implication: If UF = ℕ, then { } = ℕ.
Regards, WM
On 20.02.2025 03:23, Richard Damon wrote:
On 2/19/25 11:57 AM, WM wrote:
Note, your subject line uses the word you mean, "necessary", but you
ignore the fact that a set of necessary elements doesn't need to exist.
Assume a set of sufficient FISONs. |ℕ \ {1, 2, 3, ..., n}| = ℵo is true for all FISONs. That contradicts the assumption.
Regards, WM
Am 18.02.2025 um 18:14 schrieb Jim Burns:
For inductive sets with multiple.inductive.subsets,
it's inadequate to prove that a subset is inductive
in order to conclude that subset is the whole set.
The [set of all] FISONs [is]
an inductive set with only one inductive subset.
The subset of {F} [holding all omissibles]
is inductive.
Showing that a subset is inductive
is called proof by induction.
That is what I do.
{F} is the only.inductive.subset of {F}.
In the case of {F}, a proof by induction shows
that any subset of {F} with that property is {F},
because that subset can't be anything else..
That reasoning is silent about
whether the _set_ (not its elements) has A(k).
The set without any element is empty.
The elements are defined by induction
in order to guarantee the infinite set.
Um aber die Existenz "unendlicher" Mengen zu sichern,
bedürfen wir noch des folgenden ... Axioms.
[Zermelo: Untersuchungen über
die Grundlagen der Mengenlehre I, S. 266]
On 19.02.2025 19:58, Jim Burns wrote:
Not proven for the set {F} of FISONs,
which is not a FISON.
1)
Induction covers all elements of
an infinite inductive set.
F(1) ∈ F und F(n) ∈ F ==> F(n+1) ∈ F
describes the infinite inductive set F of FISONs.
2) Subtraction all
FISONs {1, 2, 3, ..., n} satisfying
|ℕ \ {1, 2, 3, ..., n}| = ℵo
from F leaves the empty set.
These two well-established arguments prove my case:
UF = ℕ ==> Ø = ℕ.
The sum of any two natural numbers
is a natural number.
The "sum" of all the natural numbers,
for any reasonable definition of that,
will be larger than any natural number,
and not a natural number.
Like the product.
Reason is the potential infinity of definable numbers.
Nevertheless:
Zermelo proves
the existence of an inductive *set*
by induction.
AXIOM I 'extensionality' means
proving each FISON is omissible is no
proving {F} is omissible.
Why do you think the set has the same property as
its elements?
Nevertheless:
Zermelo proves
the existence of an inductive *set*
by induction.
You keep using that word.
I do not think it means what you think it means.
Am Thu, 20 Feb 2025 11:26:13 +0100 schrieb WM:
Subtraction of all elements leaves the empty set.Substitute: UF = X != N. Does it follow that X = {}?
UF = ℕ ==> Ø = ℕ.
Is that because every natural is finite?The set of all natural numbers which can be defined is
finite. infinitesimally smaller than ℕ.
WM <[email protected]> wrote:
The set of all natural numbers which can be defined is (potentially in-)
finite. infinitesimally smaller than ℕ.
The set of all natural numbers is N, by definition. N is infinite.
Lose the useless qulifications. They're not helping anybody.
Proof: If UF = ℕ is assumed, then F(1) can be omitted without changing
the union of the remainder.
I can accept this, though it needs a bit more rigour. Under what
conditions can a FISON be omitted from a union of them without changing
that union? Only when there is a subsequent FISON in the union.
And if F(n) can be omitted without changing this union, then also
F(n+1) can be omitted without changing this union.
This is questionable,indeed. What precisely is the nature of the relationship between F(n) and F(n+1) which allows this?
You have failed to prove this inductive step.
Again, you can omit F(n+1) when and only when there is a bigger FISON in
the union.
That makes the omitted FISONs the inductive collection of all FISONs
and proves the implication: If UF = ℕ, then { } = ℕ.
You have entirely failed to prove that. A problem is that you are
defining members of a set only by their relationship to other members.
On finishing your alleged induction, that relationship no longer holds -
You end up with a set of FISONs none of which can be omitted, since
there isn't a bigger FISON in the complementary set of remainders. So
the whole mechanism collapses like a bubble bursting.
A large part of the problem is that you are using FISONs as though they
were some sort of primitive.
See how long ago it is since you actually
defined what you mean by FISON on this group, if you ever have.
On 20.02.2025 03:23, Richard Damon wrote:
Assume a set of sufficient FISONs.
|ℕ \ {1, 2, 3, ..., n}| = ℵo
is true for all FISONs.
That contradicts the assumption.
That contradicts the assumption.
On 2/20/2025 5:32 AM, WM wrote:
On 20.02.2025 03:23, Richard Damon wrote:
Assume a set of sufficient FISONs.
== Assume S ⊆ {F} exists such that
⋃S is the only.inductive.subset of ⋃S
|ℕ \ {1, 2, 3, ..., n}| = ℵo
is true for all FISONs.
Yes.
⎛ For any two FISONs {i:i≤j} and {i:i≤k},
⎜ their sum {i:i≤j+k} is a FISON.
That contradicts the assumption.
What is the assumption?
What is the contradiction?
On 20.02.2025 15:30, Jim Burns wrote:
AXIOM I 'extensionality' means
proving each FISON is omissible is no
proving {F} is omissible.
∀n ∈ ℕ: n+1 ∈ ℕ.
Together with 1 ∈ ℕ this defines the set ℕ.
(*)
Addition of all numbers defined by (*)
to the empty set
is tantamount to
addition of ℕ to the empty set.
Subtraction of all numbers defined by (*)
from ℕ
is tantamount to
subtraction of ℕ from ℕ.
Homework:
Prove the same for FISONs or v. Neumann ordinals.
On 20.02.2025 15:30, Jim Burns wrote:Absolutely not. That would be {N}, which is not equal to N = {1, 2,
AXIOM I 'extensionality' means proving each FISON is omissible is no∀n ∈ ℕ: n+1 ∈ ℕ. Together with 1 ∈ ℕ this defines the set ℕ. (*)
proving {F} is omissible.
Addition of all numbers defined by (*) to the empty set is tantamount to addition of ℕ to the empty set.
On 20.02.2025 20:05, Jim Burns wrote:
On 2/20/2025 5:32 AM, WM wrote:
Assume a set of sufficient FISONs.
== Assume S ⊆ {F} exists such that
⋃S is the only.inductive.subset of ⋃S
|ℕ \ {1, 2, 3, ..., n}| = ℵo
is true for all FISONs.
Yes.
⎛ For any two FISONs {i:i≤j} and {i:i≤k},
⎜ their sum {i:i≤j+k} is a FISON.
Yes.
That contradicts the assumption.
What is the assumption?
The assumption is the existence of S.
What is the contradiction?
The contradiction is that
induction proves every FISON useless
and therefore S not existing.
On 2/20/2025 2:27 PM, WM wrote:
On 20.02.2025 20:05, Jim Burns wrote:
On 2/20/2025 5:32 AM, WM wrote:
Assume a set of sufficient FISONs.
== Assume S ⊆ {F} exists such that
⋃S is the only.inductive.subset of ⋃S
|ℕ \ {1, 2, 3, ..., n}| = ℵo
is true for all FISONs.
Yes.
⎛ For any two FISONs {i:i≤j} and {i:i≤k},
⎜ their sum {i:i≤j+k} is a FISON.
Yes.
⎛ For any two FISONs {i:i≤j} and {i:i≤k},
⎜ their sum {i:i≤j+k} is a FISON.
⎜
⎜ For each {i:i≤k} ⊆ ⋃S, there is
⎜ a larger {i:j+1≤i≤k+1} ⊆ ⋃S\{i:i≤j}
Are you denying that {i:j+1≤i≤k+1} exists?
Are you denying that {i:j+1≤i≤k+1} is
larger than {i:i≤k} ?
⎜ ⋃S\{i:i≤j} is not.smaller than ⋃S
⎜
⎜ ⋃S\{i:i≤j} ⊆ ⋃S
⎜
⎝ ⋃S\{i:i≤j} is not.larger than ⋃S
That contradicts the assumption.
What is the assumption?
The assumption is the existence of S.
What is the contradiction?
The contradiction is that
induction proves every FISON useless
Each FISON being uselessᵂᴹ (not.last) is
a consequence of ⋃S
being the only.inductive.subset of ⋃S
And vice versa.
What are the TWO statements which
contradict each other?
and therefore S not existing.
On 20.02.2025 17:03, Jim Burns wrote:
Why do you think the set has the same property as
its elements?
That is not true in general but can be proven in this case. See my other posting. Or see Frege: When all trees of a forest haven been burned,
then the forest has been burned.
Nevertheless:
Zermelo proves
the existence of an inductive *set*
by induction.
You keep using that word.
I do not think it means what you think it means.
Then read it and ponder over it until you recognize that he said it:
The elements are defined by induction in order to guarantee the infinite
set. Um aber die Existenz "unendlicher" Mengen zu sichern, bedürfen wir
noch des folgenden ... Axioms. [Zermelo: Untersuchungen über die
Grundlagen der Mengenlehre I, S. 266]
Regards, WM
In order to secure the existance of "infinite" sets, we still need the following axioms.
There is no mention of "induction" in the statement.
Am Thu, 20 Feb 2025 18:50:27 +0100 schrieb WM:
∀n ∈ ℕ: n+1 ∈ ℕ. Together with 1 ∈ ℕ this defines the set ℕ. (*)Absolutely not.
Addition of all numbers defined by (*) to the empty set is tantamount to
addition of ℕ to the empty set.
That would be {N}
On 2/20/2025 12:50 PM, WM wrote:
On 20.02.2025 15:30, Jim Burns wrote:
AXIOM I 'extensionality' means
proving each FISON is omissible is no
proving {F} is omissible.
∀n ∈ ℕ: n+1 ∈ ℕ.
Together with 1 ∈ ℕ this defines the set ℕ.
(*)
You (WM) have left out that
ℕ is the only.inductive.subset of ℕ
S⊆ℕ ∧ 1∈S ∧ ∀n∈S:n+1∈S ⇒ S=ℕ
(**)
Addition of all numbers defined by (*)
to the empty set
is tantamount to addition of ℕ to the empty set.
{**} prevents the addition of extra elements.
Subtraction of all numbers defined by (*)
from ℕ
is tantamount to subtraction of ℕ from ℕ.
Homework:
Prove the same for FISONs or v. Neumann ordinals.
(!) Have you (WM) started reading my proofs?
For the sets of all (finite) FISONs and
of all finite von Neumann ordinals,
(**) is satisfied as a consequence of
the finitude of their elements.
On 2/20/2025 2:27 PM, WM wrote:
Are you denying that {i:j+1≤i≤k+1} exists?
Are you denying that {i:j+1≤i≤k+1} is
larger than {i:i≤k} ?
That contradicts the assumption.
What is the assumption?
The assumption is the existence of S.
What is the contradiction?
The contradiction is that
induction proves every FISON useless
Each FISON being uselessᵂᴹ (not.last)
is
a consequence of ⋃S
being the only.inductive.subset of ⋃S
And vice versa.
What are the TWO statements which
contradict each other?
On 21.02.2025 02:18, Richard Damon wrote:Yet not the element Z.
In order to secure the existance of "infinite" sets, we still need theAlmost correct, but only singular: we still need the following axiom.
following axioms.
The following axiom is the axiom of induction. The set Z contains the
There is no mention of "induction" in the statement.
empty set as an element and with every element a also the element {a}.
On 21.02.2025 02:18, Richard Damon wrote:
In order to secure the existance of "infinite" sets, we still need the
following axioms.
Almost correct, but only singular: we still need the following axiom.
There is no mention of "induction" in the statement.
The following axiom is the axiom of induction. The set Z contains the
empty set as an element and with every element a also the element {a}.
Regards, WM
On 20.02.2025 20:46, Jim Burns wrote:
On 2/20/2025 12:50 PM, WM wrote:
On 20.02.2025 15:30, Jim Burns wrote:
AXIOM I 'extensionality' means
proving each FISON is omissible is no
proving {F} is omissible.
∀n ∈ ℕ: n+1 ∈ ℕ.
Together with 1 ∈ ℕ this defines the set ℕ.
(*)
You (WM) have left out that
ℕ is the only.inductive.subset of ℕ
ℕ_def to be precise.
S⊆ℕ ∧ 1∈S ∧ ∀n∈S:n+1∈S ⇒ S=ℕ
(**)
Addition of all numbers defined by (*)
to the empty set
is tantamount to addition of ℕ to the empty set.
{**} prevents the addition of extra elements.
No extra elements are available and
no extra elements shall be added.
Subtraction of all numbers defined by (*)
from ℕ
is tantamount to subtraction of ℕ from ℕ.
Homework:
Prove the same for FISONs or v. Neumann ordinals.
(!) Have you (WM) started reading my proofs?
Why should I? I discuss my proof.
For the sets of all (finite) FISONs and
of all finite von Neumann ordinals,
(**) is satisfied as a consequence of
the finitude of their elements.
S = ℕ_def
Am Fri, 21 Feb 2025 10:39:46 +0100 schrieb WM:
On 21.02.2025 02:18, Richard Damon wrote:Yet not the element Z.
In order to secure the existance of "infinite" sets, we still need theAlmost correct, but only singular: we still need the following axiom.
following axioms.
The following axiom is the axiom of induction. The set Z contains the
There is no mention of "induction" in the statement.
empty set as an element and with every element a also the element {a}.
On 2/21/25 4:39 AM, WM wrote:
On 21.02.2025 02:18, Richard Damon wrote:
In order to secure the existance of "infinite" sets, we still need
the following axioms.
Almost correct, but only singular: we still need the following axiom.
Read what you wrote:
Axioms.
There is no mention of "induction" in the statement.
The following axiom is the axiom of induction. The set Z contains the
empty set as an element and with every element a also the element {a}.
I will admit that I haven't read the paper in the German, but I have
studied the general concepts of the theory, and "induction" is not the
basis of the existance of the infinite set, at least not in the final
works.
On 2/21/2025 4:19 AM, WM wrote:
Homework:
Prove the same for FISONs or v. Neumann ordinals.
(!) Have you (WM) started reading my proofs?
Why should I? I discuss my proof.
Why are you asking me for proofs you won't read?
For the sets of all (finite) FISONs and
of all finite von Neumann ordinals,
(**) is satisfied as a consequence of
the finitude of their elements.
S = ℕ_def
ℕ_def = ℕ′
S = {i:A(i)}
⎛ A(1) ∧ ∀n∈ℕ′:A(n)⇒A(n+1) ⇒
⎜ {i:A(i)} ∈ {S″⊆ℕ′:inductive.S″}
⎜
⎜ {S″⊆ℕ′:inductive.S″} = {ℕ′}
⎜ ⇐ all and only finites are in ℕ′
⎜
⎜ ( {ℕ′} ≠ ℕ′ )
⎜
⎜ {i:A(i)} ∈ {S″⊆ℕ′:inductive.S″} ∧
⎜ {S″⊆ℕ′:inductive.S″} = {ℕ′} ⇒
⎜ {i:A(i)} = ℕ′
⎜
⎜ {i:A(i)} = ℕ′ ∧
⎜ ∀k ∈ {i:A(i)}: A(k) ⇒
⎝ ∀k ∈ ℕ′: A(k)
Proof by induction for ℕ′
⇐ all and only finites are in ℕ′
On 21.02.2025 11:37, joes wrote:Z is not a set containing Z; Z cannot be "removed" from Z.
Am Fri, 21 Feb 2025 10:39:46 +0100 schrieb WM:Zermelo claims to secure the existence of set Z.
On 21.02.2025 02:18, Richard Damon wrote:Yet not the element Z.
In order to secure the existance of "infinite" sets, we still needAlmost correct, but only singular: we still need the following axiom.
the following axioms.
The following axiom is the axiom of induction. The set Z contains the
There is no mention of "induction" in the statement.
empty set as an element and with every element a also the element {a}.
This set can be handled, for instance removed from Z. A set containing Z
is not useful in this context.
On 21.02.2025 17:14, Jim Burns wrote:
On 2/21/2025 4:19 AM, WM wrote:
On 20.02.2025 20:46, Jim Burns wrote:
For the sets of all (finite) FISONs and
of all finite von Neumann ordinals,
(**) is satisfied as a consequence of
the finitude of their elements.
S = ℕ_def
ℕ_def = ℕ′
S = {i:A(i)}
⎛ A(1) ∧ ∀n∈ℕ′:A(n)⇒A(n+1) ⇒
⎜ {i:A(i)} ∈ {S″⊆ℕ′:inductive.S″}
⎜
⎜ {S″⊆ℕ′:inductive.S″} = {ℕ′}
⎜ ⇐ all and only finites are in ℕ′
⎜
⎜ ( {ℕ′} ≠ ℕ′ )
We need not consider {ℕ′}.
⎜ {i:A(i)} ∈ {S″⊆ℕ′:inductive.S″} ∧
⎜ {S″⊆ℕ′:inductive.S″} = {ℕ′} ⇒
⎜ {i:A(i)} = ℕ′
⎜
⎜ {i:A(i)} = ℕ′ ∧
⎜ ∀k ∈ {i:A(i)}: A(k) ⇒
⎝ ∀k ∈ ℕ′: A(k)
Proof by induction for ℕ′
⇐ all and only finites are in ℕ′
And all together are the set ℕ′.
On 21.02.2025 11:37, joes wrote:
Am Fri, 21 Feb 2025 10:39:46 +0100 schrieb WM:
On 21.02.2025 02:18, Richard Damon wrote:
In order to secure the existance of "infinite" sets,
we still need the following axioms.
Almost correct, but only singular:
we still need the following axiom.
There is no mention of "induction" in the statement.
The following axiom is the axiom of induction.
The set Z contains the empty set as an element and
with every element a also the element {a}.
Yet not the element Z.
Zermelo claims to secure the existence of set Z.
This set can be handled,
for instance removed from Z.
A set containing Z
is not useful in this context.
Regards, WM
Le 03/02/2025 à 16:11, crank Wolfang Mückenheim aka WM a écrit: [idiotic nonsense]
Anyone able to claim such a fallacy shouldn't be allowed to be put in
front of students, in any country.
Le 03/02/2025 à 18:22, WM a écrit : [idiotic nonsense]
Name ONE mathematician who supports your idiotic claim. One.
On 21.02.2025 13:18, Richard Damon wrote:
On 2/21/25 4:39 AM, WM wrote:
On 21.02.2025 02:18, Richard Damon wrote:
In order to secure the existance of "infinite" sets, we still need
the following axioms.
Almost correct, but only singular: we still need the following axiom.
Read what you wrote:
Axioms.
You wrote it.
There is no mention of "induction" in the statement.
The following axiom is the axiom of induction. The set Z contains the
empty set as an element and with every element a also the element {a}.
I will admit that I haven't read the paper in the German, but I have
studied the general concepts of the theory, and "induction" is not the
basis of the existance of the infinite set, at least not in the final
works.
Either you have not understood the final works or they are irrelevant.
Regards, WM
Am Fri, 21 Feb 2025 18:17:32 +0100 schrieb WM:
Zermelo claims to secure the existence of set Z.Z is not a set containing Z; Z cannot be "removed" from Z.
This set can be handled, for instance removed from Z. A set containing Z
is not useful in this context.
Since that is the earlier work, and not the later refined work, maybe
you should update your studies. (Note this creates Z, not N)
Note, as I understand it, that initial Zermelo Set Theory didn't even
HAVE "induction",
but its last axiom was the Axiom of Infinity that say
that there exists in this domain the set Z that contains the null set as
an element and is so constituted that to each of its element a, there corresponds a further element of the form {a}, in other words with each
of its elements a, it also contains the corresponding set {a} as an
element.
Note, this is NOT the statement of induction.
On 20.02.2025 17:03, Jim Burns wrote:How do you prove it without induction?
Why do you think the set has the same property as its elements?That is not true in general but can be proven in this case.
On 20.02.2025 03:23, Richard Damon wrote:It... no? N \ UA = {}.
On 2/19/25 11:57 AM, WM wrote:
Note, your subject line uses the word you mean, "necessary", but youAssume a set of sufficient FISONs. |ℕ \ {1, 2, 3, ..., n}| = ℵo is true for all FISONs. That contradicts the assumption.
ignore the fact that a set of necessary elements doesn't need to exist.
Am Sat, 22 Feb 2025 11:12:17 +0100 schrieb WM:
On 21.02.2025 19:45, joes wrote:But Z \ {Z} = Z, because Z !e Z.
Am Fri, 21 Feb 2025 18:17:32 +0100 schrieb WM:Of course removing all elements from Z produces the empty set.
Zermelo claims to secure the existence of set Z.Z is not a set containing Z; Z cannot be "removed" from Z.
This set can be handled, for instance removed from Z. A set containing >>>> Z is not useful in this context.
On 19.02.2025 18:19, joes wrote:Yeah yeah, given the premise, which you don't accept. It would be
Am Wed, 19 Feb 2025 18:07:56 +0100 schrieb WM:
Learn what an implication is.Proven however is this: UF = ℕ ⟹ Ø = ℕ.No. That would mean UF = {}, which you can’t seriously believe.
X is clearly an inductive set, like N. Your wrong argumentIf you think UF = X != N, then you need to accept U{} = X.No. I don't know the relation between a FISON and X.
Am Thu, 20 Feb 2025 19:03:39 +0100 schrieb WM:
On 20.02.2025 12:46, joes wrote:Ok. Why does it follow for N only?
Am Thu, 20 Feb 2025 11:26:13 +0100 schrieb WM:No.
Subtraction of all elements leaves the empty set.Substitute: UF = X != N. Does it follow that X = {}?
UF = ℕ ==> Ø = ℕ.
Am Thu, 20 Feb 2025 11:29:26 +0100 schrieb WM:
On 19.02.2025 18:19, joes wrote:
X is clearly an inductive set, like N.If you think UF = X != N, then you need to accept U{} = X.No. I don't know the relation between a FISON and X.
On 21.02.2025 19:45, joes wrote:But Z \ {Z} = Z, because Z !e Z.
Am Fri, 21 Feb 2025 18:17:32 +0100 schrieb WM:
Of course removing all elements from Z produces the empty set.Zermelo claims to secure the existence of set Z.Z is not a set containing Z; Z cannot be "removed" from Z.
This set can be handled, for instance removed from Z. A set containing
Z is not useful in this context.
However,
all sets Z which satisfy Infinityᶻᶠᶜ
HAVE A SUBSET which
is usable in a proof.by.inductionᶻᶠᶜ
On 20.02.2025 12:46, joes wrote:Ok. Why does it follow for N only?
Am Thu, 20 Feb 2025 11:26:13 +0100 schrieb WM:
No.Subtraction of all elements leaves the empty set.Substitute: UF = X != N. Does it follow that X = {}?
UF = ℕ ==> Ø = ℕ.
--No, it is because ∀n ∈ U(F): |ℕ \ {1, 2, 3, ..., n}| = ℵo.Is that because every natural is finite?The set of all natural numbers which can be defined is finite.
infinitesimally smaller than ℕ.
WM explained on 2/22/2025 :
On 22.02.2025 02:05, Richard Damon wrote:
Note, as I understand it, that initial Zermelo Set Theory didn't even
HAVE "induction",
You don't understand it.
He understands it better than you do apparently. Zermelo's theory didn't
have transfinite induction.
but its last axiom was the Axiom of Infinity that say that there
exists in this domain the set Z that contains the null set as an
element and is so constituted that to each of its element a, there
corresponds a further element of the form {a}, in other words with
each of its elements a, it also contains the corresponding set {a} as
an element.
That is induction.
That is the successor function, it doesn't extend to the transfinite.
Note, this is NOT the statement of induction.
AI Overview
No, Zermelo's set theory by itself does not explicitly include the
principle of finite induction;
Key points:
On 21.02.2025 19:45, joes wrote:
Am Fri, 21 Feb 2025 18:17:32 +0100 schrieb WM:
Zermelo claims to secure the existence of set Z.Z is not a set containing Z; Z cannot be "removed" from Z.
This set can be handled, for instance removed from Z. A set containing Z >>> is not useful in this context.
Of course removing all elements from Z produces the empty set.
Regards, WM
WM formulated on Saturday :
On 22.02.2025 12:14, FromTheRafters wrote:
WM explained on 2/22/2025 :
On 22.02.2025 02:05, Richard Damon wrote:
Note, as I understand it, that initial Zermelo Set Theory didn't
even HAVE "induction",
You don't understand it.
He understands it better than you do apparently. Zermelo's theory
didn't have transfinite induction.
Who claimed that?
Certainly not you.
First element and the successor function is induction.
First element and successor function is 'inductive hypothesis' not
induction.
On 22.02.2025 13:13, FromTheRafters wrote:
WM formulated on Saturday :
On 22.02.2025 12:14, FromTheRafters wrote:
WM explained on 2/22/2025 :
On 22.02.2025 02:05, Richard Damon wrote:
Note, as I understand it, that initial Zermelo Set Theory didn't
even HAVE "induction",
You don't understand it.
He understands it better than you do apparently. Zermelo's theory
didn't have transfinite induction.
Who claimed that?
Certainly not you.
Only stupids could do so.
First element and the successor function is induction.
First element and successor function is 'inductive hypothesis' not
induction.
Nonsense. The inductive step is proved in my example from |ℕ \ {1, 2,
3, ..., n}| = ℵo, i.e., by several axioms, and in Zermelo's example by a single axiom.
Regards, WM
Peano's successors are not induction.
Induction is the axiom that lets your prove that a set contains the set
of Natural Numbers. It isn't a "construction" technique.
On 22.02.2025 11:45, joes wrote:Why can you not induce over an inductive set?
Am Thu, 20 Feb 2025 11:29:26 +0100 schrieb WM:
On 19.02.2025 18:19, joes wrote:
If X is defined as the true UF (I had overlooked that), then we cannotX is clearly an inductive set, like N.If you think UF = X != N, then you need to accept U{} = X.No. I don't know the relation between a FISON and X.
apply |X \ {1, 2, 3, ..., n}| = ℵo and then U{} = X does not follow.
X is a FISON like every finite union of FISONs.No, X is an obviously infinite union.
Am Sat, 22 Feb 2025 12:10:12 +0100 schrieb WM:
On 22.02.2025 11:45, joes wrote:Why can you not induce over an inductive set?
Am Thu, 20 Feb 2025 11:29:26 +0100 schrieb WM:If X is defined as the true UF (I had overlooked that), then we cannot
On 19.02.2025 18:19, joes wrote:
X is clearly an inductive set, like N.If you think UF = X != N, then you need to accept U{} = X.No. I don't know the relation between a FISON and X.
apply |X \ {1, 2, 3, ..., n}| = ℵo and then U{} = X does not follow.
X is a FISON like every finite union of FISONs.No, X is an obviously infinite union.
On 22.02.2025 14:23, joes wrote:Those don't have FISes.
Am Sat, 22 Feb 2025 12:10:12 +0100 schrieb WM:Dark numbers.
On 22.02.2025 11:45, joes wrote:Why can you not induce over an inductive set?
Am Thu, 20 Feb 2025 11:29:26 +0100 schrieb WM:If X is defined as the true UF (I had overlooked that), then we cannot
On 19.02.2025 18:19, joes wrote:
X is clearly an inductive set, like N.If you think UF = X != N, then you need to accept U{} = X.No. I don't know the relation between a FISON and X.
apply |X \ {1, 2, 3, ..., n}| = ℵo and then U{} = X does not follow.
Why should there?There is no actually infinite set of FISONs because there are never two actually infinite consecutive sets in ℕ.X is a FISON like every finite union of FISONs.No, X is an obviously infinite union.
On 2/22/25 7:35 AM, WM wrote:
The inductive step is proved in my example from |ℕ \ {1, 2,
3, ..., n}| = ℵo, i.e., by several axioms, and in Zermelo's example by
a single axiom.
Which just proves that no specific FISON is individually REQURED to make
the set of Natural Numbers.
After serious thinking WM wrote :
On 22.02.2025 13:15, Richard Damon wrote:
Peano's successors are not induction.
Induction is the axiom that lets your prove that a set contains the
set of Natural Numbers. It isn't a "construction" technique.
Induction is the feature, proven or claimed, that an element exists in
the set and with any element also its successor.
No,
Am Sat, 22 Feb 2025 15:07:08 +0100 schrieb WM:
On 22.02.2025 14:23, joes wrote:
Why should there?There is no actually infinite set of FISONs because there are never twoX is a FISON like every finite union of FISONs.No, X is an obviously infinite union.
actually infinite consecutive sets in ℕ.
On 22.02.2025 15:57, joes wrote:I don't get it. What even should the second set be? You are arguing
Am Sat, 22 Feb 2025 15:07:08 +0100 schrieb WM:
On 22.02.2025 14:23, joes wrote:
Try to name the first number of the second actually infinite setWhy should there?There is no actually infinite set of FISONs because there are neverX is a FISON like every finite union of FISONs.No, X is an obviously infinite union.
two actually infinite consecutive sets in ℕ.
following upon the first actually infinite set. Then you will
understand.
Am Sat, 22 Feb 2025 19:11:09 +0100 schrieb WM:
On 22.02.2025 15:57, joes wrote:I don't get it. What even should the second set be?
Am Sat, 22 Feb 2025 15:07:08 +0100 schrieb WM:Try to name the first number of the second actually infinite set
On 22.02.2025 14:23, joes wrote:
Why should there?There is no actually infinite set of FISONs because there are neverX is a FISON like every finite union of FISONs.No, X is an obviously infinite union.
two actually infinite consecutive sets in ℕ.
following upon the first actually infinite set. Then you will
understand.
On 22.02.2025 19:19, joes wrote:So why is the union of inf. many FISONs finite?
Am Sat, 22 Feb 2025 19:11:09 +0100 schrieb WM:All FISONs have an actually infinite set of dark numbers as successors:
On 22.02.2025 15:57, joes wrote:I don't get it. What even should the second set be?
Am Sat, 22 Feb 2025 15:07:08 +0100 schrieb WM:Try to name the first number of the second actually infinite set
On 22.02.2025 14:23, joes wrote:
Why should there?There is no actually infinite set of FISONs because there are neverX is a FISON like every finite union of FISONs.No, X is an obviously infinite union.
two actually infinite consecutive sets in ℕ.
following upon the first actually infinite set. Then you will
understand.
∀n ∈ U(F): |ℕ \ {1, 2, 3, ..., n}| = ℵo. This set differs for every FISON but is not less than ℵo for ay FISON.
On 22.02.2025 15:03, Richard Damon wrote:Haven't you agreed that omitting everything does change the union?
On 2/22/25 7:35 AM, WM wrote:If every FISON can be omitted, ten nothing remains for a sufficient set. Because if there was any sufficient set, it would have a first FISON.
The inductive step is proved in my example from |ℕ \ {1, 2,Which just proves that no specific FISON is individually REQURED to
3, ..., n}| = ℵo, i.e., by several axioms, and in Zermelo's example by >>> a single axiom.
make the set of Natural Numbers.
On 21.02.2025 20:40, Jim Burns wrote:
However,
all sets Z which satisfy Infinityᶻᶠᶜ
HAVE A SUBSET which
is usable in a proof.by.inductionᶻᶠᶜ
Zermelo calls that set Z_0 or set of numbers.
We can remove all numbers
from Z_0 and produce the empty set.
Homework: Show the same for the set of FISONs.
On 22.02.2025 15:03, Richard Damon wrote:
On 2/22/25 7:35 AM, WM wrote:
The inductive step is proved in my example from |ℕ \ {1, 2, 3, ...,
n}| = ℵo, i.e., by several axioms, and in Zermelo's example by a
single axiom.
Which just proves that no specific FISON is individually REQURED to
make the set of Natural Numbers.
If every FISON can be omitted, ten nothing remains for a sufficient set. Because if there was any sufficient set, it would have a first FISON.
Regards, WM
Am Sat, 22 Feb 2025 19:32:16 +0100 schrieb WM:
All FISONs have an actually infinite set of dark numbers as successors:So why is the union of inf. many FISONs finite?
∀n ∈ U(F): |ℕ \ {1, 2, 3, ..., n}| = ℵo. This set differs for every >> FISON but is not less than ℵo for any FISON.
Am Sat, 22 Feb 2025 19:02:12 +0100 schrieb WM:
If every FISON can be omitted, then nothing remains for a sufficient set.Haven't you agreed that omitting everything does change the union?
Because if there was any sufficient set, it would have a first FISON.
WM pretended :
Induction is the feature, proven or claimed, that an element exists
in the set and with any element also its successor.
No,
Can't you understand English?
Yes, quite well.
With any (here individuality is implied) element also its (here
immediate is implied) successor.
This is proven by me:
No it isn't,
On 2/22/25 1:02 PM, WM wrote:
If every FISON can be omitted, then nothing remains for a sufficient
set. Because if there was any sufficient set, it would have a first
FISON.
The fact that none of them are individually requred doesn't mean you
can't use them.
I guess you do think that we can't factor 36, since none of the factors
are requires, so all can be omitted from the set you can use.
On 2/22/25 9:04 AM, WM wrote:
Induction is the feature, proven or claimed, that an element exists inWhere do you get that from?
the set and with any element also its successor.
The induction rule is the method used to
prove that a statement P(n) is true for every natural number, by showing
that P(0) is true, and that if P(n) is true, the P(n+1) must be true.
Your claim is just the opposite of induction.
No FISON is infinite, but the set of all of them is
Natural Number is infinte, but the set of all of them, the Natural
Numbers is.
On 2/22/2025 5:17 AM, WM wrote:
On 21.02.2025 20:40, Jim Burns wrote:
However,
all sets Z which satisfy Infinityᶻᶠᶜ
HAVE A SUBSET which
is usable in a proof.by.inductionᶻᶠᶜ
Zermelo calls that set Z_0 or set of numbers.
The usable SUBSET ℕⁱᵒᵒⁱˢˢ of Zermelo's asserted Z
is its.own.only.inductive.sub.set (iooiss).
Our sets do not change.
I'm pretty sure your ℕ_def = ℕⁱᵒᵒⁱˢˢ
Our sets do not change.
⎛ You (WM) say 'inductive S' means
⎜ ∀k:S∋k⇒S∋k+1 ∧ S∋1
⎜ Then
⎜ ℕ₁ⁱᵒᵒⁱˢˢ = {1,2,3,...}
⎜
⎜ We (matheologists) say 'inductive S' means
⎜ ∀k:S∋k⇒S∋k+1 ∧ S∋0
⎜ Then
⎜ ℕ₀ⁱᵒᵒⁱˢˢ = {0,1,2,...}
For each two meanings 'inductiveₓ' and 'inductiveᵥ',
We can remove all numbers from Z_0 and produce the empty set.
We can remove all numbers
from {137} and produce the empty set.
{137} ≠ 137
Homework: Show the same for the set of FISONs.
Am Sat, 22 Feb 2025 18:46:25 +0000 schrieb joes:No, there are infinitely many FISONs. Why not?
Am Sat, 22 Feb 2025 19:32:16 +0100 schrieb WM:There are only finitely many FISONs. Therefore their union is finite
On 22.02.2025 19:19, joes wrote:So why is the union of inf. many FISONs finite?
Am Sat, 22 Feb 2025 19:11:09 +0100 schrieb WM:All FISONs have an actually infinite set of dark numbers as
On 22.02.2025 15:57, joes wrote:I don't get it. What even should the second set be?
Am Sat, 22 Feb 2025 15:07:08 +0100 schrieb WM:
On 22.02.2025 14:23, joes wrote:
Try to name the first number of the second actually infinite setWhy should there?There is no actually infinite set of FISONs because there areX is a FISON like every finite union of FISONs.No, X is an obviously infinite union.
never two actually infinite consecutive sets in ℕ.
following upon the first actually infinite set. Then you will
understand.
succdessors ∀n ∈ U(F): |ℕ \ {1, 2, 3, ..., n}| = ℵo. This set differs
for every FISON but is not less than ℵo for ay FISON.
too.
On 22.02.2025 19:21, joes wrote:
Am Sat, 22 Feb 2025 19:02:12 +0100 schrieb WM:
If every FISON can be omitted, then nothing remains for a sufficientHaven't you agreed that omitting everything does change the union?
set.
Because if there was any sufficient set, it would have a first FISON.
Omitting everything does change the real union but not the assumed union.
Regards, WM
On 22.02.2025 19:21, joes wrote:OMFG. Nobody is saying that the union of the elements of the empty set
Am Sat, 22 Feb 2025 19:02:12 +0100 schrieb WM:
Omitting everything does change the real union but not the assumedIf every FISON can be omitted, then nothing remains for a sufficientHaven't you agreed that omitting everything does change the union?
set.
Because if there was any sufficient set, it would have a first FISON.
union.
On 22.02.2025 21:29, Jim Burns wrote:
On 2/22/2025 5:17 AM, WM wrote:
We can remove all numbers from Z_0
and produce the empty set.
We can remove all numbers from {137}
and produce the empty set.
{137} ≠ 137
If we remove all numbers from {137}
we produce { },
if we remove all numbers from 137
nothing remains.
It is a matter of taste
whether the empty set is nothing.
The usable SUBSET ℕⁱᵒᵒⁱˢˢ of Zermelo's asserted Z
is its.own.only.inductive.sub.set (iooiss).
Our sets do not change.
I'm pretty sure your ℕ_def = ℕⁱᵒᵒⁱˢˢ
Our sets do not change.
Yes, except that
Zermelo starts with zero or empty set.
Please stop your silly labeling.
Zermelo calls it Z_0.
For each two meanings
'inductiveₓ' and 'inductiveᵥ',
That is not a significant difference.
Homework: Show the same for the set of FISONs.
No, there are infinitely many FISONs. Why not?
On 2/23/2025 5:34 AM, WM wrote:
On 22.02.2025 21:29, Jim Burns wrote:
On 2/22/2025 5:17 AM, WM wrote:
We can remove all numbers from Z_0
and produce the empty set.
We can remove all numbers from {137}
and produce the empty set.
{137} ≠ 137
If we remove all numbers from {137}
we produce { },
if we remove all numbers from 137
nothing remains.
It is a matter of taste
whether the empty set is nothing.
Nothing is what is in {}.
{{}} ≠ {}
Nothing is not what is in {{}}.
The set of all FISONs with
each omissible FISON omitted
is {F}\{F} = {}
The set of all FISONs with
the set of omissible FISONs omitted
is {F}\{{F}} = {F}
Homework: Show the same for the set of FISONs.
{{}} ≠ {}
On 23.02.2025 14:55, Jim Burns wrote:No. As we've seen, you cannot omit all FISONs.
On 2/23/2025 5:34 AM, WM wrote:Right. UF = ℕ ==> Ø = ℕ.
On 22.02.2025 21:29, Jim Burns wrote:Nothing is what is in {}.
On 2/22/2025 5:17 AM, WM wrote:
If we remove all numbers from {137} we produce { },We can remove all numbers from Z_0 and produce the empty set.We can remove all numbers from {137}
and produce the empty set.
{137} ≠ 137
if we remove all numbers from 137 nothing remains.
It is a matter of taste whether the empty set is nothing.
{{}} ≠ {}
Nothing is not what is in {{}}.
The set of all FISONs with each omissible FISON omitted is {F}\{F} = {}
Indeed, because {F} is not an element of F, which is why I'm wonderingThe set of all FISONs with the set of omissible FISONs omitted isThere is no reason to consider {{F}} at all.
{F}\{{F}} = {F}
We omit all F(n) which amounts to remove F.No, it amounts to the elements of F, not those of {F}.
Like all natural numbers amount to ℕ (not {ℕ})Like all FISONs amount to F, not {F}.
Neumann would like to have a word with you.Wrong.{{}} ≠ {}Homework: Show the same for the set of FISONs.
Am Sun, 23 Feb 2025 15:43:48 +0100 schrieb WM:
Right. UF = ℕ ==> Ø = ℕ.No. As we've seen, you cannot omit all FISONs.
We omit all F(n) which amounts to remove F.No, it amounts to the elements of F, not those of {F}.
Like all natural numbers amount to ℕ (not {ℕ})Like all FISONs amount to F, not {F}.
Neumann would like to have a word with you.Wrong.{{}} ≠ {}Homework: Show the same for the set of FISONs.
Am Sun, 23 Feb 2025 18:08:25 +0100 schrieb WM:
On 23.02.2025 16:37, joes wrote:All. All FISONs can not be omitted, only a finite number,
Am Sun, 23 Feb 2025 15:43:48 +0100 schrieb WM:Which one cannot?
Right. UF = ℕ ==> Ø = ℕ.No. As we've seen, you cannot omit all FISONs.
On 23.02.2025 16:37, joes wrote:All. All FISONs can not be omitted, only a finite number,
Am Sun, 23 Feb 2025 15:43:48 +0100 schrieb WM:
Which one cannot?Right. UF = ℕ ==> Ø = ℕ.No. As we've seen, you cannot omit all FISONs.
And cutting this.Indeed, because {F} is not an element of F, which is why I'm wonderingThe set of all FISONs with the set of omissible FISONs omitted isThere is no reason to consider {{F}} at all.
{F}\{{F}} = {F}
why you keep talking about it.
No, the set F is not {F}. F is the set *of its elements*. F is not anAll elements F(n) of F are F.We omit all F(n) which amounts to remove F.No, it amounts to the elements of F, not those of {F}.
Thus F \ {F} = F. It doesn't even make sense to remove F from F.So it is.Like all natural numbers amount to ℕ (not {ℕ})Like all FISONs amount to F, not {F}.
Am Sun, 23 Feb 2025 17:26:27 +0000 schrieb joes:Induction does not "produce" any infinite number.
Am Sun, 23 Feb 2025 18:08:25 +0100 schrieb WM:Induction produces ℕ, not only a finite number.
On 23.02.2025 16:37, joes wrote:All. All FISONs can not be omitted, only a finite number,
Am Sun, 23 Feb 2025 15:43:48 +0100 schrieb WM:Which one cannot?
Right. UF = ℕ ==> Ø = ℕ.No. As we've seen, you cannot omit all FISONs.
though it doesn't matter which. If you omit all, you get a different
set than when omitting any one.
And cutting this.Indeed, because {F} is not an element of F, which is why I'mThe set of all FISONs with the set of omissible FISONs omitted isThere is no reason to consider {{F}} at all.
{F}\{{F}} = {F}
wondering why you keep talking about it.
No, the set F is not {F}. F is the set *of its elements*. F is not anAll elements F(n) of F are F.We omit all F(n) which amounts to remove F.No, it amounts to the elements of F, not those of {F}.>
element.
Thus F \ {F} = F. It doesn't even make sense to remove F from F.So it is.Like all natural numbers amount to ℕ (not {ℕ})Like all FISONs amount to F, not {F}.
Therefore your posts are irrelevant and will no longer be read.--
On 23.02.2025 14:55, Jim Burns wrote:
On 2/23/2025 5:34 AM, WM wrote:
On 22.02.2025 21:29, Jim Burns wrote:
On 2/22/2025 5:17 AM, WM wrote:
{{}} ≠ {}
Wrong.
We can remove all numbers from Z_0
and produce the empty set.
We can remove all numbers from {137}
and produce the empty set.
{137} ≠ 137
If we remove all numbers from {137}
we produce { },
if we remove all numbers from 137
nothing remains.
It is a matter of taste
whether the empty set is nothing.
Nothing is what is in {}.
{{}} ≠ {}
Nothing is not what is in {{}}.
The set of all FISONs with
each omissible FISON omitted
is {F}\{F} = {}
Right. UF = ℕ ==> Ø = ℕ.
The set of all FISONs with
the set of omissible FISONs omitted
is {F}\{{F}} = {F}
There is no reason to consider {{F}} at all.
We omit all F(n) which amounts to remove F.
Like all natural numbers amount to ℕ (not {ℕ})
Homework: Show the same for the set of FISONs.
Am Sun, 23 Feb 2025 18:33:51 +0100 schrieb WM:
Induction produces ℕ, not only a finite number.Induction does not "produce" any infinite number.
On 2/23/2025 9:43 AM, WM wrote:
There is no reason to consider {{F}} at all.
There is reason, but
only for people wanting to be correct.
Regards, WMWe omit all F(n) which amounts to remove F.
Like all natural numbers amount to ℕ (not {ℕ})
On 23.02.2025 19:34, Jim Burns wrote:
On 2/23/2025 9:43 AM, WM wrote:
There is no reason to consider {{F}} at all.
There is reason, but
only for people wanting to be correct.
Peano, Zermelo, or v. Neumann
Peano, Zermelo, or v. Neumann create ℕ
as well as the set F of all FISONs
by induction over the members
for use in set theory
without being what you erroneously call correct.
We omit all F(n) which amounts to remove F.
Like all natural numbers amount to ℕ (not {ℕ})
On 23.02.2025 19:34, Jim Burns wrote:
On 2/23/2025 9:43 AM, WM wrote:
There is no reason to consider {{F}} at all.
There is reason, but
only for people wanting to be correct.
Peano, Zermelo, or v. Neumann create ℕ as well as the set F of all
FISONs by induction over the members for use in set theory without being
what you erroneously call correct.
Regards, WMWe omit all F(n) which amounts to remove F.
Like all natural numbers amount to ℕ (not {ℕ})
On 2/23/2025 2:32 PM, WM wrote:
On 23.02.2025 19:34, Jim Burns wrote:
On 2/23/2025 9:43 AM, WM wrote:
There is no reason to consider {{F}} at all.
There is reason, but
only for people wanting to be correct.
Peano, Zermelo, or v. Neumann
...agree that {{F}} ≠ {F}
Peano, Zermelo, or v. Neumann create ℕ
Peano, Zermelo, and v. Neumann assert axioms
from which the existence of ℕ follows
in a finite.sequence of not.first.false claims.
as well as the set F of all FISONs
by induction over the members
for use in set theory
without being what you erroneously call correct.
A proof.by.induction shows that
some set,
such as the set {x:A(x)} of x such that A(x),
is inductive.
The conclusion of a proof.by.induction
is that {x:A(x)} is the whole set.
However,
not just any "whole set" is reliable here.
It must be a whole set such that
knowing {x:A(x)} is inductive
narrows
which set {x:A(x)} can be
to one set: that whole set.
We omit all F(n) which amounts to remove F.
Like all natural numbers amount to ℕ (not {ℕ})
Each natural number is in the domain of ST+F
ℕ is not in the domain of ST+F
On 2/24/2025 9:59 AM, WM wrote:
On 23.02.2025 23:03, Jim Burns wrote:
On 2/23/2025 2:32 PM, WM wrote:
On 23.02.2025 19:34, Jim Burns wrote:
On 2/23/2025 9:43 AM, WM wrote:
There is no reason to consider {{F}} at all.
There is reason, but
only for people wantcing to be correct.
Peano, Zermelo, or v. Neumann
...agree that {{F}} ≠ {F}
and also that 3 ≠ pi.
<WM<JB>>
{{}} ≠ {}
Wrong.
These axioms can be applied to show that
all FISONs can be removed.
{1,2}\{1,{2}} = {2}
The set of finite ordinals after v. Neuman
is undoubtedly such a set.
The set of finite ordinals after v. Neumann
is not a finite set.
A claim for each of its elements is
silent about the set.
Claims about each natural number
are silent about ℕ.
On 23.02.2025 23:03, Jim Burns wrote:
On 2/23/2025 2:32 PM, WM wrote:
On 23.02.2025 19:34, Jim Burns wrote:
On 2/23/2025 9:43 AM, WM wrote:
There is no reason to consider {{F}} at all.
There is reason, but
only for people wantcing to be correct.
Peano, Zermelo, or v. Neumann
...agree that {{F}} ≠ {F}
and also that 3 ≠ pi.
{{}} ≠ {}
Wrong.
Peano, Zermelo, or v. Neumann create ℕ
Peano, Zermelo, and v. Neumann assert axioms
from which the existence of ℕ follows
in a finite.sequence of not.first.false claims.
These axioms can be applied to show that
all FISONs can be removed.
as well as the set F of all FISONs
by induction over the members
for use in set theory
without being what you erroneously call correct.
A proof.by.induction shows that
some set,
such as the set {x:A(x)} of x such that A(x),
is inductive.
The conclusion of a proof.by.induction
is that {x:A(x)} is the whole set.
However,
not just any "whole set" is reliable here.
It must be a whole set such that
knowing {x:A(x)} is inductive
narrows
which set {x:A(x)} can be
to one set: that whole set.
The set of finite ordinals after v. Neuman
is undoubtedly such a set.
We omit all F(n) which amounts to remove F.
Like all natural numbers amount to ℕ (not {ℕ})
Each natural number is in the domain of ST+F
ℕ is not in the domain of ST+F
Only all FISONs = natural numbers are
the matter of my proof.
According to Zermelo they make up the set ℕ.
On 24.02.2025 17:40, FromTheRafters wrote:
WM brought next idea :
These axioms can be applied to show that
all FISONs can be removed.
Not in a system where sets don't change.
Addition and subtraction of sets are a common techniques.
Thereby sets are changed.
It is possible
but not useful to express this ponderously
by fixed sets.
WM brought next idea :
These axioms can be applied to show that all FISONs can be removed.
Not in a system where sets don't change.
On 23.02.2025 23:03, Jim Burns wrote:Nope. Any number of them can be removed. Indeed, removing all
On 2/23/2025 2:32 PM, WM wrote:and also that 3 ≠ pi.
On 23.02.2025 19:34, Jim Burns wrote:...agree that {{F}} ≠ {F}
On 2/23/2025 9:43 AM, WM wrote:
Peano, Zermelo, or v. NeumannThere is no reason to consider {{F}} at all.There is reason, but only for people wanting to be correct.
These axioms can be applied to show that all FISONs can be removed.
Peano, Zermelo, or v. Neumann create ℕPeano, Zermelo, and v. Neumann assert axioms from which the existence
of ℕ follows in a finite.sequence of not.first.false claims.
The set N is not a natural number that the proof concerns.The set of finite ordinals after v. Neuman is undoubtedly such a set.as well as the set F of all FISONs by induction over the members forA proof.by.induction shows that some set,
use in set theory without being what you erroneously call correct.
such as the set {x:A(x)} of x such that A(x), is inductive.
The conclusion of a proof.by.induction is that {x:A(x)} is the whole
set.
However, not just any "whole set" is reliable here.
It must be a whole set such that knowing {x:A(x)} is inductive narrows
which set {x:A(x)} can be to one set: that whole set.
Only all FISONs = natural numbers are the matter of my proof. AccordingEach natural number is in the domain of ST+F ℕ is not in the domain ofWe omit all F(n) which amounts to remove F.
Like all natural numbers amount to ℕ (not {ℕ})
ST+F
to Zermelo they make up the set ℕ.
On 24.02.2025 17:36, Jim Burns wrote:Yes, you are wrong trying to talk about N \ {N}.
On 2/24/2025 9:59 AM, WM wrote:It is wrong to apply this in the present framework.
On 23.02.2025 23:03, Jim Burns wrote:<WM<JB>>
On 2/23/2025 2:32 PM, WM wrote:
On 23.02.2025 19:34, Jim Burns wrote:
On 2/23/2025 9:43 AM, WM wrote:
and also that 3 ≠ pi....agree that {{F}} ≠ {F}Peano, Zermelo, or v. NeumannThere is no reason to consider {{F}} at all.There is reason, but only for people wantcing to be correct.
{{}} ≠ {}
That is your nonsense. Try and formalise your shit.Only such nonsense available?These axioms can be applied to show that all FISONs can be removed.{1,2}\{1,{2}} = {2}
That has no bearing. Properties don’t transfer from elements to their set.You are wrong. The existence of the set is guaranteed by elements whichThe set of finite ordinals after v. Neuman is undoubtedly such a set.The set of finite ordinals after v. Neumann is not a finite set.
A claim for each of its elements is silent about the set.
are defined by induction. Note that induction has been invented for
proofs concerning infinitely many element.
We are not disputing the existence.Claims about each natural number are silent about ℕ.Claims about the existence of all natural numbers are claims about the exitstence of ℕ.
On 24.02.2025 17:36, Jim Burns wrote:
On 2/24/2025 9:59 AM, WM wrote:
On 23.02.2025 23:03, Jim Burns wrote:
<WM<JB>>
{{}} ≠ {}
Wrong.
It is wrong to apply this
in the present framework.
Zermelo creates all natural numbers by induction
and by that guarantees the existence of the set ℕ.
These axioms can be applied to show that
all FISONs can be removed.
{1,2}\{1,{2}} = {2}
Only such nonsense available?
Claims about the existence of all natural numbers
are claims about the exitstence of ℕ.
Am Mon, 24 Feb 2025 15:59:18 +0100 schrieb WM:
These axioms can be applied to show that all FISONs can be removed.Nope. Any number of them can be removed. Indeed, removing all
changes the union, as you have admitted.
Only all FISONs = natural numbers are the matter of my proof. AccordingThe set N is not a natural number
to Zermelo they make up the set ℕ.
that the proof concerns.
On 2/24/2025 12:17 PM, WM wrote:
On 24.02.2025 17:40, FromTheRafters wrote:
WM brought next idea :
These axioms can be applied to show that
all FISONs can be removed.
Not in a system where sets don't change.
Addition and subtraction of sets are a common techniques.
Thereby sets are changed.
Thereby a relationship between unchanging sets
is described.
It is possible
but not useful to express this ponderously by fixed sets.
Mathematical sets and
mathematical objects in general
do not change,
Am Mon, 24 Feb 2025 18:10:26 +0100 schrieb WM:
On 24.02.2025 17:36, Jim Burns wrote:
On 2/24/2025 9:59 AM, WM wrote:
On 23.02.2025 23:03, Jim Burns wrote:
Yes, you are wrong trying to talk about N \ {N}.It is wrong to apply this in the present framework.{{}} ≠ {}
That is your nonsense.Only such nonsense available?These axioms can be applied to show that all FISONs can be removed.{1,2}\{1,{2}} = {2}
Properties don’t transfer from elements to their set.
We are not disputing the existence.Claims about each natural number are silent about ℕ.Claims about the existence of all natural numbers are claims about the
exitstence of ℕ.
On 24.02.2025 19:01, joes wrote:There are more FISONs than any natural number: infinitely many.
Am Mon, 24 Feb 2025 15:59:18 +0100 schrieb WM:
There are not more than any number.These axioms can be applied to show that all FISONs can be removed.Nope. Any number of them can be removed. Indeed, removing all changes
the union, as you have admitted.
You pretend to remove the whole set.Of course. Nobody said so.Only all FISONs = natural numbers are the matter of my proof.The set N is not a natural number
According to Zermelo they make up the set ℕ.
No, you are wrong.that the proof concerns.But Zermelo claims to produce the set ℕ. Is he wrong?
Hint: In fact he produces ℕ_def.
On 2/24/2025 12:10 PM, WM wrote:
In the present framework,
you (WM) confuse a claim about each FISON in {F}
with a claim about {F}.
Zermelo creates all natural numbers by induction
and by that guarantees the existence of the set ℕ.
I guarantee that Zermelo was a finite being
and that, as such, he did not perform any supertask.
The existence of the set which
is its.own.only.inductive.subset
is proven from Zermelo's axioms.
We call that set ℕ.
{1,2}\{1,{2}} = {2}
Only such nonsense available?
I'll grant you that it's trivial.
You (WM) have made it necessary to cover this.
On 24.02.2025 19:25, Jim Burns wrote:Yes you do talk about removing {F}.
On 2/24/2025 12:10 PM, WM wrote:
In the present framework,I never talked about {F}.
you (WM) confuse a claim about each FISON in {F} with a claim about
{F}.
--I'll grant you that it's trivial.{1,2}\{1,{2}} = {2}Only such nonsense available?
You (WM) have made it necessary to cover this.
Am Tue, 25 Feb 2025 10:02:47 +0100 schrieb WM:
On 24.02.2025 19:01, joes wrote:There are more FISONs than any natural number: infinitely many.
Am Mon, 24 Feb 2025 15:59:18 +0100 schrieb WM:There are not more than any number.
These axioms can be applied to show that all FISONs can be removed.Nope. Any number of them can be removed. Indeed, removing all changes
the union, as you have admitted.
You pretend to remove the whole set.Of course. Nobody said so.Only all FISONs = natural numbers are the matter of my proof.The set N is not a natural number
According to Zermelo they make up the set ℕ.
But Zermelo claims to produce the set ℕ. Is he wrong?No, you are wrong.
Hint: In fact he produces ℕ_def.
Am Tue, 25 Feb 2025 10:49:34 +0100 schrieb WM:
On 24.02.2025 19:25, Jim Burns wrote:Yes you do talk about removing {F}.
On 2/24/2025 12:10 PM, WM wrote:I never talked about {F}.
In the present framework,
you (WM) confuse a claim about each FISON in {F} with a claim about
{F}.
On 25.02.2025 10:17, joes wrote:Immediate contradiction: there is no largest natural.
Am Tue, 25 Feb 2025 10:02:47 +0100 schrieb WM:Yes potential infinity. There are more FISONs than any fixed natural
On 24.02.2025 19:01, joes wrote:There are more FISONs than any natural number: infinitely many.
Am Mon, 24 Feb 2025 15:59:18 +0100 schrieb WM:There are not more than any number.
These axioms can be applied to show that all FISONs can be removed.Nope. Any number of them can be removed. Indeed, removing all changes
the union, as you have admitted.
number, but the number of FISONs is a natural number
because the sequenceIt does, actually, converge on N.
{1}
{2, 1}
{3, 2, 1}
...
has no limit.
Further induction produces no actual infinity.WDYM "further"? Induction goes up to infinity.
Huh?I remove by induction all natural numbers. What of ℕ remains in your opinion?You pretend to remove the whole set.Of course. Nobody said so.Only all FISONs = natural numbers are the matter of my proof.The set N is not a natural number
According to Zermelo they make up the set ℕ.
I didn't. The *elements* are finite.You claimed yourself that induction produces only a finite number of elements. Zermelo used induction.But Zermelo claims to produce the set ℕ. Is he wrong?No, you are wrong.
Hint: In fact he produces ℕ_def.
On 24.02.2025 19:25, Jim Burns wrote:
On 2/24/2025 12:10 PM, WM wrote:
In the present framework,
you (WM) confuse a claim about each FISON in {F}
with a claim about {F}.
I never talked about {F}.
Zermelo creates all natural numbers by induction
and by that guarantees the existence of the set ℕ.
I guarantee that Zermelo was a finite being
and that, as such, he did not perform any supertask.
Therefore he used induction.
The existence of the set which
is its.own.only.inductive.subset
is proven from Zermelo's axioms.
We call that set ℕ.
Defined by induction.
{1,2}\{1,{2}} = {2}
Only such nonsense available?
I'll grant you that it's trivial.
You (WM) have made it necessary to cover this.
No.Your massive misunderstanding shows up above. I never used {ℕ} or {F}.
Regards, WM
Am Tue, 25 Feb 2025 10:55:35 +0100 schrieb WM:
Immediate contradiction: there is no largest natural.There are more FISONs than any natural number: infinitely many.Yes potential infinity. There are more FISONs than any fixed natural
number, but the number of FISONs is a natural number
because the sequenceIt does, actually, converge on N.
{1}
{2, 1}
{3, 2, 1}
...
has no limit.
Further induction produces no actual infinity.WDYM "further"? Induction goes up to infinity.
Huh?I remove by induction all natural numbers. What of ℕ remains in yourYou pretend to remove the whole set.Of course. Nobody said so.Only all FISONs = natural numbers are the matter of my proof.The set N is not a natural number
According to Zermelo they make up the set ℕ.
opinion?
I didn't. The *elements* are finite.You claimed yourself that induction produces only a finite number ofBut Zermelo claims to produce the set ℕ. Is he wrong?No, you are wrong.
Hint: In fact he produces ℕ_def.
elements. Zermelo used induction.
joes presented the following explanation :
WDYM "further"? Induction goes up to infinity.
With von Neumann's construction of ordinals it can be extended to
transfinite ordinal numbers. Zermelo's construction cannot be extended
this way.
On 25.02.2025 11:37, joes wrote:Impossible. The set of FISONs does not change.
Am Tue, 25 Feb 2025 10:55:35 +0100 schrieb WM:
There is no largest FISON either. Nevertheless, the number of FISONs isImmediate contradiction: there is no largest natural.There are more FISONs than any natural number: infinitely many.Yes potential infinity. There are more FISONs than any fixed natural
number, but the number of FISONs is a natural number
a natural number, although not fixed.
Those what? The naturals are not sequences.Not those which satisfy ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.because the sequence {1}It does, actually, converge on N.
{2, 1}
{3, 2, 1}
...
has no limit.
Whatever. Induction does not include infinity, although it doesFurther I state that induction does not produce actual infinity.Further induction produces no actual infinity.WDYM "further"? Induction goes up to infinity.
You cannot remove a non-natural number of elements, such as all of them.I remove by induction all natural numbers. What of ℕ remains in yourYou pretend to remove the whole set.Of course. Nobody said so.Only all FISONs = natural numbers are the matter of my proof.The set N is not a natural number
According to Zermelo they make up the set ℕ.
opinion?
The other way around: the number of finite elements is infinite.The elements n are finite and contain the number of elements {1, 2, 3,I didn't. The *elements* are finite.You claimed yourself that induction produces only a finite number ofBut Zermelo claims to produce the set ℕ. Is he wrong?No, you are wrong.
Hint: In fact he produces ℕ_def.
elements. Zermelo used induction.
..., n} which also are finite.
On 24.02.2025 19:00, Jim Burns wrote:
On 2/24/2025 12:17 PM, WM wrote:
It is possible
but not useful to express this ponderously
by fixed sets.
Mathematical sets and
mathematical objects in general
do not change,
(although they can represent change in various ways).
The set of citizens changes
(a set that is mathematically describable).
The set of known prime numbers changes
(a mathematical set).
Addition and subtraction of sets are
a common techniques.
Thereby sets are changed.
Thereby a relationship between unchanging sets
is described.
That is the clumsy description.
In order to be existing sets must be created
by men or by God.
On 25.02.2025 11:37, joes wrote:
Am Tue, 25 Feb 2025 10:55:35 +0100 schrieb WM:
Immediate contradiction: there is no largest natural.There are more FISONs than any natural number: infinitely many.Yes potential infinity. There are more FISONs than any fixed natural
number, but the number of FISONs is a natural number
There is no largest FISON either. Nevertheless, the number of FISONs is
a natural number, although not fixed.
because the sequenceIt does, actually, converge on N.
{1}
{2, 1}
{3, 2, 1}
...
has no limit.
Not those which satisfy ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
Further induction produces no actual infinity.WDYM "further"? Induction goes up to infinity.
Further I state that induction does not produce actual infinity.
Huh?I remove by induction all natural numbers. What of ℕ remains in yourYou pretend to remove the whole set.Of course. Nobody said so.Only all FISONs = natural numbers are the matter of my proof.The set N is not a natural number
According to Zermelo they make up the set ℕ.
opinion?
That is not a natural number.
I didn't. The *elements* are finite.You claimed yourself that induction produces only a finite number ofBut Zermelo claims to produce the set ℕ. Is he wrong?No, you are wrong.
Hint: In fact he produces ℕ_def.
elements. Zermelo used induction.
The elements n are finite and contain the number of elements {1, 2,
3, ..., n} which also are finite.
Regards, WM
On 27.02.2025 19:19, Jim Burns wrote:
On 2/27/2025 5:45 AM, WM wrote:
On 26.02.2025 23:17, Jim Burns wrote:
This next bit you (WM) might like, for a change.
It looks like the pseudo.induction.rule which
you have been trying to use.
It is induction.
This is what you (WM) have called induction:
⎛ Each inductive predicate A
No, I call induction
a very restricted number of predicates.
I prefer Wikipedia:
∀P( P(1) /\ ∀k(P(k) ==> P(k+1)) ==> ∀n (P(n)).
If A(n) is useless for UA = ℕ,
then A(n+1) us useless too.
No reason to extend this simple concept.
I do it order to avoid the following waffle:
What that version of 'induction' seems to say
is false if it's read literally.
It's false that
each inductive predicate is true.without.exception
_in each domain without exception_
Z₀ is a subset of a set Z holding 0 and all the {a}
By the same induction
I prove UF = ℕ ==> Ø = ℕ.
What you use to prove that is
∀n:Aᴺ(n) ⇒ A(ℕ)
That is how Zermelo guarantees Z₀.
That's not induction.
It seems to follow from confusion over
the difference between a set and its elements.
There is no difference in some cases like these:
When all n are added by induction to the empty set,
then we have constructed ℕ.
When all n are subtrated by induction from ℕ
then we have created the empty set.
Do you agree?
On 2/27/2025 5:45 AM, WM wrote:
On 26.02.2025 23:17, Jim Burns wrote:
This next bit you (WM) might like, for a change.
It looks like the pseudo.induction.rule which
you have been trying to use.
It is induction.
This is what you (WM) have called induction:
⎛ Each inductive predicate A
What that version of 'induction' seems to say
is false if it's read literally.
It's false that
each inductive predicate is true.without.exception
_in each domain without exception
Z₀ is a subset of a set Z holding 0 and all the {a}
By the same induction
I prove UF = ℕ ==> Ø = ℕ.
What you use to prove that is
∀n:Aᴺ(n) ⇒ A(ℕ)
That's not induction.
It seems to follow from confusion over
the difference between a set and its elements.
On 27.02.2025 19:19, Jim Burns wrote:Absolutely not. He does not prove Z e Z.
On 2/27/2025 5:45 AM, WM wrote:
On 26.02.2025 23:17, Jim Burns wrote:
That is how Zermelo guarantees Z₀.By the same induction I prove UF = ℕ ==> Ø = ℕ.What you use to prove that is ∀n:Aᴺ(n) ⇒ A(ℕ)
There totally is a difference between N and each of its elements.That's not induction.There is no difference in some cases like these:
It seems to follow from confusion over the difference between a set and
its elements.
When all n are added by induction to the empty set, then we have
constructed ℕ.
When all n are subtrated by induction from ℕ then we have created the empty set.
Am Thu, 27 Feb 2025 20:50:11 +0100 schrieb WM:
That is how Zermelo guarantees Z₀.Absolutely not. He does not prove Z e Z.
On 2/27/2025 2:50 PM, WM wrote:
On 27.02.2025 19:19, Jim Burns wrote:
On 2/27/2025 5:45 AM, WM wrote:
On 26.02.2025 23:17, Jim Burns wrote:
This next bit you (WM) might like, for a change.
It looks like the pseudo.induction.rule which
you have been trying to use.
It is induction.
This is what you (WM) have called induction:
⎛ Each inductive predicate A
No, I call induction
a very restricted number of predicates.
I prefer Wikipedia:
∀P (P(1) /\ ∀k(P(k) ==> P(k+1)) ==> ∀n (P(n)).
</WM>
Date: Thu, 6 Feb 2025 17:55:57 +0100
Message-ID: <vo2pit$31hlr$[email protected]>
If A(n) is useless for UA = ℕ,
then A(n+1) us useless too.
No reason to extend this simple concept.
You extend ∀n:Aᴺ(n) to Aᴺ(ℕ)
but you only claim it, you don't justify it.
I do it order to avoid the following waffle:
How very Orwellian of you.
I justify my claims. Doing so is 'waffle'.
You don't. Abstaining from doing so is
'mathematics' and 'logic' and 'geometry'.
What that version of 'induction' seems to say
is false if it's read literally.
It's false that
each inductive predicate is true.without.exception
_in each domain without exception_
Z₀ is a subset of a set Z holding 0 and all the {a}
By the same induction
I prove UF = ℕ ==> Ø = ℕ.
What you use to prove that is
∀n:Aᴺ(n) ⇒ A(ℕ)
That is how Zermelo guarantees Z₀.
Zermello's Infinity guarantees a superset Z of Z₀
From Z, it follows,
by Powerset and by Separation,
that Z₀ exists.
∀n:Aᴺ(n) ⇒ Aᴺ(ℕ)
is your fantasy.
You would find your posts greatly improved
by criticizing (if you can) _our_ reasoning,
There is no difference in some cases like these:
When all n are added by induction to the empty set,
then we have constructed ℕ.
When we have shown that there is
the intersection of all inductive subsets of
an inductive set,
then we have constructed ℕ.
In this context,
a 'construction' is a proof of existence.
When all n are subtracted by induction from ℕ
then we have created the empty set.
Do you agree?
I am trying to reach some expressions
with which I can answer you and be understood.
I'm not there yet.
WM formulated the question :
On 27.02.2025 19:43, FromTheRafters wrote:
Generative AI is experimental.
Much better informed than you.
Are you saying it is wrong?
On 27.02.2025 21:41, Jim Burns wrote:
On 2/27/2025 2:50 PM, WM wrote:
On 27.02.2025 19:19, Jim Burns wrote:
This is what you (WM) have called induction:
⎛ Each inductive predicate A is true.without.exception.
⎜ ∀ᵖʳᵉᵈA: A(0) ∧ ∀k:A(k)⇒A(k+1) ⇒
⎝ ∀n:A(n)
(1)
No, I call induction
a very restricted number of predicates.
I prefer Wikipedia:
∀P (P(1) /\ ∀k(P(k) ==> P(k+1)) ==> ∀n (P(n)).
Correct.
But not necessary in its generality
for my purpose.
If A(n) is useless for UA = ℕ,
then A(n+1) us useless too.
No reason to extend this simple concept.
You extend ∀n:Aᴺ(n) to Aᴺ(ℕ)
but you only claim it, you don't justify it.
I use Zermelo's approach
without wich there is no set theory.
By the same induction
I prove UF = ℕ ==> Ø = ℕ.
What you use to prove that is
∀n:Aᴺ(n) ⇒ A(ℕ)
That is how Zermelo guarantees Z₀.
Zermello's Infinity guarantees a superset Z of Z₀
How is that accomplished?
From Z, it follows,
by Powerset and by Separation,
that Z₀ exists.
∀n:Aᴺ(n) ⇒ Aᴺ(ℕ)
is your fantasy.
It is Zermelo's approach.
You would find your posts greatly improved
by criticizing (if you can) _our_ reasoning,
You deny Zermelo's approach.
His Z is ensured by induction.
When we have shown that there is
the intersection of all inductive subsets of
an inductive set,
then we have constructed ℕ.
We don't even need the intersection
if we reduce Zermelo's approach to
Lorenzen's approach:
I is a natural number, and
if x is a natural numbers then x+1 is a natural number.
On 2/27/2025 5:01 PM, WM wrote:
Zermelo's approach does not extend ∀n:Aᴺ(n) to Aᴺ(ℕ)
Zermello's Infinity guarantees a superset Z of Z₀
How is that accomplished?
By Zermelo's approach:
Z ⇒ 𝒫(Z)
It's not.even.wrong.
And it's not Zermelo's approach,
Z
His Z is ensured by induction.
Nope.
We don't even need the intersection
if we reduce Zermelo's approach to
Lorenzen's approach:
I is a natural number, and
if x is a natural numbers then x+1 is a natural number.
Consider Robinson arithmetic.
Proofs by induction are unreliable in Robinson arithmetic.
Wherever you got Lorenzen's approach from,
send it back.
On 28.02.2025 01:00, Jim Burns wrote:Nonono. The set is not added to itself.
On 2/27/2025 5:01 PM, WM wrote:
Zermelo's approach does not extend ∀n:Aᴺ(n) to Aᴺ(ℕ)It does. Zermelo says it, and it is easy to prove it:
Adding all natural numbers established the set ℕ.
No, the set of naturals is not a natural.Wherever you got Lorenzen's approach from, send it back.Then you admit to stand outside of mathematics. Lorenzen uses the same induction as Cantor, Dedekind, Peano, Schmidt, Zermelo, or v. Neumann. Addition of all natnumbers results in the set of all natnumbers.
Am Fri, 28 Feb 2025 10:12:13 +0100 schrieb WM:
On 28.02.2025 01:00, Jim Burns wrote:Nonono. The set is not added to itself.
On 2/27/2025 5:01 PM, WM wrote:It does. Zermelo says it, and it is easy to prove it:
Zermelo's approach does not extend ∀n:Aᴺ(n) to Aᴺ(ℕ)
Adding all natural numbers established the set ℕ.
[...]
No, the set of naturals is not a natural.Wherever you got Lorenzen's approach from, send it back.Then you admit to stand outside of mathematics. Lorenzen uses the same
induction as Cantor, Dedekind, Peano, Schmidt, Zermelo, or v. Neumann.
Addition of all natnumbers results in the set of all natnumbers.
On 28.02.2025 11:34, joes wrote:Yes, but that is what you and Jim wrote above.
Am Fri, 28 Feb 2025 10:12:13 +0100 schrieb WM:The set is established (not added) by all its elements.
On 28.02.2025 01:00, Jim Burns wrote:Nonono. The set is not added to itself.
On 2/27/2025 5:01 PM, WM wrote:It does. Zermelo says it, and it is easy to prove it:
Zermelo's approach does not extend ∀n:Aᴺ(n) to Aᴺ(ℕ)
Adding all natural numbers established the set ℕ.
Yes, its elements, not itself.The set ℕ is constructed by adding all its numbers or FISONs:No, the set of naturals is not a natural.Wherever you got Lorenzen's approach from, send it back.Then you admit to stand outside of mathematics. Lorenzen uses the same
induction as Cantor, Dedekind, Peano, Schmidt, Zermelo, or v. Neumann.
Addition of all natnumbers results in the set of all natnumbers.
ℕ = {1, 2, 3, ...}
= {1} U {2} U {3} U ...
= {1} U {1, 2} U {1, 2, 3} U ...
On 27.02.2025 05:06, Richard Damon wrote:
And where does that use the term "Induction"
He does not use the term but the matter: { } and with a also {a}.
Regards, WM
On 28.02.2025 01:00, Jim Burns wrote:
On 2/27/2025 5:01 PM, WM wrote:
Zermelo's approach
does not extend ∀n:Aᴺ(n) to Aᴺ(ℕ)
It does.
Zermelo says it,
and it is easy to prove it:
Adding all natural numbers established the set ℕ.
Zermello's Infinity guarantees
a superset Z of Z₀
How is that accomplished?
By Zermelo's approach:
Z ⇒ 𝒫(Z)
How is Z accomplished?
It's not.even.wrong.
And it's not Zermelo's approach,
Z
Where does he get is Z from?
His Z is ensured by induction.
Nope.
{ } and a ==> {a}.
then x+1 is a natural number.When we have shown that there is
the intersection of all inductive subsets of
an inductive set,
then we have constructed ℕ.
We don't even need the intersection
if we reduce Zermelo's approach to
Lorenzen's approach:
I is a natural number, and
if x is a natural numbers
Consider Robinson arithmetic.
No.
Proofs by induction are unreliable
in Robinson arithmetic.
Irrelevant.
On 2/27/25 5:53 AM, WM wrote:
On 27.02.2025 05:06, Richard Damon wrote:
And where does that use the term "Induction"
He does not use the term but the matter: { } and with a also {a}.
So, you admit that you are just misquoting the source,
Yes, at any finite point in the iteration, the set build so far has a
maximum element, but when we let the process complete (even though we
can't do it) the result changes to not have a maximum element.
On 2/27/25 2:50 PM, WM wrote:
No, I call induction a very restricted number of predicates.
And thus admit that what you call induction isn't what everyone else
does,
On 2/28/2025 4:12 AM, WM wrote:
On 28.02.2025 01:00, Jim Burns wrote:
On 2/27/2025 5:01 PM, WM wrote:
Zermelo's approach
does not extend ∀n:Aᴺ(n) to Aᴺ(ℕ)
It does.
Eppur si muove.
Zermelo says it,
Nope.
and it is easy to prove it:
Adding all natural numbers established the set ℕ.
We are finite beings. We do not do that.
How is Z accomplished?
Z is NOT accomplishedᵂᴹ.
Zermelo describes the Z in the discussion
{ } and a ==> {a}.
True of Z because,
when Zermelo describes Z,
Zermelo describes such a set.
Z being such a set is not induction.
Induction proves inductive a subset of
a set which is its.own.only.inductive.subset.
An inductive proof only proves about
a set which is its.own.only.inductive.subset,
like Z₀ and like ℕ, perhaps not like Z
Proofs by induction are unreliable
in Robinson arithmetic.
Irrelevant.
What you (WM) think is a proof by induction
is unreliable. But you don't care?
On 28.02.2025 16:52, Jim Burns wrote:Zermelo doesn't say that Z is in Z.
On 2/28/2025 4:12 AM, WM wrote:[spam]
On 28.02.2025 01:00, Jim Burns wrote:Eppur si muove.
On 2/27/2025 5:01 PM, WM wrote:
Zermelo's approach does not extend ∀n:Aᴺ(n) to Aᴺ(ℕ)It does.
Zermelo says it,Nope.
They do reach N in the limit.Therefore we use FISONs without approaching ℕ.and it is easy to prove it:We are finite beings. We do not do that.
Adding all natural numbers established the set ℕ.
On 28.02.2025 15:46, Richard Damon wrote:Yes, it can. You are simply incapable of conceiving this.
On 2/27/25 5:53 AM, WM wrote:
On 27.02.2025 05:06, Richard Damon wrote:
Yes, at any finite point in the iteration, the set build so far has aThe process cannot be completed with FISONs. Each one is finite, and so
maximum element, but when we let the process complete (even though we
can't do it) the result changes to not have a maximum element.
is their potentially infinite number. That does never change.
On 28.02.2025 16:52, Jim Burns wrote:
An inductive proof only proves about
a set which is its.owcn.only.inductive.subset,
like Z₀ and like ℕ, perhaps not like Z
Z contains many inductive subsets.
Am Fri, 28 Feb 2025 18:04:33 +0100 schrieb WM:
Zermelo doesn't say that Z is in Z.
Therefore we use FISONs without approaching ℕ.They do reach N in the limit.
On 2/28/2025 12:04 PM, WM wrote:
On 28.02.2025 16:52, Jim Burns wrote:
An inductive proof only proves about
a set which is its.owcn.only.inductive.subset,
like Z₀ and like ℕ, perhaps not like Z
Z contains many inductive subsets.
(We don't know whether Z=Z₀ or Z≠Z₀)
Consider this WM.inductive.proof:
Am Fri, 28 Feb 2025 17:44:17 +0100 schrieb WM:
On 28.02.2025 15:46, Richard Damon wrote:
Yes, it can.Yes, at any finite point in the iteration, the set build so far has aThe process cannot be completed with FISONs. Each one is finite, and so
maximum element, but when we let the process complete (even though we
can't do it) the result changes to not have a maximum element.
is their potentially infinite number. That does never change.
On 2/28/25 11:44 AM, WM wrote:
No, Zermelo uses induction. I did not say that he uses the term
induction.
No, what you describe is NOT "induction".
The process cannot be completed with FISONs. Each one is finite, andIt can not be completed with a finite number of them. It can be
so is their potentially infinite number. That does never change.
completed with the complete set of them.
Am Sat, 01 Mar 2025 12:58:28 +0100 schrieb WM:
On 28.02.2025 18:25, joes wrote:Taking a limit "bridges the gap".
Am Fri, 28 Feb 2025 18:04:33 +0100 schrieb WM:But he says that the infinite Z exists because of his induction.
Zermelo doesn't say that Z is in Z.
No. ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo. Note for all! The limitTherefore we use FISONs without approaching ℕ.They do reach N in the limit.
cannot exist without bridging this infinite gap.
On 28.02.2025 20:25, Jim Burns wrote:
On 2/28/2025 12:04 PM, WM wrote:
On 28.02.2025 16:52, Jim Burns wrote:
An inductive proof only proves about
a set which is its.own.only.inductive.subset,
like Z₀ and like ℕ, perhaps not like Z
Z contains many inductive subsets.
(We don't know whether Z=Z₀ or Z≠Z₀)
Z contains every set a.
Z_0 contains only 0, {0}, {{0}}, ...
Consider this WM.inductive.proof:
It is sufficient
It is sufficient
for my purpose to consider
0, {0}, {{0}}, ...
which is an infinite set produced by induction.
On 3/1/25 6:58 AM, WM wrote:
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo. Note for all! The limit >> cannot exist without bridging this infinite gap.But what is "n" in the above definition?
Note, the value AT the limit, and the values appraching the limit of
things can be different.
On 3/1/25 9:31 AM, WM wrote:
On 28.02.2025 20:45, Richard Damon wrote:
On 2/28/25 11:44 AM, WM wrote:
No, Zermelo uses induction. I did not say that he uses the term
induction.
No, what you describe is NOT "induction".
Correct your qualifiers. Or look it up.
F(1) ∈ F and F(n) ∈ F ==> F(n+1) ∈ F describes the infinite inductive >> set F of FISONs.
You have a source that uses the term "Induction" for the recursive
iteration that builds the set?
Processes using defined individuals like FISONs cannot surpass a
finite set. FISONs are finite. Their number will never be greater than
finite.
{1}
{2, 1}
{3, 2, 1}
...
Nope, the axion of Induction (which needs the axiom of infinity) says
that the process DOES complete.
Since F(0) exists, and the existance of F(n) says that F(n+1) exist, the axion of induction says that F(m) exist for ALL m a member of the
Natural Numbers, and thus the set of FISION is infinite,
On 3/1/2025 7:51 AM, WM wrote:
Z_0 contains only 0, {0}, {{0}}, ...
Z₀ = {0,{0},{{0}},...} is
the only subset of Z₀ which
holds 0 and, for each a, holds {a}
In narrower Z₀,
inductivity identifies a unique set.
On 01.03.2025 18:47, Jim Burns wrote:
On 3/1/2025 7:51 AM, WM wrote:
Z_0 contains only 0, {0}, {{0}}, ...
Z₀ = {0,{0},{{0}},...} is
the only subset of Z₀ which
holds 0 and, for each a, holds {a}
Z₀ is defined by induction.
Likewise UF is defined by induction.
ℕ \ F(1) \ F(2) \ F(3) \ ... = ℵo
= ℕ \ (F(1) U F(2) U F(3) U ...) = ℵo.
In narrower Z₀,
inductivity identifies a unique set.
So it is.
On 03/01/2025 12:10 PM, Jim Burns wrote:
Our inductive conclusions are justifiedly universal
because they are in narrower Z₀
not because of any supertask.
Like Zeno's that show motion is impossible?
On 03/01/2025 04:00 PM, Jim Burns wrote:
On 3/1/2025 5:55 PM, Ross Finlayson wrote:
On 03/01/2025 12:10 PM, Jim Burns wrote:
Our inductive conclusions are justifiedly universal
because they are in narrower Z₀
not because of any supertask.
Like Zeno's that show motion is impossible?
Let's play "Where's Zeno?"
A property P is pre.inductive iff,
if each set in set 𝒞 of sets has P,
then its intersection ⋂𝒞 has P
∀S∈𝒞:P(𝒞) ⇒ P(⋂𝒞)
For example,
'holding 7,11,13' is pre.inductive.
If each set of a set 𝒞 of sets holds 7,11,13
then ⋂𝒞 holds 7,11,13.
Another example is 'being inductive'.
Consider a set Z such that P(Z)
for pre.inductive property P
The intersection ⋂𝒫ᴾ(Z) of subsets having P
has P.
The only subset of ⋂𝒫ᴾ(Z) which has P
is ⋂𝒫ᴾ(Z)
⎛ Assume, for {i:A(i)} ⊆ ⋂𝒫ᴾ(Z)
⎜ P{i:A(i)}
⎜
⎜ There is only one subset of ⋂𝒫ᴾ(Z) which has P
⎜ {i:A(i)} = ⋂𝒫ᴾ(Z)
⎜
⎜ By definition of {i:A(i)}
⎜ ∀n ∈ {i:A(i)}: A(n)
⎜
⎜ {i:A(i)} = ⋂𝒫ᴾ(Z)
⎜
⎝ ∀n ∈ ⋂𝒫ᴾ(Z): A(n)
Therefore,
if P{i:A(i)}
then ∀n ∈ ⋂𝒫ᴾ(Z): A(n)
For
P(S) ⇔ 0∈S ∧ ∀k∈⋂𝒫ᴾ(Z):k∈S⇒k+1∈S
and
ℕ = ⋂𝒫ᴾ(Z):
A(0) ∧ ∀k∈ℕ:A(k)⇒A(k+1) ⇒ ∀n∈ℕ:A(n)
Where's Zeno?
He's standing right next to you.
Yet, you must've crossed the bridge.
He yells loud, "you are on the other side".
The geometric series you'd have is "complete", then?
I.e., attaching your cases to the geometric series.
⋂𝒫ᴾ(Z) = ℕ+
WM presented the following explanation :
Processes using defined individuals like FISONs cannot surpass a
finite set. FISONs are finite. Their number will never be greater than
finite.
{1}
{2, 1}
{3, 2, 1}
...
Those are still not FISONs.
Am Sat, 01 Mar 2025 19:09:25 +0100 schrieb WM:
On 01.03.2025 17:25, Richard Damon wrote:
On 3/1/25 9:31 AM, WM wrote:
On 28.02.2025 20:45, Richard Damon wrote:
On 2/28/25 11:44 AM, WM wrote:
You can't argue with induction if you don't believe in it.> Nope, the axion of Induction (which needs the axiom of infinity) saysProcesses using defined individuals like FISONs cannot surpass a
finite set. FISONs are finite. Their number will never be greater than >>>> finite.
{1}
{2, 1}
{3, 2, 1}
...
> that the process DOES complete.
It is in error.
Not finite.Since F(0) exists, and the existance of F(n) says that F(n+1) exist,Not actually infinite.
the axion of induction says that F(m) exist for ALL m a member of the
Natural Numbers, and thus the set of FISION is infinite,
On 3/1/2025 1:28 PM, WM wrote:
On 01.03.2025 18:47, Jim Burns wrote:
On 3/1/2025 7:51 AM, WM wrote:
Z_0 contains only 0, {0}, {{0}}, ...
Z₀ = {0,{0},{{0}},...} is
the only subset of Z₀ which
holds 0 and, for each a, holds {a}
Z₀ is defined by induction.
inductive(Z)
inductive(Z) :⇔ 0∈Z ∧ ∀a:a∈Z⇒{a}∈Z
Z₀ is the emptiest (<EZ>"einfachste"?)
inductive set.
inductive(W) ⇒ Z₀ ⊆ W
inductive(Z₀)
Z₀ is not constructed by supertask.
Z₀ = ⋂𝒫ⁱⁿᵈ(Z)
Likewise UF is defined by induction.
ℕ \ F(1) \ F(2) \ F(3) \ ... = ℵo
= ℕ \ (F(1) U F(2) U F(3) U ...) = ℵo.
In narrower Z₀,
inductivity identifies a unique set.
So it is.
Thank you.
{S⊆Z₀:inductive.S} = {Z₀} ∧
inductive{i:A(i)} ⇒
{i:A(i)} ∈ {SsZ₀:inductive.S} = {Z₀} ⇒
Z₀ = {i:A(i)}
Z₀ = {i:A(i)} ∧
∀n ∈ {i:A(i)}: A(n) ⇒
∀n ∈ Z₀: A(n)
Our inductive conclusions are justifiedly universal
because they are in narrower Z₀
not because of any supertask.
On 01.03.2025 17:25, Richard Damon wrote:
On 3/1/25 6:58 AM, WM wrote:
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo. Note for all! The limit >>> cannot exist without bridging this infinite gap.But what is "n" in the above definition?
It is a natural number that we can represent in principle by making as
many strokes.
Note, the value AT the limit, and the values appraching the limit of
things can be different.
The cannot differ by a fixed quantity like ℵo.
Regards, WM
On 01.03.2025 17:25, Richard Damon wrote:
On 3/1/25 9:31 AM, WM wrote:
On 28.02.2025 20:45, Richard Damon wrote:
On 2/28/25 11:44 AM, WM wrote:
No, Zermelo uses induction. I did not say that he uses the term
induction.
No, what you describe is NOT "induction".
Correct your qualifiers. Or look it up.
F(1) ∈ F and F(n) ∈ F ==> F(n+1) ∈ F describes the infinite inductive >>> set F of FISONs.
You have a source that uses the term "Induction" for the recursive
iteration that builds the set?
∀P(P(0) /\ ∀k(P(k) ==> P(k+1)) ==> ∀n (P(n)))
Wikipedia
Processes using defined individuals like FISONs cannot surpass a
finite set. FISONs are finite. Their number will never be greater
than finite.
{1}
{2, 1}
{3, 2, 1}
...
Nope, the axion of Induction (which needs the axiom of infinity) says
that the process DOES complete.
It is in error.
Since F(0) exists, and the existance of F(n) says that F(n+1) exist,
the axion of induction says that F(m) exist for ALL m a member of the
Natural Numbers, and thus the set of FISION is infinite,
Not actually infinite.
Regards, WM
On 01.03.2025 21:10, Jim Burns wrote:
On 3/1/2025 1:28 PM, WM wrote:
On 01.03.2025 18:47, Jim Burns wrote:
On 3/1/2025 7:51 AM, WM wrote:
Z_0 contains only 0, {0}, {{0}}, ...
Z₀ = {0,{0},{{0}},...} is
the only subset of Z₀ which
holds 0 and, for each a, holds {a}
Z₀ is defined by induction.
inductive(Z)
inductive(Z) :⇔ 0∈Z ∧ ∀a:a∈Z⇒{a}∈Z
Z₀ is the emptiest (<EZ>"einfachste"?)
inductive set.
The simplest example.
inductive(W) ⇒ Z₀ ⊆ W
inductive(Z₀)
Z₀ is not constructed by supertask.
Inductive Z₀:
{ } ∈ Z₀, and
if {{{...{{{ }}}...}}} with n curly brackets ∈ Z₀
then {{{...{{{ }}}...}}} with n+1 curly brackets ∈ Z₀.
Z₀ = ⋂𝒫ⁱⁿᵈ(Z)
Also true.
Our inductive conclusions are justifiedly universal
because they are in narrower Z₀
not because of any supertask.
Induction abbreviates a supertask.
If 1 then 2, if 2 then 3, and so on.
But supertasks will never pass through the dark numbers.
They can only extend the defined numbers without end,
never crossing the infinitely larger domain of dark numbers
- if such exist at all!
On 3/1/25 1:09 PM, WM wrote:
On 01.03.2025 17:25, Richard Damon wrote:
On 3/1/25 9:31 AM, WM wrote:
On 28.02.2025 20:45, Richard Damon wrote:
On 2/28/25 11:44 AM, WM wrote:
No, Zermelo uses induction. I did not say that he uses the term
induction.
No, what you describe is NOT "induction".
Correct your qualifiers. Or look it up.
F(1) ∈ F and F(n) ∈ F ==> F(n+1) ∈ F describes the infinite
inductive set F of FISONs.
You have a source that uses the term "Induction" for the recursive
iteration that builds the set?
∀P(P(0) /\ ∀k(P(k) ==> P(k+1)) ==> ∀n (P(n)))
Wikipedia
And what SET did that build? P was a statement about a relationship.
You can try to claim that the system it creates is
inconsistent, but to do so you need to show the error using the system
as defined.
On 3/1/25 1:18 PM, WM wrote:
On 01.03.2025 17:25, Richard Damon wrote:
On 3/1/25 6:58 AM, WM wrote:
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo. Note for all! The limitBut what is "n" in the above definition?
cannot exist without bridging this infinite gap.
It is a natural number that we can represent in principle by making as
many strokes.
SO, Which Natural Number can it not be that makes N_def different than N?
Note, the value AT the limit, and the values appraching the limit of
things can be different.
The cannot differ by a fixed quantity like ℵo.
Sure a limit can. What says it can't?
WM brought next idea :
On 01.03.2025 19:38, FromTheRafters wrote:
WM presented the following explanation :
Processes using defined individuals like FISONs cannot surpass a
finite set. FISONs are finite. Their number will never be greater
than finite.
{1}
{2, 1}
{3, 2, 1}
...
Those are still not FISONs.
They are. The order in curly brackets does not matter.
AI Chatbot
https://poe.com/chat/370x9xpc7krhbconr8j
Assistant
Poe
The term "FISON" stands for "finite initial segment of natural numbers."
A FISON is typically defined as a finite set of the form {0,1,2,…,n} for some non-negative integer n.
The set {4,3,2,1} does not meet this definition for a couple of reasons:
Order: A FISON must include all natural numbers starting from 0 up
to n in ascending order. The set {4,3,2,1} is not in ascending order and
does not include 0.
Completeness: A FISON must include all natural numbers less than or equal to its maximum element. In this case, the maximum element is 4,
but the set is missing 0.
Therefore, {4,3,2,1} is not a FISON. A valid FISON that includes up to 4 would be {0,1,2,3,4}.
On 3/2/2025 4:48 AM, WM wrote:
On 01.03.2025 21:10, Jim Burns wrote:
On 3/1/2025 1:28 PM, WM wrote:
On 01.03.2025 18:47, Jim Burns wrote:
On 3/1/2025 7:51 AM, WM wrote:
Z_0 contains only 0, {0}, {{0}}, ...
Z₀ = {0,{0},{{0}},...} is
the only subset of Z₀ which
holds 0 and, for each a, holds {a}
Z₀ is defined by induction.
inductive(Z)
inductive(Z) :⇔ 0∈Z ∧ ∀a:a∈Z⇒{a}∈Z
Z₀ is the emptiest (<EZ>"einfachste"?)
inductive set.
The simplest example.
I think Zermelo might be using 'einfachste==simplest'
in the same way that I'm using 'emptiest==leerste':
as a pointer to the unique extremum of
all sets with property P -- where P isn't size.
Z₀ is the simplest, the emptiest, but not the smallest.
(Z₀ has same.sized inductive supersets.)
Simplicity is tricky, where membership is not,
so I will continue say Z₀ is emptiest.
Inductive Z₀:
{ } ∈ Z₀, and
if {{{...{{{ }}}...}}} with n curly brackets ∈ Z₀
then {{{...{{{ }}}...}}} with n+1 curly brackets ∈ Z₀.
I wish to draw a distinction between
two senses of the verb 'construct'
'construct[make]' and 'construct[know]'.
Only 'construct[make]' is used for roads and bridges.
This is the more.commonly.used sense by far.
Only 'construct[know]' is used for abstract entities
such as ℕ and ℝ and ℂ.
We construct[know] Z₀ by finite.length proof.
Writing down and checking a finite.length proof
is not a supertask.
We do not construct[make] Z₀
In addition to Z₀ being an abstract object
and thus no more construct[make]able than {},
construct[make]ing Z₀ would require a supertask,
and we are finite non.supertaskers.
Induction abbreviates a supertask.
If 1 then 2, if 2 then 3, and so on. But supertasks will never pass
through the dark numbers.
A claim about an indefinite element of Z₀ = ⋂𝒫ⁱⁿᵈ(Z)
cannot have a counter.example outside of Z₀
That is the source of a matheologian's certainty.
They can only extend the defined numbers without end,
never crossing the infinitely larger domain of dark numbers
- if such exist at all!
Even if dark numbers exist, √2 remains irrational.
On 02.03.2025 13:22, Richard Damon wrote:Ah no, you're shifting the quantifier there. There are many sets of
On 3/1/25 1:09 PM, WM wrote:Induction is the nucleus: P(0) /\ ∀k(P(k) ==> P(k+1)
On 01.03.2025 17:25, Richard Damon wrote:
On 3/1/25 9:31 AM, WM wrote:∀P(P(0) /\ ∀k(P(k) ==> P(k+1)) ==> ∀n (P(n)))
On 28.02.2025 20:45, Richard Damon wrote:You have a source that uses the term "Induction" for the recursive
On 2/28/25 11:44 AM, WM wrote:Correct your qualifiers. Or look it up.
No, Zermelo uses induction. I did not say that he uses the termNo, what you describe is NOT "induction".
induction.
F(1) ∈ F and F(n) ∈ F ==> F(n+1) ∈ F describes the infinite
inductive set F of FISONs.
iteration that builds the set?
Wikipedia
P gives a relationship, for instance "is element of F". The universal quantifier ∀ proves that all FISONs belong to the set that can be
And what SET did that build? P was a statement about a relationship.
removed.
--You can try to claim that the system it creates is inconsistent, but toI did: UF = ℕ ==> Ø = ℕ.
do so you need to show the error using the system as defined.
On 02.03.2025 18:32, Jim Burns wrote:
On 3/2/2025 4:48 AM, WM wrote:
Induction abbreviates a supertask.
If 1 then 2, if 2 then 3, and so on.
But supertasks will never pass through
the dark numbers.
A claim about an indefinite element of Z₀ = ⋂𝒫ⁱⁿᵈ(Z)
cannot have a counter.example outside of Z₀
That is the source of a matheologian's certainty.
It has, namely ω, ω/2, etc.
They can only extend the defined numbers without end,
never crossing the infinitely larger domain of dark numbers
- if such exist at all!
Even if dark numbers exist, √2 remains irrational.
Yes.
But it has no decimal representation.
On 02.03.2025 13:22, Richard Damon wrote:
On 3/1/25 1:09 PM, WM wrote:
On 01.03.2025 17:25, Richard Damon wrote:
On 3/1/25 9:31 AM, WM wrote:
On 28.02.2025 20:45, Richard Damon wrote:
On 2/28/25 11:44 AM, WM wrote:
No, Zermelo uses induction. I did not say that he uses the term
induction.
No, what you describe is NOT "induction".
Correct your qualifiers. Or look it up.
F(1) ∈ F and F(n) ∈ F ==> F(n+1) ∈ F describes the infinite
inductive set F of FISONs.
You have a source that uses the term "Induction" for the recursive
iteration that builds the set?
∀P(P(0) /\ ∀k(P(k) ==> P(k+1)) ==> ∀n (P(n)))
Wikipedia
Induction is the nucleus: P(0) /\ ∀k(P(k) ==> P(k+1)
And what SET did that build? P was a statement about a relationship.
P gives a relationship, for instance "is element of F". The universal quantifier ∀ proves that all FISONs belong to the set that can be removed.
You can try to claim that the system it creates is inconsistent, but
to do so you need to show the error using the system as defined.
I did: UF = ℕ ==> Ø = ℕ.
Regards, WM
WM expressed precisely :
Two mistakes. FISONs have been created by Virgil and me. Therefore we
know what their definition is. They start with 1. Ascending order is
not a feature of sets but of sequences.
Yes, but your notation of {3,2,1} is not a sequence but a set.
A
sequence would be (3,2,1). You write sets and try to treat them as
segments of sequences.
On 3/2/2025 1:52 PM, WM wrote:
On 02.03.2025 18:32, Jim Burns wrote:
On 3/2/2025 4:48 AM, WM wrote:
Induction abbreviates a supertask.
If 1 then 2, if 2 then 3, and so on.
But supertasks will never pass through
the dark numbers.
A claim about an indefinite element of Z₀ = ⋂𝒫ⁱⁿᵈ(Z)
cannot have a counter.example outside of Z₀
That is the source of a matheologian's certainty.
It has, namely ω, ω/2, etc.
ω is outside Z₀
ω cannot be a counter.example to
a claim about an indefinite element of Z₀
ω/2 is outside Z₀ and outside the ordinals.
ω/2 cannot be a counter.example to
a claim about an indefinite element of Z₀, or to
a claim about an indefinite ordinal.
They can only extend the defined numbers without end,
never crossing the infinitely larger domain of dark numbers
- if such exist at all!
Even if dark numbers exist, √2 remains irrational.
Yes.
But it has no decimal representation.
√2 splits all finite decimal representations.
Each is < √2 or > √2
Therefore,
two such decimal.splitting.points √2 and √2′
don't exist.
On 3/2/25 1:05 PM, WM wrote:
SO, Which Natural Number can it not be that makes N_def different
than N?
They cannot be expressed, neither by stroks nor by digits. Their
existence can only be proved: UF = ℕ ==> Ø = ℕ.
So, what numbers can't be expressed?
What is the highest expressable number?
If there isn't one, why not?
This is the flaw in your logic, you think there are two different
classes of the infinite set of Natural Numbers, one "defined" that can't
have a highest (as what keeps us from defining the next number) and then
you need to invent something higher to hide the fact that your first set
is actually an infinite set of all the Natural Numbers,
There are dark numbers collected.Note, the value AT the limit, and the values appraching the limit
of things can be different.
They cannot differ by a fixed quantity like ℵo.
Sure a limit can. What says it can't?
Yes, when we talk about "in the limit of completing the set" it is a different sort of operation than the normal mathematics limit,
as we get
infinities that of course never change "value". That is why we don't
write it as a normal mathematics limit. N is not limit n-> inf of F(n),
Am Sun, 02 Mar 2025 18:51:05 +0100 schrieb WM:
On 02.03.2025 13:22, Richard Damon wrote:Ah no, you're shifting the quantifier there.
On 3/1/25 1:09 PM, WM wrote:Induction is the nucleus: P(0) /\ ∀k(P(k) ==> P(k+1)
On 01.03.2025 17:25, Richard Damon wrote:
On 3/1/25 9:31 AM, WM wrote:∀P(P(0) /\ ∀k(P(k) ==> P(k+1)) ==> ∀n (P(n)))
On 28.02.2025 20:45, Richard Damon wrote:You have a source that uses the term "Induction" for the recursive
On 2/28/25 11:44 AM, WM wrote:Correct your qualifiers. Or look it up.
No, Zermelo uses induction. I did not say that he uses the term >>>>>>>> induction.No, what you describe is NOT "induction".
F(1) ∈ F and F(n) ∈ F ==> F(n+1) ∈ F describes the infinite
inductive set F of FISONs.
iteration that builds the set?
Wikipedia
P gives a relationship, for instance "is element of F". The universal
And what SET did that build? P was a statement about a relationship.
quantifier ∀ proves that all FISONs belong to the set that can be
removed.
There are many sets of
FISONs that can be removed, one set per natural, containing all
FISONS of numbers less than that natural (also more sets of non-
contiguous FISONs). But those are all finite.
The set of sets of
FISONs that can be removed together does not contain the set of
all FISONs (although it does contain the infinite sets of the odd
or even FISONs).
On 3/2/25 12:51 PM, WM wrote:
No, it means the set of REQUIRED FISONs is empty, not that we can't use
a set of FISONs to make N.
I did: UF = ℕ ==> Ø = ℕ.
But you didn't. You showed that there are no specific FISON requried to
be in that set.
On 02.03.2025 21:28, Jim Burns wrote:No. Z_0 is equivalent to N.
On 3/2/2025 1:52 PM, WM wrote:You are right. I argued above concerning Cantor's actually infinite ℕ.
On 02.03.2025 18:32, Jim Burns wrote:ω is outside Z₀ ω cannot be a counter.example to a claim about an
On 3/2/2025 4:48 AM, WM wrote:
It has, namely ω, ω/2, etc.Induction abbreviates a supertask.A claim about an indefinite element of Z₀ = ⋂𝒫ⁱⁿᵈ(Z)
If 1 then 2, if 2 then 3, and so on.
But supertasks will never pass through the dark numbers.
cannot have a counter.example outside of Z₀ That is the source of a
matheologian's certainty.
indefinite element of Z₀
It has the undefinable elements ω/2, ω/10, ω/20 inside and ω outside. Z₀ does not contain undefinable elements. It contains simply all numbers that have FISONs (like v. Neumann constructs the FISONs directly).
sqrt(2) has an infinite decimal representation, like all reals. It justω/2 is outside Z₀ and outside the ordinals. ω/2 cannot be aRight. All in Z₀ is definable.
counter.example to a claim about an indefinite element of Z₀, or to a
claim about an indefinite ordinal.
∀n ∈ Z₀: |ℕ \ {1, 2, 3, ..., n}| = ℵo
Yes.
√2 splits all finite decimal representations. Each is < √2 or > √2Yes. But it has no decimal representation.They can only extend the defined numbers without end, never crossing >>>>> the infinitely larger domain of dark numbers - if such exist at all!Even if dark numbers exist, √2 remains irrational.
And in the same way that Z_0 doesn't contain infinite elements, the setTherefore, two such decimal.splitting.points √2 and √2′ don't exist.Yes, but the topic is this: In exactly the same way as Z₀ is constructed
by its elements, the set of removable FISONs is constructed by its
elements, namely by induction.
On 02.03.2025 20:31, joes wrote:He can't. The *set of* all FISONs can't be removed.
Am Sun, 02 Mar 2025 18:51:05 +0100 schrieb WM:That is allowed if Zermelo csn do it.
On 02.03.2025 13:22, Richard Damon wrote:Ah no, you're shifting the quantifier there.
On 3/1/25 1:09 PM, WM wrote:Induction is the nucleus: P(0) /\ ∀k(P(k) ==> P(k+1)
On 01.03.2025 17:25, Richard Damon wrote:
On 3/1/25 9:31 AM, WM wrote:∀P(P(0) /\ ∀k(P(k) ==> P(k+1)) ==> ∀n (P(n)))
On 28.02.2025 20:45, Richard Damon wrote:You have a source that uses the term "Induction" for the recursive >>>>>> iteration that builds the set?
On 2/28/25 11:44 AM, WM wrote:Correct your qualifiers. Or look it up.
No, Zermelo uses induction. I did not say that he uses the term >>>>>>>>> induction.No, what you describe is NOT "induction".
F(1) ∈ F and F(n) ∈ F ==> F(n+1) ∈ F describes the infinite >>>>>>> inductive set F of FISONs.
Wikipedia
P gives a relationship, for instance "is element of F". The universal
And what SET did that build? P was a statement about a relationship.
quantifier ∀ proves that all FISONs belong to the set that can be
removed.
A quantifier shift is never a valid deduction (even though the resultant sentence may be true otherwise).There are many sets of FISONs that can be removed, one set per natural,Of course all FISONs are finite. All FISONs of Zermelo's set are finite.
containing all FISONS of numbers less than that natural (also more sets
of non-contiguous FISONs). But those are all finite.
By induction Zermelo produces the set. Quantifier shift, obviously
allowed by Zermelo. There is no other way to construct an infinite set.
And in the same way Z_0 doesn't contain the set of all elements.The set of sets of FISONs that can be removed together does not containIn exactly the same way as Z₀ is constructed by its elements, the set of removable FISONs is constructed by its elements.
the set of all FISONs (although it does contain the infinite sets of
the odd or even FISONs).
on 3/3/2025, WM supposed :
On 02.03.2025 20:28, FromTheRafters wrote:
WM expressed precisely :
Two mistakes. FISONs have been created by Virgil and me. Therefore
we know what their definition is. They start with 1. Ascending order
is not a feature of sets but of sequences.
You didn't create FISONs, you created a acronym the parts of which were already defined. What sense does a 'segment' mean if the elements have
no rank?
Am Mon, 03 Mar 2025 09:57:47 +0100 schrieb WM:
Z₀ does not contain undefinable elements. It contains simply all numbers >> that have FISONs (like v. Neumann constructs the FISONs directly).No. Z_0 is equivalent to N.
sqrt(2) has an infinite decimal representation, like all reals. It just doesn't have the period (0).
And in the same way that Z_0 doesn't contain infinite elements, the setTherefore, two such decimal.splitting.points √2 and √2′ don't exist. >> Yes, but the topic is this: In exactly the same way as Z₀ is constructed >> by its elements, the set of removable FISONs is constructed by itselements, namely by induction.
of removable sets of FISONs doesn't.
Am Mon, 03 Mar 2025 09:48:16 +0100 schrieb WM:
He can't.Ah no, you're shifting the quantifier there.That is allowed if Zermelo can do it.
By induction Zermelo produces the set. Quantifier shift, obviouslyA quantifier shift is never a valid deduction (even though the resultant sentence may be true otherwise).
allowed by Zermelo. There is no other way to construct an infinite set.
And in the same way Z_0 doesn't contain the set of all elements.The set of sets of FISONs that can be removed together does not containIn exactly the same way as Z₀ is constructed by its elements, the set of >> removable FISONs is constructed by its elements.
the set of all FISONs (although it does contain the infinite sets of
the odd or even FISONs).
WM formulated the question :
On 03.03.2025 12:58, joes wrote:
Am Mon, 03 Mar 2025 09:48:16 +0100 schrieb WM:
He can't.Ah no, you're shifting the quantifier there.That is allowed if Zermelo can do it.
But it has been accepted that with all defined elements the set is
defined.
By induction Zermelo produces the set. Quantifier shift, obviously
allowed by Zermelo. There is no other way to construct an infinite set. >>> A quantifier shift is never a valid deduction (even though the resultant >>> sentence may be true otherwise).
Zermelo has no other deduction. The set is constructed by its elements.
And in the same way Z_0 doesn't contain the set of all elements.The set of sets of FISONs that can be removed together does notIn exactly the same way as Z₀ is constructed by its elements, the
contain
the set of all FISONs (although it does contain the infinite sets of >>>>> the odd or even FISONs).
set of
removable FISONs is constructed by its elements.
Of course not. It is the set of all elements.
Discrete elements.
On 02.03.2025 21:28, Jim Burns wrote:
On 3/2/2025 1:52 PM, WM wrote:
On 02.03.2025 18:32, Jim Burns wrote:
On 3/2/2025 4:48 AM, WM wrote:
Induction abbreviates a supertask.
If 1 then 2, if 2 then 3, and so on.
But supertasks will never pass through
the dark numbers.
They can only
extend the defined numbers without end,
never crossing the infinitely larger domain of
dark numbers - if such exist at all!
Even if dark numbers exist, √2 remains irrational.
Yes.
But it has no decimal representation.
√2 splits all finite decimal representations.
Each is < √2 or > √2
Yes.
[...]
Therefore,
two such decimal.splitting.points √2 and √2′
don't exist.
Yes, but the topic is this:
In exactly the same way as Z₀
is constructed by its elements,
the set of removable FISONs
is constructed by its elements,
namely by induction.
On 03.03.2025 12:54, joes wrote:Like that.
Am Mon, 03 Mar 2025 09:57:47 +0100 schrieb WM:
How that? By induction we prove for every element n ∈ Z₀:Z₀ does not contain undefinable elements. It contains simply allNo. Z_0 is equivalent to N.
numbers that have FISONs (like v. Neumann constructs the FISONs
directly).
|ℕ \ {1, 2, 3, ..., n}| = ℵo
That is not a natural position.sqrt(2) has an infinite decimal representation, like all reals. It justThere is no actually infinite decimal representation. It would have ω digits. What digit would be at position ω/2?
doesn't have the period (0).
No, finite sets of them are removable (including sets containing onlyOf course not. Only finite FISONs are existing and removable.And in the same way that Z_0 doesn't contain infinite elements, the setTherefore, two such decimal.splitting.points √2 and √2′ don't exist. >>> Yes, but the topic is this: In exactly the same way as Z₀ isconstructed by its elements, the set of removable FISONs is
constructed by its elements, namely by induction.
of removable sets of FISONs doesn't.
If all are removed, none remains. UF = ℕ ==> Ø = ℕ.That "all" you want to remove is not in the set.
On 03.03.2025 12:58, joes wrote:No, the set is not contained in itself.
Am Mon, 03 Mar 2025 09:48:16 +0100 schrieb WM:
But it has been accepted that with all defined elements the set isHe can't.Ah no, you're shifting the quantifier there.That is allowed if Zermelo can do it.
defined.
Zermelo doesn't even need to shift quantifiers.Zermelo has no other deduction. The set is constructed by its elements.By induction Zermelo produces the set. Quantifier shift, obviouslyA quantifier shift is never a valid deduction (even though the
allowed by Zermelo. There is no other way to construct an infinite
set.
resultant sentence may be true otherwise).
And the set of removable sets of FISONs is just that and doesn't containOf course not. It is the set of all elements.And in the same way Z_0 doesn't contain the set of all elements.The set of sets of FISONs that can be removed together does notIn exactly the same way as Z₀ is constructed by its elements, the set
contain the set of all FISONs (although it does contain the infinite
sets of the odd or even FISONs).
of removable FISONs is constructed by its elements.
On 03.03.2025 18:01, FromTheRafters wrote:
Discrete elements.
Yes, of course:
{ }, {{ }}, {{{ }}}, ...
This is the sequence of numbers.
According to Zermelo its existence guarantees
the set of numbers Z₀ = {{ }, {{ }}, {{{ }}}, ...}.
On 3/3/2025 3:57 AM, WM wrote:
Zermelo's ℕ and Cantor's ℕ are the same
up to isomorphism.
On 3/3/2025 3:57 AM, WM wrote:
Yes, but the topic is this:
In exactly the same way as Z₀
is constructed by its elements,
the set of removable FISONs
is constructed by its elements,
namely by induction.
The real numbers are too many
to be constructed[made] by induction,
even if construct[make]ing.by.induction
were a thing.
On 03.03.2025 01:42, Richard Damon wrote:
On 3/2/25 1:05 PM, WM wrote:
SO, Which Natural Number can it not be that makes N_def different
than N?
They cannot be expressed, neither by stroks nor by digits. Their
existence can only be proved: UF = ℕ ==> Ø = ℕ.
So, what numbers can't be expressed?
Dark numbers.
What is the highest expressable number?
That does not exist because with n also n+1 is expressable.
If there isn't one, why not?
We call that phenomenon potential infinity.
This is the flaw in your logic, you think there are two different
classes of the infinite set of Natural Numbers, one "defined" that
can't have a highest (as what keeps us from defining the next number)
and then you need to invent something higher to hide the fact that
your first set is actually an infinite set of all the Natural Numbers,
It can be proved. All infiite sets as produced by Peano, Zermelo, or v. Neumann are potentially infinite, namely produced by inductive reasoning
with subsequent quantifier exchange. In exactly the same way as Z₀ is constructed by its elements, the set of removable FISONs is constructed
by its elements. The result is Z₀ = ℕ ==> Ø = ℕ.
There are dark numbers collected.Note, the value AT the limit, and the values appraching the limit
of things can be different.
They cannot differ by a fixed quantity like ℵo.
Sure a limit can. What says it can't?
Yes, when we talk about "in the limit of completing the set" it is a
different sort of operation than the normal mathematics limit,
It adds the dark numbers to the inductive set.
as we get infinities that of course never change "value". That is why
we don't write it as a normal mathematics limit. N is not limit n->
inf of F(n),
Right. You try to collect the dark numbers without mentioning it.
Regards, WM
On 03.03.2025 20:26, Jim Burns wrote:No, he doesn't claim infinitely large naturals.
On 3/3/2025 3:57 AM, WM wrote:
Zermelo's ℕ and Cantor's ℕ are the same up to isomorphism.No. Zermelo uses induction which never goes beyond a finite number of
finite numbers.
Cantor claims a number of finite numbers which is larger than all finite numbers.
On 3/3/25 4:10 AM, WM wrote:
They are just an artifact that your logic blew up when N_def became
infinite, blinding you to the truth.
What is the highest expressable number?
That does not exist because with n also n+1 is expressable.
So, why isn't N_Def the same as N?
We call that phenomenon potential infinity.
WHich is just infinity,
On 3/3/25 3:57 AM, WM wrote:
I argued above concerning Cantor's actually infinite ℕ.
It has the undefinable elements ω/2, ω/10, ω/20 inside and ω outside.
No it doesn't. Is 1/2 a Natural Number?
Am Mon, 03 Mar 2025 22:56:46 +0100 schrieb WM:
On 03.03.2025 20:26, Jim Burns wrote:
On 3/3/2025 3:57 AM, WM wrote:
No, he doesn't claim infinitely large naturals.Zermelo's ℕ and Cantor's ℕ are the same up to isomorphism.No. Zermelo uses induction which never goes beyond a finite number of
finite numbers.
Cantor claims a number of finite numbers which is larger than all finite
numbers.
On 3/3/25 4:20 AM, WM wrote:
Every set of FISONs which is assumed to be ℕ must have a first
element. No FISON can serve as such. That is proven by induction.
But the claim isn't that some FISON is N, it is that some infinite set
of FISONs, none individually required, will union to N.
On 3/3/2025 4:56 PM, WM wrote:
The topic is this:
In exactly the same way as Z₀ is constructed by its elements,
the set of removable FISONs
is constructed by its elements,
namely by induction.
⎜ During a typical Gish gallop,
⎜ the galloper confronts an opponent with a rapid series
⎜ of specious arguments,
On 03.03.2025 20:26, Jim Burns wrote:
Zermelo's ℕ and Cantor's ℕ are the same
up to isomorphism.
No.
Zermelo uses
induction which never goes beyond
a finite number of finite numbers.
Cantor claims
a number of finite numbers which is
larger than all finite numbers.
But that is not the topic.
The topic is this:
In exactly the same way as
Z₀ is constructed by its elements,
the set of removable FISONs
is constructed by its elements,
namely by induction.
On 04.03.2025 08:39, joes wrote:Yes. That is not a natural number.
Am Mon, 03 Mar 2025 22:56:46 +0100 schrieb WM:Cantor claims that the number of natural numbers is a quantity larger
On 03.03.2025 20:26, Jim Burns wrote:No, he doesn't claim infinitely large naturals.
On 3/3/2025 3:57 AM, WM wrote:
Zermelo's ℕ and Cantor's ℕ are the same up to isomorphism.No. Zermelo uses induction which never goes beyond a finite number of
finite numbers.
Cantor claims a number of finite numbers which is larger than all
finite numbers.
than any finite number.
"Die Zahl ℵ₀ ist größer als jede endliche Zahl"
On 04.03.2025 02:05, Richard Damon wrote:Wrong. That is exactly what induction does.
On 3/3/25 4:10 AM, WM wrote:
They are just an artifact that your logic blew up when N_def becameBy induction the number of numbers can never become an actually infinite quantity, larger than all finite numbers.
infinite, blinding you to the truth.
Of course. What's your point?Every FISON contains only a finite set of numbers. ℕ is more.So, why isn't N_Def the same as N?What is the highest expressable number?That does not exist because with n also n+1 is expressable.
Am Tue, 04 Mar 2025 10:07:43 +0100 schrieb WM:
On 04.03.2025 02:05, Richard Damon wrote:Wrong. That is exactly what induction does.
On 3/3/25 4:10 AM, WM wrote:By induction the number of numbers can never become an actually infinite
They are just an artifact that your logic blew up when N_def became
infinite, blinding you to the truth.
quantity, larger than all finite numbers.
WM explained on 3/4/2025 :
On 04.03.2025 02:05, Richard Damon wrote:
On 3/3/25 4:20 AM, WM wrote:
Every set of FISONs which is assumed to be ℕ must have a first
element. No FISON can serve as such. That is proven by induction.
But the claim isn't that some FISON is N, it is that some infinite
set of FISONs, none individually required, will union to N.
If there is a set, then it has a first element.
Wrong, neither the emptyset
On 04.03.2025 13:20, joes wrote:Same difference.
Am Tue, 04 Mar 2025 10:07:43 +0100 schrieb WM:No. It gets larger than every fixed natnumber. It never becomes larger
On 04.03.2025 02:05, Richard Damon wrote:Wrong. That is exactly what induction does.
On 3/3/25 4:10 AM, WM wrote:By induction the number of numbers can never become an actually
They are just an artifact that your logic blew up when N_def became
infinite, blinding you to the truth.
infinite quantity, larger than all finite numbers.
than all natnumbers.
On 04.03.2025 11:42, Jim Burns wrote:
On 3/3/2025 4:56 PM, WM wrote:
On 03.03.2025 20:26, Jim Burns wrote:
On 3/3/2025 3:57 AM, WM wrote:
On 02.03.2025 21:28, Jim Burns wrote:
On 3/2/2025 1:52 PM, WM wrote:
On 02.03.2025 18:32, Jim Burns wrote:
On 3/2/2025 4:48 AM, WM wrote:
Induction abbreviates a supertask.
If 1 then 2, if 2 then 3, and so on.
But supertasks will never pass through
the dark numbers.
A claim about an indefinite element of
Z₀ = ⋂𝒫ⁱⁿᵈ(Z)
cannot have a counter.example outside of Z₀
That is the source of a matheologian's certainty.
It has, namely ω, ω/2, etc.
ω is outside Z₀
ω cannot be a counter.example to
a claim about an indefinite element of Z₀
You are right.
I argued above concerning
Cantor's actually infinite ℕ.
Zermelo's ℕ and Cantor's ℕ are the same
up to isomorphism.
No.
Zermelo uses
induction which never goes beyond
a finite number of finite numbers.
Cantor claims
a number of finite numbers which is
larger than all finite numbers.
Yes.
Zermelo's ℕ and Cantor's ℕ are bracketed by
the union of FISONs and
the intersection of inductive subsets.
The union of FISONs and
the intersection of inductive subsets
are the same set.
⎜ During a typical Gish gallop,
⎜ the galloper confronts an opponent with a rapid series
⎜ of specious arguments,
Here is only *one* argument standing for a long while.
On 3/4/2025 5:49 AM, WM wrote:
Here is only *one* argument standing for a long while.
The union ⋃{F} of FISONs and
the intersection ⋂𝒫ⁱⁿᵈ of inductive subsets
are the same set.
Your (WM's) darkᵂᴹ numbers
are bracketed by ⋂𝒫ⁱⁿᵈ and ⋃{F}
⎛ ⋂𝒫ⁱⁿᵈ = ⋃{F}
⎜ Mehdi Hasan, a British journalist, suggests using
⎜ three steps to beat the Gish gallop:
On 04.03.2025 17:43, Jim Burns wrote:
On 3/4/2025 5:49 AM, WM wrote:
Here is only *one* argument standing for a long while.
The union ⋃{F} of FISONs and
the intersection ⋂𝒫ⁱⁿᵈ of inductive subsets
are the same set.
Correct.
You need not the intersection however because
Z₀ can also be defined by
{ } ∈ Z₀, and
if {{{...{{{ }}}...}}} with n curly brackets ∈ Z₀
then {{{...{{{ }}}...}}} with n+1 curly brackets ∈ Z₀.
Your (WM's) darkᵂᴹ numbers
are bracketed by ⋂𝒫ⁱⁿᵈ and ⋃{F}
No, they are following.
Every induction surpasses every finite number
but never all finite numbers.
It is always a finite number.
Cantor's ℕ however is larger:
"Unter einem A.-U. ist dagegen ein Quantum zu verstehen,
das einerseits nicht veränderlich,
sondern vielmehr in allen seinen Teilen fest und bestimmt,
eine richtige Konstante ist,
zugleich aber andrerseits jede endliche Größe
derselben Art an Größe übertrifft.
Als Beispiel führe ich die Gesamtheit,
den Inbegriff aller endlichen ganzen positiven Zahlen an;
diese Menge ist ein Ding für sich und bildet,
ganz abgesehen von der natürlichen Folge
der dazu gehörigen Zahlen,
ein in allen Teilen festes, bestimmtes Quantum,
... das offenbar größer zu nennen ist als jede endliche Anzahl.
Therefore ℕ is more than UF.
By induction like Zermelo we find:
∀n ∈ U(F): |ℕ \ {1, 2, 3, ..., n}| = ℵo.
⎛ ⋂𝒫ⁱⁿᵈ = ⋃{F}
True.
⎜ Mehdi Hasan, a British journalist, suggests using
⎜ three steps to beat the Gish gallop:
Am Tue, 04 Mar 2025 20:56:17 +0100 schrieb WM:
Every induction surpasses every finite numberNo. That is why you use induction instead of directly proving the finite number of sentences.
but never all finite numbers. It is always a finite number.
On 04.03.2025 17:43, Jim Burns wrote:No. That is why you use induction instead of directly proving the finite
On 3/4/2025 5:49 AM, WM wrote:
Correct. You need not the intersection however because Z₀ can also be defined by { } ∈ Z₀, and if {{{...{{{ }}}...}}} with n curly brackets ∈ Z₀ then {{{...{{{ }}}...}}} with n+1 curly brackets ∈ Z₀.Here is only *one* argument standing for a long while.The union ⋃{F} of FISONs and the intersection ⋂𝒫ⁱⁿᵈ of inductive
subsets are the same set.
No, they are following. Every induction surpasses every finite number
Your (WM's) darkᵂᴹ numbers are bracketed by ⋂𝒫ⁱⁿᵈ and ⋃{F}
but never all finite numbers. It is always a finite number.
Cantor's N however is larger: [...]What we don't find: |N \ U{F(n): n e N}| = Aleph_0,
Therefore ℕ is more than UF.
By induction like Zermelo we find:
∀n ∈ U(F): |ℕ \ {1, 2, 3, ..., n}| = ℵo.
That is called sealioning, I believe.⎛ ⋂𝒫ⁱⁿᵈ = ⋃{F}True.
⎜ Mehdi Hasan, a British journalist, suggests using ⎜ three steps toI need only one step: I delete your gibberish without addressing it and
beat the Gish gallop:
put the topic.
On 3/4/2025 2:56 PM, WM wrote:
On 04.03.2025 17:43, Jim Burns wrote:
On 3/4/2025 5:49 AM, WM wrote:
Here is only *one* argument standing for a long while.
The union ⋃{F} of FISONs and
the intersection ⋂𝒫ⁱⁿᵈ of inductive subsets
are the same set.
Correct.
Proper supersets
of ⋂𝒫ⁱⁿᵈ contain extra elements.
You (WM) have previously said that
your (WM's) ℕ doesn't have extra elements.
You need not the intersection however because
Z₀ can also be defined by
{ } ∈ Z₀, and
if {{{...{{{ }}}...}}} with n curly brackets ∈ Z₀
then {{{...{{{ }}}...}}} with n+1 curly brackets ∈ Z₀.
No.
For Z, not Z₀, ∀a: a ∈ Z ⇒ {a} ∈ Z
Z₀ is the simplest inductive subset of Z.
But you say "n+1 curly brackets".
Plus, "finitely.many" eliminates your darkᵂᴹ numbers.
ω/2 ω/10 and ω/20 follow ω?
Every induction surpasses every finite number but never all finite
numbers.
It is always a finite number.
Cantor's ℕ however is larger:
"Unter einem A.-U. ist dagegen ein Quantum zu verstehen,
das einerseits nicht veränderlich,
sondern vielmehr in allen seinen Teilen fest und bestimmt,
eine richtige Konstante ist,
zugleich aber andrerseits jede endliche Größe
derselben Art an Größe übertrifft. Als Beispiel führe ich die Gesamtheit,
den Inbegriff aller endlichen ganzen positiven Zahlen an;
diese Menge ist ein Ding für sich und bildet,
ganz abgesehen von der natürlichen Folge
der dazu gehörigen Zahlen,
ein in allen Teilen festes, bestimmtes Quantum,
... das offenbar größer zu nennen ist als jede endliche Anzahl.
⎛ "By an A.-U. (that abbreviates actual infinity), on the other hand, we understand
⎜ a quantum that is on the one hand not variable,
⎜ but rather fixed and determined in all its parts,
⎜ a true constant,
⎜ but at the same time
⎜ on the other hand exceeds in size
⎜ any finite quantity of the same kind.
⎜ As an example, I cite
⎜ the totality, the sum of all finite whole positive numbers;
⎜ this set is a thing in itself and, quite apart from
⎜ the natural sequence of the numbers belonging to it,
⎜ forms a quantum that is fixed and determined in all its parts,
⎝ ... which can obviously be called larger than any finite number.
Therefore ℕ is more than UF.
By induction like Zermelo we find:
∀n ∈ U(F): |ℕ \ {1, 2, 3, ..., n}| = ℵo.
∀n ∈ ⋃{F}: |⋃{F}\{1,2,3,...,n}| = ℵ₀
⎛ Assume ⋃{F} is finite.
On 04.03.2025 02:05, Richard Damon wrote:
On 3/3/25 4:10 AM, WM wrote:
They are just an artifact that your logic blew up when N_def became
infinite, blinding you to the truth.
By induction the number of numbers can never become an actually infinite quantity, larger than all finite numbers.
What is the highest expressable number?
That does not exist because with n also n+1 is expressable.
So, why isn't N_Def the same as N?
Every FISON contains only a finite set of numbers. ℕ is more.
We call that phenomenon potential infinity.
WHich is just infinity,
Cantor denies your claim.
"Nevertheless the transfinite cannot be considered a subsection of what
is usually called 'potentially infinite'. Because the latter is not
(like every individual transfinite and in general everything due to an
'idea divina') determined in itself, fixed, and unchangeable, but a
finite in the process of change, having in each of its current states a finite size; like, for instance, the temporal duration since the
beginning of the world, which, when measured in some time-unit, for
instance a year, is finite in every moment, but always growing beyond
all finite limits, without ever becoming really infinitely large." [G. Cantor, letter to I. Jeiler (13 Oct 1895)]
Here he is right.
Regards, WM
On 05.03.2025 10:24, joes wrote:No.
Am Tue, 04 Mar 2025 20:56:17 +0100 schrieb WM:
For the last time: Understand or shut up!Every induction surpasses every finite number but never all finiteNo. That is why you use induction instead of directly proving the
numbers. It is always a finite number.
finite number of sentences.
If UF = ℕ, then there is at least one subset of FISONs producing ℕ. Such a set has a first element which is not conpletely useless.
On 04.03.2025 23:05, Jim Burns wrote:No. N *is* UF.
On 3/4/2025 2:56 PM, WM wrote:We talk about subsets!
On 04.03.2025 17:43, Jim Burns wrote:Proper supersets
On 3/4/2025 5:49 AM, WM wrote:
Correct.Here is only *one* argument standing for a long while.The union ⋃{F} of FISONs and the intersection ⋂𝒫ⁱⁿᵈ of inductive
subsets are the same set.
of ⋂𝒫ⁱⁿᵈ contain extra elements.ℕ is a proper superset of ⋃F. It contains all the dark natural numbers. ⋃F contains only defined natural numbers.
You (WM) have previously said that your (WM's) ℕ doesn't have extra
elements.
Please specify?Ha. Caught red handed. Yes there is no mathematical possibility toYou need not the intersection however because Z₀ can also be definedNo.
by { } ∈ Z₀, and if {{{...{{{ }}}...}}} with n curly brackets ∈ Z₀ >>> then {{{...{{{ }}}...}}} with n+1 curly brackets ∈ Z₀.
contradict my claim.
Gee, I wonder.For Z, not Z₀, ∀a: a ∈ Z ⇒ {a} ∈ Z Z₀ is the simplest inductive subsetWhat does Z₀ contain that is not in my inductive description?
of Z.
And N.But you say "n+1 curly brackets".You can also say: 0 and if n ∈ ℕ then n+1 ∈ ℕ. That is only another notation of number n.
Plus, "finitely.many" eliminates your darkᵂᴹ numbers.a then {a} means precisely as many as n then n+1.
ω/2 ω/10 and ω/20 follow ω?No they follow Z₀ and UF.
Neither does N.Every induction surpasses every finite number but never all finite
numbers. It is always a finite number.
Cantor's ℕ however is larger:
⎛ "By an A.-U. (that abbreviates actual infinity), on the other hand,
we understand ⎜ a quantum that is on the one hand not variable,
⎜ but rather fixed and determined in all its parts,
⎜ a true constant,
⎜ but at the same time ⎜ on the other hand exceeds in size ⎜ any finite
quantity of the same kind.
⎜ As an example, I cite ⎜ the totality, the sum of all finite whole
positive numbers;
⎜ this set is a thing in itself and, quite apart from ⎜ the natural
sequence of the numbers belonging to it,
⎜ forms a quantum that is fixed and determined in all its parts,
⎝ ... which can obviously be called larger than any finite number.
No. ⋃{F} is wrong because F is the set of FISONs. ⋃{F} = F.Therefore ℕ is more than UF.∀n ∈ ⋃{F}: |⋃{F}\{1,2,3,...,n}| = ℵ₀
By induction like Zermelo we find:
∀n ∈ U(F): |ℕ \ {1, 2, 3, ..., n}| = ℵo.
UF does not contain an n that is larger than all n by ℵ₀.
Is it finite or not?⎛ Assume ⋃{F} is finite.UF is (potentially in-) finite. It is constructed by induction which is variable and therefore has no largest number but never reaches a
completion.
On 04.03.2025 23:05, Jim Burns wrote:
ω/2 ω/10 and ω/20 follow ω?
No they follow Z₀ and UF.
On 3/4/25 4:07 AM, WM wrote:
Wrong, Cantor shows that the number of Natural Numbers generated by the iterative method of, we have 0, and for every number we have its
successor, is not one of those finite numbers, but is another number Aleph0
We call that phenomenon potential infinity.
WHich is just infinity,
Cantor denies your claim.
"Nevertheless the transfinite cannot be considered a subsection of
what is usually called 'potentially infinite'. Because the latter is
not (like every individual transfinite and in general everything due
to an 'idea divina') determined in itself, fixed, and unchangeable,
but a finite in the process of change, having in each of its current
states a finite size; like, for instance, the temporal duration since
the beginning of the world, which, when measured in some time-unit,
for instance a year, is finite in every moment, but always growing
beyond all finite limits, without ever becoming really infinitely
large." [G. Cantor, letter to I. Jeiler (13 Oct 1895)]
Here he is right.
Which doesn't mean what you think it means.
He is pointing out that these "transfinite" concepts aren't part of the infinite set built by iteration (the "potential infinity") but is beyond
it.
We SEE the "potentially infinite" via a process, where each step is
finite, but the final result of it *IS* an infinite thing.
None of the members of N are themselves infinite, but the set itself is.
On 3/5/2025 4:22 AM, WM wrote:No.
On 04.03.2025 23:05, Jim Burns wrote:
ω/2 ω/10 and ω/20 follow ω?
No they follow Z₀ and UF.
ω = ⋃{F}
ω is the first non.finite ordinal.Yes.
ω is the set of all and only finite ordinals.That is another use of ω.
F = {F(1), F(2), ..., F(n), ...}
UF = {1, 2, 3, ...}
Am 05.03.2025 um 20:26 schrieb Jim Burns:
On 3/5/2025 4:22 AM, WM wrote:
On 04.03.2025 23:05, Jim Burns wrote:
ω/2 ω/10 and ω/20 follow ω?
No they follow Z₀ and UF.
ω = ⋃{F}
No.
F = {F(1), F(2), ..., F(n), ...}
UF = {1, 2, 3, ...}
∀n ∈ ℕ: ω/n > n.
ω is the first non.finite ordinal.
Yes.
ω is the set of all and only finite ordinals.
That is another use of ω.
On 3/5/2025 4:28 PM, WM wrote:
Am 05.03.2025 um 20:26 schrieb Jim Burns:
On 3/5/2025 4:22 AM, WM wrote:
On 04.03.2025 23:05, Jim Burns wrote:
ω/2 ω/10 and ω/20 follow ω?
No they follow Z₀ and UF.
ω = ⋃{F}
No.
F = {F(1), F(2), ..., F(n), ...}
UF = {1, 2, 3, ...}
∀n ∈ ℕ: ω/n > n.
I think that you (WM) are using a different dictionary.
definableᵂᴹ: finiteⁿᵒᵗᐧᵂᴹ ( #[0,x)ᵒʳᵈ < #[x+1)ᵒʳᵈ )
darkᵂᴹ: finiteⁿᵒᵗᐧᵂᴹ ( #[0,x)ᵒʳᵈ < #[x+1)ᵒʳᵈ )
matheologicalᵂᴹ: infiniteⁿᵒᵗᐧᵂᴹ ( #[0,x)ᵒʳᵈ = #[x+1)ᵒʳᵈ )
ω is the first non.finite ordinal.
Yes.
ω is the set of all and only finite ordinals.
That is another use of ω.
It is von Neumann's use of ω
Am 05.03.2025 um 20:26 schrieb Jim Burns:I fail to see a difference.
On 3/5/2025 4:22 AM, WM wrote:
On 04.03.2025 23:05, Jim Burns wrote:
No.ω = ⋃{F}ω/2 ω/10 and ω/20 follow ω?No they follow Z₀ and UF.
F = {F(1), F(2), ..., F(n), ...}
UF = {1, 2, 3, ...}
I think that you (WM) are using a different dictionary.
{ } ∈ Z₀, and
if {{{...{{{ }}}...}}} with n curly brackets ∈ Z₀
then {{{...{{{ }}}...}}} with n+1 curly brackets ∈ Z₀.
No.
Am Wed, 05 Mar 2025 22:05:04 +0100 schrieb WM:
Therefore iterartion fails to produce actual infinity.As an element, but not as the number of elements (=the size of the set).
Wrong. Of course induction "is infinite". UF is not empty.
Am 05.03.2025 um 13:25 schrieb Richard Damon:Nope.
On 3/4/25 4:07 AM, WM wrote:
Wrong, Cantor shows that the number of Natural Numbers generated by theBelow you contradict yourself.
iterative method of, we have 0, and for every number we have its
successor, is not one of those finite numbers, but is another number
Aleph0
He doesn't say what you say.He means what he says.We call that phenomenon potential infinity.WHich is just infinity,
Cantor denies your claim.
"Nevertheless the transfinite cannot be considered a subsection of
what is usually called 'potentially infinite'. Because the latter is
not (like every individual transfinite and in general everything due
to an 'idea divina') determined in itself, fixed, and unchangeable,
but a finite in the process of change, having in each of its current
states a finite size; like, for instance, the temporal duration since
the beginning of the world, which, when measured in some time-unit,
for instance a year, is finite in every moment, but always growing
beyond all finite limits, without ever becoming really infinitely
large." [G. Cantor, letter to I. Jeiler (13 Oct 1895)]
Here he is right.
Which doesn't mean what you think it means.
As an element, but not as the number of elements (=the size of the set).He is pointing out that these "transfinite" concepts aren't part of theTherefore iterartion fails to produce actual infinity.
infinite set built by iteration (the "potential infinity") but is
beyond it.
Speak for yourself. You are evidently unable to understand limits andWe SEE the "potentially infinite" via a process, where each step isThere is no final result. You are unable to understand infinity.
finite, but the final result of it *IS* an infinite thing.
Because the latter is a finite in the process of change, having in eachNo, infinity isn't finite and doesn't change.
of its current without ever becoming really infinitely large.
Wrong. Of course induction "is infinite". UF is not empty.None of the members of N are themselves infinite, but the set itselfNot by recursion or induction! Therefore UF is a proper subset of ℕ. UF
is.
= ℕ ==> Ø = ℕ
{ } ∈ Z₀, and
if {{{...{{{ }}}...}}} with n curly brackets ∈ Z₀
then {{{...{{{ }}}...}}} with n+1 curly brackets ∈ Z₀.
Am 06.03.2025 um 02:52 schrieb Jim Burns:
I think that you (WM) are using a different dictionary.
I bthink that you (JB) should correct your statement:
<</JB<WM>>>
You need not the intersection however because
Z₀ can also be defined by
{ } ∈ Z₀, and
if {{{...{{{ }}}...}}} with n curly brackets ∈ Z₀
then {{{...{{{ }}}...}}} with n+1 curly brackets ∈ Z₀.
No.
{ } ∈ Z₀, and
if {{{...{{{ }}}...}}} with n curly brackets ∈ Z₀
then {{{...{{{ }}}...}}} with n+1 curly brackets ∈ Z₀.
No.
Regards, WM
Am 06.03.2025 um 10:06 schrieb joes:
Am Wed, 05 Mar 2025 22:05:04 +0100 schrieb WM:
Therefore iterartion fails to produce
actual infinity.
As an element, but not as
the number of elements (=the size of the set).
The number of elements is an element
for every element produced by induction.
Wrong.
Of course induction "is infinite".
UF is not empty.
Induction is potentially infinite.
Never an actually infinite set is porduced.
Am Wed, 05 Mar 2025 22:28:18 +0100 schrieb WM:
F = {F(1), F(2), ..., F(n), ...}I fail to see a difference.
UF = {1, 2, 3, ...}
On 3/6/2025 3:54 AM, WM wrote:
Am 06.03.2025 um 02:52 schrieb Jim Burns:
I think that you (WM) are using a different dictionary.
I think that you (JB) should correct your statement:
<<JB<WM>>>
<</JB<WM>>>
You need not the intersection however because
Z₀ can also be defined by
{ } ∈ Z₀, and
if {{{...{{{ }}}...}}} with n curly brackets ∈ Z₀
then {{{...{{{ }}}...}}} with n+1 curly brackets ∈ Z₀.
No.
No BECAUSE
that doesn't DEFINE Z₀
What does your inductive description permit in Z₀
which should not be permitted in Z₀ ?
Anything.
Your description permits Bob, as long as
{Bob}, {{Bob}}, {{{Bob}}}, ... are also in Z₀
Am 06.03.2025 um 09:54 schrieb WM: [...]
{ } ∈ Z₀, and
if {{{...{{{ }}}...}}} with n curly brackets ∈ Z₀
then {{{...{{{ }}}...}}} with n+1 curly brackets ∈ Z₀.
Right. So in math we would rather write
| {} ∈ Z₀, and for all x: if x ∈ Z₀ then {x} ∈ Z₀
(a more or less equivalent statement).
On 3/6/2025 4:15 AM, WM wrote:
Am 06.03.2025 um 10:06 schrieb joes:
Am Wed, 05 Mar 2025 22:05:04 +0100 schrieb WM:
Therefore iteration fails to produce
actual infinity.
As an element, but not as
the number of elements (=the size of the set).
We do NOT construct[make] sets.
We construct[know] sets.
The number of elements is an element
for every element produced by induction.
For each element in a set which
is its.own.only.inductive.subset,
that element's set of priors has
fuller.by.one sets which are larger.
Induction is potentially infinite.
Proof.by.induction completes
On 06.03.2025 12:08, Jim Burns wrote:
On 3/6/2025 3:54 AM, WM wrote:
Am 06.03.2025 um 02:52 schrieb Jim Burns:
<<JB<WM>>>
<</JB<WM>>>
You need not the intersection however because
Z₀ can also be defined by
{ } ∈ Z₀, and
if {{{...{{{ }}}...}}} with n curly brackets ∈ Z₀
then {{{...{{{ }}}...}}} with n+1 curly brackets ∈ Z₀.
No.
No BECAUSE
that doesn't DEFINE Z₀
It defines Z₀ precisely.
What does your inductive description permit in Z₀
which should not be permitted in Z₀ ?
Anything.
Your description permits Bob, as long as
{Bob}, {{Bob}}, {{{Bob}}}, ... are also in Z₀
Induction excludes Bob.
Induction excludes Bob.
It defines Z₀ precisely.
He is not in the empty set and
not in the set containing the empty set and
not in any set with more curly brackets.
Am 05.03.2025 um 22:28 schrieb WM:
F = {F(1), F(2), ..., F(n), ...}
with F(n) := {m e IN : m <= n} (n e IN).
UF = {1, 2, 3, ...}
Exactly! Schön, dass Du das nun endlich erkennst, Mückenheim.
On 06.03.2025 14:47, Jim Burns wrote:Infinite sets are.
On 3/6/2025 4:15 AM, WM wrote:Without their construction/proof we don't know whether infinite sets
Am 06.03.2025 um 10:06 schrieb joes:We do NOT construct[make] sets.
Am Wed, 05 Mar 2025 22:05:04 +0100 schrieb WM:
Therefore iteration fails to produce actual infinity.As an element, but not as the number of elements (=the size of the
set).
We construct[know] sets.
exist at all. Um aber die Existenz "unendlicher" Mengen zu sichern,
bedürfen wir noch des folgenden ... Axioms. [Zermelo: Untersuchungen
über die Grundlagen der Mengenlehre I, S. 266] The elements are defined
by induction in order to guarantee the existence of infinite sets.
Never a set of elements is constructed which is larger than every finite number.The number of elements is an element for every element produced byFor each element in a set which is its.own.only.inductive.subset, that
induction.
element's set of priors has fuller.by.one sets which are larger.
If it didn't, it wouldn't work.never!Induction is potentially infinite.Proof.by.induction completes
On 06.03.2025 14:47, Jim Burns wrote:
On 3/6/2025 4:15 AM, WM wrote:
Am 06.03.2025 um 10:06 schrieb joes:
Am Wed, 05 Mar 2025 22:05:04 +0100 schrieb WM:
Therefore iteration fails to produce
actual infinity.
As an element, but not as
the number of elements (=the size of the set).
We do NOT construct[make] sets.
We construct[know] sets.
Without their construction/proof
we don't know whether infinite sets exist at all.
Um aber die Existenz "unendlicher" Mengen zu sichern,
bedürfen wir noch des folgenden ... Axioms.
[Zermelo: Untersuchungen über die Grundlagen
der Mengenlehre I, S. 266]
The elements are defined by induction
in order to guarantee the existence of infinite sets.
On 3/6/2025 12:15 PM, WM wrote:
Constructions[proofs] construct[prove.to.exist]
infinite sets,
but those constructions[proofs] aren't
endless iterations.
This proof is finite:
⎛ Infinity: inductive Z exists.
⎜ PowerSet: 𝒫(Z) exists.
⎜ Separation: 𝒫ⁱⁿᵈ(Z) exists.
⎝ Separation: ⋂𝒫ⁱⁿᵈ(Z) exists.
We know Z₀ exists
Um aber die Existenz "unendlicher" Mengen zu sichern, bedürfen wir
noch des folgenden ... Axioms.
Those axioms ensure the existence of Z₀
but not the way that you (WM) think they ensure it.
[Zermelo: Untersuchungen über die Grundlagen
der Mengenlehre I, S. 266]
The elements are defined by induction
in order to guarantee the existence of infinite sets.
Am Thu, 06 Mar 2025 18:15:01 +0100 schrieb WM:
Never a set of elements is constructed which is larger than every finiteInfinite sets are.
number.
If it didn't, it wouldn't work.never!Induction is potentially infinite.Proof.by.induction completes
On 3/6/2025 12:05 PM, WM wrote:
On 06.03.2025 12:08, Jim Burns wrote:
On 3/6/2025 3:54 AM, WM wrote:
Am 06.03.2025 um 02:52 schrieb Jim Burns:
<<JB<WM>>>
<</JB<WM>>>
You need not the intersection however because
Z₀ can also be defined by
{ } ∈ Z₀, and
if {{{...{{{ }}}...}}} with n curly brackets ∈ Z₀
then {{{...{{{ }}}...}}} with n+1 curly brackets ∈ Z₀.
No.
No BECAUSE
that doesn't DEFINE Z₀
It defines Z₀ precisely.
It does not exclude Bob.
In order to exclude Bob,
some clause must state or imply Bob isn't in Z₀
Induction says what's IN Z₀
More description is needed.
What does your inductive description permit in Z₀
which should not be permitted in Z₀ ?
Anything.
Your description permits Bob, as long as
{Bob}, {{Bob}}, {{{Bob}}}, ... are also in Z₀
Induction excludes Bob.
I generalize 'end segment':
An unreliable argument is unacceptable.
Induction excludes Bob.
Induction alone does not exclude Bob.
Induction alone does not exclude Bob.
On 06.03.2025 21:23, joes wrote:Inductive sets are of course larger than any finite number.
Am Thu, 06 Mar 2025 18:15:01 +0100 schrieb WM:
Not inductive sets.Never a set of elements is constructed which is larger than everyInfinite sets are.
finite number.
--It constructs and works for all potentially infinite sets.If it didn't, it wouldn't work.never!Induction is potentially infinite.Proof.by.induction completes
On 06.03.2025 20:03, Jim Burns wrote:
On 3/6/2025 12:05 PM, WM wrote:
On 06.03.2025 12:08, Jim Burns wrote:
On 3/6/2025 3:54 AM, WM wrote:
Am 06.03.2025 um 02:52 schrieb Jim Burns:
<<JB<WM>>>
<</JB<WM>>>
You need not the intersection however
because
Z₀ can also be defined by
{ } ∈ Z₀, and
if {{{...{{{ }}}...}}} with n curly brackets ∈ Z₀
then {{{...{{{ }}}...}}} with n+1 curly brackets ∈ Z₀.
No.
No BECAUSE
that doesn't DEFINE Z₀
It defines Z₀ precisely.
It does not exclude Bob.
In order to exclude Bob,
some clause must state or imply Bob isn't in Z₀
No.
The first element is the empty set.
It does not contain Bob.
All following elements are
the empty set equipped with further curlybrackets.
Induction says what's IN Z₀
More description is needed.
Therefore there is no Bob.
What does your inductive description permit in Z₀
which should not be permitted in Z₀ ?
Anything.
Your description permits Bob, as long as
{Bob}, {{Bob}}, {{{Bob}}}, ... are also in Z₀
Induction excludes Bob.
On 3/7/2025 4:23 AM, WM wrote:
You assume that
only
{} and its curly.bracket.followers are in Z₀
You didn't say that, above.
Instead, you said pretty much the opposite, above,
saying that intersection (making 'only') isn't needed.
The description ∀a: a ↦ {a} of a set
says
that each thing in the set
has its Zermelo.sequence in the set.
{}, {{}}, {{{}}}, ...
is
the Zermelo.sequence of {}
Bob, {Bob}, {{Bob}}, ...
is the Zermelo.sequence of Bob.
Kevin, {Kevin}, {{Kevin}}, ...
is the Zermelo.sequence of Kevin.
⎛ Everything in a Zermelo.sequence
⎜ has its Zermelo.sequence
⎜ contained in the over.sequence.
Zermelo's Axiom of Infinite asserts
the existence of ONE OF those or similar sets
with AT LEAST only {}, {{}}, {{{}}}, ...
That indefinite set is referred to as Z
Induction says what's IN Z₀
More description is needed.
Therefore there is no Bob.
Induction alone,
On 07.03.2025 16:08, Jim Burns wrote:
On 3/7/2025 4:23 AM, WM wrote:
You assume that
only
{} and its curly.bracket.followers are in Z₀
That is what
the intersection of all Zermelo-inductive sets
produces.
<<JB<WM>>>
<</JB<WM>>>
You need not the intersection however
because
Z₀ can also be defined by
{ } ∈ Z₀, and
if {{{...{{{ }}}...}}} with n curly brackets ∈ Z₀
then {{{...{{{ }}}...}}} with n+1 curly brackets ∈ Z₀.
No.
You didn't say that, above.
Instead, you said pretty much the opposite, above,
saying that intersection (making 'only') isn't needed.
I said that
the intersection isn't needed when
instead of Zermelo's approach
{} and its curly bracket followers are used.
The description ∀a: a ↦ {a} of a set
says
that each thing in the set
has its Zermelo.sequence in the set.
{}, {{}}, {{{}}}, ...
is
the Zermelo.sequence of {}
It is what Zermelo calls
the sequence of numbers.
Bob, {Bob}, {{Bob}}, ...
is the Zermelo.sequence of Bob.
Kevin, {Kevin}, {{Kevin}}, ...
is the Zermelo.sequence of Kevin.
⎛ Everything in a Zermelo.sequence
⎜ has its Zermelo.sequence
⎜ contained in the over.sequence.
That is true but irrelevant when
only {} and its curly bracket followers are used.
Zermelo's Axiom of Infinite asserts
the existence of ONE OF those or similar sets
with AT LEAST only {}, {{}}, {{{}}}, ...
That indefinite set is referred to as Z
And {} and its curly bracket followers
is referred to as Z₀.
Induction says what's IN Z₀
More description is needed.
Therefore there is no Bob.
Induction alone,
yiels {} and its curly bracket followers.
And now try to find a difference to
F(1) and F(n) --> F(n+1).
On 3/7/2025 11:07 AM, WM wrote:
On 07.03.2025 16:08, Jim Burns wrote:
On 3/7/2025 4:23 AM, WM wrote:
You assume that
only
{} and its curly.bracket.followers are in Z₀
That is what
the intersection of all Zermelo-inductive sets
produces.
<<JB<WM>>>
<</JB<WM>>>
You need not the intersection however
because
Z₀ can also be defined by
{ } ∈ Z₀, and
if {{{...{{{ }}}...}}} with n curly brackets ∈ Z₀
then {{{...{{{ }}}...}}} with n+1 curly brackets ∈ Z₀.
No.
You didn't say that, above.
Instead, you said pretty much the opposite, above,
saying that intersection (making 'only') isn't needed.
I said that
the intersection isn't needed when
instead of Zermelo's approach
{} and its curly bracket followers are used.
I said (way back when) that
maybe you (WM) are sneaking ℕ in through the back door.
When you have described what you mean by
an indefinite curly.bracket.follower of {}
(which you haven't done yet),
you will find that you have described
an indefinite natural number.
How do you know which set?
Zermelo's Axiom of Infinite asserts
the existence of ONE OF those or similar sets
with AT LEAST only {}, {{}}, {{{}}}, ...
That indefinite set is referred to as Z
And {} and its curly bracket followers
is referred to as Z₀.
Can you say what those are without intersection?
I suspect it can be said,
but I don't know how, right now,
and I'm sure that you (WM) haven't said how.
Induction says what's IN Z₀
More description is needed.
Therefore there is no Bob.
Induction alone,
yields {} and its curly bracket followers.
You have just now added another clause to
being inductive.
On 07.03.2025 21:11, Jim Burns wrote:
On 3/7/2025 11:07 AM, WM wrote:
On 07.03.2025 16:08, Jim Burns wrote:
On 3/7/2025 4:23 AM, WM wrote:
You assume that
only
{} and its curly.bracket.followers are in Z₀
That is what
the intersection of all Zermelo-inductive sets
produces.
You need not the intersection however<</JB<WM>>>
because
Z₀ can also be defined by
{ } ∈ Z₀, and
if {{{...{{{ }}}...}}} with n curly brackets ∈ Z₀
then {{{...{{{ }}}...}}} with n+1 curly brackets ∈ Z₀.
No.
You didn't say that, above.
Instead, you said pretty much the opposite, above,
saying that intersection (making 'only') isn't needed.
I said that
the intersection isn't needed when
instead of Zermelo's approach
{} and its curly bracket followers are used.
I said (way back when) that
maybe you (WM) are sneaking ℕ in through the back door.
When you have described what you mean by
an indefinite curly.bracket.follower of {}
(which you haven't done yet),
you will find that you have described
an indefinite natural number.
I do not use indefinite followers
but followers with n curly brackets.
It can be read above.
How do you know which set?
From that unique description:
{ } ∈ Z₀, and
if {{{...{{{ }}}...}}} with n curly brackets ∈ Z₀
then {{{...{{{ }}}...}}} with n+1 curly brackets ∈ Z₀.
I suspect it can be said,
but I don't know how, right now,
and I'm sure that you (WM) haven't said how.
Is it so bewildering to
replace "1" and "if n ∈ ℕ then n+1 ∈ ℕ"
by Zermelo's notation?
On 3/8/2025 3:45 AM, WM wrote:
You need not the intersection however
because
Z₀ can also be defined by
{ } ∈ Z₀, and
if {{{...{{{ }}}...}}} with n curly brackets ∈ Z₀
then {{{...{{{ }}}...}}} with n+1 curly brackets ∈ Z₀.
Zermelo's Axiom of Infinity describes Z
Z ∋ {} ∧ ∀a: Z ∋ a ⇒ Z ∋ {a}
Multiple sets satisfy that unique description.
That's an indefinite description.
If only
{{{...{{{ }}}...}}} with finitely.many curly brackets
each with immediate.predecessor ⋃{{{...{{{ }}}...}}}
and with {} being one of its priors
are in Z₀
then, yes, that is a definite description.
And, yes, that seems to be what you meant.
If you ever want to make your descriptions clearer,
you won't be getting complaints from me.
I suspect it can be said,
but I don't know how, right now,
and I'm sure that you (WM) haven't said how.
Is it so bewildering to
replace "1" and "if n ∈ ℕ then n+1 ∈ ℕ"
by Zermelo's notation?
If n ranges over transfinite ordinals,
then your ℕ is unreliable in proof.by.induction.
It's not what Zermelo, von Neumann, or Cantor
mean by ℕ
If n ranges over only finite ordinals,
then
you didn't explicitly say that,
which explicitness is the reason for writing,
and,
if you had explicitly said that,
you should have said somewhere
what a finite ordinal is,
On 08.03.2025 12:58, Jim Burns wrote:
On 3/8/2025 3:45 AM, WM wrote:
You need not the intersection however
because
Z₀ can also be defined by
{ } ∈ Z₀, and
if {{{...{{{ }}}...}}} with n curly brackets ∈ Z₀
then {{{...{{{ }}}...}}} with n+1 curly brackets ∈ Z₀.
Zermelo's Axiom of Infinity describes Z
Z ∋ {} ∧ ∀a: Z ∋ a ⇒ Z ∋ {a}
Multiple sets satisfy that unique description.
That's an indefinite description.
My description is definite.
If only
{{{...{{{ }}}...}}} with finitely.many curly brackets
each with immediate.predecessor ⋃{{{...{{{ }}}...}}}
and with {} being one of its priors
are in Z₀
then, yes, that is a definite description.
And, yes, that seems to be what you meant.
If you ever want to make your descriptions clearer,
you won't be getting complaints from me.
That is exactly what I meant.
And that's also the induction of my argument
UF = ℕ ==> Ø = ℕ.
If n ranges over only finite ordinals,
then
you didn't explicitly say that,
Induction ranges over finite ordinals.
which explicitness is the reason for writing,
and,
if you had explicitly said that,
you should have said somewhere
what a finite ordinal is,
That is not under discussion here.
n is usually denoting
a natural number.
FISONs are finite by definition.
On 3/8/2025 9:09 AM, WM wrote:
On 08.03.2025 12:58, Jim Burns wrote:
On 3/8/2025 3:45 AM, WM wrote:
You need not the intersection however
because
Z₀ can also be defined by
{ } ∈ Z₀, and
if {{{...{{{ }}}...}}} with n curly brackets ∈ Z₀
then {{{...{{{ }}}...}}} with n+1 curly brackets ∈ Z₀.
Zermelo's Axiom of Infinity describes Z
Z ∋ {} ∧ ∀a: Z ∋ a ⇒ Z ∋ {a}
Multiple sets satisfy that unique description.
That's an indefinite description.
My description is definite.
Your description is only definite because
ℕ ∋ n is definite.
What ℕ is,
the interesting part of your description of Z₀,
must be found elsewhere.
If only
{{{...{{{ }}}...}}} with finitely.many curly brackets
each with immediate.predecessor ⋃{{{...{{{ }}}...}}}
and with {} being one of its priors
are in Z₀
then, yes, that is a definite description.
And, yes, that seems to be what you meant.
If you ever want to make your descriptions clearer,
you won't be getting complaints from me.
That is exactly what I meant.
And that's also the induction of my argument
UF = ℕ ==> Ø = ℕ.
A proof by induction
(a proof by this.inductive.subset.is.the.whole.set)
is only reliable for
a set which is its.own.only.inductive subset.
ℕ is its.own.only.inductive subset.
Z₀ is its.own.only.inductive subset.
⋃{F} union of the set {F} of all FISONs,
is its.own.only.inductive subset.
For any set W which
is its.own.only.inductive subset (ℕ,Z₀,⋃{F},...),
∀j∈W:∃k∈W: j < k
{F} is its.own.only.inductive.subset.
No FISON in {F} is F(ω-1)
you should have said somewhere
what a finite ordinal is,
That is not under discussion here.
That has been under discussion for decades.
I think that these decades of discussion
have been, in large part, you assigning
different meanings to 'finite', etc.
and matheologians (among whom I place myself)
trying to discern what your meanings are.
Here's my best guess:
definableᵂᴹ == finite
darkᵂᴹ == finite == big
n is usually denoting
a natural number.
Do we mean the same by 'natural number'?
FISONs are finite by definition.
Do we mean the same by 'finite'?
On 08.03.2025 20:27, Jim Burns wrote:
On 3/8/2025 9:09 AM, WM wrote:
On 08.03.2025 12:58, Jim Burns wrote:
you should have said somewhere
what a finite ordinal is,
That is not under discussion here.
That has been under discussion for decades.
I think that these decades of discussion
have been, in large part, you assigning
different meanings to 'finite', etc.
and matheologians (among whom I place myself)
trying to discern what your meanings are.
Here's my best guess:
definableᵂᴹ == finite
[...]
darkᵂᴹ == finite == big
[...]
n is usually denoting
a natural number.
Do we mean the same by 'natural number'?
There are two different meanings:
All positive integers having FISONs or
all positive integers.
FISONs are finite by definition.
Do we mean the same by 'finite'?
A natural number n is finite.
It is an integer between 0 and ω: 0 < n < ω.
Sometimes 0 is included, never ω is included.
My description is definite.
Your description is only definite because
ℕ ∋ n is definite.
No, my description is definite
because
every n can be obtained by addition of 1's
(or of curly brackets).
You even avoid hearing what we mean.
On 09.03.2025 17:26, Jim Burns wrote:
On 3/8/2025 2:27 PM, Jim Burns wrote:
Here's my best guess:
definableᵂᴹ == finiteⁿᵒᵗᐧᵂᴹ == #A<#Aᣕᵇ
darkᵂᴹ == finiteⁿᵒᵗᐧᵂᴹ == big and #A<#Aᣕᵇ
matheologicalᵂᴹ == infiniteⁿᵒᵗᐧᵂᴹ == #A=#Aᣕᵇ
You even avoid hearing what we mean.
I am interested in
the difference
that you see between
Z₀ defined by { } ∈ Z₀, and
if {{{...{{{ }}}...}}} with n curly brackets ∈ Z₀
then {{{...{{{ }}}...}}} with n+1 curly brackets ∈ Z₀
and
The set of FISONs failing to have
the union ℕ defied by induction:
|ℕ \ {1}| = ℵo, and
if |ℕ \ {1, 2, 3, ..., n}| = ℵo
then |ℕ \ {1, 2, 3, ..., n+1}| = ℵo.
On 09.03.2025 17:26, Jim Burns wrote:
You even avoid hearing what we mean.I am interested in the difference that you see between
Z₀ defined by { } ∈ Z₀, and if {{{...{{{ }}}...}}} with n curly brackets
∈ Z₀ then {{{...{{{ }}}...}}} with n+1 curly brackets ∈ Z₀
and
The set of FISONs failing to have the union ℕ defied by induction:
|ℕ \ {1}| = ℵo, and if |ℕ \ {1, 2, 3, ..., n}| = ℵo then |ℕ \ {1, 2, 3,
..., n+1}| = ℵo.
On 3/9/2025 3:13 PM, WM wrote:
You even avoid hearing what we mean.
I am interested in
the difference
that you see between
Z₀ defined by { } ∈ Z₀, and
if {{{...{{{ }}}...}}} with n curly brackets ∈ Z₀
then {{{...{{{ }}}...}}} with n+1 curly brackets ∈ Z₀
and
The set of FISONs failing to have
the union ℕ defied by induction:
|ℕ \ {1}| = ℵo, and
if |ℕ \ {1, 2, 3, ..., n}| = ℵo
then |ℕ \ {1, 2, 3, ..., n+1}| = ℵo.
For the second,
I have elsewhere provided you with
the description of a FISON
For the first,
you aren't using
any definition which fills
a role comparable to "FISON'.
I can see that you aren't because
you think you have defined 'finite' there,
somehow (by a darkᵂᴹ definition?),
and because
you have said explicitly that you don't need
a definition elsewhere.
You (WM) think you don't need to say
what natural number is,
even where you clearly have taken the term
and used it in your own, private way.
Am Sun, 09 Mar 2025 20:13:53 +0100 schrieb WM:
On 09.03.2025 17:26, Jim Burns wrote:
You even avoid hearing what we mean.I am interested in the difference that you see between
Z₀ defined by { } ∈ Z₀, and if {{{...{{{ }}}...}}} with n curly brackets
∈ Z₀ then {{{...{{{ }}}...}}} with n+1 curly brackets ∈ Z₀
and
The set of FISONs failing to have the union ℕ defied by induction:
|ℕ \ {1}| = ℵo, and if |ℕ \ {1, 2, 3, ..., n}| = ℵo then |ℕ \ {1, 2,
3,
..., n+1}| = ℵo.
How you can view the first set not to have the same union (by Neumann's equivalence) as the the second one would be most welcome.
Am Sun, 09 Mar 2025 20:13:53 +0100 schrieb WM:
I am interested in the difference that you see between
Z₀ defined by { } ∈ Z₀, and if {{{...{{{ }}}...}}} with n curly brackets
∈ Z₀ then {{{...{{{ }}}...}}} with n+1 curly brackets ∈ Z₀
and
The set of FISONs failing to have the union ℕ defied by induction:
|ℕ \ {1}| = ℵo, and if |ℕ \ {1, 2, 3, ..., n}| = ℵo then |ℕ \ {1, 2,
3,
..., n+1}| = ℵo.
How you can view the first set not to have the same union (by Neumann's equivalence) as the the second one would be most welcome.
Am Sun, 09 Mar 2025 20:13:53 +0100 schrieb WM:
I am interested in the difference that you see between
Z₀ defined by { } ∈ Z₀, and if {{{...{{{ }}}...}}} with n curly brackets
∈ Z₀ then {{{...{{{ }}}...}}} with n+1 curly brackets ∈ Z₀
and
The set of FISONs failing to have the union ℕ defied by induction:
|ℕ \ {1}| = ℵo, and if |ℕ \ {1, 2, 3, ..., n}| = ℵo then |ℕ \ {1, 2, 3,
..., n+1}| = ℵo.
How you can view the first set not to have the same union (by Neumann's equivalence) as the the second one would be most welcome.
On 10.03.2025 00:11, Jim Burns wrote:
On 3/9/2025 3:13 PM, WM wrote:
</JB>Here's my best guess:
definableᵂᴹ == finiteⁿᵒᵗᐧᵂᴹ == #A<#Aᣕᵇ
darkᵂᴹ == finiteⁿᵒᵗᐧᵂᴹ == big and #A<#Aᣕᵇ
matheologicalᵂᴹ == infiniteⁿᵒᵗᐧᵂᴹ == #A=#Aᣕᵇ
You even avoid hearing what we mean.
I am interested in
the difference
that you see between
Z₀ defined by { } ∈ Z₀, and
if {{{...{{{ }}}...}}} with n curly brackets ∈ Z₀
then {{{...{{{ }}}...}}} with n+1 curly brackets ∈ Z₀
and
The set of FISONs failing to have
the union ℕ defied by induction:
|ℕ \ {1}| = ℵo, and
if |ℕ \ {1, 2, 3, ..., n}| = ℵo
then |ℕ \ {1, 2, 3, ..., n+1}| = ℵo.
For the second,
I have elsewhere provided you with
the description of a FISON
Where?
For the first,
you aren't using
any definition which fills
a role comparable to "FISON'.
Wrong.
{{ }} = {1},
{{{ }}} = {1, 2},
{{{{ }}}} = {1, 2, 3}.
I can see that you aren't because
you think you have defined 'finite' there,
All FISONs are finite.
somehow (by a darkᵂᴹ definition?),
and because
you have said explicitly that you don't need
a definition elsewhere.
Before you recognized that
you have run out of counterarguments
you never doubted that definition.
You (WM) think you don't need to say
what natural number is,
even where you clearly have taken the term
and used it in your own, private way.
I use definable natumbers
as everybody knows how to use them.
On 3/10/2025 4:42 AM, WM wrote:
For example, here:
⎛ A FISON is linearly ordered,
⎜ begins at 0, ends at a FISON.end,
That description isn't the usual description.
I'd guess the Peano axioms are the usual.
{{ }} = {1},
{{{ }}} = {1, 2},
{{{{ }}}} = {1, 2, 3}.
You've defined 3 numbers and stopped.
Peano and Zermelo and Cantor and I
describe infinitely.many.
On 3/10/2025 1:20 PM, WM wrote:
On 10.03.2025 17:31, Jim Burns wrote:
On 3/10/2025 4:42 AM, WM wrote:
For example, here:
⎛ A FISON is linearly ordered,
⎜ begins at 0, ends at a FISON.end,
That is sufficient.
The real interval [0,x] is linearly ordered,
begins at 0, and,
if [0,x] is a FISON, ends at a FISON.end.
A real interval isn't what you mean by 'FISON'
is it?
Peano and Zermelo and Cantor and I
describe infinitely.many.
Yes, without end.
The descriptions end.
What is described is without end.
Not.first.false claims are true
about what does not end
because _the descriptions_ end (are finite).
Because _the claims_ are finitely.many,
if there is a false claim,
then there is a first.false claim.
On 10.03.2025 17:31, Jim Burns wrote:
On 3/10/2025 4:42 AM, WM wrote:
For example, here:
⎛ A FISON is linearly ordered,
⎜ begins at 0, ends at a FISON.end,
That is sufficient.
And that is what I use, except 0.
Peano and Zermelo and Cantor and I
describe infinitely.many.
Yes, without end.
On 10.03.2025 18:47, Jim Burns wrote:
On 3/10/2025 1:20 PM, WM wrote:
On 10.03.2025 17:31, Jim Burns wrote:
For example, here:
⎛ A FISON is linearly ordered,
⎜ begins at 0, ends at a FISON.end,
That is sufficient.
The real interval [0,x] is linearly ordered,
begins at 0, and,
if [0,x] is a FISON, ends at a FISON.end.
Why can't you stay with the topic?!
Peano and Zermelo and Cantor and I
describe infinitely.many.
Yes, without end.
The descriptions end.
The sets don't.
Because _the claims_ are finitely.many,
if there is a false claim,
then there is a first.false claim.
On 10.03.2025 17:31, Jim Burns wrote:
For example, here:
⎛ A FISON is linearly ordered,
⎜ begins at 0, ends at a FISON.end,
That is sufficient.
And that is what I use, except 0.
That description isn't the usual description.
I'd guess the Peano axioms are the usual.
It is irrelevant what you prefer.
Zermelo, Peano, v. Neumann.
All use the same induction.
⎝ Therefore, this a property with no exceptions.
is
reliable.
On 10.03.2025 20:37, Jim Burns wrote:
⎝ Therefore, this a property with no exceptions.
is reliable.
Do you really expect that your gobbledegook is of interest???
Either explain the following or spare your efforts. No one will read that.
Z₀ is defined by induction: { } ∈ Z₀, and if {{{...{{{ }}}...}}} with n
curly brackets ∈ Z₀ then {{{...{{{ }}}...}}} with n+1 curly brackets ∈ Z₀.
Likewise the removable UF is defined by induction.
ℕ \ F(1), and if ℕ \ F(1) \ F(2) \ F(3) \ ... \ F(n) = ℵo
then ℕ \ F(1) \ F(2) \ F(3) \ ... \ F(n+1) = ℵo.
What is the difference in your opinion?
Regards, WM
And that is what I use, except 0.
Then don't use that. It's insufficient.
That description isn't the usual description.
I'd guess the Peano axioms are the usual.
It is irrelevant what you prefer.
Zermelo, Peano, v. Neumann.
All use the same induction.
For FISONs _as I describe them_
⎛ A FISON is linearly ordered,
⎜ begins at 0, ends at a FISON.end, and,
⎜ for each split,
⎜ its foresplit ends at i or is empty and
⎜ its hindsplit begins at j or is empty,
⎝ i and j such that i+1 = j
the proof.by.induction
⎛ This is an inductive property.
⎝ Therefore, this a property with no exceptions.
is
reliable.
⎛ Assume otherwise.
⎜ Assume
⎜ A(k) is inductive on FISON.numbers
⎜ A(0) ∧ ∀k:∃⁽ꟳ⁾F′∋k: A(k)⇒A(k+1)
⎜ and
⎜ FISON.number 𝔊 exists such that ¬A(𝔊)
⎜
⎜ {F} ∋ F(𝔊) = [0,𝔊] ∋ 𝔊
⎜
⎜ From property A(k)
⎜ define sweep A{≤k}
⎜ A{≤k} ⇔ ∀j≤k: A(j)
⎜
⎜ A{≤k+1} ⇔ A{≤k} ∧ A(k+1)
⎜
⎜ Sweep A{≤k} determines a split of [0,𝔊]
⎜ F′ = {i:A{≤i}} ∋ 0
⎜ H′ = {j:¬A{≤j}} ∋ 𝔊
⎜ F′ ᵉᵃᶜʰ<ᵉᵃᶜʰ H′
⎜
⎜ [0,𝔊] is a FISON.
⎜ F′ holds last i′
⎜ H′ holds first j′
⎜ i′+1 = j′
⎜
⎜ From the definition of F′ and H′
⎜ A{≤i′} ∧ ¬A{≤i′+1}
⎜
⎜ A{≤i′} ∧ ¬(A{≤i′} ∧ A(i'+1))
⎜
⎜ A{≤i′} ∧ ¬A(i'+1)
⎜
⎜ A(i′) ∧ ¬A(i'+1)
⎜
⎜ ¬(A(i′)⇒A(i′+1))
⎜
⎜ However,
⎜ ∀k:∃⁽ꟳ⁾F′∋k: A(k)⇒A(k+1)
⎜
⎜ A(i′)⇒A(i′+1)
⎝ Contradiction.
Therefore,
the proof.by.induction on FISON.numbers
⎛ This is an inductive property.
⎝ Therefore, this a property with no exceptions.
is
reliable.
WM <[email protected]> wrote:
Z₀ is defined by induction: {} ∈ Z₀, and if {{{...{{{}}}...}}} with n >> curly brackets ∈ Z₀ then {{{...{{{}}}...}}} with n+1 curly brackets ∈ Z₀.
WM <[email protected]> wrote:
ℕ \ F(1), and if [for any n e ℕ] ℕ \ F(1) \ F(2) \ F(3) \ ... \ F(n) = ℵo
then ℕ \ F(1) \ F(2) \ F(3) \ ... \ F(n+1) = ℵo.
WM <[email protected]> wrote:
ℕ \ F(1), and if [for any n e ℕ] ℕ \ F(1) \ F(2) \ F(3) \ ... \ F(n) >>> = ℵo
then ℕ \ F(1) \ F(2) \ F(3) \ ... \ F(n+1) = ℵo.
Daraus kann man "per Induktion" folgern:
An e IN: ℕ \ F(1) \ F(2) \ F(3) \ ... \ F(n) = ℵo.
Daraus folgt aber NICHT:
ℕ \ F(1) \ F(2) \ F(3) \ ... = ℵo.
WM <[email protected]> wrote:
On 10.03.2025 20:37, Jim Burns wrote:
⎝ Therefore, this a property with no exceptions.
is reliable.
Do you really expect that your gobbledegook is of interest???
I find it of interest, yes. It refutes _your_ gobbledegook in a way that
you cannot answer.
I thought you (WM) might be interested to hear that
we can prove that proofs.by.induction are reliable
for FISONs.
Am 10.03.2025 um 22:04 schrieb Alan Mackenzie:
WM <[email protected]> wrote:
Z₀ is defined by induction: {} ∈ Z₀, and if {{{...{{{}}}...}}} with n >>> curly brackets ∈ Z₀ then {{{...{{{}}}...}}} with n+1 curly brackets ∈ >>> Z₀.
Actually, Z₀ is NOT "defined" (WM) by that.
But for Z₀ the following holds:
{} ∈ Z₀
and
Ax e Z₀: {x} e Z₀
[Ax(x e Z₀ -> {x} e Z₀]
We don't "count" "curly brackets" in this context.
On 10.03.2025 00:11, Jim Burns wrote:This is wrong as written. Perhaps you mean equivalences, but they are
On 3/9/2025 3:13 PM, WM wrote:
Where?I am interested in the difference that you see between
Z₀ defined by { } ∈ Z₀, and if {{{...{{{ }}}...}}} with n curly
brackets ∈ Z₀ then {{{...{{{ }}}...}}} with n+1 curly brackets ∈ Z₀ >>> and
The set of FISONs failing to have the union ℕ defied by induction:
|ℕ \ {1}| = ℵo, and if |ℕ \ {1, 2, 3, ..., n}| = ℵo then |ℕ \ {1, 2,
3, ..., n+1}| = ℵo.
For the second,
I have elsewhere provided you with the description of a FISON
For the first,Wrong. {{ }} = {1}, {{{ }}} = {1, 2}, {{{{ }}}} = {1, 2, 3}.
you aren't using any definition which fills a role comparable to
"FISON'.
Not an argument. What's the definition?I can see that you aren't because you think you have defined 'finite'All FISONs are finite.
there,
somehow (by a darkᵂᴹ definition?),Before you recognized that you have run out of counterarguments you
and because you have said explicitly that you don't need a definition
elsewhere.
never doubted that definition.
You can't hide your personal usage behind "everybody knows". If every-You (WM) think you don't need to say what natural number is,I use definable natumbers as everybody knows how to use them.
even where you clearly have taken the term and used it in your own,
private way.
On 10.03.2025 09:30, joes wrote:No, they are not finite. You can't believe Z_0 to be "complete" in
Am Sun, 09 Mar 2025 20:13:53 +0100 schrieb WM:
I am interested in the difference that you see between
Z₀ defined by { } ∈ Z₀, and if {{{...{{{ }}}...}}} with n curly
brackets ∈ Z₀ then {{{...{{{ }}}...}}} with n+1 curly brackets ∈ Z₀ >>> and
The set of FISONs failing to have the union ℕ defied by induction:
|ℕ \ {1}| = ℵo, and if |ℕ \ {1, 2, 3, ..., n}| = ℵo then |ℕ \ {1, 2,
3, ..., n+1}| = ℵo.
How you can view the first set not to have the same union (by Neumann's
equivalence) as the the second one would be most welcome. In any case,
the second set does not exist according to your contradictory specifi-
cation. They are clearly isomorphic.
That is no my opinion but Jim Burns' opinion.
Both sets are identical. Both are potentially infinite collections.
Am Mon, 10 Mar 2025 09:42:05 +0100 schrieb WM:
Wrong. {{ }} = {1}, {{{ }}} = {1, 2}, {{{{ }}}} = {1, 2, 3}.This is wrong as written. Perhaps you mean equivalences, but they are certainly not equal.
Am Mon, 10 Mar 2025 16:20:58 +0100 schrieb WM:
On 10.03.2025 09:30, joes wrote:
Am Sun, 09 Mar 2025 20:13:53 +0100 schrieb WM:
I am interested in the difference that you see between
Z₀ defined by { } ∈ Z₀, and if {{{...{{{ }}}...}}} with n curly
brackets ∈ Z₀ then {{{...{{{ }}}...}}} with n+1 curly brackets ∈ Z₀
and
The set of FISONs failing to have the union ℕ defied by induction:
|ℕ \ {1}| = ℵo, and if |ℕ \ {1, 2, 3, ..., n}| = ℵo then |ℕ \ {1, 2,
3, ..., n+1}| = ℵo.
How you can view the first set not to have the same union (by Neumann's
equivalence) as the the second one would be most welcome. In any case,
the second set does not exist according to your contradictory specifi-
cation. They are clearly isomorphic.
That is no my opinion but Jim Burns' opinion.
Both sets are identical. Both are potentially infinite collections.
No, they are not finite.
You can't believe Z_0 to be "complete" in
your sense if you don't think the second set is (and accept that they
are equivalent).
On 10.03.2025 23:07, Jim Burns wrote:
I thought you (WM) might be interested to hear that
we can prove that proofs.by.induction are reliable
for FISONs.
That is trivial.
I am interested in the difference you see between
Zermelo's Z₀ defined or ensurede
Zermelo's Z₀ defined or ensurede by induction:
{ } ∈ Z₀,
and if
{{{...{{{ }}}...}}} with n curly brackets ∈ Z₀
then
{{{...{{{ }}}...}}} with n+1 curly brackets ∈ Z₀.
and
the the set F of removable FISONs
defined or ensured by induction.
ℕ \ F(1) = ℵo,
and if
ℕ \ F(1) \ F(2) \ F(3) \ ... \ F(n) = ℵo
then
ℕ \ F(1) \ F(2) \ F(3) \ ... \ F(n+1) = ℵo.
On 3/11/2025 4:28 AM, WM wrote:
On 10.03.2025 23:07, Jim Burns wrote:
I thought you (WM) might be interested to hear that
we can prove that proofs.by.induction are reliable
for FISONs.
That is trivial.
Sadly,
you (WM) are wrong to think you know
what that sentence means.
I am interested in the difference you see between
Zermelo's Z₀ defined or ensurede
The difference between 'defined' and 'ensured'
is that
Zermelo defines Z (Z₀ is after)
to be
an inductive set, and so Z is inductive,
but that definition doesn't ensure
that Z exists.
Zermelo's Z₀ defined or ensurede by induction:
{ } ∈ Z₀,
and if
{{{...{{{ }}}...}}} with n curly brackets ∈ Z₀
then
{{{...{{{ }}}...}}} with n+1 curly brackets ∈ Z₀.
You (WM) are defining your own (not Zermelo's) Z₀ᵂᴹ
Perhaps you feel that Zermelo's definition is
more complicated than needed.
Perhaps it displeases you that
Zermelo's isn't vague enough
to have room for your darkᵂᴹ numbers.
When we get see what we're intended to understand,
we get to see that Z₀ᵂᴹ doesn't hold darkᵂᴹ numbers.
and
the the set F of removable FISONs
defined or ensured by induction.
ℕ \ F(1) = ℵo,
and if
ℕ \ F(1) \ F(2) \ F(3) \ ... \ F(n) = ℵo
then
ℕ \ F(1) \ F(2) \ F(3) \ ... \ F(n+1) = ℵo.
|A| < ℵ₀ ∧ |B| = ℵ₀ ⇒ |B\A| = |B| = ℵ₀
On 11.03.2025 12:03, joes wrote:I mean, the sets on the left all contain only one element.
Am Mon, 10 Mar 2025 09:42:05 +0100 schrieb WM:
These are different languages for the same notion.Wrong. {{ }} = {1}, {{{ }}} = {1, 2}, {{{{ }}}} = {1, 2, 3}.This is wrong as written. Perhaps you mean equivalences, but they are
certainly not equal.
The words are different, the numbers are the same.
Am Tue, 11 Mar 2025 18:04:45 +0100 schrieb WM:
On 11.03.2025 12:03, joes wrote:I mean, the sets on the left all contain only one element.
Am Mon, 10 Mar 2025 09:42:05 +0100 schrieb WM:These are different languages for the same notion.
Wrong. {{ }} = {1}, {{{ }}} = {1, 2}, {{{{ }}}} = {1, 2, 3}.This is wrong as written. Perhaps you mean equivalences, but they are
certainly not equal.
The words are different, the numbers are the same.
On 11.03.2025 18:11, Jim Burns wrote:
to be
an inductive set, and so Z is inductive,
but that definition doesn't ensure
that Z exists.
He defines Z by induction in order to ensure
the existence of an infinite set.
|A| < ℵ₀ ∧ |B| = ℵ₀ ⇒ |B\A| = |B| = ℵ₀
|Z₀| < ℵ₀
|Z₀| < ℵ₀ ∧ |ℕ| = ℵ₀ ⇒ |ℕ\Z₀| = |ℕ| = ℵ₀
|UF| < ℵ₀ ∧ |ℕ| = ℵ₀ ⇒ |ℕ\UF| = |ℕ| = ℵ₀
UF = ℕ ==> Ø = ℕ
On 3/11/2025 2:01 PM, WM wrote:
Zermelo defines Z to be an inductive set.
Defining Z₀ to be inductive is insufficient
to prove, from a subset ⊆ Z₀ being inductive,
that the subset = the whole set Z₀
|A| < ℵ₀ ∧ |B| = ℵ₀ ⇒ |B\A| = |B| = ℵ₀
|Z₀| < ℵ₀
No.
Z₀ is inductive.
On 11.03.2025 20:25, Jim Burns wrote:
On 3/11/2025 2:01 PM, WM wrote:
Zermelo defines Z to be an inductive set.
in order to ensure the existence of
an infinite or inductive set.
in order to ensure the existence of
an infinite or inductive set.
On 3/11/2025 5:28 PM, WM wrote:
On 11.03.2025 20:25, Jim Burns wrote:
On 3/11/2025 2:01 PM, WM wrote:
Zermelo defines Z to be an inductive set.
in order to ensure the existence of
an infinite or inductive set.
No.
There is a definition and there is an axiom.
They are different kinds of things.
The definition of Z states that
Z is inductive.
in order to ensure the existence of
an infinite or inductive set.
A definition doesn't ensure existence.
That's what an axiom does.
On 11.03.2025 18:19, joes wrote:No, natural numbers are not equal to sets of them (yes, they
Am Tue, 11 Mar 2025 18:04:45 +0100 schrieb WM:The sets on the right-hand side are FISONs = natural numbers.
On 11.03.2025 12:03, joes wrote:I mean, the sets on the left all contain only one element.
Am Mon, 10 Mar 2025 09:42:05 +0100 schrieb WM:These are different languages for the same notion.
Wrong. {{ }} = {1}, {{{ }}} = {1, 2}, {{{{ }}}} = {1, 2, 3}.This is wrong as written. Perhaps you mean equivalences, but they are
certainly not equal.
The words are different, the numbers are the same.
On 11.03.2025 23:51, Jim Burns wrote:
On 3/11/2025 5:28 PM, WM wrote:
On 11.03.2025 20:25, Jim Burns wrote:
On 3/11/2025 2:01 PM, WM wrote:
Zermelo defines Z to be an inductive set.
in order to ensure the existence of
an infinite or inductive set.
No.
Um aber die Existenz "unendlicher" Mengen zu sichern,
bedürfen wir noch des folgenden ... Axioms.
[Zermelo: Untersuchungen über die Grundlagen
der Mengenlehre I, S. 266]
There is a definition and there is an axiom.
Where are they?
Please quote a definition
which is outside of the axiom.
They are different kinds of things.
And you are unable to understand the connection
between these two things.
Is it the language?
Or is it a general deficit?
The definition of Z states that
Z is inductive.
Please quote this definition which
is not the axiom.
in order to ensure the existence of
an infinite or inductive set.
A definition doesn't ensure existence.
That's what an axiom does.
This axiom
This axiom ensures the sequence of numbers
by the same induction
which ensures the dark numbers:
which ensures the dark numbers:
{} ∈ Z₀, and for all x: if x ∈ Z₀ then {x} ∈ Z₀
ℕ \ F(1) = ℵo,
and if
ℕ \ F(1) \ F(2) \ F(3) \ ... \ F(n) = ℵo
then
ℕ \ F(1) \ F(2) \ F(3) \ ... \ F(n+1) = ℵo.
On 3/12/2025 5:03 AM, WM wrote:
On 11.03.2025 23:51, Jim Burns wrote:
On 3/11/2025 5:28 PM, WM wrote:
On 11.03.2025 20:25, Jim Burns wrote:
On 3/11/2025 2:01 PM, WM wrote:
Zermelo defines Z to be an inductive set.
in order to ensure the existence of
an infinite or inductive set.
No.
Um aber die Existenz "unendlicher" Mengen zu sichern,
bedürfen wir noch des folgenden ... Axioms.
[Zermelo: Untersuchungen über die Grundlagen der Mengenlehre I, S. 266]
...which isn't
/ Um aber die Existenz "unendlicher" Mengen zu sichern,
\ bedürfen wir noch des folgenden ... Definition.
There is a definition and there is an axiom.
Where are they?
Please quote a definition
⎛ Menge Z, welche die Nullmenge als Element enthält und
⎜ so beschaffen ist, daß jedem ihrer Elemente a
⎝ ein weiteres Element der Form {a} entspricht
how are your darkᵂᴹ numbers ensured?
{} ∈ Z₀, and for all x: if x ∈ Z₀ then {x} ∈ Z₀
ℕ \ F(1) = ℵo,
and if
ℕ \ F(1) \ F(2) \ F(3) \ ... \ F(n) = ℵo
then
ℕ \ F(1) \ F(2) \ F(3) \ ... \ F(n+1) = ℵo.
On 12.03.2025 11:54, Jim Burns wrote:
On 3/12/2025 5:03 AM, WM wrote:
On 11.03.2025 23:51, Jim Burns wrote:
There is a definition and there is an axiom.
Where are they?
Please quote a definition
⎛ Menge Z, welche die Nullmenge als Element enthält und
⎜ so beschaffen ist, daß jedem ihrer Elemente a
⎝ ein weiteres Element der Form {a} entspricht
That is just the induction.
how are your darkᵂᴹ numbers ensured?
Here:
Fron Zermelo's induction
{} ∈ Z₀, and for all x: if x ∈ Z₀ then {x} ∈ Z₀
my induction is ensured
ℕ \ F(1) = ℵo,
and if
ℕ \ F(1) \ F(2) \ F(3) \ ... \ F(n) = ℵo
then
ℕ \ F(1) \ F(2) \ F(3) \ ... \ F(n+1) = ℵo.
On 3/12/2025 9:04 AM, WM wrote:
⎛ Menge Z, welche die Nullmenge als Element enthält und
⎜ so beschaffen ist, daß jedem ihrer Elemente a
⎝ ein weiteres Element der Form {a} entspricht
⎛ set Z, which contains the zero set as an element and
⎜ is such that each of its elements a
⎝ corresponds to another element of the form {a}
That is just the induction.
A definition answers "What is Z ?"
An axiom answers "Does Z exist?"
The axiom does not create an inductive set.
Z₀ -- defined as the emptiest inductive set --
doesn't hold darkᵂᴹ {} and
doesn't hold any visibleᵂᴹ x and darkᵂᴹ {x}
{darkᵂᴹ} = Z₀\{visibleᵂᴹ} = {}
my induction is ensured
Your induction is not ensured by
your declaration that it's ensured.
How is your induction ensured?
ℕ \ F(1) = ℵo,
and if
ℕ \ F(1) \ F(2) \ F(3) \ ... \ F(n) = ℵo
then
ℕ \ F(1) \ F(2) \ F(3) \ ... \ F(n+1) = ℵo.
WM wrote on 3/12/2025 :
Inductive sets do not contain dark numbers.
Is the set of natural numbers an inductive set?
WM wrote :
On 12.03.2025 21:12, FromTheRafters wrote:
WM wrote on 3/12/2025 :
Inductive sets do not contain dark numbers.
Is the set of natural numbers an inductive set?
The set of definable natural numbers
[...]
That is not what I asked.
I happen to know that the naturals are exemplary of the
smallest inductive set.
WM formulated on Thursday :
On 13.03.2025 10:59, FromTheRafters wrote:
I happen to know that the naturals are exemplary of the smallest
inductive set.
The naturals defined by Cantor have more elements than any natural
number can measure. Inductive sets never can leave the domain measured
by natural numbers.
So, now you are saying that the superset of the naturals has cardinality aleph_zero and as such is only countably infinite
but bigger than the
set of naturals?
On 13.03.2025 10:59, FromTheRafters wrote:No contradiction there.
I happen to know that the naturals are exemplary of the smallestThe naturals defined by Cantor have more elements than any natural
inductive set.
number can measure. Inductive sets never can leave the domain measured
by natural numbers.
Am 05.03.2025 um 13:25 schrieb Richard Damon:
On 3/4/25 4:07 AM, WM wrote:
Wrong, Cantor shows that the number of Natural Numbers generated by
the iterative method of, we have 0, and for every number we have its
successor, is not one of those finite numbers, but is another number
Aleph0
Below you contradict yourself.
We call that phenomenon potential infinity.
WHich is just infinity,
Cantor denies your claim.
"Nevertheless the transfinite cannot be considered a subsection of
what is usually called 'potentially infinite'. Because the latter is
not (like every individual transfinite and in general everything due
to an 'idea divina') determined in itself, fixed, and unchangeable,
but a finite in the process of change, having in each of its current
states a finite size; like, for instance, the temporal duration since
the beginning of the world, which, when measured in some time-unit,
for instance a year, is finite in every moment, but always growing
beyond all finite limits, without ever becoming really infinitely
large." [G. Cantor, letter to I. Jeiler (13 Oct 1895)]
Here he is right.
Which doesn't mean what you think it means.
He means what he says.
He is pointing out that these "transfinite" concepts aren't part of
the infinite set built by iteration (the "potential infinity") but is
beyond it.
Therefore iterartion fails to produce actual infinity.
We SEE the "potentially infinite" via a process, where each step is
finite, but the final result of it *IS* an infinite thing.
There is no final result. You are unable to understand infinity.
Because the latter is a finite in the process of change, having in each
of its current without ever becoming really infinitely large.
None of the members of N are themselves infinite, but the set itself is.
Not by recursion or induction! Therefore UF is a proper subset of ℕ.
UF = ℕ ==> Ø = ℕ
Regards, WM
On 3/5/25 4:05 PM, WM wrote:
Therefore iteration fails to produce actual infinity.
Where do you get that?
The memeber of the set we iterate in are all
finite, but the resultant set is infinite itself.
On 04.03.2025 08:39, joes wrote:
Am Mon, 03 Mar 2025 22:56:46 +0100 schrieb WM:
On 03.03.2025 20:26, Jim Burns wrote:No, he doesn't claim infinitely large naturals.
On 3/3/2025 3:57 AM, WM wrote:
Zermelo's ℕ and Cantor's ℕ are the same up to isomorphism.No. Zermelo uses induction which never goes beyond a finite number of
finite numbers.
Cantor claims a number of finite numbers which is larger than all finite >>> numbers.
Cantor claims that the number of natural numbers is a quantity larger
than any finite number.
"Die Zahl ℵ₀ ist größer als jede endliche Zahl"
"Unter einem A.-U. ist dagegen ein Quantum zu verstehen, das einerseits
nicht veränderlich, sondern vielmehr in allen seinen Teilen fest und bestimmt, eine richtige Konstante ist, zugleich aber andrerseits jede endliche Größe derselben Art an Größe übertrifft. Als Beispiel führe ich
die Gesamtheit, den Inbegriff aller endlichen ganzen positiven Zahlen
an; diese Menge ist ein Ding für sich und bildet, ganz abgesehen von der natürlichen Folge der dazu gehörigen Zahlen, ein in allen Teilen festes, bestimmtes Quantum, ein , das offenbar größer zu nennen ist
als jede endliche Anzahl.
Sowas wird durch Induktion niemals erreicht.
Gruß, WM
Regards, WM
But the number of Natural Numbers isn't itself a Natural Number, so that doesn't contradict the statement.
On 11.04.2025 04:03, Richard Damon wrote:No, it is impossible if you only use a finite number of naturals. You
On 3/5/25 4:05 PM, WM wrote:
Therefore iteration fails to produce actual infinity.Where do you get that?
The memeber of the set we iterate in are all finite, but the resultantIt is impossible to iterate the definable terms such that the remainedr
set is infinite itself.
is less than infinity. The never iterated Terms are almost all terms.
Am Fri, 11 Apr 2025 14:43:59 +0200 schrieb WM:
It is impossible to iterate the definable terms such that the remainderNo,
is less than infinity. The never iterated Terms are almost all terms.
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