Le 07/04/2024 à 13:16, Richard Damon a écrit :
On 4/7/24 4:32 AM, WM wrote:
Le 06/04/2024 à 22:03, Richard Damon a écrit :
On 4/6/24 3:40 PM, WM wrote:
Le 06/04/2024 à 15:58, Richard Damon a écrit :
On 4/6/24 9:55 AM, WM wrote:
That mapping is Cantor's proposal. But for every other mapping,
the O's would also remain. All O's! It is th lossless exchange
which proves it.
Cantor's proposal is between members of two distinct sets.
No. He does not specify that. And there is no reason to do so,
except that it can be used to contradict the ridiculous nonsense
that there are as many fractions as prime numbers.y
But he DOES, as he talks about the two SETS of numbers that are
matched up.
One set and its subset. Dedekind: A system S is said to be /infinite/
if it is similar to a real part of itself. To consider them as two
sets does not change the numbers of elements.
But does affect your logic of pairing.
No. Since there are precisely as many natnumbers n as natnumber
fractions n/1, nothing is affected. The only effect is that the Os can
be proven to remain the same number in every step. This is true in all mappings but more easily seen in mine.
So, With infinite sets, a proper subset CAN be the same size as its
parent.
Impossible.
You are just PROVING you don't understand how infinity works,
I understand that a crowd of fools has been tricked by Cantor.
Regards, WM
On 4/7/24 9:23 AM, WM wrote:
So, With infinite sets, a proper subset CAN be the same size as its
parent.
Impossible.
Nope, PROVEN.
Since the DEFINITION of "Same Size" is the ability to make a 1-to-1
mapping between the sets.
Do you want to claim that two sets that you can match EVERY DISTINCT
element of one to a UNIQUE DISTINCT ELEMENT of the other are NOT the
same size?
and we can build such a mapping between the set of natural Numbers (N)
with the set of even Numbers (E).
Since for ALL elements n, a member of the Natural Numbers, there exists
an element e, a member of tghe Even Nubers, such that the value of e is
twice the value of n (e = 2n)
EVERY element of N is mapped to a DISTINCT element of E.
Try to find an exception
Le 07/04/2024 à 19:56, Richard Damon a écrit :
On 4/7/24 9:23 AM, WM wrote:
So, With infinite sets, a proper subset CAN be the same size as its
parent.
Impossible.
Nope, PROVEN.
Proven impossble with my matrix,
Since the DEFINITION of "Same Size" is the ability to make a 1-to-1
mapping between the sets.
Do you want to claim that two sets that you can match EVERY DISTINCT
element of one to a UNIQUE DISTINCT ELEMENT of the other are NOT the
same size?
and we can build such a mapping between the set of natural Numbers (N)
with the set of even Numbers (E).
Only handwaving by "and so on"
Since for ALL elements n, a member of the Natural Numbers, there
exists an element e, a member of tghe Even Nubers, such that the value
of e is twice the value of n (e = 2n)
EVERY element of N is mapped to a DISTINCT element of E.
Try to find an exception
In all cases there are infinitely many exceptions.
∀n ∈ ℕ_applied: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
Regards, WM
On 4/8/24 9:44 AM, WM wrote:
Le 07/04/2024 à 19:56, Richard Damon a écrit :
On 4/7/24 9:23 AM, WM wrote:
So, With infinite sets, a proper subset CAN be the same size as its
parent.
Impossible.
Nope, PROVEN.
Proven impossble with my matrix,
Nope, since you matrix doesn't follow the required form.
and we can build such a mapping between the set of natural Numbers (N)
with the set of even Numbers (E).
Only handwaving by "and so on"
In all cases there are infinitely many exceptions.
∀n ∈ ℕ_applied: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
I didn't say "N_applied", I said N.
Le 09/04/2024 à 01:22, Richard Damon a écrit :
On 4/8/24 9:44 AM, WM wrote:
Le 07/04/2024 à 19:56, Richard Damon a écrit :
On 4/7/24 9:23 AM, WM wrote:
So, With infinite sets, a proper subset CAN be the same size as
its parent.
Impossible.
Nope, PROVEN.
Proven impossble with my matrix,
Nope, since you matrix doesn't follow the required form.
It does precilsely.
and we can build such a mapping between the set of natural Numbers
(N) with the set of even Numbers (E).
Only handwaving by "and so on"
In all cases there are infinitely many exceptions.
∀n ∈ ℕ_applied: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
I didn't say "N_applied", I said N.
But what you can use belongs to ℕ_applied. Otherwise show a natural
number that completes the bijection, i.e., which has not infinitely many pairings on front.
Regards, WM
Nope, you can use ALL of the Natural Numbers.
If you can only "use" N_applied, your system doesn't actually HAVE the Natural numbers
On 4/9/24 8:16 AM, WM wrote:
Nope, because ONE set is not TWO Sets.
I didn't say "N_applied", I said N.
But what you can use belongs to ℕ_applied. Otherwise show a natural
number that completes the bijection, i.e., which has not infinitely many
pairings on front.
Nope, you can use ALL of the Natural Numbers.
You "Complete" the bijection by showing the infinite sets map one to one
by the formula of the bijection.
Le 10/04/2024 à 01:06, Richard Damon a écrit :
On 4/9/24 8:16 AM, WM wrote:
Nope, because ONE set is not TWO Sets.
In the set ℚ there are as many indices n/1 as are indices n in ℕ. If indexing all fractions was possible, it was possible with indices n/1.
But it isn't.
You can use only a small minority because almost all remain unused:
I didn't say "N_applied", I said N.
But what you can use belongs to ℕ_applied. Otherwise show a natural
number that completes the bijection, i.e., which has not infinitely
many pairings on front.
Nope, you can use ALL of the Natural Numbers.
∀n ∈ ℕ_used: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
You "Complete" the bijection by showing the infinite sets map one to
one by the formula of the bijection.
I show that Cantor's bijection fails.
Regards, WM
Le 10/04/2024 à 01:06, Richard Damon a écrit :
Nope, you can use ALL of the Natural Numbers.
If you can only "use" N_applied, your system doesn't actually HAVE the
Natural numbers
If you have used all numbers, none is remaining.
Use a number with less than ℵo successors for the start.
Regards, WM
On 4/10/24 4:14 PM, WM wrote:
Le 10/04/2024 à 01:06, Richard Damon a écrit :
On 4/9/24 8:16 AM, WM wrote:
Nope, because ONE set is not TWO Sets.
In the set ℚ there are as many indices n/1 as are indices n in ℕ. If
indexing all fractions was possible, it was possible with indices n/1.
But it isn't.
You can use only a small minority because almost all remain unused:
I didn't say "N_applied", I said N.
But what you can use belongs to ℕ_applied. Otherwise show a natural
number that completes the bijection, i.e., which has not infinitely
many pairings on front.
Nope, you can use ALL of the Natural Numbers.
∀n ∈ ℕ_used: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
Which ones were unused by e = 2*n?
Except that there *IS NO* Natural Number with less than ℵo successors,
The infinite set is inexhaustible by finite operations.
Le 11/04/2024 à 01:05, Richard Damon a écrit :
On 4/10/24 4:14 PM, WM wrote:
Le 10/04/2024 à 01:06, Richard Damon a écrit :
Nope, you can use ALL of the Natural Numbers.
You can use only a small minority because
almost all remain unused:
∀n ∈ ℕ_used: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
Which ones were unused by e = 2*n?
Those of {1, 2, 3, ...} with less than ℵo successors.
Those of {1, 2, 3, ...} with less than ℵo successors.
Le 11/04/2024 à 01:04, Richard Damon a écrit :
Except that there *IS NO* Natural Number with less than ℵo successors,
That is wrong if/because *all* natural numbers have no successors.
The infinite set is inexhaustible by finite operations.
Then enumerating of infinite sets is impossible because all indices are mapped by finite operations with no exception.
Regards, WM
Le 11/04/2024 à 01:05, Richard Damon a écrit :
On 4/10/24 4:14 PM, WM wrote:
Le 10/04/2024 à 01:06, Richard Damon a écrit :
On 4/9/24 8:16 AM, WM wrote:
Nope, because ONE set is not TWO Sets.
In the set ℚ there are as many indices n/1 as are indices n in ℕ. If >>> indexing all fractions was possible, it was possible with indices
n/1. But it isn't.
You can use only a small minority because almost all remain unused:
I didn't say "N_applied", I said N.
But what you can use belongs to ℕ_applied. Otherwise show a natural >>>>> number that completes the bijection, i.e., which has not infinitely
many pairings on front.
Nope, you can use ALL of the Natural Numbers.
∀n ∈ ℕ_used: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
Which ones were unused by e = 2*n?
Those of {1, 2, 3, ...} with less than ℵo successors.
Regards, WM
On 4/11/2024 8:07 AM, WM wrote:
Which ones were unused by e = 2*n?
Those of {1, 2, 3, ...} with less than ℵo successors.
{1,2,3,…} is
the least.upper.bound of all finiteⁿᵒᵗᐧᵂᴹ.{1,2,3,…,k}
For each m ∈ {1,2,3,…}:
( for each k ∈ {1,2,3,…}
m+k is a successor of m in {1,2,3,…} )
On 4/11/24 8:03 AM, WM wrote:
Le 11/04/2024 à 01:04, Richard Damon a écrit :
Except that there *IS NO* Natural Number with less than ℵo successors,
That is wrong if/because *all* natural numbers have no successors.
How do you expect to index an infihite set with finite operations.
Le 11/04/2024 à 19:36, Jim Burns a écrit :
On 4/11/2024 8:07 AM, WM wrote:
Which ones were unused by e = 2*n?
Those of {1, 2, 3, ...} with less than ℵo successors.
{1,2,3,…} is
the least.upper.bound of all finiteⁿᵒᵗᐧᵂᴹ.{1,2,3,…,k}
Yes, but many natnumbers will never be used.
For each m ∈ {1,2,3,…}:
( for each k ∈ {1,2,3,…}
m+k is a successor of m in {1,2,3,…} )
Consider the set {1, 2, 3, ..., ω} and multiply every element by 2 with
the result {2, 4, 6, ..., ω*2}. What elements fall between ω and ω*2?
What size has the interval between N*2 and ω*2?
On 4/11/24 8:07 AM, WM wrote:
Le 11/04/2024 à 01:05, Richard Damon a écrit :
On 4/10/24 4:14 PM, WM wrote:
Le 10/04/2024 à 01:06, Richard Damon a écrit :
On 4/9/24 8:16 AM, WM wrote:
Nope, because ONE set is not TWO Sets.
In the set ℚ there are as many indices n/1 as are indices n in ℕ. If >>>> indexing all fractions was possible, it was possible with indices
n/1. But it isn't.
You can use only a small minority because almost all remain unused:
I didn't say "N_applied", I said N.
But what you can use belongs to ℕ_applied. Otherwise show a natural >>>>>> number that completes the bijection, i.e., which has not infinitely >>>>>> many pairings on front.
Nope, you can use ALL of the Natural Numbers.
∀n ∈ ℕ_used: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
Which ones were unused by e = 2*n?
Those of {1, 2, 3, ...} with less than ℵo successors.
But none of them have less than ℵo successors,
because if have less, the
all have less since earlier ones have only finitely more,
Le 12/04/2024 à 00:32, Richard Damon a écrit :
On 4/11/24 8:03 AM, WM wrote:
Le 11/04/2024 à 01:04, Richard Damon a écrit :
Except that there *IS NO* Natural Number with less than ℵo successors, >>>That is wrong if/because *all* natural numbers have no successors.
{1, 2, 3, ...} there are no successors. But every useable number has ∀n
∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo successors.
How do you expect to index an infihite set with finite operations.
I don't. Matheologians do.
Regards, WM
Le 12/04/2024 à 03:16, Richard Damon a écrit :
On 4/11/24 8:07 AM, WM wrote:
Le 11/04/2024 à 01:05, Richard Damon a écrit :
On 4/10/24 4:14 PM, WM wrote:
Le 10/04/2024 à 01:06, Richard Damon a écrit :
On 4/9/24 8:16 AM, WM wrote:
Nope, because ONE set is not TWO Sets.
In the set ℚ there are as many indices n/1 as are indices n in ℕ. >>>>> If indexing all fractions was possible, it was possible with
indices n/1. But it isn't.
You can use only a small minority because almost all remain unused:
I didn't say "N_applied", I said N.
But what you can use belongs to ℕ_applied. Otherwise show a
natural number that completes the bijection, i.e., which has not >>>>>>> infinitely many pairings on front.
Nope, you can use ALL of the Natural Numbers.
∀n ∈ ℕ_used: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
Which ones were unused by e = 2*n?
Those of {1, 2, 3, ...} with less than ℵo successors.
But none of them have less than ℵo successors,
None of them which can be used as individuals have less than ℵo successors.
because if have less, the all have less since earlier ones have only
finitely more,
No, there are infinitely many natural numbers. Only between the usable numbers there a finite difference can be calculated, because the
infinitely many follow upon them and are dark.
Regards, WM
On 4/12/24 9:40 AM, WM wrote:
Le 12/04/2024 à 00:32, Richard Damon a écrit :
On 4/11/24 8:03 AM, WM wrote:
Le 11/04/2024 à 01:04, Richard Damon a écrit :
Except that there *IS NO* Natural Number with less than ℵo successors, >>>>That is wrong if/because *all* natural numbers have no successors.
{1, 2, 3, ...} there are no successors. But every useable number has ∀n
∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo successors.
Why do you say there are no succesors. As you say, for EVERY Natural
Number there are ℵo successors
Even for all the numbers in the ..., as it NEVER ENDS and thus EVERY
value in it still has ℵo successors.
How do you expect to index an infihite set with finite operations.
I don't. Matheologians do.
Then why do you talk about a set of Number you "don't believe" in?
You can't talk about "Cantor" and his bijections, if you don't accept
the rules he discussed his theories in.
WM schrieb:
Consider the set {1, 2, 3, ..., ω} and multiply every element by 2 with
the result {2, 4, 6, ..., ω*2}. What elements fall between ω and ω*2?
{w+1, w+2, w+3, ...,}
What size has the interval between N*2 and ω*2?
N*2 is not a number, so there is no interval between it and w*2
Le 12/04/2024 à 15:56, Tom Bola a écrit :
WM schrieb:
Consider the set {1, 2, 3, ..., ω} and multiply every element by 2 with >>> the result {2, 4, 6, ..., ω*2}. What elements fall between ω and ω*2?
{w+1, w+2, w+3, ...,}
No, all elements emergeing from doubling have larger distances than 1.
What size has the interval between N*2 and ω*2?
N*2 is not a number, so there is no interval between it and w*2
N*2 is a set having elements but not including w*2. So there is a
distance.
On 4/12/24 9:45 AM, WM wrote:
There are no "Natural Numbers" with less than ℵo successors.
Le 12/04/2024 à 15:56, Tom Bola a écrit :
WM schrieb:
Consider the set {1, 2, 3, ..., ω} and multiply every element by 2 with >>> the result {2, 4, 6, ..., ω*2}. What elements fall between ω and ω*2?
{w+1, w+2, w+3, ...}
No, all elements emergeing from doubling have larger distances than 1.
What size has the interval between N*2 and ω*2?
N*2 is not a number, so there is no interval between it and w*2
N*2 is a set having elements but not including w*2. So there is a
distance.
WM schrieb:
Le 12/04/2024 à 15:56, Tom Bola a écrit :
WM schrieb:
Consider the set {1, 2, 3, ..., ω} and multiply every element by 2 with >>>> the result {2, 4, 6, ..., ω*2}. What elements fall between ω and ω*2? >>>{w+1, w+2, w+3, ...}
No, all elements emergeing from doubling have larger distances than 1.
What size has the interval between N*2 and ω*2?
N*2 is not a number, so there is no interval between it and w*2
N*2 is a set having elements but not including w*2. So there is a
distance.
This is wrong because there is a distance to any element of that set.
But you probably are meaning the distance between the set limit of IN which is w and w*2
Le 12/04/2024 à 16:40, Tom Bola a écrit :
WM schrieb:
Le 12/04/2024 à 15:56, Tom Bola a écrit :
WM schrieb:
Consider the set {1, 2, 3, ..., ω} and multiply every element by 2
with the result {2, 4, 6, ..., ω*2}. What elements fall between ω
and ω*2?
{w+1, w+2, w+3, ...}
No, all elements emergeing from doubling have larger distances than 1.
What size has the interval between N*2 and ω*2?
N*2 is not a number, so there is no interval between it and w*2
N*2 is a set having elements but not including w*2. So there is a
distance.
This is wrong because there is a distance to any element of that set.
But you probably are meaning the distance between the set limit of IN
which is w and w*2
I am meaning the distance between N*2 and ω*2 after multiplication.
Before, the distance between N and ω is zero.
Regards, WM
Le 12/04/2024 à 15:58, Richard Damon a écrit :
On 4/12/24 9:40 AM, WM wrote:
Le 12/04/2024 à 00:32, Richard Damon a écrit :
On 4/11/24 8:03 AM, WM wrote:
Le 11/04/2024 à 01:04, Richard Damon a écrit :
Except that there *IS NO* Natural Number with less than ℵo
successors,
That is wrong if/because *all* natural numbers have no successors.
{1, 2, 3, ...} there are no successors. But every useable number has
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo successors.
Why do you say there are no succesors. As you say, for EVERY Natural
Number there are ℵo successors
No, only for every definable number. Between ℕ and ω there is nothing.
Even for all the numbers in the ..., as it NEVER ENDS and thus EVERY
value in it still has ℵo successors.
Beyond ℕ there is none.
How do you expect to index an infihite set with finite operations.
I don't. Matheologians do.
Then why do you talk about a set of Number you "don't believe" in?
I believe in the complete set which ends before ω.
You can't talk about "Cantor" and his bijections, if you don't accept
the rules he discussed his theories in.
He erred.
Regards, WM
Le 12/04/2024 à 16:40, Tom Bola a écrit :
WM schrieb:
Le 12/04/2024 à 15:56, Tom Bola a écrit :
WM schrieb:
Consider the set {1, 2, 3, ..., ω} and multiply every element by 2 with >>>>> the result {2, 4, 6, ..., ω*2}. What elements fall between ω and ω*2? >>>>{w+1, w+2, w+3, ...}
No, all elements emergeing from doubling have larger distances than 1.
What size has the interval between N*2 and ω*2?
N*2 is not a number, so there is no interval between it and w*2
N*2 is a set having elements but not including w*2. So there is a
distance.
This is wrong because there is a distance to any element of that set.
But you probably are meaning the distance between the set limit of IN which >> is w and w*2
I am meaning the distance between N*2 and ω*2 after multiplication.
Le 12/04/2024 à 16:02, Richard Damon a écrit :
On 4/12/24 9:45 AM, WM wrote:
There are no "Natural Numbers" with less than ℵo successors.
Self-contradiction: These successors are natural numbers but can't be
used because they can't be used up.
Regards, WM
WM schrieb:
Le 12/04/2024 à 16:40, Tom Bola a écrit :
WM schrieb:
Le 12/04/2024 à 15:56, Tom Bola a écrit :
WM schrieb:
Consider the set {1, 2, 3, ..., ω} and multiply every element by 2 with >>>>>> the result {2, 4, 6, ..., ω*2}. What elements fall between ω and ω*2? >>>>>{w+1, w+2, w+3, ...}
No, all elements emergeing from doubling have larger distances than 1. >>>>>
What size has the interval between N*2 and ω*2?
N*2 is not a number, so there is no interval between it and w*2
N*2 is a set having elements but not including w*2. So there is a
distance.
This is wrong because there is a distance to any element of that set.
But you probably are meaning the distance between the set limit of IN which >>> is w and w*2
I am meaning the distance between N*2 and ω*2 after multiplication.
Yes, that is the set after multiplication:
{0, 2, 4, 6, ..., w, w+1, w+2, w+3, ..., w*2}
On 4/12/24 10:44 AM, WM wrote:
"Distance" between a set and a value is not a defined term.
Le 12/04/2024 à 16:57, Richard Damon a écrit :
On 4/12/24 10:44 AM, WM wrote:
"Distance" between a set and a value is not a defined term.
You are wrong. Distance between (0, 10] and 11 is ?
Regards, WM
Le 12/04/2024 à 16:57, Richard Damon a écrit :
On 4/12/24 10:44 AM, WM wrote:
"Distance" between a set and a value is not a defined term.
You are wrong. Distance between (0, 10] and 11 is ?
Regards, WM
Le 12/04/2024 à 17:00, Tom Bola a écrit :
WM schrieb:
Le 12/04/2024 à 16:40, Tom Bola a écrit :
WM schrieb:
Le 12/04/2024 à 15:56, Tom Bola a écrit :
WM schrieb:
Consider the set {1, 2, 3, ..., ω} and multiply every element by 2 with
the result {2, 4, 6, ..., ω*2}. What elements fall between ω and ω*2?
{w+1, w+2, w+3, ...}
No, all elements emergeing from doubling have larger distances than 1. >>>>>>
What size has the interval between N*2 and ω*2?
N*2 is not a number, so there is no interval between it and w*2
N*2 is a set having elements but not including w*2. So there is a
distance.
This is wrong because there is a distance to any element of that set.
But you probably are meaning the distance between the set limit of IN which
is w and w*2
I am meaning the distance between N*2 and ω*2 after multiplication.
Yes, that is the set after multiplication:
{0, 2, 4, 6, ..., w, w+1, w+2, w+3, ..., w*2}
Why are the distances below ω 2 but beyond ω 1?
On 4/12/24 11:22 AM, WM wrote:
Le 12/04/2024 à 16:57, Richard Damon a écrit :
On 4/12/24 10:44 AM, WM wrote:
"Distance" between a set and a value is not a defined term.
You are wrong. Distance between (0, 10] and 11 is ?
Not a defined term for NUMBERS.
WM schrieb:
Le 12/04/2024 à 17:00, Tom Bola a écrit :
WM schrieb:
Le 12/04/2024 à 16:40, Tom Bola a écrit :
WM schrieb:
Le 12/04/2024 à 15:56, Tom Bola a écrit :
WM schrieb:
Consider the set {1, 2, 3, ..., ω} and multiply every element by 2 with
the result {2, 4, 6, ..., ω*2}. What elements fall between ω and ω*2?
{w+1, w+2, w+3, ...}
No, all elements emergeing from doubling have larger distances than 1. >>>>>>>
What size has the interval between N*2 and ω*2?
N*2 is not a number, so there is no interval between it and w*2
N*2 is a set having elements but not including w*2. So there is a
distance.
This is wrong because there is a distance to any element of that set. >>>>> But you probably are meaning the distance between the set limit of IN which
is w and w*2
I am meaning the distance between N*2 and ω*2 after multiplication.
Yes, that is the set after multiplication:
{0, 2, 4, 6, ..., w, w+1, w+2, w+3, ..., w*2}
Why are the distances below ω 2 but beyond ω 1?
This is the union of the image from IN under f(n)=2n and
the "elements fall between ω and ω*2" that you wanted above
The image of IN under f(n)=2n and w is still {0, 2, 4, 6, ..., w*2}
Le 12/04/2024 à 17:47, Richard Damon a écrit :
On 4/12/24 11:22 AM, WM wrote:
Le 12/04/2024 à 16:57, Richard Damon a écrit :
On 4/12/24 10:44 AM, WM wrote:
"Distance" between a set and a value is not a defined term.
You are wrong. Distance between (0, 10] and 11 is ?
Not a defined term for NUMBERS.
It is 1 in geometry, in mathematics, and in set theory: The minimum
length to bridge.
Regards, WM
Le 12/04/2024 à 17:51, Tom Bola a écrit :
WM schrieb:
Le 12/04/2024 à 17:00, Tom Bola a écrit :
WM schrieb:
Le 12/04/2024 à 16:40, Tom Bola a écrit :
WM schrieb:
Le 12/04/2024 à 15:56, Tom Bola a écrit :
WM schrieb:
Consider the set {1, 2, 3, ..., ω} and multiply every element by 2 with
the result {2, 4, 6, ..., ω*2}. What elements fall between ω and ω*2?
{w+1, w+2, w+3, ...}
No, all elements emergeing from doubling have larger distances than 1. >>>>>>>>
N*2 is a set having elements but not including w*2. So there is a >>>>>>> distance.What size has the interval between N*2 and ω*2?
N*2 is not a number, so there is no interval between it and w*2 >>>>>>>
This is wrong because there is a distance to any element of that set. >>>>>> But you probably are meaning the distance between the set limit of IN which
is w and w*2
I am meaning the distance between N*2 and ω*2 after multiplication.
Yes, that is the set after multiplication:
{0, 2, 4, 6, ..., w, w+1, w+2, w+3, ..., w*2}
Why are the distances below ω 2 but beyond ω 1?
This is the union of the image from IN under f(n)=2n and
the "elements fall between ω and ω*2" that you wanted above
I wanted the image of 1, 2, 3, ..., ω under multiplication by 2.
The image of IN under f(n)=2n and w is still {0, 2, 4, 6, ..., w*2}
Yes, but where is ω in this sequence?
Le 11/04/2024 à 19:36, Jim Burns a écrit :
On 4/11/2024 8:07 AM, WM wrote:
Which ones were unused by e = 2*n?
Those of {1, 2, 3, ...} with less than ℵo successors.
{1,2,3,…} is
the least.upper.bound of all finiteⁿᵒᵗᐧᵂᴹ.{1,2,3,…,k}
Yes,
but many natnumbers will never be used.
For each m ∈ {1,2,3,…}:
( for each k ∈ {1,2,3,…}
m+k is a successor of m in {1,2,3,…} )
Consider the set {1, 2, 3, ..., ω}
and multiply every element by 2
with the result {2, 4, 6, ..., ω*2}.
What elements fall between ω and ω*2?
What size has the interval between ℕ*2 and ω*2?
If you were right, then
no elements would fall between ω and ω*2
but all new elements
(larger than all in {1,2,3,…})
would stay in {1,2,3,…}
while the intervall between ω and ω*2
would be infinite and empty.
On 4/12/2024 7:40 AM, WM wrote:
Le 12/04/2024 à 16:02, Richard Damon a écrit :
On 4/12/24 9:45 AM, WM wrote:
There are no "Natural Numbers" with less than ℵo successors.
Self-contradiction: These successors are natural numbers but can't be
used because they can't be used up.
How many successors does say, 42 have?
Infinite!
How many predecessors does 3 have?
...
?
This is a finite view of an infinite process. As soon as I say three, I
am in a finite realm, so to speak.
On 4/12/2024 6:56 AM, Tom Bola wrote:
WM schrieb:
Le 11/04/2024 à 19:36, Jim Burns a écrit :
On 4/11/2024 8:07 AM, WM wrote:
Which ones were unused by e = 2*n?
Those of {1, 2, 3, ...} with less than ℵo successors.
{1,2,3,…} is
the least.upper.bound of all finiteⁿᵒᵗᐧᵂᴹ.{1,2,3,…,k}
Yes, but many natnumbers will never be used.
For each m ∈ {1,2,3,…}:
( for each k ∈ {1,2,3,…}
m+k is a successor of m in {1,2,3,…} )
Consider the set {1, 2, 3, ..., ω} and multiply every element by 2 with >>> the result {2, 4, 6, ..., ω*2}. What elements fall between ω and ω*2?
{w+1, w+2, w+3, ...,}
What size has the interval between N*2 and ω*2?
N*2 is not a number, so there is no interval between it and w*2
Why does WM _continue_ to think that infinity is some "really" huge
number? We have tried to help him out with this...
Le 12/04/2024 à 18:58, Richard Damon a écrit :
No, "Set Theory" doesn't talk about "differences" based on the values
of the elements, because that is outside of the domain of Set Theory.
Learn about ordered sets and well-ordered sets. For a start look here https://en.wikipedia.org/wiki/Partially_ordered_set.
Regrads, WM
Le 12/04/2024 à 20:57, Jim Burns a écrit :
On 4/12/2024 9:18 AM, WM wrote:
usedᵂᴹ or not.usedᵂᴹ
not.exists
All individually usable numbers satisfy
∀n ∈ ℕ_ind: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
All collectively useable numbers satisfy
|ℕ \ {1, 2, 3, ...}| = 0
What elements fall between ω and ω*2?
ω+1 ω+2 ...
What elements of {1, 2, 3, ..., ω}*2 fall between ω and ω*2?
Their distances must be 2.
What size has the interval between ℕ*2 and ω*2?
I take you to be asking how many ordinals are
after each of {2,4,6,...} and before ω+ω
No, the question is: What elements of {1, 2, 3, ..., ω}*2 fall between ω and ω*2?
No doubled ordinal is between ω and ω+ω
Then {1, 2, 3, ..., ω}*2 increases the gap between {1, 2, 3, ...} and ω from 0 to infinity. Very bumpy. Only a factor 2 is possible when doubling.
Regards, WM
On 4/12/2024 9:18 AM, WM wrote:
usedᵂᴹ or not.usedᵂᴹ
not.exists
What elements fall between ω and ω*2?
ω+1 ω+2 ...
What size has the interval between ℕ*2 and ω*2?
I take you to be asking how many ordinals are
after each of {2,4,6,...} and before ω+ω
No doubled ordinal is between ω and ω+ω
No, "Set Theory" doesn't talk about "differences" based on the values of
the elements, because that is outside of the domain of Set Theory.
Le 12/04/2024 à 18:58, Richard Damon a écrit :
No, "Set Theory" doesn't talk about "differences" based on the values
of the elements, because that is outside of the domain of Set Theory.
Learn about ordered sets and well-ordered sets. For a start look here https://en.wikipedia.org/wiki/Partially_ordered_set. Regrads, WM
Le 12/04/2024 à 20:57, Jim Burns a écrit :
usedᵂᴹ or not.usedᵂᴹ
not.exists
k, n in {1,2,3,…} and
k⁺¹ not.in {1,2,3,…} or
n⁺¹ not.in {1,2,3,…} or
k+n not.in {1,2,3,…} or
k⋅n not.in {1,2,3,…}
All individually usable numbers satisfy
∀n ∈ ℕ_ind: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
All collectively useable numbers satisfy
|ℕ \ {1, 2, 3, ...}| = 0
What elements fall between ω and ω*2?
ω+1 ω+2 ...
What elements of {1, 2, 3, ..., ω}*2
fall between ω and ω*2?
Their distances must be 2.
What elements of {1, 2, 3, ..., ω}*2
fall between ω and ω*2?
On 4/13/24 8:23 AM, WM wrote:
Le 12/04/2024 à 18:58, Richard Damon a écrit :
No, "Set Theory" doesn't talk about "differences" based on the values
of the elements, because that is outside of the domain of Set Theory.
Learn about ordered sets and well-ordered sets. For a start look here
https://en.wikipedia.org/wiki/Partially_ordered_set.
Which is MATH THEORY, not SET THEORY (which is a SPECIFIC subset of
math, DIFFERENT than Order theory).
On 4/13/24 8:35 AM, WM wrote:
No, the question is: What elements of {1, 2, 3, ..., ω}*2 fall between ω >> and ω*2?
None.
On 4/13/2024 8:35 AM, WM wrote:
What elements of {1, 2, 3, ..., ω}*2
fall between ω and ω*2?
Their distances must be 2.
Why 2 ?
For example, 0-1 isn't an ordinal.
if n < ω
then 2⋅n < ω
Le 13/04/2024 à 15:05, Richard Damon a écrit :
On 4/13/24 8:35 AM, WM wrote:
No, the question is: What elements of {1, 2, 3, ..., ω}*2 fall
between ω and ω*2?
None.
That is not possible in a linear operation like doubling.
Regards, WM
Le 13/04/2024 à 14:59, Richard Damon a écrit :
On 4/13/24 8:23 AM, WM wrote:
Le 12/04/2024 à 18:58, Richard Damon a écrit :
No, "Set Theory" doesn't talk about "differences" based on the
values of the elements, because that is outside of the domain of Set
Theory.
Learn about ordered sets and well-ordered sets. For a start look here
https://en.wikipedia.org/wiki/Partially_ordered_set.
Which is MATH THEORY, not SET THEORY (which is a SPECIFIC subset of
math, DIFFERENT than Order theory).
Do you know the expression ordinal number? It is related to order. Here
are the first chapters from Cantor's fundamental essay:
Beiträge zur Begründung der transfiniten Mengenlehre ........................................ 282
§ 1. Der Mächtigkeitsbegriff oder die Kardinalzahl ........................................ 282
§ 2. Das "Größer" und "Kleiner" bei Mächtigkeiten.......................................... 284
§ 3. Die Addition und Multiplikation von Mächtigkeiten ..................................... 285
§ 4. Die Potenzierung von Mächtigkeiten ............................................... 287
VII § 5. Die endlichen Kardinalzahlen ............................................... 289
§ 6. Die kleinste transfinite Kardinalzahl Alef-null ......................................... 292
§ 7. Die Ordnungstypen einfach geordneter Mengen....................................... 296
§ 8. Addition und Multiplikation von Ordnungstypen ....................................... 301
§ 9. Der Ordnungstypus der Menge R aller rationalen Zahlen, die
größer als 0 und kleiner als 1 sind, in
ihrer natürlichen Rangordnung ............................................. 303
§ 10. Die in einer transfiniten geordneten Menge enthaltenen Fundamentalreihen .................. 307
§ 11. Der Ordnungstypus der Linearkontinuums X ...................................... 310
§ 12. Die wohlgeordneten Mengen .................................................... 312
§ 13. Die Abschnitte wohlgeordneter Mengen ............................................ 314
§ 14. Die Ordnungszahlen wohlgeordneter Mengen ......
Note that Ordnung means order.
Last warning: If you dare again to waste my time by your stupid
ignorance I will never again talk to you.
Regards, WM
Le 13/04/2024 à 21:16, Jim Burns a écrit :
if n < ω
then 2⋅n < ω
That is impossible because
doubling is a linear operation.
Le 13/04/2024 à 21:16, Jim Burns a écrit :
On 4/13/2024 8:35 AM, WM wrote:
What elements of {1, 2, 3, ..., ω}*2
fall between ω and ω*2?
Their distances must be 2.
Why 2 ?
Doubling of ordinals
Every Natural Number has aleph_0 numbers above it, one of which will be double it.
Thus, doubling ALL the Natural Numbers still gives you
results that are all Natural Numbers.
WM schrieb:
Le 13/04/2024 à 21:16, Jim Burns a écrit :
On 4/13/2024 8:35 AM, WM wrote:
What elements of {1, 2, 3, ..., ω}*2
fall between ω and ω*2?
Their distances must be 2.
Why 2 ?
Doubling of ordinals
The smallest ordinals behind all n in IN possible
would be the w's in something like
{0, 1, 2, 3, ..., w, w+1, w+2, w+3, ... w+w}
which, under f(n)=n*2, has the image
{0, 2, 4, 6, ..., w*2, w*2+2, w*2+4, w*2+4, ... w*2+w*2}
On 4/14/2024 3:12 PM, WM wrote:
Le 13/04/2024 à 21:16, Jim Burns a écrit :
if n < ω
then 2⋅n < ω
That is impossible because
doubling is a linear operation.
You (WM) have decided that
ω is like all the numbers n < ω
Whatever it might mean
to put ω and 1 on the same line,
If n is a number
different.in.size from its nearest.neighbors,
then 2⋅n is a number
different.in.size from its nearest.neighbors.
If n is a number less than
the least.upper.bound of numbers
different.in.size from their nearest.neighbors,
then 2⋅n is a number less than
the least.upper.bound of numbers
different.in.size from their nearest.neighbors.
If n < ω
then 2⋅n < ω
Le 14/04/2024 à 22:17, Tom Bola a écrit :
WM schrieb:
Le 13/04/2024 à 21:16, Jim Burns a écrit :
On 4/13/2024 8:35 AM, WM wrote:
What elements of {1, 2, 3, ..., ω}*2
fall between ω and ω*2?
Their distances must be 2.
Why 2 ?
Doubling of ordinals
The smallest ordinals behind all n in IN possible
would be the w's in something like
{0, 1, 2, 3, ..., w, w+1, w+2, w+3, ... w+w}
which, under f(n)=n*2, has the image
{0, 2, 4, 6, ..., w*2, w*2+2, w*2+4, w*2+4, ... w*2+w*2}
Right! But ω remains like 6 remains.
Doubling all cannot be
absorbed by existing natnumbers.
On 4/15/2024 6:00 AM, Tom Bola wrote:
WM drivels:
Le 14/04/2024 à 22:17, Tom Bola a écrit :
WM schrieb:
Le 13/04/2024 à 21:16, Jim Burns a écrit :
On 4/13/2024 8:35 AM, WM wrote:
What elements of {1, 2, 3, ..., ω}*2
fall between ω and ω*2?
Their distances must be 2.
Why 2 ?
Doubling of ordinals
The smallest ordinals behind all n in IN possible
would be the w's in something like
{0, 1, 2, 3, ..., w, w+1, w+2, w+3, ... w+w}
which, under f(n)=n*2, has the image
{0, 2, 4, 6, ..., w*2, w*2+2, w*2+4, w*2+4, ... w*2+w*2}
Right! But ω remains like 6 remains.
No - w in the domain is related by the function to w*2,
same as every other element in the domain:
0, 1, 2, 3, ..., w, w+1, w+2, w+3, ... w+w
| | | | ||| | | | | |
0, 2, 4, 6, ..., w*2, w*2+2, w*2+4, w*2+6, ... w*2+w*2
Anyhow - your "doubling" idea is childish and idiotic bullshit, as always.
Ditto!
Le 14/04/2024 à 22:39, Jim Burns a écrit :
If n is a number
different.in.size from its nearest.neighbors,
then 2⋅n is a number
different.in.size from its nearest.neighbors.
If n is a number less than
the least.upper.bound of numbers
different.in.size from their nearest.neighbors,
then 2⋅n is a number less than
the least.upper.bound of numbers
different.in.size from their nearest.neighbors.
That is wrong if
all natnumbers are present already such that
no further natnumbers fits below ω.
Le 14/04/2024 à 22:39, Jim Burns a écrit :
On 4/14/2024 3:12 PM, WM wrote:
Le 13/04/2024 à 21:16, Jim Burns a écrit :
if n < ω
then 2⋅n < ω
That is impossible because
doubling is a linear operation.
You (WM) have decided that
ω is like all the numbers n < ω
Cantor has decided, that ω is an ordinal which can be counted and passed
by counting (Hilbert).
Whatever it might mean
to put ω and 1 on the same line,
It is Cantor's number classes. See Transfinity p. 42.
If n is a number
different.in.size from its nearest.neighbors,
then 2*n is a number
different.in.size from its nearest.neighbors.
If n is a number less than
the least.upper.bound of numbers
different.in.size from their nearest.neighbors,
then 2*n is a number less than
the least.upper.bound of numbers
different.in.size from their nearest.neighbors.
That is wrong if all natnumbers are present already such that no further natnumbers fits below ω.
If n < ω
then 2*n < ω
Le 14/04/2024 à 21:23, Richard Damon a écrit :
Every Natural Number has aleph_0 numbers above it, one of which will
be double it.
All natnumbers have no natnumbers following them. Doubling all cannot be absorbed by existing natnumbers.
Thus, doubling ALL the Natural Numbers still gives you results that
are all Natural Numbers.
Impossible, since all are doubled.
Regards, WM
On 4/15/2024 1:13 PM, Tom Bola wrote:
Chris M. Thomasson drivels:
On 4/15/2024 6:00 AM, Tom Bola wrote:
WM drivels:Ditto!
Le 14/04/2024 à 22:17, Tom Bola a écrit :
WM schrieb:
Le 13/04/2024 à 21:16, Jim Burns a écrit :
On 4/13/2024 8:35 AM, WM wrote:
What elements of {1, 2, 3, ..., ω}*2
fall between ω and ω*2?
Their distances must be 2.
Why 2 ?
Doubling of ordinals
The smallest ordinals behind all n in IN possible
would be the w's in something like
{0, 1, 2, 3, ..., w, w+1, w+2, w+3, ... w+w}
which, under f(n)=n*2, has the image
{0, 2, 4, 6, ..., w*2, w*2+2, w*2+4, w*2+4, ... w*2+w*2}
Right! But ω remains like 6 remains.
No - w in the domain is related by the function to w*2,
same as every other element in the domain:
0, 1, 2, 3, ..., w, w+1, w+2, w+3, ... w+w
| | | | ||| | | | | |
0, 2, 4, 6, ..., w*2, w*2+2, w*2+4, w*2+6, ... w*2+w*2
Anyhow - your "doubling" idea is childish and idiotic bullshit, as always. >>>
ROFL - I haven't got any "doubling" idea for a list of the IN u {w}.
Where was I disagreeing with you?
WM drivels:
Le 14/04/2024 à 22:17, Tom Bola a écrit :
WM schrieb:
Le 13/04/2024 à 21:16, Jim Burns a écrit :
On 4/13/2024 8:35 AM, WM wrote:
What elements of {1, 2, 3, ..., ω}*2
fall between ω and ω*2?
Their distances must be 2.
Why 2 ?
Doubling of ordinals
The smallest ordinals behind all n in IN possible
would be the w's in something like
{0, 1, 2, 3, ..., w, w+1, w+2, w+3, ... w+w}
which, under f(n)=n*2, has the image
{0, 2, 4, 6, ..., w*2, w*2+2, w*2+4, w*2+4, ... w*2+w*2}
Right! But ω remains like 6 remains.
No - w in the domain is related by the function to w*2,
same as every other element in the domain:
0, 1, 2, 3, ..., w, w+1, w+2, w+3, ... w+w
| | | | ||| | | | | |
0, 2, 4, 6, ..., w*2, w*2+2, w*2+4, w*2+6, ... w*2+w*2
Anyhow - your "doubling" idea is childish and idiotic bullshit, as always.
WM
Doubling all cannot be
absorbed by existing natnumbers.
There is no "doubling" since functions are relations between 2 fixed sets.
0, 1, 2, 3, ..., w,
| | | | ||| |
0, 2, 4, 6, ..., w*2,
With Dedekind, all infinite sets like IN are having proper subsets and have therefore real supersets so there is any required "place" in infinite sets.
On 4/15/2024 7:59 AM, WM wrote:
If
[0,n) is different.in.size
from [0,n+1) and [0,n-1)
then
[0,n+n) is different.in.size
from [0,n+n+1) and [0,n+n-1)
Would you (WM) like me to explain that to you?
Name a Natural Number that can not be doubled!!!
Thus, doubling ALL the Natural Numbers still gives you results that
are all Natural Numbers.
Impossible, since all are doubled.
POSSIBLE!!!!
That is the "miracle" of INFINITY.
On 15/04/2024 14:00, Tom Bola wrote:
0, 1, 2, 3, ..., w, w+1, w+2, w+3, ... w+w
| | | | ||| | | | | |
0, 2, 4, 6, ..., w*2, w*2+2, w*2+4, w*2+6, ... w*2+w*2
Should be:
0, 1, 2, 3, ..., w, w+1, w+2, w+3, ... w+w
| | | | ||| | | | | |
0, 2, 4, 6, ..., w*2, w*2+1, w*2+2, w*2+3, ... w*2+w*2
Anyhow - your "doubling" idea is childish and idiotic bullshit, as always. >>
..of course. :)
Le 15/04/2024 à 15:00, Tom Bola a écrit :
0, 1, 2, 3, ..., w,Where is the position of ω in the lower row?
| | | | ||| |
0, 2, 4, 6, ..., w*2,
Le 15/04/2024 à 23:06, Jim Burns a écrit :
On 4/15/2024 7:59 AM, WM wrote:
If
[0,n) is different.in.size
from [0,n+1) and [0,n-1)
then
[0,n+n) is different.in.size
from [0,n+n+1) and [0,n+n-1)
Would you (WM) like me to explain that to you?
It is enough if you explain where ω is in the lower line:
0, 1, 2, 3, ..., w
| | | | ||| |
0, 2, 4, 6, ..., w*2
Le 16/04/2024 à 02:35, Mike Terry a écrit :
On 15/04/2024 14:00, Tom Bola wrote:
0, 1, 2, 3, ..., w, w+1, w+2, w+3, ... w+w
| | | | ||| | | | | |
0, 2, 4, 6, ..., w*2, w*2+2, w*2+4, w*2+6, ... w*2+w*2
Should be:
0, 1, 2, 3, ..., w, w+1, w+2, w+3, ... w+w
| | | | ||| | | | | |
0, 2, 4, 6, ..., w*2, w*2+1, w*2+2, w*2+3, ... w*2+w*2
No. (ω+1)*2 = ω*2 + 2
Anyhow - your "doubling" idea is childish and idiotic bullshit, as always. >>>
..of course. :)
What ordinal number cannot be doubled? Where is ω in the lowwr row?
Le 16/04/2024 à 02:35, Mike Terry a écrit :
On 15/04/2024 14:00, Tom Bola wrote:
0, 1, 2, 3, ..., w, w+1, w+2, w+3, ... w+w
| | | | ||| | | | | |
0, 2, 4, 6, ..., w*2, w*2+2, w*2+4, w*2+6, ... w*2+w*2
Should be:
0, 1, 2, 3, ..., w, w+1, w+2, w+3, ... w+w
| | | | ||| | | | | |
0, 2, 4, 6, ..., w*2, w*2+1, w*2+2, w*2+3, ... w*2+w*2
No. (ω+1)*2 = ω*2 + 2
0, 1, 2, 3, ..., w,Where is the position of ω in the lower row?
| | | | ||| |
0, 2, 4, 6, ..., w*2,
w is not in the image = { 0, 2, 4, 6, ..., w*2, ... w*2+w*2 }
The same with the members of { 1, 2, 3, ... } who aren't in the image.
On 16/04/2024 16:05, WM wrote:
Le 16/04/2024 à 02:35, Mike Terry a écrit :
On 15/04/2024 14:00, Tom Bola wrote:
0, 1, 2, 3, ..., w, w+1, w+2, w+3, ... w+w
| | | | ||| | | |
| |
0, 2, 4, 6, ..., w*2, w*2+2, w*2+4, w*2+6, ... w*2+w*2
Should be:
0, 1, 2, 3, ..., w, w+1, w+2, w+3, ... w+w
| | | | ||| | | |
| |
0, 2, 4, 6, ..., w*2, w*2+1, w*2+2, w*2+3, ... w*2+w*2
No. (ω+1)*2 = ω*2 + 2
No, you need to learn how ordinal arithmetic works:
Le 16/04/2024 à 18:09, Tom Bola a écrit :
0, 1, 2, 3, ..., w,Where is the position of ω in the lower row?
| | | | ||| |
0, 2, 4, 6, ..., w*2,
w is not in the image = { 0, 2, 4, 6, ..., w*2, ... w*2+w*2 }
But ω is smaller than w*2
The same with the members of { 1, 2, 3, ... } who aren't in the image.
2 is. But also the places of all odd ordinals can be defined by all
smaller and all larger even ordinals.
Le 16/04/2024 à 18:38, Mike Terry a écrit :
On 16/04/2024 16:05, WM wrote:I see: 1 + ω = ω.
Le 16/04/2024 à 02:35, Mike Terry a écrit :
On 15/04/2024 14:00, Tom Bola wrote:
0, 1, 2, 3, ..., w, w+1, w+2, w+3, ... w+w
| | | | ||| | | |
| |
0, 2, 4, 6, ..., w*2, w*2+2, w*2+4, w*2+6, ... w*2+w*2
Should be:
0, 1, 2, 3, ..., w, w+1, w+2, w+3, ... w+w
| | | | ||| | | |
| |
0, 2, 4, 6, ..., w*2, w*2+1, w*2+2, w*2+3, ... w*2+w*2
No. (ω+1)*2 = ω*2 + 2
No, you need to learn how ordinal arithmetic works:
Nevertheless the question remains where in the second row is ω located, doesn't it?.
Le 16/04/2024 à 18:38, Mike Terry a écrit :
On 16/04/2024 16:05, WM wrote:I see: 1 + ω = ω.
Le 16/04/2024 à 02:35, Mike Terry a écrit :
On 15/04/2024 14:00, Tom Bola wrote:
0, 1, 2, 3, ..., w, w+1, w+2, w+3, ... w+w
| | | | ||| | | | | |
0, 2, 4, 6, ..., w*2, w*2+2, w*2+4, w*2+6, ... w*2+w*2
Should be:
0, 1, 2, 3, ..., w, w+1, w+2, w+3, ... w+w
| | | | ||| | | | | |
0, 2, 4, 6, ..., w*2, w*2+1, w*2+2, w*2+3, ... w*2+w*2
No. (ω+1)*2 = ω*2 + 2
No, you need to learn how ordinal arithmetic works:
Nevertheless the question remains where in the second row is ω located, doesn't it?.
Regards, WM
Regards, WM
Le 16/04/2024 à 00:38, Richard Damon a écrit :
Name a Natural Number that can not be doubled!!!All nameable numbers belong to a potentially infinite collection, a very small initial segment of ℕ.
Thus, doubling ALL the Natural Numbers still gives you results that
are all Natural Numbers.
Impossible, since all are doubled.
POSSIBLE!!!!
That is the "miracle" of INFINITY.
No, it is the incongruence of thinking.
For potential infinity it is true. But for actual infinity it is wrong.
Regards, WM
WM schrieb:
Nevertheless the question remains where in the second row is ω located,
doesn't it?.
NOPE - because w is not in the IMAGE of your f(ord) = 2*ord
On 4/16/24 10:59 AM, WM wrote:
Note, no "Natural Number" is actually infinite, that distinction falls
on omega. But the SET of the Natural Numbers is Actually Infinite in
Size, having a size of Aleph_0.
Does that mean the Natural Numbers themselves are individually only "Potentially Infinite" but the set of them is "Actually Infinite" by
your definitions?
WM schrieb:
The same with the members of { 1, 2, 3, ... } who aren't in the image.
2 is. But also the places of all odd ordinals can be defined by all
smaller and all larger even ordinals.
Nein, sondern es ist das gleiche wie oben: 0,1,2,3,... sind im Urbild
ABER 0,2,4,6,... sind im Bild deiner Funktion.
On 4/16/24 2:36 PM, WM wrote:
Le 16/04/2024 à 18:38, Mike Terry a écrit :Why does it need to be there?
On 16/04/2024 16:05, WM wrote:I see: 1 + ω = ω.
Le 16/04/2024 à 02:35, Mike Terry a écrit :
On 15/04/2024 14:00, Tom Bola wrote:
0, 1, 2, 3, ..., w, w+1, w+2, w+3, ... w+w >>>>>> | | | | ||| | | | | |
0, 2, 4, 6, ..., w*2, w*2+2, w*2+4, w*2+6, ... w*2+w*2
Should be:
0, 1, 2, 3, ..., w, w+1, w+2, w+3, ... w+w
| | | | ||| | | | | |
0, 2, 4, 6, ..., w*2, w*2+1, w*2+2, w*2+3, ... w*2+w*2
No. (ω+1)*2 = ω*2 + 2
No, you need to learn how ordinal arithmetic works:
Nevertheless the question remains where in the second row is ω located,
doesn't it?.
THe two set/series are of the same size even though the bottom misses
the "odd" values.
Le 16/04/2024 à 21:48, Tom Bola a écrit :
WM schrieb:
ω*2 is present. Therefore ω or the ordinals next to it must be localized below.Nevertheless the question remains where in the second row is ω located, >>> doesn't it?.
NOPE - because w is not in the IMAGE of your f(ord) = 2*ord
WM drivels bullshit again and again, as always:
Le 16/04/2024 à 21:48, Tom Bola a écrit :
WM schrieb:ω*2 is present. Therefore ω or the ordinals next to it must be localized >> below.
Nevertheless the question remains where in the second row is ω located, >>>> doesn't it?.
NOPE - because w is not in the IMAGE of your f(ord) = 2*ord
Also, 2, 4, 6, ... are present in the image but not 1, 2, 3, ...
Le 17/04/2024 à 21:38, Tom Bola a écrit :
WM drivels bullshit again and again, as always:
Le 16/04/2024 à 21:48, Tom Bola a écrit :
WM schrieb:ω*2 is present. Therefore ω or the ordinals next to it must be localized >>> below.
Nevertheless the question remains where in the second row is ω located, >>>>> doesn't it?.
NOPE - because w is not in the IMAGE of your f(ord) = 2*ord
Also, 2, 4, 6, ... are present in the image but not 1, 2, 3, ...
... there is no ordinal between IN and ω.
...
Le 17/04/2024 à 01:27, Richard Damon a écrit :
On 4/16/24 10:59 AM, WM wrote:
Note, no "Natural Number" is actually infinite, that distinction falls
on omega. But the SET of the Natural Numbers is Actually Infinite in
Size, having a size of Aleph_0.
The size of ℕ is |ω|. That means ℕ extends on the ordinal line from 0 to ω. By multiplying every natural number the extendion is doubled. That is mathematics. Every contrary opinion is foolish.
Does that mean the Natural Numbers themselves are individually only
"Potentially Infinite" but the set of them is "Actually Infinite" by
your definitions?
The visible natural numbers are potentially infinite. The set ℕ is
assumed to be actualy infinite. This cannot be known let alone be
proven. But we can assume it and draw conclusions. One of them is that nothing fits between all natural numbers and ω.
Regards, WM
Le 15/04/2024 à 23:06, Jim Burns a écrit :
On 4/15/2024 7:59 AM, WM wrote:
That is wrong if
all natnumbers are present already such that
no further natnumbers fits below ω.
If
[0,n) is different.in.size
from [0,n+1) and [0,n-1)
then
[0,n+n) is different.in.size
from [0,n+n+1) and [0,n+n-1)
Would you (WM) like me to explain that to you?
It is enough if you explain
where ω is in the lower line:
0, 1, 2, 3, ..., w
| | | | ||| | 0, 2, 4, 6, ..., w*2
By doubling the number of elements remains the same, but the populated interval is (0, ω2) with ω amidst.
On 4/17/24 2:49 PM, WM wrote:
Le 17/04/2024 à 01:27, Richard Damon a écrit :
On 4/16/24 10:59 AM, WM wrote:
Note, no "Natural Number" is actually infinite, that distinction falls
on omega. But the SET of the Natural Numbers is Actually Infinite in
Size, having a size of Aleph_0.
The size of ℕ is |ω|. That means ℕ extends on the ordinal line from 0 to
ω. By multiplying every natural number the extendion is doubled. That is
mathematics. Every contrary opinion is foolish.
Nope, the size of ℕ, that is |ℕ| is aleph_0. ω is an ORDINAL number (showing order), not a cardinal number (showing size). This means that
the first transfinite ordinal above the set ℕ would be the value ω. So yes, as you say below, there are no "finite" numbers between the set ℕ
and the value ω, but since ℕ has no "highest" member there is no "predecessor" to ω, just as there is no predecessor to 0 in the Natural Numbers.
Note, there are some extended transfinite systems that do put
transfinite numbers in that gap. (just like the Rationals put numbers
between 0 and 1 which are consecutive in the Natural Numbers).
Does that mean the Natural Numbers themselves are individually only
"Potentially Infinite" but the set of them is "Actually Infinite" by
your definitions?
The visible natural numbers are potentially infinite. The set ℕ is
assumed to be actualy infinite. This cannot be known let alone be
proven. But we can assume it and draw conclusions. One of them is that
nothing fits between all natural numbers and ω.
There is no defined set of "visible natural numbers" except in your own broken logic. ALL Natural numbers fit your definition of "visible", but
your logic can not handle that this set is actually infinite in size.
The set ℕ IS ACTUALLY infinite in size, as BY DEFINITION, there is no finite number that can count all its members, there is no "highest" member.
Refusing to beleive the definition, just proves that you logic system
can not handle infinite sets.
Le 17/04/2024 à 21:38, Tom Bola a écrit :
WM drivels bullshit again and again, as always:
Le 16/04/2024 à 21:48, Tom Bola a écrit :
WM schrieb:ω*2 is present. Therefore ω or the ordinals next to it must be
Nevertheless the question remains where in the second row is ω
located, doesn't it?.
NOPE - because w is not in the IMAGE of your f(ord) = 2*ord
localized below.
Also, 2, 4, 6, ... are present in the image but not 1, 2, 3, ...
If you accept set theory, then you have to accept too that there is no ordinal between ℕ and ω. The interval populated by ℕ is (0, ω). By doubling the number of elements remains the same, but the populated
interval is (0, ω2) with ω amidst.
I do not claim that ω is in the image, but it is amidst the interval.
That proves that doubled numbers surpassed it.
Regards, WM
On 4/8/24 9:44 AM, WM wrote:
Le 07/04/2024 à 19:56, Richard Damon a écrit :
On 4/7/24 9:23 AM, WM wrote:
So, With infinite sets, a proper subset CAN be the same size as
its parent.
Impossible.
Nope, PROVEN.
Proven impossble with my matrix,
Nope, since you matrix doesn't follow the required form.
All you have done is proven that YOUR logic can yield different
contractory results depending on which valid path you follow.
That means that YOUR logic system is proven INCONSISTENT, and thus
BLOWN UP.
Le 17/04/2024 à 01:29, Richard Damon a écrit :
On 4/16/24 2:36 PM, WM wrote:
Le 16/04/2024 à 18:38, Mike Terry a écrit :Why does it need to be there?
On 16/04/2024 16:05, WM wrote:I see: 1 + ω = ω.
Le 16/04/2024 à 02:35, Mike Terry a écrit :
On 15/04/2024 14:00, Tom Bola wrote:
0, 1, 2, 3, ..., w, w+1, w+2, w+3, ... w+w >>>>>>> | | | | ||| | | | | |
0, 2, 4, 6, ..., w*2, w*2+2, w*2+4, w*2+6, ... w*2+w*2
Should be:
0, 1, 2, 3, ..., w, w+1, w+2, w+3, ... w+w >>>>>> | | | | ||| | | | | |
0, 2, 4, 6, ..., w*2, w*2+1, w*2+2, w*2+3, ... w*2+w*2
No. (ω+1)*2 = ω*2 + 2
No, you need to learn how ordinal arithmetic works:
Nevertheless the question remains where in the second row is ω
located, doesn't it?.
Because ω is smaller than ω2.
THe two set/series are of the same size even though the bottom misses
the "odd" values.
The two sets are of same number of elements because every elements is doubled. But after multiplication these elements cover an interval twice
as large as before. The end is ω2. Therefore ω or its neighbours are located in the midst of the interval.
Regards, WM
On 4/17/24 2:52 PM, WM wrote:
Le 17/04/2024 à 01:29, Richard Damon a écrit :
On 4/16/24 2:36 PM, WM wrote:
Le 16/04/2024 à 18:38, Mike Terry a écrit :Why does it need to be there?
On 16/04/2024 16:05, WM wrote:I see: 1 + ω = ω.
Le 16/04/2024 à 02:35, Mike Terry a écrit :
On 15/04/2024 14:00, Tom Bola wrote:
0, 1, 2, 3, ..., w, w+1, w+2, w+3, ... w+w >>>>>>>> | | | | ||| | | | | |
0, 2, 4, 6, ..., w*2, w*2+2, w*2+4, w*2+6, ... w*2+w*2
Should be:
0, 1, 2, 3, ..., w, w+1, w+2, w+3, ... w+w >>>>>>> | | | | ||| | | | | |
0, 2, 4, 6, ..., w*2, w*2+1, w*2+2, w*2+3, ... w*2+w*2
No. (ω+1)*2 = ω*2 + 2
No, you need to learn how ordinal arithmetic works:
Nevertheless the question remains where in the second row is ω
located, doesn't it?.
Because ω is smaller than ω2.
So, the bottom row is missing all the odd natural numbers. why can't it
skip all the odd multiples of ω too?
THe two set/series are of the same size even though the bottom misses
the "odd" values.
The two sets are of same number of elements because every elements is
doubled. But after multiplication these elements cover an interval twice
as large as before. The end is ω2. Therefore ω or its neighbours are
located in the midst of the interval.
Nope, unless by "midst" you mean that one is above it and the rest below it.
Note, your original set wasn't "equally spaced" as the ω at the end is
an infinite distance from the rest of the set, so you can't use equal
spacing logic on the second set.
On 4/17/24 2:49 PM, WM wrote:
The size of ℕ is |ω|. That means ℕ extends on the ordinal line from 0 to
ω. By multiplying every natural number the extendion is doubled. That is
mathematics. Every contrary opinion is foolish.
Nope, the size of ℕ, that is |ℕ| is aleph_0. ω is an ORDINAL number
(showing order), not a cardinal number (showing size). This means that
the first transfinite ordinal above the set ℕ would be the value ω. So yes, as you say below, there are no "finite" numbers between the set ℕ
and the value ω, but since ℕ has no "highest" member there is no "predecessor" to ω, just as there is no predecessor to 0 in the Natural Numbers.
Richard Damon <[email protected]> writes:
That means that YOUR logic system is proven INCONSISTENT, and thus
BLOWN UP.
And that is where the intrigue lies.
On 4/17/24 4:30 PM, WM wrote:
I do not claim that ω is in the image, but it is amidst the interval.
That proves that doubled numbers surpassed it.
Only the ω doubled passed it. The rest stayed below ω, and no natural number doubled isn't a natural number.
On 4/16/2024 10:33 AM, WM wrote:
It is enough if you explain
where ω is in the lower line:
0, 1, 2, 3, ..., w
| | | | ||| |
0, 2, 4, 6, ..., w*2
Note, your original set wasn't "equally spaced" as the ω at the end is
an infinite distance from the rest of the set,
so you can't use equal
spacing logic on the second set.
Also, there is no spacing defined in sets (like image and preimage)
except N, Z, Q, R is used, which is not the case here because the
(ordered) collection of ordinal numbers and limit ordinals do are
neither existing nor defined on an equidistant scale (or row of points).
Le 18/04/2024 à 01:27, Phil Carmody a écrit :
Richard Damon <[email protected]> writes:
That means that YOUR logic system is proven INCONSISTENT, and thus
BLOWN UP.
And that is where the intrigue lies.
My logic system says: ω is amidst the interval (0, ω2) because in the
image of the function f(x) = x*2
Le 18/04/2024 à 01:26, Richard Damon a écrit :
On 4/17/24 4:30 PM, WM wrote:
I do not claim that ω is in the image, but it is amidst the interval.
That proves that doubled numbers surpassed it.
Only the ω doubled passed it. The rest stayed below ω, and no natural
number doubled isn't a natural number.
You are wrong. ω is amidst the interval (0, ω2) because
in the image there are as many ordinals in (ω, ω2) as in (0, ω)
Le 18/04/2024 à 01:30, Richard Damon a écrit :
Note, your original set wasn't "equally spaced" as the ω at the end is
an infinite distance from the rest of the set,
Wrong. The infinite distance between 0 and ω is realized [LOL, this is
not physics] by only natural numbers.
Note that there are infinitely many natural numbers and not the
least space between them and ω.
so you can't use equal spacing logic on the second set.
ω is amidst the interval (0, ω2)
because in the image there are as many ordinals in (ω, ω2) as in (0, ω).
Le 18/04/2024 à 01:41, Tom Bola a écrit :
Also, there is no spacing defined in sets (like image and preimage)
except N, Z, Q, R is used, which is not the case here because the
(ordered) collection of ordinal numbers and limit ordinals do are
neither existing nor defined on an equidistant scale (or row of points).
ω is amidst the interval (0, ω2) because in the image
Clown WM drivels bullshit:
Le 18/04/2024 à 01:30, Richard Damon a écrit :
Note, your original set wasn't "equally spaced" as the ω at the end is
an infinite distance from the rest of the set,
Wrong. The infinite distance between 0 and ω is realized [LOL, this is
not physics] by only natural numbers.
Note that there are infinitely many natural numbers and not the
least space between them and ω.
Idiot, there is no "space" in IN between each pairs of n and n+1, either.
You are phatasizing idiotic bullshit all day and night long.
so you can't use equal spacing logic on the second set.
But yes, our math defines that all n in IN are before w.
ω is amidst the interval (0, ω2)
because in the image there are as many ordinals in (ω, ω2) as in (0, ω).
This has nothing to do with your "doubling" bullshit because the image is {0,2,4,6,...,w*2,...} under f(0)=2*o of the domain
{0,1,2,3,...,w,...}.
Get that already, imbecile clown.
On 18/04/2024 16:29, Tom Bola wrote:
This has nothing to do with your "doubling" bullshit because the image is
{0,2,4,6,...,w*2,...} under f(o)=2*o of the domain
That should be f(o) - o*2
Le 18/04/2024 à 00:50, Jim Burns a écrit :
On 4/16/2024 10:33 AM, WM wrote:
Le 15/04/2024 à 23:06, Jim Burns a écrit :
On 4/15/2024 7:59 AM, WM wrote:
That is wrong if
all natnumbers are present already such that
no further natnumbers fits below ω.
It is enough if you explain
where ω is in the lower line:
0, 1, 2, 3, ..., w
| | | | ||| | 0, 2, 4, 6, ..., w*2
Please answer the question.
Le 18/04/2024 à 00:50, Jim Burns a écrit :
On 4/16/2024 10:33 AM, WM wrote:
Le 15/04/2024 à 23:06, Jim Burns a écrit :
On 4/15/2024 7:59 AM, WM wrote:
That is wrong if
all natnumbers are present already such that
no further natnumbers fits below ω.
It is enough if you explain
where ω is in the lower line:
0, 1, 2, 3, ..., w
| | | | ||| | 0, 2, 4, 6, ..., w*2
Please answer the question.
It is enough if you explain
where ω is in the lower line:
0, 1, 2, 3, ..., w
| | | | ||| | 0, 2, 4, 6, ..., w*2
Le 18/04/2024 à 00:59, Richard Damon a écrit :
On 4/17/24 2:49 PM, WM wrote:
The size of ℕ is |ω|. That means ℕ extends on the ordinal line from 0 >>> to ω. By multiplying every natural number the extendion is doubled.
That is mathematics. Every contrary opinion is foolish.
Nope, the size of ℕ, that is |ℕ| is aleph_0. ω is an ORDINAL number
on the ordinal axis beyond all natural numbers with nothing else before it. ℵo = |ℕ| = |ω|.
(showing order), not a cardinal number (showing size). This means that
the first transfinite ordinal above the set ℕ would be the value ω. So
yes, as you say below, there are no "finite" numbers between the set ℕ
and the value ω, but since ℕ has no "highest" member there is no
"predecessor" to ω, just as there is no predecessor to 0 in the
Natural Numbers.
Nonsense.
ω follows upon all natural numbers. There is nothing between them and ω.
Regards, WM
Le 18/04/2024 à 01:27, Phil Carmody a écrit :
Richard Damon <[email protected]> writes:
That means that YOUR logic system is proven INCONSISTENT, and thus
BLOWN UP.
And that is where the intrigue lies.
My logic system says: ω is amidst the interval (0, ω2) because in the
image of the function f(x) = x*2 there are as many ordinals in (ω, ω2)
as in (0, ω).
Regards, WM
Le 18/04/2024 à 01:26, Richard Damon a écrit :
On 4/17/24 4:30 PM, WM wrote:
I do not claim that ω is in the image, but it is amidst the interval.
That proves that doubled numbers surpassed it.
Only the ω doubled passed it. The rest stayed below ω, and no natural
number doubled isn't a natural number.
You are wrong. ω is amidst the interval (0, ω2) because in the image
there are as many ordinals in (ω, ω2) as in (0, ω).
Regards, WM
On 4/18/2024 10:54 AM, WM wrote:
It is enough if you explain
where ω is in the lower line:
∀κ < ω: k⋅2 < ω
∀κ ≥ ω: k⋅2 > ω
On 4/18/2024 10:54 AM, WM wrote:0, 1, 2, 3, ..., w = X
Le 18/04/2024 à 00:50, Jim Burns a écrit :
On 4/16/2024 10:33 AM, WM wrote:
Le 15/04/2024 à 23:06, Jim Burns a écrit :
On 4/15/2024 7:59 AM, WM wrote:
That is wrong if
all natnumbers are present already such that
no further natnumbers fits below ω.
It is enough if you explain
where ω is in the lower line:
Please answer the question.
| Set X fits set Y
| X and Y are same.sized
| X [=] Y
is a claim that
a function exists 1.to.1.onto.from.to:#
On 4/18/24 11:05 AM, WM wrote:
ω is amidst the interval (0, ω2) because in the image
there are as many ordinals in (ω, ω2) as in (0, ω).
But all those ordinals are transfinite ordinals, and none are the value
of double a finite Natural Number.
On 4/18/24 10:59 AM, WM wrote:
ω follows upon all natural numbers. There is nothing between them and ω.
Right, and any Natural Number * 2 is a Natural Number, so less than ω.
Your logic can't handle the fact that the set of Natural Numbers is unbounded on the high side, so it doesn't understand that.
On 4/18/24 11:08 AM, WM wrote:
Le 18/04/2024 à 01:27, Phil Carmody a écrit :
Richard Damon <[email protected]> writes:
That means that YOUR logic system is proven INCONSISTENT, and thus
BLOWN UP.
And that is where the intrigue lies.
My logic system says: ω is amidst the interval (0, ω2) because in the
image of the function f(x) = x*2 there are as many ordinals in (ω, ω2)
as in (0, ω).
And if your logic doesn't understand that the "ordinals" in (ω, ω*2) are different than those of (1, 2, 3, ...) it is just broken.
Any finite number *2 will be less than ω.
Le 19/04/2024 à 00:09, Jim Burns a écrit :
On 4/18/2024 10:54 AM, WM wrote:
Le 18/04/2024 à 00:50, Jim Burns a écrit :
On 4/16/2024 10:33 AM, WM wrote:
Le 15/04/2024 à 23:06, Jim Burns a écrit :
On 4/15/2024 7:59 AM, WM wrote:
That is wrong if
all natnumbers are present already such that
no further natnumbers fits below ω.
It is enough if you explain
where ω is in the lower line:
0, 1, 2, 3, ..., w = X
| | | | ||| | 0, 2, 4, 6, ..., w*2 = Y
Please answer the question.
| Set X fits set Y
Yes both sets have the same number of elements.
But the interval covered by Y is twice as large as
that covered by X.
| X and Y are same.sized
| X [=] Y
is a claim that
a function exists 1.to.1.onto.from.to:
| ∃f: X → Y: ¬∃x≠x′:f(x)=f(x′) ∧ f(X)=Y
The question was: Where in the second line sits ω?
Sorry, I could not find the symbol ω in your text.
I assume you deny to answer because
the answer would prove you wrong.
The question was: Where in the second line sits ω?
0, 1, 2, 3, ..., w = X
| | | | ||| | 0, 2, 4, 6, ..., w*2 = Y
On 4/19/2024 8:20 AM, WM wrote:
Le 19/04/2024 à 00:35, Richard Damon a écrit :
On 4/18/24 11:05 AM, WM wrote:
ω is amidst the interval (0, ω2) because in the image there are as
many ordinals in (ω, ω2) as in (0, ω).
But all those ordinals are transfinite ordinals, and none are the
value of double a finite Natural Number.
You are in error. Counting goes like this: 1, 2, 3, ..., ω, ω+1, ω+2, .. >> You simply pass ω although no known natural number k+1 will reach ω. But >> by multiplication, which goes faster, you cannot pass ω?
Huh?
any_natural_number * 2 = another_natural_number
These natural numbers are already in the set of all natural numbers.
On 4/19/2024 8:37 AM, WM wrote:
Le 19/04/2024 à 00:42, Richard Damon a écrit :
On 4/18/24 10:59 AM, WM wrote:
Right, and any Natural Number * 2 is a Natural Number, so less than ω.
ω follows upon all natural numbers. There is nothing between them and ω. >>
If all elements of the set {1, 2, 3, ...} are doubled and nevertheless
remain below ω, then you have created new natural numbers which have not
been doubled. Hence you have not doubled all natural numbers. But that
is what has to be done and, according to actual infinity, can be done.
Your logic can't handle the fact that the set of Natural Numbers is
unbounded on the high side, so it doesn't understand that.
Your logic can't double all natural numbers such that none below ω is
missing. You create always new natural numbers. They have not been
doubled.
Multiplying any natural number by two gives a natural number that is
already in the set of all natural numbers.
Le 19/04/2024 à 00:09, Jim Burns a écrit :
On 4/18/2024 10:54 AM, WM wrote:
It is enough if you explain
where ω is in the lower line:
0, 1, 2, 3, ..., ω
| | | | ||| |
0, 2, 4, 6, ..., ω*2
∀κ < ω: k⋅2 < ω
That means
the space between ω and ω*2 remains empty of
poducts 2k, and
not all natural numbers have been doubled
because new products have been inserted below ω.
∀κ ≥ ω: k⋅2 > ω
But is it also less than ω*2?
Are all products {ω, ω+1, ω+2, ω+3, ...}*2
less than ω*2?
Le 19/04/2024 à 00:35, Richard Damon a écrit :
On 4/18/24 11:05 AM, WM wrote:
ω is amidst the interval (0, ω2) because in the image there are as
many ordinals in (ω, ω2) as in (0, ω).
But all those ordinals are transfinite ordinals, and none are the
value of double a finite Natural Number.
You are in error. Counting goes like this: 1, 2, 3, ..., ω, ω+1, ω+2, .. You simply pass ω although no known natural number k+1 will reach ω. But
by multiplication, which goes faster, you cannot pass ω?
Regards, WM
Le 19/04/2024 à 00:42, Richard Damon a écrit :
On 4/18/24 10:59 AM, WM wrote:
ω follows upon all natural numbers. There is nothing between them and ω.
Right, and any Natural Number * 2 is a Natural Number, so less than ω.
If all elements of the set {1, 2, 3, ...} are doubled and nevertheless
remain below ω, then you have created new natural numbers which have not been doubled. Hence you have not doubled all natural numbers. But that
is what has to be done and, according to actual infinity, can be done.
Your logic can't handle the fact that the set of Natural Numbers is
unbounded on the high side, so it doesn't understand that.
Your logic can't double all natural numbers such that none below ω is missing. You create always new natural numbers. They have not been
doubled.
Regards, WM
Le 19/04/2024 à 00:40, Richard Damon a écrit :
On 4/18/24 11:08 AM, WM wrote:
Le 18/04/2024 à 01:27, Phil Carmody a écrit :
Richard Damon <[email protected]> writes:
That means that YOUR logic system is proven INCONSISTENT, and thus
BLOWN UP.
And that is where the intrigue lies.
My logic system says: ω is amidst the interval (0, ω2) because in the
image of the function f(x) = x*2 there are as many ordinals in (ω,
ω2) as in (0, ω).
And if your logic doesn't understand that the "ordinals" in (ω, ω*2)
are different than those of (1, 2, 3, ...) it is just broken.
Of course they are different since all those of (1, 2, 3, ...) have been doubled and therefore cannot remain the same set.
Any finite number *2 will be less than ω.
Not if all those which could emerge from doubling are doubled too.
Regards, WM
Le 19/04/2024 à 00:35, Richard Damon a écrit :
On 4/18/24 11:05 AM, WM wrote:
ω is amidst the interval (0, ω2) because in the image there are as
many ordinals in (ω, ω2) as in (0, ω).
But all those ordinals are transfinite ordinals, and none are the
value of double a finite Natural Number.
You are in error. Counting goes like this: 1, 2, 3, ..., ω, ω+1, ω+2, .. You simply pass ω although no known natural number k+1 will reach ω. But
by multiplication, which goes faster, you cannot pass ω?
Regards, WM
On 4/19/2024 4:36 PM, Richard Damon wrote:
On 4/19/24 11:20 AM, WM wrote:
Le 19/04/2024 à 00:35, Richard Damon a écrit :
On 4/18/24 11:05 AM, WM wrote:
ω is amidst the interval (0, ω2) because in the image there are as >>>>> many ordinals in (ω, ω2) as in (0, ω).
But all those ordinals are transfinite ordinals, and none are the
value of double a finite Natural Number.
You are in error. Counting goes like this: 1, 2, 3, ..., ω, ω+1, ω+2, >>> .. You simply pass ω although no known natural number k+1 will reach
ω. But by multiplication, which goes faster, you cannot pass ω?
Regards, WM
Nope, "counting individual numbers" NEVER gets to ω.
ω is what you get to when you go BEYOND just "counting", which is why
your bounded logic can't handle it, as it needs to count to things.
ω isn't just some "big" number, but a step above what you get by
counting and moving into dealing with infinite sets.
Perhaps WM thinks that ω is a finite boundary for the natural numbers.
Then, I can see where he thinks it must be some really big, hyper huge, natural number. Strange!
Chris M. Thomasson schrieb:
Multiplying any natural number by two gives a natural number that is
already in the set of all natural numbers.
WM knows that definition but he doesn't ACCEPT it in his own math
since 30++ years (which he doesn't mention this difference) because
he feels that the used logic in today's mathematics is very wrong.
On 4/19/2024 11:05 AM, WM wrote:
0, 1, 2, 3, ..., w = X
| | | | ||| |
0, 2, 4, 6, ..., w*2 = Y
I answer and you refuse to see it because
ω isn't what you (WM) want ω to be.
ω isn't in Y
On 4/19/24 11:37 AM, WM wrote:
Your logic can't double all natural numbers such that none below ω is
missing. You create always new natural numbers. They have not been
doubled.
Nope. The whole countable infinity was always there
On 4/19/2024 11:15 AM, WM wrote:
Le 19/04/2024 à 00:09, Jim Burns a écrit :
On 4/18/2024 10:54 AM, WM wrote:
It is enough if you explain
where ω is in the lower line:
0, 1, 2, 3, ..., ω
| | | | ||| |
0, 2, 4, 6, ..., ω*2
∀κ < ω: k⋅2 < ω
That means
the space between ω and ω*2 remains empty of
poducts 2k, and
not all natural numbers have been doubled
because new products have been inserted below ω.
Between ω and ω⋅2 is empty of
products k⋅2 from k < ω
Nothing is inserted anywhere.
Each finite even is finite and below ω
and is double an ordinal finite and below ω
Le 20/04/2024 à 00:24, Jim Burns a écrit :
On 4/19/2024 11:15 AM, WM wrote:
Le 19/04/2024 à 00:09, Jim Burns a écrit :
On 4/18/2024 10:54 AM, WM wrote:
It is enough if you explain
where ω is in the lower line:
0, 1, 2, 3, ..., ω
| | | | ||| |
0, 2, 4, 6, ..., ω*2
∀κ < ω: k⋅2 < ω
That means
the space between ω and ω*2 remains empty of
poducts 2k, and
not all natural numbers have been doubled
because new products have been inserted below ω.
Between ω and ω⋅2 is empty of
products k⋅2 from k < ω
How do the ordinal numbers ω+1, ω+2, ... come into being?
Nothing is inserted anywhere.
Each finite even is finite and below ω
and is double an ordinal finite and below ω
Then that one directly before ω is not multiplied. Or it is not
existing. But what exists directly before ω?
Regards, WM
Nope, "counting individual numbers" NEVER gets to ω.
On 4/19/24 11:39 AM, WM wrote:
Not if all those which could emerge from doubling are doubled too.
Yep, since you can't "use up" an infinte set that way.
On 4/20/24 12:31 PM, WM wrote:
Then that one directly before ω is not multiplied. Or it is not
existing. But what exists directly before ω?
But there isn't one "directly before" ω, as you don't seem to understand what the ... means. The ... contains a set with no upper bound.
The ... shows a sequence of a countably infinite number of digits, with
no end. You pass it when you do a Transfinite step in your counting.
Le 20/04/2024 à 01:32, Richard Damon a écrit :
On 4/19/24 11:37 AM, WM wrote:
Your logic can't double all natural numbers such that none below ω is
missing. You create always new natural numbers. They have not been
doubled.
Nope. The whole countable infinity was always there
Doubling doubles. The interval (0, ω) is doubled to (0, ω2).
Regards, WM
Le 20/04/2024 à 01:36, Richard Damon a écrit :
Nope, "counting individual numbers" NEVER gets to ω.
Then you will never get to ω and never count a countable set. Then no
set is countable.
Regards, WM
Le 20/04/2024 à 01:38, Richard Damon a écrit :
On 4/19/24 11:39 AM, WM wrote:
Not if all those which could emerge from doubling are doubled too.
Yep, since you can't "use up" an infinte set that way.
It must be possible to double all elements. Or you cannot complete the counting of the fractions.
Regards, WM
Le 20/04/2024 à 18:40, Richard Damon a écrit :
On 4/20/24 12:31 PM, WM wrote:
Then that one directly before ω is not multiplied. Or it is not
existing. But what exists directly before ω?
But there isn't one "directly before" ω, as you don't seem to
understand what the ... means. The ... contains a set with no upper
bound.
What is before ω?
The ... shows a sequence of a countably infinite number of digits,
with no end. You pass it when you do a Transfinite step in your counting.
Cantor claims that all natural numbers can be passed by bijecting them
with elements of coubtable sets in finite steps. Nothing remains
untouched before ω.
Regards, WM
WM presented the following explanation :
Le 19/04/2024 à 22:57, Tom Bola a écrit :
Chris M. Thomasson schrieb:
Multiplying any natural number by two gives a natural number that is
already in the set of all natural numbers.
WM knows that definition but he doesn't ACCEPT it in his own math since
30++ years (which he doesn't mention this difference) because he feels that >>> the used logic in today's mathematics is very wrong.
I accept that, but it excludes the idea that ω follows upon all natural
numbers with no gap in between. But this idea is what I presently
investigate.
What is a gap in the ordinals?
On 4/20/24 12:39 PM, WM wrote:
Doubling doubles. The interval (0, ω) is doubled to (0, ω2).
First, if we are talking the set of NATURAL Numbers, we don't normally
talk about "intervals" like this, as (0, ω) would be the set {1, 2, 3,
...} and when doubled would be the set {2, 4, 6, ...} which isn't really
an "inteval" any more as it skips values.
WM presented the following explanation :
I accept that, but it excludes the idea that ω follows upon all natural
numbers with no gap in between. But this idea is what I presently
investigate.
What is a gap in the ordinals?
You can't say something is actually "impossible" if it is only
impossible when you artificially restrict yourself from doing thing that
you can actually do.
On 4/20/24 12:42 PM, WM wrote:
It must be possible to double all elements. Or you cannot complete the
counting of the fractions.
Of course you can double all the elements, they just double into a
subset of themselves.
Le 20/04/2024 à 19:05, Richard Damon a écrit :
On 4/20/24 12:39 PM, WM wrote:
Doubling doubles. The interval (0, ω) is doubled to (0, ω2).
First, if we are talking the set of NATURAL Numbers, we don't normally
talk about "intervals" like this, as (0, ω) would be the set {1, 2, 3,
...} and when doubled would be the set {2, 4, 6, ...} which isn't
really an "inteval" any more as it skips values.
The interval (0, ω2) is covered by {2, 4, 6, ...}.
Regards, WM
Le 20/04/2024 à 19:09, Richard Damon a écrit :
You can't say something is actually "impossible" if it is only
impossible when you artificially restrict yourself from doing thing
that you can actually do.
Counting/mapping of all natnumbers is only possible if counting/mapping
of all natnumbers is possible. You deny it and endorse it simultaneously.
Regards, WM
Le 20/04/2024 à 19:17, Richard Damon a écrit :
On 4/20/24 12:42 PM, WM wrote:
It must be possible to double all elements. Or you cannot complete
the counting of the fractions.
Of course you can double all the elements, they just double into a
subset of themselves.
Impossible.
Regards, WM
Le 20/04/2024 à 19:42, FromTheRafters a écrit :
What is a gap in the ordinals?It is a not existing ordinal between two ordinals.
It is a not existing natural number next to ω, for instance.
the idea that ω follows upon all natural numbers with no gap in between.
Le 20/04/2024 à 19:17, Richard Damon a écrit :
On 4/20/24 12:42 PM, WM wrote:
Impossible.It must be possible to double all elements.
Of course you can double all the elements, they just double into a
subset of themselves.
Le 20/04/2024 à 00:24, Jim Burns a écrit :
Then that one directly before ω
Each finite even is finite and below ω and is double an ordinal finite and below ω
is not multiplied.
Or it is not existing.
But what exists directly before ω?
Le 19/04/2024 à 21:16, Jim Burns a écrit :
On 4/19/2024 11:05 AM, WM wrote:
0, 1, 2, 3, ..., w = X
| | | | ||| |
0, 2, 4, 6, ..., w*2 = Y
I answer and you refuse to see it because
ω isn't what you (WM) want ω to be.
I assume that it is the first transfinite number,
following directly upon all natural numbers.
0, 1, 2, 3, ..., w = X
| | | | ||| |
0, 2, 4, 6, ..., w*2 = Y
If you can't or won't understand that,
you need not answer.
ω isn't in Y
Wrong.
On 4/20/24 2:38 PM, WM wrote:
Le 20/04/2024 à 19:09, Richard Damon a écrit :
You can't say something is actually "impossible" if it is only
impossible when you artificially restrict yourself from doing thing
that you can actually do.
Counting/mapping of all natnumbers is only possible if counting/mapping
of all natnumbers is possible. You deny it and endorse it simultaneously.
And Mapping is possible,
Mapping and finitely counting are DIFFERENT operations.
On 4/20/24 2:31 PM, WM wrote:
The interval (0, ω2) is covered by {2, 4, 6, ...}.But that interval would be actually { 1, 2, 3, 4, 5, 6, ..., ω, ω+1,
ω+2, ...}
so, the interval notation is just incorrect as it includes members that aren't in the actual resutls.
So yes, the results are IN the interval, but are NOT THE interval.
On 4/20/24 2:40 PM, WM wrote:
But True.Of course you can double all the elements, they just double into a
subset of themselves.
Impossible.
It may seem impossible to you,
Le 20/04/2024 à 20:49, Richard Damon a écrit :
On 4/20/24 2:40 PM, WM wrote:
But True.Of course you can double all the elements, they just double into a
subset of themselves.
Impossible.
It may seem impossible to you,
The original set is in (0, ω). The doubled set
Le 20/04/2024 à 20:47, Richard Damon a écrit :
On 4/20/24 2:31 PM, WM wrote:
The interval (0, ω2) is covered by {2, 4, 6, ...}.
But that interval would be actually { 1, 2, 3, 4, 5, 6, ..., ω, ω+1,
ω+2, ...}
so, the interval notation is just incorrect as it includes members
that aren't in the actual resutls.
So yes, the results are _in_ the interval, but are NOT THE interval.
The results cannot be compressed to the interval (0, ω) of the set { 1,
2, 3, ...}. This shows that new numbers are generated by multiplication.
Am 22.04.2024 um 16:25 schrieb WM:
Le 20/04/2024 à 20:49, Richard Damon a écrit :
On 4/20/24 2:40 PM, WM wrote:
But True.Of course you can double all the elements, they just double into a
subset of themselves.
Impossible.
It may seem impossible to you,
The original set is in (0, ω). The doubled set
is in (0, ω) too (since it is in the original set).
Hint: {2*n : n e IN} c IN.
Alle geraden (natürlichen) Zahlen sind natürliche Zahlen. Erzählst Du Deinen Studenten an der THA etwas anderes?
Proof: Let k in {2*n : n e IN}. Then there is an n in IN such that k =
2*n. Since IN is closed unter multiplication and 2 is in IN too, 2*n in
IN and hence k in IN. Hence {2*n : n e IN} c IN. qed
See: https://www.cuemath.com/questions/which-of-the-following-sets-are-closed-under-multiplication-select-all-that-apply/
WM wrote :
What is a gap in the ordinals?
It is a not existing ordinal between two ordinals. It is a not existing
natural number next to ω, for instance.
Sets do not have not existing elements.
There is no "one directly before ω",
Hence there is no finite ordinal "directly before ω".
Am 20.04.2024 um 20:34 schrieb WM:
Le 20/04/2024 à 19:42, FromTheRafters a écrit :
What is a gap in the ordinals?It is a not existing ordinal between two ordinals.
Aha, dann gibt es also zwischen allen Ordinalzahlen "gaps".
It is a not existing natural number next to ω, for instance.
Ja, solche natürlichen Zahlen gibt es in der Tat nicht.
the idea that ω follows upon all natural numbers with no gap in between.
Hint: There is no "gap in between" in the following sense:
ω is (provably) the smallest ordinal AFTER (large than) all finite
ordinals.
Alle geraden (natürlichen) Zahlen sind natürliche Zahlen. Erzählst Du Deinen Studenten an der THA etwas anderes?
Proof: Let k in {2*n : n e IN}. Then there is an n in IN such that k =
2*n. Since IN is closed unter multiplication and 2 is in IN too, 2*n in
IN and hence k in IN. Hence {2*n : n e IN} c IN. qed
On 4/20/2024 12:23 PM, WM wrote:
Le 19/04/2024 à 21:16, Jim Burns a écrit :
On 4/19/2024 11:05 AM, WM wrote:
0, 1, 2, 3, ..., w = X
| | | | ||| |
0, 2, 4, 6, ..., w*2 = Y
I answer and you refuse to see it because
ω isn't what you (WM) want ω to be.
I assume that it is the first transfinite number,
following directly upon all natural numbers.
I answer and you refuse to see
what the first transfinite ordinal is.
Le 21/04/2024 à 18:32, Moebius a écrit :
Alle geraden (natürlichen) Zahlen sind natürliche Zahlen. Erzählst Du
Deinen Studenten an der THA etwas anderes?
No, you are right. But not all doubling results in even natural numbers.
Proof: Let k in {2*n : n e IN}. Then there is an n in IN such that k =Proof: The original set ℕ fits into (0, ω). The doubled set doesn't.
2*n. Since IN is closed unter multiplication and 2 is in IN too, 2*n
in IN and hence k in IN. Hence {2*n : n e IN} c IN. qed
there is an [...] ordinal ω-1.
Hence there is [especially] no finite ordinal "directly before ω"
You have never heard of it.
That does not make or even prove it not existent.
Le 21/04/2024 à 00:44, FromTheRafters a écrit :
WM wrote :
What is a gap in the ordinals?
It is a not existing ordinal between two ordinals. It is a not
existing natural number next to ω, for instance.
Sets do not have not existing elements.
What is ω - 1?
The original set fits in (0, ω). The doubled set doesn't.
[nonsense deleted]
If all are existing, why are not all countable?
<bla bla> What else should prevent counting?
Le 20/04/2024 à 01:36, Richard Damon a écrit :
Then you will never get to ω
Nope, "counting individual numbers" NEVER gets to ω.
and never count a countable set.
Then no set is countable.
[saudummen Scheißdreck gelöscht]
Am 22.04.2024 um 17:35 schrieb WM:
The original set fits in (0, ω). The doubled set doesn't.
It does.
Le 21/04/2024 à 20:08, Jim Burns a écrit :
On 4/20/2024 12:23 PM, WM wrote:
Le 19/04/2024 à 21:16, Jim Burns a écrit :
On 4/19/2024 11:05 AM, WM wrote:
0, 1, 2, 3, ..., w = X
| | | | ||| |
0, 2, 4, 6, ..., w*2 = Y
I answer and you refuse to see it because
ω isn't what you (WM) want ω to be.
I assume that it is the first transfinite number,
following directly upon all natural numbers.
I answer and you refuse to see
what the first transfinite ordinal is.
The original set fits in (0, ω).
The doubled set doesn't.
A /countable set/ S (as defined in the context of set theory) is
NOT "countable" because it can be counted (and/or there is someone who
can count its elements), but because there is an injective function f
such that f maps IN onto S.
On 4/22/2024 11:35 AM, WM wrote:
The original set fits in (0, ω).0, 1, 2, 3, ..., w = X
| | | | ||| |
0, 2, 4, 6, ..., w*2 = Y
The doubled set doesn't.
⦅0,ω+ω⦆ fits ⦅0,ω⦆
A /countable set/ S (as defined in the context of set theory) is
NOT "countable" because it can be counted (and/or there is someone who
can count its elements), but because there is an injective function f
such that f maps IN onto S.
Le 22/04/2024 à 19:27, Moebius a écrit :
A /countable set/ S (as defined in the context of set theory) is NOTWhy can't it be counted?
"countable" because it can be counted (and/or there is someone who can
count its elements), but because there is an injective function f such
that f maps IN onto S.
Le 22/04/2024 à 18:08, Moebius a écrit :
Am 22.04.2024 um 17:35 schrieb WM:[Dummschwatz]
The original set fits in (0, ω). The doubled set doesn't.
It does.
Am 22.04.2024 um 19:53 schrieb WM:
Le 22/04/2024 à 19:27, Moebius a écrit :
A /countable set/ S (as defined in the context of set theory) is NOTWhy can't it be counted?
"countable" because it can be counted (and/or there is someone who
can count its elements), but because there is an injective function f
such that f maps IN onto S.
Because we finite beings (except of Chuch Norris of course) cannot count infinitely many object, that's why, Mückenheim.
⦅0,ω+ω⦆ fits ⦅0,ω⦆
However, deleting context courts confusion,
as you have noticed.
Am 22.04.2024 um 19:32 schrieb Jim Burns:
⦅0,ω+ω⦆ fits ⦅0,ω⦆
Def:
(u, v) := {x e ORD : u < x < v}
(with x,y e ORD)
Then ⦅0,ω⦆ c ⦅0,ω+ω⦆, but not ⦅0,ω+ω⦆ c ⦅0,ω⦆.
After all, ω < ω+1 < ω+2 < ... < ω+ω.
So with "I fits J",
you don't mean "I fits in J",
but rather "J fits in I", I guess.
Right?
A commen term in this context is "contains".
Hence ⦅0,ω+ω⦆ contains ⦅0,ω⦆.
Am 22.04.2024 um 21:36 schrieb Jim Burns:
However, deleting context courts confusion,
as you have noticed.
That's why I tend to use (i.e. stick to)
standard notation and common notions
(in this context). :-)
Le 21/04/2024 à 00:44, FromTheRafters a écrit :
WM wrote :
What is a gap in the ordinals?
It is a not existing ordinal between two ordinals. It is a not
existing natural number next to ω, for instance.
Sets do not have not existing elements.
What is ω - 1?
Regards, WM
Le 21/04/2024 à 18:07, Moebius a écrit :
Am 20.04.2024 um 20:34 schrieb WM:
Le 20/04/2024 à 19:42, FromTheRafters a écrit :
What is a gap in the ordinals?It is a not existing ordinal between two ordinals.
Aha, dann gibt es also zwischen allen Ordinalzahlen "gaps".
More precisely: A gap on the ordinal axis is a not existing ordinal in distance 1 from an existing ordinal. Example: Between -oo and 0 there us
a huge gap.
It is a not existing natural number next to ω, for instance.
Ja, solche natürlichen Zahlen gibt es in der Tat nicht.
Below ω there is a huge gap or there are dark ordinals.
the idea that ω follows upon all natural numbers with no gap inbetween.
Hint: There is no "gap in between" in the following sense:
ω is (provably) the smallest ordinal AFTER (large than) all finite
ordinals.
If so, then at ω-n there are gaps.
Regards, WM
Le 20/04/2024 à 20:47, Richard Damon a écrit :
On 4/20/24 2:31 PM, WM wrote:
The interval (0, ω2) is covered by {2, 4, 6, ...}.But that interval would be actually { 1, 2, 3, 4, 5, 6, ..., ω, ω+1,
ω+2, ...}
so, the interval notation is just incorrect as it includes members
that aren't in the actual resutls.
So yes, the results are IN the interval, but are NOT THE interval.
The results cannot be compressed to the interval (0, ω) of the set { 1,
2, 3, ...}. This shows that new numbers are generated by multiplication.
Regards, WM
Moebius schrieb:
Am 22.04.2024 um 21:36 schrieb Jim Burns:
However, deleting context courts confusion,
as you have noticed.
That's why I tend to use (i.e. stick to) standard notation and common
notions (in this context).
Folks and people are different, the more in societies with more plurality.
Le 20/04/2024 à 20:48, Richard Damon a écrit :
On 4/20/24 2:38 PM, WM wrote:
Le 20/04/2024 à 19:09, Richard Damon a écrit :
You can't say something is actually "impossible" if it is only
impossible when you artificially restrict yourself from doing thing
that you can actually do.
Counting/mapping of all natnumbers is only possible if
counting/mapping of all natnumbers is possible. You deny it and
endorse it simultaneously.
And Mapping is possible,
Mapping of all is only possible if all are existing. If all are
existing, why are not all countable? Because many are dark.
Mapping and finitely counting are DIFFERENT operations.
Agreed. Mapping ℕ --> ℕ is possible, even if almost all numbers cannot
be counted to because they are dark. What else should prevent counting?
Regards, WM
Le 21/04/2024 à 20:08, Jim Burns a écrit :
On 4/20/2024 12:23 PM, WM wrote:
Le 19/04/2024 à 21:16, Jim Burns a écrit :
On 4/19/2024 11:05 AM, WM wrote:
0, 1, 2, 3, ..., w = X
| | | | ||| |
0, 2, 4, 6, ..., w*2 = Y
I answer and you refuse to see it because
ω isn't what you (WM) want ω to be.
I assume that it is the first transfinite number,
following directly upon all natural numbers.
I answer and you refuse to see
what the first transfinite ordinal is.
The original set fits in (0, ω). The doubled set doesn't.
Regards, WM
Le 20/04/2024 à 20:49, Richard Damon a écrit :
On 4/20/24 2:40 PM, WM wrote:
But True.Of course you can double all the elements, they just double into a
subset of themselves.
Impossible.
It may seem impossible to you,
The original set is in (0, ω). The doubled set cannot be compressed to
(0, ω). This disproves your unquestioning faith.
Regards, WM
Le 21/04/2024 à 18:32, Moebius a écrit :
Alle geraden (natürlichen) Zahlen sind natürliche Zahlen. Erzählst Du
Deinen Studenten an der THA etwas anderes?
No, you are right. But not all doubling results in even natural numbers.
Proof: Let k in {2*n : n e IN}. Then there is an n in IN such that k =
2*n. Since IN is closed unter multiplication and 2 is in IN too, 2*n
in IN and hence k in IN. Hence {2*n : n e IN} c IN. qed
Proof: The original set ℕ fits into (0, ω). The doubled set doesn't.
Regards, WM
On 4/22/2024 3:40 PM, Moebius wrote:
Am 22.04.2024 um 21:36 schrieb Jim Burns:
However, deleting context courts confusion,
as you have noticed.
That's why I tend to use (i.e. stick to)
standard notation and common notions
(in this context). :-)
Since you find that that works for you,
I would not dream of asking you to do otherwise.
My impression of WM is he is someone who
thinks that they know what these common notions are,
and who doesn't know.
When I tried (many times) to explain to WM
common notions in common notation,
what he read is not what I wrote,
not even if I wrote it a hundred times.
So, now I try to find uncommon ways to write them.
But the notions themselves are standard, I think.
I'm not an idiot, but I am not Dedekind or Frege
or Erdős, either.
At most, I am offering a different emphasis,
it seems to me. And the emphasis previously
encountered by WM didn't take, so why not?
Am 22.04.2024 um 21:44 schrieb Tom Bola:
Moebius schrieb:
Am 22.04.2024 um 21:36 schrieb Jim Burns:
However, deleting context courts confusion,
as you have noticed.
That's why I tend to use (i.e. stick to)
standard notation and common notions
(in this context).
Folks and people are different,
the more in societies with more plurality.
Agree.
But, if I may say so,
your idiosyncratic notation and notions
make it difficult to follow
your [JB's?] proofs and arguments.
Usually, mathematicians (and many other people)
try to communicate their ideas as good as the can.
I'm mean it's a common communication strategy
(with the aim to be understood).
Just a comment.
Lit.: https://courses.media.mit.edu/2004spring/mas966/Peirce%201878%20Make%20Ideas%20Clear.pdf
Le 22/04/2024 à 19:32, Jim Burns a écrit :
On 4/22/2024 11:35 AM, WM wrote:
0, 1, 2, 3, ..., w = X
| | | | ||| |
0, 2, 4, 6, ..., w*2 = Y
The original set fits in (0, ω).
The doubled set doesn't.
⦅0,ω+ω⦆ fits ⦅0,ω⦆
No. ω+ω = ω*2.
WM has brought this to us :
Le 21/04/2024 à 00:44, FromTheRafters a écrit :
WM wrote :
What is a gap in the ordinals?
It is a not existing ordinal between two ordinals. It is a not existing >>>> natural number next to ω, for instance.
Sets do not have not existing elements.
What is ω - 1?
An absurdity.
Yes, there is sort of a gap below ω as you can only get to ω via a
"hyper" step, not a normal step, like from 1 to 2.
There are no Ordinals in that gap,
But none of them are "Dark", because all are individually usable and nameable.
We can't name ALL of them at once, but can name ANY of them we want.
On the other hand, "in principle" it might be "possible" - if you
believe in/accept supertasks.
See: https://en.wikipedia.org/wiki/Supertask
Note that set theory does not rely on supertasks. That's why we use a
set theoretic definition for the notion /countable set/. See above.
Hint: On this basis we can prove for example (a) that the set of
rational numbers (Q) is a countable set
and (b) that the set of real
numbers (R) is an uncountable set.
On 4/22/24 10:15 AM, WM wrote:
The results cannot be compressed to the interval (0, ω) of the set { 1,Of course they can be compressed into the interval (0, ω), as every
2, 3, ...}. This shows that new numbers are generated by multiplication.
finite number n < ω, when doubled results in a finite number 2n which is also < ω.
Name one that can't be!
Your logic just blows you up because it doesn't understand the nature of unbounded infinite sets.
You invent "dark" numbers, because you can't look at the nature of the unbounded because you are blinded by your finite logic that has blown
your logic to smiterines because you used it when it can't work.
Le 23/04/2024 à 00:58, Richard Damon a écrit :
Yes, there is sort of a gap below ω as you can only get to ω via a
"hyper" step, not a normal step, like from 1 to 2.
How many normal steps covers a hyper step?
There are no Ordinals in that gap,
What is in that gap?
On 4/22/2024 11:02 AM, WM wrote:
Le 22/04/2024 à 19:27, Moebius a écrit :
A /countable set/ S (as defined in the context of set theory) is NOT
"countable" because it can be counted (and/or there is someone who can
count its elements), but because there is an injective function f such
that f maps IN onto S.
Why can't it be counted?
Its unbounded.
Le 23/04/2024 à 01:01, Richard Damon a écrit :
On 4/22/24 10:15 AM, WM wrote:
The results cannot be compressed to the interval (0, ω) of the set { 1, >>> 2, 3, ...}. This shows that new numbers are generated by multiplication. >>>Of course they can be compressed into the interval (0, ω), as every
finite number n < ω, when doubled results in a finite number 2n which is
also < ω.
Try to map the closed interval [0, ω]*2 = [0, ω*2].
If [0, ω) --> [0, ω) and ω*2 --> ω*2,
then ω*2 is the only image point in (ω, ω*2].
Infinitely points many remain empty.
Crippled mathematics. Ugly. Inacceptable.
On 4/22/2024 1:56 PM, WM wrote:
Le 22/04/2024 à 19:32, Jim Burns a écrit :
On 4/22/2024 11:35 AM, WM wrote:
0, 1, 2, 3, ..., w = X
| | | | ||| |
0, 2, 4, 6, ..., w*2 = Y
The original set fits in (0, ω).
The doubled set doesn't.
⦅0,ω+ω⦆ fits ⦅0,ω⦆
No. ω+ω = ω*2.
Yes. ω+ω = ω⋅2
⦅0,ω+ω⦆ fits ⦅0,ω⦆
⦅0,ω⋅2⦆ fits ⦅0,ω⦆
ω+i ⟼ 2⋅i
WM schrieb:
Le 23/04/2024 à 00:58, Richard Damon a écrit :
Yes, there is sort of a gap below ω as you can only get to ω via a
"hyper" step, not a normal step, like from 1 to 2.
How many normal steps covers a hyper step?
It's the same with dimensions - one cannot "cover" length with width.
There are no Ordinals in that gap,
What is in that gap?
The same that is between each pair of n and n+1 in IN: nothing.
WM was thinking very hard :
Le 22/04/2024 à 20:02, FromTheRafters a écrit :
WM has brought this to us :
Le 21/04/2024 à 00:44, FromTheRafters a écrit :
WM wrote :
What is a gap in the ordinals?
It is a not existing ordinal between two ordinals. It is a not existing >>>>>> natural number next to ω, for instance.
Sets do not have not existing elements.
What is ω - 1?
An absurdity.
Not more nor less than ω.
What is the value of ω?
Le 23/04/2024 à 20:47, Tom Bola a écrit :
WM schrieb:
Le 23/04/2024 à 00:58, Richard Damon a écrit :
Yes, there is sort of a gap below ω as you can only get to ω via a
"hyper" step, not a normal step, like from 1 to 2.
How many normal steps covers a hyper step?
It's the same with dimensions - one cannot "cover" length with width.
There are no Ordinals in that gap,
What is in that gap?
The same that is between each pair of n and n+1 in IN: nothing.
Why so much nothing? Why does the interval [0, ω] not collapse?
Le 23/04/2024 à 01:01, Richard Damon a écrit :
On 4/22/24 10:15 AM, WM wrote:
The results cannot be compressed to the interval (0, ω) of the set { 1, >>> 2, 3, ...}. This shows that new numbers are generated by multiplication. >>>Of course they can be compressed into the interval (0, ω), as every
finite number n < ω, when doubled results in a finite number 2n which is
also < ω.
Try to map the closed interval [0, ω]*2 = [0, ω*2].
WM schrieb:
Le 23/04/2024 à 01:01, Richard Damon a écrit :
On 4/22/24 10:15 AM, WM wrote:
The results cannot be compressed to the interval (0, ω) of the set { 1, >>>> 2, 3, ...}. This shows that new numbers are generated by multiplication. >>>>Of course they can be compressed into the interval (0, ω), as every
finite number n < ω, when doubled results in a finite number 2n which is >>> also < ω.
Try to map the closed interval [0, ω]*2 = [0, ω*2].
f =
(0, 0),
(1, 2),
(2, 4),
(3, 6),
(..., ...),
(w, w*2)
On 4/22/24 10:15 AM, WM wrote:
The results cannot be compressed to the interval (0, ω) of the set { 1,Of course they can be compressed into the interval (0, ω), as every
2, 3, ...}. This shows that new numbers are generated by multiplication.
finite number n < ω, when doubled results in a finite number 2n which is also < ω.
Le 23/04/2024 à 05:51, Jim Burns a écrit :
On 4/22/2024 1:56 PM, WM wrote:
Le 22/04/2024 à 19:32, Jim Burns a écrit :
On 4/22/2024 11:35 AM, WM wrote:
0, 1, 2, 3, ..., w = X
| | | | ||| |
0, 2, 4, 6, ..., w*2 = Y
The original set fits in (0, ω).
The doubled set doesn't.
⦅0,ω+ω⦆ fits ⦅0,ω⦆
No. ω+ω = ω*2.
Yes. ω+ω = ω⋅2
⦅0,ω+ω⦆ fits ⦅0,ω⦆
⦅0,ω⋅2⦆ fits ⦅0,ω⦆
Use closed intervals: [0, ω]*2 = [0, ω*2]
If [0, ω) --> [0, ω) and ω*2 --> ω*2,
then ω*2 is the only image point in
the interval (ω, ω*2].
Infinitely many points remain empty.
Ugly mathematics.
ω+i ⟼ 2⋅i
ω+i is not mapped.
Le 23/04/2024 à 21:45, Tom Bola a écrit :
WM schrieb:
Le 23/04/2024 à 01:01, Richard Damon a écrit :
On 4/22/24 10:15 AM, WM wrote:
The results cannot be compressed to the interval (0, ω) of the set { 1, >>>>> 2, 3, ...}. This shows that new numbers are generated by multiplication. >>>>>Of course they can be compressed into the interval (0, ω), as every
finite number n < ω, when doubled results in a finite number 2n which is >>>> also < ω.
Try to map the closed interval [0, ω]*2 = [0, ω*2].
f =
(0, 0),
(1, 2),
(2, 4),
(3, 6),
(..., ...),
(w, w*2)
Indeed. When n reaches ω, 2n reaches ω2.
On 4/23/2024 11:47 AM, Tom Bola wrote:
WM schrieb:
Le 23/04/2024 à 00:58, Richard Damon a écrit :
Yes, there is sort of a gap below ω as you can only get to ω via a
"hyper" step, not a normal step, like from 1 to 2.
How many normal steps covers a hyper step?
It's the same with dimensions - one cannot "cover" length with width.
A hyper step wrt the reals can be as simple as 0 + 1 because there is an infinity between 0 and 1?
There are no Ordinals in that gap,
What is in that gap?
The same that is between each pair of n and n+1 in IN: nothing.
On 4/23/2024 12:13 PM, WM wrote:
Le 23/04/2024 à 20:57, FromTheRafters a écrit :
WM was thinking very hard :
Le 22/04/2024 à 20:02, FromTheRafters a écrit :
WM has brought this to us :
Le 21/04/2024 à 00:44, FromTheRafters a écrit :
WM wrote :
What is a gap in the ordinals?
It is a not existing ordinal between two ordinals. It is a not >>>>>>>> existing natural number next to ω, for instance.
Sets do not have not existing elements.
What is ω - 1?
An absurdity.
Not more nor less than ω.
What is the value of ω?
It cannot be expressed in units as a finite initial segment of ℕ.
So, you were always missing your largest natural number from the get go?
You are a moron? WM, sure seems to think that there _is_ a largest
natural number, he just cannot see it, so, well, it must be dark. lol!
Your largest natural number is already in the set of all natural numbers!
Le 22/04/2024 à 20:02, FromTheRafters a écrit :
WM has brought this to us :
Le 21/04/2024 à 00:44, FromTheRafters a écrit :
WM wrote :
What is a gap in the ordinals?
It is a not existing ordinal between two ordinals.
It is a not existing natural number next to ω,
for instance.
Sets do not have not existing elements.
What is ω - 1?
An absurdity.
An absurdity.
Not more nor less than ω.
Le 23/04/2024 à 01:01, Richard Damon a écrit :
On 4/22/24 10:15 AM, WM wrote:
The results cannot be compressed to the interval (0, ω) of the set {Of course they can be compressed into the interval (0, ω), as every
1, 2, 3, ...}. This shows that new numbers are generated by
multiplication.
finite number n < ω, when doubled results in a finite number 2n which
is also < ω.
Try to map the closed interval [0, ω]*2 = [0, ω*2].
If [0, ω) --> [0, ω) and ω*2 --> ω*2, then ω*2 is the only image point in (ω, ω*2]. Infinitely points many remain empty. Crippled mathematics. Ugly. Inacceptable.
Regards, WM
Name one that can't be!
Your logic just blows you up because it doesn't understand the nature
of unbounded infinite sets.
You invent "dark" numbers, because you can't look at the nature of the
unbounded because you are blinded by your finite logic that has blown
your logic to smiterines because you used it when it can't work.
Le 23/04/2024 à 00:58, Richard Damon a écrit :
Yes, there is sort of a gap below ω as you can only get to ω via a
"hyper" step, not a normal step, like from 1 to 2.
How many normal steps covers a hyper step?
There are no Ordinals in that gap,
What is in that gap?
Regards, WM
Le 22/04/2024 à 20:02, FromTheRafters a écrit :
WM has brought this to us :
Le 21/04/2024 à 00:44, FromTheRafters a écrit :
WM wrote :
What is a gap in the ordinals?
It is a not existing ordinal between two ordinals. It is a not
existing natural number next to ω, for instance.
Sets do not have not existing elements.
What is ω - 1?
An absurdity.
Not more nor less than ω.
Regards, WM
Le 23/04/2024 à 01:01, Richard Damon a écrit :
On 4/22/24 10:15 AM, WM wrote:
The results cannot be compressed to the interval (0, ω) of the set {Of course they can be compressed into the interval (0, ω), as every
1, 2, 3, ...}. This shows that new numbers are generated by
multiplication.
finite number n < ω, when doubled results in a finite number 2n which
is also < ω.
Try to map the closed interval [0, ω]*2 = [0, ω*2].
If [0, ω) --> [0, ω) and ω*2 --> ω*2, then ω*2 is the only image point in (ω, ω*2]. Infinitely many points remain empty. Crippled mathematics. Ugly. Inacceptable.
Regards, WM
Le 23/04/2024 à 21:45, Tom Bola a écrit :
WM schrieb:
Le 23/04/2024 à 01:01, Richard Damon a écrit :
On 4/22/24 10:15 AM, WM wrote:
The results cannot be compressed to the interval (0, ω) of the setOf course they can be compressed into the interval (0, ω), as every
{ 1, 2, 3, ...}. This shows that new numbers are generated by
multiplication.
finite number n < ω, when doubled results in a finite number 2n
which is also < ω.
Try to map the closed interval [0, ω]*2 = [0, ω*2].
f = (0, 0),
(1, 2),
(2, 4),
(3, 6),
(..., ...),
(w, w*2)
Indeed. When n reaches ω, 2n reaches ω2.
Regards, WM
Le 23/04/2024 à 01:04, Richard Damon a écrit :
But none of them are "Dark", because all are individually usable andThen you can name all. Simply use always only half of the remaining time.
nameable.
We can't name ALL of them at once, but can name ANY of them we want.
Regards, WM
WM pretended :
Le 23/04/2024 à 20:57, FromTheRafters a écrit :
WM was thinking very hard :
Le 22/04/2024 à 20:02, FromTheRafters a écrit :
WM has brought this to us :
Le 21/04/2024 à 00:44, FromTheRafters a écrit :
WM wrote :
What is a gap in the ordinals?
It is a not existing ordinal between two ordinals. It is a not >>>>>>>> existing natural number next to ω, for instance.
Sets do not have not existing elements.
What is ω - 1?
An absurdity.
Not more nor less than ω.
What is the value of ω?
It cannot be expressed in units as a finite initial segment of ℕ.
It is a rank, not a value.
WM schrieb:
Le 23/04/2024 à 21:45, Tom Bola a écrit :
WM schrieb:
Le 23/04/2024 à 01:01, Richard Damon a écrit :
On 4/22/24 10:15 AM, WM wrote:
The results cannot be compressed to the interval (0, ω) of the set { 1, >>>>>> 2, 3, ...}. This shows that new numbers are generated by multiplication. >>>>>>Of course they can be compressed into the interval (0, ω), as every >>>>> finite number n < ω, when doubled results in a finite number 2n which is >>>>> also < ω.
Try to map the closed interval [0, ω]*2 = [0, ω*2].
f =
(0, 0),
(1, 2),
(2, 4),
(3, 6),
(..., ...),
(w, w*2)
Indeed. When n reaches ω, 2n reaches ω2.
I forgot the surrounding brackets {}...
f = {
(0, 0),
(1, 2),
(2, 4),
(3, 6),
(..., ...),
(w, w*2) }
If you want something else *you* are (the one) to define it.
On 4/23/2024 2:57 PM, WM wrote:
Le 23/04/2024 à 05:51, Jim Burns a écrit :
On 4/22/2024 1:56 PM, WM wrote:
Le 22/04/2024 à 19:32, Jim Burns a écrit :
On 4/22/2024 11:35 AM, WM wrote:
0, 1, 2, 3, ..., w = X
| | | | ||| |
0, 2, 4, 6, ..., w*2 = Y
The original set fits in (0, ω).
The doubled set doesn't.
⦅0,ω+ω⦆ fits ⦅0,ω⦆
No. ω+ω = ω*2.
Yes. ω+ω = ω⋅2
⦅0,ω+ω⦆ fits ⦅0,ω⦆
⦅0,ω⋅2⦆ fits ⦅0,ω⦆
Use closed intervals: [0, ω]*2 = [0, ω*2]
No.
1 ∉ ⟦0,ω⟧ᣔ×2
1 ∈ ⟦0,ω⋅2⟧
⟦0,ω⟧ᣔ×2 ≠ ⟦0,ω⋅2⟧
If [0, ω) --> [0, ω) and ω*2 --> ω*2,
then ω*2 is the only image point in
the interval (ω, ω*2].
Infinitely many points remain empty.
Ordinals in ⟦0,ω⦆ double to ordinals in ⟦0,ω⦆
⟦0,ω⦆ᣔ×2 ⊆ ⟦0,ω⦆
Arithmetic of the familiar.
ω in ⟦ω,ω⋅2⟧ doubles to ω⋅2 in ⟦ω,ω⋅2⟧
Arithmetic of the less.familiar.
Ugly mathematics.
de gustibus non disputandum est
ω+i ⟼ 2⋅i
ω+i is not mapped.
It is mapped in front of your eyes, sic: '⟼'
not your doubling.map, but
the map which show that ⦅0,ω⋅2⦆ fits ⦅0,ω⦆
⟦ω,ω×2⦆ maps to {even<ω}
⟦0,ω⦆ maps to {odd<ω}
⟦0,ω×2⦆ maps to ⟦0,ω⦆
On 4/23/2024 2:06 PM, WM wrote:
Le 22/04/2024 à 20:02, FromTheRafters a écrit :
What is ω - 1?
An absurdity.
ω-1 is that which has anything and everything
following in a finite not.first.false claim.sequence:
Rumpelstiltskin spins straw into gold,
But if ω - 1 existed, it would be before ω, and thus less than it.
On 4/23/24 3:51 PM, WM wrote:
Indeed. When n reaches ω, 2n reaches ω2.
But n, the finite number, never reaches ω.
Richard Damon wrote :
Of course, if you logic tries to make that transfinite value finite because >> that is all it can handle, your system just blows up in your face.
IMO, it might be best not to think of it as a value at all.
On 4/23/24 2:35 PM, WM wrote:
Le 23/04/2024 à 00:58, Richard Damon a écrit :
Yes, there is sort of a gap below ω as you can only get to ω via a
"hyper" step, not a normal step, like from 1 to 2.
How many normal steps covers a hyper step?
Aleph_0 (and infinte value).
There are no Ordinals in that gap,
What is in that gap?
Nothing.
the next number below ω
Le 24/04/2024 à 03:50, Richard Damon a écrit :
But if ω - 1 existed, it would be before ω, and thus less than it.
It is a natnumber, but dark.
On 4/23/24 3:34 PM, WM wrote:
Le 23/04/2024 à 01:01, Richard Damon a écrit :
On 4/22/24 10:15 AM, WM wrote:
The results cannot be compressed to the interval (0, ω) of the set {Of course they can be compressed into the interval (0, ω), as every
1, 2, 3, ...}. This shows that new numbers are generated by
multiplication.
finite number n < ω, when doubled results in a finite number 2n which
is also < ω.
Try to map the closed interval [0, ω]*2 = [0, ω*2].
If [0, ω) --> [0, ω) and ω*2 --> ω*2, then ω*2 is the only image point >> in (ω, ω*2]. Infinitely many points remain empty. Crippled mathematics.
Ugly. Inacceptable.
Why?
[0, ω]*2 = { [0, w), ω } *2 = {[0, w), ω*2} since the Natural numbers (what [0, ω) represents) are closed under multiplication.
The fact that a mixed set of two different classes of ordinals ends up
with two different classes of ordinals isn't surprizing.
The fact that there are "gaps" in the result isn't surprising, as we see
the same gaps in the finite part of the set:
0, 1, 2, 3 ... *2 => 0, 2, 4, 6, ...
What is between ω and ω*2?
Le 24/04/2024 à 04:01, Richard Damon a écrit :
[0, ω]*2 = { [0, w), ω } *2 = {[0, ω), ω*2} since the Natural numbers
(what [0, ω) represents) are closed under multiplication.
Impossible if (0, ω) is completely filled.
Every 2n is larger than n.
You are right. 1 belongs not in [0, ω]*2 and of ω we don't know it.
Ordinals in ⟦0,ω⦆ double to ordinals in ⟦0,ω⦆
⟦0,ω⦆ᣔ×2 ⊆ ⟦0,ω⦆
Doubling a set of natnumbers yields elements which are not in the set.
for every set {1, 2, 3, ..., n} doubling extends the set.
Am 24.04.2024 um 19:05 schrieb WM:
the next number below ω
Man hat es Dir schon zig-mal erklärt, Mückenheim:
Es gibt keine "next number below ω".
Wie dumm kann ein Mensch eigentlich sein?
Le 24/04/2024 à 03:52, Richard Damon a écrit :
On 4/23/24 2:35 PM, WM wrote:
Le 23/04/2024 à 00:58, Richard Damon a écrit :
you can only get to ω via a "hyper" step, not a normal step, like from 1 to 2.
How many normal steps covers a hyper step?
aleph_0
There is no ordinal between the finite numbers and ω.
But something requiring ℵo steps. Looks like ordinals.Ja, die ordinals/natnums between the number n and ω (for each and every
Le 23/04/2024 à 21:55, Jim Burns a écrit :
Arithmetic of the familiar.
Nevertheless it is wrong because
for every set {1, 2, 3, ..., n}
doubling extends the set.
All infinitely many natnumbers are finite n.
ω+i ⟼ 2⋅i
ω+i is not mapped.
It is mapped in front of your eyes, sic: '⟼'
not your doubling.map, but
the map which show that ⦅0,ω⋅2⦆ fits ⦅0,ω⦆
You claim that ω+3 = 2*3?
⟦ω,ω×2⦆ maps to {even<ω}
⟦0,ω⦆ maps to {odd<ω}
⟦0,ω×2⦆ maps to ⟦0,ω⦆
No.
How can it be 100% completely filled when its unbounded?
Le 24/04/2024 à 02:03, Jim Burns a écrit :
On 4/23/2024 2:06 PM, WM wrote:
Le 22/04/2024 à 20:02, FromTheRafters a écrit :
What is ω - 1?
An absurdity.
ω-1 is that which has anything and everything
following in a finite not.first.false claim.sequence:
Rumpelstiltskin spins straw into gold,
Is there an ordinal between ℕ and ω? If not, then the next number below
ω is a natural.
Regards, WM
Then I can say that (XYZ) + .5 is "inside" an interval, but its not representational wrt 0, 1, 2, 3, 4, 5, ect... ?
Any good, or total crap?
Am 24.04.2024 um 22:36 schrieb Chris M. Thomasson:
I can say that, well let create a symbol... (XYZ) hold all of the
natural numbers. Therefore (XYZ) + 1 is already in (XYZ), fair enough?
I have to admit that I don't know what you are talking about.
We have the symbol "IN" to denote the set of (all) natural numbers.
Now: For all n e IN: n + 1 e IN.
Is that what you mean? :-)
How about:
"I can say that, well let introduce a symbol... [that denotes the set of
all natural number, say "IN".] [The set] IN hold all of the natural
numbers. Therefore n + 1 (with n in IN) is already in IN, fair enough?"
?
Yeah, fair enough. :-)
Though it's not about the symbol (say "IN"), but about the set (it
denotes).
I can say that, well let create a symbol... (XYZ) hold all of the
natural numbers. Therefore (XYZ) + 1 is already in (XYZ), fair enough?
1, 1.25, 1.5, 1.75, 2
I this finite section, 1 and 2 are natural, and the other three are not.
However, they [all] are within the set of natural numbers
but three of them cannot be directly represented.
Although, they can be indexed using natural numbers, zero aside for a
moment:
[0] = 1
[1] = 1.25
[2] = 1.5
[3] = 1.75
[4] = 2
There is an infinity [of rational numbers] between 1 and 2... They can be indexed using naturals?
;^) Just having some fun with numbers here.
Deeper down the rabbit hole. There is an infinity between 1 and 1.25...
Am 24.04.2024 um 19:01 schrieb WM:
What is between ω and ω*2?
Was wohl, Du Hirni? Wenn wir von Ordinalzahlen reden, dann gilt:
ω < ω+1 < ω+2 < ... < ω+ω = ω*2
Die Menge der Ordinalzahlen, die zwischen ω und ω*2 liegen, ist also
{o e ORD : ω < o < ω*2} = {ω+1, ω+2, ω+2, ...}.
Übrigens enthält diese Menge aleph_0 Ordinalzahlen
for every set {1, 2, 3, ..., n} doubling extends the set.
Ja, Mückenheim, gut beobachtet. Dem hat auch noch nie jemand widersprochen.
Es gibt nur kein n e IN, so dass {1, 2, 3, ..., n} = IN ist.
Mit anderen Worten: Es gibt keine größte natürliche Zahl.
Le 24/04/2024 à 19:40, Moebius a écrit :
Am 24.04.2024 um 19:01 schrieb WM:
What is between ω and ω*2?
Was wohl, Du Hirni? Wenn wir von Ordinalzahlen reden, dann gilt:
ω < ω+1 < ω+2 < ... < ω+ω = ω*2
Die Menge der Ordinalzahlen, die zwischen ω und ω*2 liegen, ist also
{o e ORD : ω < o < ω*2} = {ω+1, ω+2, ω+2, ...}.
Übrigens enthält diese Menge aleph_0 Ordinalzahlen
every second of which is in [...] (0, ω)*2.
WM explained :
Le 24/04/2024 à 12:49, FromTheRafters a écrit :
Richard Damon wrote :
Of course, if you logic tries to make that transfinite value finiteIMO, it might be best not to think of it as a value at all.
because that is all it can handle, your system just blows up in your face. >>>
Cantor called it a number.
A number representing an order type.
On 4/24/2024 12:57 PM, WM wrote:
Le 23/04/2024 à 21:55, Jim Burns a écrit :
Arithmetic of the familiar.
Nevertheless it is wrong because
for every set {1, 2, 3, ..., n}
doubling extends the set.
Doubling doesn't extend it to or beyond ω
The arithmetic of ω is not the familiar arithmetic.
It would also raise the question of
who is responsible for putting you (WM)
in front of a classroom of students.
On 4/24/2024 10:16 AM, WM wrote:
Le 24/04/2024 à 04:01, Richard Damon a écrit :[...]
On 4/23/24 3:34 PM, WM wrote:
Le 23/04/2024 à 01:01, Richard Damon a écrit :
On 4/22/24 10:15 AM, WM wrote:
The results cannot be compressed to the interval (0, ω) of the set >>>>>> { 1, 2, 3, ...}. This shows that new numbers are generated byOf course they can be compressed into the interval (0, ω), as every >>>>> finite number n < ω, when doubled results in a finite number 2n
multiplication.
which is also < ω.
Try to map the closed interval [0, ω]*2 = [0, ω*2].
If [0, ω) --> [0, ω) and ω*2 --> ω*2, then ω*2 is the only image
point in (ω, ω*2]. Infinitely many points remain empty. Crippled
mathematics. Ugly. Inacceptable.
Why?
Continuity.
[0, ω]*2 = { [0, w), ω } *2 = {[0, w), ω*2} since the Natural numbers >>> (what [0, ω) represents) are closed under multiplication.
Impossible if (0, ω) is completely filled. Every 2n is larger than n.
How can it be 100% completely filled when its unbounded?
On 4/24/24 1:06 PM, WM wrote:
Le 24/04/2024 à 03:50, Richard Damon a écrit :
But if ω - 1 existed, it would be before ω, and thus less than it.
It is a natnumber, but dark.
Can't be, because ALL Natural Numbers have a successor that are Natural Numbers
On 4/24/24 1:05 PM, WM wrote:
Is there an ordinal between ℕ and ω? If not, then the next number below >> ω is a natural.
Nope, because "next below" is not an operation defined for ω, just like
it isn't for 0.
The ONLY numbers below ω are the Natural Numbers, but "next" doesn't apply.
On 4/24/24 1:16 PM, WM wrote:
Le 24/04/2024 à 04:01, Richard Damon a écrit :
On 4/23/24 3:34 PM, WM wrote:
Le 23/04/2024 à 01:01, Richard Damon a écrit :
On 4/22/24 10:15 AM, WM wrote:
The results cannot be compressed to the interval (0, ω) of the set >>>>>> { 1, 2, 3, ...}. This shows that new numbers are generated byOf course they can be compressed into the interval (0, ω), as every >>>>> finite number n < ω, when doubled results in a finite number 2n
multiplication.
which is also < ω.
Try to map the closed interval [0, ω]*2 = [0, ω*2].
If [0, ω) --> [0, ω) and ω*2 --> ω*2, then ω*2 is the only image
point in (ω, ω*2]. Infinitely many points remain empty. Crippled
mathematics. Ugly. Inacceptable.
Why?
Continuity.
Who says we have that sort of "Continuity"?
Am 25.04.2024 um 21:40 schrieb WM:
Le 24/04/2024 à 19:40, Moebius a écrit :
Am 24.04.2024 um 19:01 schrieb WM:
What is between ω and ω*2?
Was wohl, Du Hirni? Wenn wir von Ordinalzahlen reden, dann gilt:
ω < ω+1 < ω+2 < ... < ω+ω = ω*2
Die Menge der Ordinalzahlen, die zwischen ω und ω*2 liegen, ist also
{o e ORD : ω < o < ω*2} = {ω+1, ω+2, ω+2, ...}.
Übrigens enthält diese Menge aleph_0 Ordinalzahlen
every second of which is in [...] (0, ω)*2.
None of which is in (0, ω)*2, since (0, ω)*2 c (0, ω)
Le 25/04/2024 à 21:47, Moebius a écrit :
Am 25.04.2024 um 21:40 schrieb WM:
Le 24/04/2024 à 19:40, Moebius a écrit :
Am 24.04.2024 um 19:01 schrieb WM:
What is between ω and ω*2?
Was wohl, Du Hirni? Wenn wir von Ordinalzahlen reden, dann gilt:
ω < ω+1 < ω+2 < ... < ω+ω = ω*2
Die Menge der Ordinalzahlen, die zwischen ω und ω*2 liegen, ist also >>>>
{o e ORD : ω < o < ω*2} = {ω+1, ω+2, ω+2, ...}.
Übrigens enthält diese Menge aleph_0 Ordinalzahlen
every second of which is in [...] (0, ω)*2.
None of which is in (0, ω)*2, since (0, ω)*2 c (0, ω)
Nonsense.
The complete set is enlarged by doubling.
If ω is existing, then something is next.
what about:
[0] = 0
[1] = 1/3
[2] = 2/3
[3] = 1
.(3) for 1/3 is just how base 10 represents 1/3. Think about hittinga cycle during any long division in base 10. We can stop the process at
what about:
[0] = 0
[1] = 1/3
[2] = 2/3
[3] = 1
.(3) for 1/3 is just how base 10 represents 1/3. Think about hitting a
cycle during any long division in base 10. We can stop the process at
any detected "cycle"... Once we hit a cycle its a rational. Fair enough?
Or did I fuck up again?
Le 24/04/2024 à 21:42, Jim Burns a écrit :
On 4/24/2024 12:57 PM, WM wrote:
Le 23/04/2024 à 21:55, Jim Burns a écrit :
Arithmetic of the familiar.
Nevertheless it is wrong because
for every set {1, 2, 3, ..., n}
doubling extends the set.
Doubling doesn't extend it to or beyond ω
If all smaller numbers are doubled,
then there is no place for
the doubled numbers below ω.
If all smaller numbers are doubled,
then there is no place for
the doubled numbers below ω.
The arithmetic of ω is not
the familiar arithmetic.
The arithmetic of doubling produces
other numbers than the doubled ones.
ω+i ⟼ 2⋅i
ω+i is not mapped.
It is mapped in front of your eyes, sic: '⟼'
not your doubling.map, but
the map which show that ⦅0,ω⋅2⦆ fits ⦅0,ω⦆
You claim that ω+3 = 2*3?
I defined a map for which ω+3 ⟼ 2⋅3
It would explain a lot of your (WM's) posts,
if you don't know what a function/map is,
what familiar addition is,
what familiar multiplication is.
It would also raise the question of
who is responsible for putting you (WM)
in front of a classroom of students.
Those are experts which
a disappearing Bob cannot estimate.
On 4/25/2024 4:03 PM, WM wrote:
Le 24/04/2024 à 21:42, Jim Burns a écrit :
On 4/24/2024 12:57 PM, WM wrote:
Le 23/04/2024 à 21:55, Jim Burns a écrit :
Arithmetic of the familiar.
Nevertheless it is wrong because
for every set {1, 2, 3, ..., n}
doubling extends the set.
Doubling doesn't extend it to or beyond ω
If all smaller numbers are doubled,
then there is no place for
the doubled numbers below ω.
"Infinite" does not mean "humongous".
finite ⟺ can be counted to from 0
finite ⟺ below ω
If n is below ω
then n can be counted to from 0
then n⋅2 can be counted to from n
then n⋅2 can be counted to from 0 through n
then n⋅2 is below ω
If n is below w
then n⋅2 is below w
If all smaller numbers are doubled,
then there is no place for
the doubled numbers below ω.
The arithmetic of ω is not
the familiar arithmetic.
The arithmetic of doubling produces
other numbers than the doubled ones.
The arithmetic of doubling
numbers which can be counted to
produces only
doubled numbers which can be counted to.
"Infinite" does not mean "humongous".
ω+i ⟼ 2⋅i
ω+i is not mapped.
It is mapped in front of your eyes, sic: '⟼'
not your doubling.map, but
the map which show that ⦅0,ω⋅2⦆ fits ⦅0,ω⦆
You claim that ω+3 = 2*3?
I defined a map for which ω+3 ⟼ 2⋅3
It would explain a lot of your (WM's) posts,
if you don't know what a function/map is,
what familiar addition is,
what familiar multiplication is.
It would also raise the question of
who is responsible for putting you (WM)
in front of a classroom of students.
Those are experts which
a disappearing Bob cannot estimate.
Experts who think, as you (WM) do,
that ω+3 ⟼ 2⋅3 means ω+3 = 2⋅3
That is a radical new use of the word "expert".
Le 25/04/2024 à 01:13, Richard Damon a écrit :
On 4/24/24 1:06 PM, WM wrote:
Le 24/04/2024 à 03:50, Richard Damon a écrit :
But if ω - 1 existed, it would be before ω, and thus less than it.
It is a natnumber, but dark.
Can't be, because ALL Natural Numbers have a successor that are
Natural Numbers
all visible ones.
Regards, WM
Le 25/04/2024 à 01:13, Richard Damon a écrit :
On 4/24/24 1:05 PM, WM wrote:
Is there an ordinal between ℕ and ω? If not, then the next number
below ω is a natural.
Nope, because "next below" is not an operation defined for ω, just
like it isn't for 0.
Defined or not, there is something on the way from 1 to ω. Only natural numbers or empty steps.
The ONLY numbers below ω are the Natural Numbers, but "next" doesn't
apply.
If ω is existing, then something is next.
Regards, WM
Le 24/04/2024 à 20:08, FromTheRafters a écrit :
WM explained :
Le 24/04/2024 à 12:49, FromTheRafters a écrit :
Richard Damon wrote :
Of course, if you logic tries to make that transfinite value finite
because that is all it can handle, your system just blows up in
your face.
IMO, it might be best not to think of it as a value at all.
Cantor called it a number.
A number representing an order type.
like 2 or 5.
Regards, WM
Le 25/04/2024 à 21:47, Moebius a écrit :
Am 25.04.2024 um 21:40 schrieb WM:
Le 24/04/2024 à 19:40, Moebius a écrit :
Am 24.04.2024 um 19:01 schrieb WM:
What is between ω and ω*2?
Was wohl, Du Hirni? Wenn wir von Ordinalzahlen reden, dann gilt:
ω < ω+1 < ω+2 < ... < ω+ω = ω*2
Die Menge der Ordinalzahlen, die zwischen ω und ω*2 liegen, ist also >>>>
{o e ORD : ω < o < ω*2} = {ω+1, ω+2, ω+2, ...}.
Übrigens enthält diese Menge aleph_0 Ordinalzahlen
every second of which is in [...] (0, ω)*2.
None of which is in (0, ω)*2, since (0, ω)*2 c (0, ω)
Nonsense.
The complete set is enlarged by doubling.
Regards, WM
On 4/25/24 4:15 PM, WM wrote:
Le 25/04/2024 à 01:13, Richard Damon a écrit :
On 4/24/24 1:05 PM, WM wrote:
Is there an ordinal between ℕ and ω? If not, then the next number
below ω is a natural.
Nope, because "next below" is not an operation defined for ω, just
like it isn't for 0.
Defined or not, there is something on the way from 1 to ω. Only natural
numbers or empty steps.
But there is no "step" to ω from the value before it, only the hyper
step from the full set of finite numbers.
The ONLY numbers below ω are the Natural Numbers, but "next" doesn't
apply.
If ω is existing, then something is next.
next above, yes, ω+1
next below, no, the problem is the set below it is unbounded, and thus doesn't have a "highest" value in it.
Richard Damon schrieb:
On 4/25/24 4:15 PM, WM wrote:
Le 25/04/2024 à 01:13, Richard Damon a écrit :
On 4/24/24 1:05 PM, WM wrote:
Is there an ordinal between ℕ and ω? If not, then the next number >>>>> below ω is a natural.
Nope, because "next below" is not an operation defined for ω, just
like it isn't for 0.
Defined or not, there is something on the way from 1 to ω. Only natural >>> numbers or empty steps.
But there is no "step" to ω from the value before it, only the hyper
step from the full set of finite numbers.
The ONLY numbers below ω are the Natural Numbers, but "next" doesn't
apply.
If ω is existing, then something is next.
next above, yes, ω+1
next below, no, the problem is the set below it is unbounded, and thus
doesn't have a "highest" value in it.
Better say something like "the case" but don't say "the problem", because WM takes this as an admission that today's set theory is "unlogical" and wrong. ;)
On 4/25/2024 4:03 PM, WM wrote:
If all smaller numbers are doubled,
then there is no place for
the doubled numbers below ω.
If n is below ω
then n can be counted to from 0
then n⋅2 can be counted to from n
I defined a map for which ω+3 ⟼ 2⋅3
After serious thinking WM wrote :
Le 24/04/2024 à 20:08, FromTheRafters a écrit :
WM explained :
Le 24/04/2024 à 12:49, FromTheRafters a écrit :
Richard Damon wrote :
Of course, if you logic tries to make that transfinite value finite >>>>>> because that is all it can handle, your system just blows up in your >>>>>> face.
IMO, it might be best not to think of it as a value at all.
Cantor called it a number.
A number representing an order type.
like 2 or 5.
No, like:
If ω is existing, then something is next.
next above, yes, ω+1
next below, no,
the problem is the set below it is unbounded,
and thus
doesn't have a "highest" value in it.
Le 26/04/2024 à 04:03, Richard Damon a écrit :
If ω is existing, then something is next.
next above, yes, ω+1
next below, no,
I disagree.
the problem is the set below it is unbounded,
The set is bounded by ω.
and thus doesn't have a "highest" value in it.
The value is there but invisible.
Regards, WM
On 4/26/24 7:14 AM, Tom Bola wrote:
Richard Damon schrieb:
On 4/25/24 4:15 PM, WM wrote:
Le 25/04/2024 à 01:13, Richard Damon a écrit :
On 4/24/24 1:05 PM, WM wrote:
Is there an ordinal between ℕ and ω? If not, then the next number >>>>>> below ω is a natural.
Nope, because "next below" is not an operation defined for ω, just
like it isn't for 0.
Defined or not, there is something on the way from 1 to ω. Only natural >>>> numbers or empty steps.
But there is no "step" to ω from the value before it, only the hyper
step from the full set of finite numbers.
The ONLY numbers below ω are the Natural Numbers, but "next" doesn't >>>>> apply.
If ω is existing, then something is next.
next above, yes, ω+1
next below, no, the problem is the set below it is unbounded, and thus
doesn't have a "highest" value in it.
Better say something like "the case" but don't say "the problem", because WM >> takes this as an admission that today's set theory is "unlogical" and wrong. >> ;)
But the "Problem" is just in his logic. HE is the one that want there to
be a predecessor to ω, today's set theory doesn't make that claim. There
is ORDER between the parts of the set, but not a "sequence"
There *IS* an "unlogic" in the statement, but only because HE tries to
make something that isn't.
On 4/25/24 3:51 PM, WM wrote:
Le 24/04/2024 à 20:08, FromTheRafters a écrit :
WM explained :
Le 24/04/2024 à 12:49, FromTheRafters a écrit :
Richard Damon wrote :
Of course, if you logic tries to make that transfinite value finite >>>>>> because that is all it can handle, your system just blows up in
your face.
IMO, it might be best not to think of it as a value at all.
Cantor called it a number.
A number representing an order type.
like 2 or 5.
No, because 2 or 5 were FINITE numbers, not transfinite numbers.
On 4/26/24 10:41 AM, WM wrote:
The set is bounded by ω.
But ω isn't in the set, so it can't be the upper bound of the set that
is below ω
Le 26/04/2024 à 16:49, Richard Damon a écrit :
On 4/26/24 10:41 AM, WM wrote:
The set is bounded by ω.
But ω isn't in the set, so it can't be the upper bound of the set that
is below ω
Learn the meaning of upper bound. ω*2 is an upper bound too.
Fact is that below ω there are natnumbers and above there are none.
Hence they cease below the bound ω.
Regards, WM
Le 26/04/2024 à 01:11, Jim Burns a écrit :
On 4/25/2024 4:03 PM, WM wrote:
If all smaller numbers are doubled,
then there is no place for
the doubled numbers below ω.
If n is below ω
then n can be counted to from 0
then n⋅2 can be counted to from n
That is true for definable numbers
but not for the last numbers before ω.
Le 26/04/2024 à 04:08, Richard Damon a écrit :
On 4/25/24 3:51 PM, WM wrote:
Le 24/04/2024 à 20:08, FromTheRafters a écrit :
WM explained :
Le 24/04/2024 à 12:49, FromTheRafters a écrit :
Richard Damon wrote :
Of course, if you logic tries to make that transfinite value
finite because that is all it can handle, your system just blows >>>>>>> up in your face.
IMO, it might be best not to think of it as a value at all.
Cantor called it a number.
A number representing an order type.
like 2 or 5.
No, because 2 or 5 were FINITE numbers, not transfinite numbers.
Learn what an order type is according to Cantor.
Regards, WM
On 4/26/24 1:02 PM, WM wrote:
Le 26/04/2024 à 16:49, Richard Damon a écrit :
On 4/26/24 10:41 AM, WM wrote:
The set is bounded by ω.
But ω isn't in the set, so it can't be the upper bound of the set that
is below ω
Learn the meaning of upper bound. ω*2 is an upper bound too.
But since it isn't in the set, you can't use it as the value to "step
back" to.
Note, that value isn't "of the set that is below ω".
Fact is that below ω there are natnumbers and above there are none.
Hence they cease below the bound ω.
Yes, but it can't be the "Upper Bound" used to step back to from ω.
There is not "Upper Bound" for the Natural Numbers *IN* the Natural
Numbers to be the "last" value to step back to.
Thus, your logic about ω-1 breaks. The value does not exist, as there
can not be a "transfinite" value below ω, by its definition, and ω-1
can't be a Natural Number, as then it would be the "last" Natural
Number, but that set doesn't have a "last" member, as its "Upper Bound"
is outside the set, as it is a "Unbounded" set.
On 4/25/2024 1:06 PM, WM wrote:
Le 24/04/2024 à 22:07, "Chris M. Thomasson" a écrit :
On 4/24/2024 10:16 AM, WM wrote:
Le 24/04/2024 à 04:01, Richard Damon a écrit :[...]
On 4/23/24 3:34 PM, WM wrote:
Le 23/04/2024 à 01:01, Richard Damon a écrit :
On 4/22/24 10:15 AM, WM wrote:
The results cannot be compressed to the interval (0, ω) of the >>>>>>>> set { 1, 2, 3, ...}. This shows that new numbers are generated by >>>>>>>> multiplication.Of course they can be compressed into the interval (0, ω), as
every finite number n < ω, when doubled results in a finite number >>>>>>> 2n which is also < ω.
Try to map the closed interval [0, ω]*2 = [0, ω*2].
If [0, ω) --> [0, ω) and ω*2 --> ω*2, then ω*2 is the only image >>>>>> point in (ω, ω*2]. Infinitely many points remain empty. Crippled >>>>>> mathematics. Ugly. Inacceptable.
Why?
Continuity.
[0, ω]*2 = { [0, w), ω } *2 = {[0, w), ω*2} since the Natural
numbers (what [0, ω) represents) are closed under multiplication.
Impossible if (0, ω) is completely filled. Every 2n is larger than n.
How can it be 100% completely filled when its unbounded?
Ask Cantor or Bolzano. It can also be completely counted - according to
Cantor.
How can you completely count an infinite unbounded set? You have some
issues WM... Deal with it. Your folly is not our problem... ;^o
On 4/26/24 5:48 PM, Tom Bola wrote:
Chris M. Thomasson schrieb:
On 4/25/2024 1:06 PM, WM wrote:
Le 24/04/2024 à 22:07, "Chris M. Thomasson" a écrit :
On 4/24/2024 10:16 AM, WM wrote:
Le 24/04/2024 à 04:01, Richard Damon a écrit :
On 4/23/24 3:34 PM, WM wrote:
Le 23/04/2024 à 01:01, Richard Damon a écrit :
On 4/22/24 10:15 AM, WM wrote:
The results cannot be compressed to the interval (0, ω) of the >>>>>>>>>> set { 1, 2, 3, ...}. This shows that new numbers are generated by >>>>>>>>>> multiplication.Of course they can be compressed into the interval (0, ω), as >>>>>>>>> every finite number n < ω, when doubled results in a finite number >>>>>>>>> 2n which is also < ω.
Try to map the closed interval [0, ω]*2 = [0, ω*2].
If [0, ω) --> [0, ω) and ω*2 --> ω*2, then ω*2 is the only image >>>>>>>> point in (ω, ω*2]. Infinitely many points remain empty. Crippled >>>>>>>> mathematics. Ugly. Inacceptable.
Why?
Continuity.
Impossible if (0, ω) is completely filled. Every 2n is larger than n. >>>>> [...]
[0, ω]*2 = { [0, w), ω } *2 = {[0, w), ω*2} since the Natural >>>>>>> numbers (what [0, ω) represents) are closed under multiplication. >>>>>>
How can it be 100% completely filled when its unbounded?
Ask Cantor or Bolzano. It can also be completely counted - according to >>>> Cantor.
How can you completely count an infinite unbounded set? You have some
issues WM... Deal with it. Your folly is not our problem... ;^o
WM takes himself to be a "finitist" - he denies that there is somewhere
in the universe any infinity "realized" (judging that math is happening
in the mental space), so he combats this idea and exacts "the existence"
of a first real number right to zero and a last finite number left to w
altogether with any idea of inifinite objects elsewhere in the universe.
But then he wants to talk about the transfinite numbers, which just can
not exist in the finite domain he tries to live in.
He seems to accept that mathematics does generate "infinite" numbers,
but then wants them to still act like the finite numbers he understands,
and complains about how they seem to break all the rules,
so everything needs to be come "dark" to hide the problems.
Chris M. Thomasson schrieb:
On 4/25/2024 1:06 PM, WM wrote:
Le 24/04/2024 à 22:07, "Chris M. Thomasson" a écrit :
On 4/24/2024 10:16 AM, WM wrote:
Le 24/04/2024 à 04:01, Richard Damon a écrit :
On 4/23/24 3:34 PM, WM wrote:
Le 23/04/2024 à 01:01, Richard Damon a écrit :
On 4/22/24 10:15 AM, WM wrote:
The results cannot be compressed to the interval (0, ω) of the >>>>>>>>> set { 1, 2, 3, ...}. This shows that new numbers are generated by >>>>>>>>> multiplication.Of course they can be compressed into the interval (0, ω), as >>>>>>>> every finite number n < ω, when doubled results in a finite number >>>>>>>> 2n which is also < ω.
Try to map the closed interval [0, ω]*2 = [0, ω*2].
If [0, ω) --> [0, ω) and ω*2 --> ω*2, then ω*2 is the only image >>>>>>> point in (ω, ω*2]. Infinitely many points remain empty. Crippled >>>>>>> mathematics. Ugly. Inacceptable.
Why?
Continuity.
[0, ω]*2 = { [0, w), ω } *2 = {[0, w), ω*2} since the Natural
numbers (what [0, ω) represents) are closed under multiplication.
Impossible if (0, ω) is completely filled. Every 2n is larger than n. >>>> [...]
How can it be 100% completely filled when its unbounded?
Ask Cantor or Bolzano. It can also be completely counted - according to
Cantor.
How can you completely count an infinite unbounded set? You have some
issues WM... Deal with it. Your folly is not our problem... ;^o
WM takes himself to be a "finitist" - he denies that there is somewhere
in the universe any infinity "realized" (judging that math is happening
in the mental space), so he combats this idea and exacts "the existence"
of a first real number right to zero and a last finite number left to w altogether with any idea of inifinite objects elsewhere in the universe.
On 4/26/24 1:02 PM, WM wrote:
Le 26/04/2024 à 16:49, Richard Damon a écrit :
On 4/26/24 10:41 AM, WM wrote:
The set is bounded by ω.
But ω isn't in the set, so it can't be the upper bound of the set that
is below ω
Learn the meaning of upper bound. ω*2 is an upper bound too.
But since it isn't in the set, you can't use it as the value to "step
back" to.
Note, that value isn't "of the set that is below ω".
Fact is that below ω there are natnumbers and above there are none.
Hence they cease below the bound ω.
Yes, but it can't be the "Upper Bound" used to step back to from ω.
There is not "Upper Bound" for the Natural Numbers *IN* the Natural
Numbers to be the "last" value to step back to.
Thus, your logic about ω-1 breaks. The value does not exist,
On 4/26/2024 10:37 AM, WM wrote:
Le 26/04/2024 à 01:11, Jim Burns a écrit :
On 4/25/2024 4:03 PM, WM wrote:
If all smaller numbers are doubled,
then there is no place for
the doubled numbers below ω.
If n is below ω
then n can be counted to from 0
then n⋅2 can be counted to from n
That is true for definable numbers
but not for the last numbers before ω.
If any number below n canNOT be counted to from 0
then n itself canNOT be counted to from 0
Thus,
each number which CAN be counted to from 0
is not above
any number which canNOT be counted to from 0
By definition,
ω is between
numbers which CAN be counted to from 0 and
numbers which canNOT be counted to from 0
Imagine being someone who denies that definition of ω
Because the following isn't a claim about ω
you (the denier) should still admit:
if n can be counted to from 0
then n⋅2 can be counted to from n
then n⋅2 can be counted to from 0 (through n)
If ω exists as defined,
then doubling never crosses ω
(from CAN to canNOT)
Even if ω doesn't exist as defined,
then doubling never crosses
_where ω would be if ω existed_
(from CAN to canNOT)
ω is NOT a simply.humongous.instance of
the numbers 0 1 2 3 ...
ω marks a boundary between domains with
different descriptions (CAN and canNOT).
Imagine being someone who denies that
ω marks that boundary.
With or without the marker,
the domains (CAN and canNOT) remain
the domains (CAN and canNOT).
On 4/26/2024 10:37 AM, WM wrote:
Le 26/04/2024 à 01:11, Jim Burns a écrit :
On 4/25/2024 4:03 PM, WM wrote:
If all smaller numbers are doubled,
then there is no place for
the doubled numbers below ω.
If n is below ω
then n can be counted to from 0
then n⋅2 can be counted to from n
That is true for definable numbers
but not for the last numbers before ω.
If any number below n canNOT be counted to from 0
then n itself canNOT be counted to from 0
Thus,
each number which CAN be counted to from 0
is not above
any number which canNOT be counted to from 0
By definition,
ω is between
numbers which CAN be counted to from 0 and
numbers which canNOT be counted to from 0
Imagine being someone who denies that definition of ω
Because the following isn't a claim about ω
you (the denier) should still admit:
if n can be counted to from 0
then n⋅2 can be counted to from n
then n⋅2 can be counted to from 0 (through n)
If ω exists as defined,
then doubling never crosses ω
(from CAN to canNOT)
On 4/26/24 1:06 PM, WM wrote:
Le 26/04/2024 à 04:08, Richard Damon a écrit :Maybe you should. "Order Types" are not "Numbers" but properties of how
On 4/25/24 3:51 PM, WM wrote:
Le 24/04/2024 à 20:08, FromTheRafters a écrit :
WM explained :
Le 24/04/2024 à 12:49, FromTheRafters a écrit :
Richard Damon wrote :
Of course, if you logic tries to make that transfinite value
finite because that is all it can handle, your system just blows >>>>>>>> up in your face.
IMO, it might be best not to think of it as a value at all.
Cantor called it a number.
A number representing an order type.
like 2 or 5.
No, because 2 or 5 were FINITE numbers, not transfinite numbers.
Learn what an order type is according to Cantor.
you order the set of numbers.
Le 26/04/2024 à 21:10, Richard Damon a écrit :
On 4/26/24 1:06 PM, WM wrote:
Le 26/04/2024 à 04:08, Richard Damon a écrit :Maybe you should. "Order Types" are not "Numbers" but properties of
On 4/25/24 3:51 PM, WM wrote:
Le 24/04/2024 à 20:08, FromTheRafters a écrit :
WM explained :
Le 24/04/2024 à 12:49, FromTheRafters a écrit :
Richard Damon wrote :
Of course, if you logic tries to make that transfinite value >>>>>>>>> finite because that is all it can handle, your system just
blows up in your face.
IMO, it might be best not to think of it as a value at all.
Cantor called it a number.
A number representing an order type.
like 2 or 5.
No, because 2 or 5 were FINITE numbers, not transfinite numbers.
Learn what an order type is according to Cantor.
how you order the set of numbers.
Your ignorance is actually unlimited.
Cantor created and defined these notions: Den Ordnungstypus einer wohlgeordneten Menge F nennen wir die ihr zukommende "Ordnungszahl".
Regards, WM
Le 26/04/2024 à 20:13, Richard Damon a écrit :
On 4/26/24 1:02 PM, WM wrote:
Le 26/04/2024 à 16:49, Richard Damon a écrit :
On 4/26/24 10:41 AM, WM wrote:
The set is bounded by ω.
But ω isn't in the set, so it can't be the upper bound of the set
that is below ω
Learn the meaning of upper bound. ω*2 is an upper bound too.
But since it isn't in the set, you can't use it as the value to "step
back" to.
The question was whether it is an upper bound. Bounds need not belong to
a set.
Note, that value isn't "of the set that is below ω".
Therefore this set ends before.
Fact is that below ω there are natnumbers and above there are none.
Hence they cease below the bound ω.
Yes, but it can't be the "Upper Bound" used to step back to from ω.
Then something else must be used.
There is not "Upper Bound" for the Natural Numbers *IN* the Natural
Numbers to be the "last" value to step back to.
Thus, your logic about ω-1 breaks. The value does not exist,
In actual infinity something is below ω.
Regards, WM
Le 26/04/2024 à 20:53, Jim Burns a écrit :
On 4/26/2024 10:37 AM, WM wrote:
That is true for definable numbers
but not for the last numbers before ω.
If any number below n canNOT be counted to from 0
then n itself canNOT be counted to from 0
Thus,
each number which CAN be counted to from 0
is not above
any number which canNOT be counted to from 0
By definition,
ω is between
numbers which CAN be counted to from 0 and
numbers which canNOT be counted to from 0
Imagine being someone who denies that definition of ω
Because the following isn't a claim about ω
you (the denier) should still admit:
if n can be counted to from 0
then n⋅2 can be counted to from n
then n⋅2 can be counted to from 0 (through n)
If ω exists as defined,
then doubling never crosses ω
(from CAN to canNOT)
Hence
not all natural numbers exist as defined
for visible numbers.
I suppose Cantor Pairing is another one that WM does not understand.
We can map back-and-forth using the pairing functions.
Nothing is lost, and [...]
4.
If n can be counted to from 0
then n⋅2 can be counted to _from n_
n⋅2 can be counted to from 0 _through n_
On 4/27/24 10:54 AM, WM wrote:
Le 26/04/2024 à 21:10, Richard Damon a écrit :
On 4/26/24 1:06 PM, WM wrote:
Le 26/04/2024 à 04:08, Richard Damon a écrit :Maybe you should. "Order Types" are not "Numbers" but properties of
On 4/25/24 3:51 PM, WM wrote:
Le 24/04/2024 à 20:08, FromTheRafters a écrit :
WM explained :
Le 24/04/2024 à 12:49, FromTheRafters a écrit :
Richard Damon wrote :
Of course, if you logic tries to make that transfinite value >>>>>>>>>> finite because that is all it can handle, your system just >>>>>>>>>> blows up in your face.
IMO, it might be best not to think of it as a value at all.
Cantor called it a number.
A number representing an order type.
like 2 or 5.
No, because 2 or 5 were FINITE numbers, not transfinite numbers.
Learn what an order type is according to Cantor.
how you order the set of numbers.
Your ignorance is actually unlimited.
Cantor created and defined these notions: Den Ordnungstypus einer
wohlgeordneten Menge F nennen wir die ihr zukommende "Ordnungszahl".
Right, and 2 or 5 are not assigned as a number for an ordinal type.
There is no set with an Ordinal Type of 2 or 5.
On 4/27/24 10:36 AM, WM wrote:
But since it isn't in the set, you can't use it as the value to "step
back" to.
The question was whether it is an upper bound. Bounds need not belong to
a set.
They do if you want to use them to step into the set, like you did.
Note, that value isn't "of the set that is below ω".
Therefore this set ends before.
Only if it HAS an "end"
On 4/27/2024 10:46 AM, WM wrote:
If n can be counted to from 0
then n⋅2 can be counted to _from n_
then n⋅2 can be counted to from 0 _through n_
If n is before ω
then n⋅2 before ω
However,
it is always true that,
if n is before ω
then n⋅2 before ω
My guess is that you (WM) feel
Am 28.04.2024 um 19:29 schrieb Jim Burns:
4.
If n can be counted to from 0
then n⋅2 can be counted to _from n_
Since n⋅2 is just short for n+n (in my book).
n+n can be counted to from 0 the following way: 0 -> 1 -> 2 -> .. -> n
and further n -> n+1 -> n+2 -> .. -> n+n (just like we already did when counting from 0 to n).*)
Hence
n⋅2 can be counted to from 0 _through n_
Indeed!
In other words,
"If n is a flying rainbow sparkle pony
then n⋅2 is a flying rainbow sparkle pony."
{1, 2, 3, ..., ω}*2
results in {2, 4, 6, ..., ω*2}
ω or ω+1 is [NOT] amidst [...]
Le 28/04/2024 à 19:29, Jim Burns a écrit :
On 4/27/2024 10:46 AM, WM wrote:
If n can be counted to from 0
then n⋅2 can be counted to _from n_
then n⋅2 can be counted to from 0 _through n_
That is true.
If n is before ω
then n⋅2 before ω
That is not true.
However,
it is always true that,
if n is before ω
then n⋅2 before ω
No.
My guess is that you (WM) feel
Whatever you guess, {1, 2, 3, ..., ω}*2 results in {2, 4, 6, ..., ω*2}.
crippled [brain]
{1, 2, 3, ..., ω}*2 results in {2, 4, 6, ..., ω*2}
ω or ω+1 is amidst
Le 28/04/2024 à 19:29, Jim Burns a écrit :
1.
We describe some things.
For our convenience,
we refer to what we describe by some label.
Whatever label we use, "natural number" or
"flying rainbow sparkle pony",
using that label uses _that description_
If n can be counted to from 0
then n⋅2 can be counted to _from n_
then n⋅2 can be counted to from 0 _through n_
That is true.
If n is before ω
then n⋅2 before ω
That is not true.
Le 27/04/2024 à 19:27, Richard Damon a écrit :
On 4/27/24 10:36 AM, WM wrote:
But since it isn't in the set, you can't use it as the value to
"step back" to.
The question was whether it is an upper bound. Bounds need not belong
to a set.
They do if you want to use them to step into the set, like you did.
I use the upper bond only as uoper bound.
Note, that value isn't "of the set that is below ω".
Therefore this set ends before.
Only if it HAS an "end"
The end is smaller than ω.
Regards, WM
Le 27/04/2024 à 19:38, Richard Damon a écrit :
On 4/27/24 10:54 AM, WM wrote:
Le 26/04/2024 à 21:10, Richard Damon a écrit :
On 4/26/24 1:06 PM, WM wrote:
Le 26/04/2024 à 04:08, Richard Damon a écrit :Maybe you should. "Order Types" are not "Numbers" but properties of
On 4/25/24 3:51 PM, WM wrote:
Le 24/04/2024 à 20:08, FromTheRafters a écrit :
WM explained :
Le 24/04/2024 à 12:49, FromTheRafters a écrit :
Richard Damon wrote :Cantor called it a number.
Of course, if you logic tries to make that transfinite value >>>>>>>>>>> finite because that is all it can handle, your system just >>>>>>>>>>> blows up in your face.
IMO, it might be best not to think of it as a value at all. >>>>>>>>>
A number representing an order type.
like 2 or 5.
No, because 2 or 5 were FINITE numbers, not transfinite numbers.
Learn what an order type is according to Cantor.
how you order the set of numbers.
Your ignorance is actually unlimited.
Cantor created and defined these notions: Den Ordnungstypus einer
wohlgeordneten Menge F nennen wir die ihr zukommende "Ordnungszahl".
Right, and 2 or 5 are not assigned as a number for an ordinal type.
Your ignorance is actually unlimited but your capability to learn is zero.
There is no set with an Ordinal Type of 2 or 5.
What are (1, 2) and (1, 2, 3, 4, 5) or (17, 28, 55, 67, 100)?
Regards, WM
If n is before ω then n⋅2 is before ω. (*)
That is not true.
Jim Burns schrieb:
ω is NOT a simply.humongous.instance of
the numbers 0 1 2 3 ...
ω marks a boundary between domains with
different descriptions (CAN and canNOT).
Imagine being someone who denies that
ω marks that boundary.
With or without the marker,
the domains (CAN and canNOT) remain
the domains (CAN and canNOT).
This is really well put!
Unfortunately,
WM is not interested in
our ideas of (our) math and logic
but in his own (mostly read up upon) ideas and
his flexible and willingly deformable "true logic"
which a "normal" person can even "feel" --
but even more is WM interested in
WHAT (we) folks CLAIM and STATE about (our) math,
more than about that math itself.
The end is smaller than ω.
Am 02.05.2024 00:46:01 Jim Burns schrieb:
I think it's possible that
WM has no objection
to the run.of.the.mill claims about
the first.upper.bound of
numbers which can be counted.to from 0
(and things like that) as long as
those claims are not made using symbols
such as ω ℕ ℵ₀ which
WM has made his mathematizedᵂᴹ claims about.
I think that WM has a very fixed idea of the
"world of math" which is fixed by nature and
not a creation of culture in the mind of men
which WM thinks is given by THE ONE real nature
and by THE ONE true logic which can ONLY be
detected and "seen" rather than defined and
built.
If n can be counted to from 0
then n⋅2 can be counted to _from n_
then n⋅2 can be counted to from 0 _through n_
That is true.
If n is before ω
then n⋅2 before ω
That is not true.
Am 02.05.2024 20:00:50 Jim Burns schrieb:
WM will say that
ω is the first infiniteᵂᴹ ordinal,
but he does NOT mean that
ω is the first infiniteⁿᵒᵗᐧᵂᴹ ordinal.
WM rejects the idea that
there is "actual infinity" which
is realized in nature
and WM rejects it also
in the any mental space because
it is "wrong logic" and idiocy
(but he tends to "allow" for
the idea of "potential infinity").
WMs philosophy is like ultrafinitistic
while he is too dense for any mathematic thinking
which he lacks to basically understand altogether)...
Le 02/05/2024 à 19:06, Moebius a écrit :
The sequence of natural numbers (all of which are smaller than ω) does
not have an end,
Then it would stretch beyond all, in particular beyond ω. For your convenience: The sequence of unit fractions would stretch beyond zero.
Le 02/05/2024 à 19:06, Moebius a écrit :
Am 30.04.2024 um 14:56 schrieb WM:
The end is smaller than ω.
The sequence of natural numbers (all of which are smaller than ω) does
not have an end,
Then it would stretch beyond all, in particular beyond ω. For your convenience: The sequence of unit fractions would stretch beyond zero.
Regards, WM
Am 30.04.2024 um 14:56 schrieb WM:
The end is smaller than ω.
The sequence of natural numbers (all of which are smaller than ω) does
not have an end,
Learn what an order type is according to Cantor.Maybe you should. "Order Types" are not "Numbers" but properties of
how you order the set of numbers.
Right, and 2 or 5 are not assigned as a number for an ordinal type.
Your ignorance is actually unlimited but your capability to learn is zero. >>
There is no set with an Ordinal Type of 2 or 5.
What are (1, 2) and (1, 2, 3, 4, 5) or (17, 28, 55, 67, 100)?
Finite sets,
Am 30.04.2024 um 15:12 schrieb WM:
If n is before ω then n⋅2 is before ω. (*)
That is not true.
Wir haben mithin also gezeigt, dass "If n is
before ω then n⋅2 is before ω" gilt. qed
On 4/30/24 8:56 AM, WM wrote:
Le 27/04/2024 à 19:27, Richard Damon a écrit :
On 4/27/24 10:36 AM, WM wrote:
But since it isn't in the set, you can't use it as the value to
"step back" to.
The question was whether it is an upper bound. Bounds need not belong
to a set.
They do if you want to use them to step into the set, like you did.
I use the upper bond only as upper bound.
But then you can't talk about using it to "step back" from omega.
Yes, ALL Natural Numbers are less than omega.
That doesn't mean there is a largest Natural Number.
In fact, the fact that the smallest Upper Bound we can find is Omega,
which is out of the set of Natural Numbers means that there is no
"Highest" Natural number.
But there isn't a specific number that is the end.
That is the problem with your logic.
I think it's possible that
WM thinks that
a mathematical claim is mathematical because
of the great certainty with which it is expressed.
I think that WM has a very fixed idea of the
"world of math" which is fixed by nature and
not a creation of culture in the mind of men
WM rejects the idea that there is "actual infinity" which is
realized in nature and WM rejects it also in the any mental space
because
My guess is that,
whether WM is aware of it or not,
he follows this line of thought:
| Infinitenessⁿᵒᵗᐧᵂᴹ is weird.
| Infinitenessⁿᵒᵗᐧᵂᴹ is wrong.
WM call various things
"actually infinite" and "potentially.infinite".
What does he mean by those terms?
NOT "What does Cantor mean? Euclid mean?"
They have no successors which must remain.
Le 02/05/2024 à 01:32, Tom Bola a écrit :
I think that WM has a very fixed idea of the
"world of math" which is fixed by nature and
not a creation of culture in the mind of men
I don't know whether actual infinity is true.
Le 03/05/2024 à 11:07, Jim Burns a écrit :
WM call various things
"actually infinite" and "potentially.infinite".
What does he mean by those terms?
NOT "What does Cantor mean? Euclid mean?"
By actual infinity I mean what Cantor started but could not stand.
By actual infinity I mean what ZFC claims but cannot maintain.
It is the idea that infinite sets are fixed such that no element can be
added or removed.
It is the idea that the natural numbers reach immediately from 0 until ω such that no number can be inserted between ℕ and ω. If every natural nunmber is increased by 1, then the set is shifted from {1, 2, 3, ...} to
{2, 3, 4, ..., ω}.
this omega is the first
infinite ordinal. It is not larger than the natural numbers, it *is*
the natural numbers.
Le 03/05/2024 à 13:48, joes a écrit :
By actual infinity I mean what Cantor started but could not stand.
By actual infinity I mean what ZFC claims but cannot maintain.
It is the idea that infinite sets are fixed such that no element can be
added or removed.
It is the idea that the natural numbers reach immediately from 0 until ω such that no number can be inserted between ℕ and ω. If every natural nunmber is increased by 1, then the set is shifted from {1, 2, 3, ...} to
{2, 3, 4, ..., ω}.
Le 03/05/2024 à 15:23, FromTheRafters a écrit :
this omega is the first
infinite ordinal. It is not larger than the natural numbers, it *is*
the natural numbers.
This idea is not the meaning assigned to ω by Cantor who invented it.
Am 03.05.2024 15:57:52 WM drivels bullshit:
Le 03/05/2024 à 11:07, Jim Burns a écrit :
WM call various things
"actually infinite" and "potentially.infinite".
What does he mean by those terms?
NOT "What does Cantor mean? Euclid mean?"
By actual infinity I mean what Cantor started but could not stand.
By actual infinity I mean what ZFC claims but cannot maintain.
It is the idea that infinite sets are fixed such that no element can be
added or removed.
It is the idea that the natural numbers reach immediately from 0 until ω
such that no number can be inserted between ℕ and ω. If every natural
nunmber is increased by 1, then the set is shifted from {1, 2, 3, ...} to
{2, 3, 4, ..., ω}.
No, nothing must be shifted - it is enough to give each element another
name like above, or e.g. "a", "b", ..., "aa", "bb", ... or whatever.
Le 03/05/2024 à 16:09, Tom Bola a écrit :
Am 03.05.2024 15:57:52 WM drivels bullshit:
Le 03/05/2024 à 11:07, Jim Burns a écrit :
WM call various things
"actually infinite" and "potentially.infinite".
What does he mean by those terms?
NOT "What does Cantor mean? Euclid mean?"
By actual infinity I mean what Cantor started but could not stand.
By actual infinity I mean what ZFC claims but cannot maintain.
It is the idea that infinite sets are fixed such that no element can be
added or removed.
It is the idea that the natural numbers reach immediately from 0 until ω >>> such that no number can be inserted between ℕ and ω. If every natural >>> nunmber is increased by 1, then the set is shifted from {1, 2, 3, ...} to >>> {2, 3, 4, ..., ω}.
No, nothing must be shifted - it is enough to give each element another
name like above, or e.g. "a", "b", ..., "aa", "bb", ... or whatever.
When 1 is added to 3, then no other name but precisely 4 is required.
omega is the first infinite ordinal. It is not larger than the natural numbers, it *is* the
natural numbers.
Am 03.05.2024 16:10:04 WM schrieb:
Le 03/05/2024 à 15:23, FromTheRafters a écrit :
this omega is the first
infinite ordinal. It is not larger than the natural numbers, it *is*
the natural numbers.
This idea is not the meaning assigned to ω by Cantor who invented it.
But IN and w are isomorphic - which is a fact, no matter whatever
"Cantor invented".
Am 03.05.2024 um 16:13 schrieb Tom Bola:
But IN and w are isomorphic - which is a fact, no matter whatever "Cantor invented".IN and ω are IDENTICAL in modern/axiomatic set theory (say ZF(C)) you fucking asshole.
Lies mal irgendwas zu dem Thema oder halt die Schnauze (gleiches würde ich auch RR empfehlen).
Am 03.05.2024 16:10:04 WM schrieb:
But IN and w are isomorphic - which is a fact, no matter whatever
"Cantor invented".
Le 03/05/2024 à 16:13, Tom Bola a écrit :
Am 03.05.2024 16:10:04 WM schrieb:
Le 03/05/2024 à 15:23, FromTheRafters a écrit :
this omega is the first
infinite ordinal. It is not larger than the natural numbers, it *is*
the natural numbers.
This idea is not the meaning assigned to ω by Cantor who invented it.
But IN and w are isomorphic - which is a fact, no matter whatever
"Cantor invented".
Isomorphic is not identical.
Further N has no morphology or structure,
hence it differes considerably from ω.
Am 03.05.2024 um 16:13 schrieb Tom Bola:
Am 03.05.2024 16:10:04 WM schrieb:is not the meaning assigned to ω by Cantor who invented it.
But IN and w are isomorphic - which is a fact, no matter whatever
"Cantor invented".
IN and ω are IDENTICAL
you fucking asshole
Le 01/05/2024 à 00:49, Richard Damon a écrit :
[...]
It is so easy:
Every findable number has ℵo successors.
Not all can be found.All natural numbers have no successors.
There are more.
Every findable number has ℵo successors.
Not all can be found.
All natural numbers have no successors.
There are more.
Every findable number has ℵo successors.
Not all can be found.All natural numbers have no successors.
There are more.
But IN and w are isomorphic - which is a fact, no matter whatever
"Cantor invented".
IN and ω are IDENTICAL in modern/axiomatic set theory (say ZF(C)) you fucking asshole.
Am 03.05.2024 um 16:13 schrieb Tom Bola:
Am 03.05.2024 16:10:04 WM schrieb:is not the meaning assigned to ω by Cantor who invented it.
But IN and w are isomorphic - which is a fact, no matter whatever
"Cantor invented".
IN and ω are IDENTICAL in modern/axiomatic set theory (say ZF(C))
On 5/3/2024 9:21 AM, WM wrote:
Le 01/05/2024 à 00:49, Richard Damon a écrit :
It is so easy:
Every findable number has ℵo successors.
Not all can be found.All natural numbers have no successors.
There are more.
For numbers j k, with j before k
NEVER is it so that
k CAN be counted.
There are no successors which unavoidably will remain.From all *definable* natnumbers (Peano-natnumbers) however we know that
j canNOT be counted.to from 0
For numbers j j+1
NEVER is it so that
j CAN be counted.to from 0 and
j+1 canNOT be counted.to from 0
Every findable number has ℵo successors.
Not all can be found.
Number i CAN be counted.to from 0
Number k CAN be counted.to from 0
For numbers j before k
NEVER is it so that
i+j CAN be counted.to from 0 and
(i+j)+1 canNOT be counted.to from 0
and
NEVER is it so that
i+j CAN be counted.to from 0 and
i+(j+1) canNOT be counted.to from 0
ALWAYS is it so that
i+k CAN be counted.to from 0
Number i CAN be counted.to from 0
For each number k which CAN be counted.to from 0
i+k _after i_ CAN be counted.to from 0
Each number i which CAN be counted.to from 0
has as many successors i+k as there are
all numbers k which CAN be counted.to from 0
Each number i which CAN be counted.to from 0
has ℵ₀ successors which CAN be counted.to from 0
All natural numbers have no successors.
There are more.
‖ is the first.upper.bound of
numbers which can be counted.to from 0
It's almost always named ω instead
but you (WM) don't accept the name ω for that
and there are no logical consequences to
naming it something else, such as ‖
ξ is after ‖
NEVER is it so that
ξ CAN be counted.to from 0
Otherwise, ‖ isn't a bound.
j is before ‖
NEVER is it so that
j canNOT be counted.to from 0
Otherwise, j, which
no number which CAN be counted.to from 0 is after
is before.‖.upper.bound of
numbers which CAN be counted.to from 0
And ‖ ISN'T the first.upper.bound of
numbers which can be counted.to from 0
Each number i which CAN be counted.to from 0
has ℵ₀ successors which CAN be counted.to from 0
Each number i before ‖
has ℵ₀ successors before ‖
----
‖ canNOT be counted.to from 0
Otherwise,
‖+1 CAN be counted.to from 0
and ‖ ISN'T a bound.
Every findable number has ℵo successors.
Not all can be found.All natural numbers have no successors.
There are more.
Each number i before ‖
has ℵ₀ successors before ‖
Each number i before ‖ and only those before
CAN be counted.to from 0
‖ canNOT be counted.to from 0
Le 03/05/2024 à 17:57, Jim Burns a écrit :
On 5/3/2024 9:21 AM, WM wrote:
It is so easy:
Every findable number has ℵo successors.
Not all can be found.All natural numbers have no successors.
There are more.
For numbers j k, with j before k
NEVER is it so that
k CAN be counted.
You can claim that only for
numbers which you can distinguish.
You can claim that only for
numbers which you can distinguish.
Le 03/05/2024 à 11:07, Jim Burns a écrit :
WM call various things
"actually infinite" and "potentially.infinite".
What does he mean by those terms?
NOT "What does Cantor mean? Euclid mean?"
By actual infinity I mean what Cantor [...]
It is the idea that infinite sets are fixed
such that no element can be added or removed.
It is the idea that
the natural numbers reach immediately from 0 until ω
If every natural nunmber is increased by 1,
then the set is shifted from {1, 2, 3, ...}
to {2, 3, 4, ..., ω}.
The increase of every number mentioned above
can only be applied collectively.
Le 03/05/2024 à 16:33, Moebius a écrit :
IN and ω are IDENTICAL in modern/axiomatic set theory (say ZF(C))
showing that <bla ba bla>
Le 03/05/2024 à 16:33, Moebius a écrit :
IN and ω are IDENTICAL in modern/axiomatic set theory (say ZF(C))
showing that <bla ba bla>
On 5/3/2024 1:03 PM, WM wrote:
You can claim that only for
numbers which you can distinguish.
We can claim
for a number which can be counted.to from 0
that each number before it
can be counted.to from 0
WM used his keyboard to write :
Try to overcome your inertia and comprehend the difference:
All natural numbers can be subtracted from the set of all real numbers
No,
Sets
still don't change, no matter how much you like to think of them as
changing.
Each number which CAN be counted.to from 0
is before
ℵ₀.many numbers which CAN be counted.to from 0
Your description of ω is a description of
a number which CAN be counted.to from 0
Am 03.05.2024 16:33:47 Moebius drivels:
But IN and w are isomorphic - which is a fact, no matter whatever
"Cantor invented".
IN and ω are IDENTICAL in modern/axiomatic set theory (say ZF(C)) you
fucking asshole.
Here is some help for you <bla bla bla>
Le 03/05/2024 à 16:33, Moebius a écrit :
IN and ω are IDENTICAL in modern/axiomatic set theory (say ZF(C))
showing that <bla ba bla>
The Isomorphiesatz von Dedekind
Le 01/05/2024 à 23:46, Moebius a écrit :
Am 30.04.2024 um 15:12 schrieb WM:
If n is before ω then n⋅2 is before ω. (*)
That is not true.
Wir haben mithin also gezeigt, dass "If n is
before ω then n⋅2 is before ω" gilt. qed
Your proof holds for all natnumbers that can be found - a tiny initial segment of ℕ.
Every findable natnumber has ℵo natsuccessors. ==> Not all can be found. All natural numbers have no natsuccessors. ==> There are more natnumbers
than can be found.
But they can only be applied collectively.
Am 03.05.2024 um 18:37 schrieb Tom Bola:
Am 03.05.2024 16:33:47 Moebius drivels:
But IN and w are isomorphic - which is a fact, no matter whatever
"Cantor invented".
IN and ω are IDENTICAL in modern/axiomatic set theory (say ZF(C))
you fucking asshole
I wonder if WM thinks that if every natural number is less than ω, then
ω must represent a largest natural number. Uggghhhh. ;^o
On 5/3/2024 7:17 AM, Moebius wrote:
Am 03.05.2024 um 15:23 schrieb FromTheRafters:
omega is the first infinite ordinal. It is not larger than the
natural numbers, it *is* the natural numbers.
Well, actually, it is larger than _each and every_ natural number.
Using symbols: An e IN: n < ω.
Using a common depiction: 0 < 1 < 2 < 3 < ... < ω.
See: https://en.wikipedia.org/wiki/Ordinal_number
On the other hand, you are right, in the context of axiomatic set
theory (if the natural numbers and the ordinals are defined due to von
Neumann) we do have IN = ω.
0 < ω
1 < ω
2 < ω
3 < ω
...
This holds true for infinity...
0 < ω
1 < ω
2 < ω
3 < ω
... ad infinitum.
ω blows WM's mind. All natural numbers are less than it. Therefore, WM thinks ω must be a largest natural number. wow. What a fool!
Knowing WM, his rope would be 100 feet long and he jumps of a 25 foot
cliff.
Safety rope :-) :-) :-)
Le 03/05/2024 à 21:32, Jim Burns a écrit :
Your description of ω is a description of
a number which CAN be counted.to from 0
Not at all! You miunderstand or are lying.
It is the idea that the natural numbers
reach immediately from 0 until ω
If every natural nunmber is increased by 1,
then the set is shifted
from {1, 2, 3, ...} to {2, 3, 4, ..., ω}.
Le 01/05/2024 à 00:49, Richard Damon a écrit :
On 4/30/24 8:56 AM, WM wrote:
Le 27/04/2024 à 19:27, Richard Damon a écrit :
On 4/27/24 10:36 AM, WM wrote:
But since it isn't in the set, you can't use it as the value to
"step back" to.
The question was whether it is an upper bound. Bounds need not
belong to a set.
They do if you want to use them to step into the set, like you did.
I use the upper bond only as upper bound.
But then you can't talk about using it to "step back" from omega.
I can't because almost all natural numbers are dark.
Yes, ALL Natural Numbers are less than omega.
Yes.
That doesn't mean there is a largest Natural Number.
In fact, the fact that the smallest Upper Bound we can find is Omega,
which is out of the set of Natural Numbers means that there is no
"Highest" Natural number.
The findable natural numbers constitute only a small initial segment of
ℕ. Every findable number has ℵo successors. But these successors are natural numbers too which can be subtracted from ℕ, such than no natural numbers remain, only collectively.
But there isn't a specific number that is the end.
There is no such number findable.
That is the problem with your logic.
That is not a problem with logic. It is only a problem for people who
cannot think farther than the definable or findable numbers reach. Intellectual inertia.
It is so easy:
Every findable number has ℵo successors. ==> Not all can be found.
All natural numbers have no successors. ==> There are more.
Regards, WM
Le 03/05/2024 à 11:07, Jim Burns a écrit :
WM call various things
"actually infinite" and "potentially.infinite".
What does he mean by those terms?
NOT "What does Cantor mean? Euclid mean?"
By actual infinity I mean what Cantor started but could not stand.
By actual infinity I mean what ZFC claims but cannot maintain.
It is the idea that infinite sets are fixed such that no element can be
added or removed.
It is the idea that the natural numbers reach immediately from 0 until ω such that no number can be inserted between ℕ and ω. If every natural nunmber is increased by 1, then the set is shifted from {1, 2, 3, ...}
to {2, 3, 4, ..., ω}.
Of course the precondition for this theory is that almost all natural
numbers are dark, only a tiny initial segment contains numbers that can
be applied individually. The increase of every number mentioned above
can only be applied collectively.
Regards, WM
Le 02/05/2024 à 01:32, Tom Bola a écrit :
I think that WM has a very fixed idea of the "world of math" which is
fixed by nature and not a creation of culture in the mind of men
I don't know whether actual infinity is true. But if so, then all
natural numbers can be subtracted from the set of all real numbers. ==>
They have no successors which must remain.
From all definable natnumbers however we know that they have ℵo successors. ==> Not all are definable. A huge, infinite subset is dark.
Regards, WM
By actual infinity I mean what Cantor started but could not stand.
By actual infinity I mean what ZFC claims but cannot maintain.
Le 03/05/2024 à 21:32, Jim Burns a écrit :
On 5/3/2024 9:57 AM, WM wrote:
It is the idea that infinite sets are fixed
such that no element can be added or removed.
It is the idea that
the natural numbers reach immediately from 0 until ω
Each number which CAN be counted.to from 0
is before
ℵ₀.many numbers which CAN be counted.to from 0
If all could be counted to,
they would not remain after every counted number.
If all could be counted to,
they would not remain after every counted number.
On 05/05/2024 10:59 AM, Jim Burns wrote:
The meaning of (1.) depends upon
'n' NOT being
the true name of any number which CAN be counted.to,
the way that "Rumpelstiltskin" is the true name of
a certain straw.into.gold.spinner.
An analogy better than "name" is "pronoun"
'n' is more like a pronoun than a name.
"It is a natural number", etc.
Variable.names are a big improvement over pronouns
because, in every natural language I'm aware of,
there are no more than a handful of pronouns,
used with many handfuls of referents, and
their distinct referents are kept distinct
by context, AKA, figuring.it.out.
Even if the figuring.out doesn't fail, a lot of work.
The expression
| x < y and y < z implies x < z
|
is a big improvement in clarity over
a paragraph of muddle with three pronouns.
x y z act like pronouns, though.
How about disambiguating quantifiers so that
something like the universal quantifier
gets disambiguated to reflect
a for-any/for-each/for-every/for-all
when it's so that
things like the Sorites/Heap or transfer principle
apply.
Similarly
the existential quantifier is often to be disambiguated
"exists", "exists-unique", "exists-plural",
these kinds of things.
English as a language has a rich variety of copulas.
Le 03/05/2024 à 21:32, Jim Burns a écrit :
Each number which CAN be counted.to from 0 is before
ℵ₀.many numbers which CAN be counted.to from 0
If all could be counted to, they would not remain after every [number which can be] counted [to].
On 05/03/2024 07:06 AM, Tom Bola wrote:
Am 03.05.2024 15:41:24 WM drivels:
Le 02/05/2024 à 01:32, Tom Bola a écrit :
I think that WM has a very fixed idea of the
"world of math" which is fixed by nature and
not a creation of culture in the mind of men
I don't know whether actual infinity is true.
Being true IN MIND is
all the Mathematicians want and need.
That there is a "theory of truth" has two schools:
"idealists" are "platonists" who
agree mathematics is discovered,
"nominalists" are "figurists" who
each say mathematics is invented.
on 5/3/2024, WM supposed :
Try to overcome your inertia and comprehend the difference:
All natural numbers can be subtracted from the set of all real numbers
No,
No? You can't find |R \ |N?
That is the construction of a "Difference Set" not some sort of
subtraction within a set which makes it smaller. Sets don't change.
On 5/3/2024 4:31 PM, WM wrote:
If all could be counted to,
they would not remain after every counted number.
From each number n which CAN be counted.to
for each number k which CAN be counted.to
more.than.k numbers which CAN be counted.to
can be reached immediately
from n to n+k+1
None of these immediate more.than.k numbers remain
after every number n which CAN be counted.to.
Am 03.05.2024 um 22:31 schrieb WM:
Le 03/05/2024 à 21:32, Jim Burns a écrit :
Each number which CAN be counted.to from 0 is before
ℵ₀.many numbers which CAN be counted.to from 0
If all could be counted to, they would not remain after every [number which can
be] counted [to].
Du bringst hier wieder einmal die
Quantoren durcheinander.
On 5/3/24 9:41 AM, WM wrote:
I don't know whether actual infinity is true. But if so, then all
natural numbers can be subtracted from the set of all real numbers. ==>
They have no successors which must remain.
From all definable natnumbers however we know that they have ℵo
successors. ==> Not all are definable. A huge, infinite subset is dark.
But the problem is you are using logic that can't actually have the set
of Natural Numbers,
You THINK you can subtract your set of "definable" numbers from the
Natual Numbers in your system,
On 5/3/24 9:21 AM, WM wrote:
Every findable number has ℵo successors. ==> Not all can be found.
All natural numbers have no successors. ==> There are more.
So, what is your actual definition of a "findable" number?
How does that definition actually differ IN ITS DEFINITION from the definition of the Natural Numbers?
On 05/05/2024 03:02 PM, Jim Burns wrote:
For my wish,
I would like everyone to be clear on what
standard.issue quantifiers and variables
mean.
I think that,
way off in that glorious future,
both you and I will be able to be
satisfactorily understood.
And what more could there be
to wish for?
Well, one might aver that extra-ordinary
universal quantifiers are merely syntactic sugar,
yet there's that in the low- and high- orders,
or the first and final, that what they would
reflect of the _effects_ of quantification,
something like
for-any A?
for-each A+
for-every A*
for-all A$
that it is so that the sputniks or extras
of the quantification in the extra-ordinary,
have that a quantifier disambiguation:
is in the syntax.
Then for the rest of it,
On 5/6/2024 2:59 PM, WM wrote:
Le 05/05/2024 à 19:59, Jim Burns a écrit :
On 5/3/2024 4:31 PM, WM wrote:
If all could be counted to,
they would not remain after every counted number.
Of those which CAN be counted.to,
each CAN be counted to.
Nevertheless almost all, namely ℵo, remain.
Of those which CAN be counted.to,
each CAN be counted to,
and
none remain after all which CAN be counted.to
Of those which CAN be counted.to,
each CAN be counted to,
and
none remain after all which CAN be counted.to,
Le 05/05/2024 à 19:59, Jim Burns a écrit :
On 5/3/2024 4:31 PM, WM wrote:
If all could be counted to,
they would not remain after every counted number.
From each number n which CAN be counted.to
for each number k which CAN be counted.to
more.than.k numbers which CAN be counted.to
can be reached immediately
from n to n+k+1
None of these immediate more.than.k numbers remain
after every number n which CAN be counted.to.
Nevertheless almost all, namely ℵo, remain.
Le 04/05/2024 à 05:18, Richard Damon a écrit :
So, what is your actual definition of a "findable" number?
A number that can be put in trichotomy with its neighbours.
So, what is your actual definition of a "findable" number?
A number that can be put in trichotomy with its neighbours.
Le 06/05/2024 à 22:16, Jim Burns a écrit :
On 5/6/2024 2:59 PM, WM wrote:
Le 05/05/2024 à 19:59, Jim Burns a écrit :
On 5/3/2024 4:31 PM, WM wrote:
If all could be counted to,
they would not remain after every counted number.
Of those which CAN be counted.to,
each CAN be counted to.
Of course.
But ∀n which can be counted to:
|ℕ \ {1, 2, 3, ..., n}| = ℵo.
Nevertheless almost all, namely ℵo, remain.
Of those which CAN be counted.to,
each CAN be counted to,
and
none remain after all which CAN be counted.to
Count to one that has
less than ℵo numbers which
cannot be counted to.
Fail.
Le 04/05/2024 à 05:18, Richard Damon a écrit :
On 5/3/24 9:21 AM, WM wrote:
Every findable number has ℵo successors. ==> Not all can be found.
All natural numbers have no successors. ==> There are more.
So, what is your actual definition of a "findable" number?
A number that can be put in trichotomy with its neighbours.
How does that definition actually differ IN ITS DEFINITION from the
definition of the Natural Numbers?
The set of all natural numbers consists mainly of dark numbers. All
numbers which can be put in trichotomy belong to a finite set
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
which is followed by an infinite set the numbers of whch cannot be distinguished and which can only be manipulated collectively.
|ℕ \ {1, 2, 3, ...}| = 0
Regards, WM
On 5/6/24 4:20 PM, WM wrote:
the numbers of which cannot be distinguished and which can only be manipulated collectively.
Le 04/05/2024 à 05:18, Richard Damon a écrit :
On 5/3/24 9:21 AM, WM wrote:
Every findable number has ℵo successors. ==> Not all can be found.
All natural numbers have no successors. ==> There are more.
So, what is your actual definition of a "findable" number?
A number that can be put in trichotomy with its neighbours.
How does that definition actually differ IN ITS DEFINITION from the
definition of the Natural Numbers?
The set of all natural numbers consists mainly of dark numbers. All
numbers which can be put in trichotomy belong to a finite set
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
which is followed by an infinite set the numbers of whch cannot be distinguished and which can only be manipulated collectively.
|ℕ \ {1, 2, 3, ...}| = 0
Regards, WM
Le 05/05/2024 à 19:59, Jim Burns a écrit :
On 5/3/2024 4:31 PM, WM wrote:
If all could be counted to,
they would not remain after every counted number.
From each number n which CAN be counted.to
for each number k which CAN be counted.to
more.than.k numbers which CAN be counted.to
can be reached immediately
from n to n+k+1
None of these immediate more.than.k numbers remain
after every number n which CAN be counted.to.
Nevertheless almost all, namely ℵo, remain.
Regards, WM
So, what is your actual definition of a "findable" number?
A number that can be put in trichotomy with its neighbours.
Also sind offenbar alle n in IN "findable". Welch Überraschung!
All which canNOT be counted.to are not.in ℕ
GIVEN Natural Number is followed by an infinite set of numbers
above it,
∀n ∈ ℕ_def: ∃^ℵo m ∈ ℕ, m > n.
∀n ∈ ℕ: ~∃^ℵo n ∈ (ℝ\ℕ).
Le 07/05/2024 à 00:11, Jim Burns a écrit :
All which canNOT be counted.to are not.in ℕ
All which canNOT be counted.to are not.in ℕ_def.
∀n ∈ ℕ_def: ∃^ℵo m ∈ ℕ, m > n.
∀n ∈ ℕ: ~∃^ℵo n ∈ (ℝ\ℕ).
ℕ_def is what everyone else calls ℕ
∀n ∈ ℕ_def: ∃^ℵo m ∈ ℕ, m > n. [WM]
∀n ∈ ℕ_def: ∃^ℵo m ∈ ℕ_def, m > n.
∀n ∈ ℕ: ~∃^ℵo n ∈ (ℝ\ℕ). [WM]
Simplified:
| ∀n ∈ ℕ: ~∃ n ∈ (ℝ\ℕ).
Does that make any sense to you?
Am 08.05.2024 um 23:55 schrieb Jim Burns:
ℕ_def is what everyone else calls ℕ
∀n ∈ ℕ_def: ∃^ℵo m ∈ ℕ, m > n. [WM]
∀n ∈ ℕ_def: ∃^ℵo m ∈ ℕ_def, m > n.
In this case, I'd prefer
∀ n ∈ ℕ: ∃^ℵo m ∈ ℕ: m > n.
Am 09.05.2024 um 02:27 schrieb Moebius:
Simplified:
| ∀n ∈ ℕ: ~∃ n ∈ (ℝ\ℕ).
Does that make any sense to you?
Maybe he just means:
~∃n ∈ ℕ: n ∈ (ℝ\ℕ))
or
∀n ∈ ℕ: ~(n ∈ (ℝ\ℕ)) .
Who knows?
Am 08.05.2024 um 23:55 schrieb Jim Burns:
Hint: This formula is
∀n ∈ ℕ: ~∃^ℵo n ∈ (ℝ\ℕ). [WM]
nonsense.
("~∃^ℵo" is not the problem.
It just means that
"there aren't countably many" ...).
Simplified:
| ∀n ∈ ℕ: ~∃ n ∈ (ℝ\ℕ).
Does that make any sense to you?
Le 07/05/2024 à 02:50, Richard Damon a écrit :
GIVEN Natural Number is followed by an infinite set of numbers above it,
Yes. Other natural numbers are existing but cannot be given.
∀n ∈ ℕ_def: ∃^ℵo m ∈ ℕ, m > n.
∀n ∈ ℕ: ~∃^ℵo n ∈ (ℝ\ℕ).
Regards, WM
∀n ∈ ℕ: ∃^ℵo m ∈ ℕ: m > n.
Am 08.05.2024 um 21:53 schrieb WM:
∀n ∈ ℕ_def: ∃^ℵo m ∈ ℕ, m > n.
Das lässt sich sogar noch etwas verbesser (ohne undefinierte "Brgiffe"
zu verwenden wie z. B. IN_def):
∀n ∈ ℕ: ∃^ℵo m ∈ ℕ, m > n.
Against your preference:
WM thinks ℕ is larger than ℕ_def
On 5/8/24 3:58 PM, WM wrote:
But N_def can be all of N
There are NO values in the set of Natural Numbers that do NOT have an infinite number of values about them.
X\ℕ n ℕ = {}
or even
X\Y n Y = {}.
What an incredible insight!
Am 08.05.2024 um 21:58 schrieb WM:
∀n ∈ ℕ_def: ∃^ℵo m ∈ ℕ: m > n.
Großartig
Daraus können wir direkt auf
∀n ∈ ℕ_def: ∃^ℵo m ∈ ℕ: 1/m < 1/n.
schließen.
Und damit auf
∃^ℵo m ∈ ℕ_def: 1/m <= x.
Mit anderen Worten:
Ax_def > 0: NUF(x_def) = ℵo.
On 5/8/2024 3:55 PM, WM wrote:
Le 07/05/2024 à 00:11, Jim Burns a écrit :
All which canNOT be counted.to are not.in ℕ
All which canNOT be counted.to are not.in ℕ_def.
And all which CAN be counted.to are in ℕ_def.
ℕ_def is the set of all and only
numbers which CAN be counted.to.
ℕ_def is what everyone else calls ℕ
|ℕ_def| = ℵ₀
Le 09/05/2024 à 03:33, Richard Damon a écrit :
On 5/8/24 3:58 PM, WM wrote:
But N_def can be all of N
No.
When removing ℕ_def, then always ℵo elements of ℕ remain.
There are NO values in the set of Natural Numbers that do NOT have an
infinite number of values about them.
When removing ℕ, then no elements of ℕ remain.
Regards, WM
Le 09/05/2024 à 17:24, Moebius a écrit : [...]
Le 09/05/2024 à 03:33, Richard Damon a écrit :
On 5/8/24 3:58 PM, WM wrote:
But N_def can be all of N
No.
Le 08/05/2024 à 23:55, Jim Burns a écrit :
On 5/8/2024 3:55 PM, WM wrote:
Le 07/05/2024 à 00:11, Jim Burns a écrit :
All which canNOT be counted.to are not.in ℕ
All which canNOT be counted.to are not.in ℕ_def.
And all which CAN be counted.to are in ℕ_def.
Yes.
ℕ_def is the set of all and only
numbers which CAN be counted.to.
ℕ_def is what everyone else calls ℕ
Because
almost everyone has not yet realized, that
ℕ contains also the natural numbers which
cannot be counted to and which
do not leave ℵ₀ successors after being removed from ℝ.
|ℕ_def| = ℵ₀
ℕ_def is a potentially infinite collection
and as such has no fxed number of elements.
We use the indefinite oo in this case.
But every n ∈ ℕ_def has ℵ₀ successors
which never vanish by counting.
They can be removed only collectively
such that nothing of ℕ remains.
On 5/10/2024 8:18 AM, WM wrote:
Le 08/05/2024 à 23:55, Jim Burns a écrit :
On 5/8/2024 3:55 PM, WM wrote:
Le 07/05/2024 à 00:11, Jim Burns a écrit :
All which canNOT be counted.to are not.in ℕ
All which canNOT be counted.to are not.in ℕ_def.
And all which CAN be counted.to are in ℕ_def.
Yes.
Thank you.
ℕ_def is the set of all and only
numbers which CAN be counted.to.
ℕ_def is what everyone else calls ℕ
ℕ contains also the natural numbers which cannot be counted to and [bla]
Weⁿᵒᵗᐧᵂᴹ use ℕ to refer to
the set of all and only
numbers which CAN be counted.to,
but we could use ℕ_def
or use ω
or use ⋃ₙ⟨⟨0…n⟩⟩
or use ♃
or use 🐎
to refer to the set of all and only
numbers which CAN be counted.to.
Whichever way we refer to it,
all which canNOT be counted.to are not.in it
and all which CAN be counted.to are in it
|ℕ_def| = ℵ₀
ℕ_def is a potentially infinite collection and <bla>
as such has no fxed number of elements.
ℕ_def ℕⁿᵒᵗᐧᵂᴹ ω ⋃ₙ⟨⟨0…n⟩⟩ ♃ 🐎 is
the set of all and only numbers which CAN be counted.to
and as such its elements are fixed,
because nothing exists which is
partly.countable.to and partly.not.countable.to.
Nothing is partly.in and partly.out.
We use the indefinite oo in this case.
Weⁿᵒᵗᐧᵂᴹ use ℵ₀ to refer to |ℕ_def|
But we could use
|ℕⁿᵒᵗᐧᵂᴹ| |ω| |⋃ₙ⟨⟨0…n⟩⟩| |♃| |🐎| to refer to |ℕ_def|
The claims weⁿᵒᵗᐧᵂᴹ make for
ℕ_def ℕⁿᵒᵗᐧᵂᴹ ω ⋃ₙ⟨⟨0…n⟩⟩ ♃ 🐎 are NOT altered by
pretending weⁿᵒᵗᐧᵂᴹ are NOT referring to
the set of all and only numbers which CAN be counted.to
But every n ∈ ℕ_def has ℵ₀ successors <bla>
Am 10.05.2024 um 20:56 schrieb Jim Burns:
ℕ_def ℕⁿᵒᵗᐧᵂᴹ ω ⋃ₙ⟨⟨0…n⟩⟩ ♃ 🐎 is
the set of all and only numbers which CAN be counted.to
and as such its elements are fixed,
because nothing exists which is
partly.countable.to and partly.not.countable.to.
Nothing is partly.in and partly.out.
At least in the context of set theory.
Talking about the set ℕⁿᵒᵗᐧᵂᴹ
in the context of set theory
is most of what I can hope to accomplish.
However,
I think a little more might be available.
I have the impression that
the ordinals which are _represented_ in set theory
are, in some sense, _prior to_ set theory or
orthogonal to set theory, at least,
with neither prior.
I'm thinking of
the von Neumann cumulative hierarchy of sets.
V₀ = ∅
Vᵦ₊₁ = 𝒫(Vᵦ)
Vᵧ = ⋃ᵝᑉᵞ Vᵦ
It seems advisable that we grant ordinals
their existence before we head down that road.
And natural numbers are finite ordinals.
I think it can be argued that
ω [...] exists
independently of set theories.
I'm pretty sure we can derive induction from that.
I'm pretty sure we can derive induction from that.
Am 11.05.2024 um 01:44 schrieb Jim Burns:
For ω
I suggest the definition:
γ < ω ⟺
∀β: 0 < β ≤ γ ⇒
∃α: 0 ≤ α < γ ∧ α+1 = β
I'm pretty sure we can derive induction from that.
Incidentally, I'm just studying some (early) papers
concerning this question.
E. Zermelo, Sur les ensembles finis et
le principe de l'induction complète (1909)
K Grelling, Die Axiome der Arithmetik mit
besonderer Berücksichtigung der
Beziehungen zur Mengenlehre (1910)
K Grelling, Mengenlehre (1924)
But all these considerations presuppose
some sort of set theory.
Am 11.05.2024 um 01:44 schrieb Jim Burns:
I think it can be argued that
ω [...] exists
independently of set theories.
Quite an interresting view.
Historical fact:
Infinite ordinals were introduced by Cantor
in the context of his "transfinite set theory".
But I get your idea (I think).
We may imagine (!)
infinitely many natural numbers
independent of set theory:
1 < 2 < 3 < ...
Of course, no set IN which
"contains" these numbers in this case.
In addition to all these numbers
we may imagine an additional "number" which
is larger than all these (natural) numbers:
1 < 2 < 3 < ... < ω.
Right?
(Actually, I sometimes didn't like the term
"infinite ordinal" in this connection.
My thought was:
"It's just an additional number." :-P)
Ax > 0: NUF(x) = ℵo.
On 5/10/24 8:34 AM, WM wrote:
When removing ℕ_def, then always ℵo elements of ℕ remain.Your problem is that N_Def isn't actually a singlely defined set.
When removing ℕ, then no elements of ℕ remain.
The set of "defined" values *IS* N, not a subset of it,
You just try to claim that not all natural numbers are definable, but
you can't actually prove it.
On 5/10/2024 8:18 AM, WM wrote:
Le 08/05/2024 à 23:55, Jim Burns a écrit :
On 5/8/2024 3:55 PM, WM wrote:
Le 07/05/2024 à 00:11, Jim Burns a écrit :
All which canNOT be counted.to are not.in ℕ
All which canNOT be counted.to are not.in ℕ_def.
And all which CAN be counted.to are in ℕ_def.
Yes.
Thank you.
ℕ_def is the set of all and only
numbers which CAN be counted.to.
ℕ_def is what everyone else calls ℕ
ℕ_def is a potentially infinite collection
and as such has no fxed number of elements.
We use the indefinite oo in this case.
Weⁿᵒᵗᐧᵂᴹ use ℵ₀ to refer to |ℕ_def|
Am 10.05.2024 um 20:56 schrieb Jim Burns:
ℕ contains also the natural numbers which cannot be counted to.
No, there aren't such "numbers" in ℕ.
Le 10/05/2024 à 16:21, Moebius a écrit :
Ax > 0: NUF(x) = ℵo.
Then there would sit ℵo unit fractions before any x > 0,
that is at x = 0.
<saudummer Scheißdreck gelöscht>
Le 10/05/2024 à 16:24, Richard Damon a écrit :
On 5/10/24 8:34 AM, WM wrote:
When removing ℕ_def, then always ℵo elements of ℕ remain.Your problem is that N_Def isn't actually a singlely defined set.
When removing ℕ, then no elements of ℕ remain.
It is not a (fixed) set but a (variable) collection.
The set of "defined" values *IS* N, not a subset of it,
All values of ℕ can be removed such that none remains.
You cannot define all natnumbers such that none remains.
You just try to claim that not all natural numbers are definable, but
you can't actually prove it.
You cannot define all natnumbers such that none remains.
Regards, WM
Am 13.05.2024 um 22:15 schrieb WM:
Le 10/05/2024 à 16:21, Moebius a écrit :
Ax > 0: NUF(x) = ℵo.
Then there would sit ℵo unit fractions before any x > 0,
Yes, that's what this formula says.
that is at x = 0.
No,. that's not "at x = 0",
That's exactly: Ax > 0: NUF(x) = ℵo.
On 5/13/24 4:19 PM, WM wrote:
Le 10/05/2024 à 16:24, Richard Damon a écrit :
On 5/10/24 8:34 AM, WM wrote:
When removing ℕ_def, then always ℵo elements of ℕ remain.Your problem is that N_Def isn't actually a singlely defined set.
When removing ℕ, then no elements of ℕ remain.
It is not a (fixed) set but a (variable) collection.
So, what does that mean your logic is saying?
The set of "defined" values *IS* N, not a subset of it,
All values of ℕ can be removed such that none remains.
You cannot define all natnumbers such that none remains.
Since ℕ is DEFINED as the set of Natural Numbers, either your term natnumbers is something different,
You just try to claim that not all natural numbers are definable, but
you can't actually prove it.
You cannot define all natnumbers such that none remains.
Since it is clear that your term "natnumbers" is something you haven't defined here, it becomes a question what is it defined as.
Le 10/05/2024 à 16:21, Moebius a écrit :
Ax > 0: NUF(x) = ℵo.
Then there would sit
ℵo unit fractions before any x > 0,
that is at x = 0.
Wrong.
No unit fraction sits at x = 0.
Try again.
On 5/13/2024 4:15 PM, WM wrote:
Then there would sit
ℵo unit fractions before any x > 0,
that is at x = 0.
No.
⎛ ¬∃x > 0: NUF(x) < ℵ₀
⎝ ∀x > 0: NUF(x) = ℵ₀
Le 14/05/2024 à 01:16, Moebius a écrit :
Am 13.05.2024 um 22:15 schrieb WM:Where is it?
Le 10/05/2024 à 16:21, Moebius a écrit :
Ax > 0: NUF(x) = ℵo.
Then there would sit ℵo unit fractions before any x > 0,
Yes, that's what this formula says.
Where can the first unit fractions exist on the real line
if none exists before 0 and none exists at 0?
Le 14/05/2024 à 01:16, Moebius a écrit :
Am 13.05.2024 um 22:15 schrieb WM:Where is it?
Le 10/05/2024 à 16:21, Moebius a écrit :
Ax > 0: NUF(x) = ℵo.
Then there would sit ℵo unit fractions before any x > 0,
Yes, that's what this formula says.
Where can the first unit fractions exist on the real line
if none exists before 0 and none exists at 0?
What [non negative real number] is before [each and every real] x > 0and not 0?
Where do the [...] ℵo unit fractions sit?
⎛ ¬∃x > 0: NUF(x) < ℵ₀
⎝ ∀x > 0: NUF(x) = ℵ₀
Don't repeat your claims.
can [you] show the point x where ℵ₀ unit fractions sit.
Where does NUF(x) change from 0 to ℵ₀?
What [real number] is before [each and every real] x > 0 and not 0?
Where do the [...] ℵo unit fractions sit?
⎛ ¬∃x > 0: NUF(x) < ℵ₀
⎝ ∀x > 0: NUF(x) = ℵ₀
Don't repeat your claims.
can [you] show the point x where ℵ₀ unit fractions sit.
Where does NUF(x) change from 0 to ℵ₀?
Le 14/05/2024 à 13:58, Jim Burns a écrit :
On 5/13/2024 4:15 PM, WM wrote:
Then there would sit
ℵo unit fractions before any x > 0,
that is at x = 0.
No.
What is before any x > 0 and not 0?
Where do the first ℵo unit fractions sit?
⎛ ¬∃x > 0: NUF(x) < ℵ₀
⎝ ∀x > 0: NUF(x) = ℵ₀
Don't repeat your claims.
They are nonsensical unless
you can show the point x where
ℵ₀ unit fractions sit.
Where does NUF(x) change from 0 to ℵ₀?
Am 14.05.2024 um 13:13 schrieb WM:
Le 14/05/2024 à 01:16, Moebius a écrit :
Am 13.05.2024 um 22:15 schrieb WM:Where is it?
Le 10/05/2024 à 16:21, Moebius a écrit :
Ax > 0: NUF(x) = ℵo.
Then there would sit ℵo unit fractions before any x > 0,
Yes, that's what this formula says.
Where is what?
Where can the first unit fractions exist on the real line
Nowhere. Since there is no such unit fraction, it can't be anywhere.
if none exists before 0 and none exists at 0?
But infinitely many unit fractions exist before each and every x > 0 (x
e IR).
Am 14.05.2024 um 14:47 schrieb WM:
can [you] show the point x where ℵ₀ unit fractions sit.
Nein, Mückenheim, den Punkt kann man nicht zeigen, da es ihn nicht gibt.
In jeder epsilon-Umgebung um 0 gibt es allerdings abzählbar unendlich
viele Stammbrüche.
Where does NUF(x) change from 0 to ℵ₀?
NUF besitzt im Punkt x = 0 eine "Unstetigkeitsstelle",
NUF(x) = 0 für x <= 0
und
NUF(x) = ℵ₀ für x > 0.
NUF(x) is not larger than 1 before NUF(x) = 1 has been realized.
NUF(x) changes from 0 to ℵ₀ in (-x′/2,x′/2)
but x′ ∉ (-x′/2,x′/2)
...and the same for each point > 0
You (WM) incorrectly think that that's incorrect,
I'd guess for the reason that
you incorrectly think that
a quantifier shift gives reliable results,
which it does not.
Every [nonempty] subset of the real line has a first element.
[...] But we are interested in NUF(x), not NUF(eps).
NUF besitzt im Punkt x = 0 eine "Unstetigkeitsstelle",
No.
in fact the case.NUF(x) = 0 für x <= 0
und
NUF(x) = ℵ₀ für x > 0.
That is
NUF(x) cannot <bla bla bla>
Le 14/05/2024 à 22:07, Jim Burns a écrit :
NUF(x) changes from 0 to ℵ₀ in (-x′/2,x′/2)
but x′ ∉ (-x′/2,x′/2)
For every definable x'.
...and the same for each point > 0
No.
∀n ∈ ℕ:
1/n - 1/(n+1) = 1/(n(n+1)) > 0 .
NUF(x) cannot assume a value > 1
before it has assumed the value 1.
You (WM) incorrectly think that that's incorrect,
I'd guess for the reason that
you incorrectly think that
a quantifier shift gives reliable results,
which it does not.
The function NUF(x) is defined at every x.
No further quantifier used.
On 5/15/2024 6:12 AM, WM wrote:
Every subset of the real line has a first element.
WM is just talking nonsense.
A finite subsection of the real line does have a first element.
Say, the
following line [segment] denoted by these two points. See, I defined it:
p0 = -42+0i
p1 = 742+0i
p0 = -4
p1 = 742
The line from p0 to p1 has a start (p0) and an end (p1).
This is a _finite_ subsection.
However, in the reals it has _infinite_ points between p0 and p1
Am 15.05.2024 um 15:30 schrieb WM:
NUF(x) cannot <bla bla bla>
NUF(x) = 0 für x e IR, x <= 0
und
NUF(x) = ℵ₀ für x e IR, x > 0.
This means: img(NUF) = {0, ℵ₀}.
I was just thinking that the line [segment from] p0 [to] p1 definitely has
a first and last element.
-4 is the first [smallest] element, and 742 is the last [largest] element [in the set]
{ -4, ..., 742 } ?
Chris M. Thomasson used his keyboard to write :Except in Mückenland of course: "Every subset of the real line has a
I was just thinking that the line p0 and p1 definitely
has a first and last element. -4 is the first element, and 742 is the
last element with regard to:
{ -4, ..., 742 }
?
That would be [-4,742] a closed interval which, as a set, has a first element. Whereas (-4, 742] as a set, does not.
Am 15.05.2024 um 15:25 schrieb WM:
in fact the case.NUF(x) = 0 für x <= 0
und
NUF(x) = ℵ₀ für x > 0.
That is
On 5/15/2024 2:13 PM, Moebius wrote:
Am 15.05.2024 um 17:40 schrieb Chris M. Thomasson:
On 5/15/2024 6:12 AM, WM wrote:
Every subset of the real line has a first element.
This claim is simply FALSE.
This is a _finite_ subsection.
Nope, but the LENGTH of this interval is finite.
I was just thinking that the line p0 and p1
definitely has a first and last element.
-4 is the first element, and
742 is the last element with regard to:
{ -4, ..., 742 }
Le 15/05/2024 à 19:30, Jim Burns a écrit :
| Assume
| ∃x > 0: NUF(x) < ℵ₀
| is correct.
|
| Assume
| x′ exists such that |⅟ℕ∩(0,x′]| < ℵ₀
|
| ⅟ℕ∩(0,x′] is totally.ordered and finite.
Not in the dark numbers.
∀x > 0: NUF(x) = ℵ₀
is correct.
Tell me where the first point
loaded with ℵ₀ unit fractions is.
On 5/15/2024 9:30 AM, WM wrote:
Le 14/05/2024 à 22:07, Jim Burns a écrit :
NUF(x) changes from 0 to ℵ₀ in (-x′/2,x′/2)
but x′ ∉ (-x′/2,x′/2)
For every definable x'.
For each x' ∈ ℝ⁺ such that,
No point x exists such that NUF(x) = 1
You (WM) say
∀x > 0: NUF(x) = ℵ₀
is incorrect.
Which is to say, you (WM) say
∃x > 0: NUF(x) < ℵ₀
is correct.
| Assume
| ∃x > 0: NUF(x) < ℵ₀
| is correct.
|
| Assume
| x′ exists such that |⅟ℕ∩(0,x′]| < ℵ₀
|
| ⅟ℕ∩(0,x′] is totally.ordered and finite.
∀x > 0: NUF(x) = ℵ₀
is correct.
Le 14/05/2024 à 15:35, Moebius a écrit :
Am 14.05.2024 um 13:13 schrieb WM:
Where can the first unit fractions exist
on the real line
Nowhere.
Since there is no such unit fraction,
it can't be anywhere.
Every subset of the real line has
a first element.
Le 15/05/2024 à 16:52, Moebius a écrit :
Am 15.05.2024 um 15:25 schrieb WM:
NUF(x) = 0 für x <= 0
und
NUF(x) = ℵ₀ für x > 0.
That is
in fact the case.
Then you can tell me
On 5/15/2024 9:12 AM, WM wrote:
Le 14/05/2024 à 15:35, Moebius a écrit :
Am 14.05.2024 um 13:13 schrieb WM:
Where can the first unit fractions exist
on the real line
Nowhere.
Since there is no such unit fraction,
it can't be anywhere.
Every subset of the real line has
a first element.
No.
| Assume ⅟ℕ∩(0,1] has first element ⅟G
|
| 0 < ½⋅⅟G < ⅟G < 2⋅⅟G
| There IS a unit.fraction ⅟k < 2⋅⅟G
| There is NOT a unit.fraction < ½⋅⅟G
|
| ⅟k < 2⋅⅟G exists
| (⅟k)/4 < (2⋅⅟G)/4
| ⅟(4⋅k) < ½⋅⅟G
| There IS a unit.fraction ⅟(4⋅k) < ½⋅⅟G
| Contradiction.
Therefore,
⅟ℕ∩(0,1] does NOT have a first element.
Am 16.05.2024 um 20:56 schrieb Jim Burns:
On 5/15/2024 9:12 AM, WM wrote:
Le 14/05/2024 à 15:35, Moebius a écrit :
Am 14.05.2024 um 13:13 schrieb WM:
Where can the first unit fractions exist
on the real line
Nowhere.
Since there is no such unit fraction,
it can't be anywhere.
Every subset of the real line has
a first element.
No.
| Assume ⅟ℕ∩(0,1] has first element ⅟G
|
| 0 < ½⋅⅟G < ⅟G < 2⋅⅟G
| There IS a unit.fraction ⅟k < 2⋅⅟G
| There is NOT a unit.fraction < ½⋅⅟G
|
| ⅟k < 2⋅⅟G exists
| (⅟k)/4 < (2⋅⅟G)/4
| ⅟(4⋅k) < ½⋅⅟G
| There IS a unit.fraction ⅟(4⋅k) < ½⋅⅟G
| Contradiction.
Therefore,
⅟ℕ∩(0,1] does NOT have a first element.
How about?
Let S = {x e IR : 0 < x}.
Then S is a (nomempty) subset of IR.
Assume there is a first/smallest element in S.
Let r be this element.
Then r e IR and 0 < r and
hence r/2 e IR and 0 < r/2.
Hence r/2 in S.
But r/2 < r.
Contradiction!
Hence S does not have a first/smallest Element.
This disproves WM's claim. [ ]
It seems that
WM likes to assert false statements.
I grabbed a proof I had laying around
and re.purposed it,
so that's my excuse.
But yours is better.
Am 16.05.2024 um 20:15 schrieb WM:
Le 15/05/2024 à 16:52, Moebius a écrit :
Am 15.05.2024 um 15:25 schrieb WM:
NUF(x) = 0 für x <= 0
und
NUF(x) = ℵ₀ für x > 0.
That is
in fact the case.
Then you can tell me the coordinate x where the first ℵ₀ unit fractions sit.
No, I can't.
Le 17/05/2024 à 02:21, Jim Burns a écrit :
On 5/16/2024 7:00 PM, Moebius wrote:
Am 16.05.2024 um 20:56 schrieb Jim Burns:
On 5/15/2024 9:12 AM, WM wrote:
Every subset of the real line has
a first element.
I grabbed a proof I had laying around
and re.purposed it,
so that's my excuse.
But yours is better.
Between each pair of unit fractions,
there is a finite distance.
To gather ℵ₀ unit fractions,
ℵ₀ finite distances are required.
Their sum is not zero.
Therefore the statement
NUF(x) = ℵ₀ für x > 0
is wrong.
Wrt to unit fractions, 1/1, 1/2, 1/3, 1/4, ...
There is no smallest unit fraction. However, is it fair to say that
there is a largest unit fraction at 1/1?
On 5/17/2024 12:08 PM, Jim Burns wrote:
On 5/17/2024 10:08 AM, WM wrote:
Le 17/05/2024 à 02:21, Jim Burns a écrit :
On 5/16/2024 7:00 PM, Moebius wrote:
Am 16.05.2024 um 20:56 schrieb Jim Burns:
On 5/15/2024 9:12 AM, WM wrote:
Every subset of the real line has
a first element.
Wrt to unit fractions, 1/1, 1/2, 1/3, 1/4, ...
There is no smallest unit fraction.
However, is it fair to say that
there is a largest unit fraction at 1/1?
I made some notes on Cantor
pairing here:
https://youtu.be/XkwgJt5bxKI
On 5/17/2024 5:24 PM, Chris M. Thomasson wrote:
I made some notes on Cantor pairing here:
https://youtu.be/XkwgJt5bxKI
Very pretty.
Chris M. Thomasson presented the following explanation :
There is no smallest unit fraction. However, is it fair to say that there is >> a largest unit fraction at 1/1?
Sure,
Note that in Mückenhausen there is a smallest unit fraction.
On 5/17/2024 10:08 AM, WM wrote:
Le 17/05/2024 à 02:21, Jim Burns a écrit :
On 5/16/2024 7:00 PM, Moebius wrote:
Am 16.05.2024 um 20:56 schrieb Jim Burns:
On 5/15/2024 9:12 AM, WM wrote:
Between each pair of unit fractions,
there is a finite distance.
Before each unit fraction ⅟n,
there is a smaller unit fraction ⅟(n+1).
Each unit fraction is not first.
The unit.fraction.subset of the real line
does NOT have a first element.
To gather ℵ₀ unit fractions,
ℵ₀ finite distances are required.
Their sum is not zero.
The greatest lower bound of d > 0 is 0
which does NOT contradict |(x,x+d)| ≮ ℵ₀
Therefore the statement
NUF(x) = ℵ₀ für x > 0
is wrong.
Le 17/05/2024 à 21:08, Jim Burns a écrit :
On 5/17/2024 10:08 AM, WM wrote:
Le 17/05/2024 à 02:21, Jim Burns a écrit :
On 5/16/2024 7:00 PM, Moebius wrote:
Am 16.05.2024 um 20:56 schrieb Jim Burns:
On 5/15/2024 9:12 AM, WM wrote:
Between each pair of unit fractions,
there is a finite distance.
Before each unit fraction ⅟n,
there is a smaller unit fraction ⅟(n+1).
These two conditions cannot be satisfied simultaneously.
Each unit fraction is not first.
The unit.fraction.subset of the real line
does NOT have a first element.
This condition violates this condition:
To gather ℵ₀ unit fractions,
ℵ₀ finite distances are required.
Le 17/05/2024 à 21:08, Jim Burns a écrit :
On 5/17/2024 10:08 AM, WM wrote:
Between each pair of unit fractions,
there is a finite distance.
Before each unit fraction ⅟n,
there is a smaller unit fraction ⅟(n+1).
These two conditions
cannot be satisfied simultaneously.
On 5/17/2024 3:33 PM, Moebius wrote:
Am 17.05.2024 um 23:24 schrieb Chris M. Thomasson:
Wrt to unit fractions, 1/1, 1/2, 1/3, 1/4, ...
There is no smallest unit fraction. However, is it fair to say that
there is a largest unit fraction at 1/1?
Sure/natürlich. Since this is the case. :-P
Proof: u is a unit fraction, iff (by definition of /unit fraction/)
there's a natural number n such that u = 1/n. This means that 1/1 is a
unit fraction. Moreover: Since for all natural numbers n: 1 <= n, and
hence 1/n <= 1/1, each and every unit fraction is <= the unit fraction
1/1. Hence 1/1 is indeed the largest unit fraction. [ ]
Actually, we may depict the/all unit fraction the following way:
1/1 > 1/2 > 1/3 > 1/4 > ...
Note that in Mückenhausen there is a smallest unit fraction. :-P
Wild! I still wonder why he thinks there is a smallest unit fraction.
A troll, or does he _really_ believe that nonsense? Humm...
The fact that we cannot find one, in his mind, merely proves his point.
on 5/19/2024, Moebius supposed :
Am 20.05.2024 um 00:04 schrieb FromTheRafters:
The fact that we cannot find one,
in his mind, merely proves his point.
We usually aren't able to "find"
something which does not exist
(except in Mückenland of course).
Yes, that's true,
how many times has he written something akin to
"Try to find 'A', fail, therefore dark numbers."
Between each pair of unit fractions,
there is a finite distance.
Strange way of thinking. For some reason it reminds me of a tree falling
in a forest...
Yes, that's true, how many times has he written something akin to "Try
to find 'A', fail, therefore dark numbers."
Keep in mind that unit fractions 1/x are not dense, say, wrt the reals?
Le 20/05/2024 à 11:28, FromTheRafters a écrit :
Yes, that's true, how many times has he written something akin to "Try
to find 'A', fail, therefore dark numbers."
Between each pair of unit fractions, there is a finite distance.
Hence not <bla bla bla>
Le 20/05/2024 à 22:15, "Chris M. Thomasson" a écrit :
Keep in mind that unit fractions 1/x are not dense, say, wrt the reals?
That is the essence of my proof.
On 5/20/2024 1:26 PM, Chris M. Thomasson wrote:
On 5/20/2024 1:22 PM, WM wrote:
Le 20/05/2024 à 22:15, "Chris M. Thomasson" a écrit :
That is the essence of my proof.
Keep in mind that unit fractions 1/x are not dense, say, wrt the reals? >>>
How does you proof get into realms where you seem to say there
actually is a "largest" natural number, so to speak?
WM, have you ever taken a look at the following book(s)?
https://youtu.be/rVtHrgdcvZA
For some reason, when I think of you, I think of this! Wow. ;^)
On 5/20/2024 2:02 PM, Moebius wrote:
Am 20.05.2024 um 22:27 schrieb Chris M. Thomasson:
On 5/20/2024 1:26 PM, Chris M. Thomasson wrote:
On 5/20/2024 1:22 PM, WM wrote:
Le 20/05/2024 à 22:15, "Chris M. Thomasson" a écrit :
Keep in mind that unit fractions 1/x are not dense, say, wrt the
reals?
That is the essence of my proof.
How does you proof get into realms where you seem to say there
actually is a "largest" natural number, so to speak?
WM, have you ever taken a look at the following book(s)?
https://youtu.be/rVtHrgdcvZA
For some reason, when I think of you, I think of this! Wow. ;^)
Little wonder, the authors were clearly inspired by Mückenheim!
All the numbers printed in these books are clearly "visible"!
The numbers just for men (69) part still cracks me up! Big time. LOL!!!
Not even two unit fractions can satisfy the condition to sit beforeany x > 0.
Not even two unit fractions can satisfy the condition to sit
before any x > 0.
Le 19/05/2024 à 22:28, Jim Burns a écrit :
Between each pair of unit fractions,
there is a finite distance.
Hence
not even two unit fractions
can satisfy the condition to
sit before any x > 0.
Therefore
Ax > 0: NUF(x) = ℵo is wrong.
Jim Burns laid this down on his screen :
While I wait for the horse to sing,
I amuse myself with Unicode and with
proving things I've never doubted but which
are pleasant to have laid out neatly.
Have you found
a Unicode mathematical symbol for
'humongous' yet? Maybe
not.yet.first.false for
the only potentially infinite? :)
Am 20.05.2024 um 22:12 schrieb WM:
Not even two unit fractions can satisfy the condition to sit
before any x > 0.
Ax e IR, x > 0: Eq,q e {1/n : n e IN}, p =/= q: p,q < x (I)
und
Eq,q e {1/n : n e IN}, p =/= q: x e IR, x > 0: p,q < x (II)
zu verstehen.
Hint: (I) ist wahr, (II) ist falsch.
On 5/20/2024 4:08 PM, WM wrote:
Le 19/05/2024 à 22:28, Jim Burns a écrit :
Between each pair of unit fractions,
there is a finite distance.
Hence
not even two unit fractions
can satisfy the condition to
sit before any x > 0.
For any x > 0
more.than.2 unit.fractions
sit before x
among them are ⅟⌊(1+⅟x)⌋ ⅟⌊(2+⅟x)⌋ ⅟⌊(3+⅟x)⌋
zero unit fractions sit before any x > 0
Le 21/05/2024 à 00:25, Moebius a écrit :
Am 20.05.2024 um 22:12 schrieb WM:
Not even two unit fractions can satisfy the condition to sit before
any x > 0.
Ax e IR, x > 0: Eq,q e {1/n : n e IN}, p =/= q: p,q < x (I)
und
Eq,q e {1/n : n e IN}, p =/= q: x e IR, x > 0: p,q < x (II)
zu verstehen.
Hint: (I) ist wahr, (II) ist falsch.
Quantoren[blindheit]
"Ax > 0: NUF(x) = ℵo" is right.
Le 21/05/2024 à 17:50, Jim Burns a écrit :
On 5/20/2024 4:08 PM, WM wrote:
Le 19/05/2024 à 22:28, Jim Burns a écrit :
Between each pair of unit fractions,
there is a finite distance.
Hence
not even two unit fractions
can satisfy the condition to
sit before any x > 0.
For any x > 0
more.than.2 unit.fractions
sit before x
among them are ⅟⌊(1+⅟x)⌋ ⅟⌊(2+⅟x)⌋ ⅟⌊(3+⅟x)⌋
Of course,
for any x that you can name,
there are ℵo smaller unit fractions.
Ax_def > 0: NUF(x_def) = ℵo is right.
zero unit fractions sit before any x > 0
So it is!
Therefore Ax > 0: NUF(x) = ℵo is wrong.
On 5/21/2024 12:55 PM, WM wrote:
for any x > 0
more.than.any.k<ℵ₀ unit.fractions
sit before x
among them are ⅟⌊(1+sₓ/rₓ)⌋ to ⅟⌊(k+1+sₓ/rₓ)⌋
Ax_def > 0: NUF(x_def) = ℵo is right.
∀x ∈ ℝ: x > 0 ⇒ NUF(x) = ℵ₀
zero unit fractions sit before any x > 0
So it is!
Therefore Ax > 0: NUF(x) = ℵo is wrong.
No. Your "therefore" is a quantifier shift.
If there is no unit fraction smaller than all x > 0, then there is an x > 0 preventing this.
That is
Zero unit fractions sit before [each and every] x > 0.
What prevents this?
That can only be accomplished <Hirnrotz>"That can be accomplished" by the fact that each and every unit fraction
Le 22/05/2024 à 01:27, Jim Burns a écrit :
ℝ is ℚ and points between non.∅ splits of ℚ
for any x > 0
that you can determine
for any x > 0
more.than.any.k<ℵ₀ unit.fractions
sit before x
among them are ⅟⌊(1+sₓ/rₓ)⌋ to ⅟⌊(k+1+sₓ/rₓ)⌋
Ax_def > 0: NUF(x_def) = ℵo is right.
∀x ∈ ℝ: x > 0 ⇒ NUF(x) = ℵ₀
If
there is no unit fraction smaller than all x > 0,
then
there is an x > 0 preventing this.
That is merciless logic.
Am 22.05.2024 um 12:43 schrieb WM:
If there is no unit fraction smaller than all x > 0, then there is an x > 0 >> preventing this.
That is
nonsense.
Hint: ~Eu e {1/n : n e IN}: Ax > 0: u < x.
In other words:
Au e {1/n : n e IN}: Ex > 0: x <= u.
THIS DOES NOT MEAN or IMPLY
Ex > 0: Au e {1/n : n e IN}: x <= u.
"Concluding" the latter from the former is a logical fallacy known as "quantifier shift".
Am 22.05.2024 um 12:43 schrieb WM:
Zero unit fractions sit before [each and every] x > 0.
Using symbols:
~Eu e {1/n : n e IN}: Ax > 0: u < x.
What prevents this?
"What prevents this", ist der Umstand, dass auch jeder Stammbruch eine
reelle Zahl > 0 ist. Daher kann kein Stammbruch kleiner als _alle_
reellen Zahlen > 0 sein (denn dazu müsste er kleiner als er selbst sein).
"That can be accomplished" by the fact that each and every unit fraction
is a real number > 0. Hence no unit fraction can be smaller than _each
and every_ real number > 0.
On 5/22/2024 6:43 AM, WM wrote:
Le 22/05/2024 à 01:27, Jim Burns a écrit :
ℝ is ℚ and points between non.∅ splits of ℚ
for any x > 0
that you can determine
For any x > 0 in ℚ or between a non.∅ split of ℚ
more.than.any.k<ℵ₀ unit.fractions
sit before x
among them are ⅟⌊(1+sₓ/rₓ)⌋ to ⅟⌊(k+1+sₓ/rₓ)⌋
0 < rₓ/sₓ < x
If
there is no unit fraction smaller than all x > 0,
then
there is an x > 0 preventing this.
There is no x > 0 smaller than all unit fractions.
¬∃ᴿx > 0: ∀¹ᐟᴺ ⅟k: x ≤ ⅟k
| Assume otherwise.
| Assume x¹ᐟᴺ > 0: ∀¹ᐟᴺ ⅟k: x¹ᐟᴺ ≤ ⅟k
Le 22/05/2024 à 17:48, Jim Burns a écrit :
There is no x > 0 smaller than all unit fractions.
¬∃ᴿx > 0: ∀¹ᐟᴺ ⅟k: x ≤ ⅟k
There is an x >= 0 smaller than all unit fractions.
On 5/22/2024 1:57 PM, WM wrote:
Le 22/05/2024 à 17:48, Jim Burns a écrit :
There is no x > 0 smaller than all unit fractions.
¬∃ᴿx > 0: ∀¹ᐟᴺ ⅟k: x ≤ ⅟k
There is an x >= 0 smaller than all unit fractions.
| Assume you are correct.
| Assume that there is
| an x ≥ 0 smaller than all unit fractions.
|
| There are points 2⋅b¹ᐟᴺ > ½⋅b¹ᐟᴺ > 0
| such that
| Unit.fraction ⅟k < 2⋅b¹ᐟᴺ
| No unit.fraction < ½⋅b¹ᐟᴺ
|
| ⅟k < 2⋅b¹ᐟᴺ
| (⅟k)/4 < (2⋅b¹ᐟᴺ)/4
| ⅟(4⋅k) < ½⋅b¹ᐟᴺ
| Unit.fraction ⅟(4⋅k) < ½⋅b¹ᐟᴺ
| Contradiction.
Therefore,
you are not correct.
Le 22/05/2024 à 20:58, Jim Burns a écrit :
On 5/22/2024 1:57 PM, WM wrote:
Le 22/05/2024 à 17:48, Jim Burns a écrit :
There is no x > 0 smaller than all unit fractions.
¬∃ᴿx > 0: ∀¹ᐟᴺ ⅟k: x ≤ ⅟k
There is an x >= 0 smaller than all unit fractions.
Therefore,
you are not correct.
I am. It is 0.
But your "proofs" are nonsense.
Disprove this:
Between ℵo unit fractions
there are at least ℵo real numbers x.
For them NUF(x) = ℵo is wrong.
On 5/22/2024 3:13 PM, WM wrote:
But your "proofs" are nonsense.
Le 22/05/2024 à 17:48, Jim Burns a écrit :
There is no x > 0 smaller than all unit fractions.
¬∃ᴿx > 0: ∀¹ᐟᴺ ⅟k: x ≤ ⅟k
There is an x >= 0 smaller than all unit fractions.
Therefore Ax > 0: NUF(x) = ℵo is wrong.
Why do you never address this fact?
On 5/22/2024 3:13 PM, WM wrote:
Le 22/05/2024 à 20:58, Jim Burns a écrit :
On 5/22/2024 1:57 PM, WM wrote:
Le 22/05/2024 à 17:48, Jim Burns a écrit :
There is no x > 0 smaller than all unit fractions.
¬∃ᴿx > 0: ∀¹ᐟᴺ ⅟k: x ≤ ⅟k
There is an x >= 0 smaller than all unit fractions.
Therefore,
you are not correct.
I am. It is 0.
0 is the greatest lower bound of unit fractions.
Disprove this:
Between ℵo unit fractions
there are at least ℵo real numbers x.
For them NUF(x) = ℵo is wrong.
For any x > 0
there are more.than.any.k<ℵ₀ unit.fractions < x
Am 22.05.2024 um 19:57 schrieb WM:
Therefore Ax > 0: NUF(x) = ℵo is wrong.
Why do you never address this fact?
Weil es saudummer Scheißdreck ist
Le 22/05/2024 à 22:48, Jim Burns a écrit :
On 5/22/2024 3:13 PM, WM wrote:
Disprove this:
Between ℵo unit fractions
there are at least ℵo real numbers x.
For them NUF(x) = ℵo is wrong.
For any x > 0
there are more.than.any.k<ℵ₀ unit.fractions < x
If you are right,
If you are right,
then there is a contradiction,
since I am right with absolute certainty.
Hence we have to find a way
to satisfy both statements:
JB:
For any x > 0
there are ℵ₀ smaller unit fractions.
WM:
Between two unit fractions
there are ℵo real numbers x.
I have shown the way: Dark numbers.
In accordance with:
There is no unit fraction smaller than all x > 0,
and even in accordance with
For any unit fraction there are ℵ₀ smaller real x > 0.
Note that
points on the real axis are fixed and
not subject to quantifier nonsense.
Le 22/05/2024 à 22:48, Jim Burns a écrit :
[...]
Hence
we have to find a way to satisfy both statements:
JB:
For any x > 0
there are ℵ₀ smaller unit fractions.
WM:
Between two unit fractions
there are ℵo real numbers x.
I have shown the way: Dark numbers.
In accordance with:
There is no unit fraction smaller than all x > 0,
and even in accordance with
For any unit fraction
there are ℵ₀ smaller real x > 0.
Note that points on the real axis are fixed and
not subject to quantifier nonsense.
WM has brought this to us :
Le 22/05/2024 à 22:48, Jim Burns a écrit :
On 5/22/2024 3:13 PM, WM wrote:
Le 22/05/2024 à 20:58, Jim Burns a écrit :
On 5/22/2024 1:57 PM, WM wrote:
Le 22/05/2024 à 17:48, Jim Burns a écrit :
There is no x > 0 smaller than all unit fractions.
¬∃ᴿx > 0: ∀¹ᐟᴺ ⅟k: x ≤ ⅟k
There is an x >= 0 smaller than all unit fractions.
Therefore,
you are not correct.
I am. It is 0.
0 is the greatest lower bound of unit fractions.
It is an x >= 0 smaller than all unit fractions. Therefore I a correct.
Disprove this:
Between ℵo unit fractions there are at least ℵo real numbers x.
For them NUF(x) = ℵo is wrong.
For any x > 0
there are more.than.any.k<ℵ₀ unit.fractions < x
If you are right, then there is a contradiction, since I am right with
absolute certainty.
Hence we have to find a way to satisfy both statements:
WM: Between two unit fractions there are ℵo real numbers x.
FTR: Between any two unique unit fractions there are more than ℵo real numbers x.
Deal with it!
On 5/23/2024 8:10 AM, WM wrote:
WM:
Between two unit fractions
there are ℵo real numbers x.
Between any two ⅟m < ⅟n
there are more.than.any.k<ℵ₀ real.points
I have shown the way: Dark numbers.
Darkᵂᴹ numbers in ℚ and between splits of ℚ
which are between 0 and all unit fractions
do not exist, neither darklyᵂᴹ nor visiblyᵂᴹ
We can know that they don't exist by starting with
that description and then making not.first.false
claims until we get to a contradiction.
Also true:
There is no x > 0 smaller than all unit fractions.
and even in accordance with
For any unit fraction there are ℵ₀ smaller real x > 0.
Also true:
For any x > 0 there are ℵ₀ smaller unit fractions.
Note that
points on the real axis are fixed and
not subject to quantifier nonsense.
On 5/23/2024 8:10 AM, WM wrote:
There is no unit fraction smaller than all x > 0, [A]
Also true:
There is no x > 0 smaller than all unit fractions. [B]
Note that points on the real axis are fixed and
not subject to quantifier nonsense.
How variables and quantifiers work and
how you (WM) think variables and quantifiers work
are different.
Le 25/05/2024 à 19:23, Jim Burns a écrit :
On 5/23/2024 8:10 AM, WM wrote:
There is no unit fraction smaller than all x > 0, [A]
Also true:
There is no x > 0 smaller than all unit fractions. [B]
Note that
points on the real axis are fixed and
can be subdivided into two sets, namely
the set of unit fractions and
the set of positive non-unit fractions.
If A is true, then there is
a positive non-unit fraction smaller than
all unit fractions.
If B is true, then there is
a unit fraction smaller than
all positive non-unit fraction.
There is no third alternative.
But there is no decision possible.
Le 23/05/2024 à 21:52, Jim Burns a écrit :
On 5/23/2024 8:10 AM, WM wrote:
I have shown the way: Dark numbers.
Darkᵂᴹ numbers in ℚ and between splits of ℚ
which are between 0 and all unit fractions
do not exist, neither darklyᵂᴹ nor visiblyᵂᴹ
What is closer to zero,
a unit fraction or a not unit fraction?
We can know that they don't exist by starting with
that description and then making not.first.false
claims until we get to a contradiction.
The contradiction is
∀x ∈ (0, 1]: NUF(x) = ℵo
because the unit fractions are x ∈ (0, 1].
They cannot sit at a single point x,
hence the statememt is false.
Also true:
There is no x > 0 smaller than all unit fractions.
That implies that
there is a unit fractions smaller than
all other x > 0.
and even in accordance with
For any unit fraction there are ℵ₀ smaller real x > 0.
Also true:
For any x > 0 there are ℵ₀ smaller unit fractions.
Impossible because
the unit fractions cannot be smaller than themselves.
On 5/26/2024 2:53 PM, WM wrote:
What is closer to zero,
a unit fraction or a not unit fraction?
Which is larger,
an even number or an odd number?
Neither.
Show your work.
On 5/26/2024 3:15 PM, WM wrote:
Le 25/05/2024 à 19:23, Jim Burns a écrit :
On 5/23/2024 8:10 AM, WM wrote:
There is no unit fraction smaller than all x > 0, [A]
Also true:
There is no x > 0 smaller than all unit fractions. [B]
Note that
points on the real axis are fixed and
can be subdivided into two sets, namely
the set of unit fractions and
the set of positive non-unit fractions.
Agreed.
If A is true, then there is
a positive non-unit fraction smaller than
all unit fractions.
No.
| Assume that claim is correct.
| Assume that x > 0 is smaller than
| all unit.fractions.
|
| However,
| ⅟⌊(1+⅟x)⌋ is a unit.fraction < x
| Contradiction.
That claim is incorrect.
If B is true, then there is
a unit fraction smaller than
all positive non-unit fraction.
No.
| Assume that claim is correct.
| Assume that ⅟n is smaller than
| all positive non.unit.fractions.
|
| However,
| ⅟(n+π) is a non.unit.fraction < ⅟n
| Contradiction.
That claim is incorrect.
There is no third alternative.
Consider ⅟⌊(1+⅟x)⌋ and ⅟(n+π)
The decision is that not all subsets
of unit.fractions or of non.unit.fractions
have two ends.
WM was thinking very hard :
Le 23/05/2024 à 19:44, FromTheRafters a écrit :
WM has brought this to us :
Le 22/05/2024 à 22:48, Jim Burns a écrit :
On 5/22/2024 3:13 PM, WM wrote:
Le 22/05/2024 à 20:58, Jim Burns a écrit :
On 5/22/2024 1:57 PM, WM wrote:
Le 22/05/2024 à 17:48, Jim Burns a écrit :
There is no x > 0 smaller than all unit fractions.
¬∃ᴿx > 0: ∀¹ᐟᴺ ⅟k: x ≤ ⅟k
There is an x >= 0 smaller than all unit fractions.
Therefore,
you are not correct.
I am. It is 0.
0 is the greatest lower bound of unit fractions.
It is an x >= 0 smaller than all unit fractions. Therefore I a correct. >>>>
Disprove this:
Between ℵo unit fractions there are at least ℵo real numbers x. >>>>>> For them NUF(x) = ℵo is wrong.
For any x > 0
there are more.than.any.k<ℵ₀ unit.fractions < x
If you are right, then there is a contradiction, since I am right with >>>> absolute certainty.
Hence we have to find a way to satisfy both statements:
WM: Between two unit fractions there are ℵo real numbers x.
FTR: Between any two unique unit fractions there are more than ℵo real >>> numbers x.
Deal with it!
This confirms that there are ℵo real numbers too.
At least that 'many'.
Le 26/05/2024 à 23:44, Jim Burns a écrit :
On 5/26/2024 2:53 PM, WM wrote:
What is closer to zero,
a unit fraction or a not unit fraction?
Which is larger,
an even number or an odd number?
Neither.
No decision possible because
almost all numbers are dark.
Show your work.
And, by "Show your work", I do not mean
"Repeat your unsupported claim endlessly".
Points on the real axis are fixed and
can be subdivided into two sets.
The real axis and all point sets in it
have an end at or before zero.
Le 26/05/2024 à 23:21, Jim Burns a écrit :
On 5/26/2024 3:15 PM, WM wrote:
Le 25/05/2024 à 19:23, Jim Burns a écrit :
On 5/23/2024 8:10 AM, WM wrote:
There is no unit fraction smaller than all x > 0,
[A]
Also true:
There is no x > 0 smaller than all unit fractions.
[B]
Note that
points on the real axis are fixed and
can be subdivided into two sets, namely
the set of unit fractions and
the set of positive non-unit fractions.
Agreed.
If A is true, then there is
a positive non-unit fraction smaller than
all unit fractions.
No.
| Assume that claim is correct.
| Assume that x > 0 is smaller than
| all unit.fractions.
|
| However,
| ⅟⌊(1+⅟x)⌋ is a unit.fraction < x
| Contradiction.
Not possible with dark number x.
Not possible with dark number x.
That claim is incorrect.
for visible numbers.
If B is true, then there is
a unit fraction smaller than
all positive non-unit fraction.
No.
| Assume that claim is correct.
| Assume that ⅟n is smaller than
| all positive non.unit.fractions.
|
| However,
| ⅟(n+π) is a non.unit.fraction < ⅟n
| Contradiction.
Not possible with dark number n.
That claim is incorrect.
for visible numbers.
There is no third alternative.
Consider ⅟⌊(1+⅟x)⌋ and ⅟(n+π)
I consider that
points on the real axis are fixed and
can be subdivided into two sets.
The decision is that not all subsets
of unit.fractions or of non.unit.fractions
have two ends.
The real axis and all point sets in it
have an end at or before zero.
On 5/27/2024 10:03 AM, WM wrote:
Note that
points on the positive real axis are fixed and
can be subdivided into two sets, namely
the set of unit fractions and
the set of positive non-unit fractions.
The [positive] real axis and all point sets in it
have an end at or before zero.
The positive real axis ℝ⁺
does not hold a visibleᵂᴹ lower end.
Not all non.∅.subsets of visibleᵂᴹ ℝ⁺ hold two ends.
ℝ⁺ doesn't.
For each darkᵂᴹ.number.holding superset ℝ⁺∪𝔻ᵂᴹ
not all non.∅.subsets of ℝ⁺∪𝔻ᵂᴹ hold two ends.
ℝ⁺ doesn't.
Holding two ends is insufficient for
a set to be finite.
On 5/27/2024 10:09 AM, WM wrote:
Points on the real axis are fixed and
can be subdivided into two sets.
Agreed.
The real axis and all point sets in it
have an end at or before zero.
Is repeating your unsupported claim endlessly
the full extent of your work?
WM laid this down on his screen :
If a set of real numbers has no elements below 0, then it ends above or at 0.
That is not a claim but a logical conclusion.
Zero is *STILL* not in your set.
Le 27/05/2024 à 19:46, Jim Burns a écrit :
On 5/27/2024 10:09 AM, WM wrote:
Points on the real axis are fixed and
can be subdivided into two sets.
Agreed.
The real axis and all point sets in it
have an end at or before zero.
Is repeating your unsupported claim endlessly
the full extent of your work?
If a set of real numbers has no elements below 0,
then it ends
above or at 0.
That is not a claim
but a logical conclusion.
If you disagree
you are not worth to receive any answer
from me or another intelligent person.
On 5/27/2024 4:10 PM, WM wrote:
If a set of real numbers has no elements below 0,
then it ends
Support your claim.
Feeling reallyreallyreally sure
isn't logic.
Finite sequences of not.first.false claims
are logic.
[usual bullshit]
By the way: Transfinity has been published as a book. https://www.elivapress.com/en/book/book-9877032691/
Le 28/05/2024 à 21:42, Jim Burns a écrit :
On 5/27/2024 4:10 PM, WM wrote:
If a set of real numbers has no elements below 0,
then it ends
Support your claim.
The set has ended where no elements follow.
Feeling reallyreallyreally sure
The set is not where no elements are.
If the set was before,
then it has ended meanwhile.
isn't logic.
Finite sequences of not.first.false claims
are logic.
If your logic denies this,
then your first claim is wrong already.
By the way: Transfinity has been published as a book. https://www.elivapress.com/en/book/book-9877032691/
My next book is under preparation.
It will appear next year:
Evidence for dark numbers.
Le 29/05/2024 à 19:11, WM a écrit :
By the way: Transfinity has been published as a book.
https://www.elivapress.com/en/book/book-9877032691/
Well-known vanity press
Le 29/05/2024 à 19:30, Python a écrit :
Le 29/05/2024 à 19:11, WM a écrit :
By the way: Transfinity has been published as a book.
https://www.elivapress.com/en/book/book-9877032691/
Well-known vanity press
To all who may be interested:
My book has been published without any costs for me, but I will receive Royalties. Eliva Press has a very good editor who reacts fast and
detailled. That's a good reason to publish with Eliva Press. I can only recommend it.
@ Python (real name is known but without any interest): Most authors
there are professors. No environment for you.
On 5/29/2024 1:11 PM, WM wrote:
Le 28/05/2024 à 21:42, Jim Burns a écrit :
On 5/27/2024 4:10 PM, WM wrote:
If a set of real numbers has no elements below 0,
then it ends
Support your claim.
The set has ended where no elements follow.
Consider a set B of real numbers such that
real number x exists such that
x ≤ each element of B
Are you (WM) claiming that
⎛ B.element b′ exists such that
⎝ b′ ≤ each element of B
?
If, yes, that is your claim,
then please support that claim.
The set is not where no elements are.
If the set was before,
then it has ended meanwhile.
Are you claiming that
the set holds an element ≤ each element?
If, yes, you claim that,
then please support that claim.
isn't logic.
Finite sequences of not.first.false claims
are logic.
On 5/29/2024 10:11 AM, WM wrote:
[...]> Whether actually infinite sets exist is unknown
The natural numbers are infinite, this
is just one example...
It happens that WM formulated :
By the way: Transfinity has been published as a book.
https://www.elivapress.com/en/book/book-9877032691/
My next book is under preparation. It will appear next year: Evidence for
dark numbers. Here is the
Preface
This book contains the collection of my writings on dark numbers most of
which have been published already here and there in the internet. Although I >> was a strong opponent of Cantor's actual infinity, an internet discussion in >> 2018 [1] has changed my mind in that without actual infinity the real axis >> would have gaps. That is my reason for accepting it and investigating its
consequences.
Whether actually infinite sets exist is unknown and, like the existence of >> God, cannot be proven; it can only be assumed as an axiom. But if actually >> infinite sets exist then dark elements are unavoidable. Almost all elements >> of infinite sets are dark: They cannot be manipulated as individuals but only
as sets or collections. That is the essence of this book.
How thick is it? My desk wobbles.
Le 29/05/2024 à 20:29, Jim Burns a écrit :
On 5/29/2024 1:11 PM, WM wrote:
Le 28/05/2024 à 21:42, Jim Burns a écrit :
On 5/27/2024 4:10 PM, WM wrote:
If a set of real numbers has no elements below 0,
then it ends
Support your claim.
The set has ended where no elements follow.
Consider a set B of real numbers such that
real number x exists such that
x ≤ each element of B
Are you (WM) claiming that
⎛ B.element b′ exists such that
⎝ b′ ≤ each element of B
?
The answer is no/yes.
b is impossible to find.
b exists.
If, yes, that is your claim,
then please support that claim.
I did so already using unit fractions.
NUF(x) = 1 between x = 0 and x = 1/10^10^10^100000.
This is proven by the mathematics of unit fractions
∀n ∈ ℕ: 1/n - 1/(n+1) = 1/(n(n+1)) > 0 .
Never two appear simultaneously.
The set is not where no elements are.
If the set was before,
then it has ended meanwhile.
Are you claiming that
the set holds an element ≤ each element?
If, yes, you claim that,
then please support that claim.
Bob!
isn't logic.
Finite sequences of not.first.false claims
are logic.
Your sequences of not.first.false claims leads to
the result that elements can disappear by exchange.
You should recognize that this is impossible.
Therefore,
if your chain contains only correct conclusions,
then your start must be wrong.
if your chain contains only correct conclusions,
then your start must be wrong.
On 5/26/2024 3:15 PM, WM wrote:
Note that
points on the real axis are fixed and
can be subdivided into two sets, namely
the set of unit fractions and
the set of positive non-unit fractions.
Agreed.
Am 26.05.2024 um 23:21 schrieb Jim Burns:
On 5/26/2024 3:15 PM, WM wrote:
Note that
points on the real axis are fixed and
can be subdivided into two sets, namely
the set of unit fractions and
the set of positive non-unit fractions.
Agreed.
So there are no non-positive real numbers
in your book?
NUF(x) = 1 between x = 0 and x = 1/10^10^10^100000.
No, NUF(x) ≠ 1
| Assume you are correct.
| Assume NUF(x) = 1 for
| 0 < x < 1/10^10^10^100000.
|
| unit fraction ⅟n exists: 0 < ⅟n < x
| no unit fraction exists between 0 and ⅟n
|
| However,
| unit fraction ⅟(n+1) exists: 0 < ⅟(n+1) < ⅟n
| Contradiction.
Therefore, NUF(x) ≠ 1
After more definitions and more not.first.false claims,
we get to disappearances of Bob and so forth.
if your chain contains only correct conclusions,
then your start must be wrong.
Either one of EAX is wrong
⎛ E. The empty set ∅ exists.
⎜ A. For sets x and y, adjunct x⨭y = x∪{y} exists.
⎝ X. Two equi.membered sets are equal sets.
or you are wrong.
Le 30/05/2024 à 22:24, Jim Burns a écrit :
On 5/30/2024 3:44 AM, WM wrote:
NUF(x) = 1 between x = 0 and x = 1/10^10^10^100000.
No, NUF(x) ≠ 1
That would imply that
not between all unit fractions distances existed.
Therefore it is wrong.
| Assume you are correct.
| Assume NUF(x) = 1 for
| 0 < x < 1/10^10^10^100000.
|
| unit fraction ⅟n exists: 0 < ⅟n < x
| no unit fraction exists between 0 and ⅟n
|
| However,
| unit fraction ⅟(n+1) exists: 0 < ⅟(n+1) < ⅟n
| Contradiction.
Yes, if all n had successors,
there would be a contradiction.
if your chain contains only correct conclusions,
then your start must be wrong.
Either one of EAX is wrong
⎛ E. The empty set ∅ exists.
⎜ A. For sets x and y, adjunct x⨭y = x∪{y} exists.
⎝ X. Two equi.membered sets are equal sets.
or you are wrong.
Ask colleagues
(without pointing to our discussion)
whether they agree that
in the course of exchanging elements,
infinitely many elements can disappear.
Ask further whether
in the accumulation point of the sequence (1/n)
infinitely many unit fractions
can populate one and the same point.
On 5/29/2024 10:11 AM, WM wrote:
The natural numbers are infinite,
WM is a moron?
Yes, if all n had successors, <bla>
On 5/31/2024 8:00 AM, Moebius wrote:
Am 26.05.2024 um 23:21 schrieb Jim Burns:
On 5/26/2024 3:15 PM, WM wrote:
Note that
points on the real axis are fixed and
can be subdivided into two sets, namely
the set of unit fractions and
the set of positive non-unit fractions.
Agreed.
So there are no non-positive real numbers
in your book?
In context,
I could see that WM was referring to
the positive real axis.
I didn't <bla>Hint: "Discussions" with WM eventually lead to brain damage.
On 5/31/2024 1:15 PM, WM wrote:
Le 30/05/2024 à 22:24, Jim Burns a écrit :
On 5/30/2024 3:44 AM, WM wrote:
NUF(x) = 1 between x = 0 and x = 1/10^10^10^100000.
No, NUF(x) ≠ 1
That would imply that
not between all unit fractions distances existed.
No.
x > ⅟n > ⅟(n+1) > 0
Yes, if all n had successors,
there would be a contradiction.
For each n countable.to from 0
n+1 is countable.to from n
n+1 is countable.to (through n) from 0
Each countable.to n has a countable.to successor.
There is a contradiction.
Ask colleagues
(without pointing to our discussion)
whether they agree that
in the course of exchanging elements,
infinitely many elements can disappear.
Ask further whether
in the accumulation point of the sequence (1/n)
infinitely many unit fractions
can populate one and the same point.
Even better:
Hint: "Mit Ausnahme von Vertretern des Ultrafinitismus wird die Peano-Arithmetik in der Mathematik allgemein als korrekte und
konsistente Charakterisierung der natürlichen Zahlen anerkannt." (Wikipedia)
On 5/31/2024 4:35 PM, Moebius wrote:
Am 29.05.2024 um 21:05 schrieb Chris M. Thomasson:Touche. There are infinitely many natural numbers? Any better? ;^)
On 5/29/2024 10:11 AM, WM wrote:
The natural numbers are infinite,
Well, actually, not the natural numbers themselves but _the set_ of
natural numbers [is infinite].
Le 31/05/2024 à 21:15, Jim Burns a écrit :
On 5/31/2024 1:15 PM, WM wrote:
Ask colleagues
(without pointing to our discussion)
whether they agree that
in the course of exchanging elements,
infinitely many elements can disappear.
Ask further whether
in the accumulation point of the sequence (1/n)
infinitely many unit fractions
can populate one and the same point.
Even better:
Ask whether
∀ᴿ⁺y ∃ᴿ⁺x≠y: x<y implies
∃ᴿ⁺x ∀ᴿ⁺y≠x: x<y
No.
On 6/1/2024 11:15 AM, WM wrote:
Le 31/05/2024 à 21:15, Jim Burns a écrit :
On 5/31/2024 1:15 PM, WM wrote:
Ask colleagues
(without pointing to our discussion)
whether they agree that
in the course of exchanging elements,
infinitely many elements can disappear.
Ask further whether
in the accumulation point of the sequence (1/n)
infinitely many unit fractions
can populate one and the same point.
Even better:
Ask whether
∀ᴿ⁺y ∃ᴿ⁺x≠y: x<y implies
∃ᴿ⁺x ∀ᴿ⁺y≠x: x<y
No.
"No", what?
⎜ ∀ᴿ⁺y ∃ᴿ⁺x≠y: x<y implies
⎝ ∃ᴿ⁺x ∀ᴿ⁺y≠xv : x<y
Le 03/06/2024 à 10:57, Jim Burns a écrit :
⎜ ∀ᴿ⁺y ∃ᴿ⁺x≠y: x<y implies
⎝ ∃ᴿ⁺x ∀ᴿ⁺y≠xv : x<y
No this is not implied but
independently proven in Evidence for Dark Numbers,
prepublished chapter 4.2:
We assume that
all points on the [positive] real axis are fixed and
can be subdivided into two sets, namely
the set of unit fractions and
the set of positive non-unit fractions.
For visible numbers we have two
statements both of which are true:
[A]
There is no unit fraction smaller than
all positive non-unit fractions.
[B]
There is no positive non-unit fraction smaller than
all unit fractions
If A is true for dark numbers too,
then there is a positive non-unit fraction
smaller than all unit fractions.
If B is true for dark numbers too,
then there is a unit fraction
smaller than all positive non-unit fractions.
There is only one objection:
Not all subsets of unit fractions or
of non-unit fractions have two ends.
But this is dismissed by the fact that
the positive real axis and all point sets in it
have an end at or before zero.
On 6/3/2024 7:58 AM, WM wrote:
Le 03/06/2024 à 10:57, Jim Burns a écrit :
⎜ ∀ᴿ⁺y ∃ᴿ⁺x≠y: x<y implies
⎝ ∃ᴿ⁺x ∀ᴿ⁺y≠xv : x<y
No this is not implied but
independently proven in Evidence for Dark Numbers,
prepublished chapter 4.2:
We assume that
all points on the [positive] real axis are fixed and
can be subdivided into two sets, namely
the set of unit fractions and
the set of positive non-unit fractions.
2.
Or we can assume instead that
ℕ⁺ holds all.and.only numbers countable.to by.1 from.0
ℚ⁺ holds all.and.only ratios of numbers in ℕ⁺
Under assumption (2.)
[A] and [B] are provable for all of ⅟ℕ and ℝ⁺\⅟ℕ
If A is true for dark numbers too,
then there is a positive non-unit fraction
smaller than all unit fractions.
Equivalent to:
If there is no positive non.unit.fraction
smaller than all unit fraction,
then A is false for darkᵂᴹ numbers too
And then, by (2.), no darkᵂᴹ numbers are in ℝ⁺\⅟ℕ
If B is true for dark numbers too,
then there is a unit fraction
smaller than all positive non-unit fractions.
Equivalent to:
If there is no unit fraction
smaller than all positive non.unit.fractions,
then B is false for darkᵂᴹ numbers too
And then, by (2.), no darkᵂᴹ numbers are in ⅟ℕ
There is only one objection:
Versions of ℕ⁺ ℚ⁺ and ℝ⁺ which hold darkᵂᴹ numbers
are provably not the (2.) version.
Whatever is proved or claimed or hallucinated about
some other version is not a claim about
the (2.) version.
Not all subsets of unit fractions or
of non-unit fractions have two ends.
Pick a non.two.ended subset. 'Bye, Bob.
But this is dismissed by the fact that
the positive real axis and all point sets in it
have an end at or before zero.
You're too late. Bob's gone.
Le 03/06/2024 à 20:34, Jim Burns a écrit :
On 6/3/2024 7:58 AM, WM wrote:
Le 03/06/2024 à 10:57, Jim Burns a écrit :
⎜ ∀ᴿ⁺y ∃ᴿ⁺x≠y: x<y implies
⎝ ∃ᴿ⁺x ∀ᴿ⁺y≠xv : x<y
No this is not implied but
independently proven in Evidence for Dark Numbers,
prepublished chapter 4.2:
We assume that
all points on the [positive] real axis are fixed and
can be subdivided into two sets, namely
the set of unit fractions and
the set of positive non-unit fractions.
2.
Or we can assume instead that
ℕ⁺ holds all.and.only numbers countable.to by.1 from.0
ℚ⁺ holds all.and.only ratios of numbers in ℕ⁺
ℝ⁺ holds all of ℚ⁺ and all.and.only
points x between open.foresplits Fₓ and ℚ⁺\Fₓ of ℚ⁺
with no points zero distance apart
and
⅟ℕ holds all.and.only reciprocals of numbers in ℕ⁺
ℝ⁺\⅟ℕ holds all.and only the others in ℝ⁺
and
ℝ⁺ is the positive real axis.
For visible numbers
we have two statements both of which are true:
[A]
There is no unit fraction smaller than
all positive non-unit fractions.
[B]
There is no positive non-unit fraction smaller than
all unit fractions
Under assumption (2.)
[A] and [B] are provable for all of ⅟ℕ and ℝ⁺\⅟ℕ
Hence assumption (2) contradicts logic.
ℕ⁺ holds all.and.only numbers countable.to
by.1 from.0
ℚ⁺ holds all.and.only ratios of numbers in ℕ⁺
ℝ⁺ holds all of ℚ⁺ and all.and.only
points x between open.foresplits Fₓ and ℚ⁺\Fₓ of ℚ⁺
with no points zero distance apart
Only one of the two complementary sets
can and must contain the first point.
Not all subsets of unit fractions or
of non-unit fractions have two ends.
Pick a non.two.ended subset. 'Bye, Bob.
But this is dismissed by the fact that
the positive real axis and
all point sets in it
have an end at or before zero.
You're too late. Bob's gone.
Not in a mathematics based upon logic.
On 6/3/2024 3:50 PM, WM wrote:
Assumption (2.) describes
objects in our familiar arithmetic.
For each n in ℕ⁺
n countable.to from.0
n+1 is countable.to from.n
n+1 is countable.to from.0 through.n
n+1 is in ℕ⁺
n is not larger than all numbers in ℕ⁺
𝔼 is the subset of even numbers in ℕ⁺
𝕆 is the subset of odd numbers in ℕ⁺
⎛ There is no number in 𝔼 larger than
⎝ all numbers in 𝕆
⎛ There is no number in 𝕆 larger than
⎝ all numbers in 𝔼
Only one of the two complementary sets
can and must contain the first point.
Why?
Responding "Logic" or "Mathematics" is dodging.
Le 30/05/2024 à 22:24, Jim Burns a écrit :Nein, Du psychotischer Spinner, das impliziert (*) NICHT.
NUF(x) = 1 between x = 0 and x = 1/10^10^10^100000.
No, NUF(x) ≠ 1 (*)
That would imply that not between all unit fractions distances existed.
Le 04/06/2024 à 04:07, Jim Burns a écrit :
Assumption (2.) describes
objects in our familiar arithmetic.
That is true
That is truea.
because
our familiar arithmetic is based upon
potential infinity.
Only in finished infinity
dark numbers are required.
our familiar arithmetic is based upon
potential infinity.
For each n in ℕ⁺
n countable.to from.0
n+1 is countable.to from.n
n+1 is countable.to from.0 through.n
n+1 is in ℕ⁺
n is not larger than all numbers in ℕ⁺
And in particular all n have ℵo dark successors.
And in particular all n have ℵo dark successors.
𝔼 is the subset of even numbers in ℕ⁺
𝕆 is the subset of odd numbers in ℕ⁺
⎛ There is no number in 𝔼 larger than
⎝ all numbers in 𝕆
⎛ There is no number in 𝕆 larger than
⎝ all numbers in 𝔼
Dark numbers do not unveil their mystery.
Assumption 2.
ℕ⁺ holds all.and.only numbers countable.to by.1 from.0
We agree if
darkᵂᴹ numbers are countable.to by.1 from.0.
Le 04/06/2024 à 23:31, Jim Burns a écrit :
ℕ⁺ holds all.and.only numbers countable.to by.1 from.0
Of course. But that is a small minority because
For all counted n: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
Le 04/06/2024 à 23:31, Jim Burns a écrit :
On 6/4/2024 10:10 AM, WM wrote:
Le 04/06/2024 à 04:07, Jim Burns a écrit :
Assumption (2.) describes
objects in our familiar arithmetic.
That is true
Thank you.
Assumption 2.
ℕ⁺ holds all.and.only
numbers countable.to by.1 from.0
Of course.
But that is a small minority
because you cannot count all:
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
Whatever you count, most numbers remain uncounted.
ℵo is much larger than
the collection |ℕ_def| of all counted n.
We agree if
darkᵂᴹ numbers are countable.to by.1 from.0.
They remain uncounted.
Proof:
For all counted n:
|ℕ \ {1, 2, 3, ..., n}| = ℵo.
Don't you believe this statement?
Or don't you understand it?
For all counted n:
|ℕ \ {1, 2, 3, ..., n}| = ℵo.
Am 05.06.2024 um 12:43 schrieb WM:
Le 04/06/2024 à 23:31, Jim Burns a écrit :
ℕ⁺ holds all.and.only numbers countable.to by.1 from.0
Of course. But that is a small minority because
No, Mückenheim, that are ALL natural numbers.
Proof by induction: 1 is a numbers countable to by 1 from 0 (since 0+1 =
1). If n is a number countable to by 1 from 0 then n+1 s a number
countable to by 1 from 0 (since n+1 = n+1). qed
Am 05.06.2024 um 12:43 schrieb WM:
For all counted n: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
Keine Ahnung, was Du hier wieder mit Deinen "counted n" willst, Mückenheim.
Tatsächlich gilt für ALLE natürlichen Zahlen n: |ℕ \ {1, 2, 3, ..., n}| = ℵo; egal ob "counted" oder nicht "counted",
Le 05/06/2024 à 13:21, Moebius a écrit :
Am 05.06.2024 um 12:43 schrieb WM:
Le 04/06/2024 à 23:31, Jim Burns a écrit :
ℕ⁺ holds all.and.only numbers countable.to by.1 from.0
Of course. But that is a small minority because
No, Mückenheim, that are ALL natural numbers.
A religious belief, contradicted by mathematics: <bla bla bla>
On 6/5/2024 6:43 AM, WM wrote:
Le 04/06/2024 à 23:31, Jim Burns a écrit :
On 6/4/2024 10:10 AM, WM wrote:
Le 04/06/2024 à 04:07, Jim Burns a écrit :
Assumption (2.) describes
objects in our familiar arithmetic.
That is true
Thank you.
Assumption 2.
ℕ⁺ holds all.and.only
numbers countable.to by.1 from.0
Of course.
Assumption 2 in detail.
For all counted n:
|ℕ \ {1, 2, 3, ..., n}| = ℵo.
not.exists n ∈ ℕ⁺: |ℕ⁺\{n≥}| ≠ ℵ₀
What do you (WM) think that proves?
Le 05/06/2024 à 17:08, Moebius a écrit :
Tatsächlich gilt für ALLE natürlichen Zahlen n: |ℕ \ {1, 2, 3, ...,
n}| = ℵo; egal ob "counted" oder nicht "counted",
ℵo will always remain uncounted
and, contrary to Cantor's claim, uncountable.
What do you (WM) think that proves?
See:https://m.media-amazon.com/images/M/MV5BZDQwMjNiMTQtY2UwYy00NjhiLTk0ZWEtZWM5ZWMzNGFjNTVkXkEyXkFqcGdeQXVyMTQxNzMzNDI@._V1_.jpg
Le 05/06/2024 à 21:28, Jim Burns a écrit :
On 6/5/2024 6:43 AM, WM wrote:
Le 04/06/2024 à 23:31, Jim Burns a écrit :
On 6/4/2024 10:10 AM, WM wrote:
Le 04/06/2024 à 04:07, Jim Burns a écrit :
Assumption (2.) describes
objects in our familiar arithmetic.
That is true
Thank you.
Assumption 2.
ℕ⁺ holds all.and.only
numbers countable.to by.1 from.0
Of course.
Assumption 2 in detail.
does not contradict that
contrary to Cantor's claim
most natural numbers are uncountable,
although with n
also n^n^n is countable.
Am 05.06.2024 um 21:28 schrieb Jim Burns:
What do you (WM) think that proves?
Do you REALLY think that
he read you exposition in full?
See:https://m.media-amazon.com/images/M/MV5BZDQwMjNiMTQtY2UwYy00NjhiLTk0ZWEtZWM5ZWMzNGFjNTVkXkEyXkFqcGdeQXVyMTQxNzMzNDI@._V1_.jpg
Am 05.06.2024 um 23:46 schrieb Jim Burns:
maybe the horse will sing.
Sure, and hell will freeze.
Am 06.06.2024 um 01:08 schrieb Moebius:
Am 05.06.2024 um 23:46 schrieb Jim Burns:
maybe the horse will sing.
Sure, and hell will freeze.
I mean: You're just as crazy as Mückenheim.
See: https://m.media-amazon.com/images/M/MV5BZDQwMjNiMTQtY2UwYy00NjhiLTk0ZWEtZWM5ZWMzNGFjNTVkXkEyXkFqcGdeQXVyMTQxNzMzNDI@._V1_.jpg
maybe the horse will sing.
Am 06.06.2024 um 01:11 schrieb Moebius:
Am 06.06.2024 um 01:08 schrieb Moebius:
Am 05.06.2024 um 23:46 schrieb Jim Burns:
maybe the horse will sing.
Sure, and hell will freeze.
I mean: You're just as crazy as Mückenheim.
See:
https://m.media-amazon.com/images/M/MV5BZDQwMjNiMTQtY2UwYy00NjhiLTk0ZWEtZWM5ZWMzNGFjNTVkXkEyXkFqcGdeQXVyMTQxNzMzNDI@._V1_.jpg
Jim: "I know that I'm crazy."
Fine. :-)
Le 05/06/2024 à 23:31, Jim Burns a écrit :
Darkᵂᴹ numbers do not exist in ℕ⁺
That is <bla bla bla>
Darkᵂᴹ numbers do not exist in ℕ⁺
Am 06.06.2024 um 15:27 schrieb WM:
Le 05/06/2024 à 23:31, Jim Burns a écrit :
Darkᵂᴹ numbers do not exist in ℕ⁺
That is <bla bla bla>
In the context of _set theory_ ℕ is an _infinite_ set (due to Cantor).
Theorem: ∀n ∈ ℕ: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
Mit Deinem psychotischen Scheißdreck gehst Du [daher] besser zum Psychiater, Mückenheim.
Am 06.06.2024 um 15:27 schrieb WM:
Le 05/06/2024 à 23:31, Jim Burns a écrit :
Darkᵂᴹ numbers do not exist in ℕ⁺
That is <bla bla bla>
In the context of _set theory_ ℕ is an _infinite_ set (due to Cantor).
Theorem: ∀n ∈ ℕ: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
Mit Deinem psychotischen Scheißdreck gehst Du [daher] besser zumPsychiater, Mückenheim.
most natural numbers are uncountable,
although with n also n^n^n is countable.
most natural numbers are uncountable,
although with n also n^n^n is countable.
Le 05/06/2024 à 23:31, Jim Burns a écrit :
Assumption (2.) describes
objects in our familiar arithmetic.
That is true
Thank you.
Assumption 2.
ℕ⁺ holds all.and.only
numbers countable.to by.1 from.0
Of course.
Darkᵂᴹ numbers do not exist in ℕ⁺
That is potential infinity.
It is not Cantor's actual infinity.
∀n ∈ ℕ⁺: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
It is not Cantor's actual infinity.
Therefore,
each k ∈ ℕ⁺ is ℵ₀.followed.in.ℕ⁺
It is not Cantor's actual infinity.
It is our familiar arithmetic.
Am 05.06.2024 um 22:39 schrieb WM:
most natural numbers are uncountable,
Nonsense.
On 6/5/2024 1:39 PM, WM wrote:
ℵo natural numbers will remain uncounted forever.
uncounted implies that they are finite?
Le 06/06/2024 à 20:12, Moebius a écrit :
Am 05.06.2024 um 22:39 schrieb WM:
most natural numbers are uncountable,
Nonsense.
Fact: <nonsense>
Le 06/06/2024 à 20:12, Moebius a écrit :
Am 05.06.2024 um 22:39 schrieb WM:
most natural numbers are uncountable,
Nonsense.
<bla>
Am 07.06.2024 um 13:01 schrieb WM:
Le 06/06/2024 à 20:12, Moebius a écrit :
Am 05.06.2024 um 22:39 schrieb WM:
most natural numbers are uncountable,
NO natural number is "uncountable"
Proof (by induction): 1 is "countable" (at least in my book). If n is "countable", then n+1 is "countable" too (again, at least in my book).
Hence all natural numbers are "countable".
Le 06/06/2024 à 20:35, Jim Burns a écrit :
Therefore,
each k ∈ ℕ⁺ is ℵ₀.followed.in.ℕ⁺
You define some individual number and
many are following.
They are following even upon
the last number that you ever have defined.
Therefore they are undefined by you.
That is the difference between ℕ_def and ℕ.
It is not Cantor's actual infinity.
It is our familiar arithmetic.
Of course.
Most natural numbers are undefined
by you and be anyone else.
It is not Cantor's actual infinity.
It is our familiar arithmetic.
Of course.
Do you agree that
your (WM's) claims are not about
our familiar arithmetic?
Most natural numbers are undefined
by you and by anyone else.
⎛ Each number in ℕ⁺ has a successor.
⎜ Each nonzero number in ℕ⁺ has a predecessor.
⎝ Each nonempty subset of ℕ⁺ holds a first number.
Every element of the natural numbers is countable (finite) and
the entire set is countably infinite.
Le 07/06/2024 à 20:00, Jim Burns a écrit :
On 6/6/2024 4:14 PM, WM wrote:
You define some individual number and
many are following.
I give a description of an individual number j in ℕ⁺
-- a description which does not distinguish between
different numbers in ℕ⁺
Most natural numbers are undefined
by you and by anyone else.
⎛ Each number in ℕ⁺ has a successor.
⎜ Each nonzero number in ℕ⁺ has a predecessor.
⎝ Each nonempty subset of ℕ⁺ holds a first number.
That is no definition of a number
but a description of a set.
But every element of ℕ⁺ can be defined.
But every element of ℕ⁺ can be defined.
What the hell are you on man?See: https://en.wikipedia.org/wiki/Psychosis
Chris M. Thomasson has brought this to us :
[1, 2, 3, 4] is finite. [1, 2, 3, ...] Is infinite... ;^)
Curly brackets for sets, square brackets for intervals.
What the hell are you on man?
Dementia.
if ℕ⁺\Defble is not.empty
then
ℕ⁺\Defble holds first j ∈ ℕ⁺
j ∉ Defble
j-1 ∈ Defble
After decades of saying that there are 'undefinable' and hence
'undefined' natural numbers which are "dark", he recently stated that
the natural numbers are all defined
On 6/7/2024 2:00 PM, Jim Burns wrote:
[...]
Give me a natural number. Say 42, well:
{ 42, 42 + 1, 42 + 1 + 1, 42 + 1 + 1 + 1, ... }
They are all
in the set of natural numbers and are
able to be defined all at one?
Fair enough?
{ 42, 42 + 1, 42 + 1 + 1, 42 + 1 + 1 + 1, ... }
Le 07/06/2024 à 23:00, Jim Burns a écrit :
I give a description of an individual number j in ℕ⁺
-- a description which does not distinguish between
different numbers in ℕ⁺
⎛ Each number in ℕ⁺ has a successor.
⎜ Each nonzero number in ℕ⁺ has a predecessor.
⎝ Each nonempty subset of ℕ⁺ holds a first number.
if ℕ⁺\Defble is not.empty
then
ℕ⁺\Defble holds first j ∈ ℕ⁺
j ∉ Defble
j-1 ∈ Defble
That is your error.
If j ∈ Defble then j^j^j ∈ Defble.
Nevertheless j^j^j^j^j is finite, but
there are ℵo undefinable natural numbers.
This could only be disproved by defining them.
But they will never be defined
A natural number can be
definable and not defined.
On 6/8/2024 4:20 PM, Moebius wrote:
Am 08.06.2024 um 20:16 schrieb Jim Burns:
A natural number can be
definable and not defined.
Infinitely many even numbers are divisible by 2.
Excellent example! I'm stealing this.
divisible = able to be divided
divisible ≠ divided
No need for performing [a] division
for all these numbers though
(except in Mückenmath of course).
Am 08.06.2024 um 20:16 schrieb Jim Burns:
A natural number can be
definable and not defined.
Infinitely many even numbers are divisible by 2.
No need for performing the division
for all these numbers though
(except in Mückenmath of course).
On 6/8/2024 10:25 AM, Jim Burns wrote:
On 6/8/2024 1:34 AM, Chris M. Thomasson wrote:
Give me a natural number. Say 42, well:
{ 42, 42 + 1, 42 + 1 + 1, 42 + 1 + 1 + 1, ... }
They are all
in the set of natural numbers and are
able to be defined all at one?
Fair enough?
The question is
what counts as able to be defined?
A number that WM can see for _himself_?
Visually and/or mentally?
I don't know.
WM is strange for sure.
If WM does not know about the number,
then he insists that we call it dark?
Wow.
Just not 100% sure on that...
Jim Burns laid this down on his screen :
A natural number can be
definable and not defined.
But can you have a non-empty set of undefined natural numbers?
Am 09.06.2024 um 00:47 schrieb Jim Burns:
On 6/8/2024 4:20 PM, Moebius wrote:
Am 08.06.2024 um 20:16 schrieb Jim Burns:
A natural number can be
definable and not defined.
Infinitely many even numbers
are divisible by 2.
Excellent example! I'm stealing this.
N/p. Had to think for a moment
to find this helpful example.
divisible = able to be divided
divisible ≠ divided
Right.
No need for performing [a] division
for all these numbers though
(except in Mückenmath of course).
Jim Burns laid this down on his screen :
On 6/8/2024 8:42 AM, WM wrote:
But they will never be defined
Irrelevant.
Numbers between 0 and Avogadroᴬᵛᵒᵍᵃᵈʳᵒ
will never be all defined.
Numbers between 0 and Avogadroᴬᵛᵒᵍᵃᵈʳᵒ
are all definable.
An apple can be
edible and not eaten.
A tree falling in the forest can be
audible and not heard.
A natural number can be
definable and not defined.
But can you have
a non-empty set of undefined natural numbers?
Distinguishable but not distinguished?
Discernible yet not discerned?
One should be able to discern
each and every element
at least enough to decide on membership.
Ghost naturals might exist somewhere,
but are not a subset of the naturals.
WM seems to just 'not get' real numbers.
One should be able to discern
each and every element
at least enough to decide on membership.
The question is
what counts as able to be defined?
On 6/8/2024 8:42 AM, WM wrote:
If j ∈ Defble then j^j^j ∈ Defble.
Nevertheless j^j^j^j^j is finite, but
there are ℵo undefinable natural numbers.
No.
There are 0 first undefinable natural numbers.
There are 0 undefinable natural numbers.
This could only be disproved by defining them.
No.
It has been disproved by
giving a description of an individual number j in ℕ⁺
Am 08.06.2024 um 20:16 schrieb Jim Burns:
A natural number can be
definable and not defined.
Infinitely many even numbers are divisible by 2. No need for performing
the division for all these numbers though
Jim Burns laid this down on his screen :
A natural number can be
definable and not defined.
But can you have a non-empty set of undefined natural numbers?
Distinguishable but not distinguished? Discernible yet not discerned?
One should be able to discern each and every element at least enough to decide on membership.
Le 08/06/2024 à 19:25, Jim Burns a écrit :
The question is
what counts as able to be defined?
Proposal:
Every [sum or] product or power or power tower of defined numbers is or can become defined.
Every quotient like ω/1000000 is and remains undefined.
ω/2, ω/3, ω/10, ... are
Le 08/06/2024 à 20:16, Jim Burns a écrit :
On 6/8/2024 8:42 AM, WM wrote:
If j ∈ Defble then j^j^j ∈ Defble.
Nevertheless j^j^j^j^j is finite, but there are ℵo undefinable
natural numbers.
No.
Then define natural numbers with [...] no undefinable successors.
"One" can't perform a supertask.
That is the crack through which
WM's rhetoric enters.
Am 10.06.2024 um 15:26 schrieb WM:
Proposal:
Every [sum or] product or power or power tower of defined numbers is or can >> become defined.
i.e. is /definable/.
Then each and every natural number is definable.
Le 10/06/2024 à 15:57, Moebius a écrit :
Am 10.06.2024 um 15:26 schrieb WM:
Proposal:
Every [sum or] product or power or power tower of defined numbers is
or can become defined.
i.e. is /definable/.
Then each and every natural number is definable.
every number [that] can [be] define has ℵo successors.
Le 08/06/2024 à 19:25, Jim Burns a écrit :
The question is
what counts as able to be defined?
Proposal:
Every product or power or power tower of
defined numbers
is or can become defined.
Every quotient like ω/1000000
is and remains undefined.
Proposal:
Every product or power or power tower of
defined numbers
is or can become defined.
Every quotient like ω/1000000
is and remains undefined.
Hint: Each and every natural number has ℵo successors.
On 6/10/2024 9:33 AM, WM wrote:
Le 08/06/2024 à 20:16, Jim Burns a écrit :
On 6/8/2024 8:42 AM, WM wrote:
If j ∈ Defble then j^j^j ∈ Defble.
Nevertheless j^j^j^j^j is finite, but
there are ℵo undefinable natural numbers.
No.
Then define natural numbers with
less or no undefinable successors.
Proposal 1.
j is only collectively.
Proof:
You can replace it by every individual
like 4711 or 10^10^10000.
Whatever we learn
by finite only not.first.false claim.sequence about j
we learn about 4711 and 10^10^10000.
On 6/10/2024 9:26 AM, WM wrote:
Le 08/06/2024 à 19:25, Jim Burns a écrit :
The question is
what counts as able to be defined?
Proposal:
Every product or power or power tower of
defined numbers
is or can become defined.
Every quotient like ω/1000000
is and remains undefined.
Proposal 1.
Definitions are only
statements of _what the definer means_
Without evidence to the contrary,
the definer is presumed to be
honest and aware of what
Since ω/1000000 not.exists,
the properties ω/1000000 has not.matters.
Le 08/06/2024 à 20:16, Jim Burns a écrit :
On 6/8/2024 8:42 AM, WM wrote:
If j ∈ Defble then j^j^j ∈ Defble.
Nevertheless j^j^j^j^j is finite, but
there are ℵo undefinable natural numbers.
No.
Then define natural numbers with
less or no undefinable successors.
There are 0 first undefinable natural numbers.
There are 0 undefinable natural numbers.
This could only be disproved by defining them.
No.
It has been disproved by
giving a description of an individual number j in ℕ⁺
j is only collectively.
Proof:
You can replace it by every individual
like 4711 or 10^10^10000.
Le 08/06/2024 à 22:20, Moebius a écrit :
Am 08.06.2024 um 20:16 schrieb Jim Burns:
A natural number can be
definable and not defined.
Infinitely many even numbers are divisible by 2.
No need for performing the division for
all these numbers though
But we can perform it in every case we desire.
(We cannot desire it in every case however.)
| However, by 2020, most style guides accepted
| the singular 'they' as a personal pronoun.
Le 10/06/2024 à 19:09, Jim Burns a écrit :
On 6/10/2024 9:26 AM, WM wrote:
Le 08/06/2024 à 19:25, Jim Burns a écrit :
The question is
what counts as able to be defined?
Proposal:
Every product or power or power tower of
defined numbers
is or can become defined.
Every quotient like ω/1000000
is and remains undefined.
Proposal 1.
Definitions are only
statements of _what the definer means_
Without evidence to the contrary,
the definer is presumed to be
honest and aware of what
what he means!
define natural numbers with [...] no undefinable successors.
Such numbers exist
because ℕ \ {1, 2, 3, ...} does not leave successors.
On 6/10/2024 12:33 PM, WM wrote [essentially]:
You can't because
every number you can define has ℵo successors.
Le 09/06/2024 à 18:38, Jim Burns a écrit :
"One" can't perform a supertask.
That is the crack through which
WM's rhetoric enters.
You need not perform a supertask!
You need only define
You need only define
*one* natural number that has
less than almost all ℵo natural numbers following!
You can't because
every number you can define has ℵo successors.
That means almost all are undefinable or dark.
Le 10/06/2024 à 18:45, Moebius a écrit :
Hint: Each and every natural number has ℵo successors.
ℕ\{1, 2, 3, ...}
Where are they?
If ω exists, then it is a dark number too.
Are you dark? Well, in WM your real name?
On 6/10/2024 1:50 PM, FromTheRafters wrote:
Moebius has brought this to us :
Am 10.06.2024 um 21:18 schrieb Jim Burns:
| However, by 2020, most style guides accepted
| the singular 'they' as a personal pronoun.
Whatever, it's idiotic crap.
Yep, now we're stuck with trying to figure out the rules for "whom"
and "who". Whom are they, or who is they?
For some damn reason this makes me think of a scene from a movie called Idiocracy:
Why come you don't have a tattoo?
https://youtu.be/tFfTludf0SU?t=88
Wow!
On 6/10/2024 1:14 PM, Moebius wrote:
/0/ is the smallest natural number with no undefinable successors
(after all, _all_ of its ℵo "successors" are "definable", since all
natural numbers are "definable").
Indeed! WM says what about 1, well you only said zero... 1 must be dark?
Crap, or really bad? What did I miss here?
On 6/10/2024 3:47 PM, Moebius wrote:
Am 11.06.2024 um 00:33 schrieb Chris M. Thomasson:
Crap, or really bad? What did I miss here?
Math? :-)
I was just thinking about,
[1, 2, 3, 4, ...]
applied to:
[1/2, 2/2, 3/2, 4/2, ...]
creates
[.5, 1, 1.5, 2, ...]
Refining to all the naturals we get:
[1, 2, ...]
Make any sense?
On 6/10/2024 3:27 PM, Moebius wrote:
Am 11.06.2024 um 00:22 schrieb Chris M. Thomasson:
If ω exists, then it is a dark number too.
Are you dark? Well, in WM your real name?
"is"?
Yup. God damn typos! Sorry. Shit. I need to read my quick typing before
I click send. Sorry again.
Am 11.06.2024 um 01:02 schrieb Chris M. Thomasson:
On 6/10/2024 3:27 PM, Moebius wrote:
Am 11.06.2024 um 00:22 schrieb Chris M. Thomasson:
If ω exists, then it is a dark number too.
Are you dark? Well, in WM your real name?
"is"?
Yup. God damn typos! Sorry. Shit. I need to read my quick typing
before I click send. Sorry again.
N/p. It's a great feature of _natural language_ that the menaing of "statements" (often) can be deduced by taking into consideration the "context".
Make any sense?
Sure.
On 6/10/2024 4:28 PM, Moebius wrote:
Am 11.06.2024 um 01:20 schrieb Chris M. Thomasson:
Make any sense?
Sure.
Like most of your ideas.
Never considered to study math?
I am mostly a skilled programmer that enjoys interesting math, in a
sense?
Sorry about my highly non conventional ways. I tend to learn
enough math to make my ideas just work the way I envision them.
Well, most of the time.
I can create some hard problems... I can teach myself
a lot of things. This comes at a price, where its difficult for me to
consult with people like you, or more of the math types that expect a
certain format and such. Sorry about that. ;^o
I have a lot of experience in programming and math, but parts of it are highly unconventional. Strange notation and such. Ahhh, the programmer
side in me.
I only started out with fractals around early 2013. So, not that long
ago! I was a hard core multi thread sync guy before that.
Shit happens! ;^o
I'd suggest:
[...]
and (of course)
- A. Tarski, Introduction to Logic and to the Methodology of Deductive Sciences
A rather little book.
On 6/10/2024 4:09 PM, Moebius wrote:
What do you "mean" be "refining"? :-P
I guess you mean some sort of ""reduction"" (or "filtering").
Actually, I've never heard about this (in a math context).
But it certainly can be defined.
Let's call this operation "nat".
Then
nat [.5, 1, 1.5, 2, ...] = [1, 2, ...] .
And, say,
nat [.5, 2, 1.5, 7, ...] = [2, 7, ...] .
14/2 is seven, a natural? Filtering out all of the non-natural numbers 7
is in there as 14/2?
Fair enough, or just as crazy as WM? Maybe worse?
Am 11.06.2024 um 03:28 schrieb Moebius:
I'd suggest:
[...]
and (of course)
- A. Tarski, Introduction to Logic and to the Methodology of Deductive
Sciences
A rather little book.
The value-to-effort ratio might be rather high.
On 6/10/2024 1:45 PM, WM wrote:
Le 10/06/2024 à 19:09, Jim Burns a écrit :
| This use of singular 'they' had emerged by the 14th century,
| about a century after the plural 'they' It has been
| commonly employed in everyday English ever since and
| has gained currency in official contexts. Singular 'they'
| has been criticised since the mid-18th century by
| _prescriptive commentators_ who consider it an error.
| Its continued use in modern standard English has become
| more common and formally accepted with the move toward
| gender-neutral language. Some early-21st-century style guides
| described it as colloquial and less appropriate in
| formal writing. However, by 2020, most style guides accepted
| the singular 'they' as a personal pronoun.
each natural number, defined or undefined,
has ℵ₀ followers.
After all, each and every natural number has ℵo successors.
Le 10/06/2024 à 21:18, Jim Burns a écrit :
| This use of singular 'they' had emerged by the 14th century,widerlich.
| about a century after the plural 'they' It has been
| commonly employed in everyday English ever since and
| has gained currency in official contexts. Singular 'they'
| has been criticised since the mid-18th century by
| _prescriptive commentators_ who consider it an error.
| Its continued use in modern standard English has become
| more common and formally accepted with the move toward
| gender-neutral language. Some early-21st-century style guides
| described it as colloquial and less appropriate in
| formal writing. However, by 2020, most style guides accepted
| the singular 'they' as a personal pronoun.
Le 10/06/2024 à 22:29, Moebius a écrit :
After all, each and every natural number has ℵo successors.
No.
ℕ \ {1, 2, 3, ...} = ? Where are the followers?
Le 10/06/2024 à 22:16, Jim Burns a écrit :
each natural number [...] has ℵ₀ followers.
No.
ℕ \ {1, 2, 3, ...} = ? Where are the followers?
Am 11.06.2024 um 16:44 schrieb WM:
Le 10/06/2024 à 22:16, Jim Burns a écrit :
each natural number [...] has ℵ₀ followers.
No.
Yes.
ℕ \ {1, 2, 3, ...} = ? Where are the followers?
Was hat diese "Frage" mit dem oben von Jim Gesagten zu tun?
Hinweis: ℕ = {1, 2, 3, ...}. Daher ist ℕ \ {1, 2, 3, ...} = ℕ \ ℕ = { }.#
Nochmal die Behauptung (von Jim) ist, dass jedes Element in ℕ (also jede der natürlichen Zahlen 1, 2, 3, ...) ℵ₀ Nachfolger besitzt.
In der Tat gilt für alle n e ℕ: card(ℕ \ {1, 2, .., n}) = ℵ₀.
Nochmal: meine Behauptung ist, dass jedes Element in ℕ (also jede der natürlichen Zahlen 1, 2, 3, ...) ℵ₀ Nachfolger besitzt.
In der Tat gilt für alle n e ℕ: card(ℕ \ {1, 2, .., n}) = ℵ₀.
Le 11/06/2024 à 16:57, Moebius a écrit :
Am 11.06.2024 um 16:44 schrieb WM:
Le 10/06/2024 à 22:16, Jim Burns a écrit :
each natural number [...] has ℵ₀ followers.
No.
Yes.
ℕ \ {1, 2, 3, ...} = ? Where are the followers?
Was hat diese "Frage" mit dem oben von Jim Gesagten zu tun?
Hier wird [nichts] gezeigt [...]
Le 11/06/2024 à 16:59, Moebius a écrit :
Nochmal: meine Behauptung ist, dass jedes Element in ℕ (also jede der
natürlichen Zahlen 1, 2, 3, ...) ℵ₀ Nachfolger besitzt.
In der Tat gilt für alle n e ℕ: card(ℕ \ {1, 2, .., n}) = ℵ₀.
Du brichst hier immer ab.
On 6/11/2024 12:41 PM, Chris M. Thomasson wrote:
Therefore:
nat [-.5, 8/2, 14/3, 5/2, .6(6), 4/2, -7, 5.2, 0+0i, ...] = [4, 2, ...]
The nat operation seems logical to me. :^)
Or, if somebody desires to treat zero as a natural number the result
would be, wrt the 0+0i element,
[4, 2, 0, ...]
On 6/10/2024 6:28 PM, Moebius wrote:
and (of course)
- A. Tarski, Introduction to Logic and to the Methodology of Deductive Sciences
A rather little book.
:^) Thanks Man. :^)
Am 11.06.2024 um 20:53 schrieb WM:
Le 11/06/2024 à 16:57, Moebius a écrit :
Am 11.06.2024 um 16:44 schrieb WM:
Le 10/06/2024 à 22:16, Jim Burns a écrit :
each natural number [...] has ℵ₀ followers.
No.
Yes.
ℕ \ {1, 2, 3, ...} = ? Where are the followers?
Was hat diese "Frage" mit dem oben von Jim Gesagten zu tun?
Hier wird gezeigt
On 6/11/2024 7:44 AM, WM wrote:
Le 10/06/2024 à 22:16, Jim Burns a écrit :
each natural number, defined or undefined,
has ℵ₀ followers.
No. ℕ \ {1, 2, 3, ...} = ? Where are the followers?
Are you referring to:
[1/1, 2/2, 3/3, 4/4, ...]
Am 11.06.2024 um 20:56 schrieb WM:
Le 11/06/2024 à 16:59, Moebius a écrit :
Nochmal: meine Behauptung ist, dass jedes Element in ℕ (also jede der
natürlichen Zahlen 1, 2, 3, ...) ℵ₀ Nachfolger besitzt.
In der Tat gilt für alle n e ℕ: card(ℕ \ {1, 2, .., n}) = ℵ₀.
Du brichst hier immer ab.
Whatever, Mückenheim.
Meine Antwort war in der Tat etwas länger:
---------------------
Hinweis: ℕ = {1, 2, 3, ...}. Daher ist ℕ \ {1, 2, 3, ...} = ℕ \ ℕ = { }.
Bekanntlich enthält { } keine Elemente, daher ist die Frage nach iw. Nachfolgern in { } unsinnig. Andererseits gilt natürlich das jede natürliche Zahl in { } ℵ₀ Nachfolger besitzt. 🙂
Nochmal: Meine Behauptung ist, dass jedes Element in ℕ (also jede der natürlichen Zahlen 1, 2, 3, ...) ℵ₀ 'Nachfolger' besitzt.
In der Tat gilt für alle n e ℕ: card(ℕ \ {1, 2, .., n}) = ℵ₀.
Le 11/06/2024 à 22:22, Moebius a écrit :
Am 11.06.2024 um 20:53 schrieb WM:
Le 11/06/2024 à 16:57, Moebius a écrit :
Am 11.06.2024 um 16:44 schrieb WM:
Le 10/06/2024 à 22:16, Jim Burns a écrit :
each natural number [...] has ℵ₀ followers.
No.
Yes.
ℕ \ {1, 2, 3, ...} = ? Where are the followers?
Was hat diese "Frage" mit dem oben von Jim Gesagten zu tun?
Hier wird gezeigt
Le 11/06/2024 à 22:30, Moebius a écrit :
Am 11.06.2024 um 20:56 schrieb WM:
Le 11/06/2024 à 16:59, Moebius a écrit :
Nochmal: meine Behauptung ist, dass jedes Element in ℕ (also jede
der natürlichen Zahlen 1, 2, 3, ...) ℵ₀ Nachfolger besitzt.
In der Tat gilt für alle n e ℕ: card(ℕ \ {1, 2, .., n}) = ℵ₀.
Du brichst hier immer ab.
Whatever, Mückenheim.
Meine Antwort war in der Tat etwas länger:
---------------------
Hinweis: ℕ = {1, 2, 3, ...}. Daher ist ℕ \ {1, 2, 3, ...} = ℕ \ ℕ = { }.
Bekanntlich enthält { } keine Elemente, daher ist die Frage nach iw.
Nachfolgern in { } unsinnig. Andererseits gilt natürlich das jede
natürliche Zahl in { } ℵ₀ Nachfolger besitzt. 🙂
Nochmal: Meine Behauptung ist, dass jedes Element in ℕ (also jede der
natürlichen Zahlen 1, 2, 3, ...) ℵ₀ 'Nachfolger' besitzt.
In der Tat gilt für alle n e ℕ: card(ℕ \ {1, 2, .., n}) = ℵ₀.
<nonsense deleted>
"Zu jedem n in ℕ gibt es ℵ₀ Elemente in ℕ, die größer als n sind"
On 6/11/2024 10:44 AM, WM wrote:
each natural number, defined or undefined,
has ℵ₀ followers.
No.
ℕ \ {1, 2, 3, ...} = ?
Where are the followers?
Define
(Presume it is true that, when I say … , I mean … )
ℕ \ {1, 2, 3, ...} = ?
Where are the followers?
ℕ\{0,1,2,…} = ∅
Define
Each natural number, defined or undefined,
has ℵ₀ followers.
Le 12/06/2024 à 20:18, Jim Burns a écrit :
On 6/11/2024 10:44 AM, WM wrote:
Le 10/06/2024 à 22:16, Jim Burns a écrit :
each natural number, defined or undefined,
has ℵ₀ followers.
No.
ℕ \ {1, 2, 3, ...} = ?
Where are the followers?
Define
(Presume it is true that, when I say … , I mean … )
Simply name one.
ℕ \ {1, 2, 3, ...} = ?
Where are the followers?
ℕ\{0,1,2,…} = ∅
So there are no followers?
Define
No. Simply answer.
Le 10/06/2024 à 22:16, Jim Burns a écrit :
each natural number, defined or undefined,
has ℵ₀ followers.
No.
each natural number, defined or undefined,
has ℵ₀ followers.
No.
ℕ \ {1, 2, 3, ...} = ?
Where are the followers?
each natural number, defined or undefined,
has ℵ₀ followers.
No.
ℕ \ {1, 2, 3, ...} = ?
Where are the followers?
each natural number, defined or undefined,
has ℵ₀ followers.
On 6/12/2024 2:27 PM, WM wrote:
Le 12/06/2024 à 20:18, Jim Burns a écrit :
ℕ \ {1, 2, 3, ...} = ?
Where are the followers?
ℕ\{0,1,2,…} = ∅
So there are no followers?
there is no j ∈ ℕ after ℕ
ℕ is the minimal inductive meta.set.
for each j ∈ ℕ
there are ℵ₀.many followers in ℕ
Are there infinitely many "dark" numbers?
Is there only a finite number of "light" numbers?
WM formulated on Wednesday :
Le 12/06/2024 à 14:29, Moebius a écrit :
"Zu jedem n in ℕ gibt es ℵ₀ Elemente in ℕ, die größer als n >>> sind"
Wenn jedes n aus ℕ entfernt wird, bleiben also noch größere? Welche sind >> das?
You don't remove them from a set, you remove them from consideration in
being a member of a new set.
On 6/12/2024 1:38 PM, WM wrote:
Yes, it is impossible to find a last one before ω.
Indeed. You are correct here. There is no last dark number?
On 6/12/2024 1:51 PM, Chris M. Thomasson wrote:
On 6/12/2024 1:38 PM, WM wrote:
Yes, it is impossible to find a last one before ω.
Indeed. You are correct here. There is no last dark number?
There are infinitely many dark natural numbers, aka, we have not seen
(aka, calculated, "printed them out") them yet? There is no largest dark number, right?
Le 12/06/2024 à 20:54, Jim Burns a écrit :
On 6/12/2024 2:27 PM, WM wrote:
Le 12/06/2024 à 20:18, Jim Burns a écrit :
On 6/11/2024 10:44 AM, WM wrote:
ℕ \ {1, 2, 3, ...} = ?
Where are the followers?
ℕ\{0,1,2,…} = ∅
So there are no followers?
there is no j ∈ ℕ after ℕ
i.e., after all natural numbers.
That means:
If every number is subtracted,
then no successors remain.
If only definable numbersa are
subtracted, then successors remain.
ℕ is the minimal inductive meta.set.
ℕ is all natural numbers.
Not more and not less.
WM laid this down on his screen :
Le 12/06/2024 à 22:41, FromTheRafters a écrit :
WM formulated on Wednesday :
Le 12/06/2024 à 14:29, Moebius a écrit :
"Zu jedem n in ℕ gibt es ℵ₀ Elemente in ℕ,
die größer als n sind"
Wenn jedes n aus ℕ entfernt wird,
bleiben also noch größere?
Welche sind das?
You don't remove them from a set,
you remove them from consideration
in being a member of a new set.
I remove them from the set.
Sets still don't change.
Le 12/06/2024 à 14:29, Moebius a écrit :
"Zu jedem n in ℕ gibt es ℵ₀ Elemente in ℕ, die größer als n sind"
Wenn jedes [Element] aus ℕ entfernt wird, bleiben [...]
Chris M. Thomasson explained :
{0, 1, 2, 3, ...}
WM artifically removes say, 2:
{0, 1, 3, ...}.
There are still infinitely many numbers left... :^)
Of the roster forms you posted above, only the first suggests a unique pattern to go with the ellipsis.
Sets [...] don't change.
{ 0, 1, 2, 3, ... }
WM artifically removes say, 2:
{ 0, 1, 3, ... }
There are still infinitely many numbers left... :^)
and ω/10000, let alone to find those between ω/10000 and ω.
i.e., after all natural numbers. That means:
If every number is subtracted, then no successors remain.
On 6/12/2024 4:51 PM, FromTheRafters wrote:
Sets still don't change.
Yes.
If sets changed, [...]
Am 12.06.2024 22:33:18 WM schrieb:
If every number is subtracted, then no successors remain. [WM]
Moebius wrote :
Am 13.06.2024 um 01:37 schrieb FromTheRafters:
Chris M. Thomasson explained :
{0, 1, 2, 3, ...}
WM artifically removes say, 2:
{0, 1, 3, ...}.
There are still infinitely many numbers left... :^)
Of the roster forms you posted above, only the first suggests a
unique pattern to go with the ellipsis.
C'mon. That's lame, man. :-P
Okay, then what is the obvious pattern?
To me, the triangulars come to mind, and there are infinitely many of
them indeed. Could also be the naturals with only the two not being considered which is evident in context. [...]
I hope you see that the roster form should not be ambiguous.
On 6/12/2024 5:37 PM, FromTheRafters wrote:
Moebius submitted this idea :
But Chris EXPLAINED what he MEANS with "{0, 1, 3, ...}":
Yes, and <bla bla bla>
|"WM artifically removes say, 2 [from {0, 1, 2, 3, ...}]: {0, 1, 3,
...}."
See? Clear enough to figure out the meaning of "{0, 1, 3, ...}" IN
THIS (THE PRESENT) CONTEXT.
<bla bla bla>
Raising hand... Chris, you fucked up again wrt notation... ;^o
ZAPPPPP!!! YIKES!!!
The unit fractions are separated.
NUF(0) = 0. NUF(1) = ℵo.
The increase happens.
Es geht hier aber um den trivialen mengentheoretischen Sacherhalt:
"Zu jedem n in ℕ gibt es ℵ₀ Elemente in ℕ, die größer als n sind"
On 6/12/2024 2:03 PM, Chris M. Thomasson wrote:
On 6/12/2024 1:58 PM, WM wrote:
Le 12/06/2024 à 22:51, "Chris M. Thomasson" a écrit :
On 6/12/2024 1:38 PM, WM wrote:
Yes, it is impossible to find a last one before ω.
Indeed. You are correct here. There is no last dark number?
There is one (think of the first unit fraction) but as it is and
remains dark, it cannot be caught.
I thought there were infinitely many dark natural numbers? If so, how
can there possibly be a largest dark number?
You told me that there are indeed infinitely many dark natural numbers, right?
On 6/12/2024 4:51 PM, FromTheRafters wrote:
WM laid this down on his screen :
I remove them from the set.
Sets still don't change.
Yes.
On 6/12/2024 4:33 PM, WM wrote:
Le 12/06/2024 à 20:54, Jim Burns a écrit :
On 6/12/2024 2:27 PM, WM wrote:
Le 12/06/2024 à 20:18, Jim Burns a écrit :
On 6/11/2024 10:44 AM, WM wrote:
ℕ \ {1, 2, 3, ...} = ?
Where are the followers?
ℕ\{0,1,2,…} = ∅
So there are no followers?
there is no j ∈ ℕ after ℕ
i.e., after all natural numbers.
There is no natural number after (≥)
all natural numbers.
That means:
If every number is subtracted,
then no successors remain.
If every number is deleted,
then every number is deleted.
If only definable numbers are
subtracted, then successors remain.
Only if some natural number is undefinable.
If any natural number is undefinable, then
the first undefinable has a definable predecessor.
No undefinable has a definable predecessor.
No natural number is undefinable.
After all definables are deleted from ℕ
no successors (no anything) remain in ℕ
By 'ℕ' I mean the minimal inductive meta.set.
Le 13/06/2024 à 01:33, Moebius a écrit :
Es geht hier aber um den trivialen mengentheoretischen Sacherhalt:
"Zu jedem n in ℕ gibt es ℵ₀ Elemente in ℕ, die größer als n sind"
Where are they when all n e ℕ are removed?
on 6/12/2024, Chris M. Thomasson supposed :
On 6/12/2024 5:37 PM, FromTheRafters wrote:
Moebius submitted this idea :
Am 13.06.2024 um 02:00 schrieb FromTheRafters:
Moebius wrote :
Am 13.06.2024 um 01:37 schrieb FromTheRafters:
Chris M. Thomasson explained :
{0, 1, 2, 3, ...}
WM artifically removes say, 2:
{0, 1, 3, ...}.
There are still infinitely many numbers left... :^)
Of the roster forms you posted above, only the first suggests a
unique pattern to go with the ellipsis.
C'mon. That's lame, man. :-P
Okay, then what is the obvious pattern?
Yeah, "{0, 1, 3, ...}" (as such) does not suggest a pattern, agree.
But it does, and not the one he intended.
But Chris EXPLAINED what he MEANS with "{0, 1, 3, ...}":
Yes, and it was obvious when he used square brackets earlier that he
meant a set, yet he was corrected. How else does one learn?
"WM artifically removes say, 2 [from {0, 1, 2, 3, ...}]: {0, 1, 3,...}."
See? Clear enough to figure out the meaning of "{0, 1, 3, ...}" IN
THIS (THE PRESENT) CONTEXT.
Calm down, his "wrong" brackets were addressed also even though in
context he was understood.
Raising hand... Chris, you fucked up again wrt notation... ;^o
ZAPPPPP!!! YIKES!!!
Sorry... ;^o
I'll try not to help anymore.
dark numbers have no discernible order
Fact is
Le 12/06/2024 à 23:47, Jim Burns a écrit :
On 6/12/2024 4:51 PM, FromTheRafters wrote:
WM laid this down on his screen :
I remove them from the set.
Sets still don't change.
Yes.
The set of americam citizens changes permanently.
On 6/12/2024 4:33 PM, WM wrote:
If every number is subtracted,
then no successors remain.
If only definable numbers are subtracted,
then successors remain.
ℕ is all natural numbers.
Not more and not less.
for each j ∈ ℕ
there are ℵ₀.many followers in ℕ
Is ℕ more than every j ∈ ℕ?
What is it more?
Le 12/06/2024 à 20:54, Jim Burns a écrit :
On 6/12/2024 2:27 PM, WM wrote:
Le 12/06/2024 à 20:18, Jim Burns a écrit :
ℕ\{0,1,2,…} = ∅
So there are no followers?
there is no j ∈ ℕ after ℕ
i.e., after all natural numbers.
That means:
If every number is subtracted,
then no successors remain.
If only definable numbers are subtracted,
then successors remain.
If every number is subtracted,
then no successors remain.
If only definable numbers are subtracted,
then successors remain.
ℕ is the minimal inductive meta.set.
ℕ is all natural numbers.
Not more and not less.
for each j ∈ ℕ
there are ℵ₀.many followers in ℕ
Is ℕ more than every j ∈ ℕ?
What is it more?
Le 12/06/2024 à 23:12, Jim Burns a écrit :
If any natural number is undefinable, then
the first undefinable has a definable predecessor.
That is your error.
The definable numbers are definable and
have definable successors.
You will never get into the dark numbers by
counting or defining.
On 6/13/2024 6:55 AM, WM wrote:
Le 12/06/2024 à 23:12, Jim Burns a écrit :
If any natural number is undefinable, then
the first undefinable has a definable predecessor.
Am 12.06.2024 22:33:18 WM schrieb:
i.e., after all natural numbers. That means:
If every number is subtracted, then no successors remain.
Thats absolutely the same as:
"If every number is noticed, then no successors remain. "
Am 13.06.2024 um 02:14 schrieb Tom Bola:
Am 12.06.2024 22:33:18 WM schrieb:
If every number is subtracted, then no successors remain. [WM]
Eine wirklich bemerkenswerte Erkenntnis!
Am 12.06.2024 um 23:01 schrieb WM:
The unit fractions are separated.
An,m e IN: n =/= m <-> 1/n =/= 1/m.
NUF(0) = 0. NUF(1) = ℵo.
Ja. Mehr noch:
Ax e IR, x > 0: NUF(x) = ℵo
Google mal nach "Häufungspunkt".
The increase happens.
Jawohl! It's just that "the increase" actually is a _jump_.
Le 13/06/2024 à 02:26, Moebius a écrit :
Am 13.06.2024 um 02:14 schrieb Tom Bola:
Am 12.06.2024 22:33:18 WM schrieb:
If every number is subtracted, then no successors remain. [WM]
Eine wirklich bemerkenswerte Erkenntnis!
It contradicts your claim that every number has ℵo successors.
On 6/12/2024 4:33 PM, WM wrote:
there is no j ∈ ℕ after ℕ
i.e., after all natural numbers.
there is no
natural number after (≥) all natural numbers.
I have placed proposal 2 (sets: E A X) and
proposal 3 (meta.sets: E A X C MX) before you.
They have different consequences.
If each finite von Neumann ordinal
is deleted from
the set of finite von Neumann ordinals,
nothing is left in the result.
We finite beings cannot perform a supertask,
which is what it takes to count all
finite von Neumann ordinals.
However,
it is not a supertask
Am 13.06.2024 um 12:48 schrieb WM:
dark numbers have no discernible order
Definiere erst mal den Begriff "discernible order",
Le 13/06/2024 à 04:37, Moebius a écrit :
Am 12.06.2024 um 23:01 schrieb WM:
The unit fractions are separated.
An,m e IN: n =/= m <-> 1/n =/= 1/m.
NUF(0) = 0. NUF(1) = ℵo.
Ja. Mehr noch:
Ax e IR, x > 0: NUF(x) = ℵo
Wrong.
Google mal nach "Häufungspunkt".
Am 13.06.2024 um 14:22 schrieb Jim Burns:
On 6/12/2024 4:33 PM, WM wrote:
If every number is subtracted,
then no successors remain.
If only definable numbers are subtracted,
then successors remain.
Observation: If "definable numbers" is replaced by "finitely many
numbers"
| If every number is subtracted,
| then no successors remain.
| If only finitely many numbers are subtracted
| then [infinitely many] successors remain.
Le 13/06/2024 à 14:22, Jim Burns a écrit :
there is no
natural number after (≥) all natural numbers.
But there are many after all definable natural numbers.
On 6/13/2024 6:55 AM, WM wrote:
Le 12/06/2024 à 23:12, Jim Burns a écrit :
If any natural number is undefinable, then
the first undefinable has a definable predecessor.
That is your error.
The definable numbers are definable and
have definable successors.
The minimal inductive set contains
all and only finite von Neumann ordinals.
You will never get into the dark numbers by
counting or defining.
There is no final finite von Neumann ordinal.
By 'natural number' I mean
'finite von Neumann ordinal'.
By ℕ I mean
minimal inductive set.
Le 13/06/2024 à 14:50, Moebius a écrit :
Am 13.06.2024 um 14:22 schrieb Jim Burns:
On 6/12/2024 4:33 PM, WM wrote:
If every number is subtracted,
then no successors remain.
If only definable numbers are subtracted,
then successors remain.
Observation: If "definable numbers" is replaced by "finitely many
numbers"
By definition definable implies existence of a FISON.
Am 13.06.2024 um 15:53 schrieb WM:
Le 13/06/2024 à 02:26, Moebius a écrit :
Am 13.06.2024 um 02:14 schrieb Tom Bola:
Am 12.06.2024 22:33:18 WM schrieb:
If every number is subtracted, then no successors remain. [WM]
Eine wirklich bemerkenswerte Erkenntnis!
It contradicts your claim that every number has ℵo successors.
No,
Le 13/06/2024 à 16:32, Moebius a écrit :
"In der Analysis ist ein Häufungspunkt einer Menge anschaulich ein
Punkt, der unendlich viele Punkte der Menge in seiner Nähe hat."
But never two unit fractions at the same coordinate.
"In der Analysis ist ein Häufungspunkt einer Menge anschaulich ein
Punkt, der unendlich viele Punkte der Menge in seiner Nähe hat."
The minimal inductive set contains
all and only finite von Neumann ordinals.
Yes. I call it ℕ_def.
By ℕ I mean [the]
minimal inductive set.
That is what I call ℕ_def.
Every number has ℵo [numbers as] successors.
If every number is subtracted [nothing] remain[s].
Two simple steps.
Am 13.06.2024 um 16:28 schrieb WM:
Le 13/06/2024 à 14:22, Jim Burns a écrit :
there is no
natural number after (≥) all natural numbers.
But there are many after all definable natural numbers.
Woanders hast Du geschrieben:
"ω cannot be connected by a FISON to zero. This however is required by
my definition of /definable/."
Dann lass Dir gesagt sein, dass JEDE natürliche Zahl (per def) "is
connected by a FISON to zero".
Tatsächlich SIND die natürlichen Zahlen
(nach von Neumann) FISONS
So, nein, there are NO natural numbers after all "definable" natural
numbers (since ALL natural numbers are "definable").
Btw. Hast Du hier nicht "definable" mit "defined" verwechselt?
Le 13/06/2024 à 16:40, Moebius a écrit :
Am 13.06.2024 um 16:28 schrieb WM:
Le 13/06/2024 à 14:22, Jim Burns a écrit :
there is no
natural number after (≥) all natural numbers.
But there are many after all definable natural numbers.
Woanders hast Du geschrieben:
"ω cannot be connected by a FISON to zero. This however is required by
my definition of /definable/."
Dann lass Dir gesagt sein, dass JEDE natürliche Zahl (per def) "is
connected by a FISON to zero".
No.
Tatsächlich SIND die natürlichen Zahlen (nach von Neumann) FISONS
Yes.
So, nein, there are NO natural numbers after all "definable" natural
numbers (since ALL natural numbers are "definable").
Btw. Hast Du hier nicht "definable" mit "defined" verwechselt?
There are defined numbers, a subcollection of definable numbers, a subcollection of all natural numbers.
Am 13.06.2024 um 16:39 schrieb WM:
Le 13/06/2024 à 14:50, Moebius a écrit :
Am 13.06.2024 um 14:22 schrieb Jim Burns:
On 6/12/2024 4:33 PM, WM wrote:
If every number is subtracted,
then no successors remain.
If only definable numbers are subtracted,
then successors remain.
Observation: If "definable numbers" is replaced by "finitely many
numbers"
By definition definable implies existence of a FISON.
Ich glaube, Du verwechselst hier gerade "definable" mit "defined".
Aber natürlich impliziert "defined" "definable". :-)
(Denn wenn eine Zahl nicht "definable" wäre, könnte sie auch nicht "defined" sein.)
Hinweis: Da die natürlichen Zahlen nach von Neumann FISONs SIND, sind
diese also auch "definable" bzw. "defined".
Damit ist dann aber klar, dass die Behauptung "If only definable numbers
are subtracted, then successors remain." falsch ist.
"Richtig" dagegen wäre die Behauptung: "If only finitely many numbers
are subtracted, then successors remain."
Am 13.06.2024 um 16:56 schrieb WM:
Le 13/06/2024 à 16:32, Moebius a écrit :
"In der Analysis ist ein Häufungspunkt einer Menge anschaulich ein
Punkt, der unendlich viele Punkte der Menge in seiner Nähe hat."
But never two unit fractions at the same coordinate.
Hat das je jemand behauptet
Am 13.06.2024 um 16:55 schrieb WM:
Every number has ℵo successors.
If every number is subtracted ℵo successors remain.
Two simple steps.
In der Tat.
Le 13/06/2024 à 16:53, Moebius a écrit :
Am 13.06.2024 um 16:39 schrieb WM:
Le 13/06/2024 à 14:50, Moebius a écrit :
Am 13.06.2024 um 14:22 schrieb Jim Burns:
On 6/12/2024 4:33 PM, WM wrote:
If every number is subtracted,
then no successors remain.
If only definable numbers are subtracted,
then successors remain.
Observation: If "definable numbers" is replaced by "finitely many
numbers"
By definition definable implies existence of a FISON.
Ich glaube, Du verwechselst hier gerade "definable" mit "defined".
Aber natürlich impliziert "defined" "definable". :-)
(Denn wenn eine Zahl nicht "definable" wäre, könnte sie auch nicht
"defined" sein.)
Hinweis: Da die natürlichen Zahlen nach von Neumann FISONs SIND, sind
diese also auch "definable" bzw. "defined".
Of course.
Damit ist dann aber klar, dass die Behauptung "If only definable
numbers are subtracted, then successors remain." falsch ist.
No.
"Richtig" dagegen wäre die Behauptung: "If only finitely many numbers
are subtracted, then successors remain."
All FISONs or v. Neumann ordinals [in IN] are finite.
Theorem: More than finitely many finite initial segments cannot be merged.
Proof: This is caused by the pigeonhole principle and the definition
"finite initial segment". If each of the first n positive integers has a unary representation in form of a string, like ooooo, that is shorter
than n then, by the pigeonhole principle, there must be two different positive integers defined by the same unary representation.
Clearly this is absurd.
Le 13/06/2024 à 17:02, Moebius a écrit :
Am 13.06.2024 um 16:55 schrieb WM:
Every number has ℵo successors.
If every number is subtracted ℵo successors remain.
Two simple steps.
In der Tat.
Am 13.06.2024 um 17:13 schrieb WM:
Le 13/06/2024 à 17:02, Moebius a écrit :
Am 13.06.2024 um 16:55 schrieb WM:
Every number has ℵo successors.
If every number is subtracted ℵo successors remain.
Two simple steps.
Offensichtlich verstehst Du insbesondere die Bedeutung des Wortes "alle" (bzw. 'every') nicht.
Am 13.06.2024 um 17:01 schrieb WM:
Le 13/06/2024 à 16:40, Moebius a écrit :
Am 13.06.2024 um 16:28 schrieb WM:
Le 13/06/2024 à 14:22, Jim Burns a écrit :
there is no
natural number after (≥) all natural numbers.
But there are many after all definable natural numbers.
Woanders hast Du geschrieben:
"ω cannot be connected by a FISON to zero. This however is required
by my definition of /definable/."
Dann lass Dir gesagt sein, dass JEDE natürliche Zahl (per def) "is
connected by a FISON to zero".
No.
Doch, doch, denn:
Tatsächlich SIND die natürlichen Zahlen (nach von Neumann) FISONS
Yes.
D. h. (für jede natürliche Zahl n > 0:) die natürliche Zahl n ist durch die FISON {0, ..., n-1} mit 0 "verbunden". (Sonderfall n = 0, da ist die FISON {}.)
Also:
So, nein, there are NO natural numbers after all "definable" natural
numbers (since ALL natural numbers are "definable").
Hinweis: Richtig wäre z. B.
| Every number has ℵo successors.
Am 13.06.2024 um 17:55 schrieb WM:
Daher würde ich es lieber so formulieren (um das klar zu stellen):
| Every number has ℵo numbers as successors.
| If only numbers having ℵo successors are removed (i.e. _all_ numbers)
| then nothing remains.
nothing remains.
Le 13/06/2024 à 17:48, Moebius a écrit :
Hinweis: Richtig wäre z. B.
| Every number has ℵo successors.
Vorschlag zur Logik:
If only numbers having ℵo successors are removed,
then ℵo successors remain.
Solltest Du das auch ablehnen, ist keine sinnvolle Diskussion mit Dir möglich.
Am 13.06.2024 um 18:04 schrieb Moebius:
Am 13.06.2024 um 17:55 schrieb WM:
Daher würde ich es lieber so formulieren (um das klar zu stellen):
| Every number has ℵo numbers as successors.
| If only numbers having ℵo successors are removed (i.e. _all_ numbers)
| then nothing remains.
Hier anhand einiger Beispiele im Detail erklärt: [...]
D. h. _alle_ Elemente in IN haben ℵo Elemente in IN als "Nachfolger".
Wenn man nun die Differenzmenge von IN und der Menge {n e IN : n hat ℵo Elemente in IN als "Nachfolger"} betrachtet, dann ist das IN \ IN = { }.
nothing remains.
Am 13.06.2024 um 17:55 schrieb WM:
Vorschlag zur Logik:
If only numbers having ℵo successors are removed,
then ℵo successors remain.
Nein, dann remains NO successor, da bekanntlich auch die successors
numbers sind;
Daher würde ich es lieber so formulieren (um das klar zu stellen):
| If only numbers having ℵo successors are removed
Le 13/06/2024 à 16:28, Moebius a écrit :
[...]
Every number has ℵo successors.
If every number is subtracted the successors remain.
Two simple steps.
Le 13/06/2024 à 18:04, Moebius a écrit :
Am 13.06.2024 um 17:55 schrieb WM:
Vorschlag zur Logik:
If only numbers having ℵo successors are removed, then ℵo successors >>> remain.
Nein, dann remains NO successor, da bekanntlich auch die successors
numbers sind;
aber keine, die ℵo successors haben.
Daher würde ich es lieber so formulieren (um das klar zu stellen):
On 6/13/2024 7:59 AM, Moebius wrote:
Am 13.06.2024 um 16:45 schrieb WM:[...]
The minimal inductive set contains
all and only finite von Neumann ordinals.
Yes. I call it ℕ_def.
Mathematians usually call it "ℕ" or "ω".
Please try to use standard notation (to avoid confusion).
I am blushing now. I need some help with standard notation, read study
and get helped by people like you, ect... ;^o
Le 13/06/2024 à 15:16, Jim Burns a écrit :
The minimal inductive set contains
all and only finite von Neumann ordinals.
Yes. I call it ℕ_def.
You will never get into the dark numbers
by counting or defining.
There is no final finite von Neumann ordinal.
Correct.
That is the reason why
you cannot leave this collection.
By 'natural number' I mean
'finite von Neumann ordinal'.
That implies the existence of a FISON
and hence definable number.
Induction means existence of FISONs.
By ℕ I mean
minimal inductive set.
That is what I call ℕ_def.
By induction we prove that
ℵo numbers of ℕ remain before ω.
Am 13.06.2024 um 20:03 schrieb WM:
Le 13/06/2024 à 18:04, Moebius a écrit :
Am 13.06.2024 um 17:55 schrieb WM:
Vorschlag zur Logik:
If only numbers having ℵo successors are removed, then ℵo successors >>>> remain.
Nein, dann remains NO successor, da bekanntlich auch die successors
numbers sind;
aber keine, die ℵo successors haben.
Da widersprichst Du Dir gerade selbst, da Du im ersten Satz sagtest:
"Every number has ℵo numbers as successors."
Du scheinst innerhalb kürzester Zeit zu vergessen, was Du eben noch
gesagt hattest.
On 6/13/2024 7:59 AM, Moebius wrote:
Please, Mückenheim, try to use standard notation (to avoid confusion).
I am blushing now. I need some help with standard notation, read study
and get helped by people like you, ect... ;^o
Thanks.
Am 13.06.2024 um 21:13 schrieb Chris M. Thomasson:
On 6/13/2024 7:59 AM, Moebius wrote:
Please, Mückenheim, try to use standard notation (to avoid confusion).
I am blushing now. I need some help with standard notation, read study
and get helped by people like you, ect... ;^o
Thanks.
There's a difference between a reasonable guy not using standard
notation (because he never formally learned it) and a CRANK who
deliberately (or unconsciously) AVOIDS using standard notation.
On 6/13/2024 7:56 AM, WM wrote:
But never two unit fractions at the same coordinate.
Well, here are two unit fractions:
A = 1/2
B = 1/2
On 6/13/2024 10:55 AM, WM wrote:
If every number which has ℵ₀ successors
is deleted,
then none remain.
Explanations should be as simple as possible,
but no simpler.
Am 13.06.2024 um 20:03 schrieb WM:
Le 13/06/2024 à 18:04, Moebius a écrit :
Am 13.06.2024 um 17:55 schrieb WM:
Vorschlag zur Logik:
If only numbers having ℵo successors are removed, then ℵo successors >>>> remain.
Nein, dann remains NO successor, da bekanntlich auch die successors
numbers sind;
aber keine, die ℵo successors haben.
Da widersprichst Du Dir gerade selbst, da Du im ersten Satz sagtest:
"Every number has ℵo numbers as successors."
On 6/13/2024 3:48 AM, WM wrote:
Fact is: If we assume the existence of ω at the ordinal line, then
something must exist before, either dark numbers or nothing. There is no
third alternative. Or can you imagine one?
Let be imagine something that was dark:
A = 1024^42426969
B = A + 1
A and B were already in the set of natural numbers, right?
Natürlich hat jeder
der ℵo successors (einer Zahl) wieder ℵo successors.
Le 13/06/2024 à 21:39, Moebius a écrit :
Am 13.06.2024 um 20:03 schrieb WM:
Le 13/06/2024 à 18:04, Moebius a écrit :
Am 13.06.2024 um 17:55 schrieb WM:
Vorschlag zur Logik:
If only numbers having ℵo successors are removed, then
ℵo successors remain.
Nein, dann remains NO successor, da bekanntlich auch die successors
numbers sind;
aber keine, die ℵo successors haben.
Not every number has ℵo numbers as successors.
If all numbers which have ℵo successors are deleted from ℕ, then ℵo successors remain in ℕ
Le 13/06/2024 à 20:25, Jim Burns a écrit :
On 6/13/2024 10:55 AM, WM wrote:
Every number has ℵo successors.
If every number which hasn't ℵ₀ successors
is deleted,
then none are deleted.
If every number is subtracted
the successors remain.
If every number which has ℵ₀ successors
is deleted,
then none remain.
Try to think better.
The successors are not deleted.
Every number has ℵo successors.
Explanations should be as simple as possible,
but no simpler.
Here is the simplest possible explanation:
If all numbers which have ℵo successors
are deleted from ℕ,
Every number has ℵo successors.
then ℵo successors remain in ℕ
- in every step
and, if existing, in the limit.
But every definable number is deleted.
None remains.
Every number has ℵo successors.
Le 13/06/2024 à 23:52, Moebius a écrit :
Note that [...] the successors remain by definition.
On 6/14/2024 4:34 AM, WM wrote:
Le 13/06/2024 à 20:25, Jim Burns a écrit :
On 6/13/2024 10:55 AM, WM wrote:
Every number has ℵo successors.
Yes.
You:
Every number has ℵo successors.
If all numbers which have ℵo successors
are deleted from ℕ,
...and
Every number has ℵo successors.
then ℵo successors remain in ℕ
...then successors which haven't ℵ₀ successors
remain in the new not.ℕ set.
That is 0 successors.
Nun hat jede Zahl in IN unendlich viele Nachfolger (in IN). Wenn mal
also eine Teilmenge von IN betrachtet, die keine natürlichen Zahlen enthält, die unendlich viele Nachfolger (in IN) besitzen, dann ist diese Teilmenge leer.
Am 14.06.2024 um 10:48 schrieb WM:
Offenbar versuchst Du gerade wieder einen unsinnigen Scheißdreck, den Du formuliert hast, unter den Teppich zu kehren.
Du hattest geschrieben:
"Every number has ℵo successors." (*)
If all numbers which have ℵo successors are deleted from ℕ, then ℵo successors remain in ℕ
Das steht im Widerspruch zu (*).
Le 14/06/2024 à 13:17, Moebius a écrit :
Am 14.06.2024 um 10:48 schrieb WM:
Offenbar versuchst Du gerade wieder einen unsinnigen Scheißdreck, den
Du formuliert hast, unter den Teppich zu kehren.
Du hattest geschrieben:
"Every number has ℵo successors." (*)
No, you are a liar.
Every number has ℵo successors.
If every number is subtracted the successors remain.
Two simple steps.
Regards, WM
If all numbers which have ℵo successors are deleted from ℕ, then ℵo
successors remain in ℕ
Das steht im Widerspruch zu (*).
Of course. (*) is wrong.
Le 14/06/2024 à 18:34, Moebius a écrit :
Nun hat jede Zahl in IN unendlich viele Nachfolger (in IN). Wenn man
also eine Teilmenge von IN betrachtet, die keine natürlichen Zahlen
enthält, die unendlich viele Nachfolger (in IN) besitzen, dann ist
diese Teilmenge leer.
Note that
Am 14.06.2024 um 18:29 schrieb WM:
Le 14/06/2024 à 13:17, Moebius a écrit :
Du hattest geschrieben:
"Every number has ℵo successors." (*)
No, you are a liar.
So, so. Du hast als NICHT am 13.06. um 16:55 Uhr
| Every number has ℵo successors.
| If every number is subtracted the successors remain.
| Two simple steps.
geschrieben?
Welche natürliche Zahl hat denn NICHT "ℵo successors"? Kannst Du so eine angeben/nennen
und für sie BEWEISEN, dass die Aussage stimmt? Oder
kannst Du wenigstens beweisen, dass es so eine natürliche Zahl GIBT?
WM: If every number is subtracted, then no successors remain.
If every number is subtracted the successors remain.
Le 14/06/2024 à 16:56, Jim Burns a écrit :
On 6/14/2024 4:34 AM, WM wrote:
Le 13/06/2024 à 20:25, Jim Burns a écrit :
On 6/13/2024 10:55 AM, WM wrote:
Every number has ℵo successors.
Yes.
No.
Here is the context:
WM:
If every number is subtracted,
then no successors remain.
FF:
Eine wirklich bemerkenswerte Erkenntnis!
WM:
It contradicts your claim that
every number has ℵo successors.
FF:
No,
WM (Proof by contradiction):
Every number has ℵo successors.
If every number is subtracted
the successors remain.
You:
Every number has ℵo successors.
In a proof by contradiction.
Note that
only the numbers with successors are deleted,
the successors remain by definition.
So, the following sequence is never dark?
________________
A = 0
A = A + 1 = 1
A = A + 1 = 2
A = A + 1 = 3
... (on and on) ...
________________
On 6/14/2024 1:41 AM, WM wrote:So let's read his statement the following way: "But never two
Le 13/06/2024 à 20:56, "Chris M. Thomasson" a écrit :
On 6/13/2024 7:56 AM, WM wrote:
But never two unit fractions at the same coordinate.
Well, here are two unit fractions:
A = 1/2
B = 1/2
No that is only one. Please stop to be silly.
How many items do I have here:
A = 1/2
B = 1/2
?
On 6/14/2024 1:41 AM, WM wrote:So let's read his statement the following way: "But never two
Le 13/06/2024 à 20:56, "Chris M. Thomasson" a écrit :
On 6/13/2024 7:56 AM, WM wrote:
But never two unit fractions at the same coordinate.
Well, here are two unit fractions:
A = 1/2
B = 1/2
No that is only one. Please stop to be silly.
How many items do I have here:
A = 1/2
B = 1/2
?
I hope you see that the roster form should not be ambiguous.
Am 13.06.2024 um 02:37 schrieb FromTheRafters:
I hope you see that
the roster form should not be ambiguous.
Sure.
{0, 1, 2, 3, ...} \ {2} = {0, 1, 3, ...} ,
or, if you like:
{0, 1, 2, 3, ...} \ {2} = {0, 1, 3, 4, ...} .
:-P
It's often context which avoids ambiguity.
On the other had, I'd agree that the meaning of, say,
"{0, 1, 2, 3, ...}"
or
"{1, 2, 3, ...}"
is unambiguous by convention.
On 6/14/2024 6:24 PM, Moebius wrote:
Am 13.06.2024 um 02:37 schrieb FromTheRafters:
I hope you see that
the roster form should not be ambiguous.
Sure.
{0, 1, 2, 3, ...} \ {2} = {0, 1, 3, ...} ,
or, if you like:
{0, 1, 2, 3, ...} \ {2} = {0, 1, 3, 4, ...} .
:-P
It's often context which avoids ambiguity.
On the other had, I'd agree that the meaning of, say,
"{0, 1, 2, 3, ...}"
or
"{1, 2, 3, ...}"
is unambiguous by convention.
Unambiguous by almost.universal convention.
Almost.
Consider a poster who says {0,1,2,3,...} ≠ ℕ
that ℕ\{0,1,2,3,...} holds featherbedding elements
which sneer at mere induction.
An unconventional ℕ
I see two options.
1.
Treat the poster approximately the same as
the guy on the corner telling everyone passing by
that his cat is God. Distance. Distance is good.
2.
Use unconventional not.misunderstood notation
to convey the conventional ideas, and
why the convention is what it is.
There is an implicit assumption here
that the poster does not think his cat is God.
I have taken option 2,
although I appreciate the point of view
of those taking option 1.
I mention my choice in order to answer questions
which you seem to have about whether
I think my cat is God.
Please, do not worry. I do not have a cat.
On 5/2/2024 10:06 AM, Moebius wrote:
Am 30.04.2024 um 14:56 schrieb WM:
The end is smaller than ω.
The sequence of natural numbers (all of which are smaller than ω) does
not have an end, Du Depp!
Fwiw, here how I generally write my recursive functions. Say the natural numbers, including zero here:
r[0] = 0
r[n + 1] = r[n] + 1
Lets expand that for a couple of iterations:
r[0] = 0
r[1] = r[0] + 1 = 1
r[2] = r[1] + 1 = 2
r[3] = r[2] + 1 = 3
...
[n] where n is an index
vs:
r_0 = 0
r_(n+1) = r_n + 1
Both seem okay to me, as in I can understand both of them.
On 6/14/2024 12:39 PM, WM wrote:
WM (Proof by contradiction):
[Assume:] Every number has ℵo successors.
If every number is subtracted the successors remain.
Am 14.06.2024 um 20:52 schrieb Jim Burns:
On 6/14/2024 12:39 PM, WM wrote:
Just seen here:
"number(s)" (WM) seems to refer to "natural number(s)" in this context.
WM (Proof by contradiction):
[Assume:] Every number has ℵo successors.
Actually, we do not have to assume that, since it can be proved (in the context of mathematics/set theory).
An e IN: card({m e IN : m > n}) = ℵo.
If every number is subtracted the successors remain.
Huh?! Just a silly (psychotic) claim. If _every_ number "is subtracted" (based on "the set of numbers+their successors"), then NO numbers (and
hence no successors) "remain" [in the new/resulting set]. (After all,
the successors of any number are numbers too.*)
What did WM prove here? That he's a complete idiot?
_____________________________________________
*) An e IN: {m e IN : m > n} c IN.
Am 14.06.2024 um 20:52 schrieb Jim Burns:
How long can you
listen to this before you start to freak out? lol.
https://youtu.be/kdCJunw_Jgg
How long can you listen to this before you start to freak out? lol
https://youtu.be/kdCJunw_Jgg
Am 14.06.2024 um 20:52 schrieb Jim Burns:
What did WM prove here? That he's a complete idiot?
"But never two
(different) unit fractions at the same coordinate."
Well, it's hard to see why he feels the need to express this triviality.
On 6/14/2024 12:39 PM, WM wrote:
every number in {i:∀₃Xᴬ⤾⁺¹₀:X∋i}
has ℵ₀ successors in {i:∀₃Xᴬ⤾⁺¹₀:X∋i}
Note that
only the numbers with successors are deleted,
the successors remain by definition.
sⱼ∘f:{0<} → {j⁺¹<} is 1.to.1
Concerning his claim "But never two (different) unit fractions at the
same coordinate.": it is so trivial that it's hard to state/express it properly.
Unit fractions are real numbers. And "two" unit fractions (and hence
real numbers) a, b are either identical (a = b) or not (a =/= b). That's
all.
Well, my last shot: If a and b are two _different_ unit fractions, then there's no real number c such that a = c = b (though -of course- a and b
are real numbers).
On 6/16/2024 7:21 AM, WM wrote:
Le 14/06/2024 à 23:43, Moebius a écrit:
"But never two (different) unit fractions at the same coordinate." [WM]
Well, it's hard to see why he feels the need to express this triviality.
is the reason: Before every x > 0 there are many unit fractions.
It is wrong
since there would be no different x to distinguish them.
Am 16.06.2024 um 21:41 schrieb Moebius:
Concerning his claim "But never two (different) unit fractions at the
same coordinate.": it is so trivial that it's hard to state/express it
properly.
Unit fractions are real numbers. And "two" unit fractions (and hence
real numbers) a, b are either identical (a = b) or not (a =/= b).
That's all.
Well, my last shot: If a and b are two _different_ unit fractions,
then there's no real number c such that a = c = b (though -of course-
a and b are real numbers).
To put it in simple words: /different/ means /not the same/.
On 5/2/2024 10:06 AM, Moebius wrote:
[...]
Fwiw,
here how I generally write my recursive functions.
Say the natural numbers, including zero here:
r[0] = 0
r[n + 1] = r[n] + 1
Lets expand that for a couple of iterations:
r[0] = 0
r[1] = r[0] + 1 = 1
r[2] = r[1] + 1 = 2
r[3] = r[2] + 1 = 3
...
[n] where n is an index
vs:
r_0 = 0
r_(n+1) = r_n + 1
Both seem okay to me, as in
I can understand both of them.
Am 15.06.2024 um 03:06 schrieb Jim Burns:
I see two options.
1.
Treat the poster approximately the same as
the guy on the corner telling everyone passing by
that his cat is God. Distance. Distance is good.
Agree.
After all, *my* cat is God
(and there is only ONE God, as is well known).
I speculate that WM doesn't do declarative programming, at all.
| For all real numbers x > 0: there are infinitely many unit fractions u
such that u < x [true]
| There are infinitely many unit fractions u such that for all real
numbers x > 0: u < x [false]
Unit fractions are real numbers.
-- Larry Niven, "The Subject Is Closed"
Am 17.06.2024 um 20:31 schrieb Jim Burns:
-- Larry Niven, "The Subject Is Closed"
A good read?
On 6/17/2024 6:00 PM, Moebius wrote:
Am 17.06.2024 um 20:31 schrieb Jim Burns:
-- Larry Niven, "The Subject Is Closed"
A good read?
Golden Age of Science Fiction, seriously.
Am 16.06.2024 um 21:51 schrieb Moebius:
Am 16.06.2024 um 21:41 schrieb Moebius:
Concerning his claim
"But never two (different) unit fractions at
the same coordinate.":
it is so trivial that
it's hard to state/express it properly.
Unit fractions are real numbers.
And "two" unit fractions
(and hence real numbers) a, b are
either identical (a = b) or not (a =/= b).
That's all.
Well, my last shot:
If a and b are two _different_ unit fractions,
then there's no real number c
such that a = c = b
(though -of course- a and b are real numbers).
To put it in simple words:
/different/ means /not the same/.
So _different_ unit fractions
CAN'T BE "at the same coordinate"
(i.e. equal to the same real number).
Le 14/06/2024 à 20:52, Jim Burns a écrit :
every number in {i:∀₃Xᴬ⤾⁺¹₀:X∋i}
has ℵ₀ successors in {i:∀₃Xᴬ⤾⁺¹₀:X∋i}
Note that
only the numbers with successors are deleted,
the successors remain by definition.
sⱼ∘f:{0<} → {j⁺¹<} is 1.to.1
If only numbers having ℵo successors are removed
and only as long as ℵo successors remain,
then ℵo successors remain and
every definable number is removed.
You cannot defina a number that remains.
But ℵo successors remain
So _different_ unit fractions
CAN'T BE "at the same coordinate"
(i.e. equal to the same real number).
Yes.
However, WM gives 'fraction' and 'rational'
different meanings.
1/2 2/4 3/6 4/8 ...
are all different _fractions_ corresponding to
the same _rational_ at the same point.
_As rationals_
whatever we can say about 1/2
we can say about 2/4
Conventionally. By definition.
_As fractions_
in Cantor's sequence,
the index of 1/2 is 2
the index of 2/4 is 12
We can say different things.
If all numbers having ℵ₀ followers are removed
then no numbers remain.
There is no first number remaining.
There is no number remaining.
We can _define_ a number that remains;
however,
the _existence_ of the number defined
leads to self.contradiction
Am 17.06.2024 um 19:46 schrieb Jim Burns:
I speculate that WM doesn't do declarative programming, at all.
As far as I can tell, WM never has done ANY programming.
Le 18/06/2024 à 19:12, Jim Burns a écrit :
If all numbers having ℵ₀ followers are removed
then no numbers remain.
Note that the ℵ₀ followers are not removed.
There is no first number remaining.
Correct.
There is no number remaining.
Wrong.
We can _define_ a number that remains;
No.
however,
the _existence_ of the number defined
leads to self.contradiction
Therefor you cannot define a number that remains.
Le 17/06/2024 à 23:53, Moebius a écrit :
Am 17.06.2024 um 19:46 schrieb Jim Burns:
I speculate that
WM doesn't do declarative programming,
at all.
As far as I can tell,
WM never has done ANY programming.
Fortran and TI59.
Programs run for several days.
On 6/18/2024 4:16 PM, WM wrote:
Le 17/06/2024 à 23:53, Moebius a écrit :
Am 17.06.2024 um 19:46 schrieb Jim Burns:
I speculate that
WM doesn't do declarative programming,
at all.
As far as I can tell,
WM never has done ANY programming.
Fortran and TI59.
Programs run for several days.
Oh, you poor, poor thing.
https://www.youtube.com/watch?v=yfJXd0rSCqo
My (JB's) first was ALGOL which is lovely.
My second was COBOL which I shudder to remember.
a = 5 elephants
a = b
b must equal 5 elephants. However, if we add in something, say:
a = 5 elephants
b = 5 cats
a = b?
Well, 5 = 5, but what they define is different...
Is that what you are getting at?
On 6/18/2024 3:45 PM, Moebius wrote:
Am 19.06.2024 um 00:21 schrieb Chris M. Thomasson:[...]
Your "approach" will work "out of the box" in physics:
:^)
Then I think of where:
a = 1/2 + 1/2 = 1
b = 1/4 + 1/4 + 1/2 = 1
a = b = 1
Then, think of b being the complex number:
b = 1+0i
a = b = 1
but a is unit fractions and b is a complex number.
They are equal as in able to be plotted in the same place on the real
line [in the coplex plane], but use different means to get to the same number?
Fair enough?
I will courageously assert: it depends.
On 6/17/2024 5:04 PM, Jim Burns wrote:
On 6/16/2024 6:52 PM, Moebius wrote:
Am 16.06.2024 um 21:51 schrieb Moebius:
To put it in simple words:
/different/ means /not the same/.
We ask
what can we say (or not.say) about a and b
if they are (or are.not) the same?
a = 5 elephants
a = b
b must equal 5 elephants.
However, if we add in something, say:
a = 5 elephants
b = 5 cats
a = b?
Well, 5 = 5, but what they define is different...
Is that what you are getting at?
a = 5 elephants
b = 5 cats
a = b?
Each unit fraction is a real number, but not all real numbers are unit fractions... Fair enough? ;^)
If the set of numbers.remaining
does not hold a first element,
then the set of numbers.remaining
is the empty set.
WM wrote on 6/19/2024 :
Le 18/06/2024 à 23:06, Jim Burns a écrit :
If the set of numbers.remaining
does not hold a first element,
then the set of numbers.remaining
is the empty set.
That is your big mistake! Start to count, continue, continue, continue, .. . >> What you can determine that you can count. The set of not counted numbers
remains infinite. But you cannot determine a first element.
Of course you can.
Le 18/06/2024 à 23:06, Jim Burns a écrit :
For subsets of the minimal inductive meta.set
{i:∀₃Xᴬ⤾⁺¹₀:X∋i}
only the empty set does not hold a first element.
If the set of numbers.remaining
does not hold a first element,
then the set of numbers.remaining
is the empty set.
That is your big mistake!
For subsets of the minimal inductive meta.set
{i:∀₃Xᴬ⤾⁺¹₀:X∋i}
only the empty set does not hold a first element.
If the set of numbers.remaining [in ⋂{Xᴬ⤾⁺¹₀}]
does not hold a first element [in ⋂{Xᴬ⤾⁺¹₀}],
then the set of numbers.remaining [in ⋂{Xᴬ⤾⁺¹₀}]
is the empty set.
That is your big mistake!
Start to count, continue, continue, continue,.. .
What you can determine that you can count.
The set of not counted numbers remains infinite.
But you cannot determine a first element.
All your following waffle is worthless,
because it violates this fundamental truth.
Simply try it instead of "proving"
counterfactual nonense.
On 6/19/2024 2:37 PM, WM wrote:
If the set of numbers.remaining [in ⋂{Xᴬ⤾⁺¹₀}]
does not hold a first element [in ⋂{Xᴬ⤾⁺¹₀}],
then the set of numbers.remaining [in ⋂{Xᴬ⤾⁺¹₀}]
is the empty set.
That is your big mistake!
For subsets of the union ⋃{⟨0…n⟩} of FISONs
only the empty set does not hold a first element.
[1]
Le 21/06/2024 à 02:18, Jim Burns a écrit :
On 6/19/2024 2:37 PM, WM wrote:
Le 18/06/2024 à 23:06, Jim Burns a écrit :
If the set of numbers.remaining [in ⋂{Xᴬ⤾⁺¹₀}]
does not hold a first element [in ⋂{Xᴬ⤾⁺¹₀}],
then the set of numbers.remaining [in ⋂{Xᴬ⤾⁺¹₀}]
is the empty set.
That is your big mistake!
For subsets of the union ⋃{⟨0…n⟩} of FISONs
only the empty set does not hold a first element.
[1]
That is true.
Since from every FISON F(n) = {1,2,3,...,n}
we know by definition that
it is neither sufficient nor necessary
to make the union of FISONs ℕ,
we can remove it from the union and
find
∪{F(1),F(2),F(3),...} = ℕ ==> ∪{ } = ℕ .
Among the FISONs of ℕ there is not,
in any enumeration,
a first one that is required to yield
the union ℕ.
Usually this is apologized by the fact,
that even in
{1, 2} ∪ {2, 3} ∪ {3, 1} = {1, 2, 3}
(*)
it is impossible to find a first set which
cannot be omitted from the union
to yield {1, 2, 3}.
But this argument fails.
It is not a set of sets which is subject to
Cantor's theorem B
(every embodiment of different numbers of
the first and the second number class
has a smallest number)
but only every set of ordinal numbers.
Therefore
we always have to enumerate the sets.
In case of FISONs this is simple.
We apply the natural order:
{1,2,3,...,n} --> n.
Of course
every other enumeration would also do.
In case of the sets (*)
we can use the written order from left to right.
Then the first set not to be omitted is {2, 3}
because after having omitted {1, 2} already,
2 would then be missing in the union.
Every other order is possible and
has a first set which cannot be omitted.
Le 21/06/2024 à 02:18, Jim Burns a écrit :
On 6/19/2024 2:37 PM, WM wrote:
Le 18/06/2024 à 23:06, Jim Burns a écrit :
If the set of numbers.remaining [in ⋂{Xᴬ⤾⁺¹₀}]
does not hold a first element [in ⋂{Xᴬ⤾⁺¹₀}],
then the set of numbers.remaining [in ⋂{Xᴬ⤾⁺¹₀}]
is the empty set.
That is your big mistake!
For subsets of the union ⋃{⟨0…n⟩} of FISONs
only the empty set does not hold a first element.
[1]
That is true.
Since from every FISON F(n) = {1,2,3,...,n}
we know by definition that
it is neither sufficient nor necessary
to make the union of FISONs ℕ,
we can remove it from the union and
find
∪{F(1),F(2),F(3),...} = ℕ ==> ∪{ } = ℕ .
Among the FISONs of ℕ there is not,
in any enumeration,
a first one that is required to yield
the union ℕ.
Usually
On 06/20/2024 05:18 PM, Jim Burns wrote:
We can learn this in our 13.7×10⁹.year.old universe.
You know about 30 years about
it was only 13.4, billion years old.
Imagine the simplest sort of argument that
deduction immediately solves, yet induction
never does.
"Not ultimately untrue."
If you reject that, then remember that
the usual Axiom of Infinity isn't just
naming the second constant of ZF after
the empty set: it's also a restriction
of comprehension because naive comprehension
makes it extra-ordinary.
However,
if all FISONs are removed,
no remaining FISON is a superset.
The convention is that ⋃{} = {}
But, either way, ⋃{} ≠ ℕ
Among the FISONs of ℕ there is not,
in any enumeration,
a first one that is required to yield
the union ℕ.
For each FISON, another FISON is proper superset. ∀ᶠⁱˢᵒⁿ⟨0…j⟩ ∃ᶠⁱˢᵒⁿ⟨0…k⟩≠⟨0…j⟩: ⟨0…j⟩ ⊂ ⟨0…k⟩
Therefore,
there is NO FISON superset all FISONs.
On 6/21/2024 2:59 AM, WM wrote:
∪{F(1),F(2),F(3),...} = ℕ ==> ∪{ } = ℕ .
⋃{} requires nuance.
Vacuously, everything is in each set in {}
Which makes ⋃{} the universal class.
Which is weird.
The convention is that ⋃{} = {}
But, either way, ⋃{} ≠ ℕ
All of which is irrelevant to ⋃{}
which is always {}
Le 21/06/2024 à 21:13, Jim Burns a écrit :
However,
if all FISONs are removed,
no remaining FISON is a superset.
The set of necerssary FISONs,
if existing [nonempty], must have a smallest element
- according to Cantor.
But it has not.
Therefore there is no necessary FISON.
We cannot find a necessary FISON because
each one covers only a tiny subset of ℕ.
ℵo Elements are missing.
That is not expalint by quantifier magic
but by mathematical facts:
Dark numbers.
Therefore,
there is NO FISON superset all FISONs.
But every FISON is a very, very proper subset:
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
That statement covers all FISONs.
Le 22/06/2024 à 13:58, Jim Burns a écrit :
[...]
The set of FISONs has a smallest element, {1}.
The set of FISONs covering the first 100 natnumbers
has a smallest element, namely {1, 2 3, ..., 100}.
The set of FISONs
covering the natural numbers has no first element.
On the contrary we can prove that every FISON fails.
Only
a very fanatic matheologian like Franz Fritsche
can claim, that
the union of FISONs is infinitely larger than
every FISON [each FISON].
According to Cantor's theorem B
a set of necessary FISONs to be unioned does not exist.
Therefore we have: UF(n) = ℕ ⟹ U{ } = ℕ.
On 6/22/2024 8:11 AM, WM wrote:
if existing [nonempty], must have a smallest element
- according to Cantor.
But it has not.
Therefore there is no necessary FISON.
Yes.
No FISON is necessary ⇔
Each FISON is unnecessary ⇐
For each FISON, a proper.superset.FISON exists.
We cannot find a necessary FISON because
each one covers only a tiny subset of ℕ.
Yes.
ℵo Elements are missing.
No.
Each FISON is in {FISON}
FISON.union U{FISON} and
minimal.inductive ⋂{inductive} are
both inductive and
both well.ordered.
Infinity is not finite.
But every FISON is a very, very proper subset:
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
That statement covers all FISONs.
U{FISON} = ⋂{inductive}
∀n ∈ U{FISON}: |U{FISON}\{1,2,3,...,n}| = ℵ₀
On 6/22/2024 12:31 PM, WM wrote:
Your (WM's) "argument" is that it is fanatic matheology
to claim that ⋃{FISON} is an IISON
an Infinite.Initial.Segment.Of.Naturals.
According to Cantor's theorem B
a set of necessary FISONs to be unioned does not exist.
According to proposal 3 == "∅, x∪{y}, {y:P(x)} exist"
a set of FISONS.to.be.unioned exists
Removing a single FISON not.changes the union.
Removing all FISONs is not removing a single FISON
and changes the union.
Therefore we have: UF(n) = ℕ ⟹ U{ } = ℕ.
Do you (WM) have a math.like reason?
On 6/22/2024 5:11 AM, WM wrote:
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
That statement covers all FISONs.
An infinite set of FISON's? Sorry for my notation here, but it should
get the point across:
[ { 1 }, { 1, 2 }, { 1, 2, 3 }, { 1, 2, 3, 4}, ... ]
?
Le 22/06/2024 à 19:34, Jim Burns a écrit :
On 6/22/2024 8:11 AM, WM wrote:
But every FISON is a very, very proper subset:
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
That statement covers all FISONs.
U{FISON} = ⋂{inductive}
∀n ∈ U{FISON}: |U{FISON}\{1,2,3,...,n}| = ℵ₀
Each FISON is Union of all predecessors.
U{FISON} is a FISON.
No FISON has ℵ₀ elements.
No union of FISONs has ℵ₀ elements.
On 6/22/2024 4:12 PM, WM wrote:
Yes,
U{FISON} is the union of all its predecessors.
No,
U{FISON} is not a FISON.
U{FISON} is the union of all its predecessors.
U{FISON} is not a FISON. It is not finite.
No FISON has ℵ₀ elements.
U{FISON} is not a FISON. It is not finite.
Le 22/06/2024 à 21:09, "Chris M. Thomasson" a écrit :
On 6/22/2024 5:11 AM, WM wrote:
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
That statement covers all FISONs.
An infinite set of FISON's?
Sorry for my notation here, but
it should get the point across:
[ { 1 }, { 1, 2 }, { 1, 2, 3 }, { 1, 2, 3, 4}, ... ]
?
Every existing [nonempty] set of ordinals has
a first or smallest element.
The set of FISONs necessary to yield ℕ would be
a set of ordinals.
It is well defined since
for every FISON
we can determine whether it is necessary.
It has no smallest element. ==>
It is empty.
IF ℕ is the union of FISONs,
THEN ℕ is empty.
ℕ is not empty.
On 6/22/2024 4:23 PM, WM wrote:
IF ℕ is the union of FISONs,
THEN ℕ is empty.
ℕ is not empty.
IF ℕ is the union of necessary FISONs,
THEN ℕ is empty.
ℕ is not empty.
ℕ is not the union of necessary FISONs,
⎛ No single FISON is necessary.
⎜ Even! No _finite_ FISON.subset is necessary.
⎝ The whole infinite FISON.set is necessary.
On 6/22/2024 1:23 PM, WM wrote:
Every existing set of ordinals has a first or smallest element. The setA infinite sequence of FISONS:
of FISONs necessary to yield ℕ would be a set of ordinals. It is well
defined since for every FISON we can determine whether it is necessary.
It has no smallest element. ==> It is empty.
IF ℕ is the union of FISONs, THEN ℕ is empty. ℕ is not empty.
WM laid this down on his screen :
Le 22/06/2024 à 19:34, Jim Burns a écrit :
On 6/22/2024 8:11 AM, WM wrote:
if existing [nonempty], must have a smallest element - according
to Cantor.
But it has not.
Therefore there is no necessary FISON.
Yes.
No FISON is necessary ⇔
Each FISON is unnecessary ⇐
For each FISON, a proper.superset.FISON exists.
The set has no first element. ==> The set is empty.
If ℕ is union of FISONs, then ℕ is empty.
The set of integers is empty? Say it ain't so!
Le 22/06/2024 à 23:04, Jim Burns a écrit :
U{FISON} is the union of all its predecessors.
U{FISON} is not a FISON. It is not finite.
The union cannot be larger than all its FISONs.
Yes,
U{FISON} is the union of all its predecessors.
No,
U{FISON} is not a FISON.
How can adding a FISON
change the character of the union?
Every existing [nonempty] set of ordinals has
a first or smallest element.
The set of FISONs necessary to yield ℕ would be
[an empty] set of ordinals.
It is well defined since
for every FISON we can determine
whether it is necessary.
It has no smallest element.
It is empty.
On 6/22/2024 5:30 PM, WM wrote:
The union cannot be larger than all its FISONs.
The union cannot be smaller than any FISON.
Each FISON is smaller than another FISON,
Yes,
the set of FISONs necessary to yield ℕ is empty.
U{FISON} is the union of the set of
both necessary and unnecessary FISONs
WM laid this down on his screen :
The set has no first element. ==> The set is empty.
If ℕ is union of FISONs, then ℕ is empty.
The set of integers is empty? Say it ain't so!
Le 23/06/2024 à 19:58, Jim Burns a écrit :
U{FISON} is the union of the set of
both necessary and unnecessary FISONs
The unnecessary FISONs can be removed.
The union cannot be larger than all its FISONs.
The union cannot be smaller than any FISON.
Agreed.
Each FISON is smaller than another FISON,
Therefore the union is a FISON.
The sequence of FISONs is potentially infinite.
There is no last one.
The end is evolving.
It runs through ℕ
but never covers a substantial part.
Yes,
the set of FISONs necessary to yield ℕ is empty.
That proves that
ℕ is more than all FISONs.
According to Cantor's theorem B
the set of necesary FISONs must have
a smallest element.
This proves that ℕ is not the union of FISONs.
On 6/23/2024 4:04 PM, WM wrote:
Le 23/06/2024 à 19:58, Jim Burns a écrit :
U{FISON} is the union of the set of
both necessary and unnecessary FISONs
The unnecessary FISONs can be removed.
Your definition of 'unnecessary' is that
an unnecessaryᵂᴹ FISON when removed
leaves the union of remaining FISONs unchanged.
One unnecessaryᵂᴹ FISON can be removed.
One.
{FISON} is inductive and well.ordered.
Because well.ordered,
Any necessaryᵂᴹ FISON in {FISON}
is last in {FISON}
Because inductive,
No FISON is last in {FISON}
Any _one_ FISON can be removed and
leave the union of remaining FISONs unchanged.
Any _one_ FISON is unnecessaryᵂᴹ
Removing all of those which
singly don't change the union
changes the union, ⋃{}≠ℕ
It is not a contradiction.
Compare to:
Suppose I have 5 elephants, and
an unnecessaryᴶᴮ elephant is one when removed
leaves 4 elephants.
Any _one_ elephant can be removed and
leave 4 elephants.
Any _one_ elephant is unnecessaryᴶᴮ
Removing all which
singly leave 4
doesn't leave 4
It is not a contradiction.
It is arithmetic.
The union cannot be larger than all its FISONs.
The union cannot be smaller than any FISON.
Agreed.
Each FISON is smaller than another FISON,
Therefore the union is a FISON.
Therefore the union is NOT a FISON.
It runs through ℕ
but never covers a substantial part.
Related:
∀ᴿx>0: NUF(x) = ℵ₀
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