On 08.11.2024 14:09, Mikko wrote:
On 2024-11-07 13:21:42 +0000, WM said:
On 07.11.2024 10:22, Mikko wrote:
On 2024-11-06 17:55:15 +0000, WM said:
On 06.11.2024 16:04, Mikko wrote:
On 2024-11-06 10:01:21 +0000, WM said:
I leave ε = 1. No shrinking. Every point outside of the intervals is >>>>>>> nearer to an endpoint than to the contents.
This discussion started with message that clearly discussed limits when >>>>>> ε approaches 0. The case ε = 1 was only about a specific unimportant >>>>>> question.
When ε approaches 0 then the measure of the real axis is, according to >>>>> Cantor's results, 0. That shows that his results are wrong.
It is not the measure of the real axis but the set of rationals. The
real axis more than just the rationals. The irrationals are also a
part of the real axis.
But not between irrational points.
Real axis contains both real and irrational numbers and nothing else.
Between any two points of the real axis there are both rational and
irrational points.
If Cantors enumeration of the rationals is complete, then all rationals
are in the sequence 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1,
1/5, 2/4, 3/3, 4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ... and none is outside.
Therefore also irrational numbers cannot be there.
Of course this is wrong.
It proves that not all rational numbers are countable and in the sequence.
On 2024-11-08 16:30:23 +0000, WM said:
If Cantors enumeration of the rationals is complete, then all rationals
are in the sequence 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1,
1/5, 2/4, 3/3, 4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ... and none
is outside.
All positive rationals quite obviously are in the sequence. Non-positive rationals are not.
Therefore also irrational numbers cannot be there.
That is equally obvious.
Of course this is wrong.
You may call it wrong but that's the way they are.
It proves that not all rational numbers are countable and in the
sequence.
Calling a truth wrong does not prove anything.
On 2024-11-09 21:30:47 +0000, WM said:
On 09.11.2024 15:03, Mikko wrote:
On 2024-11-08 16:30:23 +0000, WM said:
If Cantors enumeration of the rationals is complete, then all rationals >>>> are in the sequence 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2,
4/1, 1/5, 2/4, 3/3, 4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ... and
none is outside.
All positive rationals quite obviously are in the sequence. Non-positive >>> rationals are not.
Therefore also irrational numbers cannot be there.
That is equally obvious.
Of course this is wrong.
You may call it wrong but that's the way they are.
The measure of all intervals J(n) = [n - √2/10, n + √2/10] is smaller
than 3.
Maybe, maybe not, depending on what is all n.
If all n is all reals then
the measure of their union is infinite.
On 2024-11-09 21:30:47 +0000, WM said:rationals
On 09.11.2024 15:03, Mikko wrote:
On 2024-11-08 16:30:23 +0000, WM said:
If Cantors enumeration of the rationals is complete, then all
4/1, 1/5, 2/4, 3/3, 4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ... andare in the sequence 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2,
Non-positive
All positive rationals quite obviously are in the sequence.
smaller than 3.rationals are not.
Therefore also irrational numbers cannot be there.
That is equally obvious.
Of course this is wrong.
You may call it wrong but that's the way they are.
The measure of all intervals J(n) = [n - √2/10, n + √2/10] is
Maybe, maybe not, depending on what is all n.
If all n is all reals then
the measure of their union is infinite.
On 09.11.2024 15:03, Mikko wrote:
On 2024-11-08 16:30:23 +0000, WM said:
If Cantors enumeration of the rationals is complete, then all rationals
are in the sequence 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1,
1/5, 2/4, 3/3, 4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ... and none is >>> outside.
All positive rationals quite obviously are in the sequence. Non-positive
rationals are not.
Therefore also irrational numbers cannot be there.
That is equally obvious.
Of course this is wrong.
You may call it wrong but that's the way they are.
The measure of all intervals J(n) = [n - √2/10, n + √2/10] is smaller than 3.
On 10.11.2024 11:20, Mikko wrote:
On 2024-11-09 21:30:47 +0000, WM said:
On 09.11.2024 15:03, Mikko wrote:
On 2024-11-08 16:30:23 +0000, WM said:
If Cantors enumeration of the rationals is complete, then all rationals >>>>> are in the sequence 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, >>>>> 1/5, 2/4, 3/3, 4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ... and none is >>>>> outside.
All positive rationals quite obviously are in the sequence. Non-positive >>>> rationals are not.
Therefore also irrational numbers cannot be there.
That is equally obvious.
Of course this is wrong.
You may call it wrong but that's the way they are.
The measure of all intervals J(n) = [n - √2/10, n + √2/10] is smaller than 3.
Maybe, maybe not, depending on what is all n.
It is, as usual, all natural numbers.
On 10.11.2024 11:20, Mikko wrote:
On 2024-11-09 21:30:47 +0000, WM said:
4/1, 1/5, 2/4, 3/3, 4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ... andOn 09.11.2024 15:03, Mikko wrote:
On 2024-11-08 16:30:23 +0000, WM said:
If Cantors enumeration of the rationals is complete, then all rationals
are in the sequence 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2,
none is outside.
smaller than 3.
All positive rationals quite obviously are in the sequence. Non-positive
rationals are not.
Therefore also irrational numbers cannot be there.
That is equally obvious.
Of course this is wrong.
You may call it wrong but that's the way they are.
The measure of all intervals J(n) = [n - √2/10, n + √2/10] is
Maybe, maybe not, depending on what is all n.
It is, as usual, all natural numbers.
If all n is all reals then
the measure of their union is infinite.
But n is all naturals as you could have found out yourself, by the measure < 3.
On 2024-11-10 10:54:02 +0000, WM said:
The measure of the set of all those intervals is infinite.
Between the intervals J(n) and (Jn+1) there are infinitely many rational
and irrational numbers but no hatural numbers.
On 2024-11-10 10:54:02 +0000, WM said:
The measure of all intervals J(n) = [n - √2/10, n + √2/10] is
smaller than 3.
Maybe, maybe not, depending on what is all n.
It is, as usual, all natural numbers.
The measure of the interval J(n) is √2/5, which is roghly 0,28.
The measure of the set of all those intervals is infinite.
Between the intervals J(n) and (Jn+1) there are infinitely many rational
and irrational numbers but no hatural numbers.
On 11/05/2024 02:29 AM, Mikko wrote:
Geometry is only another language for the same thing.
Another language is an unnecessary complication that only reeasls
an intent to deceive.
No, the meaning is clear. Of course, because some intevals overlap,
you should have specified what exacly you mean by "nearer". But as
ε shriks the overlappings disappear and the distance between any
two intevals approaches the distance between their centers we may
define distance between the intervals as the distance between their
endpoints even wne ε > 0.
On 06.11.2024 03:48, Ross Finlayson wrote:
On 11/05/2024 02:29 AM, Mikko wrote:
Geometry is only another language for the same thing.
Another language is an unnecessary complication that only reeasls
an intent to deceive.
It is a clearer language.
No, the meaning is clear. Of course, because some intevals overlap,
you should have specified what exacly you mean by "nearer". But as
ε shriks the overlappings disappear and the distance between any
two intevals approaches the distance between their centers we may
define distance between the intervals as the distance between their
endpoints even wne ε > 0.
I leave ε = 1. No shrinking. Every point outside of the intervals is
nearer to an endpoint than to the contents.
On 2024-11-06 10:01:21 +0000, WM said:
I leave ε = 1. No shrinking. Every point outside of the intervals is
nearer to an endpoint than to the contents.
This discussion started with message that clearly discussed limits when
ε approaches 0. The case ε = 1 was only about a specific unimportant question.
On 06.11.2024 16:04, Mikko wrote:
On 2024-11-06 10:01:21 +0000, WM said:
I leave ε = 1. No shrinking. Every point outside of the intervals is
nearer to an endpoint than to the contents.
This discussion started with message that clearly discussed limits when
ε approaches 0. The case ε = 1 was only about a specific unimportant
question.
When ε approaches 0 then the measure of the real axis is, according to Cantor's results, 0. That shows that his results are wrong.
But the important question is also covered by ε = 1. The measure of the
real axis is, according to Cantor's results, less than 3. That shows
that his results are wrong.
On 2024-11-06 17:55:15 +0000, WM said:
On 06.11.2024 16:04, Mikko wrote:
On 2024-11-06 10:01:21 +0000, WM said:
I leave ε = 1. No shrinking. Every point outside of the intervals is
nearer to an endpoint than to the contents.
This discussion started with message that clearly discussed limits when
ε approaches 0. The case ε = 1 was only about a specific unimportant
question.
When ε approaches 0 then the measure of the real axis is, according to
Cantor's results, 0. That shows that his results are wrong.
It is not the measure of the real axis but the set of rationals. The
real axis more than just the rationals. The irrationals are also a
part of the real axis.
But the important question is also covered by ε = 1. The measure of
the real axis is, according to Cantor's results, less than 3. That
shows that his results are wrong.
No, that is not Cantor's result,
On 07.11.2024 10:22, Mikko wrote:
On 2024-11-06 17:55:15 +0000, WM said:
On 06.11.2024 16:04, Mikko wrote:
On 2024-11-06 10:01:21 +0000, WM said:
I leave ε = 1. No shrinking. Every point outside of the intervals is >>>>> nearer to an endpoint than to the contents.
This discussion started with message that clearly discussed limits when >>>> ε approaches 0. The case ε = 1 was only about a specific unimportant >>>> question.
When ε approaches 0 then the measure of the real axis is, according to
Cantor's results, 0. That shows that his results are wrong.
It is not the measure of the real axis but the set of rationals. The
real axis more than just the rationals. The irrationals are also a
part of the real axis.
But not between irrational points.
But the important question is also covered by ε = 1. The measure of the >>> real axis is, according to Cantor's results, less than 3. That shows
that his results are wrong.
No, that is not Cantor's result,
It is Cantor's result that all rationals are countable, hence inside my intervals.
But we can use the following estimation that should convince everyone:
Use the intervals I(n) = [n - sqrt(2)/2^n, n + sqrt(2)/2^n]. Since n
and q_n can be in bijection, these intervals are sufficient to cover
all q_n. That means by clever reordering them you can cover the whole positive axis except "boundaries".
And an even more suggestive approximation:
Replace the I(n) by intervals J(n) = [n - 1/10, n + 1/10].
On 2024-11-07 13:21:42 +0000, WM said:
On 07.11.2024 10:22, Mikko wrote:
On 2024-11-06 17:55:15 +0000, WM said:
On 06.11.2024 16:04, Mikko wrote:
On 2024-11-06 10:01:21 +0000, WM said:
I leave ε = 1. No shrinking. Every point outside of the intervals >>>>>> is nearer to an endpoint than to the contents.
This discussion started with message that clearly discussed limits
when
ε approaches 0. The case ε = 1 was only about a specific unimportant >>>>> question.
When ε approaches 0 then the measure of the real axis is, according
to Cantor's results, 0. That shows that his results are wrong.
It is not the measure of the real axis but the set of rationals. The
real axis more than just the rationals. The irrationals are also a
part of the real axis.
But not between irrational points.
Real axis contains both real and irrational numbers and nothing else.
Between any two points of the real axis there are both rational and irrational points.
It is Cantor's result that all rationals are countable, hence inside
my intervals.
That is but what you said above is not.
But we can use the following estimation that should convince everyone:
Use the intervals I(n) = [n - sqrt(2)/2^n, n + sqrt(2)/2^n]. Since n
and q_n can be in bijection, these intervals are sufficient to cover
all q_n. That means by clever reordering them you can cover the whole
positive axis except "boundaries".
Depends on the type of n.
And an even more suggestive approximation:
Replace the I(n) by intervals J(n) = [n - 1/10, n + 1/10].
Likewise.
J(n) = [n - 1/10, n + 1/10]
--------_1_--------_2_--------_3_--------_4_--------_5_--------_...
On 2024-11-11 11:33:52 +0000, WM said:
Between the intervals J(n) and (Jn+1) there are infinitely many rational >>> and irrational numbers but no hatural numbers.
Therefore infinitely many natural numbers must become centres of
intervals, if Cantor was right. But that is impossible.
Where did Cantor say otherwise?
On 11.11.2024 12:15, Mikko wrote:
On 2024-11-10 10:54:02 +0000, WM said:
The measure of all intervals J(n) = [n - √2/10, n + √2/10] is smaller than 3.
Maybe, maybe not, depending on what is all n.
It is, as usual, all natural numbers.
The measure of the interval J(n) is √2/5, which is roghly 0,28.
Agreed, I said smaller than 3.
The measure of the set of all those intervals is infinite.
The density or relative measure is √2/5. By shifting intervals this
density cannot grow. Therefore the intervals cannot cover the real
axis, let alone infinitely often.
Between the intervals J(n) and (Jn+1) there are infinitely many rational
and irrational numbers but no hatural numbers.
Therefore infinitely many natural numbers must become centres of
intervals, if Cantor was right. But that is impossible.
On 12.11.2024 14:45, Mikko wrote:
On 2024-11-11 11:33:52 +0000, WM said:
Between the intervals J(n) and (Jn+1) there are infinitely many rational >>>> and irrational numbers but no hatural numbers.
Therefore infinitely many natural numbers must become centres of
intervals, if Cantor was right. But that is impossible.
Where did Cantor say otherwise?
Cantor said that all rationals are within the sequence and hence within
all intervals. I prove that rationals are in the complement.
On 2024-11-12 13:59:24 +0000, WM said:
Cantor said that all rationals are within the sequence and hence
within all intervals. I prove that rationals are in the complement.
He said that about his sequence and his intervals. Infinitely many of them are in intervals that do not overlap with any of your J(n).
On 13.11.2024 11:39, Mikko wrote:
On 2024-11-12 13:59:24 +0000, WM said:
Cantor said that all rationals are within the sequence and hence within
all intervals. I prove that rationals are in the complement.
He said that about his sequence and his intervals. Infinitely many of them >> are in intervals that do not overlap with any of your J(n).
The intervals J(n) = [n - 1/10, n + 1/10] cover the relative measure
1/5 of ℝ+. By translating them to match Cantor's intervals they cover
ℝ+ infinitely often. This is impossible. Therefore set theorists must discard geometry.
On 2024-11-13 16:14:02 +0000, WM said:
On 13.11.2024 11:39, Mikko wrote:
On 2024-11-12 13:59:24 +0000, WM said:
Cantor said that all rationals are within the sequence and hence
within all intervals. I prove that rationals are in the complement.
He said that about his sequence and his intervals. Infinitely many of
them
are in intervals that do not overlap with any of your J(n).
The intervals J(n) = [n - 1/10, n + 1/10] cover the relative measure
1/5 of ℝ+. By translating them to match Cantor's intervals they cover
ℝ+ infinitely often. This is impossible. Therefore set theorists must
discard geometry.
The intervals J(n) are what they are. Translated intervals are not the same intervals. The properties of the translated set depend on how you translate.
For example, if you translate them to J'(n) = (n/100 - 1/10, n/100 + 1/10) then the translated intervals J'(n) wholly cover the postive side of the
real line.
On 14.11.2024 10:17, Mikko wrote:
On 2024-11-13 16:14:02 +0000, WM said:
On 13.11.2024 11:39, Mikko wrote:
On 2024-11-12 13:59:24 +0000, WM said:
Cantor said that all rationals are within the sequence and hence within >>>>> all intervals. I prove that rationals are in the complement.
He said that about his sequence and his intervals. Infinitely many of them >>>> are in intervals that do not overlap with any of your J(n).
The intervals J(n) = [n - 1/10, n + 1/10] cover the relative measure
1/5 of ℝ+. By translating them to match Cantor's intervals they cover
ℝ+ infinitely often. This is impossible. Therefore set theorists must
discard geometry.
The intervals J(n) are what they are. Translated intervals are not the same >> intervals. The properties of the translated set depend on how you translate.
No. Covering by intervals is completely independent of their
individuality and therefore of their order.
Therefore you can either believe in set theory or in geometry. Both contradict each other.
For example, if you translate them to J'(n) = (n/100 - 1/10, n/100 + 1/10) >> then the translated intervals J'(n) wholly cover the postive side of the
real line.
By shuffling the same set of intervals which do not cover ℝ+ in the
initial configuration, it is impossible to cover more. That's geometry.
On 2024-11-14 10:34:52 +0000, WM said:
No. Covering by intervals is completely independent of their
individuality and therefore of their order.
Translated intervals are not the same as the original ones. Not only their order but also their positions can be different as demonstrated by your example and mine, too.
Therefore you can either believe in set theory or in geometry. Both
contradict each other.
Geometry cannot contradict set theory because there is nothing both
could say. But this discussion is about set theory so geometry is not relevant.
By shuffling the same set of intervals which do not cover ℝ+ in the
initial configuration, it is impossible to cover more. That's geometry.
So what part of ℝ+ is not covered by my J'?
On 15.11.2024 11:43, Mikko wrote:
On 2024-11-14 10:34:52 +0000, WM said:
No. Covering by intervals is completely independent of their
individuality and therefore of their order.
Translated intervals are not the same as the original ones. Not only their >> order but also their positions can be different as demonstrated by your
example and mine, too.
If the do not cover the whole figure in their initial order, then they
cannot do so in any other order.
On 2024-11-15 12:00:43 +0000, WM said:
On 15.11.2024 11:43, Mikko wrote:
On 2024-11-14 10:34:52 +0000, WM said:
No. Covering by intervals is completely independent of their
individuality and therefore of their order.
Translated intervals are not the same as the original ones. Not only
their
order but also their positions can be different as demonstrated by your
example and mine, too.
If they do not cover the whole figure in their initial order, then they
cannot do so in any other order.
So you want to retract your claims that involve another order?
On 15.11.2024 11:43, Mikko wrote:
Translated intervals are not
the same as the original ones.
Not only their order
but also their positions can be different
as demonstrated by your example and mine, too.
If the do not cover the whole figure
in their initial order,
then they cannot do so in any other order.
Since according to Cantor's formula
the smaller parts of ℝ+ are frequently covered,
in the larger parts much gets uncovered.
Every definable rational is covered.
That is called potential infinity.
On 11/15/2024 7:00 AM, WM wrote:
On 15.11.2024 11:43, Mikko wrote:
Translated intervals are not
the same as the original ones.
Not only their order
but also their positions can be different
as demonstrated by your example and mine, too.
If they do not cover the whole figure
in their initial order,
then they cannot do so in any other order.
Sets for which that is true
are finite sets.
⎝ And our sets do not change.
On 16.11.2024 20:30, Jim Burns wrote:
On 11/15/2024 7:00 AM, WM wrote:
On 15.11.2024 11:43, Mikko wrote:
Translated intervals are not
the same as the original ones.
Not only their order
but also their positions can be different
as demonstrated by your example and mine, too.
If they do not cover the whole figure
in their initial order,
then they cannot do so in any other order.
Sets for which that is true
are finite sets.
Sets for which that is true are
sets which obey geometrical rules.
⎝ And our sets do not change.
Therefore
the set of intervals cannot grow.
On 11/16/2024 2:54 PM, WM wrote:
Therefore
the set of intervals cannot grow.
An infinite set can match a proper superset
without growing.
Because it is infinite.
On 16.11.2024 10:21, Mikko wrote:
On 2024-11-15 12:00:43 +0000, WM said:
On 15.11.2024 11:43, Mikko wrote:
On 2024-11-14 10:34:52 +0000, WM said:
No. Covering by intervals is completely independent of their
individuality and therefore of their order.
Translated intervals are not the same as the original ones. Not only their >>>> order but also their positions can be different as demonstrated by your >>>> example and mine, too.
If they do not cover the whole figure in their initial order, then they
cannot do so in any other order.
So you want to retract your claims that involve another order?
My claim is the obvious truth that the intervals [n - 1/10, n + 1/10]
in every order do not cover the positive real line, let alone
infinitely often.
Regards, WM
The intervals J(n) = [n - 1/10, n + 1/10] cover the relative measure
1/5 of ℝ+. By translating them to match Cantor's intervals they cover
ℝ+ infinitely often. This is impossible. Therefore set theorists must discard geometry.
On 2024-11-16 19:42:22 +0000, WM said:
On 16.11.2024 10:21, Mikko wrote:So you regard invalid what you said on 2024-11-13 16:14:02 +0000:
On 2024-11-15 12:00:43 +0000, WM said:
On 15.11.2024 11:43, Mikko wrote:
On 2024-11-14 10:34:52 +0000, WM said:
No. Covering by intervals is completely independent of their
individuality and therefore of their order.
Translated intervals are not the same as the original ones. Not
only their
order but also their positions can be different as demonstrated by
your
example and mine, too.
If they do not cover the whole figure in their initial order, then
they cannot do so in any other order.
So you want to retract your claims that involve another order?
My claim is the obvious truth that the intervals [n - 1/10, n + 1/10]
in every order do not cover the positive real line, let alone
infinitely often.
The intervals J(n) = [n - 1/10, n + 1/10] cover the relative measure
1/5 of ℝ+. By translating them to match Cantor's intervals they cover
ℝ+ infinitely often.
This is impossible.
Therefore set theorists must
discard geometry.
There you translate them so that not only their order but also their positions change. But You also rejected my J' intervals without giving
any reason that does not reject your "translated" intervals.
On 2024-11-17 10:29:31 +0000, WM said:
Your J'(n) = (n/100 - 1/10, n/100 + 1/10) are 100 times more than mine.
For every reordering of a finite subset of my intervals J(n) the
relative covering remains constant, namely 1/5.
The analytical limit proves that the constant sequence 1/5, 1/5, 1/5,
... has limit 1/5. This is the relative covering of the infinite set
and of every reordering.
My J'(n) are your J(n) translated much as your translated J(n) except
that they are not re-ordered.
My J'(n) are as numerous as your J(n): there is one of each for every
natural number n.
Each my J'(n) has the same size as your corresponding J(n): 1/5.
One more similarity is that neither is relevant to the subject.
Your J'(n) = (n/100 - 1/10, n/100 + 1/10) are 100 times more than mine.
For every reordering of a finite subset of my intervals J(n) the
relative covering remains constant, namely 1/5.
The analytical limit proves that the constant sequence 1/5, 1/5, 1/5,
... has limit 1/5. This is the relative covering of the infinite set
and of every reordering.
On 17.11.2024 13:28, Mikko wrote:
On 2024-11-17 10:29:31 +0000, WM said:
Your J'(n) = (n/100 - 1/10, n/100 + 1/10) are 100 times more than mine.
For every reordering of a finite subset of my intervals J(n) the
relative covering remains constant, namely 1/5.
The analytical limit proves that the constant sequence 1/5, 1/5, 1/5,
... has limit 1/5. This is the relative covering of the infinite set
and of every reordering.
My J'(n) are your J(n) translated much as your translated J(n) except
that they are not re-ordered.
My J'(n) are as numerous as your J(n): there is one of each for every
natural number n.
There are 100 intervals for each natural number.
This can be proven by bijecting J'(100n) and J(n). My intervals are
then exhausted, yours are not.
Each my J'(n) has the same size as your corresponding J(n): 1/5.
One more similarity is that neither is relevant to the subject.
Only if you believe in matheology and resist mathematics.
Geometry says that your intervals cover the real line, my do not.
On 2024-11-17 12:46:29 +0000, WM said:
There are 100 intervals for each natural number.
This can be proven by bijecting J'(100n) and J(n). My intervals are
then exhausted, yours are not.
Irrelevant.
In mathematics unproven claims do not count.
On 16.11.2024 23:36, Jim Burns wrote:
On 11/16/2024 2:54 PM, WM wrote:
Therefore
the set of intervals cannot grow.
An infinite set can match a proper superset
without growing.
But with shrinking.
When it matches
first itself and then a proper subset,
then it has decreased.
The set of even numbers has fewer elements
than the set of integers.
Because it is infinite.
The interval [0, 1] is infinite because
it can be split into infinitely many subsets.
But its measure remains constant.
There is no reason except naivety
to believe that the intervals [n - 1/10, n + 1/10]
could cover the real line infinitely often.
On 18.11.2024 10:58, Mikko wrote:
On 2024-11-17 12:46:29 +0000, WM said:
There are 100 intervals for each natural number.
This can be proven by bijecting J'(100n) and J(n). My intervals are
then exhausted, yours are not.
Irrelevant.
Very relevant.
In mathematics unproven claims do not count.
Geometry is only another language of mathematics.
On 2024-11-18 14:29:40 +0000, WM said:
On 18.11.2024 10:58, Mikko wrote:
On 2024-11-17 12:46:29 +0000, WM said:
There are 100 intervals for each natural number.
This can be proven by bijecting J'(100n) and J(n). My intervals are
then exhausted, yours are not.
Irrelevant.
Very relevant.
It is not relevant if no relevancy is shown.
In mathematics unproven claims do not count.
Geometry is only another language of mathematics.
Therefore unproven claims don't count in geometry.
The set of even numbers is
a proper subset of the set of integers,
AND
the set of even numbers can match
the set of integers without either set changing.
On 18.11.2024 20:22, Jim Burns wrote:
An infinite set
can match some proper supersets without growing and
can match some proper subsets without shrinking.
Sets which can't aren't infinite.
The set of even numbers is
a proper subset of the set of integers,
AND
the set of even numbers can match
the set of integers without either set changing.
That implies that
our well-known intervals
our well-known intervals can
cover the real line or
reduce the average covering to 1/1000000000.
But every finite translation of
any finite subset of intervals maintains
the relative covering 1/5.
If the infinite set has
the relative covering 1 (or more or less),
then you claim that
the sequence 1/5, 1/5, 1/5, ... has
limit 1 (or more or less).
So you deny analysis or / and geometry.
On 11/19/2024 6:01 AM, WM wrote:
That implies that
our well-known intervals
Sets with different intervals are different.
Our sets do not change.
Sets of our well.known.intervals
can match some proper supersets without growing
Relative covering isn't measure.
You haven't defined 'relative covering'.
Giving examples isn't a definition.
I claim that there are functions f:ℝ→ℝ
such that
⟨ f(⅟1) f(⅟2) f(⅟3) ... ⟩ =
⟨ ⅟5 ⅟5 ⅟5 ... ⟩
and f(0) = 1
So you deny analysis or / and geometry.
I deny what you think analysis and geometry are.
I accept infinite sets
and discontinuous functions
What is it you (WM) accuse infinite sets of,
other than not being finite?
Note:
An infinite set
can match some proper supersets without growing
On 19.11.2024 17:26, Jim Burns wrote:
On 11/19/2024 6:01 AM, WM wrote:
That implies that
our well-known intervals
Sets with different intervals are different.
Our sets do not change.
The intervals before and after shifting are not different. Only their positions are.
On 2024-11-20 11:42:15 +0000, WM said:
The intervals before and after shifting are not different. Only their
positions are.
The intervals are different. A shifted interval contains a different
set of numbers.
On 21.11.2024 10:16, Mikko wrote:
On 2024-11-20 11:42:15 +0000, WM said:
The intervals before and after shifting are not different. Only their
positions are.
The intervals are different. A shifted interval contains a different
set of numbers.
Consider this simplified argument. Let every unit interval after a
natural number n which is divisible by 10 be coloured black: (10n,
10n+1]. All others are white. Is it possible to shift the black
intervals so that the whole real axis becomes black?
On 2024-11-21 10:21:40 +0000, WM said:
On 21.11.2024 10:16, Mikko wrote:
On 2024-11-20 11:42:15 +0000, WM said:
The intervals before and after shifting are not different. Only
their positions are.
The intervals are different. A shifted interval contains a different
set of numbers.
Consider this simplified argument. Let every unit interval after a
natural number n which is divisible by 10 be coloured black: (10n,
10n+1]. All others are white. Is it possible to shift the black
intervals so that the whole real axis becomes black?
Yes. Shift the interval (10n, 10n+1) to (n/2, n/2+1).
On 21.11.2024 10:16, Mikko wrote:
On 2024-11-20 11:42:15 +0000, WM said:
On 19.11.2024 17:26, Jim Burns wrote:
On 11/19/2024 6:01 AM, WM wrote:
That implies that
our well-known intervals
Sets with different intervals are different.
Our sets do not change.
The intervals before and after shifting are not different.
Only their positions are.
The intervals are different.
A shifted interval contains a different set of numbers.
The intervals before and after shifting are not different.
Only their positions are.
The intervals are different.
A shifted interval contains a different set of numbers.
Consider this simplified argument.
Let every unit interval after
a natural number n which is divisible by 10
be coloured black: (10n, 10n+1].
All others are white.
Is it possible to shift the black intervals
Is it possible to shift the black intervals
so that the whole real axis becomes black?
No.
Although there are infinitely many black intervals,
the white intervals will remain in the majority.
For every finite distance (0, 10n)
the relative covering is precisely 1/10,
whether or not the intervals have been moved or
remain at their original sites.
That means the function decribing this,
1/10, 1/10, 1/10, ...
has limit 1/10.
That is the quotient of
the infinity of black intervals and
the infinity of all intervals.
On 11/21/2024 5:21 AM, WM wrote:
That means the function describing this,
1/10, 1/10, 1/10, ...
has limit 1/10.
That is the quotient of
the infinity of black intervals and
the infinity of all intervals.
The Paradox of the Discontinuous Function
(not a paradox):
lim.⟨ rc(1), rc(2), rc(3), ... ⟩ ≠
rc( lim.⟨ 1, 2, 3, ... ⟩ )
You (WM) do not "believe in"
proper.superset.matching sets
discontinuous functions
On 21.11.2024 16:39, Jim Burns wrote:
On 11/21/2024 5:21 AM, WM wrote:
That means the function describing this,
1/10, 1/10, 1/10, ...
has limit 1/10.
That is the quotient of
the infinity of black intervals and
the infinity of all intervals.
The Paradox of the Discontinuous Function
(not a paradox):
It is a paradox
that only 1/10 of the real line is covered
for every finite interval (0, n]
but all is covered completely in the limit.
By what is it covered,
after all n have been proved unable?
lim.⟨ rc(1), rc(2), rc(3), ... ⟩ ≠
rc( lim.⟨ 1, 2, 3, ... ⟩ )
You (WM) do not "believe in"
proper.superset.matching sets
discontinuous functions
There is no reason to believe in magic.
But if you do, then
all Cantor-bijections can fail as well
"in the infinite".
Then mathematics is insufficient
to determine limits.
On 11/21/2024 11:24 AM, WM wrote:
By what is it covered,
after all n have been proved unable?
⎛ n ↦ i/j ↦ n
⎜
⎜ (i+j) := ⌈(2⋅n+¼)¹ᐟ²+½⌉
⎜ i := n-((i+j)-1)⋅((i+j)-2)/2
⎜ j := (i+j)-i
⎜
⎝ (i+j-1)⋅(i+j-2)/2+i = n
There is no reason to believe in magic.
But if you do, then
all Cantor-bijections can fail as well
"in the infinite".
Then mathematics is insufficient
to determine limits.
I am not enough of a scholar to know
that this is true of _all_ mathematics, but
I know that much knowledge of infinity,
including what I'm most familiar with,
is grounded in the _finite_
Here, I DON'T refer to finite numbers, etc.
I refer to finite sequences of CLAIMS,
each of which is true.or.not.first.false.
On 11/21/2024 2:21 PM, WM wrote:
On 21.11.2024 19:54, Jim Burns wrote:
On 11/21/2024 11:24 AM, WM wrote:
By what is it covered,
after all n have been proved unable?
⎛ n ↦ i/j ↦ n
⎜
⎜ (i+j) := ⌈(2⋅n+¼)¹ᐟ²+½⌉
⎜ i := n-((i+j)-1)⋅((i+j)-2)/2
⎜ j := (i+j)-i
⎜
⎝ (i+j-1)⋅(i+j-2)/2+i = n
That is not an answer.
You (WM) see it as "indistinguishable from magic".
That's a shame for your students.
The _description_ is completed.
It's right there.
On 21.11.2024 19:54, Jim Burns wrote:
On 11/21/2024 11:24 AM, WM wrote:
By what is it covered,
after all n have been proved unable?
⎛ n ↦ i/j ↦ n
⎜
⎜ (i+j) := ⌈(2⋅n+¼)¹ᐟ²+½⌉
⎜ i := n-((i+j)-1)⋅((i+j)-2)/2
⎜ j := (i+j)-i
⎜
⎝ (i+j-1)⋅(i+j-2)/2+i = n
That is not an answer.
Further it is only valid for
the first numbers which are followed by
almost all numbers.
Never completed.
On 21.11.2024 11:59, Mikko wrote:
On 2024-11-21 10:21:40 +0000, WM said:
On 21.11.2024 10:16, Mikko wrote:
On 2024-11-20 11:42:15 +0000, WM said:
The intervals before and after shifting are not different. Only their >>>>> positions are.
The intervals are different. A shifted interval contains a different
set of numbers.
Consider this simplified argument. Let every unit interval after a
natural number n which is divisible by 10 be coloured black: (10n,
10n+1]. All others are white. Is it possible to shift the black
intervals so that the whole real axis becomes black?
Yes. Shift the interval (10n, 10n+1) to (n/2, n/2+1).
For every finite (0, n] the relative covering remains f(n) = 1/10, independent of shifting. The constant sequence has limit 1/10.
On 2024-11-21 11:03:28 +0000, WM said:
For every finite (0, n] the relative covering remains f(n) = 1/10,
independent of shifting. The constant sequence has limit 1/10.
That is irrelevant to your question whether the whole interval becomes
black if the shifted intervals (n/2, n/2+1) are painted black.
And that question is irrelevant to the topic specified on the subject line.
On 21.11.2024 22:46, Jim Burns wrote:
On 11/21/2024 2:21 PM, WM wrote:
On 21.11.2024 19:54, Jim Burns wrote:
On 11/21/2024 11:24 AM, WM wrote:
By what is it covered,
after all n have been proved unable?
⎛ n ↦ i/j ↦ n
⎜
⎜ (i+j) := ⌈(2⋅n+¼)¹ᐟ²+½⌉
⎜ i := n-((i+j)-1)⋅((i+j)-2)/2
⎜ j := (i+j)-i
⎜
⎝ (i+j-1)⋅(i+j-2)/2+i = n
That is not an answer.
Further it is only valid for
the first numbers which are followed by
almost all numbers.
Never completed.
The _description_ is completed.
It's right there.
The description of the set
not of all its elements.
On 11/21/2024 4:57 PM, WM wrote:For instance, not all indices of the endsegments can be counted to.
The _description_ is completed.
It's right there.
The description of the set
not of all its elements.
The description is sufficient in order to
finitely.investigate infinitely.many.
----
We have spent a lot of pixels discussing FISONs,
finite initial segments of naturals.
However,
here I consider FISOCs,
finite initial segments of claims.
FISOCs share a useful property with FISONs,
they are well.ordered.
If any claim has a property,
then some claim has that property first.
If any claim is written in Comic Sans,
then some claim is in Comic Sans first.
If any claim is false,
then some claim is false first.
Consider a specific FISOC with
a description of what.we.are.considering, broadly.
⎛⎛ ℕ⁺ holds numbers countable.to from.1
⎜⎜ ℚ⁺ holds ratios of numbers in ℕ⁺
⎜⎝ ℝ⁺ holds points between splits of ℚ⁺
⎜ Further claims about elements of ℕ⁺ ℚ⁺ ℝ⁺
⎝ which are each true.or.not.first.false
Broadly speaking,
claims can be true and can be false.
Broadly speaking,
the initial ℕ⁺.ℚ⁺.ℝ⁺ claims can be false
about some three sets or other.
In those broader after.false instances,
the following not.first.false claims
are not.first.false
whether they are true or they are false.
In the broader after.false instances,
the following not.first.false claims
are NOT an answer.
However,
more narrowly,
for what.we.are.considering,
the initial ℕ⁺.ℚ⁺.ℝ⁺ claims are true.
On 22.11.2024 19:51, Jim Burns wrote:⎛ (i+j) := ⌈(2⋅n+¼)¹ᐟ²+½⌉
On 11/21/2024 4:57 PM, WM wrote:On 11/22/2024 1:51 PM, Jim Burns wrote:
On 11/21/2024 4:57 PM, WM wrote:
On 21.11.2024 22:46, Jim Burns wrote:
The _description_ is completed.
It's right there.
The description of the set
not of all its elements.
The description
is sufficient in order to
finitely.investigate infinitely.many.
For instance,
not all indices of the endsegments
can be counted to.
Remember:
The intersection of all endsegments is empty,
but the intersection of
endsegments which can be counted to
is infinite.
Note that every endsegment loses only one number.
Therefore there must
exist infinitely many finite endsegments.
On 22.11.2024 09:42, Mikko wrote:
On 2024-11-21 11:03:28 +0000, WM said:
For every finite (0, n] the relative covering remains f(n) = 1/10,
independent of shifting. The constant sequence has limit 1/10.
That is irrelevant to your question whether the whole interval becomes
black if the shifted intervals (n/2, n/2+1) are painted black.
It is relevant by three reasons:
1) The limit of the sequence f(n) of relative coverings in (0, n] is
1/10, not 1. Therefore the relative covering 1 would contradict
analysis.
2) Since for all intervals (0, n] the relative covering is 1/10, the additional blackies must be taken from the nowhere.
3) Since a shifted blacky leaves a white unit interval where it has
left, the white must remain such that the whole real axis can never
become black.
On 2024-11-22 10:53:32 +0000, WM said:
On 22.11.2024 09:42, Mikko wrote:
On 2024-11-21 11:03:28 +0000, WM said:
For every finite (0, n] the relative covering remains f(n) = 1/10,
independent of shifting. The constant sequence has limit 1/10.
That is irrelevant to your question whether the whole interval becomes
black if the shifted intervals (n/2, n/2+1) are painted black.
It is relevant by three reasons:
1) The limit of the sequence f(n) of relative coverings in (0, n] is
1/10, not 1. Therefore the relative covering 1 would contradict analysis.
2) Since for all intervals (0, n] the relative covering is 1/10, the
additional blackies must be taken from the nowhere.
3) Since a shifted blacky leaves a white unit interval where it has
left, the white must remain such that the whole real axis can never
become black.
You say that it is relevant but you don't show how that is relevant
to the fact that there is no real number between the intervals (n/2, n/2+1) that is not a part of at least one of those intervals.
On 11/22/2024 4:30 PM, WM wrote:
Remember:
The intersection of all endsegments is empty,
but the intersection of
endsegments which can be counted to
is infinite.
No one should "remember" that.
It is incorrect.
Note that every endsegment loses only one number.
Finite cardinalities can lose one number.
Therefore there must
exist infinitely many finite endsegments.
On 22.11.2024 23:56, Jim Burns wrote:
On 11/22/2024 4:30 PM, WM wrote:
[...][...]
∀k ∈ ℕ_def:
∩{E(1), E(2), ..., E(k)} = E(k),
|E(k)| = ℵ₀
For all endsegments:
∀k ∈ ℕ:
|E(k+1)| = |E(k)| - 1
Therefore there must
exist infinitely many finite endsegments.
Remember:
The intersection of all endsegments is empty,
but the intersection of
endsegments which can be counted to
is infinite.
No one should "remember" that.
It is incorrect.
∀k ∈ ℕ_def:
∩{E(1), E(2), ..., E(k)} = E(k),
|E(k)| = ℵ₀
∩{E(1), E(2), ...} = { }.
On 11/23/2024 3:54 AM, WM wrote:
∀k ∈ ℕ_def:
∩{E(1), E(2), ..., E(k)} = E(k),
|E(k)| = ℵ₀
For all endsegments:
∀k ∈ ℕ:
|E(k+1)| = |E(k)| - 1
Perhaps what you intend to say is
⎛ For all post.definable end.segments
⎜ ∀k ∈ (ℕ\ℕ_def):
⎝ |E(k+1)| = |E(k)| - 1
Perhaps what you intend to say is
⎛ There are post.definable end.segments
⎝ which have finite cardinalities.
I nearly agreed with that,
because our ℕ = ℕ_def
However,
your mention of 'E(k+1)' implies
⎛ ∀k ∈ (ℕ\ℕ_def):
⎝ (ℕ\ℕ_def) ∋ k+1
For _your_ end.segments,
k ↦ k+1 : one.to.one
E(k) → E(k+1) : one.to.one
|E(k)| ≤ |E(k+1)|
Also,
E(k) ⊇ E(k+1)
|E(k)| ≥ |E(k+1)|
|E(k)| = |E(k+1)|
Remember:
The intersection of all endsegments is empty,
but the intersection of
endsegments which can be counted to
is infinite.
No one should "remember" that.
It is incorrect.
∀k ∈ ℕ_def:
∩{E(1), E(2), ..., E(k)} = E(k),
|E(k)| = ℵ₀
∩{E(1), E(2), ...} = { }.
⎜ ℕ_def ∋ 0
⎜ ℕ_def ⊆ S ⇐ S ∋ 0 ∧ ∀k ∈ S: S ∋ k+1
⎝ Each end.segment of ℕ_def is countable.to.
⋂{E(k):k∈ℕ_def} = {}
because
∀j ∈ ℕ_def:
j ∉ E(j+1) ∈ {E(k):k∈ℕ_def}
j ∉ ⋂{E(k):k∈ℕ_def}
The end.segment.intersection is empty because
each end.segment of ℕ_def is countable.past.What is countable.past?
On 23.11.2024 16:33, Jim Burns wrote:
For _your_ end.segments,
k ↦ k+1 : one.to.one
E(k) → E(k+1) : one.to.one
|E(k)| ≤ |E(k+1)|
No.
|E(k)| ≥ |E(k+1)| = |E(k)| - 1.
On 11/23/2024 12:23 PM, WM wrote:
|E(k)| ≥ |E(k+1)| = |E(k)| - 1.
E(k) ⊇ E(k+1)
|E(k)| ≥ |E(k+1)|
doesn't contradict
|E(k)| ≤ |E(k+1)|
Together,
|E(k)| = |E(k+1)|
and
|E(k+1)| doesn't lose one number.
|E(k)| = |E(k+1)| is infinite.
----
Do you (WM) object to
k ↦ k+1 : one.to.one
On 23.11.2024 22:20, Jim Burns wrote:
Do you (WM) object to
k ↦ k+1 : one.to.one
I don't know what that waffle should mean.
E(k) ⊇ E(k+1)
|E(k)| ≥ |E(k+1)|
doesn't contradict
|E(k)| ≤ |E(k+1)|
It does.
Together,
|E(k)| = |E(k+1)|
and
|E(k+1)| doesn't lose one number.
Spare your nonsense.
On 23.11.2024 09:07, Mikko wrote:
On 2024-11-22 10:53:32 +0000, WM said:
On 22.11.2024 09:42, Mikko wrote:
On 2024-11-21 11:03:28 +0000, WM said:
For every finite (0, n] the relative covering remains f(n) = 1/10,
independent of shifting. The constant sequence has limit 1/10.
That is irrelevant to your question whether the whole interval becomes >>>> black if the shifted intervals (n/2, n/2+1) are painted black.
It is relevant by three reasons:
1) The limit of the sequence f(n) of relative coverings in (0, n] is
1/10, not 1. Therefore the relative covering 1 would contradict
analysis.
2) Since for all intervals (0, n] the relative covering is 1/10, the
additional blackies must be taken from the nowhere.
3) Since a shifted blacky leaves a white unit interval where it has
left, the white must remain such that the whole real axis can never
become black.
You say that it is relevant but you don't show how that is relevant
to the fact that there is no real number between the intervals (n/2, n/2+1) >> that is not a part of at least one of those intervals.
Because that has nothing to do with the topic under discussion. See
points 1, 2, and 3. They are to be discussed.
On 11/23/2024 4:39 PM, WM wrote:
On 23.11.2024 22:20, Jim Burns wrote:
Do you (WM) object to
k ↦ k+1 : one.to.one
I don't know what that waffle should mean.
k ↦ k+1 means the successor operation.
'One.to.one' means that,
if j≠k then j+1≠k+1
different numbers have different successors.
I am claiming that
different numbers have different successors.
E(k) ⊇ E(k+1)
|E(k)| ≥ |E(k+1)|
doesn't contradict
|E(k)| ≤ |E(k+1)|
It does.
Then you (WM) also don't know
what a contradiction is.
On 2024-11-23 08:49:18 +0000, WM said:
It is relevant by three reasons:
1) The limit of the sequence f(n) of relative coverings in (0, n] is
1/10, not 1. Therefore the relative covering 1 would contradict
analysis.
2) Since for all intervals (0, n] the relative covering is 1/10, the
additional blackies must be taken from the nowhere.
3) Since a shifted blacky leaves a white unit interval where it has
left, the white must remain such that the whole real axis can never
become black.
You say that it is relevant but you don't show how that is relevant
to the fact that there is no real number between the intervals (n/2,
n/2+1)
that is not a part of at least one of those intervals.
Because that has nothing to do with the topic under discussion. See
points 1, 2, and 3. They are to be discussed.
The subject line specifies that the discussion should be about Cantor's enumeration of the rational numbers.
OP specifies that the discussion shall be baout the sequence of
itnrevals
The 1, 2, and 3 above are not relevant to the topic sepcified by the
subject line and OP.
On 23.11.2024 23:10, Jim Burns wrote:
On 11/23/2024 4:39 PM, WM wrote:
On 23.11.2024 22:20, Jim Burns wrote:
Do you (WM) object to
k ↦ k+1 : one.to.one
I don't know what that waffle should mean.
k ↦ k+1 means the successor operation.
'One.to.one' means that,
if j≠k then j+1≠k+1
different numbers have different successors.
I am claiming that
different numbers have different successors.
Ok.
E(k) ⊇ E(k+1)
|E(k)| ≥ |E(k+1)|
doesn't contradict
|E(k)| ≤ |E(k+1)|
It does.
Then you (WM) also don't know
what a contradiction is.
|E(k)| ≤ |E(k+1)| is wrong.
On 11/24/2024 9:05 AM, WM wrote:
On 23.11.2024 23:10, Jim Burns wrote:
On 11/23/2024 4:39 PM, WM wrote:
On 23.11.2024 22:20, Jim Burns wrote:
Do you (WM) object to
k ↦ k+1 : one.to.one
I don't know what that waffle should mean.
k ↦ k+1 means the successor operation.
'One.to.one' means that,
if j≠k then j+1≠k+1
different numbers have different successors.
I am claiming that
different numbers have different successors.
Ok.
"Ok, I understand you"
or
"Ok, different numbers have different successors"
?
What we mean by
|E(k)| ≤ |E(k+1)|
is that
there is a one.to.one function
from E(k) to E(k+1)
The successor operation, for example.
So, there is.
So, |E(k)| ≤ |E(k+1)| isn't wrong.
On 24.11.2024 20:26, Jim Burns wrote:
What we mean by
|E(k)| ≤ |E(k+1)|
is that
there is a one.to.one function
from E(k) to E(k+1)
The successor operation, for example.
What I mean is the fact that
∀k ∈ ℕ: |E(k+1)| = |E(k)| - 1
whereas Cantor's ℵo is a very unsharp measure.
So, there is.
So, |E(k)| ≤ |E(k+1)| isn't wrong.
Cantor's nonsense has many faces.
It i not suitable for serious maths.
On 11/24/2024 2:42 PM, WM wrote:
On 24.11.2024 20:26, Jim Burns wrote:
What we mean by
|E(k)| ≤ |E(k+1)|
is that
there is a one.to.one function
from E(k) to E(k+1)
The successor operation, for example.
What I mean is the fact that
∀k ∈ ℕ: |E(k+1)| = |E(k)| - 1
whereas Cantor's ℵo is a very unsharp measure.
Finite cardinalities can change by 1.
Infinite cardinalities are larger than
each finite cardinality,
and cannot change by 1.
∀k ∈ ℕ: |E(k+1)| = |E(k)| - 1
ℕ is the set of each and only finite cardinalities.
|ℕ| isn't a finite cardinality.
|ℕ| cannot change by 1.
|ℕ| can change by 1.
Cantor's nonsense has many faces.
It i not suitable for serious maths.
Cardinalities which cannot change by 1
do not change by 1 when they're called unserious.
On 24.11.2024 21:17, Jim Burns wrote:
On 11/24/2024 2:42 PM, WM wrote:
On 24.11.2024 20:26, Jim Burns wrote:
What we mean by
|E(k)| ≤ |E(k+1)|
is that
there is a one.to.one function
from E(k) to E(k+1)
The successor operation, for example.
What I mean is the fact that
∀k ∈ ℕ: |E(k+1)| = |E(k)| - 1
whereas Cantor's ℵo is a very unsharp measure.
Finite cardinalities can change by 1.
Endsegmentes can change by 1 element.
Therefore their number of elements can change by 1.
On 24.11.2024 13:38, Mikko wrote:
On 2024-11-23 08:49:18 +0000, WM said:
It is relevant by three reasons:
1) The limit of the sequence f(n) of relative coverings in (0, n] is >>>>> 1/10, not 1. Therefore the relative covering 1 would contradict
analysis.
2) Since for all intervals (0, n] the relative covering is 1/10, the >>>>> additional blackies must be taken from the nowhere.
3) Since a shifted blacky leaves a white unit interval where it has
left, the white must remain such that the whole real axis can never
become black.
You say that it is relevant but you don't show how that is relevant
to the fact that there is no real number between the intervals (n/2, n/2+1)
that is not a part of at least one of those intervals.
Because that has nothing to do with the topic under discussion. See
points 1, 2, and 3. They are to be discussed.
The subject line specifies that the discussion should be about Cantor's
enumeration of the rational numbers.
OP specifies that the discussion shall be baout the sequence of
itnrevals
That is a mistake. Should read:
[q_n - ε*sqrt(2)/2^n, q_n + ε*sqrt(2)/2^n].
The 1, 2, and 3 above are not relevant to the topic sepcified by the
subject line and OP.
My last example contradicts a simpler bijection, namely that between
all natural numbers and all natural numbers divisible by 10: Let every
unit interval on the real axis after a number 10n carry a black hat.
Then it should be possible to cover all intervals with black hats.
On 11/24/2024 3:56 PM, WM wrote:
Endsegments can change by 1 element.
Therefore their number of elements can change by 1.
Yes,
each end.segment.set can change by 1 element
However,
for each end.segment.set E(k)
for each finite cardinality j
there is a larger.than.j subset E(k)\E(k+j+1)
For each end.segment.set E(k)
for each finite cardinality j
j is not the cardinality of E(k)
On 2024-11-24 14:01:15 +0000, WM said:
On 24.11.2024 13:38, Mikko wrote:
On 2024-11-23 08:49:18 +0000, WM said:
It is relevant by three reasons:
1) The limit of the sequence f(n) of relative coverings in (0, n]
is 1/10, not 1. Therefore the relative covering 1 would contradict >>>>>> analysis.
2) Since for all intervals (0, n] the relative covering is 1/10,
the additional blackies must be taken from the nowhere.
3) Since a shifted blacky leaves a white unit interval where it
has left, the white must remain such that the whole real axis can
never become black.
You say that it is relevant but you don't show how that is relevant
to the fact that there is no real number between the intervals
(n/2, n/2+1)
that is not a part of at least one of those intervals.
Because that has nothing to do with the topic under discussion. See
points 1, 2, and 3. They are to be discussed.
The subject line specifies that the discussion should be about Cantor's
enumeration of the rational numbers.
OP specifies that the discussion shall be baout the sequence of
itnrevals
That is a mistake. Should read:
[q_n - ε*sqrt(2)/2^n, q_n + ε*sqrt(2)/2^n].
OK but the following applies to that, too:
The 1, 2, and 3 above are not relevant to the topic sepcified by the
subject line and OP.
My last example contradicts a simpler bijection, namely that between
all natural numbers and all natural numbers divisible by 10: Let every
unit interval on the real axis after a number 10n carry a black hat.
Then it should be possible to cover all intervals with black hats.
What does "contradicts" in "contradicts a simpler bijection"?
On 24.11.2024 22:33, Jim Burns wrote:
On 11/24/2024 3:56 PM, WM wrote:
Endsegments can change by 1 element.
Therefore their number of elements can change by 1.
Yes,
each end.segment.set can change by 1 element
However,
for each end.segment.set E(k)
for each finite cardinality j
there is a larger.than.j subset E(k)\E(k+j+1)
Finite cardinalities belong to dark endsegments.
Not every dark endsegment has a successor.
For each end.segment.set E(k)
for each finite cardinality j
j is not the cardinality of E(k)
The endsegments
only can have an empty intersection
if there are endsegments with 3, 2, 1, 0 elements.
On 11/25/2024 8:52 AM, WM wrote:
Finite cardinalities belong to dark endsegments.
Finite cardinals can change by 1
Each end.segment Eᶠⁱⁿ(k) of the finite.cardinalities ℕᶠⁱⁿ
holds a countable.to.from.0 least.element
The endsegments
only can have an empty intersection
if there are endsegments with 3, 2, 1, 0 elements.
The end.segments
can only have a non.empty intersection
if there is an element which is in each end.segment.
On 25.11.2024 09:43, Mikko wrote:
On 2024-11-24 14:01:15 +0000, WM said:
On 24.11.2024 13:38, Mikko wrote:
On 2024-11-23 08:49:18 +0000, WM said:
It is relevant by three reasons:
1) The limit of the sequence f(n) of relative coverings in (0, n] is >>>>>>> 1/10, not 1. Therefore the relative covering 1 would contradict
analysis.
2) Since for all intervals (0, n] the relative covering is 1/10, the >>>>>>> additional blackies must be taken from the nowhere.
3) Since a shifted blacky leaves a white unit interval where it has >>>>>>> left, the white must remain such that the whole real axis can never >>>>>>> become black.
You say that it is relevant but you don't show how that is relevant >>>>>> to the fact that there is no real number between the intervals (n/2, n/2+1)
that is not a part of at least one of those intervals.
Because that has nothing to do with the topic under discussion. See
points 1, 2, and 3. They are to be discussed.
The subject line specifies that the discussion should be about Cantor's >>>> enumeration of the rational numbers.
OP specifies that the discussion shall be baout the sequence of
itnrevals
That is a mistake. Should read:
[q_n - ε*sqrt(2)/2^n, q_n + ε*sqrt(2)/2^n].
OK but the following applies to that, too:
The 1, 2, and 3 above are not relevant to the topic sepcified by the
subject line and OP.
My last example contradicts a simpler bijection, namely that between
all natural numbers and all natural numbers divisible by 10: Let every
unit interval on the real axis after a number 10n carry a black hat.
Then it should be possible to cover all intervals with black hats.
What does "contradicts" in "contradicts a simpler bijection"?
The simple example contradicts a bijection between the two sets
described above.
On 2024-11-25 14:38:13 +0000, WM said:
The simple example contradicts a bijection between the two sets
described above.
What does "contradicts a bijection" mean?
On 26.11.2024 10:09, Mikko wrote:
On 2024-11-25 14:38:13 +0000, WM said:
It shows that the mapping claimed to be a bijection is not a bijection.The simple example contradicts a bijection between the two sets
described above.
What does "contradicts a bijection" mean?
Regards, WM
On 11/26/24 6:07 AM, WM wrote:
On 26.11.2024 10:09, Mikko wrote:
On 2024-11-25 14:38:13 +0000, WM said:It shows that the mapping claimed to be a bijection is not a bijection.
The simple example contradicts a bijection between the two sets
described above.
What does "contradicts a bijection" mean?
Where did you do that with the ACTUAL bijection, and not just your
strawman "equivalent".
Which element of which infinite set did not participate in the bijection?
On 26.11.2024 14:07, Richard Damon wrote:
On 11/26/24 6:07 AM, WM wrote:
On 26.11.2024 10:09, Mikko wrote:
On 2024-11-25 14:38:13 +0000, WM said:It shows that the mapping claimed to be a bijection is not a bijection.
The simple example contradicts a bijection between the two sets
described above.
What does "contradicts a bijection" mean?
Where did you do that with the ACTUAL bijection, and not just your
strawman "equivalent".
An actual bijection is assumed: Consider the black hats at every 10 n
and white hats at all other numbers n. It is possible to shift the black
hats such that every interval (0, n] is completely covered by black
hats. There is no first n discernible that cannot be covered by black
hat. But the origin of each used black hat larger than n is now covered
by a white hat. Without deleting all white hats it is not possible to
cover all n by black hats. But deleting white hats is prohibited by
logic. Exchanging can never delete one of the exchanged elements.
Which element of which infinite set did not participate in the bijection?
That is the crucial point! The white hats remain as long as logic is
valid. But their carriers cannot be found. They are dark.
Regards, WM
On 26.11.2024 15:50, Richard Damon wrote:
On 11/26/24 8:58 AM, WM wrote:
An actual bijection is assumed: Consider the black hats at every 10 n
and white hats at all other numbers n. It is possible to shift the
black hats such that every interval (0, n] is completely covered by
black hats. There is no first n discernible that cannot be covered by
black hat. But the origin of each used black hat larger than n is now
covered by a white hat. Without deleting all white hats it is not
possible to cover all n by black hats. But deleting white hats is
prohibited by logic. Exchanging can never delete one of the exchanged
elements.
But a bijection is NOT a set to itself, but between two sets.
The set of natural numbers divisible by 10 and the set of natural
numbers are two different sets.
The white hats aren't on the "OTHER" numbers, (unless you bijection is
from numbers zero mod 10 to number non-zero mod 10), but ALL numbers
have a white hat, and every tenth has a black hat, and the bijection
shows we can pair every black hat to a white hat.
Let all numbers n have white hats and in addition the 10n have black hats. Claim: We can cover every white hat by a black hat. But when we take a
black hat from 20 to give it to 2, then 20 has a white hat, and so on.
How can the white hats disappear?
Which element of which infinite set did not participate in the
bijection?
That is the crucial point! The white hats remain as long as logic is
valid. But their carriers cannot be found. They are dark.
And nothing in the logic says that white hats go away.
Nothing in logic allows that. But Cantor claims it erroneously.
Regards, WM
On 11/26/24 8:58 AM, WM wrote:
An actual bijection is assumed: Consider the black hats at every 10 n
and white hats at all other numbers n. It is possible to shift the
black hats such that every interval (0, n] is completely covered by
black hats. There is no first n discernible that cannot be covered by
black hat. But the origin of each used black hat larger than n is now
covered by a white hat. Without deleting all white hats it is not
possible to cover all n by black hats. But deleting white hats is
prohibited by logic. Exchanging can never delete one of the exchanged
elements.
But a bijection is NOT a set to itself, but between two sets.
The white hats aren't on the "OTHER" numbers, (unless you bijection is
from numbers zero mod 10 to number non-zero mod 10), but ALL numbers
have a white hat, and every tenth has a black hat, and the bijection
shows we can pair every black hat to a white hat.
Which element of which infinite set did not participate in the
bijection?
That is the crucial point! The white hats remain as long as logic is
valid. But their carriers cannot be found. They are dark.
And nothing in the logic says that white hats go away.
On 26.11.2024 06:58, Jim Burns wrote:
On 11/25/2024 8:52 AM, WM wrote:
Finite cardinalities belong to dark endsegments.
Finite cardinals can change by 1
Yes.
The last endsegments have 3, 2, 1, 0 elements.
Each end.segment Eᶠⁱⁿ(k) of the finite.cardinalities ℕᶠⁱⁿ
holds a countable.to.from.0 least.element
No.
All elements of finite endsegments
(and almost all of infinite endsegments)
are dark.
Each end.segment Eᶠⁱⁿ(k) of the finite.cardinalities ℕᶠⁱⁿ
holds a countable.to.from.0 least.element
No.
The endsegments
only can have an empty intersection
if there are endsegments with 3, 2, 1, 0 elements.
The end.segments
can only have a non.empty intersection
if there is an element which is in each end.segment.
That is the case for every non-empty endsegment
before all elements are lost.
Otherwise
two endsegments with different elements
must exist.
That is impossible by inclusion monotony.
Finite cardinals can change by 1
Yes.
The last endsegments have 3, 2, 1, 0 elements.
On 26.11.2024 19:49, Jim Burns wrote:
There are no last end.segments of ℕᶠⁱⁿ
There are no finitely.sized end segments of ℕᶠⁱⁿ
There are no finite cardinals common to
each end.segment of ℕᶠⁱⁿ
That is a contradiction.
If there are no common numbers,
then all numbers must have been lost.
But then no numbers are remaining.
Then
there are finite endsegments because
∀k ∈ ℕ: |E(k+1)| = |E(k)| - 1.
There are no last end.segments of ℕᶠⁱⁿ
There are no finitely.sized end segments of ℕᶠⁱⁿ
There are no finite cardinals common to
each end.segment of ℕᶠⁱⁿ
WM <[email protected]> wrote:
And nothing in the logic says that white hats go away.
Nothing in logic allows that. But Cantor claims it erroneously.
Cantor NEVER had the white hats go away, he just showed you could match
every number with a black hat
On 26.11.2024 18:24, Richard Damon wrote:
WM <[email protected]> wrote:
And nothing in the logic says that white hats go away.
Nothing in logic allows that. But Cantor claims it erroneously.
Cantor NEVER had the white hats go away, he just showed you could match
every number with a black hat
That means all white hats must disappear under black hats. That means
the white and black must be exchanged until no white remains. That is impossible.
Regards, WM
On 26.11.2024 10:09, Mikko wrote:
On 2024-11-25 14:38:13 +0000, WM said:
It shows that the mapping claimed to be a bijection is not a bijection.The simple example contradicts a bijection between the two sets
described above.
What does "contradicts a bijection" mean?
On 11/26/2024 2:15 PM, WM wrote:
On 26.11.2024 19:49, Jim Burns wrote:
There are no last end.segments of ℕᶠⁱⁿ
There are no finitely.sized end segments of ℕᶠⁱⁿ
There are no finite cardinals common to
each end.segment of ℕᶠⁱⁿ
That is a contradiction.
It contradicts ℕᶠⁱⁿ being finite, nothing else.
If there are no common numbers,
then all numbers must have been lost.
But then no numbers are remaining.
Yes.
Each finite.cardinal k is countable.past to
k+1 which indexes
Eᶠⁱⁿ(k+1) which doesn't hold
k which is not common to
all end segments.
Each finite.cardinal k is not.in
the intersection of all end segments,
the set of elements common to all end.segments,
which is empty.
No numbers are remaining.
Then there are finite endsegments because
∀k ∈ ℕ: |E(k+1)| = |E(k)| - 1.
For each cardinal.which.can.change.by.1 j
WM <[email protected]> wrote:
On 26.11.2024 18:24, Richard Damon wrote:
WM <[email protected]> wrote:
And nothing in the logic says that white hats go away.
Nothing in logic allows that. But Cantor claims it erroneously.
Cantor NEVER had the white hats go away, he just showed you could match
every number with a black hat
That means all white hats must disappear under black hats. That means
the white and black must be exchanged until no white remains. That is
impossible.
Why is there a white hat under the black hat?
They are two *DIFFERENT* sets, one with the number {1, 2, 3, 4, …} each with a white hat, and one with the numbers {10, 20, 30, 40, …} each with a black hat. Just because we gave 10 a white hat in the first set doesn’t give the 10 in the second set that hat, then are DIFFERENT sets.
We can exchange the white hat from the first set on 1 with the black hat on the second set on 10.
We then swap hats between 1st set 2 and 2nd set 20, and so on for the first set n and the second set 10n, and we see that ALL the numbers in the first set get their black hats and all the numbers in the second set get white hats, and thus we have proven that this is a bijection, and the sets use
have the same cardinality.
You are just showing you don’t understand what is happening,
On 2024-11-26 11:07:57 +0000, WM said:
On 26.11.2024 10:09, Mikko wrote:
On 2024-11-25 14:38:13 +0000, WM said:It shows that the mapping claimed to be a bijection is not a bijection.
The simple example contradicts a bijection between the two sets
described above.
What does "contradicts a bijection" mean?
If so, no bijection is contradicted.
On 27.11.2024 10:33, Mikko wrote:
On 2024-11-26 11:07:57 +0000, WM said:
On 26.11.2024 10:09, Mikko wrote:
On 2024-11-25 14:38:13 +0000, WM said:It shows that the mapping claimed to be a bijection is not a bijection.
The simple example contradicts a bijection between the two sets
described above.
What does "contradicts a bijection" mean?
If so, no bijection is contradicted.
The possibility of a bijection between the sets ℕ = {1, 2, 3, ...} and
D = {10n | n ∈ ℕ} is contradicted.
Assume that a bijection between natural numbers divisible by 10 and all natural numbers is possible. First establish a bijection between the
numbers 10n ∈ ℕ and the numbers 10n ∈ D. This can be visualized by attaching black hats to the natural numbers of the form 10n ∈ ℕ and
white hats to the remaining natural numbers. The black hats indicate
that a number n ∈ ℕ has a partner in D. Now it should be possible to shift the black hats such that all natural numbers are covered by black
hats. The mapping f(10n) is started by exchanging white hats and black
hats precisely as would have been defined without the intermediate step:
1 gets it from 10, 2 gets it from 20, 3 gets it from 30, and so on,
precisely as would be done without the intermediate first step.
However when all numbers of an interval (1, 2, 3, ..., n) are equipped
with black hats, then some black hats have been taken from outside of
the interval, from larger 10n which in turn have received white hats. If
all natural numbers are equipped with black hats, then all white hats
have disappeared. But hats cannot disappear by exchanging them.
Regards, WM
On 26.11.2024 20:32, Richard Damon wrote:
WM <[email protected]> wrote:
On 26.11.2024 18:24, Richard Damon wrote:
WM <[email protected]> wrote:
And nothing in the logic says that white hats go away.
Nothing in logic allows that. But Cantor claims it erroneously.
Cantor NEVER had the white hats go away, he just showed you could match >>>> every number with a black hat
That means all white hats must disappear under black hats. That means
the white and black must be exchanged until no white remains. That is
impossible.
Why is there a white hat under the black hat?
That is irrelevant. The black hats must cover all numbers ℕ but cannot because when black and white hats are exchanged never a white hat is
deleted.
They are two *DIFFERENT* sets, one with the number {1, 2, 3, 4, …} each
with a white hat, and one with the numbers {10, 20, 30, 40, …} each
with a
black hat. Just because we gave 10 a white hat in the first set doesn’t
give the 10 in the second set that hat, then are DIFFERENT sets.
We have the black hats from the second set {10, 20, 30, 40, …}. They
shall cover all numbers The numbers divisible by 10 from the first set ℕ
= {1, 2, 3, 4, …} have been covered at the outset.
We can exchange the white hat from the first set on 1 with the black
hat on
the second set on 10.
We can first cover all numbers 10n of the first set ℕ with black hats.
That does not increase or decrease the set of black hats.
We then swap hats between 1st set 2 and 2nd set 20, and so on for the
first
set n and the second set 10n, and we see that ALL the numbers in the
first
set get their black hats and all the numbers in the second set get white
hats, and thus we have proven that this is a bijection, and the sets use
have the same cardinality.
You are just showing you don’t understand what is happening,
Do you understand that the bijection is not disturbed in the least when
we first cover all numbers 10n of the first set ℕ with black hats. We
can proceed then precisely in the same way as you say, can't we? What is different in your opinion?
Regards, WM
On 26.11.2024 20:44, Jim Burns wrote:
On 11/26/2024 2:15 PM, WM wrote:
On 26.11.2024 19:49, Jim Burns wrote:
There are no last end.segments of ℕᶠⁱⁿ
There are no finitely.sized end segments of ℕᶠⁱⁿ
There are no finite cardinals common to
each end.segment of ℕᶠⁱⁿ
That is a contradiction.
It contradicts ℕᶠⁱⁿ being finite, nothing else.
It contradicts inclusion monotony.
If there are no common numbers,
then all numbers must have been lost.
But then no numbers are remaining.
Yes.
Then also
no numbers are remaining in the endsegments.
Each finite.cardinal k is countable.past to
k+1 which indexes
Eᶠⁱⁿ(k+1) which doesn't hold
k which is not common to
all end segments.
Each finite.cardinal k is not.in
the intersection of all end segments,
the set of elements common to all end.segments,
which is empty.
No numbers are remaining.
That is true.
But you claimed that every endsegment is infinite.
In an infinite endsegment
numbers are remaining.
In many infinite endsegments
infinitely many numbers are the same.
Then there are finite endsegments because
∀k ∈ ℕ: |E(k+1)| = |E(k)| - 1.
For each cardinal.which.can.change.by.1 j
|Eᶠⁱⁿ(k)| is larger than j
|Eᶠⁱⁿ(k)| isn't j
|Eᶠⁱⁿ(k)| isn't any cardinal.which.can.change.by.1
|Eᶠⁱⁿ(k)| cannot change by 1
|Eᶠⁱⁿ(k+1)| = |Eᶠⁱⁿ(k)|
That does not contradict the fact that
infinite endsegments have
infinitely many numbers in common
and hence an infinite intersection.
On 11/27/24 5:57 AM, WM wrote:
You switch between a white hat being revealed from under the black hat
when it is moved from 10n to n, and the hats being switched.
The second happens when you have TWO sets of numbers,
In no cases were white hats "deleted" because in both cases we still had
an infinite set of numbers with white hats
We have the black hats from the second set {10, 20, 30, 40, …}. They
shall cover all numbers The numbers divisible by 10 from the first set
ℕ = {1, 2, 3, 4, …} have been covered at the outset.
No, that isn't the bijection being talked about.
You can't disprove a bijection by not following it.
You don't seem to understand Cantor's claim, it isn't that every
possible attempt at a bijection will succeed, it is that *IF* there is
one, the sets are equal.
We then swap hats between 1st set 2 and 2nd set 20, and so on for the
first
set n and the second set 10n, and we see that ALL the numbers in the
first
set get their black hats and all the numbers in the second set get white >>> hats, and thus we have proven that this is a bijection, and the sets use >>> have the same cardinality.
You are just showing you don’t understand what is happening,
Do you understand that the bijection is not disturbed in the least
when we first cover all numbers 10n of the first set ℕ with black
hats. We can proceed then precisely in the same way as you say, can't
we? What is different in your opinion?
Of course it is, since you can't complete the first part.
On 11/27/2024 6:04 AM, WM wrote:
However,
one makes a quantifier shift, unreliable,
to go from that to
⛔⎛ there is an end segment such that
⛔⎜ for each number (finite cardinal)
⛔⎝ the number isn't in the end segment.
Each end.segment is infinite.
Their intersection of all is empty.
These claims do not conflict.
In an infinite endsegment
numbers are remaining.
In many infinite endsegments infinitely many numbers are the same.
And the intersection of all,
which isn't any end.segment,
is empty.
On 27.11.2024 15:16, Richard Damon wrote:
On 11/27/24 5:57 AM, WM wrote:
You switch between a white hat being revealed from under the black hat
when it is moved from 10n to n, and the hats being switched.
It is irrelevant, but assume the hats being switched. Black hats for
numbers 10n, white hats for all others.
The second happens when you have TWO sets of numbers,
We have one set, namely ℕ to be covered with black hats. The natural numbers 10n are already covered. Now the black hats have to be moved to
cover all natural numbers.
In no cases were white hats "deleted" because in both cases we still
had an infinite set of numbers with white hats
But Cantor claims that all can be replaced by black hats.
We have the black hats from the second set {10, 20, 30, 40, …}. They
shall cover all numbers The numbers divisible by 10 from the first
set ℕ = {1, 2, 3, 4, …} have been covered at the outset.
No, that isn't the bijection being talked about.
This bijection comes afterwards.
You can't disprove a bijection by not following it.
You can follow it now. The number of black hats remains the same.
You don't seem to understand Cantor's claim, it isn't that every
possible attempt at a bijection will succeed, it is that *IF* there is
one, the sets are equal.
Now show that there is one.
We then swap hats between 1st set 2 and 2nd set 20, and so on for
the first
set n and the second set 10n, and we see that ALL the numbers in the
first
set get their black hats and all the numbers in the second set get
white
hats, and thus we have proven that this is a bijection, and the sets
use
have the same cardinality.
You are just showing you don’t understand what is happening,
Do you understand that the bijection is not disturbed in the least
when we first cover all numbers 10n of the first set ℕ with black
hats. We can proceed then precisely in the same way as you say, can't
we? What is different in your opinion?
Of course it is, since you can't complete the first part.
It is completed! Every number 10n starts wit a black hat.
Regards, WM
On 11/27/24 3:09 PM, WM wrote:
It is completed! Every number 10n starts wit a black hat.
Right, and it gives that hat to n, so every number gets one.
And, 10n will get back a hat from 100n, so it still have one at the end.
On 27.11.2024 22:20, Richard Damon wrote:
On 11/27/24 3:09 PM, WM wrote:
It seems so, but that is impossible. Up to every 10n, the interval 1, 2,It is completed! Every number 10n starts wit a black hat.
Right, and it gives that hat to n, so every number gets one.
And, 10n will get back a hat from 100n, so it still have one at the end.
3, ... 10n has a covering of 1/10 only. The limit of this sequence is
1/10 too. And since this is true for all natural numbers, where should additional hats come from?
Regards, WM
On 11/28/24 7:20 AM, WM wrote:
On 27.11.2024 22:20, Richard Damon wrote:
On 11/27/24 3:09 PM, WM wrote:It seems so, but that is impossible. Up to every 10n, the interval 1,
It is completed! Every number 10n starts wit a black hat.
Right, and it gives that hat to n, so every number gets one.
And, 10n will get back a hat from 100n, so it still have one at the end. >>>
2, 3, ... 10n has a covering of 1/10 only. The limit of this sequence
is 1/10 too. And since this is true for all natural numbers, where
should additional hats come from?
No, it is possilbe,
On 28.11.2024 14:01, Richard Damon wrote:
On 11/28/24 7:20 AM, WM wrote:
On 27.11.2024 22:20, Richard Damon wrote:
On 11/27/24 3:09 PM, WM wrote:It seems so, but that is impossible. Up to every 10n, the interval 1,
It is completed! Every number 10n starts wit a black hat.
Right, and it gives that hat to n, so every number gets one.
And, 10n will get back a hat from 100n, so it still have one at the
end.
2, 3, ... 10n has a covering of 1/10 only. The limit of this sequence
is 1/10 too. And since this is true for all natural numbers, where
should additional hats come from?
No, it is possilbe,
It is impossible that the sequence 1/10, 1/10, 1/10, ... has another
limit than 1/10.
Regards, WM
On 11/28/24 9:17 AM, WM wrote:No.
On 28.11.2024 14:01, Richard Damon wrote:
On 11/28/24 7:20 AM, WM wrote:
On 27.11.2024 22:20, Richard Damon wrote:
On 11/27/24 3:09 PM, WM wrote:It seems so, but that is impossible. Up to every 10n, the interval
It is completed! Every number 10n starts wit a black hat.
Right, and it gives that hat to n, so every number gets one.
And, 10n will get back a hat from 100n, so it still have one at the
end.
1, 2, 3, ... 10n has a covering of 1/10 only. The limit of this
sequence is 1/10 too. And since this is true for all natural
numbers, where should additional hats come from?
No, it is possilbe,
It is impossible that the sequence 1/10, 1/10, 1/10, ... has another
limit than 1/10.
But that makes the error of a wrong category.
The properties of the infinte set is not just the properties of the
series of the finite sets that approach it.
On 11/28/24 11:09 AM, WM wrote:
The properties of the infinte set is not just the properties of the
series of the finite sets that approach it.
Maybe. In this case it is precisely this.
Look: If for all intervals 1, 2, 3, ..., n the covering is 1/10, then
there are no natnumbers outside of all intervals and there are no hats
outside of all intervals. Therefore only a fool could believe that
infinitely many black hats were supplied after all. If you wish to be
a fool you may claim that. My students would never come down that much.
It is sort of like trying to figure out what 0^0 is.
On 28.11.2024 16:50, Richard Damon wrote:
On 11/28/24 9:17 AM, WM wrote:
On 28.11.2024 14:01, Richard Damon wrote:
On 11/28/24 7:20 AM, WM wrote:
On 27.11.2024 22:20, Richard Damon wrote:
On 11/27/24 3:09 PM, WM wrote:It seems so, but that is impossible. Up to every 10n, the interval
It is completed! Every number 10n starts wit a black hat.
Right, and it gives that hat to n, so every number gets one.
And, 10n will get back a hat from 100n, so it still have one at
the end.
1, 2, 3, ... 10n has a covering of 1/10 only. The limit of this
sequence is 1/10 too. And since this is true for all natural
numbers, where should additional hats come from?
No, it is possilbe,
It is impossible that the sequence 1/10, 1/10, 1/10, ... has another
limit than 1/10.
But that makes the error of a wrong category.No.
The properties of the infinte set is not just the properties of the
series of the finite sets that approach it.
Maybe. In this case it is precisely this.
Look: If for all intervals 1, 2, 3, ..., n the covering is 1/10, then
there are no natnumbers outside of all intervals and there are no hats outside of all intervals. Therefore only a fool could believe that
infinitely many black hats were supplied after all. If you wish to be a
fool you may claim that. My students would never come down that much.
Regards, WM
On 28.11.2024 18:17, Richard Damon wrote:
On 11/28/24 11:09 AM, WM wrote:
The properties of the infinte set is not just the properties of the
series of the finite sets that approach it.
Maybe. In this case it is precisely this.
Look: If for all intervals 1, 2, 3, ..., n the covering is 1/10, then
there are no natnumbers outside of all intervals and there are no
hats outside of all intervals. Therefore only a fool could believe
that infinitely many black hats were supplied after all. If you wish
to be a fool you may claim that. My students would never come down
that much.
It is sort of like trying to figure out what 0^0 is.
No. If for all intervals 1, 2, 3, ..., n the covering is 1/10, then
there are no natnumbers outside of all intervals and there are no hats outside of all intervals.
Regards, WM
On 29.11.2024 01:06, Richard Damon wrote:
On 11/28/24 12:50 PM, WM wrote:
If for all intervals 1, 2, 3, ..., n the covering is 1/10, then there
are no natnumbers outside of all intervals and there are no hats
outside of all intervals.
You are making the error of assuming that the infinite set is just
like a finite set that has part of it.
No. Analysis concerns infinite sequences and sets.
The problem is that the actual problem is defined on the INFINITE set,
and in that case, there ARE enough hats to cover.
No. The limit of the sequence f(n) of relative coverings in (0, n] is
1/10, not 1. Therefore the relative covering 1 would contradict analysis.
Regards, WM
On 11/28/24 12:50 PM, WM wrote:
If for all intervals 1, 2, 3, ..., n the covering is 1/10, then
there are no natnumbers outside of all intervals and there are no hats
outside of all intervals.
You are making the error of assuming that the infinite set is just like
a finite set that has part of it.
The problem is that the actual problem is defined on the INFINITE set,
and in that case, there ARE enough hats to cover.
On 11/29/24 8:44 AM, WM wrote:
On 29.11.2024 01:06, Richard Damon wrote:
On 11/28/24 12:50 PM, WM wrote:
If for all intervals 1, 2, 3, ..., n the covering is 1/10, then
there are no natnumbers outside of all intervals and there are no hats >>>> outside of all intervals.
You are making the error of assuming that the infinite set is just
like a finite set that has part of it.
No. Analysis concerns infinite sequences and sets.
You are looking at FINITE sets, and then trying to extrapolate to an
infinte set, which doesn't work.
On 11/29/24 8:44 AM, WM wrote:
On 29.11.2024 01:06, Richard Damon wrote:
On 11/28/24 12:50 PM, WM wrote:
If for all intervals 1, 2, 3, ..., n the covering is 1/10, then
there are no natnumbers outside of all intervals and there are no hats >>>> outside of all intervals.
You are making the error of assuming that the infinite set is just
like a finite set that has part of it.
No. Analysis concerns infinite sequences and sets.
You are looking at FINITE sets, and then trying to extrapolate to an
infinte set, which doesn't work.
The problem is that the actual problem is defined on the INFINITE
set, and in that case, there ARE enough hats to cover.
No. The limit of the sequence f(n) of relative coverings in (0, n] is
1/10, not 1. Therefore the relative covering 1 would contradict analysis.
And 0^x is 0, and x^0 is 1, which shows that just because you have a
constant sequence, it limit is not necessarily the final value.
On 29.11.2024 14:57, Richard Damon wrote:
On 11/29/24 8:44 AM, WM wrote:
On 29.11.2024 01:06, Richard Damon wrote:
On 11/28/24 12:50 PM, WM wrote:
If for all intervals 1, 2, 3, ..., n the covering is 1/10, then
there are no natnumbers outside of all intervals and there are no hats >>>>> outside of all intervals.
You are making the error of assuming that the infinite set is just
like a finite set that has part of it.
No. Analysis concerns infinite sequences and sets.
You are looking at FINITE sets, and then trying to extrapolate to an
infinte set, which doesn't work.
Analysis shows the way. If it differs from set theory, then one of both
is wrong.
Regards, WM
On 11/29/24 9:10 AM, WM wrote:
On 29.11.2024 14:57, Richard Damon wrote:But set theory doesn't say anything that you are saying.
On 11/29/24 8:44 AM, WM wrote:
On 29.11.2024 01:06, Richard Damon wrote:
On 11/28/24 12:50 PM, WM wrote:
If for all intervals 1, 2, 3, ..., n the covering is 1/10, then
there are no natnumbers outside of all intervals and there are no
hats
outside of all intervals.
You are making the error of assuming that the infinite set is just
like a finite set that has part of it.
No. Analysis concerns infinite sequences and sets.
You are looking at FINITE sets, and then trying to extrapolate to an
infinte set, which doesn't work.
Analysis shows the way. If it differs from set theory, then one of
both is wrong.
Nothing in set theory talks about the equivalence of properties of the infinite set to the properties of the finite sets
On 29.11.2024 14:57, Richard Damon wrote:
On 11/29/24 8:44 AM, WM wrote:
On 29.11.2024 01:06, Richard Damon wrote:
On 11/28/24 12:50 PM, WM wrote:
If for all intervals 1, 2, 3, ..., n the covering is 1/10, then
there are no natnumbers outside of all intervals and there are no hats >>>>> outside of all intervals.
You are making the error of assuming that the infinite set is just
like a finite set that has part of it.
No. Analysis concerns infinite sequences and sets.
You are looking at FINITE sets, and then trying to extrapolate to an
infinte set, which doesn't work.
Analysis is basic.
The problem is that the actual problem is defined on the INFINITE
set, and in that case, there ARE enough hats to cover.
No. The limit of the sequence f(n) of relative coverings in (0, n] is
1/10, not 1. Therefore the relative covering 1 would contradict
analysis.
And 0^x is 0, and x^0 is 1, which shows that just because you have a
constant sequence, it limit is not necessarily the final value.
The limit of the sequence 1/9, 1/9, 1/9, ... is 1/9.
Regards, WM
On 29.11.2024 16:16, Richard Damon wrote:
On 11/29/24 9:10 AM, WM wrote:
On 29.11.2024 14:57, Richard Damon wrote:But set theory doesn't say anything that you are saying.
On 11/29/24 8:44 AM, WM wrote:
On 29.11.2024 01:06, Richard Damon wrote:
On 11/28/24 12:50 PM, WM wrote:
If for all intervals 1, 2, 3, ..., n the covering is 1/10, then
there are no natnumbers outside of all intervals and there are no >>>>>>> hats
outside of all intervals.
You are making the error of assuming that the infinite set is just >>>>>> like a finite set that has part of it.
No. Analysis concerns infinite sequences and sets.
You are looking at FINITE sets, and then trying to extrapolate to an
infinte set, which doesn't work.
Analysis shows the way. If it differs from set theory, then one of
both is wrong.
Set theory says that the sets ℕ = {1, 2, 3, ...} and D = {10n | n ∈ ℕ} can be bijected by clever mapping. That means there are as many n as 10n
as black hats. That means by clever shifting the black hats they will
cover all n.
Nothing in set theory talks about the equivalence of properties of the
infinite set to the properties of the finite sets
That is the field of analysis. I believe that application of analysis provides correct results for the infinite.
Regards, WM
On 29.11.2024 19:56, Richard Damon wrote:
On 11/29/24 12:38 PM, WM wrote:
Set theory says that the sets ℕ = {1, 2, 3, ...} and D = {10n | n ∈
ℕ} can be bijected by clever mapping. That means there are as many n
as 10n as black hats. That means by clever shifting the black hats
they will cover all n.
Yep, that's what it says, and it is true.
Note, NOTHING in that statement talks about working with finite subset
of those sets,
That's why set theory is in contradiction with mathematics.
So, you are just shown to be using the logic of lies.
Analysis is no a lie.
Nothing in set theory talks about the equivalence of properties of
the infinite set to the properties of the finite sets
That is the field of analysis. I believe that application of analysis
provides correct results for the infinite.
Apparently your concept of it doesn't, because you are using broken
rules,
No, I am using a very simple and sound rule. If all hats of finite
intervals (0, n] fail to cover more than 1/10, then it is impossible to
cover more than 1/10 of the whole set ℕ because beyond all finite
intervals and all finite n, there is no supply of black hats. Even a
moderate brain should see that.
Regards, WM
maybe because you don't actually understand how to do analysis.
Sorry, starting from lies just gets you more lies.
Regards, WM
On 11/29/24 12:38 PM, WM wrote:
Set theory says that the sets ℕ = {1, 2, 3, ...} and D = {10n | n ∈ ℕ} >> can be bijected by clever mapping. That means there are as many n as
10n as black hats. That means by clever shifting the black hats they
will cover all n.
Yep, that's what it says, and it is true.
Note, NOTHING in that statement talks about working with finite subset
of those sets,
So, you are just shown to be using the logic of lies.
Nothing in set theory talks about the equivalence of properties of
the infinite set to the properties of the finite sets
That is the field of analysis. I believe that application of analysis
provides correct results for the infinite.
Apparently your concept of it doesn't, because you are using broken
rules,
Sorry, starting from lies just gets you more lies.
Regards, WM
On 29.11.2024 21:07, Richard Damon wrote:
On 11/29/24 2:28 PM, WM wrote:
Analysis is no a lie.
Bad Analysis, like what you do is.
The limit of the infinite sequence 1/9, 1/9, 1/9, ... is 1/9. Nothing is clearer than that.
No, I am using a very simple and sound rule. If all hats of finite
intervals (0, n] fail to cover more than 1/10, then it is impossible
to cover more than 1/10 of the whole set ℕ because beyond all finite
intervals and all finite n, there is no supply of black hats.
So, you admit to MAKING UP your rules based on your own ideas
This chain of arguing is irrefutable by consistent thinking.
Regards, WM
On 11/29/24 2:28 PM, WM wrote:
Analysis is no a lie.
Bad Analysis, like what you do is.
No, I am using a very simple and sound rule. If all hats of finite
intervals (0, n] fail to cover more than 1/10, then it is impossible
to cover more than 1/10 of the whole set ℕ because beyond all finite
intervals and all finite n, there is no supply of black hats.
So, you admit to MAKING UP your rules based on your own ideas
On 11/29/24 4:39 PM, WM wrote:
On 29.11.2024 21:07, Richard Damon wrote:
On 11/29/24 2:28 PM, WM wrote:
Analysis is no a lie.
Bad Analysis, like what you do is.
The limit of the infinite sequence 1/9, 1/9, 1/9, ... is 1/9. Nothing
is clearer than that.
Which doesn't actually mean anything.
The limit x-> 0 of 0^x is 0 (as it is for all x)
The limit x->0 of x^0 is 1 (as it is for all x)
But 0^0 isn't defined, even though both of the limits seem to apprach it.
You can only use the limit as the final result if it actually applies.
Since the Infinite set is actually the same kind as any of the finite
sets in the sequence, the limit doesn't apply
intervals (0, n] fail to cover more than 1/10, then it is impossible
to cover more than 1/10 of the whole set ℕ because beyond all finite >>>> intervals and all finite n, there is no supply of black hats.
So, you admit to MAKING UP your rules based on your own ideas
This chain of arguing is irrefutable by consistent thinking.
No, it is based on inconsistent thinking
On 29.11.2024 23:05, Richard Damon wrote:
On 11/29/24 4:39 PM, WM wrote:
On 29.11.2024 21:07, Richard Damon wrote:
On 11/29/24 2:28 PM, WM wrote:
Analysis is no a lie.
Bad Analysis, like what you do is.
The limit of the infinite sequence 1/9, 1/9, 1/9, ... is 1/9. Nothing
is clearer than that.
Which doesn't actually mean anything.
The limit x-> 0 of 0^x is 0 (as it is for all x)
The limit x->0 of x^0 is 1 (as it is for all x)
But 0^0 isn't defined, even though both of the limits seem to apprach it.
1/9, 1/9, 1/9, ... has the unique limit 1/9.
You can only use the limit as the final result if it actually applies.
Since the Infinite set is actually the same kind as any of the finite
sets in the sequence, the limit doesn't apply
It does.
Further>>>> I am using a very simple and sound rule. If all hats of finite
intervals (0, n] fail to cover more than 1/10, then it is
impossible to cover more than 1/10 of the whole set ℕ because
beyond all finite intervals and all finite n, there is no supply of
black hats.
So, you admit to MAKING UP your rules based on your own ideas
This chain of arguing is irrefutable by consistent thinking.
No, it is based on inconsistent thinking
That is yours.
Regards, WM
On 2024-11-27 11:10:51 +0000, WM said:
On 27.11.2024 10:33, Mikko wrote:
On 2024-11-26 11:07:57 +0000, WM said:
On 26.11.2024 10:09, Mikko wrote:If so, no bijection is contradicted.
On 2024-11-25 14:38:13 +0000, WM said:It shows that the mapping claimed to be a bijection is not a bijection. >>>
The simple example contradicts a bijection between the two sets
described above.
What does "contradicts a bijection" mean?
The possibility of a bijection between the sets ℕ = {1, 2, 3, ...}
and D = {10n | n ∈ ℕ} is contradicted.
No, it is not. You merely deny it, disregarding obvious facts.
The function
f(x) = 10 * f obviously maps every element of ℕ to a different element of
D and there is no element of D that is not 10 * f for some f so this f is
a bijection between ℕ and D.
On 27.11.2024 10:33, Mikko wrote:
On 2024-11-26 11:07:57 +0000, WM said:
On 26.11.2024 10:09, Mikko wrote:
On 2024-11-25 14:38:13 +0000, WM said:It shows that the mapping claimed to be a bijection is not a bijection.
The simple example contradicts a bijection between the two sets
described above.
What does "contradicts a bijection" mean?
If so, no bijection is contradicted.
The possibility of a bijection between the sets ℕ = {1, 2, 3, ...} and
D = {10n | n ∈ ℕ} is contradicted.
On 01.12.2024 11:17, Mikko wrote:
On 2024-11-27 11:10:51 +0000, WM said:
On 27.11.2024 10:33, Mikko wrote:
On 2024-11-26 11:07:57 +0000, WM said:
On 26.11.2024 10:09, Mikko wrote:
On 2024-11-25 14:38:13 +0000, WM said:It shows that the mapping claimed to be a bijection is not a
The simple example contradicts a bijection between the two sets
described above.
What does "contradicts a bijection" mean?
bijection.
If so, no bijection is contradicted.
The possibility of a bijection between the sets ℕ = {1, 2, 3, ...}
and D = {10n | n ∈ ℕ} is contradicted.
No, it is not. You merely deny it, disregarding obvious facts.
Obvious is that for every interval (0, n] the relative covering is 1/10,
and that there are no further black hats beyond all natnumbers n.
The function
f(x) = 10 * f obviously maps every element of ℕ to a different element of >> D and there is no element of D that is not 10 * f for some f so this f is
a bijection between ℕ and D.
It appears so. I have shown by a different example that it is wrong. The relative covering for every interval is 1/10, independent of the configuration of the hats available inside. The limit of this sequence
is 1/10.
Regards, WM
On 12/1/24 5:55 AM, WM wrote:
The relative covering for every interval is 1/10, independent of the
configuration of the hats available inside. The limit of this sequence
is 1/10.
Which just shows that you are using naive mathematics that is just inconsistant.
On 01.12.2024 13:14, Richard Damon wrote:
On 12/1/24 5:55 AM, WM wrote:
The relative covering for every interval is 1/10, independent of the
configuration of the hats available inside. The limit of this
sequence is 1/10.
Which just shows that you are using naive mathematics that is just
inconsistant.
Mathematics is consistent, set theory is not.
Regards, WM
On 01.12.2024 11:17, Mikko wrote:
On 2024-11-27 11:10:51 +0000, WM said:
On 27.11.2024 10:33, Mikko wrote:
On 2024-11-26 11:07:57 +0000, WM said:
On 26.11.2024 10:09, Mikko wrote:If so, no bijection is contradicted.
On 2024-11-25 14:38:13 +0000, WM said:It shows that the mapping claimed to be a bijection is not a bijection. >>>>
The simple example contradicts a bijection between the two sets
described above.
What does "contradicts a bijection" mean?
The possibility of a bijection between the sets ℕ = {1, 2, 3, ...} and >>> D = {10n | n ∈ ℕ} is contradicted.
No, it is not. You merely deny it, disregarding obvious facts.
Obvious is that for every interval (0, n] the relative covering is
1/10, and that there are no further black hats beyond all natnumbers n.
On 2024-12-01 10:55:15 +0000, WM said:
What does "contradicts a bijection" mean?It shows that the mapping claimed to be a bijection is not a
bijection.
If so, no bijection is contradicted.
The possibility of a bijection between the sets ℕ = {1, 2, 3, ...} >>>> and D = {10n | n ∈ ℕ} is contradicted.
No, it is not. You merely deny it, disregarding obvious facts.
Obvious is that for every interval (0, n] the relative covering is
1/10, and that there are no further black hats beyond all natnumbers n.
Irrelevant to everything quoted above.
On 12/1/24 11:50 AM, WM wrote:
On 01.12.2024 13:14, Richard Damon wrote:
On 12/1/24 5:55 AM, WM wrote:
The relative covering for every interval is 1/10, independent of the
configuration of the hats available inside. The limit of this
sequence is 1/10.
Which just shows that you are using naive mathematics that is just
inconsistant.
Mathematics is consistent, set theory is not.
But Mathematics is based on set theory,
On 01.12.2024 18:38, Richard Damon wrote:
On 12/1/24 11:50 AM, WM wrote:
On 01.12.2024 13:14, Richard Damon wrote:
On 12/1/24 5:55 AM, WM wrote:
The relative covering for every interval is 1/10, independent of
the configuration of the hats available inside. The limit of this
sequence is 1/10.
Which just shows that you are using naive mathematics that is just
inconsistant.
Mathematics is consistent, set theory is not.
But Mathematics is based on set theory,
Not at all. This is only claimed by set theorists. Mathematics is based
upon potential infinity. Merely some symbols of finite set theory like ∈ ℕ ∪ ∩ ⊆ have turned out useful.
Regards, WM
On 02.12.2024 09:41, Mikko wrote:
On 2024-12-01 10:55:15 +0000, WM said:
If so, no bijection is contradicted.What does "contradicts a bijection" mean?It shows that the mapping claimed to be a bijection is not a bijection. >>>>>>
The *claim* that a bijection is possible is disproved.
The possibility of a bijection between the sets ℕ = {1, 2, 3, ...} and
D = {10n | n ∈ ℕ} is contradicted.
No, it is not. You merely deny it, disregarding obvious facts.
Obvious is that for every interval (0, n] the relative covering is
1/10, and that there are no further black hats beyond all natnumbers n.
Irrelevant to everything quoted above.
You are unable to understand?
I understand mathematics
On 03.12.2024 11:03, Mikko wrote:
I understand mathematics
Hardly, though you may have the impression.
Proof: You cannot understand that the function
f(10n) = n from D = {10n | n ∈ ℕ} to ℕ = {1, 2, 3, ...} is not a bijection because for every initial segment {1, 2, 3, ..., n} of ℕ there are too few numbers of the form 10n that can be paired with numbers n.
Since ℕ is the union (or the limit) of the sequence of initial segments
{1, 2, 3, ..., n}, there are too few numbers of the form 10n in ℕ that
can be paired with numbers n.
Regards, WM
On 01.12.2024 18:38, Richard Damon wrote:Lol. Mathematics is axiomatised as ZFC, no mention of „potential”.
On 12/1/24 11:50 AM, WM wrote:Not at all. This is only claimed by set theorists. Mathematics is based
On 01.12.2024 13:14, Richard Damon wrote:But Mathematics is based on set theory,
On 12/1/24 5:55 AM, WM wrote:Mathematics is consistent, set theory is not.
The relative covering for every interval is 1/10, independent of the >>>>> configuration of the hats available inside. The limit of thisWhich just shows that you are using naive mathematics that is just
sequence is 1/10.
inconsistant.
upon potential infinity. Merely some symbols of finite set theory like ∈ ℕ ∪ ∩ ⊆ have turned out useful.
On 03.12.2024 11:03, Mikko wrote:Wow. For every segment there are numbers {10, 20, …, 10n}.
I understand mathematicsHardly, though you may have the impression.
Proof: You cannot understand that the function f(10n) = n from D = {10n
| n ∈ ℕ} to ℕ = {1, 2, 3, ...} is not a bijection because for every initial segment {1, 2, 3, ..., n} of ℕ there are too few numbers of the form 10n that can be paired with numbers n.
On 29.11.2024 19:56, Richard Damon wrote:Because set theory is not mathematics and not finite?
On 11/29/24 12:38 PM, WM wrote:
That's why set theory is in contradiction with mathematics.Set theory says that the sets ℕ = {1, 2, 3, ...} and D = {10n | n ∈ ℕ}Yep, that's what it says, and it is true.
can be bijected by clever mapping. That means there are as many n as
10n as black hats. That means by clever shifting the black hats they
will cover all n.
Note, NOTHING in that statement talks about working with finite subset
of those sets,
Of course they „fail”, they are a subset.No, I am using a very simple and sound rule. If all hats of finiteApparently your concept of it doesn't, because you are using brokenNothing in set theory talks about the equivalence of properties ofThat is the field of analysis. I believe that application of analysis
the infinite set to the properties of the finite sets
provides correct results for the infinite.
rules,
intervals (0, n] fail to cover more than 1/10,
Am Tue, 03 Dec 2024 12:10:53 +0100 schrieb WM:
On 03.12.2024 11:03, Mikko wrote:Wow. For every segment there are numbers {10, 20, …, 10n}.
I understand mathematicsHardly, though you may have the impression.
Proof: You cannot understand that the function f(10n) = n from D = {10n
| n ∈ ℕ} to ℕ = {1, 2, 3, ...} is not a bijection because for every
initial segment {1, 2, 3, ..., n} of ℕ there are too few numbers of the
form 10n that can be paired with numbers n.
Am Mon, 02 Dec 2024 15:52:16 +0100 schrieb WM:
On 01.12.2024 18:38, Richard Damon wrote:Lol. Mathematics is axiomatised as ZFC, no mention of „potential”.
On 12/1/24 11:50 AM, WM wrote:Not at all. This is only claimed by set theorists. Mathematics is based
On 01.12.2024 13:14, Richard Damon wrote:But Mathematics is based on set theory,
On 12/1/24 5:55 AM, WM wrote:Mathematics is consistent, set theory is not.
The relative covering for every interval is 1/10, independent of the >>>>>> configuration of the hats available inside. The limit of thisWhich just shows that you are using naive mathematics that is just
sequence is 1/10.
inconsistant.
upon potential infinity. Merely some symbols of finite set theory like ∈ >> ℕ ∪ ∩ ⊆ have turned out useful.
Am Fri, 29 Nov 2024 20:28:02 +0100 schrieb WM:
No, I am using a very simple and sound rule. If all hats of finiteOf course they „fail”, they are a subset.
intervals (0, n] fail to cover more than 1/10,
On 08.12.2024 21:29, joes wrote:
Am Mon, 02 Dec 2024 15:52:16 +0100 schrieb WM:
On 01.12.2024 18:38, Richard Damon wrote:Lol. Mathematics is axiomatised as ZFC, no mention of „potential”.
On 12/1/24 11:50 AM, WM wrote:Not at all. This is only claimed by set theorists. Mathematics is based
On 01.12.2024 13:14, Richard Damon wrote:But Mathematics is based on set theory,
On 12/1/24 5:55 AM, WM wrote:Mathematics is consistent, set theory is not.
The relative covering for every interval is 1/10, independent of the >>>>>>> configuration of the hats available inside. The limit of thisWhich just shows that you are using naive mathematics that is just >>>>>> inconsistant.
sequence is 1/10.
upon potential infinity. Merely some symbols of finite set theory like ∈ >>> ℕ ∪ ∩ ⊆ have turned out useful.
That is a lie of matheologians.
Regards, WM
On 08.12.2024 21:22, joes wrote:
Am Tue, 03 Dec 2024 12:10:53 +0100 schrieb WM:
On 03.12.2024 11:03, Mikko wrote:Wow. For every segment there are numbers {10, 20, …, 10n}.
I understand mathematicsHardly, though you may have the impression.
Proof: You cannot understand that the function f(10n) = n from D = {10n
| n ∈ ℕ} to ℕ = {1, 2, 3, ...} is not a bijection because for every >>> initial segment {1, 2, 3, ..., n} of ℕ there are too few numbers of the >>> form 10n that can be paired with numbers n.
But for every segments more are needed: {1, 2, 3, ..., 10n}
Regards, WM
The pairing is between TWO sets, not the members of a set with itself.
On 10.12.2024 13:19, Richard Damon wrote:
The pairing is between TWO sets, not the members of a set with itself.
The pairing is between the elements. Otherwise you could pair R and Q by simply claiming it.
"The infinite sequence thus defined has the peculiar property to contain
the positive rational numbers completely, and each of them only once at
a determined place." [Cantor] Note the numbers, not the set.
Regards, WM
WM <[email protected]> wrote:
On 10.12.2024 13:19, Richard Damon wrote:
The pairing is between TWO sets, not the members of a set with itself.
The pairing is between the elements. Otherwise you could pair R and Q by
simply claiming it.
"The infinite sequence thus defined has the peculiar property to contain
the positive rational numbers completely, and each of them only once at
a determined place." [Cantor] Note the numbers, not the set.
TWO different sets, not the elements of a set and some of the elements of that same set.
On 11.12.2024 01:25, Richard Damon wrote:
WM <[email protected]> wrote:
On 10.12.2024 13:19, Richard Damon wrote:
The pairing is between TWO sets, not the members of a set with itself.
The pairing is between the elements. Otherwise you could pair R and Q by >>> simply claiming it.
"The infinite sequence thus defined has the peculiar property to contain >>> the positive rational numbers completely, and each of them only once at
a determined place." [Cantor] Note the numbers, not the set.
TWO different sets, not the elements of a set and some of the elements of
that same set.
In mathematics, a set A is Dedekind-infinite (named after the German mathematician Richard Dedekind) if some proper subset B of A is
equinumerous to A. [Wikipedia].
Regards, WM
On 12/11/24 9:04 AM, WM wrote:
In mathematics, a set A is Dedekind-infinite (named after the German
mathematician Richard Dedekind) if some proper subset B of A is
equinumerous to A. [Wikipedia].
So? That isn't what Cantor was talking about in his pairings
On 12.12.2024 01:38, Richard Damon wrote:
On 12/11/24 9:04 AM, WM wrote:
In mathematics, a set A is Dedekind-infinite (named after the German
mathematician Richard Dedekind) if some proper subset B of A is
equinumerous to A. [Wikipedia].
So? That isn't what Cantor was talking about in his pairings
It is precisely this.
Regards, WM
On 10.12.2024 13:19, Richard Damon wrote:What Richard meant: do not confuse the set being mapped with the one being mapped onto.
The pairing is between TWO sets, not the members of a set with itself.
The pairing is between the elements. Otherwise you could pair R and Q by simply claiming it.
"The infinite sequence thus defined has the peculiar property to contain
the positive rational numbers completely, and each of them only once at
a determined place." [Cantor] Note the numbers, not the set.
On 12/12/24 4:53 AM, WM wrote:
On 12.12.2024 01:38, Richard Damon wrote:
On 12/11/24 9:04 AM, WM wrote:
In mathematics, a set A is Dedekind-infinite (named after the German
mathematician Richard Dedekind) if some proper subset B of A is
equinumerous to A. [Wikipedia].
So? That isn't what Cantor was talking about in his pairings
It is precisely this.
No, Cantors pairing is between two SETS, not a set and its subset.
Yes, we can call the subset a set, since it is, but then when we look at
it for the pairing, we need to be looking at its emancipated version,
not the version tied into the original set.
By your logic, *NO* set can be infinite,
as no proper subset can be
equinumerous to its superset.
Am Thu, 12 Dec 2024 10:12:26 +0100 schrieb WM:
The sequence is endless, has no end, is infinite.The end of the sequence is defined by ∀k ∈ ℕ : E(k+1) = E(k) \ {k}.
None of which are an infinite sets, so trying to take a "limit" of
combining them is just improper.
Most endsegments are infinite. But if Cantor can apply all natural
numbers as indices for his sequences, then all must leave the sequence
of endsegments. Then the sequence (E(k)) must end up empty. And there
must be a continuous staircase from E(k) to the empty set.
It makes no sense not being able to „apply” numbers. Clearly Cantor does.
The sequence IS continuous. It’s just that you misconceive of the
limit as reachable.
Am Tue, 10 Dec 2024 18:01:04 +0100 schrieb WM:
On 10.12.2024 13:19, Richard Damon wrote:What Richard meant: do not confuse the set being mapped with the one being mapped onto.
The pairing is between TWO sets, not the members of a set with itself.
The pairing is between the elements. Otherwise you could pair R and Q by
simply claiming it.
"The infinite sequence thus defined has the peculiar property to contain
the positive rational numbers completely, and each of them only once at
a determined place." [Cantor] Note the numbers, not the set.
On 11.12.2024 01:25, Richard Damon wrote:
WM <[email protected]> wrote:
On 10.12.2024 13:19, Richard Damon wrote:
The pairing is between TWO sets, not the members of a set with itself.
The pairing is between the elements. Otherwise you could pair R and Q by >>> simply claiming it.
"The infinite sequence thus defined has the peculiar property to contain >>> the positive rational numbers completely, and each of them only once at
a determined place." [Cantor] Note the numbers, not the set.
TWO different sets, not the elements of a set and some of the elements of
that same set.
In mathematics, a set A is Dedekind-infinite (named after the German mathematician Richard Dedekind) if some proper subset B of A is
equinumerous to A. [Wikipedia].
What Richard meant: do not confuse the set being mapped with the one being mapped onto.
On 12/12/2024 6:59 AM, joes wrote:
Am Tue, 10 Dec 2024 18:01:04 +0100 schrieb WM:
On 10.12.2024 13:19, Richard Damon wrote:What Richard meant: do not confuse the set being mapped with the one
The pairing is between TWO sets, not the members of a set with itself.
The pairing is between the elements. Otherwise you could pair R and Q by >>> simply claiming it.
"The infinite sequence thus defined has the peculiar property to contain >>> the positive rational numbers completely, and each of them only once at
a determined place." [Cantor] Note the numbers, not the set.
being
mapped onto.
But that's sort of what mappings are for! Aren't they?
On 2024-12-11 14:04:30 +0000, WM said:
On 11.12.2024 01:25, Richard Damon wrote:
WM <[email protected]> wrote:
On 10.12.2024 13:19, Richard Damon wrote:
The pairing is between the elements. Otherwise you could pair R and
The pairing is between TWO sets, not the members of a set with itself. >>>>
Q by
simply claiming it.
"The infinite sequence thus defined has the peculiar property to
contain
the positive rational numbers completely, and each of them only once at >>>> a determined place." [Cantor] Note the numbers, not the set.
TWO different sets, not the elements of a set and some of the
elements of
that same set.
In mathematics, a set A is Dedekind-infinite (named after the German
mathematician Richard Dedekind) if some proper subset B of A is
equinumerous to A. [Wikipedia].
Do you happen to know any set that is Dedekind-infinite?
On 12.12.2024 14:59, joes wrote:
What Richard meant: do not confuse the set being mapped with the oneI don't. D = {10n | n ∈ ℕ} is the set being mapped. The set D being mapped does not change when it is attached to the set ℕ being mapped in form of black hats.
being
mapped onto.
Regards, WM
On 12/12/24 4:57 PM, WM wrote:
D = {10n | n ∈ ℕ} is the set being mapped. The set D being
mapped does not change when it is attached to the set ℕ being mapped
in form of black hats.
And so, which element of which set didn't get mapped to a member of the
other by the defined mapping?
On 12/12/24 9:44 AM, WM wrote:
On 12.12.2024 13:26, Richard Damon wrote:
On 12/12/24 4:53 AM, WM wrote:
On 12.12.2024 01:38, Richard Damon wrote:
On 12/11/24 9:04 AM, WM wrote:
In mathematics, a set A is Dedekind-infinite (named after the
German mathematician Richard Dedekind) if some proper subset B of
A is equinumerous to A. [Wikipedia].
So? That isn't what Cantor was talking about in his pairings
It is precisely this.
No, Cantors pairing is between two SETS, not a set and its subset.
Yes, we can call the subset a set, since it is, but then when we look
at it for the pairing, we need to be looking at its emancipated
version, not the version tied into the original set.
Both is the same. In emancipated version it is not as obvious as in
the subset version.
Nope, when the subset is considered as its own independent set, the
operation you want to do isn't part of its operations.
On 12.12.2024 18:48, Mikko wrote:
On 2024-12-11 14:04:30 +0000, WM said:No, there is no such set. This is proven by my black hats = numbers of
On 11.12.2024 01:25, Richard Damon wrote:
WM <[email protected]> wrote:
On 10.12.2024 13:19, Richard Damon wrote:
The pairing is between TWO sets, not the members of a set with
itself.
The pairing is between the elements. Otherwise you could pair R and
Q by
simply claiming it.
"The infinite sequence thus defined has the peculiar property to
contain
the positive rational numbers completely, and each of them only
once at
a determined place." [Cantor] Note the numbers, not the set.
TWO different sets, not the elements of a set and some of the
elements of
that same set.
In mathematics, a set A is Dedekind-infinite (named after the German
mathematician Richard Dedekind) if some proper subset B of A is
equinumerous to A. [Wikipedia].
Do you happen to know any set that is Dedekind-infinite?
the form 10n: For every interval [1, n] the relative covering is at most 1/10. And more than all intervals are not available to supply numbers of
the form 10n.
Regards, WM
Note, the pairing is not between some elements of N that are also in D,
with other elements in N, but the elements of D and the elements on N.
You just don' understand what it means to PAIR elements of two sets.
On 13.12.2024 13:11, Richard Damon wrote:
Note, the pairing is not between some elements of N that are also in
D, with other elements in N, but the elements of D and the elements on N.
Yes all elements of D, as black hats attached to the elements 10n of ℕ, have to get attached to all elements of ℕ. There the simple shift from
10n to n (division by 10) is applied. >
You just don' understand what it means to PAIR elements of two sets.
That pairs the elements of D with the elements of ℕ. Alas, it can be
proved that for every interval [1, n] the deficit of hats amounts to at
least 90 %. And beyond all n, there are no further hats.
Regards, WM
On 12/13/24 12:00 PM, WM wrote:
On 13.12.2024 13:11, Richard Damon wrote:
Note, the pairing is not between some elements of N that are also in
D, with other elements in N, but the elements of D and the elements
on N.
Yes all elements of D, as black hats attached to the elements 10n of
ℕ, have to get attached to all elements of ℕ. There the simple shift
from 10n to n (division by 10) is applied.
No, the black hats are attached to the element of D, not N.
That pairs the elements of D with the elements of ℕ. Alas, it can be
proved that for every interval [1, n] the deficit of hats amounts to
at least 90 %. And beyond all n, there are no further hats.
But we aren't dealing with intervals of [1, n] but of the full set.
The problem is that you can't GET to "beyond all n" in the pairing, as
there are always more n to get to.
Yes, there are only 1/10th as many Black Hats as White Hats, but since
that number is Aleph_0/10, which just happens to also equal Aleph_0,
there is no "deficit" in the set of Natual Numbers.
Your logic woud say that Aleph_0/10 would be some value between
(possible dark) Natural Numbers
On 19.11.2024 10:32, Mikko wrote:
On 2024-11-18 14:29:40 +0000, WM said:
On 18.11.2024 10:58, Mikko wrote:
On 2024-11-17 12:46:29 +0000, WM said:
There are 100 intervals for each natural number.
This can be proven by bijecting J'(100n) and J(n). My intervals are
then exhausted, yours are not.
Irrelevant.
Very relevant.
It is not relevant if no relevancy is shown.
But if relevancy is only deleted, it can show up again:
Every finite translation of any finite subset of intervals J(n)
maintains the relative covering 1/5. If the infinite set has the
relative covering 1 (or more), then you claim that the sequence 1/5,
1/5, 1/5, ... has limit 1 (or more).
On 12.12.2024 18:48, Mikko wrote:
On 2024-12-11 14:04:30 +0000, WM said:No, there is no such set.
On 11.12.2024 01:25, Richard Damon wrote:
WM <[email protected]> wrote:
On 10.12.2024 13:19, Richard Damon wrote:
The pairing is between the elements. Otherwise you could pair R and Q by >>>>> simply claiming it.
The pairing is between TWO sets, not the members of a set with itself. >>>>>
"The infinite sequence thus defined has the peculiar property to contain >>>>> the positive rational numbers completely, and each of them only once at >>>>> a determined place." [Cantor] Note the numbers, not the set.
TWO different sets, not the elements of a set and some of the elements of >>>> that same set.
In mathematics, a set A is Dedekind-infinite (named after the German
mathematician Richard Dedekind) if some proper subset B of A is
equinumerous to A. [Wikipedia].
Do you happen to know any set that is Dedekind-infinite?
On 2024-11-19 11:04:08 +0000, WM said:
On 19.11.2024 10:32, Mikko wrote:
On 2024-11-18 14:29:40 +0000, WM said:
On 18.11.2024 10:58, Mikko wrote:
On 2024-11-17 12:46:29 +0000, WM said:
There are 100 intervals for each natural number.
This can be proven by bijecting J'(100n) and J(n). My intervals
are then exhausted, yours are not.
Irrelevant.
Very relevant.
It is not relevant if no relevancy is shown.
But if relevancy is only deleted, it can show up again:
Every finite translation of any finite subset of intervals J(n)
maintains the relative covering 1/5. If the infinite set has the
relative covering 1 (or more), then you claim that the sequence 1/5,
1/5, 1/5, ... has limit 1 (or more).
There is a bijection between your J and my J', where
J'(n) = (n/100 - 1/10, n/100 + 1/10): for each n there
is one interval J(n) and one interval of J'(n). Whateever
you infer from that is either an invalid inference or
a true conclusion.
On 2024-12-12 22:06:58 +0000, WM said:
No, there is no such set.In mathematics, a set A is Dedekind-infinite (named after the German
mathematician Richard Dedekind) if some proper subset B of A is
equinumerous to A. [Wikipedia].
Do you happen to know any set that is Dedekind-infinite?
The set of natural numbers, if there is any such set,
is Dedekind-infinte:
the successor function is a bijection between the set of all natural
numbers and non-zero natural numbers.
On 14.12.2024 01:03, Richard Damon wrote:
On 12/13/24 12:00 PM, WM wrote:
On 13.12.2024 13:11, Richard Damon wrote:
Note, the pairing is not between some elements of N that are also in
D, with other elements in N, but the elements of D and the elements
on N.
Yes all elements of D, as black hats attached to the elements 10n of
ℕ, have to get attached to all elements of ℕ. There the simple shift >>> from 10n to n (division by 10) is applied.
No, the black hats are attached to the element of D, not N.
They are elements of D and become attached to elements of ℕ.
That pairs the elements of D with the elements of ℕ. Alas, it can be
proved that for every interval [1, n] the deficit of hats amounts to
at least 90 %. And beyond all n, there are no further hats.
But we aren't dealing with intervals of [1, n] but of the full set.
Those who try to forbid the detailed analysis are dishonest swindlers
and tricksters and not worth to participate in scientific discussion.
The problem is that you can't GET to "beyond all n" in the pairing, as
there are always more n to get to.
If this is impossible, then also Cantor cannot use all n.
Yes, there are only 1/10th as many Black Hats as White Hats, but since
that number is Aleph_0/10, which just happens to also equal Aleph_0,
there is no "deficit" in the set of Natual Numbers.
This example proves that aleph_0 is nonsense.
Your logic woud say that Aleph_0/10 would be some value between
(possible dark) Natural Numbers
My logic says that nonsense cannot be defended by accepting just this nonsense.
Regards, WM
On 12/14/24 3:53 AM, WM wrote:+
Please refer to the simplest example I gave you on 2024-11-27:
The possibility of a bijection between the sets ℕ = {1, 2, 3, ...} and
D = {10n | n ∈ ℕ} is contradicted because for every interval (0, n]
the relative covering is not more than 1/10, and there are no further
numbers 10n beyond all natural numbers n. The sequence 1/10, 1/10,
1/10, ... has limit 1/10.
Except that we aren't dealng with the FINITE sets of {1, 2, 3, ..., n}
but for the full set of { 1, 2, 3, ... }
On 12/14/24 3:38 AM, WM wrote:
On 14.12.2024 01:03, Richard Damon wrote:
On 12/13/24 12:00 PM, WM wrote:
On 13.12.2024 13:11, Richard Damon wrote:
Note, the pairing is not between some elements of N that are also
in D, with other elements in N, but the elements of D and the
elements on N.
Yes all elements of D, as black hats attached to the elements 10n of
ℕ, have to get attached to all elements of ℕ. There the simple shift >>>> from 10n to n (division by 10) is applied.
No, the black hats are attached to the element of D, not N.
They are elements of D and become attached to elements of ℕ.
No, they are PAIR with elements of N.
There is no operatation to "Attach" sets.
That pairs the elements of D with the elements of ℕ. Alas, it can be >>>> proved that for every interval [1, n] the deficit of hats amounts to
at least 90 %. And beyond all n, there are no further hats.
But we aren't dealing with intervals of [1, n] but of the full set.
Those who try to forbid the detailed analysis are dishonest swindlers
and tricksters and not worth to participate in scientific discussion.
No, we are not forbiding "detailed" analysis
The problem is that you can't GET to "beyond all n" in the pairing,
as there are always more n to get to.
If this is impossible, then also Cantor cannot use all n.
Why can't he? The problem is in the space of the full set, not the
finite sub sets.
Yes, there are only 1/10th as many Black Hats as White Hats, but
since that number is Aleph_0/10, which just happens to also equal
Aleph_0, there is no "deficit" in the set of Natual Numbers.
This example proves that aleph_0 is nonsense.
Nope, it proves it is incompatible with finite logic.
On 14.12.2024 15:08, Richard Damon wrote:
On 12/14/24 3:53 AM, WM wrote:
+Please refer to the simplest example I gave you on 2024-11-27:
The possibility of a bijection between the sets ℕ = {1, 2, 3, ...}
and D = {10n | n ∈ ℕ} is contradicted because for every interval (0, >>> n] the relative covering is not more than 1/10, and there are no
further numbers 10n beyond all natural numbers n. The sequence 1/10,
1/10, 1/10, ... has limit 1/10.
Except that we aren't dealng with the FINITE sets of {1, 2, 3, ..., n}
but for the full set of { 1, 2, 3, ... }
Of course. It consist of all finite n. It is the union of all intervals
(0, n]. Not more!
Regards, WM
On 14.12.2024 15:08, Richard Damon wrote:
On 12/14/24 3:38 AM, WM wrote:
On 14.12.2024 01:03, Richard Damon wrote:
On 12/13/24 12:00 PM, WM wrote:
On 13.12.2024 13:11, Richard Damon wrote:
Note, the pairing is not between some elements of N that are also
in D, with other elements in N, but the elements of D and the
elements on N.
Yes all elements of D, as black hats attached to the elements 10n
of ℕ, have to get attached to all elements of ℕ. There the simple >>>>> shift from 10n to n (division by 10) is applied.
No, the black hats are attached to the element of D, not N.
They are elements of D and become attached to elements of ℕ.
No, they are PAIR with elements of N.
There is no operatation to "Attach" sets.
To put a hat on n is to attach a hat to n.
That pairs the elements of D with the elements of ℕ. Alas, it canBut we aren't dealing with intervals of [1, n] but of the full set.
be proved that for every interval [1, n] the deficit of hats
amounts to at least 90 %. And beyond all n, there are no further hats. >>>>
Those who try to forbid the detailed analysis are dishonest swindlers
and tricksters and not worth to participate in scientific discussion.
No, we are not forbiding "detailed" analysis
Then deal with all infinitely many intervals [1, n].
The problem is that you can't GET to "beyond all n" in the pairing,
as there are always more n to get to.
If this is impossible, then also Cantor cannot use all n.
Why can't he? The problem is in the space of the full set, not the
finite sub sets.
The intervals [1, n] cover the full set.
Yes, there are only 1/10th as many Black Hats as White Hats, but
since that number is Aleph_0/10, which just happens to also equal
Aleph_0, there is no "deficit" in the set of Natual Numbers.
This example proves that aleph_0 is nonsense.
Nope, it proves it is incompatible with finite logic.
There is no other logic.
Regards, WM
On 12/14/24 10:55 AM, WM wrote:
On 14.12.2024 15:08, Richard Damon wrote:
On 12/14/24 3:53 AM, WM wrote:+
Please refer to the simplest example I gave you on 2024-11-27:
The possibility of a bijection between the sets ℕ = {1, 2, 3, ...}
and D = {10n | n ∈ ℕ} is contradicted because for every interval (0, >>>> n] the relative covering is not more than 1/10, and there are no
further numbers 10n beyond all natural numbers n. The sequence 1/10,
1/10, 1/10, ... has limit 1/10.
Except that we aren't dealng with the FINITE sets of {1, 2, 3, ...,
n} but for the full set of { 1, 2, 3, ... }
Of course. It consist of all finite n. It is the union of all
intervals (0, n]. Not more!
But the set of Natural Numbers isn't built that way.
On 14.12.2024 19:53, Richard Damon wrote:
On 12/14/24 10:55 AM, WM wrote:
On 14.12.2024 15:08, Richard Damon wrote:
On 12/14/24 3:53 AM, WM wrote:+
Please refer to the simplest example I gave you on 2024-11-27:
The possibility of a bijection between the sets ℕ = {1, 2, 3, ...} >>>>> and D = {10n | n ∈ ℕ} is contradicted because for every interval >>>>> (0, n] the relative covering is not more than 1/10, and there are
no further numbers 10n beyond all natural numbers n. The sequence
1/10, 1/10, 1/10, ... has limit 1/10.
Except that we aren't dealng with the FINITE sets of {1, 2, 3, ...,
n} but for the full set of { 1, 2, 3, ... }
Of course. It consist of all finite n. It is the union of all
intervals (0, n]. Not more!
But the set of Natural Numbers isn't built that way.
You ar5e misinformed. The set of natural numbers is the union of all
FISONs F(n) = {1, 2, 3, ..., n} = (0, n].
Regards, WM
On 14.12.2024 09:41, Mikko wrote:
On 2024-11-19 11:04:08 +0000, WM said:
On 19.11.2024 10:32, Mikko wrote:
On 2024-11-18 14:29:40 +0000, WM said:
On 18.11.2024 10:58, Mikko wrote:
On 2024-11-17 12:46:29 +0000, WM said:
There are 100 intervals for each natural number.
This can be proven by bijecting J'(100n) and J(n). My intervals are >>>>>>> then exhausted, yours are not.
Irrelevant.
Very relevant.
It is not relevant if no relevancy is shown.
But if relevancy is only deleted, it can show up again:
Every finite translation of any finite subset of intervals J(n)
maintains the relative covering 1/5. If the infinite set has the
relative covering 1 (or more), then you claim that the sequence 1/5,
1/5, 1/5, ... has limit 1 (or more).
There is a bijection between your J and my J', where
J'(n) = (n/100 - 1/10, n/100 + 1/10): for each n there
is one interval J(n) and one interval of J'(n). Whateever
you infer from that is either an invalid inference or
a true conclusion.
Please refer to the simplest example I gave you on 2024-11-27:
The possibility of a bijection between the sets ℕ = {1, 2, 3, ...} and
D = {10n | n ∈ ℕ} is contradicted because for every interval (0, n] the relative covering is not more than 1/10, and there are no further
numbers 10n beyond all natural numbers n.
The sequence 1/10, 1/10, 1/10, ... has limit 1/10.
On 14.12.2024 09:52, Mikko wrote:
On 2024-12-12 22:06:58 +0000, WM said:
No, there is no such set.In mathematics, a set A is Dedekind-infinite (named after the German >>>>> mathematician Richard Dedekind) if some proper subset B of A is
equinumerous to A. [Wikipedia].
Do you happen to know any set that is Dedekind-infinite?
The set of natural numbers, if there is any such set,
If ℕ is a set, i.e. if it is complete such that all numbers can be used
for indexing sequences or in other mappings, then it can also be
exhausted such that no element remains. Then the sequence of
intersections of endsegments
E(1), E(1)∩E(2), E(1)∩E(2)∩E(3), ...
loses all content. Then, by the law
∀k ∈ ℕ : ∩{E(1), E(2), ..., E(k+1)} = ∩{E(1), E(2), ..., E(k)} \ {k}
the content must become finite.
is Dedekind-infinte:
the successor function is a bijection between the set of all natural
numbers and non-zero natural numbers.
This "bijection" appears possible but it is not.
On 12/14/24 4:34 PM, WM wrote:
On 14.12.2024 19:53, Richard Damon wrote:From where did this concept come from?
On 12/14/24 10:55 AM, WM wrote:
On 14.12.2024 15:08, Richard Damon wrote:
On 12/14/24 3:53 AM, WM wrote:+
Please refer to the simplest example I gave you on 2024-11-27:
The possibility of a bijection between the sets ℕ = {1, 2, 3, ...} >>>>>> and D = {10n | n ∈ ℕ} is contradicted because for every interval >>>>>> (0, n] the relative covering is not more than 1/10, and there are
no further numbers 10n beyond all natural numbers n. The sequence
1/10, 1/10, 1/10, ... has limit 1/10.
Except that we aren't dealng with the FINITE sets of {1, 2, 3, ...,
n} but for the full set of { 1, 2, 3, ... }
Of course. It consist of all finite n. It is the union of all
intervals (0, n]. Not more!
But the set of Natural Numbers isn't built that way.
You are misinformed. The set of natural numbers is the union of all
FISONs F(n) = {1, 2, 3, ..., n} = (0, n].
On 2024-12-14 09:50:52 +0000, WM said:
On 14.12.2024 09:52, Mikko wrote:
On 2024-12-12 22:06:58 +0000, WM said:
No, there is no such set.In mathematics, a set A is Dedekind-infinite (named after the
German mathematician Richard Dedekind) if some proper subset B of
A is equinumerous to A. [Wikipedia].
Do you happen to know any set that is Dedekind-infinite?
The set of natural numbers, if there is any such set,
If ℕ is a set, i.e. if it is complete such that all numbers can be
used for indexing sequences or in other mappings, then it can also be
exhausted such that no element remains. Then the sequence of
intersections of endsegments
E(1), E(1)∩E(2), E(1)∩E(2)∩E(3), ...
loses all content. Then, by the law
∀k ∈ ℕ : ∩{E(1), E(2), ..., E(k+1)} = ∩{E(1), E(2), ..., E(k)} \ {k}
the content must become finite.
is Dedekind-infinte:
the successor function is a bijection between the set of all natural
numbers and non-zero natural numbers.
This "bijection" appears possible but it is not.
So you say that there is a natural number that does not have a next
natural number. What number is that?
On 2024-12-14 08:53:19 +0000, WM said:
Please refer to the simplest example I gave you on 2024-11-27:
The possibility of a bijection between the sets ℕ = {1, 2, 3, ...} and
D = {10n | n ∈ ℕ} is contradicted because for every interval (0, n]
the relative covering is not more than 1/10, and there are no further
numbers 10n beyond all natural numbers n.
It is already proven that there is such bijection. What is proven cannot
be contradicted unless you can prove that 1 = 2.
The sequence 1/10, 1/10, 1/10, ... has limit 1/10.
Irrelevant as the proof of the exitence of the bijection does not
mention that sequence.
On 14.12.2024 23:45, Richard Damon wrote:
On 12/14/24 4:34 PM, WM wrote:
On 14.12.2024 19:53, Richard Damon wrote:From where did this concept come from?
On 12/14/24 10:55 AM, WM wrote:
On 14.12.2024 15:08, Richard Damon wrote:
On 12/14/24 3:53 AM, WM wrote:+
Please refer to the simplest example I gave you on 2024-11-27:
The possibility of a bijection between the sets ℕ = {1, 2,
3, ...} and D = {10n | n ∈ ℕ} is contradicted because for every >>>>>>> interval (0, n] the relative covering is not more than 1/10, and >>>>>>> there are no further numbers 10n beyond all natural numbers n.
The sequence 1/10, 1/10, 1/10, ... has limit 1/10.
Except that we aren't dealng with the FINITE sets of {1, 2,
3, ..., n} but for the full set of { 1, 2, 3, ... }
Of course. It consist of all finite n. It is the union of all
intervals (0, n]. Not more!
But the set of Natural Numbers isn't built that way.
You are misinformed. The set of natural numbers is the union of all
FISONs F(n) = {1, 2, 3, ..., n} = (0, n].
The set of FISONs is the set of natural numbers designed by v. Neumann.
https://en.wikipedia.org/wiki/Set-theoretic_definition_of_natural_numbers
Regards, WM
On 15.12.2024 12:03, Mikko wrote:
On 2024-12-14 09:50:52 +0000, WM said:
On 14.12.2024 09:52, Mikko wrote:
On 2024-12-12 22:06:58 +0000, WM said:
No, there is no such set.In mathematics, a set A is Dedekind-infinite (named after the
German mathematician Richard Dedekind) if some proper subset B of >>>>>>> A is equinumerous to A. [Wikipedia].
Do you happen to know any set that is Dedekind-infinite?
The set of natural numbers, if there is any such set,
If ℕ is a set, i.e. if it is complete such that all numbers can be
used for indexing sequences or in other mappings, then it can also be
exhausted such that no element remains. Then the sequence of
intersections of endsegments
E(1), E(1)∩E(2), E(1)∩E(2)∩E(3), ...
loses all content. Then, by the law
∀k ∈ ℕ : ∩{E(1), E(2), ..., E(k+1)} = ∩{E(1), E(2), ..., E(k)} \ {k}
the content must become finite.
is Dedekind-infinte:
the successor function is a bijection between the set of all natural
numbers and non-zero natural numbers.
This "bijection" appears possible but it is not.
So you say that there is a natural number that does not have a next
natural number. What number is that?
We cannot name dark numbers as individuals. All numbers which can be
used a individuals belong to a potentially infinite collection ℕ_def.
There is no firm end. When n belongs to ℕ_def, then also n+1 and 2n and n^n^n belong to ℕ_def. The only common property is that all the numbers belong to a finite set and have an infinite set of dark successors.
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
|ℕ \ {1, 2, 3, ...}| = 0
This is the only way to explain that the intersection of endegments
E(1), E(1)∩E(2), E(1)∩E(2)∩E(3), ...
loses all content in a sequences which allow the loss of only one number
per step:
∀k ∈ ℕ : ∩{E(1), E(2), ..., E(k+1)} = ∩{E(1), E(2), ..., E(k)} \ {k}
Regards, WM
Note, v. Neuman starts at 0, not 1,
On 12/13/24 12:00 PM, WM wrote:
On 13.12.2024 13:11, Richard Damon wrote:
Note, the pairing is not between some elements of N that are also in
D, with other elements in N, but the elements of D and the elements
on N.
Yes all elements of D, as black hats attached to the elements 10n of
ℕ, have to get attached to all elements of ℕ. There the simple shift
from 10n to n (division by 10) is applied. >
No, the black hats are attached to the element of D, not N.
But we aren't dealing with intervals of [1, n] but of the full set.
You can't "name" your dark numbers,
You already admitted that the set of definable natural numbers was in
fact the set of all natural numbers,
so there is nothing left to be dark
On 15.12.2024 13:52, Richard Damon wrote:
Note, v. Neuman starts at 0, not 1,
That is irrelevant. He uses FISONs.
Regards, WM
On 15.12.2024 13:52, Richard Damon wrote:
On 12/13/24 12:00 PM, WM wrote:
On 13.12.2024 13:11, Richard Damon wrote:
Note, the pairing is not between some elements of N that are also in
D, with other elements in N, but the elements of D and the elements
on N.
Yes all elements of D, as black hats attached to the elements 10n of
ℕ, have to get attached to all elements of ℕ. There the simple shift >>> from 10n to n (division by 10) is applied. >
No, the black hats are attached to the element of D, not N.
The black hats _are_ the element of D.
But we aren't dealing with intervals of [1, n] but of the full set.
All intervals together are the full set.
Regards, WM
On 12/15/24 2:44 PM, WM wrote:
On 15.12.2024 13:54, Richard Damon wrote:
You can't "name" your dark numbers,
because they are dark.
|ℕ \ {1, 2, 3, ...}| = 0 cannot be accomplished by visible numbers
because ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
Which just shows that the full set in infinte, and any member in it is finite, and not the last member.
On 15.12.2024 11:56, Mikko wrote:
On 2024-12-14 08:53:19 +0000, WM said:
Please refer to the simplest example I gave you on 2024-11-27:
The possibility of a bijection between the sets ℕ = {1, 2, 3, ...} and >>> D = {10n | n ∈ ℕ} is contradicted because for every interval (0, n] the >>> relative covering is not more than 1/10, and there are no further
numbers 10n beyond all natural numbers n.
It is already proven that there is such bijection. What is proven cannot
be contradicted unless you can prove that 1 = 2.
What is proven under false (self-contradictory) premises can be shown
to be false.
Here we have a limit of 1/10 from analysis and a limit of 0 from set
theory. That shows that if set theory is right, we have
1/10 = 0 ==> 1 = 0 ==> 2 = 1.
On 14.12.2024 23:45, Richard Damon wrote:
On 12/14/24 4:34 PM, WM wrote:
On 14.12.2024 19:53, Richard Damon wrote:From where did this concept come from?
On 12/14/24 10:55 AM, WM wrote:
On 14.12.2024 15:08, Richard Damon wrote:
On 12/14/24 3:53 AM, WM wrote:+
Please refer to the simplest example I gave you on 2024-11-27:
The possibility of a bijection between the sets ℕ = {1, 2, 3, ...} and
D = {10n | n ∈ ℕ} is contradicted because for every interval (0, n] the
relative covering is not more than 1/10, and there are no further >>>>>>> numbers 10n beyond all natural numbers n. The sequence 1/10, 1/10, >>>>>>> 1/10, ... has limit 1/10.
Except that we aren't dealng with the FINITE sets of {1, 2, 3, ..., n} >>>>>> but for the full set of { 1, 2, 3, ... }
Of course. It consist of all finite n. It is the union of all intervals >>>>> (0, n]. Not more!
But the set of Natural Numbers isn't built that way.
You are misinformed. The set of natural numbers is the union of all
FISONs F(n) = {1, 2, 3, ..., n} = (0, n].
The set of FISONs is the set of natural numbers designed by v. Neumann.
On 2024-12-15 11:03:27 +0000, WM said:
From where did this concept come from?Of course. It consist of all finite n. It is the union of all
intervals (0, n]. Not more!
But the set of Natural Numbers isn't built that way.
You are misinformed. The set of natural numbers is the union of all
FISONs F(n) = {1, 2, 3, ..., n} = (0, n].
The set of FISONs is the set of natural numbers designed by v. Neumann.
The construction of natural numbers as sets was presented by Cantor (apparently as a comment to Kronecker's "God made the integers, all
else is the work of man").
On 15.12.2024 12:03, Mikko wrote:
On 2024-12-14 09:50:52 +0000, WM said:
On 14.12.2024 09:52, Mikko wrote:
On 2024-12-12 22:06:58 +0000, WM said:
No, there is no such set.In mathematics, a set A is Dedekind-infinite (named after the German >>>>>>> mathematician Richard Dedekind) if some proper subset B of A is
equinumerous to A. [Wikipedia].
Do you happen to know any set that is Dedekind-infinite?
The set of natural numbers, if there is any such set,
If ℕ is a set, i.e. if it is complete such that all numbers can be used >>> for indexing sequences or in other mappings, then it can also be
exhausted such that no element remains. Then the sequence of
intersections of endsegments
E(1), E(1)∩E(2), E(1)∩E(2)∩E(3), ...
loses all content. Then, by the law
∀k ∈ ℕ : ∩{E(1), E(2), ..., E(k+1)} = ∩{E(1), E(2), ..., E(k)} \ {k}
the content must become finite.
is Dedekind-infinte:
the successor function is a bijection between the set of all natural
numbers and non-zero natural numbers.
This "bijection" appears possible but it is not.
So you say that there is a natural number that does not have a next
natural number. What number is that?
We cannot name dark numbers as individuals.
On 2024-12-15 11:33:15 +0000, WM said:
We cannot name dark numbers as individuals.
We needn't. The axioms of natural numbers ensure that every natural number has a successor,
If that is not possible then there are no
natural numbers.
On 2024-12-15 11:25:26 +0000, WM said:
Here we have a limit of 1/10 from analysis and a limit of 0 from set
theory. That shows that if set theory is right, we have
1/10 = 0 ==> 1 = 0 ==> 2 = 1.
That is the fallacy of equivocation. The limit of analysis is a different concept from the limit of set theory.
On 15.12.2024 22:14, Richard Damon wrote:
On 12/15/24 2:44 PM, WM wrote:
On 15.12.2024 13:54, Richard Damon wrote:
You can't "name" your dark numbers,
because they are dark.
|ℕ \ {1, 2, 3, ...}| = 0 cannot be accomplished by visible numbers
because ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
Which just shows that the full set in infinte, and any member in it is
finite, and not the last member.
Many members can be subtracted individually but infinitely many members cannot be subtracted individually. They are belonging to the set. They
are dark.
Regards, WM
On 12/16/24 3:59 AM, WM wrote:
On 15.12.2024 22:14, Richard Damon wrote:
On 12/15/24 2:44 PM, WM wrote:
On 15.12.2024 13:54, Richard Damon wrote:
You can't "name" your dark numbers,
because they are dark.
|ℕ \ {1, 2, 3, ...}| = 0 cannot be accomplished by visible numbers
because ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
Which just shows that the full set in infinte, and any member in it
is finite, and not the last member.
Many members can be subtracted individually but infinitely many
members cannot be subtracted individually. They are belonging to the
set. They are dark.
Sure an infinite number of members can be subtracted individually,
On 17.12.2024 00:57, Richard Damon wrote:
On 12/16/24 3:59 AM, WM wrote:
On 15.12.2024 22:14, Richard Damon wrote:
On 12/15/24 2:44 PM, WM wrote:
On 15.12.2024 13:54, Richard Damon wrote:
You can't "name" your dark numbers,
because they are dark.
|ℕ \ {1, 2, 3, ...}| = 0 cannot be accomplished by visible numbers >>>>> because ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
Which just shows that the full set in infinte, and any member in it
is finite, and not the last member.
Many members can be subtracted individually but infinitely many
members cannot be subtracted individually. They are belonging to the
set. They are dark.
Sure an infinite number of members can be subtracted individually,
An unbounded number can be subtracted individually. However, if all are subtracted individually, then a last one is subtracted. That cannot happen.
Regards, WM
On 16.12.2024 11:14, Mikko wrote:
On 2024-12-15 11:25:26 +0000, WM said:
Here we have a limit of 1/10 from analysis and a limit of 0 from set
theory. That shows that if set theory is right, we have
1/10 = 0 ==> 1 = 0 ==> 2 = 1.
That is the fallacy of equivocation. The limit of analysis is a different
concept from the limit of set theory.
The limit of analysis proves that the limit of set theory is wrong.
On 16.12.2024 11:23, Mikko wrote:
On 2024-12-15 11:33:15 +0000, WM said:
We cannot name dark numbers as individuals.
We needn't. The axioms of natural numbers ensure that every natural number >> has a successor,
The set, i.e. all numbers together, has no successor.
On 12/17/24 5:30 AM, WM wrote:
On 17.12.2024 00:57, Richard Damon wrote:
On 12/16/24 3:59 AM, WM wrote:
On 15.12.2024 22:14, Richard Damon wrote:
On 12/15/24 2:44 PM, WM wrote:
On 15.12.2024 13:54, Richard Damon wrote:
You can't "name" your dark numbers,
because they are dark.
|ℕ \ {1, 2, 3, ...}| = 0 cannot be accomplished by visible numbers >>>>>> because ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
Which just shows that the full set in infinte, and any member in it
is finite, and not the last member.
Many members can be subtracted individually but infinitely many
members cannot be subtracted individually. They are belonging to the
set. They are dark.
Sure an infinite number of members can be subtracted individually,
An unbounded number can be subtracted individually. However, if all
are subtracted individually, then a last one is subtracted. That
cannot happen.
An unbound number can't be used individually,
Am Tue, 17 Dec 2024 22:49:51 +0100 schrieb WM:
On 17.12.2024 13:34, Richard Damon wrote:He „applies“ the set of all N, as opposed to every single n.
Your logic that if it holds for all FISONs, it holds for N,Please explain what Cantor does to apply more than what I apply, namely
all n ∈ ℕ.
On 16.12.2024 11:14, Mikko wrote:
That is the fallacy of equivocation.
The limit of analysis is a different concept from
the limit of set theory.
The limit of analysis proves that
the limit of set theory is wrong.
The limit of analysis proves that
the relative covering is 1/10,
the relative non-covering is 9/10.
Set theory proves that
the relative non-covering is 0.
These numbers are answering exactly the same question.
On 17.12.2024 13:34, Richard Damon wrote:
On 12/17/24 5:30 AM, WM wrote:
On 17.12.2024 00:57, Richard Damon wrote:
On 12/16/24 3:59 AM, WM wrote:
On 15.12.2024 22:14, Richard Damon wrote:
On 12/15/24 2:44 PM, WM wrote:
On 15.12.2024 13:54, Richard Damon wrote:
You can't "name" your dark numbers,
because they are dark.
|ℕ \ {1, 2, 3, ...}| = 0 cannot be accomplished by visibleWhich just shows that the full set in infinte, and any member in
numbers because ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo. >>>>>
it is finite, and not the last member.
Many members can be subtracted individually but infinitely many
members cannot be subtracted individually. They are belonging to
the set. They are dark.
Sure an infinite number of members can be subtracted individually,
An unbounded number can be subtracted individually. However, if all
are subtracted individually, then a last one is subtracted. That
cannot happen.
An unbound number can't be used individually,
An unbounded number of members can be subtracted individually but not all.
Regards, WM
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