On 28.11.2023 16:51, Fritz Feldhase wrote:

> A ist also offensichtlich gänzlich anders definiert als B.
>
A is defined in the OP by A(0) and the replacements of the numbers of
the first column described byk = (m + n - 1)(m + n - 2)/2 + m.
B is defined by k = (m + n - 1)(m + n - 2)/2 + m.

Both are equally well defined, namely by the same formula. Only Cantor
"forgets" that the natural numbers reside in the first column already.

Regards, WM

On Tuesday, November 28, 2023 at 1:30:04 PM UTC-4, WM wrote:
> William schrieb am Dienstag, 28. November 2023 um 14:54:24 UTC+1:
> > On Tuesday, November 28, 2023 at 6:36:12 AM UTC-4, WM wrote:
> > > If A does not exist, then B does not exist.
> > Nope. B is defined. Your putative A is not defined
> Both are defined simultaneously in the OP.

Nope, as you have noted, only B is defined. Your putative A is not defined.

czwartek, 19 października 2023 o 15:04:16 UTC+2 WM napisał(a):
> 1, 1/2, 1/3, 1/4, ...
> 2, 2/2, 2/3, 2/4, ...
> 3, 3/2, 3/3, 3/4, ...
> 4, 4/2, 4/3, 4/4, ...
> 5, 5/2, 5/3, 5/4, ...
>
> Push the natnumbers of the first column (without queue) into the field of fractions and store the hit fraction always there where the natnumber has come from. Try to push the natnumbers such that all matrix positions are occupied by them. That is best done by creating a pattern like
>
> 1, 2, 4, ...
> 3, 5, 8, ...
> 6, 9, 13, ...
> ... ,
>
> According to this simple rule it is impossible, in eternity, to remove a fraction from the matrix or to attach a natnumber to a fraction.
>
> You have won as soon as you understand that the last sentence is true.
>
> Congrats, WM
Of course, you are right. If the method involves STEP-by-STEP action, then the possibility of performing an action on "ALL" elements of an infinite series or set is excluded – regardless of whether the actions are considered in time or "outside of time." The action "outside of time" is only a pretense here because the step-by-step method inherently requires minimal (delta t) one after one, in which subsequent steps are taken.

The only type of action on the elements of an infinite series or set that can encompass "ALL" elements of such a series or set is an action involving "ALL" elements simultaneously. So, "ALL" in one step, absolutely not step by step, element by element.

Acting on "ALL" elements of an infinite set simultaneously is, in practice, simply redefining the set or, perhaps more precisely, defining a NEW infinite set.

Example: "I have" an infinite set whose elements are natural numbers written in Arabic numerals:
1, 2, 3, 4, 5, 6, 7, ...
Contrary to what seems erroneously apparent to many people, and in accordance with what was described above, it is not possible to change "ALL" elements of this set step by step.
For however advanced step, i.e., any large natural number, there will always be, by nature, another number bigger by 1, and practically an infinite series of such bigger numbers, still "not handled, yet."

However, it is possible to change "ALL" elements of this set "AT ONCE."

"I take" the infinite set described above, whose elements are natural numbers and change their representation to Roman numerals:
I, II, III, IV, V, VI, VII, ...

See? I changed the qualitative parameter - the symbolic representation - of "ALL" elements of the infinite set of natural numbers :)

That's roughly everything on this topic.
Regards,
Adam

On 28.11.2023 19:36, William wrote:
> On Tuesday, November 28, 2023 at 1:30:04 PM UTC-4, WM wrote:
>> William schrieb am Dienstag, 28. November 2023 um 14:54:24 UTC+1:
>>> On Tuesday, November 28, 2023 at 6:36:12 AM UTC-4, WM wrote:
>>>> If A does not exist, then B does not exist.
>>> Nope. B is defined. Your putative A is not defined
>> Both are defined simultaneously in the OP.
>
> Nope, as you have noted, only B is defined.

I noted as you should note that every element of B is defined by the
same algorithm that defines A.

Regards, WM

On Wednesday, November 29, 2023 at 8:47:35 AM UTC-4, WM wrote:
> On 28.11.2023 19:36, William wrote:
> > On Tuesday, November 28, 2023 at 1:30:04 PM UTC-4, WM wrote:
> >> William schrieb am Dienstag, 28. November 2023 um 14:54:24 UTC+1:
> >>> On Tuesday, November 28, 2023 at 6:36:12 AM UTC-4, WM wrote:
> >>>> If A does not exist, then B does not exist.
> >>> Nope. B is defined. Your putative A is not defined
> >> Both are defined simultaneously in the OP.
> >
> > Nope, as you have noted, only B is defined.
> I noted as you should note that every element of B is defined by the
> same algorithm that defines A.

Nope. you noted there is no algorithm that defines the putative A

William schrieb am Mittwoch, 29. November 2023 um 15:49:07 UTC+1:
> On Wednesday, November 29, 2023 at 8:47:35 AM UTC-4, WM wrote:
> > On 28.11.2023 19:36, William wrote:
> > > On Tuesday, November 28, 2023 at 1:30:04 PM UTC-4, WM wrote:
> > >> William schrieb am Dienstag, 28. November 2023 um 14:54:24 UTC+1:
> > >>> On Tuesday, November 28, 2023 at 6:36:12 AM UTC-4, WM wrote:
> > >>>> If A does not exist, then B does not exist.
> > >>> Nope. B is defined. Your putative A is not defined
> > >> Both are defined simultaneously in the OP.
> > >
> > > Nope, as you have noted, only B is defined.
> > I noted as you should note that every element of B is defined by the
> > same algorithm that defines A.
> Nope. you noted there is no algorithm that defines the putative A

A is as well defined as B because in both cases every step is defined
such that no further decision is required.

Regards, WM

Fritz Feldhase schrieb am Mittwoch, 29. November 2023 um 14:36:16 UTC+1:
> On Wednesday, November 29, 2023 at 1:47:35 PM UTC+1, WM wrote:
>
> > B is defined by the
> following EXPLICIT DEFINITION,
>
> | Def.: B = (b_n,m)_(n,m e IN) with b_n,m = (m + n - 1)(m + n - 2)/2
+ m for all n,m e IN.
>
> In the context of set theory, we might even state it the following way:
>
> | Def.: B = {((n, m), (m + n - 1)(m + n - 2)/2 + m) : n,m e IN}.

Def.: A see the OP.
An algorithm is given if every step is determined and no further
decision is asked for.

Def.: of the submatrix B of A: B = {((n, m), (m + n - 1)(m + n - 2)/2 +
m) : n,m e IN}.

Therefore the submatrix B of A is defined. Further it is proven that the
matrix A contains at least one more element.

Regards, WM

On Wednesday, November 29, 2023 at 1:46:52 PM UTC-4, WM wrote:
> William schrieb am Mittwoch, 29. November 2023 um 15:49:07 UTC+1:
> > On Wednesday, November 29, 2023 at 8:47:35 AM UTC-4, WM wrote:
> > > On 28.11.2023 19:36, William wrote:
> > > > On Tuesday, November 28, 2023 at 1:30:04 PM UTC-4, WM wrote:
> > > >> William schrieb am Dienstag, 28. November 2023 um 14:54:24 UTC+1:
> > > >>> On Tuesday, November 28, 2023 at 6:36:12 AM UTC-4, WM wrote:
> > > >>>> If A does not exist, then B does not exist.
> > > >>> Nope. B is defined. Your putative A is not defined
> > > >> Both are defined simultaneously in the OP.
> > > >
> > > > Nope, as you have noted, only B is defined.
> > > I noted as you should note that every element of B is defined by the
> > > same algorithm that defines A.
> > Nope. you noted there is no algorithm that defines the putative A
> A is as well defined as B because in both cases every step

Nope. Neither B nor you putative A is defined by steps.

William schrieb am Mittwoch, 29. November 2023 um 20:27:42 UTC+1:
> On Wednesday, November 29, 2023 at 1:46:52 PM UTC-4, WM wrote:

> > A is as well defined as B because in both cases every step
> Nope. Neither B nor you putative A is defined by steps.

Mathematics is not time-dependent. If all steps are defined by the
algorithm, and if infinitely many natural numbers can be applied at all,
then this can be done like described in the OP. The only condition is
that all steps are determined and no further decisions are necessary.

Note that counting is a step-wise process like Cantor's procedure to
obtain B.

Regards, WM

On 11/29/2023 11:58 AM, WM wrote:
> William schrieb am Mittwoch, 29. November 2023 um 20:27:42 UTC+1:
> > On Wednesday, November 29, 2023 at 1:46:52 PM UTC-4, WM wrote:
>
> > > A is as well defined as B because in both cases every step
> > Nope. Neither B nor you putative A is defined by steps.
>
> Mathematics is not time-dependent. If all steps are defined by the
> algorithm, and if infinitely many natural numbers can be applied at all,
> then this can be done like described in the OP. The only condition is
> that all steps are determined and no further decisions are necessary.
>
> Note that counting is a step-wise process like Cantor's procedure to
> obtain B.

Are you the kind of person that, instead of saying 10, you say 1, 2, 3,
4, 5, 6, 7, 8, 9, 10?

On Wednesday, November 29, 2023 at 3:58:31 PM UTC-4, WM wrote:
> William schrieb am Mittwoch, 29. November 2023 um 20:27:42 UTC+1:
> > On Wednesday, November 29, 2023 at 1:46:52 PM UTC-4, WM wrote:
>
> > > A is as well defined as B because in both cases every step
> > Nope. Neither B nor you putative A is defined by steps.
> Mathematics is not time-dependent. If all steps are defined by the
> algorithm, and if infinitely many natural numbers can be applied at all,

B is not defined by "applying infinitely many natural numbers". Your putative A is not defined

William schrieb am Donnerstag, 30. November 2023 um 03:10:16 UTC+1:
> On Wednesday, November 29, 2023 at 3:58:31 PM UTC-4, WM wrote:
> > William schrieb am Mittwoch, 29. November 2023 um 20:27:42 UTC+1:
> > > On Wednesday, November 29, 2023 at 1:46:52 PM UTC-4, WM wrote:
> >
> > > > A is as well defined as B because in both cases every step
> > > Nope. Neither B nor you putative A is defined by steps.
> > Mathematics is not time-dependent. If all steps are defined by the
> > algorithm, and if infinitely many natural numbers can be applied at all,
> B is not defined by "applying infinitely many natural numbers".

It is. Name any position which is not.

> Your putative A is not defined.

Name any position of its submatrix B is as well defined as by Cantor.

Regards, WM

Fritz Feldhase schrieb am Donnerstag, 30. November 2023 um 03:22:22 UTC+1:
> On Wednesday, November 29, 2023 at 8:58:31 PM UTC+1, WM wrote:
>
> > Note that counting is a step-wise process like Cantor's procedure to obtain B.
> The ONLY 'step' "to obtain B" is its EXPLICIT DEFINITION,

That has been given in the OP. Try to find a position of B that is not defined there.

Regards, WM

On Thursday, November 30, 2023 at 11:20:35 AM UTC-4, WM wrote:

> ... its submatrix B

Your putative A (of which B is supposedly a submatrix) is not defined.

William schrieb am Donnerstag, 30. November 2023 um 16:38:24 UTC+1:
> On Thursday, November 30, 2023 at 11:20:35 AM UTC-4, WM wrote:
>
> > ... its submatrix B
>
> Your putative A (of which B is supposedly a submatrix) is not defined.

The submatrix B is defined by Cantor's formula. And it is proven that
there are more positions in the matrix which are not indexed. Therefore
the matrix of size A(0) is not completely indexd.

Regards, WM

On Thursday, November 30, 2023 at 12:17:29 PM UTC-4, WM wrote:

> ...B is defined

Correct. B is defined. However, this does not mean your putative A is defined.

William schrieb am Donnerstag, 30. November 2023 um 17:40:05 UTC+1:
> On Thursday, November 30, 2023 at 12:17:29 PM UTC-4, WM wrote:
>
> > ...B is defined
>
> Correct. B is defined. However, this does not mean your putative A is
defined.

All matrices A(0), A(1), A(2), ... have the same size. No element is
added, no element leaves. If B is defined in the OP, then it has a
smaller size because the fraction 1/2 is not in B but remains in
whatever is defined of A.

Regards, WM

On Friday, December 1, 2023 at 8:16:58 AM UTC-4, WM wrote:
>...the fraction 1/2 is not in B but remains in
> whatever is defined of A.

There is no A. A(0), A(1) , A(2) ,,,, are not "whatever is defined of A".

William schrieb am Freitag, 1. Dezember 2023 um 13:27:28 UTC+1:
> On Friday, December 1, 2023 at 8:16:58 AM UTC-4, WM wrote:
> >...the fraction 1/2 is not in B but remains in
> > whatever is defined of A.
> There is no A. A(0), A(1) , A(2) ,,,, are not "whatever is defined of A".

Let (A(n)) be the sequence A(0), A(1), A(2), ... . All existing
matrices are of same size. All elements of B are in every term A(n).
If B is completed by k = (m + n - 1)(m + n - 2)/2 + m, then also A is
completed.

Regards, WM

On Saturday, December 2, 2023 at 6:12:58 AM UTC-4, WM wrote:

> If B is completed by k = (m + n - 1)(m + n - 2)/2 + m,

Nope. B is defined by this algorithm.

> then also A is completed.

Nope. "Completed" is nonsense. The putative A does not exist.

On 02.12.2023 15:37, William wrote:
> On Saturday, December 2, 2023 at 6:12:58 AM UTC-4, WM wrote:
>
>
>> If B is completed by k = (m + n - 1)(m + n - 2)/2 + m,
>
> Nope. B is defined by this algorithm.

Then it is completed by it.

>> then also A is completed.
>
> Nope. "Completed" is nonsense.

Agreed. Like Cantor's bijection. "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." [G.
Cantor, letter to R. Lipschitz (19 Nov 1883)]
Without completeness no complete mapping.
Note: completely.

> The putative A does not exist.

It is defined by the algorithm in the OP.

Regards, WM

On 02.12.2023 15:48, Fritz Feldhase wrote:
> On Saturday, December 2, 2023 at 11:12:58 AM UTC+1, WM wrote:
>
>> If B is completed by k = (m + n - 1)(m + n - 2)/2 + m,
>
> B is not "completed

Agreed. Like Cantor's bijection. "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." [G.
Cantor, letter to R. Lipschitz (19 Nov 1883)]
Without completeness no complete mapping.

Regards, WM

On Saturday, December 2, 2023 at 11:02:18 AM UTC-4, WM wrote:
> On 02.12.2023 15:37, William wrote:
> > On Saturday, December 2, 2023 at 6:12:58 AM UTC-4, WM wrote:
> >
> >
> >> If B is completed by k = (m + n - 1)(m + n - 2)/2 + m,
> >
> > Nope. B is defined by this algorithm.
> Then it is completed by it.

> >
> > Nope. "Completed" is nonsense.
> Agreed.

??

On 02.12.2023 16:37, William wrote:
> On Saturday, December 2, 2023 at 11:02:18 AM UTC-4, WM wrote:
>> On 02.12.2023 15:37, William wrote:
>>> On Saturday, December 2, 2023 at 6:12:58 AM UTC-4, WM wrote:
>>>
>>>
>>>> If B is completed by k = (m + n - 1)(m + n - 2)/2 + m,
>>>
>>> Nope. B is defined by this algorithm.
>> Then it is completed by it.
>
>>>
>>> Nope. "Completed" is nonsense.
>> Agreed.
>
> ??

The claim that all fractions could be indexed by using all indices is
nonsense.

*If* B is defined by this algorithm, then it is completed by it.

Regards, WM

On Saturday, December 2, 2023 at 12:52:38 PM UTC-4, WM wrote:

> *If* B is defined by this algorithm, then it is completed by it.

Nope. The term "completed" is meaningless.