William schrieb am Mittwoch, 22. November 2023 um 18:41:34 UTC+1:
> On Wednesday, November 22, 2023 at 8:02:10 AM UTC-4, WM wrote:
> > On 22.11.2023 11:40, William wrote:
> > > On Wednesday, November 22, 2023 at 6:30:35 AM UTC-4, WM wrote:
> > >> On 21.11.2023 23:02, William wrote:
> > >>> On Tuesday, November 21, 2023 at 1:51:50 PM UTC-4, WM wrote:
> > >>>> On 20.11.2023 19:28, William wrote:
> > >>>>> On Monday, November 20, 2023 at 6:53:13 AM UTC-4, WM wrote:
> > >>>>>> ... every A(i) contains fractions which sit at not indexed positions,
> > >>>>>
> > >>>>> Each A(n) contains n "indexed" positions and an infinite number of "not indexed positions" for a total of an infinite number of positions.
> > >>>> Yes. The n finally indexed positions are the matrix B(n), a submatrix of
> > >>>> A(n).
> > >>>>> Your game does not tell you that the infinite number of positions in A(n) is larger than the infinite number of positions in B.
> > >>>> My game tells us (1) that the number of positions of A(n) is constant
> > >>>> for every n. Further it tells us that B(n) is a part of A(n). And (3) it
> > >>>> tells that all A(n) contain all fractions at not indexed positions.. Ony
> > >>>> a very strong belief can claim that after all positions have been
> > >>>> indexed
> > >>>
> > >>> At no step in your game have all positions been indexed.
> > >> Of course not.
> > >
> > > Hence your game cannot tell you anything "after all positions have been indexed.
> > Of course not! The game proves that it is impossible to index all
> Nope the game says nothing about "all"

A is larger than B, since B consists of the indexed position in A.
>
> B is defined, while your putative larger matrix A is not defined.

Both are defined or both are not defined. The matrix B is well-known. But B is not defined as the result of the enumeration. That is only assumed. Erroneously, as A shows.

Regards, WM

On Wednesday, November 22, 2023 at 6:13:31 PM UTC-4, Transfinity wrote:
> William schrieb am Mittwoch, 22. November 2023 um 18:41:34 UTC+1:
> > On Wednesday, November 22, 2023 at 8:02:10 AM UTC-4, WM wrote:
> > > On 22.11.2023 11:40, William wrote:
> > > > On Wednesday, November 22, 2023 at 6:30:35 AM UTC-4, WM wrote:
> > > >> On 21.11.2023 23:02, William wrote:
> > > >>> On Tuesday, November 21, 2023 at 1:51:50 PM UTC-4, WM wrote:
> > > >>>> On 20.11.2023 19:28, William wrote:
> > > >>>>> On Monday, November 20, 2023 at 6:53:13 AM UTC-4, WM wrote:
> > > >>>>>> ... every A(i) contains fractions which sit at not indexed positions,
> > > >>>>>
> > > >>>>> Each A(n) contains n "indexed" positions and an infinite number of "not indexed positions" for a total of an infinite number of positions.
> > > >>>> Yes. The n finally indexed positions are the matrix B(n), a submatrix of
> > > >>>> A(n).
> > > >>>>> Your game does not tell you that the infinite number of positions in A(n) is larger than the infinite number of positions in B.
> > > >>>> My game tells us (1) that the number of positions of A(n) is constant
> > > >>>> for every n. Further it tells us that B(n) is a part of A(n). And (3) it
> > > >>>> tells that all A(n) contain all fractions at not indexed positions. Ony
> > > >>>> a very strong belief can claim that after all positions have been
> > > >>>> indexed
> > > >>>
> > > >>> At no step in your game have all positions been indexed.
> > > >> Of course not.
> > > >
> > > > Hence your game cannot tell you anything "after all positions have been indexed.
> > > Of course not! The game proves that it is impossible to index all
> > Nope the game says nothing about "all"
> A

does not exist.

> >is larger than B, since B consists of the indexed position in A.
> >
> > B is defined, while your putative larger matrix A is not defined.
> Both are defined or both are not defined.
Nope you have agreed that there is a finite algorithm which given any position in B gives you the value at that position. No such finite algorithm exists for A.

On 23.11.2023 03:41, William wrote:
> On Wednesday, November 22, 2023 at 6:13:31 PM UTC-4, Transfinity wrote:
>> William schrieb am Mittwoch, 22. November 2023 um 18:41:34 UTC+1:
>>> On Wednesday, November 22, 2023 at 8:02:10 AM UTC-4, WM wrote:
>>>> On 22.11.2023 11:40, William wrote:
>>>>> On Wednesday, November 22, 2023 at 6:30:35 AM UTC-4, WM wrote:
>>>>>> On 21.11.2023 23:02, William wrote:
>>>>>>> On Tuesday, November 21, 2023 at 1:51:50 PM UTC-4, WM wrote:
>>>>>>>> On 20.11.2023 19:28, William wrote:
>>>>>>>>> On Monday, November 20, 2023 at 6:53:13 AM UTC-4, WM wrote:
>>>>>>>>>> ... every A(i) contains fractions which sit at not indexed positions,
>>>>>>>>>
>>>>>>>>> Each A(n) contains n "indexed" positions and an infinite number of "not indexed positions" for a total of an infinite number of positions.
>>>>>>>> Yes. The n finally indexed positions are the matrix B(n), a submatrix of
>>>>>>>> A(n).
>>>>>>>>> Your game does not tell you that the infinite number of positions in A(n) is larger than the infinite number of positions in B.
>>>>>>>> My game tells us (1) that the number of positions of A(n) is constant
>>>>>>>> for every n. Further it tells us that B(n) is a part of A(n). And (3) it
>>>>>>>> tells that all A(n) contain all fractions at not indexed positions. Ony
>>>>>>>> a very strong belief can claim that after all positions have been
>>>>>>>> indexed
>>>>>>>
>>>>>>> At no step in your game have all positions been indexed.
>>>>>> Of course not.
>>>>>
>>>>> Hence your game cannot tell you anything "after all positions have been indexed.
>>>> Of course not! The game proves that it is impossible to index all
>>> Nope the game says nothing about "all"
>> A
>
> does not exist.

All positions of B do exist in A, because I use the finite algorithm
given by Cantor. But A contains more. May it be visible or dark, it is
remaining there. By the way, note that B is only a triangle.
>
>>> is larger than B, since B consists of the indexed position in A.
>>>
>>> B is defined, while your putative larger matrix A is not defined.
>> Both are defined or both are not defined.
> Nope you have agreed that there is a finite algorithm which given any position in B gives you the value at that position.

Yes! That is the algorithm described in the OP.

> No such finite algorithm exists for A.

A(n) exists for every n that settles at its final position. Every
removed fraction is as well defined as the position of the n. But the
important fact is, that for every B(n) we know that almost all positions
of A(n) are not indexed. And if the limit B exists, then the limit A
exists and has just this property.

Regards, WM

On Thursday, November 23, 2023 at 3:57:36 AM UTC-4, WM wrote:
> Every
> removed fraction is as well defined as the position of the n.

Nope. The final position of the fraction is not given by your billiards game.
(after each step every fraction will move).

B is defined by an algorithm not a limit.
You have agreed there is no definition of the putative matrix A

On 23.11.2023 09:16, William wrote:
> On Thursday, November 23, 2023 at 3:57:36 AM UTC-4, WM wrote:
>> Every
>> removed fraction is as well defined as the position of the n.
>
>
> Nope. The final position of the fraction is not given by your billiards game.
> (after each step every fraction will move).

I said: for every step n. Not for the final position, if there is a
final matrix A and its submatrix B. But every fraction is known to stay
inside every existing matrix at a not indexed position.

>
> B is defined by an algorithm not a limit.
> You have agreed there is no definition of the putative matrix A

B is defined by the algorithm that I use, namely Cantor's. And not all
positions are indexed.

Regards, WM

On Thursday, November 23, 2023 at 4:43:24 AM UTC-4, WM wrote:

> > B is defined by an algorithm not a limit.
> > You have agreed there is no definition of the putative matrix A
> B is defined

and the putative matrix A is not defined.

On 23.11.2023 10:11, William wrote:
> On Thursday, November 23, 2023 at 4:43:24 AM UTC-4, WM wrote:
>
>>> B is defined by an algorithm not a limit.
>>> You have agreed there is no definition of the putative matrix A
>> B is defined

only as a submatrix of A
>
> and the putative matrix A is not defined.

A(n) contains every position of B(n) completely defined and it contains
all positions of fractions not indexed. That is enough to prove that not
all positions can be indexed.

Regards, WM

On Thursday, November 23, 2023 at 5:28:26 AM UTC-4, WM wrote:
> On 23.11.2023 10:11, William wrote:
> > On Thursday, November 23, 2023 at 4:43:24 AM UTC-4, WM wrote:
> >
> >>> B is defined by an algorithm not a limit.
> >>> You have agreed there is no definition of the putative matrix A
> >> B is defined
> only as a submatrix of A

Nope. There is no matrix A. There is a matrix B as B is defined by an algorithm.

On 23.11.2023 11:01, William wrote:
> On Thursday, November 23, 2023 at 5:28:26 AM UTC-4, WM wrote:
>> On 23.11.2023 10:11, William wrote:
>>> On Thursday, November 23, 2023 at 4:43:24 AM UTC-4, WM wrote:
>>>
>>>>> B is defined by an algorithm not a limit.
>>>>> You have agreed there is no definition of the putative matrix A
>>>> B is defined
>> only as a submatrix of A
>
> Nope. There is no matrix A.
There are all positions of A(0). They do not change in the course of the
sequence.

> There is a matrix B as B is defined by an algorithm.
There are all matrices A(i). They all are infinitely larger than B(i),
and all A(i) contain B(i). If all positions of A(0) could be indexed,
then this would have to be done in the course of the sequence A(i).
Because only in its course natnumbers i are attached to positions.
Therefore, whether or not A exists, for all i ∈ ℕ the enumeration lacks
infinitely many positions. And in the limit nothing is enumerated
further. Hence B covers positions of A(0), but not all. QED

Regards, WM

On Thursday, November 23, 2023 at 12:05:36 PM UTC-4, WM wrote:
> On 23.11.2023 11:01, William wrote:
> > On Thursday, November 23, 2023 at 5:28:26 AM UTC-4, WM wrote:
> >> On 23.11.2023 10:11, William wrote:
> >>> On Thursday, November 23, 2023 at 4:43:24 AM UTC-4, WM wrote:
> >>>
> >>>>> B is defined by an algorithm not a limit.
> >>>>> You have agreed there is no definition of the putative matrix A
> >>>> B is defined
> >> only as a submatrix of A
> >
> > Nope. There is no matrix A.
> There are all positions of A(0).
I.e. there is no matrix A.

On Thursday, November 23, 2023 at 12:05:36 PM UTC-4, WM wrote:
> If all positions of A(0) could be indexed,
> then this would have to be done in the course of the sequence A(i).

No for each A(i) only a finite number of the infinite number of positions or A(0) are "indexed". At no step
are all positions "indexed". Your billiards game never reaches a limit.

On 23.11.2023 18:44, William wrote:
> On Thursday, November 23, 2023 at 12:05:36 PM UTC-4, WM wrote:

>> There are all positions of A(0).
> I.e. there is no matrix A.
>

If there is no matrix B, i.e., no complete application of natnumbers,
then there is no matrix A. But what should it be good for? The
difference between natnumbers and number of matrix positions is evident
in any case.

Regards, WM

On 23.11.2023 18:55, William wrote:
> On Thursday, November 23, 2023 at 12:05:36 PM UTC-4, WM wrote:
>> If all positions of A(0) could be indexed,
>> then this would have to be done in the course of the sequence A(i).
>
>
> No for each A(i) only a finite number of the infinite number of positions or A(0) are "indexed". At no step
> are all positions "indexed".

Of course not. They are dark.

> Your billiards game never reaches a limit.

I reach what Cantor's formula k = (m + n - 1)(m + n - 2)/2 + m
facilitates, because I use this formula. All natnumbers k are applied to
matrix positions 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, ... . Not more, not
less. But all fractions are removed to other positions within the matrix
which remain without index.

Regards, WM

On Friday, November 24, 2023 at 5:48:17 AM UTC-4, WM wrote:
> On 23.11.2023 18:44, William wrote:
> > On Thursday, November 23, 2023 at 12:05:36 PM UTC-4, WM wrote:
>
> >> There are all positions of A(0).
> > I.e. there is no matrix A.
> >
> If there is no matrix B, i.e., no complete application of natnumbers,
> then there is no matrix A

Nope B is not a "complete application" it is an algorithm. B exists. A does not.

On Friday, November 24, 2023 at 5:52:37 AM UTC-4, WM wrote:
> On 23.11.2023 18:55, William wrote:
> > On Thursday, November 23, 2023 at 12:05:36 PM UTC-4, WM wrote:
> >> If all positions of A(0) could be indexed,
> >> then this would have to be done in the course of the sequence A(i).
> >
> >
> > No for each A(i) only a finite number of the infinite number of positions or A(0) are "indexed". At no step
> > are all positions "indexed".
> Of course not. They are dark.

Nope, the reason that not all positions of A(0) are "indexed" at any step is that at any step only a finite number of positions have been "indexed".

> All

Your billiards game never reaches "all". At every step only a finite number of positions have been "indexed"

William schrieb am Freitag, 24. November 2023 um 14:05:52 UTC+1:
> On Friday, November 24, 2023 at 5:48:17 AM UTC-4, WM wrote:
> > On 23.11.2023 18:44, William wrote:
> > > On Thursday, November 23, 2023 at 12:05:36 PM UTC-4, WM wrote:
> >
> > >> There are all positions of A(0).
> > > I.e. there is no matrix A.
> > >
> > If there is no matrix B, i.e., no complete application of natnumbers,
> > then there is no matrix A
> Nope B is not a "complete application" it is an algorithm. B exists.

By the belief of matheologians? B exists by the application of the first diagonal argument.

> A does not.

That part of B which "exists", exists as a submatrix of A. And A contains more, namely the position of 1/2.

Regards, WM

William schrieb am Freitag, 24. November 2023 um 14:14:24 UTC+1:
> On Friday, November 24, 2023 at 5:52:37 AM UTC-4, WM wrote:
> > On 23.11.2023 18:55, William wrote:
> > > On Thursday, November 23, 2023 at 12:05:36 PM UTC-4, WM wrote:
> > >> If all positions of A(0) could be indexed,
> > >> then this would have to be done in the course of the sequence A(i).
> > >
> > >
> > > No for each A(i) only a finite number of the infinite number of positions or A(0) are "indexed". At no step
> > > are all positions "indexed".
> > Of course not. They are dark.
> Nope, the reason that not all positions of A(0) are "indexed" at any step is that at any step only a finite number of positions have been "indexed".
>
> > All
>
> Your billiards game never reaches "all".

It reaches all of B which Cantor reches. And A contains besides B the fraction 1/2 at a position not belonging to B.

> At every step only a finite number of positions have been "indexed"

The same is true for Cantor.

Regards, WM

Fritz Feldhase schrieb am Samstag, 25. November 2023 um 05:43:44 UTC+1:
> On Friday, November 24, 2023 at 2:05:52 PM UTC+1, William wrote:
>
> > Nope B is not a "complete application" it is an algorithm.
> Actually B is a /matrix/, not an algorithm.

It is a matrix realized by an algorithm.

> Hint: We usually don't call formulas "algorithms" (in the present context).

We do. But that's irrelevant. Fact is: Only when B is realized, all natnumbers are issued. But then A contains B and in addition positions without index.

Regards, WM

On Saturday, November 25, 2023 at 6:38:37 AM UTC-4, Transfinity wrote:

> That part of B which "exists",

is B

> exists as a submatrix of A.

Your putative matrix A does not exist.

William schrieb am Samstag, 25. November 2023 um 12:50:01 UTC+1:
> On Saturday, November 25, 2023 at 6:38:37 AM UTC-4, Transfinity wrote:
>
>
> > That part of B which "exists",
> is B
> > exists as a submatrix of A.
> Your putative matrix A does not exist.

But B is produced by the algorithm described in the OP, isn't it? And in the course of this agorithm the fraction 1/2 sits at a position which is not belonging to B (but to A(0).

Regards, WM

On Saturday, November 25, 2023 at 9:16:08 AM UTC-4, Transfinity wrote:
> William schrieb am Samstag, 25. November 2023 um 12:50:01 UTC+1:
> > On Saturday, November 25, 2023 at 6:38:37 AM UTC-4, Transfinity wrote:
> >
> >
> > > That part of B which "exists",
> > is B
> > > exists as a submatrix of A.
> > Your putative matrix A does not exist.
> But B
unlike your putative A, exists.

William schrieb am Samstag, 25. November 2023 um 14:59:53 UTC+1:
> On Saturday, November 25, 2023 at 9:16:08 AM UTC-4, Transfinity wrote:
> > William schrieb am Samstag, 25. November 2023 um 12:50:01 UTC+1:
> > > On Saturday, November 25, 2023 at 6:38:37 AM UTC-4, Transfinity wrote:
> > >
> > >
> > > > That part of B which "exists",
> > > is B
> > > > exists as a submatrix of A.
> > > Your putative matrix A does not exist.
> > But B
> unlike your putative A, exists.

created by the algorithm given in the OP from the matrix A(0) which exists and contains the fraction 1/2 which never leaves but sits only at not indexed places.

Regards, WM

On Saturday, November 25, 2023 at 11:43:50 AM UTC-4, Transfinity wrote:
> William schrieb am Samstag, 25. November 2023 um 14:59:53 UTC+1:
> > On Saturday, November 25, 2023 at 9:16:08 AM UTC-4, Transfinity wrote:
> > > William schrieb am Samstag, 25. November 2023 um 12:50:01 UTC+1:
> > > > On Saturday, November 25, 2023 at 6:38:37 AM UTC-4, Transfinity wrote:
> > > >
> > > >
> > > > > That part of B which "exists",
> > > > is B
> > > > > exists as a submatrix of A.
> > > > Your putative matrix A does not exist.
> > > But B
> > unlike your putative A, exists.
> created by the algorithm given in the OP
Though you do not do so an algorithm can be defined. Unlike your putative A, B exists.

William schrieb am Samstag, 25. November 2023 um 17:40:52 UTC+1:
> On Saturday, November 25, 2023 at 11:43:50 AM UTC-4, Transfinity wrote:
> > William schrieb am Samstag, 25. November 2023 um 14:59:53 UTC+1:
> > > On Saturday, November 25, 2023 at 9:16:08 AM UTC-4, Transfinity wrote:
> > > > William schrieb am Samstag, 25. November 2023 um 12:50:01 UTC+1:
> > > > > On Saturday, November 25, 2023 at 6:38:37 AM UTC-4, Transfinity wrote:
> > > > >
> > > > >
> > > > > > That part of B which "exists",
> > > > > is B
> > > > > > exists as a submatrix of A.
> > > > > Your putative matrix A does not exist.
> > > > But B
> > > unlike your putative A, exists.
> > created by the algorithm given in the OP
> Though you do not do so an algorithm can be defined.

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 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 algorithm it is impossible, in eternity, to
remove a fraction from the matrix or to attach a natnumber to a fraction.

(1) My game tells us that the number of positions of A(n)
is constant for every n.

(2) Further it tells us that B(n) is a part of
A(n).

(3) It tells that all A(n) contain all fractions at not
indexed positions.

Regards, WM

On 11/25/2023 5:16 AM, Transfinity wrote:
> William schrieb am Samstag, 25. November 2023 um 12:50:01 UTC+1:
>> On Saturday, November 25, 2023 at 6:38:37 AM UTC-4, Transfinity wrote:
>>
>>
>>> That part of B which "exists",
>> is B
>>> exists as a submatrix of A.
>> Your putative matrix A does not exist.
>
> But B is produced by the algorithm described in the OP, isn't it? And in the course of this agorithm the fraction 1/2 sits at a position which is not belonging to B (but to A(0).

Your mind seems to be stuck on a step-by-step process, instead of saying
all...

>
> Regards, WM