On 11/9/2023 3:30 AM, Chris M. Thomasson wrote:
> On 11/9/2023 12:26 AM, Transfinity wrote:

>> [...]
>
> You do realize that
> an index can be used to create
> a unique pair, right?
> This pair can be used for
> a fraction p[0] / p[1]?
> This pair can be used to go right back to
> the unique natural number that created it.
> Its full circle:
>
> index to pair, and pair to index
>
> Nothing is lost.

Yes. Well-put.

sₖ = ⌈½+(2⋅k+¼)¹ᐟ²⌉
iₖ = k-(sₖ-1)(sₖ-2)/2
jₖ = sₖ-iₖ

sᵢⱼₖ = iₖ+jₖ
kᵢⱼₖ = iₖ+(sᵢⱼₖ-1)(sᵢⱼₖ-2)/2
sᵢⱼₖ = sₖ
kᵢⱼₖ = k

And around it goes,
index to pair to index to pair.
A snake swallowing its own tail,
in each of infinitely-many instances.

sₖᵢⱼₖ = ⌈½+(2⋅kᵢⱼₖ+¼)¹ᐟ²⌉
iₖᵢⱼₖ = kᵢⱼₖ-(sₖᵢⱼₖ-1)(sₖᵢⱼₖ-2)/2
jₖᵢⱼₖ = sₖᵢⱼₖ-iₖᵢⱼₖ
sₖᵢⱼₖ = sᵢⱼₖ = sₖ
iₖᵢⱼₖ = iₖ
jₖᵢⱼₖ = jₖ

Arithmetic.

On Thursday, November 9, 2023 at 5:29:50 PM UTC+1, Jim Burns wrote:

> If all swaps are swapped,
> then Bob is not anywhere.

Since Bob is always somewhere, never all swaps are swapped.

On 11/9/2023 1:35 PM, Fritz Feldhase wrote:
> On Thursday, November 9, 2023
> at 5:29:50 PM UTC+1, Jim Burns wrote:

>> If all swaps are swapped,
>> then Bob is not anywhere.
>
> Since Bob is always somewhere,
> never all swaps are swapped.

Either not all swaps are swapped,
or Bob is not anywhere.

If
it is a supertask to consider
infinitely-many tasks finitely,
by finitely-many finite-length claims,
then
we tremble on the edge of becoming Chuck Norris
(who has reportedly counted to infinity -- twice)
when we watch Achilles sprint after a tortoise,
or only walk across the room.

That strikes me as an extravagant shadow
for "supertask" to cast.
Perhaps we shouldn't call
making finitely-many finite-length claims
about infinitely-many tasks
"supertask".

Asking this kind of question ("Supertask is?")
commits an act of philosophy, I suspect.
But maybe you (FF) will let it go this once.

On 09.11.2023 11:22, Fritz Feldhase wrote:
> On Thursday, November 9, 2023 at 9:18:48 AM UTC+1, Transfinity wrote:
>> Fritz Feldhase schrieb am Montag, 6. November 2023 um 13:02:13 UTC+1:
>>> On Monday, November 6, 2023 at 10:15:10 AM UTC+1, Transfinity wrote:
>>>>>>
>>>>>> Cantor's attempt
>>>>>>
>>>>> is successful, it produces matrix B
>>>>>
>>>> in which step?
>>>>
>>> No steps needed for that.
>>>
>> Counting is a stepwise process.
>
> Cantor did not count the rational numbers,

But he tried and he believed that he had succeeded: "wie man sich
gewöhnlich ausdrückt, aufeinander abzählen lassen." [1883, 5. Okt.
Cantor, Brief an Wundt]

>>>
>>> Hint: You billiard
>>>
>> It produces B
>
> No, it doesn't.

It produces B because no part of B can be found that would not be produced.
>
> Hint: For each and every n e IN: A(n) =/= B.
>
Of course. A is not B but contains B. Never B alone is produced, neither
by Cantor nor by me.

Regards, WM

On 09.11.2023 19:35, Fritz Feldhase wrote:
> On Thursday, November 9, 2023 at 5:29:50 PM UTC+1, Jim Burns wrote:
>
>> If all swaps are swapped,
>> then Bob is not anywhere.
>
> Since Bob is always somewhere, never all swaps are swapped.

That is possible. But if Cantor is right

Die weite Reise, welche Herbart seiner "wandelbaren Grenze" vorschreibt,
ist eingestandenermaßen nicht auf einen endlichen Weg beschränkt; so muß
denn ihr Weg ein unendlicher, und zwar, weil er seinerseits nichts
Wandelndes, sondern überall fest ist, ein aktualunendlicher Weg sein. Es
fordert also jedes potentiale Unendliche (die wandelnde Grenze) ein
Transfinitum (den sichern Weg zum Wandeln) und kann ohne letzteres nicht
gedacht werden. [E. Zermelo: "Georg Cantor – Gesammelte Abhandlungen
mathematischen und philosophischen Inhalts", Springer, Berlin (1932) p.
393] When I first read it, I noted: wrong. But he was right in an
incredibly lucid way.

So if Cantor is right, then all swaps are swapped but Bob is present,
invisible though.

Regards, WM

On Friday, November 10, 2023 at 3:33:28 PM UTC+1, WM wrote:
> On 09.11.2023 11:22, Fritz Feldhase wrote:
> > On Thursday, November 9, 2023 at 9:18:48 AM UTC+1, Transfinity wrote:
> >> Fritz Feldhase schrieb am Montag, 6. November 2023 um 13:02:13 UTC+1:
> >>> On Monday, November 6, 2023 at 10:15:10 AM UTC+1, Transfinity wrote:
> >>>>>>
> >>>>>> Cantor's attempt
> >>>>>>
> >>>>> is successful, it produces matrix B
> >>>>>
> >>>> in which step?
> >>>>
> >>> No steps needed for that.
> >>>
> >> Counting is a stepwise process.
> >
> > Cantor did not count the rational numbers,
> >
> But he tried and

Nonsense.

Merke: Cantor war nicht so krank in der Birne wie Du.

> >>>
> >>> Hint: You billiard
> >>>
> >> It produces B
> >>
> > No, it doesn't.
> >
> It

doesn't.

> > Hint: For each and every n e IN: A(n) =/= B.
> >
> Of course. A is not B but

There is no matrix A.

On Friday, November 10, 2023 at 3:47:38 PM UTC+1, WM wrote:
> On 09.11.2023 19:35, Fritz Feldhase wrote:
> > On Thursday, November 9, 2023 at 5:29:50 PM UTC+1, Jim Burns wrote:
> >>
> >> If all swaps are swapped,
> >> then Bob is not anywhere.
> >>
> > Since Bob is always somewhere, never all swaps are swapped.
> >
> That is

a logical conclusion.

> <nonsense deleted>

On 11/10/2023 9:47 AM, WM wrote:
> On 09.11.2023 19:35, Fritz Feldhase wrote:
>> On Thursday, November 9, 2023
>> at 5:29:50 PM UTC+1, Jim Burns wrote:

>>> If all swaps are swapped,
>>> then Bob is not anywhere.
>>
>> Since Bob is always somewhere,
>> never all swaps are swapped.
>
> That is possible.
> But if Cantor is right

You mean: If all swaps are swapped

There are better ways to say what you mean.

If I said
| If Pythagoras is right,
| the square of a polygon's longest side
| equals the sum of the squares of
| the remaining sides
| there would be an implicature that,
if Pythagoras is right,
that would be true for all polygons.
We agree that it isn't, not for all.

That claim being false for some polygons
only contradicts a claim Pythagoras didn't make.

What is right about "Pythagoras is right" is
that the polygon of interest is a right triangle,
if it is, if that is what we are talking about.

If Cantor is right,
Cantor is right about the discussion being about
all swaps being swapped
(allowing for differences in GC- and WM-presentation).

Which, at last, is to say,
if all swaps are swapped.

> So if Cantor is right,
> then all swaps are swapped
> but Bob is present,
> invisible though.

For each visible ⟨p,q⟩
there is a visible swap ⟨p,q⟩↔#⟨p,q⟩
#⟨p,q⟩ := ⟨p+(p+q-1)(p+q-2)/2, 1⟩
such that
#⟨p,q⟩ is visible

If only all visible swaps are swapped,
Bob never swaps anywhere invisible.

For each visible ⟨p,q⟩
if ⟨p,q⟩ holds Bob,
then visible ⟨p,q⟩↔#⟨p,q⟩ is not swapped.

If Bob is visibly present,
then not all visible swaps are swapped.

If only all visible swaps are swapped,
Bob never swaps anywhere invisible,
and
Bob is not visibly present.

"Infinite" is not the same as
"ridiculously-large partly-darkᵂᴹ"

On 11/10/2023 9:33 AM, WM wrote:
> On 09.11.2023 11:22, Fritz Feldhase wrote:
>> On Thursday, November 9, 2023
>> at 9:18:48 AM UTC+1, Transfinity wrote:

>>> It produces B
>>
>> No, it doesn't.
>
> It produces B because
> no part of B can be found that
> would not be produced.

Each part of B is produced by finitely-many swaps.
B is not produced by finitely-many swaps.

The minimally-each-accessible
matrix-sequence ⟨ A(0) A(1) ... ⟩
only holds matrices produced by
finitely-many-swaps.
So, B is not-in ⟨ A(0) A(1) ... ⟩

Each part of B is produced by finitely-many swaps.
For each matrix-place ⟨i,j⟩
there is a finitely-indexed swap ⟨i,j⟩↔⟨kᵢⱼ,1⟩
such that,
∀k′>kᵢⱼ: A(k′)[i,j] = B[i,j]

B is not produced by finitely-many swaps.
For each finitely-indexed swap ⟨i,j⟩↔⟨kᵢⱼ,1⟩
there is a matrix-place ⟨i′,j′⟩
such that
A(kᵢⱼ)[i′,j′] ≠ B[i′,j′]

On 11/9/2023 8:29 AM, Jim Burns wrote:
> On 11/9/2023 3:30 AM, Chris M. Thomasson wrote:
>> On 11/9/2023 12:26 AM, Transfinity wrote:
>
>>> [...]
>>
>> You do realize that
>> an index can be used to create
>> a unique pair, right? This pair can be used for
>> a fraction p[0] / p[1]?
>> This pair can be used to go right back to
>> the unique natural number that created it.
>> Its full circle:
>>
>> index to pair, and pair to index
>>
>> Nothing is lost.
>
> Yes. Well-put.

Thanks Jim. :^)

Humm... Perhaps WM is so "hyper finite" that the mere idea if infinity
makes blood want to shoot out of his eyes?

>
> sₖ = ⌈½+(2⋅k+¼)¹ᐟ²⌉
> iₖ = k-(sₖ-1)(sₖ-2)/2
> jₖ = sₖ-iₖ
>
> sᵢⱼₖ = iₖ+jₖ
> kᵢⱼₖ = iₖ+(sᵢⱼₖ-1)(sᵢⱼₖ-2)/2
> sᵢⱼₖ = sₖ
> kᵢⱼₖ = k
>
> And around it goes,
> index to pair to index to pair.
> A snake swallowing its own tail,
> in each of infinitely-many instances.
>
> sₖᵢⱼₖ = ⌈½+(2⋅kᵢⱼₖ+¼)¹ᐟ²⌉
> iₖᵢⱼₖ = kᵢⱼₖ-(sₖᵢⱼₖ-1)(sₖᵢⱼₖ-2)/2
> jₖᵢⱼₖ = sₖᵢⱼₖ-iₖᵢⱼₖ
> sₖᵢⱼₖ = sᵢⱼₖ = sₖ
> iₖᵢⱼₖ = iₖ
> jₖᵢⱼₖ = jₖ
>
> Arithmetic.
>
>

On Thursday, November 9, 2023 at 4:14:43 AM UTC-4, Transfinity wrote:
> William schrieb am Montag, 6. November 2023 um 11:18:50 UTC+1:

> > Your billiard game never produces a matrix in which no fraction "sits at an indexed place" Your billiard game cannot produce B.
> What position of Cantor's matrix is missing?

No position of the matrix B, which represents Cantor's bijecition, is "missing". For each n it is easy to see that A(n) has the same positions as B but is not B. You talk about the positions of a putative matrix A. However you billiard game does not produce matrix A.

On 11/10/2023 4:12 PM, Chris M. Thomasson wrote:
> On 11/9/2023 8:29 AM, Jim Burns wrote:
>> On 11/9/2023 3:30 AM, Chris M. Thomasson wrote:
>>> On 11/9/2023 12:26 AM, Transfinity wrote:

>>>> [...]
>>>
>>> You do realize that
>>> an index can be used to create
>>> a unique pair, right? This pair can be used for
>>> a fraction p[0] / p[1]?
>>> This pair can be used to go right back to
>>> the unique natural number that created it.
>>> Its full circle:
>>>
>>> index to pair, and pair to index
>>>
>>> Nothing is lost.
>>
>> Yes. Well-put.
>
> Thanks Jim. :^)

My biggest stumbling block is that
all of this is dead-familiar to me.
I knew what you had pointed out,
but it hadn't yet occurred to me that
someone else might not know it. Duh!
So, thank you.

> Humm...
> Perhaps WM is so "hyper finite" that
> the mere idea if infinity
> makes blood want to shoot out of his eyes?

Perhaps.

My most recent theory is that
what WM thinks is infinite is basically
what we mean by finite,
but really, really, really big.

Something like that is used in quantum mechanics.
A particle is put in a box with sides of length L.
The states, energies, etc, are solved.
L is floating around in the solutions somewhere.
Then, we let L "go to infinity", after which
L _should_ become irrelevant in comparison to
other terms in the answer(s).

In other words, we let L become negligible.
And neglecting L works, sometimes.

However, hypothetical-WM,
having learned the trick of neglecting L
is unwilling to learn a second or third trick.
It is said that, when one is holding a hammer,
everything looks like a nail.
That might explain actual-WM.

>> sₖ = ⌈½+(2⋅k+¼)¹ᐟ²⌉
>> iₖ = k-(sₖ-1)(sₖ-2)/2
>> jₖ = sₖ-iₖ
>>
>> sᵢⱼₖ = iₖ+jₖ
>> kᵢⱼₖ = iₖ+(sᵢⱼₖ-1)(sᵢⱼₖ-2)/2
>> sᵢⱼₖ = sₖ
>> kᵢⱼₖ = k
>>
>> And around it goes,
>> index to pair to index to pair.
>> A snake swallowing its own tail,
>> in each of infinitely-many instances.
>>
>> sₖᵢⱼₖ = ⌈½+(2⋅kᵢⱼₖ+¼)¹ᐟ²⌉
>> iₖᵢⱼₖ = kᵢⱼₖ-(sₖᵢⱼₖ-1)(sₖᵢⱼₖ-2)/2
>> jₖᵢⱼₖ = sₖᵢⱼₖ-iₖᵢⱼₖ
>> sₖᵢⱼₖ = sᵢⱼₖ = sₖ
>> iₖᵢⱼₖ = iₖ
>> jₖᵢⱼₖ = jₖ
>>
>> Arithmetic.
>>
>>
>

On Saturday, November 11, 2023 at 12:08:59 AM UTC+1, Jim Burns wrote:
> On 11/10/2023 4:12 PM, Chris M. Thomasson wrote:
> >
> > Perhaps WM is so "hyper finite" that
> > the mere idea of infinity
> > makes blood want to shoot out of his eyes?
> >
> Perhaps.

The main problem (imho) is that he's missing a brain.

> My most recent theory is that
> what WM thinks is infinite is basically
> what we mean by finite,
> but really, really, really big.

Right. It seems that he's not able to comprehend the idea of "infinity".

Of course, little wonder, given his lacking brain.

> It is said that, when one is holding a hammer, everything looks like a nail.

Right. :-P

> That might explain actual-WM.

Yeah.

On the other hand: No brain => no mind => no comprehension.

On 11/10/2023 3:57 PM, Fritz Feldhase wrote:
> On Saturday, November 11, 2023 at 12:08:59 AM UTC+1, Jim Burns wrote:
>> On 11/10/2023 4:12 PM, Chris M. Thomasson wrote:
>>>
>>> Perhaps WM is so "hyper finite" that
>>> the mere idea of infinity
>>> makes blood want to shoot out of his eyes?
>>>
>> Perhaps.
>
> The main problem (imho) is that he's missing a brain.
>
>> My most recent theory is that
>> what WM thinks is infinite is basically
>> what we mean by finite,
>> but really, really, really big.
>
> Right. It seems that he's not able to comprehend the idea of "infinity".
>
> Of course, little wonder, given his lacking brain.
>
>> It is said that, when one is holding a hammer, everything looks like a nail.
>
> Right. :-P
>
>> That might explain actual-WM.
>
> Yeah.
>
> On the other hand: No brain => no mind => no comprehension.

I am wondering if the following could possibly cause WM to take up
drinking high proof booze everyday?

https://youtu.be/5qXSeNKXNPQ

Btw, I love fractals, and infinity is indeed involved however it does
not make me go fucking crazy! :^)

On 10.11.2023 23:36, William wrote:
> On Thursday, November 9, 2023 at 4:14:43 AM UTC-4, Transfinity wrote:
>> William schrieb am Montag, 6. November 2023 um 11:18:50 UTC+1:
>
>
>>> Your billiard game never produces a matrix in which no fraction "sits at an indexed place" Your billiard game cannot produce B.
>> What position of Cantor's matrix is missing?
>
> No position of the matrix B, which represents Cantor's bijecition, is "missing". For each n it is easy to see that A(n) has the same positions as B but is not B.

True.

> You talk about the positions of a putative matrix A. However you billiard game does not produce matrix A.

It produces all of B that Cantor produces. But it includes more than B.
That proves the impossibility to produce by Cantor's prescripition only B.

Regards, WM
>

On Saturday, November 11, 2023 at 7:44:16 AM UTC-4, WM wrote:
> On 10.11.2023 23:36, William wrote:

> > You talk about the positions of a putative matrix A. However you billiard game does not produce matrix A.
> It produces all of B that Cantor produces.

Not in a single matrix Cantor provides a bijection which can be represented by the matrix B. Your billiard game does not produce a matrix of which B is a submatrix. In particular is does not produce the putative matrix A so talking about the positions of A is meaningless,

On 11.11.2023 14:21, William wrote:
> On Saturday, November 11, 2023 at 7:44:16 AM UTC-4, WM wrote:
>> On 10.11.2023 23:36, William wrote:
>
>>> You talk about the positions of a putative matrix A. However you billiard game does not produce matrix A.
>> It produces all of B that Cantor produces.
>
> Not in a single matrix

I produce it by simply stating the rules of the game of billiards. That
is as valid as a complete mapping of all natnumbers as Cantor's mapping.
Concerning the consumption of natnumbers there is no difference that
could be shown.

> Cantor provides a bijection which can be represented by the matrix B.

So it appears, but it isimpossible as my game shows.

> Your billiard game does not produce a matrix of which B is a submatrix.

Show one natural number which is used as an index in B but not in my
game. (Of course completeness is either never attained or in both cases.
If it is attained in both cases, then also A is attained. The existence
of B implies the existence of A.)

Regards, WM

On Saturday, November 11, 2023 at 1:34:15 PM UTC-4, WM wrote:
> On 11.11.2023 14:21, William wrote:
> > On Saturday, November 11, 2023 at 7:44:16 AM UTC-4, WM wrote:
> >> On 10.11.2023 23:36, William wrote:
> >
> >>> You talk about the positions of a putative matrix A. However you billiard game does not produce matrix A.
> >> It produces all of B that Cantor produces.
> >
> > Not in a single matrix
> I produce it by simply stating the rules of the game of billiards.

Nope. those rules say that you can produce the matrices A(n). The rules of your billiard game do not give a way of producing a single matrix that has B as a submatrix. However we can specify a bijection between |N and |Nx|N. This specifies B.

On 11.11.2023 19:07, William wrote:
> On Saturday, November 11, 2023 at 1:34:15 PM UTC-4, WM wrote:
>> On 11.11.2023 14:21, William wrote:
>>> On Saturday, November 11, 2023 at 7:44:16 AM UTC-4, WM wrote:
>>>> On 10.11.2023 23:36, William wrote:
>>>
>>>>> You talk about the positions of a putative matrix A. However you billiard game does not produce matrix A.
>>>> It produces all of B that Cantor produces.
>>>
>>> Not in a single matrix
>> I produce it by simply stating the rules of the game of billiards.
>
> Nope. those rules say that you can produce the matrices A(n).

The rules say that Cantor can enumerate the fraction m/n by index k.

> The rules of your billiard game do not give a way of producing a single matrix that has B as a submatrix. However we can specify a bijection between |N and |Nx|N. This specifies B.

This specification specifies that k enumerates m/n, but for all k used
by Cantor we can prove |ℕ \ {1, 2, 3, ..., k}| = ℵo.

By the way how would you enumerate the algebraic numbers? Have you a
bijection between ℕ and the algebraics? Did Cantor or Dedekind, who
first enumerated the algebraics, have more than a stepwise process? Look
for all polynomials of height n and enumerate their roots? That will
never become complete. We must accept that a process being possible to
go on without obstacle, leads to completeness. Just like Cantor taught
us. Just like the matrices B and A.

Regards, WM

On 11.11.2023 19:07, William wrote:
> On Saturday, November 11, 2023 at 1:34:15 PM UTC-4, WM wrote:

>> I produce it by simply stating the rules of the game of billiards.
>
> Nope. those rules say that you can produce the matrices A(n). The rules of your billiard game do not give a way of producing a single matrix that has B as a submatrix. However we can specify a bijection between |N and |Nx|N. This specifies B.

The proof of equinumerosity by bijection between infinite sets, M and N,
is justified by mathematical induction: If every element of set M can be
related to one and only one corresponding element of set N and vice
versa, and if there is never an obstacle or halt(*) in this process of
assignment, then both infinite sets are in bijection.

(*) Cantor gave the necessary condition: "und es erfährt daher der aus
unsrer Regel resultierende Zuordnungsprozeß keinen Stillstand." [Cantor,
p. 239]
That means: There is no halt of the assignement process resulting from
our rule.

Can you find a halt or obstacle in the assignment process of my matrices?

Regards, WM

On Saturday, November 11, 2023 at 3:14:25 PM UTC-4, WM wrote:

> > The rules of your billiard game do not give a way of producing a single matrix that has B as a submatrix. However we can specify a bijection between |N and |Nx|N. This specifies B.
> This specification specifies that k enumerates [(m,n)]

Nope. By "k enumerates (m.n)" you mean there is a stepwise process that produces B. There is no stepwise process (like your game of billiards) that produces B. This does not mean you cannot specify B (nor does it mean that there cannot be a stepwise process involved in specifying B, it just means you cannot use *only* a stepwise process. e.g for every pair (m,n) there is an r(m,n) such that
B[m,n]=A(r(m,n))[m,n], as well as the stepwise process for producing the A(k) you need to specify the function r(m,n) )

On Saturday, November 11, 2023 at 3:23:26 PM UTC-4, WM wrote:
> Cantor gave the necessary condition: "und es erfährt daher der aus
> unsrer Regel resultierende Zuordnungsprozeß keinen Stillstand." [Cantor,
> p. 239]
> That means: There is no halt of the assignement process resulting from
> our rule.
>

Correct. Cantor gave a necessary condition, not a sufficient condition.

> Can you find a halt or obstacle in the assignment process of my matrices?

Nope. Your billiards game still does not produce a matrix such that B is a submatrix.

On 11.11.2023 21:32, William wrote:
> On Saturday, November 11, 2023 at 3:23:26 PM UTC-4, WM wrote:
>> Cantor gave the necessary condition: "und es erfährt daher der aus
>> unsrer Regel resultierende Zuordnungsprozeß keinen Stillstand." [Cantor,
>> p. 239]
>> That means: There is no halt of the assignement process resulting from
>> our rule.
>>
>
> Correct. Cantor gave a necessary condition, not a sufficient condition.

If Cantor's necessary condition is not a sufficient condition
simultaneously, then there is no sufficient condition at all and there
is no complete application of indices, because there is no other condition.

What is the sufficient condition in case of indexing the algebraic numbers?
>
>> Can you find a halt or obstacle in the assignment process of my matrices?
>
> Nope. Your billiards game still does not produce a matrix such that B is a submatrix.

Unjustified claim. Proof: You cannot find any number in B that is not
put by my matrix-language.

Regards, WM

On 11.11.2023 21:21, William wrote:
> On Saturday, November 11, 2023 at 3:14:25 PM UTC-4, WM wrote:
>
>
>>> The rules of your billiard game do not give a way of producing a single matrix that has B as a submatrix. However we can specify a bijection between |N and |Nx|N. This specifies B.
>> This specification specifies that k enumerates [(m,n)]
>
> Nope. By "k enumerates (m.n)" you mean there is a stepwise process that produces B. There is no stepwise process (like your game of billiards) that produces B.

Other ways of producing B are not existing.

>This does not mean you cannot specify B (nor does it mean that there cannot be a stepwise process involved in specifying B, it just means you cannot use *only* a stepwise process. e.g for every pair (m,n) there is an r(m,n) such that
> B[m,n]=A(r(m,n))[m,n], as well as the stepwise process for producing the A(k) you need to specify the function r(m,n) )

The function is only a shortcut describing the steps. Proof: There is no
obstacle or halt in the sequence of steps.
The function is described in the OP. It is given only in another
language. Should set theory depend on the used language?
The function for indexing the algebraic numbers for instance has been
given by Dedekind and repeated by Cantor:
Gehen wir auf die Gleichung (1), welcher eine algebraischen Zahl 
genügt und welche nach den gedachten Festsetzungen eine völlig bestimmte
ist, zurück, so möge die Summe der absoluten Beträge ihrer
Koeffizienten, vermehrt um die Zahl n - 1, wo n den Grad von  angibt,
die Höhe der Zahl  genannt und mit N bezeichnet werden; es ist also,
unter Anwendung der üblich gewordenen Bezeichnungsweise:
N = n - 1 + |a0| + |a1| + ... + |an|
Die Höhe N ist danach für jede reelle algebraische Zahl  eine
bestimmte positive ganze Zahl; umgekehrt gibt es zu jedem positiven
ganzzahligen Werte von N nur eine endliche Anzahl algebraischer reeller
Zahlen mit der Höhe N; die Anzahl derselben sei (N); es ist
beispielsweise (1) = 1; (2) = 2; (3) = 4. Es lassen sich alsdann die
Zahlen des Inbegriffes (), d. h. sämtliche algebraischen reellen Zahlen
folgendermaßen anordnen: man nehme als erste Zahl 1 die eine Zahl mit
der Höhe N = 1; lasse auf sie, der Größe nach steigend, die (2) = 2
algebraischen reellen Zahlen der Höhe N = 2 folgen, bezeichne sie mit
2, 3; an diese mögen sich die (3) = 4 Zahlen mit der Höhe N = 3,
ihrer Größe nach aufsteigend, anschließen; allgemein mögen, nachdem in
dieser Weise sämtliche Zahlen aus () bis zu einer gewissen Höhe N = N1
abgezählt und an einen bestimmten Platz gewiesen sind, die reellen
algebraischen Zahlen mit der Höhe N = N1 + 1 auf sie folgen, und zwar
der Größe nach aufsteigend; so erhält man den Inbegriff () aller
reellen algebraischen Zahlen in der Form

ω1, ω2, ... ωn,...
und kann mit Rücksicht auf diese Anordnung von der ten algebraischen
Zahl reden, wobei keine einzige aus dem Inbegriffe () vergessen ist
Use Google for translation. Only the last sentence is important:

"thus we get the epitome (ω) of all real algebraic numbers [...] and
with respect to this order we can talk about the th algebraic number
where not a single one of this epitome () has been forgotten." [E.
Zermelo: "Georg Cantor – Gesammelte Abhandlungen mathematischen und
philosophischen Inhalts", Springer, Berlin (1932) p. 116]

What is the sufficient condition here?

Regards, WM

On Sunday, November 12, 2023 at 4:33:41 AM UTC-4, WM wrote:
> On 11.11.2023 21:21, William wrote:
> > On Saturday, November 11, 2023 at 3:14:25 PM UTC-4, WM wrote:
> >
> >
> >>> The rules of your billiard game do not give a way of producing a single matrix that has B as a submatrix. However we can specify a bijection between |N and |Nx|N. This specifies B.
> >> This specification specifies that k enumerates [(m,n)]
> >
> > Nope. By "k enumerates (m.n)" you mean there is a stepwise process that produces B. There is no stepwise process (like your game of billiards) that produces B.
> Other ways of producing B are not existing.
> >This does not mean you cannot specify B (nor does it mean that there cannot be a stepwise process involved in specifying B, it just means you cannot use *only* a stepwise process. e.g for every pair (m,n) there is an r(m,n) such that
> > B[m,n]=A(r(m,n))[m,n], as well as the stepwise process for producing the A(k) you need to specify the function r(m,n) )
> The function is only a shortcut describing the steps.

Nope.

> Proof: There is no
> obstacle or halt in the sequence of steps.

So what? At no step is a matrix produdced such that B is a submatrix.