On Tuesday, October 31, 2023 at 12:43:35 PM UTC-3, WM wrote:
> William schrieb am Dienstag, 31. Oktober 2023 um 15:56:03 UTC+1:
> > On Tuesday, October 31, 2023 at 6:40:09 AM UTC-3, WM wrote:
> > > William schrieb am Montag, 30. Oktober 2023 um 22:51:21 UTC+1:
> > >
> > > > So there is a bijection between the positions of B (the elements of |Nx|N) and the natural numbers |N.
> > > There is no bijection between the positions of any A(n) [and |N]
> >
> > Note A(n) has the same positions for each n.
> A(n) is the matrix after the nth push. It has all positions which A(1), A(2), ..., and A have.
There is no matrix A

Jim Burns schrieb am Dienstag, 31. Oktober 2023 um 15:11:01 UTC+1:
> On 10/31/2023 5:20 AM, WM wrote:
> > Jim Burns schrieb am Dienstag,
> > 31. Oktober 2023 um 04:29:43 UTC+1:
> >> This is the point of
> >> Cantor's fraction-sequence,
> >> Disappearing Bob,
> >
> > and related nonsense.
> If you're claiming arithmetic is nonsense,

Disappearing Bob is nonsense.

> why do you snip my referring to that claim?

Reread the OP and find the positions of the fractions. Fail. But they are there, namely at not indexed positions nwithin the matrix, according to the rules of the game.

Regards, WM

On 10/31/2023 11:53 AM, WM wrote:
> Jim Burns schrieb am Dienstag,
> 31. Oktober 2023 um 15:11:01 UTC+1:
>> On 10/31/2023 5:20 AM, WM wrote:
>>> Jim Burns schrieb am Dienstag,
>>> 31. Oktober 2023 um 04:29:43 UTC+1:

>>>> This is the point of
>>>> Cantor's fraction-sequence,
>>>> Disappearing Bob,
>>>
>>> and related nonsense.
>>
>> If you're claiming arithmetic is nonsense,
>
> Disappearing Bob is nonsense.

If arithmetic is correct,
then Disappearing Bob is correct.

If
each natural n is at place ⟨i,j⟩
i = n-(s-1)(s-2)/2
j = s-i
s = ⌈½+(2⋅n+¼)¹ᐟ²⌉
then
each place ⟨i,j⟩ holds natural n
n = i+(s-1)(s-2)/2
s = i+j
and Bob (nee 1/2) has disappeared.

If, as you say,
Disappearing Bob is nonsense,
then
arithmetic is nonsense.

>> why do you snip my referring to that claim?
>
> Reread the OP and
> find the positions of the fractions.

Each fraction is at a ⟨row,column⟩ from
the minimally-each-accessible set ℕ

> Fail.
> But they are there, namely at
> not indexed positions n within the matrix,
> according to the rules of the game.

A fraction p/q only wanders to ⟨row,column⟩ from
the minimally-each-accessible set ℕ

Each minimally-each-accessible row/column ⟨i,j⟩
holds an index n = i+(i+j-1)(i+j-2)/2
and is not not-indexed.

If, as you say,
fractions are outside the indexed rows/columns,
they got there by means other than the game.

Arithmetic only provides wandering within
minimally-each-accessible rows/columns.

And.

After all wandering, only naturals are within
the minimally-each-accessible rows-columns.

But perhaps arithmetic is nonsense.
Make the case.

So far, the best (poor) case you have made is
the attempt to hide (eg, by snippage) that which
you are attempting to convince us of:
"arithmetic is nonsense".

Consider what it says that
you have no better case than hiding to make.

Consider the alternative:
That arithmetic is not nonsense, and
that some sets are the same "size" as
some of their proper subsets.

Jim Burns schrieb am Dienstag, 31. Oktober 2023 um 18:23:21 UTC+1:
> On 10/31/2023 11:53 AM, WM wrote:
> > Jim Burns schrieb am Dienstag,
> > 31. Oktober 2023 um 15:11:01 UTC+1:
> >> On 10/31/2023 5:20 AM, WM wrote:
> >>> Jim Burns schrieb am Dienstag,
> >>> 31. Oktober 2023 um 04:29:43 UTC+1:
>
> >>>> This is the point of
> >>>> Cantor's fraction-sequence,
> >>>> Disappearing Bob,
> >>>
> >>> and related nonsense.
> >>
> >> If you're claiming arithmetic is nonsense,
> >
> > Disappearing Bob is nonsense.
> If arithmetic is correct,
> then Disappearing Bob is correct.

The familiar arithmetic is correct in the visible domain.
Logic is correct everywhere. Lossless exchanges are lossless in logic.
>
> and Bob (nee 1/2) has disappeared

beyond visibility, not from the matrix.
>
> If, as you say,
> Disappearing Bob is nonsense,
> then
> arithmetic is nonsense.

Logic remains true. Arithmetic is based on experience in the visible domain..

> > Reread the OP and
> > find the positions of the fractions.
> Each fraction is at a ⟨row,column⟩ from
> the minimally-each-accessible set ℕ

Can't understand that word.

> If, as you say,
> fractions are outside the indexed rows/columns,
> they got there by means other than the game.

The border between visible and dark domain ist not rigid.

> After all wandering, only naturals are within
> the minimally-each-accessible rows-columns.

But the fractions remain in the matrix. How should they leave?
>
> But perhaps arithmetic is nonsense.
> Make the case.
>
> So far, the best (poor) case you have made is
> the attempt to hide (eg, by snippage) that which
> you are attempting to convince us of:
> "arithmetic is nonsense".

Arithmetic is for n also n^n^n exists. Nevertheless there are dark numbers which cannot be reached from any n.

> Consider the alternative:
> That arithmetic is not nonsense, and
> that some sets are the same "size" as
> some of their proper subsets.

Aithmetic is nonsense at the completion of the enumeration. Completion is claimed however.

Regards, WM

William schrieb am Dienstag, 31. Oktober 2023 um 16:50:54 UTC+1:
> On Tuesday, October 31, 2023 at 12:43:35 PM UTC-3, WM wrote:

> > A(n) is the matrix after the nth push. It has all positions which A(1), A(2), ..., and A have.
> There is no matrix A

Then there is no complete application of the natural nunbers. A would describe just this state.

Regards, WM

On 10/31/2023 8:53 AM, WM wrote:
> Jim Burns schrieb am Dienstag, 31. Oktober 2023 um 15:11:01 UTC+1:
>> On 10/31/2023 5:20 AM, WM wrote:
>>> Jim Burns schrieb am Dienstag,
>>> 31. Oktober 2023 um 04:29:43 UTC+1:
>>>> This is the point of
>>>> Cantor's fraction-sequence,
>>>> Disappearing Bob,
>>>
>>> and related nonsense.
>> If you're claiming arithmetic is nonsense,
>
> Disappearing Bob is nonsense.
>
>> why do you snip my referring to that claim?
>
> Reread the OP and find the positions of the fractions. Fail. But they are there, namely at not indexed positions nwithin the matrix, according to the rules of the game.

index to pair along with pair to index. Nothing is lost.

On Tuesday, October 31, 2023 at 4:02:07 PM UTC-3, WM wrote:
> William schrieb am Dienstag, 31. Oktober 2023 um 16:50:54 UTC+1:
> > On Tuesday, October 31, 2023 at 12:43:35 PM UTC-3, WM wrote:
>
> > > A(n) is the matrix after the nth push. It has all positions which A(1), A(2), ..., and A have.
> > There is no matrix A
> Then there is no complete application
meaningless word salad.

At each step n your game of billiards produces A(n). There is no step which produces A, i.e.your game of billiards never produces A.

On 10/31/2023 11:02 AM, WM wrote:
> Jim Burns schrieb am Dienstag, 31. Oktober 2023 um 18:23:21 UTC+1:
>> On 10/31/2023 11:53 AM, WM wrote:
>>> Jim Burns schrieb am Dienstag,
>>> 31. Oktober 2023 um 15:11:01 UTC+1:
>>>> On 10/31/2023 5:20 AM, WM wrote:
>>>>> Jim Burns schrieb am Dienstag,
>>>>> 31. Oktober 2023 um 04:29:43 UTC+1:
>>
>>>>>> This is the point of
>>>>>> Cantor's fraction-sequence,
>>>>>> Disappearing Bob,
>>>>>
>>>>> and related nonsense.
>>>>
>>>> If you're claiming arithmetic is nonsense,
>>>
>>> Disappearing Bob is nonsense.
>> If arithmetic is correct,
>> then Disappearing Bob is correct.
>
> The familiar arithmetic is correct in the visible domain.
> Logic is correct everywhere. Lossless exchanges are lossless in logic.

Have you heard of lossy compression?

[...]

On 10/31/2023 8:50 AM, WM wrote:
> Fritz Feldhase schrieb am Dienstag, 31. Oktober 2023 um 14:02:34 UTC+1:
>> On Tuesday, October 31, 2023 at 10:40:09 AM UTC+1, WM wrote:
>>>
>>> There is no bijection between the positions [elements] of [the matrix]
>>> 1/1, 1/2, 1/3, 1/4, ...
>>> 2/1, 2/2, 2/3, 2/4, ...
>>> 3/1, 3/2, 3/3, 3/4, ...
>>> 4/1, 4/2, 4/3, 4/4, ...
>>> 5/1, 5/2, 5/3, 5/4, ...
>>> ...
>>> and ℕ.
>
>> It is {(n/m, m + (m + n - 1)(m + n - 2)/2) : n,m e IN} = {(1/1, 1), (1/2, 2), (2/1, 3), ...}.
>
> Stay with the indexing the positions (m, n). After every push the number of natnumbers k and the number fractions p/q are constant and they all are within the matrix. After all pushes they remain in the matrix A too. But the fractions are invisible and their positions are not indexed (they are never indexed according to the rules). Your claim and

I think you need to look inside of your head to make sure you can see a
brain...

>
>> See: https://de.wikipedia.org/wiki/Cantorsche_Paarungsfunktion
>
> are refuted.
>
> Regards, WM

On 10/31/2023 2:02 PM, WM wrote:
> Jim Burns schrieb am Dienstag,
> 31. Oktober 2023 um 18:23:21 UTC+1:

>> Consider the alternative:
>> That arithmetic is not nonsense, and
>> that some sets are the same "size" as
>> some of their proper subsets.
>
> [Arithmetic] is nonsense at
> the completion of the enumeration.

Arithmetic can be described by
describing each number and each operation,
and then augmenting their description
with only not-first-false claims.

The description and augmentation are
_completely_ true of arithmetic,
because
a finite sequence of only not-first-false
can only hold true claims.

And.

Those numbers have no second end
because
that's how arithmetic is.

----
Each successor of an accessible is
a visible.

Each sum of, each product of, each power of
visibles is
a visible.

For each index n of visible ⟨i,j⟩
n = i+(i+j-2)(i+j-2)/2
and thus n is visible.

Okay, send the fractions to wander the matrix.
It won't matter in this respect:
they never step out of the visible.

And.

When all the wandering is done, and
all the indices have settled in at
all the visible rows/columns,
no non-integral fractions will be visible..

Because
infinite is not merely more-of-the-same.

> Completion is claimed however.

No unspeakably-self-equal things are claimed.

Arithmetic is claimed.

Each two visible numbers have
visible successors, a visible sum,
a visible product, a visible power.

Fractions only wander from
visible rows/columns to
visible rows/columns.

And.

When all the wandering is done,
no fractions are in any
visible row/column.

William schrieb am Dienstag, 31. Oktober 2023 um 22:22:47 UTC+1:
> On Tuesday, October 31, 2023 at 4:02:07 PM UTC-3, WM wrote:
> > William schrieb am Dienstag, 31. Oktober 2023 um 16:50:54 UTC+1:
> > > On Tuesday, October 31, 2023 at 12:43:35 PM UTC-3, WM wrote:
> >
> > > > A(n) is the matrix after the nth push. It has all positions which A(1), A(2), ..., and A have.
> > > There is no matrix A
> > Then there is no complete application
> meaningless word salad.

For the uneducated, certainly. But read Cantor: complete enumeration of the rational numbers or in modern times: Consider a complete enumeration of all programs P(i).
>
> At each step n your game of billiards produces A(n). There is no step which produces A, i.e.your game of billiards never produces A.

Then never all indices have been applied in my game. This implies that also Cantor fails when stating: "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."

Many have been forgotten before matrix A is completed.

Note: I would not deny your point. I cannot disprove it. But then you must never again claim that all fractions or algebraic numbers can be enumerated.. That would be a self-contradiction.

Regards, WM

Jim Burns schrieb am Mittwoch, 1. November 2023 um 01:28:28 UTC+1:
> On 10/31/2023 2:02 PM, WM wrote:
> > Jim Burns schrieb am Dienstag,
> > 31. Oktober 2023 um 18:23:21 UTC+1:
> >> Consider the alternative:
> >> That arithmetic is not nonsense, and
> >> that some sets are the same "size" as
> >> some of their proper subsets.
> >
> > [Arithmetic] is nonsense at
> > the completion of the enumeration.
> Arithmetic can be described by
> describing each number and each operation,

Each number and each operation have infinitely many successors. Here is Cantor. Use Google for translation. Meanwhile they are doin pretty well:
"Denn warum kann man nach jeder Grenze, die man sich in der unendlichen Ausdehnung gesetzt hat, immer wieder weiter gehen? Weil hinter jeder angebbaren Ausdehnung immer noch Ausdehnung ist."

Note: Hinter jeder *angebbaren* = after each that can be named

"dieses 'Gebiet' kann aber nicht selbst wieder etwas Veränderliches sein, da sonst jede feste Unterlage der Betrachtung fehlen würde; also ist dieses 'Gebiet' eine bestimmte aktual-unendliche Wertmenge.

The actual infinite is definite and invariable.

> Those numbers have no second end
> because
> that's how arithmetic is.

Those numbers can be completed bevcause behind them there is something else (compare the unit fractions and zero behind them).
>
> ----
> Each successor of an accessible is
> a visible.

The last unit fractions before zero are dark.
>
> Each sum of, each product of, each power of
> visibles is
> a visible.

Whatever you may argue: The fractions cannot leave the matrix but are invisible when all natnumbers have been applied.

> Okay, send the fractions to wander the matrix.
> It won't matter in this respect:
> they never step out of the visible.

In matrix A they are contained but only its submatrix B is visible.
>
> And.
>
> When all the wandering is done, and
> all the indices have settled in at
> all the visible rows/columns,
> no non-integral fractions will be visible.

So it is. But they cannot have lost A because they only change positions.
>
> Because
> infinite is not merely more-of-the-same.

Infinite is all of the same.
>
> > Completion is claimed however.
>
> No unspeakably-self-equal things are claimed.
>
> Arithmetic is claimed

and algebraic numbers: "thus we get the epitome (ω) of all real algebraic numbers [...] and with respect to this order we can talk about the nth algebraic number where not a single one of this epitome (ω) has been forgotten."
>
> Each two visible numbers have
> visible successors, a visible sum,
> a visible product, a visible power.

Yes. It is the hard task to imagine the dark successors.
>
> Fractions only wander from
> visible rows/columns to
> visible rows/columns.

Yes. It is the hard task to imagine that they nevertheless find their way to dark positions.
>
> And.
>
> When all the wandering is done,
> no fractions are in any
> visible row/column.

So it is.

Regards, WM

On Wednesday, November 1, 2023 at 11:38:01 AM UTC+1, WM wrote:

> > When all the wandering is done,

which never is the case, dumbo.

Hint: A sequence does not have an end.

On Wednesday, November 1, 2023 at 7:12:47 AM UTC-3, WM wrote:

> > At each step n your game of billiards produces A(n). There is no step which produces A, i.e.your game of billiards never produces A.
> Then never all indices have been applied in my game.

Correct. Your game does not produce A.

On 11/1/2023 6:37 AM, WM wrote:
> Jim Burns schrieb am Mittwoch,
> 1. November 2023 um 01:28:28 UTC+1:

>> [...]
>
> Whatever you may argue:
> The fractions cannot leave the matrix
> but are invisible
> when all natnumbers have been applied.
>
>> Okay, send the fractions to wander the matrix.
>> It won't matter in this respect:
>> they never step out of the visible.
>
> In matrix A they are contained
> but only its submatrix B is visible.

The fractions never step out of visible B

From visible ⟨i,j⟩ in B to ⟨k,1⟩
k = i+(s-1)(s-2)/2
s = i+j
visible ⟨k,1⟩ in B

Endlessly, within B

Any fraction from B which is in A\B
got there without having stepped there,
breaking your rules.

Visible ⟨k,1⟩ to visible ⟨i,j⟩
i = k-(s-1)(s-2)/2
j = s-i
s = ⌈½+(2⋅k+¼)¹ᐟ²⌉

The first column of B is the same "size" as
all of B

>> And.
>>
>> When all the wandering is done, and
>> all the indices have settled in at
>> all the visible rows/columns,
>> no non-integral fractions will be visible.
>
> So it is.
> But they cannot have lost A because
> they only change positions.

They cannot have entered A\B because
k = i+(i+j-1)(i+j-2)/2 is visible == in B

>> Because
>> infinite is not merely more-of-the-same.
>
> Infinite is all of the same.

Infinite X cannot be ordered with
first‖last in X
and also with,
for each split F‖H of X
last-in-F‖first-in-H

Finite A can be ordered by ≪ with
first‖last in A existing
and also with,
for each split F‖H of A
last-in-F‖first-in-H existing

I am verbing "first" "last" and "next":

≪ firsts, lasts, and nexts A
By definition, A is finite.
(Paul Stäckel 1862-1919)

----
Theorem
If ≪ firsts, lasts, and nexts A
then
(finite) A cannot match any proper subset.

If ≪ doesn't first A or doesn't last A
or doesn't next A
then
(infinite) A can match some proper subsets.

Infinite is not merely more-of-the-same.

>> Each two visible numbers have
>> visible successors, a visible sum,
>> a visible product, a visible power.
>
> Yes.
> It is the hard task
> to imagine the dark successors.

The minimally-each-visible set ℕᵐᵉᵛ of numbers
holds each visible number and
is a subset of
each set which holds each visible number.

No visible i j k are not-in ℕᵐᵉᵛ

No dark δ is in ℕᵐᵉᵛ
| Otherwise,
| ℕᵐᵉᵛ\{δ} holds each visible but
| ℕᵐᵉᵛ isn't a subset of ℕᵐᵉᵛ\{δ}

visible == in ℕᵐᵉᵛ

Each two numbers in ℕᵐᵉᵛ have
successors in ℕᵐᵉᵛ, a sum in ℕᵐᵉᵛ,
a product in ℕᵐᵉᵛ, a power in ℕᵐᵉᵛ.

For each place ⟨i,j⟩ in ℕᵐᵉᵛ×ℕᵐᵉᵛ
there is an index k in ℕᵐᵉᵛ
k = i+(s-1)(s-2)/2
s = i+j

For each index k in ℕᵐᵉᵛ
there is an index ⟨i,j⟩ in ℕᵐᵉᵛ×ℕᵐᵉᵛ
i = k-(s-1)(s-2)/2
j = s-i
s = ⌈½+(2⋅k+¼)¹ᐟ²⌉

ℕᵐᵉᵛ and ℕᵐᵉᵛ×ℕᵐᵉᵛ are the same "size".

And.

Everything in ℕᵐᵉᵛ and ℕᵐᵉᵛ×ℕᵐᵉᵛ is visible.

On Wednesday, November 1, 2023 at 6:56:12 PM UTC+1, Jim Burns wrote:

> The minimally-each-visible set ℕᵐᵉᵛ of numbers <bla>

Question: ℕᵐᵉᵛ c ℕ & ℕᵐᵉᵛ =/= ℕ?

If not, why don't you just write "ℕ" instead of "ℕᵐᵉᵛ" (like SANE persons would do)?

Trying to confuse the reader just like Mückenheim?

"visible numbes"? WTF?!

Please fuck off, together with Mückenheim!

<No answer please!>

On 11/1/2023 6:56 AM, Fritz Feldhase wrote:
> On Wednesday, November 1, 2023
> at 11:38:01 AM UTC+1, WM wrote:
>> Jim Burns schrieb am Mittwoch,
>> 1. November 2023 um 01:28:28 UTC+1:

>>> When all the wandering is done,
>
> which never is the case, dumbo.

If any place is not indexed,
then not all wandering is done.

¬P ⇒ ¬W ⊢ W ⇒ P

If all wandering is done,
then each place is indexed.

> Hint: A sequence does not have an end.

Yes,
an endless sequence does not have two ends.

The order ≪ of sequence S _nexts_ S ==
for each split F‖H of S
last-in-F‖first-in-H exists

Typically, by endless sequence, we mean
some s₀ in S starts S
but nothing in S ends S

For each split F‖H of S
≪ firsts, nexts, and lasts F and
≪ firsts, nexts, but doesn't last H

Firsting, nexting and lasting makes F finite.
Firsting, nexting but not lasting
makes H infinite.

For each split F‖H
F is finite, H is infinite.

On Wednesday, November 1, 2023 at 8:00:52 PM UTC+1, Jim Burns wrote:
> On 11/1/2023 6:56 AM, Fritz Feldhase wrote:
> >
> > | When all the wandering is done, [JB]
> >
> > which never is the case, dumbo.

Especially if we are talking about infinite "collections" AND don't allow for so called "supertasks".

See: https://en.wikipedia.org/wiki/Supertask

(Hint: In the context of mathematics/set theory we *don't* adopt supertasks, espescially not in our arguments/proofs.)

Fritz Feldhase schrieb am Mittwoch, 1. November 2023 um 11:56:41 UTC+1:
> On Wednesday, November 1, 2023 at 11:38:01 AM UTC+1, WM wrote:
>
> > > When all the wandering is done,
> which never is the case,

Then never all fractions are enumerated.
>
> Hint: A sequence does not have an end.

But it can be complete, according to Cantor.

Regards, WM

On 11/1/2023 2:36 PM, Fritz Feldhase wrote:
> On Wednesday, November 1, 2023
> at 6:56:12 PM UTC+1, Jim Burns wrote:

>> The minimally-each-visible set ℕᵐᵉᵛ
>> of numbers <bla>
>
> Question: ℕᵐᵉᵛ c ℕ & ℕᵐᵉᵛ =/= ℕ?

ℕᵐᵉᵛ = the communal ℕ
ℕᵐᵉᵛ ≠ Mückenheim's ℕ

> If not,
> why don't you just write "ℕ"
> instead of "ℕᵐᵉᵛ"
> (like SANE persons would do)?

Mückenheim responds _as though_
everyone else is using ℕ with dark numbers:
his ℕ
Even when I say otherwise.

I think that becomes more difficult to do if
I, in effect, append the definition (ᵐᵉᵛ)
every time I write ℕ

Or not. Who knows?
But that's what I'm trying to do.

> Trying to confuse the reader
> just like Mückenheim?

My sincere wish is that sane readers,
if they are confused by my notation,
will ask what I mean. As you did.

> "visible numbes"? WTF?!

Yes.
Visible, definable, accessible, tangible,
audible, olfactible. The opposite of dark.
You're new here, then?

The first number 1 is olfactible.

A number n is olfactible more broadly iff
a sequence ⟨1,..,n⟩ exists
such that
1‖n is first‖last in ⟨1,..,n⟩
and,
for each split F‖H of ⟨1,..,n⟩
last-in-F‖first-in-H exist and
are related by a successorship rule
first-in-H = last-in-F⁺¹

A number is dark if it is not that.

> Please fuck off, together with Mückenheim!
>
> <No answer please!>

Does telling someone to fuck off and then
asking them ("please") to do something
work for you (FF) very often?

On Wednesday, November 1, 2023 at 8:35:14 PM UTC+1, WM wrote:
> Fritz Feldhase schrieb am Mittwoch, 1. November 2023 um 11:56:41 UTC+1:
> > >
> > > When all the wandering is done, [JB]
> > >
> > which never is the case,
> >
> Then

Yaeh, whatever Mückenheim.

On 11/1/2023 12:46 PM, Fritz Feldhase wrote:
> On Wednesday, November 1, 2023 at 8:35:14 PM UTC+1, WM wrote:
>> Fritz Feldhase schrieb am Mittwoch, 1. November 2023 um 11:56:41 UTC+1:
>>>>
>>>> When all the wandering is done, [JB]
>>>>
>>> which never is the case,
>>>
>> Then
>
> Yaeh, whatever Mückenheim.

Afaict, WM is a hyper finite fool. Actually, it might be infinitely
moronic, but it does not know it yet... Humm...

Its all dark in the mind of WM.

William schrieb am Mittwoch, 1. November 2023 um 18:56:02 UTC+1:
> On Wednesday, November 1, 2023 at 7:12:47 AM UTC-3, WM wrote:
>
> > > At each step n your game of billiards produces A(n). There is no step which produces A, i.e.your game of billiards never produces A.
> > Then never all indices have been applied in my game.
> Correct. Your game does not produce A.

My game is closely related to Cantor's game. It produces the same as his.

Regards, WM

Jim Burns schrieb am Mittwoch, 1. November 2023 um 18:56:12 UTC+1:
> On 11/1/2023 6:37 AM, WM wrote:
> > Jim Burns schrieb am Mittwoch,
> > 1. November 2023 um 01:28:28 UTC+1:
> >> [...]
> >
> > Whatever you may argue:
> > The fractions cannot leave the matrix
> > but are invisible
> > when all natnumbers have been applied.

Simple fact by the rulesof the game. You cannot save Cantor other than by finding a deviation of my game form his.
> >
> >> Okay, send the fractions to wander the matrix.
> >> It won't matter in this respect:
> >> they never step out of the visible.
> >
> > In matrix A they are contained
> > but only its submatrix B is visible.
> The fractions never step out of visible B

Then Cantor is refuted either.
>
> Any fraction from B which is in A\B
> got there without having stepped there,
> breaking your rules.

The steps leading into A\B cannot be seen becaue A\B is dark.

> The first column of B is the same "size" as
> all of B

That is the bijection between potentially infinite collections.

Just today I received a letter where the author wrote: If bijective mappings are possible but not necessary, then in all other branches of science this would lead to refusal.

> > So it is.
> > But they cannot have lost A because
> > they only change positions.
> They cannot have entered A\B because
> k = i+(i+j-1)(i+j-2)/2 is visible == in B

Yes, for every visible k.

Regards, WM

Fritz Feldhase schrieb am Mittwoch, 1. November 2023 um 20:12:36 UTC+1:

> (Hint: In the context of mathematics/set theory we *don't* adopt supertasks, espescially not in our arguments/proofs.)

Why? Because it leads to contradictions!
If Cantor could enumerate all fractions then this could be represented by a super task because *every* pair would be visible. The supertask would cover every pair. Where should it fail? At no point! The ban of supertasks has been issued by defrauders who know all this.

Regards, WM