On 11/1/2023 2:16 PM, WM wrote:
> 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.

Cantor Pairing is real. You are a moron?

On Wednesday, November 1, 2023 at 10:16:19 PM UTC+1, WM wrote:

> My game is closely related to Cantor's [approach].

It clearly depends on it.

> It produces the same as his.

No, it doesn't, you psychotic asshole full of shit.

Your nonsense just produces shit.

Please see a psychatrist, soon!

On Wednesday, November 1, 2023 at 10:36:23 PM UTC+1, WM wrote:
> 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 they are part of philosophy, not mathematics.

See: https://en.wikipedia.org/wiki/Supertask
and: https://plato.stanford.edu/entries/spacetime-supertasks/

On Wednesday, November 1, 2023 at 6:16:19 PM UTC-3, WM wrote:
> 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. There is no last step in your game of billiards. At no step have all indices been applied.

On 11/1/2023 7:38 PM, Fritz Feldhase wrote:
> On Wednesday, November 1, 2023
> at 10:36:23 PM UTC+1, WM wrote:
>> 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 they are part of philosophy,
> not mathematics.
>
> See: https://en.wikipedia.org/wiki/Supertask
> and: https://plato.stanford.edu/entries/spacetime-supertasks/

Being part of philosophy and
not being part of mathematics
are separate propositions.

https://plato.stanford.edu/entries/philosophy-mathematics/

Philosophers have been know to use mathematics.
So have shepherds, surveyors, gatherers of small
pebbles and many other apparent non-mathematicians.

In my opinion,
when given a choice between
following the guild by-laws of mathematicians and
being correct,
pick being correct.

On Thursday, November 2, 2023 at 3:49:29 AM UTC+1, Jim Burns wrote:

nonsense.

There is no doubt: You and Mückenheim are of the same ilk.

Fuck off, you silly asshole!

On 11/1/2023 5:29 PM, WM wrote:
> Jim Burns schrieb am Mittwoch,
> 1. November 2023 um 18:56:12 UTC+1:

>> 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
> [because] A\B is dark.

We have a description true of
each swap from B.
Each of those swaps is to B
None of those swaps are to A\B

Whether the steps can or can't be _seen_
can't change what they _are_

Compare:
Let us say there are,
in our physical, observable universe,
umpteen bazillion instances of
_two things_ and of _four things_
Nearly all of them we can't see,
have never seen, and will never see.

Q. In how many cases are
two disjoint instances of two things,
combined, NOT an instance of four things?
IOW, how often is 2+2=4 false?

How do you know?
Nearly all of them can't be checked.
Are you sure?

Your answer is based on
what you know about
what has been described,
"two things" and "four things".

Your answer is not based on
what you or anyone can ever see.

Welcome to Mathematics.

>>> 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.

It can be shown in excruciating detail that
any place dark does not swap with
any place visible.
We can't see the swaps and places.
Or maybe we can.
It doesn't matter.

William schrieb am Donnerstag, 2. November 2023 um 02:35:41 UTC+1:
> On Wednesday, November 1, 2023 at 6:16:19 PM UTC-3, WM wrote:
> > 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. There is no last step in your game of billiards. At no step have all indices been applied.

But all steps are defined by the definition of the game, just like Cantor's game is defined by his definition.

Regards, WM

On Thursday, November 2, 2023 at 7:39:17 AM UTC-3, WM wrote:
> William schrieb am Donnerstag, 2. November 2023 um 02:35:41 UTC+1:
> > On Wednesday, November 1, 2023 at 6:16:19 PM UTC-3, WM wrote:
> > > 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. There is no last step in your game of billiards. At no step have all indices been applied.
> But all steps are defined by the definition of the game,

Correct. And, according to this definition, at no step have all indices been applied.

On Thursday, November 2, 2023 at 11:39:17 AM UTC+1, WM wrote:

> Cantor's [approach is based on] his definition [of equivalence].

Right. We define a function from IN x IN in IN and show that it is a bijection, thus proving equialence between IN x IN and IN.

See: https://de.wikipedia.org/wiki/Cantorsche_Paarungsfunktion

Fritz Feldhase schrieb am Donnerstag, 2. November 2023 um 00:38:17 UTC+1:
> On Wednesday, November 1, 2023 at 10:36:23 PM UTC+1, WM wrote:
> > 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 they are part of philosophy, not mathematics.

Cantor's bijections are supertasks. If you are right, then Kronenecker was right too: "I don't know what predominates in Cantor's theory – philosophy or theology, but I am sure that there is no mathematics there."

Regards, WM

Jim Burns schrieb am Donnerstag, 2. November 2023 um 03:49:29 UTC+1:

> Being part of philosophy and
> not being part of mathematics
> are separate propositions.
>
> https://plato.stanford.edu/entries/philosophy-mathematics/
>
> Philosophers have been know to use mathematics.
> So have shepherds, surveyors, gatherers of small
> pebbles and many other apparent non-mathematicians.
>
> In my opinion,
> when given a choice between
> following the guild by-laws of mathematicians and
> being correct,
> pick being correct.

Yes. FF is advocating a snooty and fraudulent fireproofwall of separation. It has been erected by the matheologial swindlers who have realized that Cantor's bijections are supertasks but must fail to reach their aim.

Regards, WM

William schrieb am Donnerstag, 2. November 2023 um 12:33:16 UTC+1:
> On Thursday, November 2, 2023 at 7:39:17 AM UTC-3, WM wrote:
> > William schrieb am Donnerstag, 2. November 2023 um 02:35:41 UTC+1:
> > > On Wednesday, November 1, 2023 at 6:16:19 PM UTC-3, WM wrote:
> > > > 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. There is no last step in your game of billiards. At no step have all indices been applied.
> > But all steps are defined by the definition of the game,
> Correct. And, according to this definition, at no step have all indices been applied.

At no step, yes, but what about: when all steps have been done? This would be Cantors definition of his enumeration and, by the way, also of his diagonal number. Before all steps have been done, it is only a rational. But all rationals can be assimilated by his list.

Regards, WM

Jim Burns schrieb am Donnerstag, 2. November 2023 um 10:46:03 UTC+1:
> On 11/1/2023 5:29 PM, WM wrote:

> >>> 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.
> It can be shown in excruciating detail that
> any place dark does not swap with
> any place visible.

May be that you are right in this point. Then there are no dark numbers and positions. Then there is no completed infinity. Then there is no transfinite set theory.

For instance: As long as Cantors diagonal number can be followed, is visible, it is a rational. Before you can enter the limit of a sequence, there are infinitely many dark terms.

∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo

If there are no dark digits, then the limit, like 1 = lim 0.999..., can only be obtained from the general behaviour. But note that Cantor's diagonal, as a pure sequence of arbitrary digits, has no limit that could be obtained from the general behaviour.

Regards, WM

Fritz Feldhase schrieb am Donnerstag, 2. November 2023 um 13:56:54 UTC+1:
> On Thursday, November 2, 2023 at 11:39:17 AM UTC+1, WM wrote:
>
> > Cantor's [approach is based on] his definition [of equivalence].
>
> Right. We define a function from IN x IN in IN and show that

every point has infinite distance from the completion. This cannot be reached as my game shows.

Regards, WM

On Thursday, November 2, 2023 at 4:22:13 PM UTC+1, WM wrote:

> Cantor's bijections are supertasks.

Mückenheim, Du hast einen schweren Dachschaden, geh doch bitte mal zum Psychater.

On Thursday, November 2, 2023 at 4:31:43 PM UTC+1, WM wrote:

> Cantor's bijections are supertasks

You don't know what you are talking about, you psychotic asshole full of shit.

Just like your buddy JB.

On 11/2/2023 5:56 AM, Fritz Feldhase wrote:
> On Thursday, November 2, 2023 at 11:39:17 AM UTC+1, WM wrote:
>
>> Cantor's [approach is based on] his definition [of equivalence].
>
> Right. We define a function from IN x IN in IN and show that it is a bijection, thus proving equialence between IN x IN and IN.
>
> See: https://de.wikipedia.org/wiki/Cantorsche_Paarungsfunktion

I can hear a whooshing sound over WM's head...

On 11/2/2023 11:46 AM, WM wrote:
> Jim Burns schrieb am Donnerstag,
> 2. November 2023 um 10:46:03 UTC+1:
>> On 11/1/2023 5:29 PM, WM wrote:

>>>>> 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.
>>
>> It can be shown in excruciating detail that
>> any place dark does not swap with
>> any place visible.
>
> May be that you are right in this point.
> Then there are no dark numbers and positions.

My claim is slightly narrower than non-existence.

Dark numbers and positions (if they exist)
are not the reason that
ℕ×ℕ and ℕ are the same "size".

The _visible_ first column matches one-to-one to
all the _visible_ rows and columns.

A better reason is that
a set S with order ≪ which
firsts and lasts and nexts S
cannot match a proper subset of S
but
a set S′ with order ≪′ which
DOESN'T first and last and next S′
CAN match a proper subset of S′

≪ firsts‖lasts S iff,
according to ≪
some first‖last of S exists in S

≪ nexts S iff,
according to ≪
for each split Fᣔ≪ᣔH of S
some last-in-F‖first-in-H exists in F‖H

We can start with
a descriptive claim about S ("first-last-next")
or a descriptive claim about S' ("otherwise")
and augment them finitely not-first-false-ly,
and arrive at claims of not-matching or matching,
which,
because that sequence of claims exists,
we'll know are true claims.

> Then there is no completed infinity.

Your completed-infinityᵂᴹ requires
things which cannot be said to self-equal.
Completed-infinityᵂᴹ will not be missed.

> Then there is no transfinite set theory.
>
> For instance:
> As long as Cantors diagonal number
> can be followed, is visible,
> it is a rational.

Cantor's unlisted anti-diagonal number
is not any of the listed numbers,
because of differences along the diagonal.

Here again,
we start with a descriptive claim about
each one of infinitely-many
| index n for real xₙ is in
| minimally-each-visible ℕᵐᵉᵛ

We augment that finitely with
only not-first-false claims about
each of infinitely-many.

We make finitely-many finite-length
claims about infinitely-many,
of which we know each claim is true because
(1) we know what we're talking about, or
(2) we know the claims are only not-first-false,
and, thus, only not-false.

> Before you can enter the limit of a sequence,
> there are infinitely many dark terms.

x is the _limit_ of sequence ⟨a₀ a₁ a₂ ...⟩
iff
each neighborhood Oₓ ∋ x holds
each of ⟨a₀ a₁ a₂ ...⟩ with
at most finitely-many exceptions.
∀Oₓ ∋ x: |⟨a₀ a₁ a₂ ...⟩\Oₓ| < ℵ₀

That says nothing about entering the limit
or whether ⟨a₀ a₁ a₂ ...⟩ ∋ x
A limit doesn't need to be in the sequence,
and, in many cases of interest,
the limit isn't in the sequence.

> ∀n ∈ ℕ_def:
> |ℕ \ {1, 2, 3, ..., n}| = ℵo

∀n ∈ ℕᵐᵉᵛ:
|ℕᵐᵉᵛ\{1,2,3,...,n}| = ℵ₀

∀⟨1,..,n⟩ ⊆ ℕᵐᵉᵛ
ℕᵐᵉᵛ = ⋂{S∈ℕᵐᵉᵛ| S ∋ 1 ∧ ∀s ∈ S ∋ s⁺¹ }

Standard order < firsts, nexts, but not-lasts ℕᵐᵉᵛ
ℕᵐᵉᵛ is infinite.

For each split F‖H of ℕᵐᵉᵛ
last-in-F‖first-in-H exist

For each split F‖H of ℕᵐᵉᵛ
< firsts, nexts, and lasts F
< firsts, nexts, but not-lasts H

For each split F‖H of ℕᵐᵉᵛ
F‖H is finite‖infinite

On Thursday, November 2, 2023 at 12:34:23 PM UTC-3, WM wrote:
> William schrieb am Donnerstag, 2. November 2023 um 12:33:16 UTC+1:
> > On Thursday, November 2, 2023 at 7:39:17 AM UTC-3, WM wrote:
> > > William schrieb am Donnerstag, 2. November 2023 um 02:35:41 UTC+1:
> > > > On Wednesday, November 1, 2023 at 6:16:19 PM UTC-3, WM wrote:
> > > > > 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. There is no last step in your game of billiards. At no step have all indices been applied.
> > > But all steps are defined by the definition of the game,
> > Correct. And, according to this definition, at no step have all indices been applied.
> At no step, yes, but what about: when all steps have been done?

This is not part of your billiard game.

On Thursday, November 2, 2023 at 4:34:23 PM UTC+1, WM wrote:

> but what about: when all steps have been done?

We don't refer to supertasks in the context of mathematical "arguments", especially not in proofs, you silly asshole.

William wrote :
> On Thursday, November 2, 2023 at 12:34:23 PM UTC-3, WM wrote:
>> William schrieb am Donnerstag, 2. November 2023 um 12:33:16 UTC+1:
>>> On Thursday, November 2, 2023 at 7:39:17 AM UTC-3, WM wrote:
>>>> William schrieb am Donnerstag, 2. November 2023 um 02:35:41 UTC+1:
>>>>> On Wednesday, November 1, 2023 at 6:16:19 PM UTC-3, WM wrote:
>>>>>> 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. There is no last step in your game of billiards. At no step have
>>>>> all indices been applied.
>>>> But all steps are defined by the definition of the game,
>>> Correct. And, according to this definition, at no step have all indices
>>> been applied.
>> At no step, yes, but what about: when all steps have been done?
>
> This is not part of your billiard game.

He doesn't have the balls to play infinite billiards, and he'd need
quite a stick too.

On 11/2/2023 11:31 AM, WM wrote:
> Jim Burns schrieb am Donnerstag,
> 2. November 2023 um 03:49:29 UTC+1:

>> [...]
>
> [...] that Cantor's bijections are supertasks

If Georg Cantor had googled "bijection",
he would have learned that
a bijection B between sets U and V is
a set B of pairs ⟨u,v⟩ ∈ U×V such that
for each u ∈ U unique vᵤ ∈ V exists: ⟨u,vᵤ⟩ ∈ B
for each v ∈ V unique uᵥ ∈ U exists: ⟨uᵥ,v⟩ ∈ B

Cantor could not have googled "bijection".
You (WM) can, though.

> but must fail to reach their aim.

The aim of a bijection B is to be
a set of pairs ⟨u,v⟩ ∈ U×V such that etc.

If B reaches that aim,
B is a bijection.

If B fails to reach that aim,
B isn't a bijection.

There is no paradox in something
not-being a bijection.
It's just that we wish to reason about
all bijections together, and so
we wish to make claims true of each bijection,
and then augment those claims.

It doesn't serve us to have
"bijections in the sweet bye and bye".
Can we make claim C about B?
Or can't we make claim C?
Knowing that is what serves us.

William schrieb am Donnerstag, 2. November 2023 um 20:30:56 UTC+1:
> On Thursday, November 2, 2023 at 12:34:23 PM UTC-3, WM wrote:
> > William schrieb am Donnerstag, 2. November 2023 um 12:33:16 UTC+1:
> > > On Thursday, November 2, 2023 at 7:39:17 AM UTC-3, WM wrote:
> > > > William schrieb am Donnerstag, 2. November 2023 um 02:35:41 UTC+1:
> > > > > On Wednesday, November 1, 2023 at 6:16:19 PM UTC-3, WM wrote:
> > > > > > 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. There is no last step in your game of billiards. At no step have all indices been applied.
> > > > But all steps are defined by the definition of the game,
> > > Correct. And, according to this definition, at no step have all indices been applied.
> > At no step, yes, but what about: when all steps have been done?
> This is not part of your billiard game.

My billiard game mimics Cantor's bijection.

Regards, WM

Fritz Feldhase schrieb am Donnerstag, 2. November 2023 um 21:12:06 UTC+1:
> On Thursday, November 2, 2023 at 4:34:23 PM UTC+1, WM wrote:
>
> > but what about: when all steps have been done?
> We don't refer to supertasks in the context of mathematical "arguments",

Then Kronecker was right: "I don't know what predominates in Cantor's theory – philosophy or theology, but I am sure that there is no mathematics there."

Regards, WM