On 11/21/2022 8:00 AM, Sergi o wrote:
> On 11/21/2022 6:39 AM, WM wrote:
>> zelos...@gmail.com schrieb am Montag, 21. November 2022 um 11:08:59
>> UTC+1:
>>> måndag 21 november 2022 kl. 10:26:55 UTC+1 skrev WM:
>>
>>>> But a much simpler example is this one. Consider the sequence of
>>>> strings
>>>> 21111111111111111111111111...
>>>> 12111111111111111111111111...
>>>> 11211111111111111111111111...
>>>> 11121111111111111111111111...
>>>> and so on, exchanging always the 2 and the 1 following next upon the 2.
>>>> Will the 2 leave the string? Why should it? There are obviously
>>>> enough places.
>>>> Nevertheless, in the final state only the string
>>>> 11111111111111111111111111...
>>>> is visible. The 2, if present yet, occupies a dark place.
>>>>
>>> Still haven't defined what "dark" even means.
>>
>> Can you see the 2 in  11111111111111111111111111... ?
>
> notion says it is all 1's
>
> if you want a 2 in it, put it in there.
[...]

Exactly! Humm... Let me try:

i[n] = floor(((1 / 9) * 10^(n+2))+1) = 1...2

where:

i[0] = floor(((1 / 9) * 10^(0+2))+1) = 12
i[1] = floor(((1 / 9) * 10^(1+2))+1) = 112
i[2] = floor(((1 / 9) * 10^(2+2))+1) = 1112
i[3] = floor(((1 / 9) * 10^(3+2))+1) = 11112
i[4] = floor(((1 / 9) * 10^(4+2))+1) = 111112
....

;^)

lol.

On 11/21/2022 7:55 PM, Chris M. Thomasson wrote:
> On 11/21/2022 8:00 AM, Sergi o wrote:
>> On 11/21/2022 6:39 AM, WM wrote:
>>> zelos...@gmail.com schrieb am Montag, 21. November 2022 um 11:08:59 UTC+1:
>>>> måndag 21 november 2022 kl. 10:26:55 UTC+1 skrev WM:
>>>
>>>>> But a much simpler example is this one. Consider the sequence of strings
>>>>> 21111111111111111111111111...
>>>>> 12111111111111111111111111...
>>>>> 11211111111111111111111111...
>>>>> 11121111111111111111111111...
>>>>> and so on, exchanging always the 2 and the 1 following next upon the 2.
>>>>> Will the 2 leave the string? Why should it? There are obviously enough places.
>>>>> Nevertheless, in the final state only the string
>>>>> 11111111111111111111111111...
>>>>> is visible. The 2, if present yet, occupies a dark place.
>>>>>
>>>> Still haven't defined what "dark" even means.
>>>
>>> Can you see the 2 in  11111111111111111111111111... ?
>>
>> notion says it is all 1's
>>
>> if you want a 2 in it, put it in there.
> [...]
>
> Exactly! Humm... Let me try:
>
> i[n] = floor(((1 / 9) * 10^(n+2))+1) = 1...2
>
> where:
>
> i[0] = floor(((1 / 9) * 10^(0+2))+1) = 12
> i[1] = floor(((1 / 9) * 10^(1+2))+1) = 112
> i[2] = floor(((1 / 9) * 10^(2+2))+1) = 1112
> i[3] = floor(((1 / 9) * 10^(3+2))+1) = 11112
> i[4] = floor(((1 / 9) * 10^(4+2))+1) = 111112
> ...
>
> ;^)
>
> lol.
>
>
>

Verified, That will do it!

i[n], the 2 inserter function, where n is a natural number, if you want your 2 at k then let n = k

(WM- note the key generator fraction, 1/9)

WM expressed precisely :
> Abstract: We will prove by means of Cantor's mapping between natural numbers
> and positive fractions that his approach to actual infinity implies the
> existence of numbers which cannot be applied as defined individuals. We will
> call them dark numbers.

You need to pair them, not compare them.

måndag 21 november 2022 kl. 20:13:37 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Montag, 21. November 2022 um 17:46:24 UTC+1:
> > måndag 21 november 2022 kl. 16:36:51 UTC+1 skrev WM:
> > > zelos...@gmail.com schrieb am Montag, 21. November 2022 um 14:43:03 UTC+1:
> > > > måndag 21 november 2022 kl. 13:39:12 UTC+1 skrev WM:
> > > > > zelos...@gmail.com schrieb am Montag, 21. November 2022 um 11:08:59 UTC+1:
> > > > > > måndag 21 november 2022 kl. 10:26:55 UTC+1 skrev WM:
> > > > >
> > > > > > > But a much simpler example is this one. Consider the sequence of strings
> > > > > > > 21111111111111111111111111...
> > > > > > > 12111111111111111111111111...
> > > > > > > 11211111111111111111111111...
> > > > > > > 11121111111111111111111111...
> > > > > > > and so on, exchanging always the 2 and the 1 following next upon the 2.
> > > > > > > Will the 2 leave the string? Why should it? There are obviously enough places.
> > > > > > > Nevertheless, in the final state only the string
> > > > > > > 11111111111111111111111111...
> > > > > > > is visible. The 2, if present yet, occupies a dark place.
> > > > > > >
> > > > > > Still haven't defined what "dark" even means.
> > > > > Can you see the 2 in 11111111111111111111111111... ?
> > > > Can you see the 9 in 11?
> > > 9 is not involved in this example but 2 is.
> > >
> > if it is ...111 there is no 2 in it.
> It is 11111111111111111111111111... But 2 has not left. The number of digits is the same in all terms of the sequence.
>
> Regards, WM
Then what you want is [1]_n 2[1]_m

Which still makes nothing dark.

Gus Gassmann schrieb am Montag, 21. November 2022 um 20:16:33 UTC+1:
> On Monday, 21 November 2022 at 14:36:16 UTC-4, WM wrote:

> > The limit is not involved when "every number p/q comes at an absolutely fixed position of a simple infinite sequence". But that is not so obvious to see as in my matrices.

> Cantor does things one way (the easy and proper way), and you insist on doing it the other way,

I do it in the same way. The only difference is that I check the indices first. The result is infinitely many O's in every term of the sequence.

Please ponder about the following dilemma:

If the infinitely many O's leave in the limit, then there is no indexing possible. If they do not leave, then there is no indexing completed.

Regards, WM

On Tuesday, November 22, 2022 at 1:10:16 PM UTC+1, WM wrote:

> The [main] difference is that

you are are a psychotic crank.

On 11/22/2022 6:10 AM, WM wrote:
> Gus Gassmann schrieb am Montag, 21. November 2022 um 20:16:33 UTC+1:
>> On Monday, 21 November 2022 at 14:36:16 UTC-4, WM wrote:
>
>>> The limit is not involved when "every number p/q comes at an absolutely fixed position of a simple infinite sequence". But that is not so obvious to see as in my matrices.
>
>> Cantor does things one way (the easy and proper way), and you insist on doing it the other way,
>
> I do it in the same way.

Liar.

>
> Please ponder about the following dilemma:

you go ponder your own bullshit.

>
> Regards, WM

On Tuesday, 22 November 2022 at 08:10:16 UTC-4, WM wrote:
> Gus Gassmann schrieb am Montag, 21. November 2022 um 20:16:33 UTC+1:
> > On Monday, 21 November 2022 at 14:36:16 UTC-4, WM wrote:
>
> > > The limit is not involved when "every number p/q comes at an absolutely fixed position of a simple infinite sequence". But that is not so obvious to see as in my matrices.
> > Cantor does things one way (the easy and proper way), and you insist on doing it the other way,
> I do it in the same way. The only difference is that ...

you are a psychotic imbecile. If there are differences, then you are *NOT* doing it the same way. That should be understandable even for you.

Gus Gassmann schrieb am Dienstag, 22. November 2022 um 19:30:00 UTC+1:
> On Tuesday, 22 November 2022 at 08:10:16 UTC-4, WM wrote:
> > Gus Gassmann schrieb am Montag, 21. November 2022 um 20:16:33 UTC+1:
> > > On Monday, 21 November 2022 at 14:36:16 UTC-4, WM wrote:
> >
> > > > The limit is not involved when "every number p/q comes at an absolutely fixed position of a simple infinite sequence". But that is not so obvious to see as in my matrices.
> > > Cantor does things one way (the easy and proper way), and you insist on doing it the other way,
> > I do it in the same way. The only difference is that ...
>
> If there are differences, then you are *NOT* doing it the same way.

Wrong. The distribution of the indices is exactly the same way, namely according to
k = (m + n - 1)(m + n - 2)/2 + m .
with the resulting sequence of fractions
1/1, 1/2, 2/1, 1/3, 2/2, 3/1, ... .

The only difference is that Cantor does not know that there are not more natnumbers than integer fractions.

Now try to solve the real problem: If the infinitely many O's leave in the limit, then there is no indexing possible. If they do not leave, then there is no indexing completed.

Regards, WM

On 11/22/2022 1:51 PM, WM wrote:
> Gus Gassmann schrieb am Dienstag, 22. November 2022 um 19:30:00 UTC+1:
>> On Tuesday, 22 November 2022 at 08:10:16 UTC-4, WM wrote:
>>> Gus Gassmann schrieb am Montag, 21. November 2022 um 20:16:33 UTC+1:
>>>> On Monday, 21 November 2022 at 14:36:16 UTC-4, WM wrote:
>>>
>>>>> The limit is not involved when "every number p/q comes at an absolutely fixed position of a simple infinite sequence". But that is not so obvious to see as in my matrices.
>>>> Cantor does things one way (the easy and proper way), and you insist on doing it the other way,
>>> I do it in the same way. The only difference is that ...
>>
>> If there are differences, then you are *NOT* doing it the same way.
>
> Wrong. The distribution of the indices is exactly the same way, namely according to
> k = (m + n - 1)(m + n - 2)/2 + m .
> with the resulting sequence of fractions
> 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, ... .
>

but *you do not use those* as you put X and O stickies on them covering up the fraction, and the natural numbers

then you have your switcharoofestival, you cant tell where what ended up where, you lose stickies and pasties, and you lose O's

everyone in this forum has rejected your idea, because it is wrong, and it is not math.

besides you do not know what 111111111... means

does 1/2 = 0.5 yet ?

> Regards, WM
>
>

On Tuesday, November 22, 2022 at 8:51:31 PM UTC+1, WM wrote:
> Gus Gassmann schrieb am Dienstag, 22. November 2022 um 19:30:00 UTC+1:
> > On Tuesday, 22 November 2022 at 08:10:16 UTC-4, WM wrote:
> > > Gus Gassmann schrieb am Montag, 21. November 2022 um 20:16:33 UTC+1:
> > > > On Monday, 21 November 2022 at 14:36:16 UTC-4, WM wrote:
> > >
> > > > > The limit is not involved when "every number p/q comes at an absolutely fixed position of a simple infinite sequence". But that is not so obvious to see as in my matrices.
> > > > Cantor does things one way (the easy and proper way), and you insist on doing it the other way,
> > > I do it in the same way. The only difference is that ...
> >
> > If there are differences, then you are *NOT* doing it the same way.
> Wrong. The distribution of the indices is exactly the same way, namely according to
> k = (m + n - 1)(m + n - 2)/2 + m .
> with the resulting sequence of fractions
> 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, ... .
>
> The only difference is that Cantor does not know that there are not more natnumbers than integer fractions.
>
> Now try to solve the real problem: If the infinitely many O's leave in the limit, then there is no indexing possible. If they do not leave, then there is no indexing completed.
>
> Regards, WM

It's amazing how dumb these people were. Cantor, Hilbert and friends
completely overlooked the dark numbers. But now the age of Mücke is upon us,
the age of the Lightning Bug, so to speak. The age of the dark number, of transparent,
iridescent, invisible, unknowable and undefinable numbers.

Hail our Great Leader, Mücke Mückenheim!!

On 11/22/2022 2:11 PM, JVR wrote:
> On Tuesday, November 22, 2022 at 8:51:31 PM UTC+1, WM wrote:
>> Gus Gassmann schrieb am Dienstag, 22. November 2022 um 19:30:00 UTC+1:
>>> On Tuesday, 22 November 2022 at 08:10:16 UTC-4, WM wrote:
>>>> Gus Gassmann schrieb am Montag, 21. November 2022 um 20:16:33 UTC+1:
>>>>> On Monday, 21 November 2022 at 14:36:16 UTC-4, WM wrote:
>>>>
>>>>>> The limit is not involved when "every number p/q comes at an absolutely fixed position of a simple infinite sequence". But that is not so obvious to see as in my matrices.
>>>>> Cantor does things one way (the easy and proper way), and you insist on doing it the other way,
>>>> I do it in the same way. The only difference is that ...
>>>
>>> If there are differences, then you are *NOT* doing it the same way.
>> Wrong. The distribution of the indices is exactly the same way, namely according to
>> k = (m + n - 1)(m + n - 2)/2 + m .
>> with the resulting sequence of fractions
>> 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, ... .
>>
>> The only difference is that Cantor does not know that there are not more natnumbers than integer fractions.
>>
>> Now try to solve the real problem: If the infinitely many O's leave in the limit, then there is no indexing possible. If they do not leave, then there is no indexing completed.
>>
>> Regards, WM
>
> It's amazing how dumb these people were. Cantor, Hilbert and friends
> completely overlooked the dark numbers. But now the age of Mücke is upon us,
> the age of the Lightning Bug, so to speak. The age of the dark number, of transparent,
> iridescent, invisible, unknowable and undefinable numbers.
>
> Hail our Great Leader, Mücke Mückenheim!!

Crypto uses most of the Dark Numbers now, and when called upon some cannot identify themselves. So, they go missing instead.

this formally dark number here:
a1075db55d416d3ca199f55b6084e2115b9345e16c5cf302fc80e9d5fbf5d4
8d0xb4bc263278d3f77a652a8d73a6bfd8ec0ba1a63923bbb4f38147fb8a943da26dcabc4c0a4709
ac3b654e7f6b3c95df727407fe10c9f3d181dcfaa2dd53da1fd90790e6f6462df7afc8ad9d90cec8
a46c7014e8e1869fa833b2905beb6c977368

was found disorientated after he was snatched from some other dudes wallet, it took months for that number to remember its name, and to tap out his
number on a table with raps, be self identified but not verified, he did find his color in the rainbow (yes WMs daffynition).

JVR was thinking very hard :
> On Tuesday, November 22, 2022 at 8:51:31 PM UTC+1, WM wrote:
>> Gus Gassmann schrieb am Dienstag, 22. November 2022 um 19:30:00 UTC+1:
>>> On Tuesday, 22 November 2022 at 08:10:16 UTC-4, WM wrote:
>>>> Gus Gassmann schrieb am Montag, 21. November 2022 um 20:16:33 UTC+1:
>>>>> On Monday, 21 November 2022 at 14:36:16 UTC-4, WM wrote:
>>>>>> The limit is not involved when "every number p/q comes at an absolutely
>>>>>> fixed position of a simple infinite sequence". But that is not so
>>>>>> obvious to see as in my matrices.
>>>>> Cantor does things one way (the easy and proper way), and you insist on
>>>>> doing it the other way,
>>>> I do it in the same way. The only difference is that ...
>>>
>>> If there are differences, then you are *NOT* doing it the same way.
>> Wrong. The distribution of the indices is exactly the same way, namely
>> according to k = (m + n - 1)(m + n - 2)/2 + m .
>> with the resulting sequence of fractions
>> 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, ... .
>>
>> The only difference is that Cantor does not know that there are not more
>> natnumbers than integer fractions.
>>
>> Now try to solve the real problem: If the infinitely many O's leave in the
>> limit, then there is no indexing possible. If they do not leave, then there
>> is no indexing completed.
>>
>> Regards, WM
>
> It's amazing how dumb these people were. Cantor, Hilbert and friends
> completely overlooked the dark numbers. But now the age of Mücke is upon us,
> the age of the Lightning Bug, so to speak. The age of the dark number, of
> transparent, iridescent, invisible, unknowable and undefinable numbers.
>
> Hail our Great Leader, Mücke Mückenheim!!

He has imagined a 'set' of numbers whose elements are included in the
set by the criterion that they have no criterion -- and thus are dark.

Genius, pure and simple.

Too bad they are useless.

On Tuesday, 22 November 2022 at 15:51:31 UTC-4, WM wrote:

....nothing but shit.

JVR schrieb am Dienstag, 22. November 2022 um 21:11:52 UTC+1:
> On Tuesday, November 22, 2022 at 8:51:31 PM UTC+1, WM wrote:
> > If the infinitely many O's leave in the limit, then there is no indexing possible. If they do not leave, then there is no indexing completed.
> >
> Cantor, Hilbert and friends
> completely overlooked the dark numbers.

Not generally. Hilbert for instance wrote: "Finally we will return to our original topic and draw the conclusions of all our investigations about the infinite. On balance the result is this: The infinite is nowhere realized; it is neither present in nature nor admissible as the foundation of our rational thinking – a remarkable harmony of being and thinking." [D. Hilbert: "Über das Unendliche", Mathematische Annalen 95 (1925) p. 190]

This excludes dark numbers. Without actual infinity there is no reason to accept dark numbers.

But if you accept actual infinity, then Cantor's bijections are disproved - or you have to find another solution for the problem: If the infinitely many O's leave in the limit, then there is no indexing possible. If they do not leave, then there is no complete indexing.

Regards, WM

On Tuesday, November 22, 2022 at 10:31:33 PM UTC+1, FromTheRafters wrote:
> JVR was thinking very hard :
> > On Tuesday, November 22, 2022 at 8:51:31 PM UTC+1, WM wrote:
> >> Gus Gassmann schrieb am Dienstag, 22. November 2022 um 19:30:00 UTC+1:
> >>> On Tuesday, 22 November 2022 at 08:10:16 UTC-4, WM wrote:
> >>>> Gus Gassmann schrieb am Montag, 21. November 2022 um 20:16:33 UTC+1:
> >>>>> On Monday, 21 November 2022 at 14:36:16 UTC-4, WM wrote:
> >>>>>> The limit is not involved when "every number p/q comes at an absolutely
> >>>>>> fixed position of a simple infinite sequence". But that is not so
> >>>>>> obvious to see as in my matrices.
> >>>>> Cantor does things one way (the easy and proper way), and you insist on
> >>>>> doing it the other way,
> >>>> I do it in the same way. The only difference is that ...
> >>>
> >>> If there are differences, then you are *NOT* doing it the same way.
> >> Wrong. The distribution of the indices is exactly the same way, namely
> >> according to k = (m + n - 1)(m + n - 2)/2 + m .
> >> with the resulting sequence of fractions
> >> 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, ... .
> >>
> >> The only difference is that Cantor does not know that there are not more
> >> natnumbers than integer fractions.
> >>
> >> Now try to solve the real problem: If the infinitely many O's leave in the
> >> limit, then there is no indexing possible. If they do not leave, then there
> >> is no indexing completed.
> >>
> >> Regards, WM
> >
> > It's amazing how dumb these people were. Cantor, Hilbert and friends
> > completely overlooked the dark numbers. But now the age of Mücke is upon us,
> > the age of the Lightning Bug, so to speak. The age of the dark number, of
> > transparent, iridescent, invisible, unknowable and undefinable numbers.
> >
> > Hail our Great Leader, Mücke Mückenheim!!
> He has imagined a 'set' of numbers whose elements are included in the
> set by the criterion that they have no criterion -- and thus are dark.
>
> Genius, pure and simple.
>
> Too bad they are useless.

What do you mean useless? They have made Mücke a Usenet celebrity.
The greatest prefosser on all of usenet. Greater even than Archie Atomicus
and the Rectum.
Actually that isn't true. He was famous already due to the fallacy of the
binary tree. This just establishes his position as primus inter pares.

FromTheRafters schrieb am Dienstag, 22. November 2022 um 22:31:33 UTC+1:

> He has imagined a 'set' of numbers whose elements are included in the
> set by the criterion that they have no criterion -- and thus are dark.
>
> Genius, pure and simple.
>
> Too bad they are useless.

They are not useless but useful to prove that the idea of equivalent infinite stes is wrong. This shows that *infinite* endsegments will not yield an empty intersection. McDuck will not become bankrupt. The Binary Tree has less paths than nodes. And so on. The most silly results of set theory are deleted.

Regards, WM

Gus Gassmann schrieb am Dienstag, 22. November 2022 um 22:55:22 UTC+1:
> On Tuesday, 22 November 2022 at 15:51:31 UTC-4, WM wrote:
>
> ...nothing

that you could discuss without showing that even your inappropriate limits will not help. A clearer contradiction of your claim is hardly possible:

If the infinitely many O's leave in the limit, then there is no indexing possible. If they do not leave, then there is no completed indexing.

Regards, WM

On 11/22/2022 4:06 PM, WM wrote:
> FromTheRafters schrieb am Dienstag, 22. November 2022 um 22:31:33 UTC+1:
>
>> He has imagined a 'set' of numbers whose elements are included in the
>> set by the criterion that they have no criterion -- and thus are dark.
>>
>> Genius, pure and simple.
>>
>> Too bad they are useless.
>
> They are not useless but useful to prove that the idea of equivalent infinite stes is wrong. This shows that *infinite* endsegments will not yield an empty intersection. McDuck will not become bankrupt. The Binary Tree has less paths than nodes. And so on. The most silly results of set theory are deleted.
>
> Regards, WM

in your darkie dreams pal, and that duck is gonna get it good...

just like framed Achilles, a loser to a cryptodira.

On 11/22/2022 4:11 PM, WM wrote:
> Gus Gassmann schrieb am Dienstag, 22. November 2022 um 22:55:22 UTC+1:
>> On Tuesday, 22 November 2022 at 15:51:31 UTC-4, WM wrote:
>>
>> ...nothing
>

<snip crap>

> Regards, WM

On 11/22/2022 2:11 PM, WM wrote:
> Gus Gassmann schrieb am Dienstag, 22. November 2022 um 22:55:22 UTC+1:
>> On Tuesday, 22 November 2022 at 15:51:31 UTC-4, WM wrote:
>>
>> ...nothing
>
> that you could discuss without showing that even your inappropriate limits will not help. A clearer contradiction of your claim is hardly possible:
>
> If the infinitely many O's leave in the limit, then there is no indexing possible. If they do not leave, then there is no completed indexing.

Any natural number, yes including zero, creates a unique pair wrt Cantor
Pairing. Nothing to miss. Nothing is dark.

On 11/22/2022 4:35 PM, Chris M. Thomasson wrote:
> On 11/22/2022 2:11 PM, WM wrote:
>> Gus Gassmann schrieb am Dienstag, 22. November 2022 um 22:55:22 UTC+1:
>>> On Tuesday, 22 November 2022 at 15:51:31 UTC-4, WM wrote:
>>>
>>> ...nothing
>>
>> that you could discuss without showing that even your inappropriate limits will not help. A clearer contradiction of your claim is hardly possible:
>>
>> If the infinitely many O's leave in the limit, then there is no indexing possible. If they do not leave, then there is no completed indexing.
>
> Any natural number, yes including zero, creates a unique pair wrt Cantor Pairing. Nothing to miss. Nothing is dark.
>

forget dark numbers, what about dark letters ?

I make e and a dark in above......

forg█t d█rk numb█rs, wh█t █bout d█rk l█tt█rs ?

here is a dark letter for cut and paste █ very useful too!

tisdag 22 november 2022 kl. 13:10:16 UTC+1 skrev WM:
> Gus Gassmann schrieb am Montag, 21. November 2022 um 20:16:33 UTC+1:
> > On Monday, 21 November 2022 at 14:36:16 UTC-4, WM wrote:
>
> > > The limit is not involved when "every number p/q comes at an absolutely fixed position of a simple infinite sequence". But that is not so obvious to see as in my matrices.
> > Cantor does things one way (the easy and proper way), and you insist on doing it the other way,
> I do it in the same way. The only difference is that I check the indices first. The result is infinitely many O's in every term of the sequence.
>
> Please ponder about the following dilemma:
>
> If the infinitely many O's leave in the limit, then there is no indexing possible. If they do not leave, then there is no indexing completed.
>
> Regards, WM
Your fucking matrix does not matter! how dumb are you!?

On Tuesday, 22 November 2022 at 18:11:39 UTC-4, WM wrote:
> Gus Gassmann schrieb am Dienstag, 22. November 2022 um 22:55:22 UTC+1:
> > On Tuesday, 22 November 2022 at 15:51:31 UTC-4, WM wrote:
> >
> > ...nothing
> that you could discuss without showing that even your inappropriate limits will not help. A clearer contradiction of your claim is hardly possible:
>
> If the infinitely many O's leave in the limit, then there is no indexing possible. If they do not leave, then there is no completed indexing.

That bullshit only shows that you have no idea how limits work, you fucking imbecile.

WM brought next idea :
> FromTheRafters schrieb am Dienstag, 22. November 2022 um 22:31:33 UTC+1:
>
>> He has imagined a 'set' of numbers whose elements are included in the
>> set by the criterion that they have no criterion -- and thus are dark.
>>
>> Genius, pure and simple.
>>
>> Too bad they are useless.
>
> They are not useless but useful to prove that the idea of equivalent infinite
> stes is wrong.

Of course it is wrong, the naturals with zero and the naturals without
zero are not the same set, they are the same size though. You would see
this if you understood what a bijection is.