On 1/27/2022 6:07 AM, WM wrote:
> Jim Burns schrieb
> am Mittwoch, 26. Januar 2022 um 19:42:39 UTC+1:
>> On 1/26/2022 5:26 AM, WM wrote:

>>> Name one that does not remain.
>> The O that starts at 1/2 in the matrix
>> does not remain in ℕ⁺⨯ℕ⁺
>
> That is a claim.
>
>> I name it _Bob_
>
> Where and when does Bob leave?
> What is his last place in the matrix?
>
>> https://www.youtube.com/watch?v=TjAg-8qqR3g&t=45s
>>
>> The O from 1/3 is named Kevin.
>> The O from 2/2 is named Stuart.
>> They also don't remain in ℕ⁺⨯ℕ⁺
>> There are more not remaining in ℕ⁺⨯ℕ⁺
>
> First answer my question.

It's disappointing that you snipped one of the
few points of agreement we have had.

Please be more careful. Someone paying less attention
might think you weren't arguing in good faith. TIA

>>>> For each O, we know where in ℕ⁺⨯{1}
>>>> the X comes from that replaces it.
>>>
>>> And we know that every X is replaced by an O.
>>
>> Right.
>> Each O swaps with a different X.
>> So there can't be more O's than X's.

Jim Burns schrieb am Donnerstag, 27. Januar 2022 um 19:01:06 UTC+1:
> On 1/27/2022 6:07 AM, WM wrote:

> > No swap removes him from the matrix.
> Right.
> One swap moves Bob within the matrix.
> Also {one swap} is a two-ended-stepping-order-able
> collection of swaps.

An infinite sequence of swaps consists of one-swaps only. You believe in an exception. There is none.

> What if the collection of swaps was NOT two-ended-
> -stepping-order-able?

Then the transpositions interchamnging X and O continue without end and without changing the numbers of X's and O's.

> >> Where is Bob after _all_ the swaps?
> Bob is out of ℕ⁺⨯{1}
> after _all_ the swaps.

No X and no O exists in any transposition. Other exist are not open. Deal with it.

> But
> each one of Bob's transpositions is followed by another
> by another transposition which picks him up again.
>
> _Because Bob has no last transposition_
> each spot at which Bob is set down is _not_
> a spot at which _all_ Bob's transpositions leave him.

But he is a conserved piece, he is preserved. He is not outside of the matrix. The only possibility is: He got lost in the dark part.
>
> Bob is not where he's been set down.
> Bob is not where he's not been set down.
> Bob is not in ℕ⁺⨯{1}

Clear evidence for being lost in the dark part.
>
> If your dark numbers are not in ℕ⁺⨯{1}

They are there too. But there they are not so clearly proved.

Regards, WM

Jim Burns schrieb am Donnerstag, 27. Januar 2022 um 22:51:52 UTC+1:
> On 1/27/2022 6:07 AM, WM wrote:

> Please be more careful. Someone paying less attention
> might think you weren't arguing in good faith.

There is no agreement required. I use facts: Bob will never be observed outside of the matrix. Bob will get out of sight within the matrix. The only explanation is dark.

By the way: Instead of changing the positions of X's and O's, we can also let them stay fixed at their positions while the fractions move around, according to Cantor's sequence. Then Bob will never get out of sight.

Regards, WM

WM schrieb am Donnerstag, 27. Januar 2022 um 22:56:29 UTC+1:
Correction
> > after _all_ the swaps.
> No X and no O exits in any transposition. Other exits are not open.

Regards, WM

On Thursday, 27 January 2022 at 17:39:01 UTC-4, WM wrote:
> horand....@gmail.com schrieb am Donnerstag, 27. Januar 2022 um 17:06:10 UTC+1:
[...]
> > Just as at every finite step there are aleph_0 X's in the matrix, there also are aleph_0 O's.
> The O's indicate not indexed fractions.

At every finite step there are aleph_0 "not indexed fractions" as well as aleph_0 X's that have not been moved yet. That's all. It is clear that you cannot comprehend this, but it is as easy as that.

horand....@gmail.com schrieb am Donnerstag, 27. Januar 2022 um 23:11:38 UTC+1:
> On Thursday, 27 January 2022 at 17:39:01 UTC-4, WM wrote:
> > horand....@gmail.com schrieb am Donnerstag, 27. Januar 2022 um 17:06:10 UTC+1:
> [...]
> > > Just as at every finite step there are aleph_0 X's in the matrix, there also are aleph_0 O's.
> > The O's indicate not indexed fractions.
> At every finite step there are aleph_0 "not indexed fractions" as well as aleph_0 X's that have not been moved yet. That's all.
That's all, in fact, because after every finite step nothing happens. Therefore most fractions remain not indexed.

Regards, WM

On 1/27/2022 4:56 PM, WM wrote:
> Jim Burns schrieb
> am Donnerstag, 27. Januar 2022 um 19:01:06 UTC+1:
>> On 1/27/2022 6:07 AM, WM wrote:

>>> No swap removes him from the matrix.
>>
>> Right.
>> One swap moves Bob within the matrix.
>> Also {one swap} is a two-ended-stepping-order-able
>> collection of swaps.
>
> An infinite sequence of swaps consists of one-swaps
> only. You believe in an exception. There is none.

Each one of Bob's swaps sets him down
somewhere in ℕ⁺⨯{1} = ⋃{{1,...,k}}⨯{1}

But
each one of Bob's swaps is followed by
another swap which picks him up again.

Bob is not where he's been set down.
Bob is not where he's not been set down.
Bob is not in ℕ⁺⨯{1}

No single swap removes Bob from the matrix.

_Because Bob has no last swap_ he can be set down on
infinitely-many spots in ℕ⁺⨯{1} and left on
none of them.

Bob-always-being-picked-up-again is not something within
the power of any single swap.

It is within the power of all the swaps for which
k = {n+d-1)*(n+d-2)/2 + n

On 1/27/2022 3:56 PM, WM wrote:
> Jim Burns schrieb am Donnerstag, 27. Januar 2022 um 19:01:06 UTC+1:
>> On 1/27/2022 6:07 AM, WM wrote:
>
>>> No swap removes him from the matrix.
>> Right.
>> One swap moves Bob within the matrix.
>> Also {one swap} is a two-ended-stepping-order-able
>> collection of swaps.
>
> An infinite sequence of swaps consists of one-swaps only.

no, it is an infinite # of swaps.

>
>> What if the collection of swaps was NOT two-ended-
>> -stepping-order-able?
>
> Then the transpositions interchamnging X and O continue without end and without changing the numbers of X's and O's.

wrong, take the limit.

>
>>>> Where is Bob after _all_ the swaps?
>> Bob is out of ℕ⁺⨯{1}
>> after _all_ the swaps.
>
> No X and no O exists in any transposition.

then transposition is meaningless

On 1/27/2022 4:02 PM, WM wrote:
> Jim Burns schrieb am Donnerstag, 27. Januar 2022 um 22:51:52 UTC+1:
>> On 1/27/2022 6:07 AM, WM wrote:
>
>> Please be more careful. Someone paying less attention
>> might think you weren't arguing in good faith.
>
<snip crap>

WM argues to confuse and mislead, he hates math, and he does not know math

> Regards, WM

On 1/27/2022 4:22 PM, WM wrote:
> horand....@gmail.com schrieb am Donnerstag, 27. Januar 2022 um 23:11:38 UTC+1:
>> On Thursday, 27 January 2022 at 17:39:01 UTC-4, WM wrote:
>>> horand....@gmail.com schrieb am Donnerstag, 27. Januar 2022 um 17:06:10 UTC+1:
>> [...]
>>>> Just as at every finite step there are aleph_0 X's in the matrix, there also are aleph_0 O's.
>>> The O's indicate not indexed fractions.
>> At every finite step there are aleph_0 "not indexed fractions" as well as aleph_0 X's that have not been moved yet. That's all.
>
> That's all, in fact, because after every finite step nothing happens.

In your math, nothing happens.

But no worries at lease you are consistent in getting the answer wrong every single time.

>
> Regards, WM

On 1/27/2022 3:47 PM, WM wrote:
> horand....@gmail.com schrieb am Donnerstag, 27. Januar 2022 um 17:06:10 UTC+1:
>> On Thursday, 27 January 2022 at 09:35:50 UTC-4, WM wrote:
>
>>>> The board is filled in with X's one step at a time.
>>> No, all X are present from the beginning. They are only re-ordered.
>> That's *NOT* true. At every step you move one X from the left column to a position somewhere else in the map.
>
> The left column belongs to the matrix. All X are present from the beginning and remain within the matrix, like the O's..
>
>>> There is no "after infinitely many steps". Hence there are infinitely many O's remaining always, in fact precisely those from the start.
>> Of *COURSE* there is "after infinitely many steps".
>
> If you like, we can model it. Nothing changes after infinitely many steps. And before there have been infinitely many replacements, tra nspositions without any loss of X or O.
>
>> They do not "leave". They are replaced by X's.
>
> No O is replaced by an X. Only their positions are interchanged.

wrong, each O is replaced by an X.

>
> Regards, WM

torsdag 27 januari 2022 kl. 14:38:14 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Donnerstag, 27. Januar 2022 um 14:24:15 UTC+1:
> > torsdag 27 januari 2022 kl. 12:32:48 UTC+1 skrev WM:
>
> > > > There is no FINITE steps, there are no steps PERIOD!
> > > But there are finite terms of the sequence --- only finite terms. Where, when, and how does the first O exit?
> >
> > >What is the frist term when an O exits? Where does it go then?
> > Irrelevant, you are obfuscating with this and trying to confuse things.
> No, I am using mathematics and logic.
> > >But there are finite terms of the sequence --- only finite terms. Where, when, and how does the first O exit?
> > You are asking "Why are all natural numbers finite but not N?"
> No, I am asking where do the O's go?
> >
> > Because N contains all of them, of course it is not finite then!
> That has nothing to do with my question.
>
> Regards, WM

>No, I am using mathematics and logic.

You aren't, stop lying

>No, I am asking where do the O's go?

Irrelevant, again all your Os and shit is irrelevant to the fact that there is a bijection

zelos...@gmail.com schrieb am Freitag, 28. Januar 2022 um 06:09:19 UTC+1:
> torsdag 27 januari 2022 kl. 14:38:14 UTC+1 skrev WM:
> > zelos...@gmail.com schrieb am Donnerstag, 27. Januar 2022 um 14:24:15 UTC+1:
> > > torsdag 27 januari 2022 kl. 12:32:48 UTC+1 skrev WM:
> >
> > > > > There is no FINITE steps, there are no steps PERIOD!
> > > > But there are finite terms of the sequence --- only finite terms. Where, when, and how does the first O exit?
> > >
> > > >What is the frist term when an O exits? Where does it go then?
> > > Irrelevant, you are obfuscating with this and trying to confuse things.
> > No, I am using mathematics and logic.
> > > >But there are finite terms of the sequence --- only finite terms. Where, when, and how does the first O exit?
> > > You are asking "Why are all natural numbers finite but not N?"
> > No, I am asking where do the O's go?
> > >
> >I am using mathematics and logic. I am asking where do the O's go?
> Irrelevant

Very relevant, because it kills set theory. Note that the O's are there and cannot exit. And if they tried, where should they settle?

> again all your Os and shit is irrelevant to the fact that there is a bijection

But the O's are relevant to prove that there is no bijection. If this proof surpasses your intellectual competence, try to use the simpler picture where the O's remain fixed at their places and the fractions try to disappear through the first column.

Regards, WM

Jim Burns schrieb am Freitag, 28. Januar 2022 um 03:13:52 UTC+1:
> On 1/27/2022 4:56 PM, WM wrote:
> > Jim Burns schrieb
> > am Donnerstag, 27. Januar 2022 um 19:01:06 UTC+1:
> >> On 1/27/2022 6:07 AM, WM wrote:
>
> >>> No swap removes him from the matrix.
> >>
> >> Right.
> >> One swap moves Bob within the matrix.
> >> Also {one swap} is a two-ended-stepping-order-able
> >> collection of swaps.
> >
> > An infinite sequence of swaps consists of one-swaps
> > only. You believe in an exception. There is none.
> Each one of Bob's swaps sets him down
> somewhere in ℕ⁺⨯{1} = ⋃{{1,...,k}}⨯{1}
> But
> each one of Bob's swaps is followed by
> another swap which picks him up again.
> Bob is not where he's been set down.
> Bob is not where he's not been set down.
> Bob is not in ℕ⁺⨯{1}
> No single swap removes Bob from the matrix.

Of course not! Where should he settle outside?
>
> _Because Bob has no last swap_ he can be set down on
> infinitely-many spots in ℕ⁺⨯{1} and left on
> none of them.

If you have problems, change the model such that the O's remain fixed while the fractions are moving. Then it should be clear to the strictest orthodox matheologian that the O's remain.

Regards, WM

sergio schrieb am Freitag, 28. Januar 2022 um 03:31:37 UTC+1:
> On 1/27/2022 3:47 PM, WM wrote:

> >> They do not "leave". They are replaced by X's.
> >
> > No O is replaced by an X. Only their positions are interchanged.
> wrong, each O is replaced by an X.

Which O is the first to exit?

Regards, WM

It happens that WM formulated :
> zelos...@gmail.com schrieb am Freitag, 28. Januar 2022 um 06:09:19 UTC+1:
>> torsdag 27 januari 2022 kl. 14:38:14 UTC+1 skrev WM:
>>> zelos...@gmail.com schrieb am Donnerstag, 27. Januar 2022 um 14:24:15
>>> UTC+1:
>>>> torsdag 27 januari 2022 kl. 12:32:48 UTC+1 skrev WM:
>>>>>> There is no FINITE steps, there are no steps PERIOD!
>>>>> But there are finite terms of the sequence --- only finite terms. Where,
>>>>> when, and how does the first O exit? What is the frist term when an O
>>>>> exits? Where does it go then?
>>>> Irrelevant, you are obfuscating with this and trying to confuse things.
>>>> No, I am using mathematics and logic.
>>>>> But there are finite terms of the sequence --- only finite terms. Where,
>>>>> when, and how does the first O exit?
>>>> You are asking "Why are all natural numbers finite but not N?" No, I am
>>>> asking where do the O's go?
>>>>
>>> I am using mathematics and logic. I am asking where do the O's go?
>> Irrelevant
>
> Very relevant, because it kills set theory. Note that the O's are there and
> cannot exit. And if they tried, where should they settle?
>
>> again all your Os and shit is irrelevant to the fact that there is a
>> bijection
>
> But the O's are relevant to prove that there is no bijection. If this proof
> surpasses your intellectual competence, try to use the simpler picture where
> the O's remain fixed at their places and the fractions try to disappear
> through the first column.

You are a fool.

On 1/28/2022 5:57 AM, WM wrote:
> Jim Burns schrieb
> am Freitag, 28. Januar 2022 um 03:13:52 UTC+1:
>> On 1/27/2022 4:56 PM, WM wrote:
>>> Jim Burns schrieb
>>> am Donnerstag, 27. Januar 2022 um 19:01:06 UTC+1:
>>>> On 1/27/2022 6:07 AM, WM wrote:

>>>>> No swap removes him from the matrix.
>>>>
>>>> Right.
>>>> One swap moves Bob within the matrix.
>>>> Also {one swap} is a two-ended-stepping-order-able
>>>> collection of swaps.
>>>
>>> An infinite sequence of swaps consists of one-swaps
>>> only. You believe in an exception. There is none.
>>
>> Each one of Bob's swaps sets him down
>> somewhere in ℕ⁺⨯{1} = ⋃{{1,...,k}}⨯{1}
>>
>> But
>> each one of Bob's swaps is followed by
>> another swap which picks him up again.
>>
>> Bob is not where he's been set down.
>> Bob is not where he's not been set down.
>> Bob is not in ℕ⁺⨯{1}
>>
>> No single swap removes Bob from the matrix.
>
> Of course not! Where should he settle outside?

Bob is not in ℕ⁺⨯{1}

We do not have global Bob-conservation,
even though we have local Bob-conservation.

The explanation is NOT dark fractions.
Bob disappears inside ℕ⁺⨯ℕ⁺
ℕ⁺ = ⋃{{1,...,k}}
Inside ℕ⁺ there are no dark numbers.
Inside ℕ⁺⨯ℕ⁺ there are no dark fractions.

The explanation IS the lack of two-ended-
-stepping-order-ability.
Wherever Bob is set down, he is picked up again.
That can't happen in a two-ended stepping-order.

The explanation is that not all collections
are like collections of sheep or of pebbles.

This difference from sheep-collections and
pebble-collections is not a disaster, but it requires
careful thought. Fortunately, you and I can
benefit from previous centuries of careful thought
about this difference.

>> _Because Bob has no last swap_ he can be set down on
>> infinitely-many spots in ℕ⁺⨯{1} and left on
>> none of them.
>
> If you have problems, change the model such that
> the O's remain fixed while the fractions are moving.
> Then it should be clear to the strictest orthodox
> matheologian that the O's remain.

Meanwhile...

<JB<WM<JB>>>
| |>> For each O, we know where in ℕ⁺⨯{1}
|>> the X comes from that replaces it.
|>
|> And we know that every X is replaced by an O.
| | Right.
| Each O swaps with a different X.
| So there can't be more O's than X's.

On Thursday, 27 January 2022 at 18:22:21 UTC-4, WM wrote:
> horand....@gmail.com schrieb am Donnerstag, 27. Januar 2022 um 23:11:38 UTC+1:
> > On Thursday, 27 January 2022 at 17:39:01 UTC-4, WM wrote:
> > > horand....@gmail.com schrieb am Donnerstag, 27. Januar 2022 um 17:06:10 UTC+1:
> > [...]
> > > > Just as at every finite step there are aleph_0 X's in the matrix, there also are aleph_0 O's.
> > > The O's indicate not indexed fractions.
> > At every finite step there are aleph_0 "not indexed fractions" as well as aleph_0 X's that have not been moved yet. That's all.
> That's all, in fact, because after every finite step nothing happens. Therefore most fractions remain not indexed.

"[A]fter every finite step" is your usual gobbledygook. Add quantifiers to the list of things you cannot comprehend, and also the fact that natural language is often ambiguous.

"After every step" only makes sense when you consider a *LIMIT*. And the limiting position is clear: Every square in the matrix has an X in it. The question of "how do they all get there?" is as idiotic as the question "how can the limit of a sequence of infinite sets ever be empty?" That, of course, is another point you will never understand.

On Friday, 28 January 2022 at 06:53:24 UTC-4, WM wrote:
[...]
> try to use the simpler picture where the O's remain fixed at their places and the fractions try to disappear through the first column.

Oh goody! Now we have fractions that are trying to disappear! Why would they want to do that? Out of shame of being associated with your nonsense?

On 1/28/2022 4:57 AM, WM wrote:
> Jim Burns schrieb am Freitag, 28. Januar 2022 um 03:13:52 UTC+1:
>> On 1/27/2022 4:56 PM, WM wrote:
>>> Jim Burns schrieb
>>> am Donnerstag, 27. Januar 2022 um 19:01:06 UTC+1:
>>>> On 1/27/2022 6:07 AM, WM wrote:
>>
>>>>> No swap removes him from the matrix.
>>>>
>>>> Right.
>>>> One swap moves Bob within the matrix.
>>>> Also {one swap} is a two-ended-stepping-order-able
>>>> collection of swaps.
>>>
>>> An infinite sequence of swaps consists of one-swaps
>>> only. You believe in an exception. There is none.
>> Each one of Bob's swaps sets him down
>> somewhere in ℕ⁺⨯{1} = ⋃{{1,...,k}}⨯{1}
>> But
>> each one of Bob's swaps is followed by
>> another swap which picks him up again.
>> Bob is not where he's been set down.
>> Bob is not where he's not been set down.
>> Bob is not in ℕ⁺⨯{1}
>> No single swap removes Bob from the matrix.
>
> Of course not! Where should he settle outside?
>>
>> _Because Bob has no last swap_ he can be set down on
>> infinitely-many spots in ℕ⁺⨯{1} and left on
>> none of them.
>
> If you have problems, change the model such that the O's remain fixed while the fractions are moving.

no, dont use X and O, use white and yellow post-it notes, and write the fraction on top first.

>Then it should be clear to the strictest orthodox matheologian that the O's remain.

all of this is not mathematics.

you use X and O's and your rubber leaky sets, because you don't know how to use equations.

it is a demonstration that you do not know math at all.

Fail.

>
> Regards, WM

On 1/28/2022 4:58 AM, WM wrote:
> sergio schrieb am Freitag, 28. Januar 2022 um 03:31:37 UTC+1:
>> On 1/27/2022 3:47 PM, WM wrote:
>
>>>> They do not "leave". They are replaced by X's.
>>>
>>> No O is replaced by an X. Only their positions are interchanged.
>> wrong, each O is replaced by an X.
>
> Which O is the first to exit?
>
> Regards, WM

did you stick the X on top of the O ?

then the O is under each X

Or did you remove the O, and then put an X there ?

if so, look in your pocket or on the floor first

On 1/28/2022 4:53 AM, WM wrote:
> zelos...@gmail.com schrieb am Freitag, 28. Januar 2022 um 06:09:19 UTC+1:
>> torsdag 27 januari 2022 kl. 14:38:14 UTC+1 skrev WM:
>>> zelos...@gmail.com schrieb am Donnerstag, 27. Januar 2022 um 14:24:15 UTC+1:
>>>> torsdag 27 januari 2022 kl. 12:32:48 UTC+1 skrev WM:
>>>
>>>>>> There is no FINITE steps, there are no steps PERIOD!
>>>>> But there are finite terms of the sequence --- only finite terms. Where, when, and how does the first O exit?
>>>>
>>>>> What is the frist term when an O exits? Where does it go then?
>>>> Irrelevant, you are obfuscating with this and trying to confuse things.
>>> No, I am using mathematics and logic.
>>>>> But there are finite terms of the sequence --- only finite terms. Where, when, and how does the first O exit?
>>>> You are asking "Why are all natural numbers finite but not N?"
>>> No, I am asking where do the O's go?
>>>>
>>> I am using mathematics and logic. I am asking where do the O's go?
>> Irrelevant
>
> Very relevant, because it kills set theory. Note that the O's are there and cannot exit. And if they tried, where should they settle?

O Ants that cannot exit.
Ants that try, but can settle anywhere

>
>> again all your Os and shit is irrelevant to the fact that there is a bijection
>
> But the O's are relevant to prove that there is no bijection.

So you failed. but chin up, if Cantor can do it, you can't.

> Regards, WM

Jim Burns schrieb am Freitag, 28. Januar 2022 um 13:47:34 UTC+1:
> On 1/28/2022 5:57 AM, WM wrote:
> > Jim Burns schrieb
> > am Freitag, 28. Januar 2022 um 03:13:52 UTC+1:
> >> On 1/27/2022 4:56 PM, WM wrote:
> >>> Jim Burns schrieb
> >>> am Donnerstag, 27. Januar 2022 um 19:01:06 UTC+1:
> >>>> On 1/27/2022 6:07 AM, WM wrote:
>
> >>>>> No swap removes him from the matrix.
> >>>>
> >>>> Right.
> >>>> One swap moves Bob within the matrix.
> >>>> Also {one swap} is a two-ended-stepping-order-able
> >>>> collection of swaps.
> >>>
> >>> An infinite sequence of swaps consists of one-swaps
> >>> only. You believe in an exception. There is none.
> >>
> >> Each one of Bob's swaps sets him down
> >> somewhere in ℕ⁺⨯{1} = ⋃{{1,...,k}}⨯{1}
> >>
> >> But
> >> each one of Bob's swaps is followed by
> >> another swap which picks him up again.
> >>
> >> Bob is not where he's been set down.
> >> Bob is not where he's not been set down.
> >> Bob is not in ℕ⁺⨯{1}
> >>
> >> No single swap removes Bob from the matrix.
> >
> > Of course not! Where should he settle outside?
>
> Bob is not in ℕ⁺⨯{1}
> We do not have global Bob-conservation,
> even though we have local Bob-conservation.
>
> The explanation is NOT dark fractions.
> Bob disappears inside ℕ⁺⨯ℕ⁺
> ℕ⁺ = ⋃{{1,...,k}}
> Inside ℕ⁺ there are no dark numbers.
> Inside ℕ⁺⨯ℕ⁺ there are no dark fractions.

There are fractions which cannot be defined. Look at this model proving dark fractions in a simple and lucid way:

Here is the eighth term of Cantor's sequence when it is realized by figures

XXXXOOOOO...
XXXOOOOOO...
XOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
OOOOOOOOO...
XOOOOOOOO...
....

Now, instead of moving X and O, let them remain fixed while the fractions

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, ...
....

are changing places.

First term: 1/1 remains at its position, indexed by 1. Second term: 1/2 receives Index 2, by changing positions with 2/1:

1/1, 2/1, 1/3, 1/4, ...
1/2, 2/2, 2/3, 2/4, ...
3/1, 3/2, 3/3, 3/4, ...
4/1, 4/2, 4/3, 4/4, ...
....

Then 2/1 receives index 3 by changing positions with 3/1
1/1, 3/1, 1/3, 1/4, ...
1/2, 2/2, 2/3, 2/4, ...
2/1, 3/2, 3/3, 3/4, ...
4/1, 4/2, 4/3, 4/4, ...
....

and so on. Since we have a bijection, it doesn't matter which sort moves. Every term of the sequence satisfies Cantor's formula k = (m + n - 1)(m + n - 2)/2 + m.

All this happens below the surface

1OOO...
2OOO...
3OOO...
4OOO...
....

which remains rigid and frozen.

Now, for every finite term you admit that all O's have a fraction. No fraction is missing, no O is empty. But in the LIMIT, there are all O's covering nothing because all fractions are in the first column. That means aleph_0 O's have passed away in a flash. You can imagine the rattle (if you think of coins in a fruit machine). This can happen only if they are dark, not identifyable indivdually.

Regards, WM

On Friday, 28 January 2022 at 15:41:34 UTC-4, WM wrote:
> Now, for every finite term you admit that all O's have a fraction. No fraction is missing, no O is empty. But in the LIMIT, there are all O's covering nothing because all fractions are in the first column.

What utter rot. The whole point is that the fractions make up the entire square. No point is left uncovered.

> That means aleph_0 O's have passed away in a flash. You can imagine the rattle (if you think of coins in a fruit machine).

Not to mention the neurons in poor Wolfi's Head. Must be tough going through life with as little comprehension as you. And you've been not learning for decades now. That's an impressive record of nonperformance.

On 1/28/2022 1:41 PM, WM wrote:
> Jim Burns schrieb am Freitag, 28. Januar 2022 um 13:47:34 UTC+1:
>> On 1/28/2022 5:57 AM, WM wrote:
>>> Jim Burns schrieb
>>> am Freitag, 28. Januar 2022 um 03:13:52 UTC+1:
>>>> On 1/27/2022 4:56 PM, WM wrote:
>>>>> Jim Burns schrieb
>>>>> am Donnerstag, 27. Januar 2022 um 19:01:06 UTC+1:
>>>>>> On 1/27/2022 6:07 AM, WM wrote:
>>
>>>>>>> No swap removes him from the matrix.
>>>>>>
>>>>>> Right.
>>>>>> One swap moves Bob within the matrix.
>>>>>> Also {one swap} is a two-ended-stepping-order-able
>>>>>> collection of swaps.
>>>>>
>>>>> An infinite sequence of swaps consists of one-swaps
>>>>> only. You believe in an exception. There is none.
>>>>
>>>> Each one of Bob's swaps sets him down
>>>> somewhere in ℕ⁺⨯{1} = ⋃{{1,...,k}}⨯{1}
>>>>
>>>> But
>>>> each one of Bob's swaps is followed by
>>>> another swap which picks him up again.
>>>>
>>>> Bob is not where he's been set down.
>>>> Bob is not where he's not been set down.
>>>> Bob is not in ℕ⁺⨯{1}
>>>>
>>>> No single swap removes Bob from the matrix.
>>>
>>> Of course not! Where should he settle outside?
>>
>> Bob is not in ℕ⁺⨯{1}
>> We do not have global Bob-conservation,
>> even though we have local Bob-conservation.
>>
>> The explanation is NOT dark fractions.
>> Bob disappears inside ℕ⁺⨯ℕ⁺
>> ℕ⁺ = ⋃{{1,...,k}}
>> Inside ℕ⁺ there are no dark numbers.
>> Inside ℕ⁺⨯ℕ⁺ there are no dark fractions.
>
> There are fractions which cannot be defined. Look at this model proving dark fractions in a simple and lucid way:
>
> Here is the eighth term of Cantor's sequence when it is realized by figures
>
> XXXXOOOOO...
> XXXOOOOOO...
> XOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> OOOOOOOOO...
> XOOOOOOOO...
> ...
>

which shows a bad X in (1,9)

> Now, instead of moving X and O, let them remain fixed while the fractions
>
> 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, ...
> ...
>

like Cantor did, that is good.

> are changing places.
>
> First term: 1/1 remains at its position, indexed by 1. Second term: 1/2 receives Index 2, by changing positions with 2/1:
>
> 1/1, 2/1, 1/3, 1/4, ...
> 1/2, 2/2, 2/3, 2/4, ...
> 3/1, 3/2, 3/3, 3/4, ...
> 4/1, 4/2, 4/3, 4/4, ...

no, there is no time to do that to an infinite matrix, nor any reason, or was your first matrix a mistake ?

> ...
>
> Then 2/1 receives index 3 by changing positions with 3/1
> 1/1, 3/1, 1/3, 1/4, ...
> 1/2, 2/2, 2/3, 2/4, ...
> 2/1, 3/2, 3/3, 3/4, ...
> 4/1, 4/2, 4/3, 4/4, ...
> ...

no, you fooling people now, you a diversionator,

>
> and so on. Since we have a bijection, it doesn't matter which sort moves. Every term of the sequence satisfies Cantor's formula k = (m + n - 1)(m + n - 2)/2 + m.

no it does not, you changed the matrix, the m,n stayed with each fraction, you have it all messed up now,

>
> All this happens below the surface

Submarine Ants

>
> 1OOO...
> 2OOO...
> 3OOO...
> 4OOO...
> ...
>

all of a sudden, this pop out ? like Magic ? you try to fool us again.

> which remains rigid and frozen.

no it does not, all you sets are rubber leaky variables

>
> Now, for every finite term you admit that all O's have a fraction.

then you made a mistake. Where/when did you put O's in ? like CheeriOs ?

No fraction is missing, no O is empty. But in the LIMIT, there are all O's covering nothing because all fractions are in the first column. That means
aleph_0 O's have passed away in a flash. You can imagine the rattle (if you think of coins in a fruit machine). This can happen only if they are dark,
not identifyable indivdually.
>
> Regards, WM

na, that is never going to work.

How about if we took a finite string, and tied it to an infinite string, then we would have an infinite string with a knot in it.
Well at least one knot, who knows how many knots are way out there.

then at least you can go cut it at k with sissors, and have your finite set again.

Ants stuck in a fruit machine
Ants that have a fraction
Ants that are frozen and ridiged
Ants that happen below the surface