On 11/27/2022 6:14 AM, WM wrote:
> Jim Burns schrieb am Samstag, 26. November 2022 um 21:06:43 UTC+1:
>> On 11/26/2022 5:37 AM, WM wrote:
>
>>> All definable steps can be discerned.
>> It is a sequence of exchanges.
>>
>> Its first exchange is ⟨1/1 1/1⟩
>> It has no last exchange.
>
> But it has only definable *FINITE* exchanges which leave the numbers of O's unchanged.

wrong, you replaced the Os with Xs or did you forget again ?

>>
>> No exchange _at all_ in the sequence
>> handles dark numbers.
>
>
> Regards, WM

On 11/27/2022 7:14 AM, WM wrote:
> Jim Burns schrieb am Samstag,
> 26. November 2022 um 21:06:43 UTC+1:
>> On 11/26/2022 5:37 AM, WM wrote:

>>> All definable steps can be discerned.
>>
>> It is a sequence of exchanges.
>> Its first exchange is ⟨1/1 1/1⟩
>> It has no last exchange.
>
> But it has only definable exchanges which
> leave the numbers of O's unchanged.

Some -- but not all -- collections have
_Bob-conservation_
https://www.youtube.com/watch?v=TjAg-8qqR3g

_Bob_ is the O which is born on the 1/2 of
the fraction-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 ...
....

The exchange ⟨1/2 2/1⟩ leaves Bob in the
matrix, but at 2/1.
-- Bob-conservation holds.

The exchanges ⟨1/2 2/1⟩ ⟨2/1 3/1⟩ leave Bob
in the matrix, but at 3/1.
-- Bob-conservation holds.

The exchanges ⟨1/2 2/1⟩ ⟨2/1 3/1⟩ ⟨3/1 6/1⟩
leave Bob in the matrix, but at 6/1.
-- Bob-conservation holds.

For each exchange in the sequence,
there is a _last_ exchange involving Bob
before that exchange. All of the
before-exchanges together leave Bob where
that last exchange leaves Bob.
Bob is _somewhere_
-- Bob-conservation holds.

However,
for each exchange in the sequence,
there is a _later_ exchange which removes
Bob from the last matrix-location he occupied
before that exchange.

After _all_ exchanges in the sequence,
Bob has been removed from any matrix location
_at all_ that he occupied at any point
-- Bob is not in the matrix
-- Bob-conservation is violated.

A bounded initial segment of exchanges
keeps Bob within a bounded initial segment
of indexes/numerators/denominators.
Bob-conservation _cannot_ be violated in
that bounded initial segment.

_All_ the exchanges in the sequence
do _not_ keep Bob within any bounded initial
segment of indexes/numerators/denominators.
Bob-conservation _can_ be violated by
_all_ the exchanges.

> But it has only definable exchanges which
> leave the numbers of O's unchanged.

Some -- but not all -- collections have
_Bob-conservation_

Bounded initial segments of the definable
exchanges have Bob-conservation.

_All_ of the definable exchanges do not
have Bob-conservation.

>> No exchange _at all_ in the sequence
>> handles dark numbers.
>
> No exchange changes the number of O's.

_All_ the exchanges together change
the number of O's.

O's ⟶ ⟨p/q k/1⟩ ⟵ no O's

On Sunday, November 27, 2022 at 7:21:35 PM UTC+1, Jim Burns wrote:

> After _all_ exchanges in the sequence,

You are talking nonsense, just like WM.

Co-Stupidity.

On Sunday, November 27, 2022 at 9:19:09 PM UTC+1, Fritz Feldhase wrote:
> On Sunday, November 27, 2022 at 7:21:35 PM UTC+1, Jim Burns wrote:
> >
> > After _all_ exchanges in the sequence,
> >
> You are talking nonsense, just like WM.
>
> Co-Stupidity.

Hint: "Nice to see that you [Jim Burns] at least have understood what others claim to be incomprehensible for average mathematicians." (WM)

Two idiots "discussing" nonsense?

See: https://is1-ssl.mzstatic.com/image/thumb/Video124/v4/6f/d4/97/6fd4975f-d2de-9b05-0b68-33491bcf0ca4/pr_source.lsr/2000x1125.webp

On 11/27/2022 2:29 PM, Fritz Feldhase wrote:
> On Sunday, November 27, 2022 at 9:19:09 PM UTC+1, Fritz Feldhase wrote:
>> On Sunday, November 27, 2022 at 7:21:35 PM UTC+1, Jim Burns wrote:
>>>
>>> After _all_ exchanges in the sequence,
>>>
>> You are talking nonsense, just like WM.
>>
>> Co-Stupidity.
>
> Hint: "Nice to see that you [Jim Burns] at least have understood what others claim to be incomprehensible for average mathematicians." (WM)

that's funny!

>
> Two idiots "discussing" nonsense?
>
> See: https://is1-ssl.mzstatic.com/image/thumb/Video124/v4/6f/d4/97/6fd4975f-d2de-9b05-0b68-33491bcf0ca4/pr_source.lsr/2000x1125.webp

lördag 26 november 2022 kl. 11:20:16 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Freitag, 25. November 2022 um 13:59:53 UTC+1:
> > fredag 25 november 2022 kl. 13:25:39 UTC+1 skrev WM:
> > > zelos...@gmail.com schrieb am Freitag, 25. November 2022 um 13:12:26 UTC+1:
> > >
> > > > Your O and X does not change the fact that I can find those two injections and thus proves you wrong.
> > >
> > > Every remaining O proves the existence of a fraction that is not indexed such that the index can be known - by definition.
> > It doesn't because you are arguing in a step by step process
> Definable indices are issued step by step.
> > WHICH IS NOT HOW FUNCTIONS FUCKING WORK!
> That means they work in undefinable ways, i.e., in the darkness.
>
> Regards, WM

Which is HORSE SHIT!

There is no "definable" or "Darkness" as you put it, it is all bullshit!

On 11/27/2022 3:29 PM, Fritz Feldhase wrote:
> On Sunday, November 27, 2022
> at 9:19:09 PM UTC+1, Fritz Feldhase wrote:
>> On Sunday, November 27, 2022
>> at 7:21:35 PM UTC+1, Jim Burns wrote:

>>> After _all_ exchanges in the sequence,
>>
>> You are talking nonsense, just like WM.
>> Co-Stupidity.

After _all_ exchanges in the sequence,
Bob has been removed from any matrix location
_at all_ that he occupied at any point

What I mean by that is:

At some point, Bob occupies pⱼ/qⱼ
⟺
there is a finite sub-sequence 𝐵𝑜𝑏ⱼ
𝐵𝑜𝑏ⱼ = ⟨ ⟨1/2,2/1⟩ ⋯ ⟨pⱼ₋₁/qⱼ₋₁,pⱼ/qⱼ⟩ ⟩
of the _Cantor_ sequence of exchanges
such that
⟨1/2, 2/1⟩ begins 𝐵𝑜𝑏ⱼ
⟨pⱼ₋₁/qⱼ₋₁, pⱼ/qⱼ⟩ ends 𝐵𝑜𝑏ⱼ
and,
for each split of 𝐵𝑜𝑏ⱼ
there is some pᵢ/qᵢ such that
⟨pᵢ₋₁/qᵢ₋₁, pᵢ/qᵢ⟩ is last-before and
⟨pᵢ/qᵢ, pᵢ₊₁/qᵢ₊₁⟩ is first-after that split.

⟨p/q, k/1⟩ is in the Cantor sequence
⟺
p and q are in ℕ⁺
k = (p+q-1)(p+q-2)/2+p

⟨p'/q', k'/1⟩ ≺ ⟨p"/q", k"/1⟩
⟺
k' < k"

at some point
Bob has been removed from pⱼ/qⱼ
⟺
at some point
Bob occupies pₖ/qₖ after occupying pⱼ/qⱼ
⟺
some 𝐵𝑜𝑏ₖ exists
which ⟨pₖ₋₁/qₖ₋₁, pₖ/qₖ⟩ ends
and which contains ⟨pⱼ/qⱼ, pⱼ₊₁/qⱼ₊₁⟩
-- the "leaving pⱼ/qⱼ" exchange

After _all_ exchanges in the sequence,
Bob has been removed from any matrix location
_at all_ that he occupied at any point
⟺
for each fraction pⱼ/qⱼ
if
𝐵𝑜𝑏ⱼ exists
which ⟨pⱼ₋₁/qⱼ₋₁, pⱼ/qⱼ⟩ ends
then
some 𝐵𝑜𝑏ₖ exists
which ⟨pₖ₋₁/qₖ₋₁, pₖ/qₖ⟩ ends
and which contains ⟨pⱼ/qⱼ, pⱼ₊₁/qⱼ₊₁⟩

----
Do you agree with this claim?
| for each fraction pⱼ/qⱼ
| if
| 𝐵𝑜𝑏ⱼ exists
| which ⟨pⱼ₋₁/qⱼ₋₁, pⱼ/qⱼ⟩ ends
| then
| some 𝐵𝑜𝑏ₖ exists
| which ⟨pₖ₋₁/qₖ₋₁, pₖ/qₖ⟩ ends
| and which contains ⟨pⱼ/qⱼ, pⱼ₊₁/qⱼ₊₁⟩

> Hint:
> "Nice to see that you [Jim Burns] at least
> have understood what others claim to be
> incomprehensible for average mathematicians."
> (WM)

I recommend against taking WM too seriously here.
He has "agreed" with me before.
Nothing ever came of it.

Suppose, though, that WM actually agrees with me.
Do you recommend that I change my positions
(many of which have proofs) in order that
hr would continue to disagree with me?

Maybe I could dissuade WM from using
first order logic correctly. Do you have
any thoughts on how to go about doing that,
if that should ever become a desirable goal?

On Monday, November 28, 2022 at 11:23:42 AM UTC+1, Jim Burns wrote:

> > Hint:
> > "Nice to see that you [Jim Burns] at least
> > have understood what others claim to be
> > incomprehensible for average mathematicians."
> > (WM)
> >
> I recommend against taking WM too seriously here.
> He has "agreed" with me before.
> Nothing ever came of it.

You don't get it, do you? He's "thinking" that you agree WITH HIM (not the other way round).

At least he's quoting you in de.sci.mathematic, claiming that you have understood his argument.

:-)

On Monday, November 28, 2022 at 11:23:42 AM UTC+1, Jim Burns wrote:

> After _all_ exchanges in the sequence, ...

Considering usual ("standard") sequences there is no such point/state etc.

Using an index set IN u {omega} (for an "extended sequence") we MIGHT "locate" "after _all_ exchanges" at the index /omega/.

But how can we DETERMINE the "outcome" of "all exchanges"?

> Bob has been removed from any matrix location

Well, so WHERE is he now? Huh?!

Gus Gassmann schrieb am Sonntag, 27. November 2022 um 14:35:29 UTC+1:
> On Sunday, 27 November 2022 at 07:57:36 UTC-4, WM wrote:
> > Gus Gassmann schrieb am Samstag, 26. November 2022 um 15:44:38 UTC+1:
> > > On Saturday, 26 November 2022 at 06:18:44 UTC-4, WM wrote:
> >
> > > > I am not interested in the limit process because it does not index any fractions in a definable way.
> > > First, there is no "limit process", but the limit of the process.
> > To take the limit is a process.
> > > And the limit of the matrix sequence you defined is the matrix of all 'X's. Your quote proves that you understand this.
> > Of course, but I understand that all definable mappings leave the number of O's unchanged. That and only that is the result of my proof.
> > > But *of course* you prefer to stop after finitely many steps
> > My proof covers all definable steps. Of course this collection is only potentially infinite,
> It is not *FINITE*, i.e, bounded by a fixed finite upper bound

Correct! But it dos not reach to a number of elements which is larger than every naturak number.

> Your set D of "definable numbers" is an *inductive set*

of course.

> hence at least as large as the set of natural numbers.

Not as large however as Cantor's ℵo-set of all natural numbers.

> But even if D were just a FISON, say,

D is the set (or a subset) of all exchanges in my example which do not change the number of O's.

Regards, WM

Jim Burns schrieb am Sonntag, 27. November 2022 um 19:21:35 UTC+1:
> On 11/27/2022 7:14 AM, WM wrote:
> > Jim Burns schrieb am Samstag,
> > 26. November 2022 um 21:06:43 UTC+1:
> >> On 11/26/2022 5:37 AM, WM wrote:
>
> >>> All definable steps can be discerned.
> >>
> >> It is a sequence of exchanges.
> >> Its first exchange is ⟨1/1 1/1⟩
> >> It has no last exchange.
> >
> > But it has only definable exchanges which
> > leave the numbers of O's unchanged.
> Some -- but not all -- collections have
> _Bob-conservation_
> https://www.youtube.com/watch?v=TjAg-8qqR3g
>
> _Bob_ is the O which is born on the 1/2 of
> the fraction-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 ...
> ...
>
> The exchange ⟨1/2 2/1⟩ leaves Bob in the
> matrix, but at 2/1.
> -- Bob-conservation holds.

Yes, at least during all definable steps.
>
> The exchanges ⟨1/2 2/1⟩ ⟨2/1 3/1⟩ leave Bob
> in the matrix, but at 3/1.
> -- Bob-conservation holds.

You have got it! You should teach a rather stupid reader in de.sci.mathematik who denies that this could be understood.
>
> The exchanges ⟨1/2 2/1⟩ ⟨2/1 3/1⟩ ⟨3/1 6/1⟩
> leave Bob in the matrix, but at 6/1.
> -- Bob-conservation holds.
> For each exchange in the sequence,
> there is a _last_ exchange involving Bob
> before that exchange. All of the
> before-exchanges together leave Bob where
> that last exchange leaves Bob.
> Bob is _somewhere_
> -- Bob-conservation holds.
>
Very good!

> However,
> for each exchange in the sequence,
> there is a _later_ exchange which removes
> Bob from the last matrix-location he occupied
> before that exchange.

That is true. Nevertheless no definable step throws it out of the matrix.
>
> After _all_ exchanges in the sequence,
> Bob has been removed from any matrix location
> _at all_ that he occupied at any point
> -- Bob is not in the matrix
> -- Bob-conservation is violated.

Agreed. But not during any definable step.
>
> A bounded initial segment of exchanges
> keeps Bob within a bounded initial segment
> of indexes/numerators/denominators.
> Bob-conservation _cannot_ be violated in
> that bounded initial segment.
>
> _All_ the exchanges in the sequence
> do _not_ keep Bob within any bounded initial
> segment of indexes/numerators/denominators.
> Bob-conservation _can_ be violated by
> _all_ the exchanges.
> > But it has only definable exchanges which
> > leave the numbers of O's unchanged.
> Some -- but not all -- collections have
> _Bob-conservation_

All dfinable steps have it. More is not claimed.
>
> Bounded initial segments of the definable
> exchanges have Bob-conservation.

Thank you for your great analyse. Only the next step goes astray.
>
> _All_ of the definable exchanges do not
> have Bob-conservation.

Then you could name that one where Bob leaves.

> > No exchange changes the number of O's.
> _All_ the exchanges together change
> the number of O's.

Yes, but no definable exchange does. And more is not claimed.

Regards, WM

Fritz Feldhase schrieb am Sonntag, 27. November 2022 um 21:19:09 UTC+1:
> On Sunday, November 27, 2022 at 7:21:35 PM UTC+1, Jim Burns wrote:
>
> > After _all_ exchanges in the sequence,
> You are talking nonsense, just like

Cantor: "then every number p/q comes at an absolutely fixed place of a simple infinite sequence."

Regards, WM

zelos...@gmail.com schrieb am Montag, 28. November 2022 um 08:05:56 UTC+1:
> lördag 26 november 2022 kl. 11:20:16 UTC+1 skrev WM:

> > Definable indices are issued step by step.
> > > WHICH IS NOT HOW FUNCTIONS FUCKING WORK!
> > That means they work in undefinable ways, i.e., in the darkness.
>
> There is no "definable" or "Darkness" as you put it, it is all bullshit!

It is the realm where step-by-step fails.
Definable indices are issued step by step.

Regards, WM

Fritz Feldhase schrieb am Montag, 28. November 2022 um 12:13:50 UTC+1:

> But how can we DETERMINE the "outcome" of "all exchanges"?

Cantor ande his disciples claim that we can.

> > Bob has been removed from any matrix location
> Well, so WHERE is he now? Huh?!

It is in the darkness, since it cannot have left the matrix. That is the only point not yet understood by JB.

Regards, WM

Fritz Feldhase schrieb am Montag, 28. November 2022 um 11:38:33 UTC+1:

> You don't get it, do you? He's "thinking" that you agree WITH HIM (not the other way round).

He obviously agrees that no O can get lost in an individual definable step.

Regards, WM

On 11/28/2022 6:25 AM, WM wrote:
> Fritz Feldhase schrieb am Sonntag, 27. November 2022 um 21:19:09 UTC+1:
>> On Sunday, November 27, 2022 at 7:21:35 PM UTC+1, Jim Burns wrote:
>>
>>> After _all_ exchanges in the sequence,
>> You are talking nonsense, just like
>
> Cantor: "then every number p/q comes at an absolutely fixed place of a simple infinite sequence."

which is his enumeration, and does not support any of your points at all.

>
> Regards, WM

On 11/28/2022 6:34 AM, WM wrote:
> Fritz Feldhase schrieb am Montag, 28. November 2022 um 12:13:50 UTC+1:
>
>> But how can we DETERMINE the "outcome" of "all exchanges"?
>
> Cantor ande his disciples claim that we can.

diversion. Cantor cant help you.

>
>>> Bob has been removed from any matrix location
>> Well, so WHERE is he now? Huh?!
>
> It is in the darkness, since it cannot have left the matrix. That is the only point not yet understood by JB.
there is no darkness, it is just you trying to fool people.

>
> Regards, WM
>
>

måndag 28 november 2022 kl. 13:29:01 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Montag, 28. November 2022 um 08:05:56 UTC+1:
> > lördag 26 november 2022 kl. 11:20:16 UTC+1 skrev WM:
>
> > > Definable indices are issued step by step.
> > > > WHICH IS NOT HOW FUNCTIONS FUCKING WORK!
> > > That means they work in undefinable ways, i.e., in the darkness.
> >
> > There is no "definable" or "Darkness" as you put it, it is all bullshit!
> It is the realm where step-by-step fails.
> Definable indices are issued step by step.
> Regards, WM
still all bullshit

Injections exists therefore a bijection exists and thus you are as always wring.

On 11/28/2022 7:22 AM, WM wrote:
> Jim Burns schrieb am Sonntag,
> 27. November 2022 um 19:21:35 UTC+1:
>> On 11/27/2022 7:14 AM, WM wrote:
>>> Jim Burns schrieb am Samstag,
>>> 26. November 2022 um 21:06:43 UTC+1:
>>>> On 11/26/2022 5:37 AM, WM wrote:

>>>>> All definable steps can be discerned.
>>>>
>>>> It is a sequence of exchanges.
>>>> Its first exchange is ⟨1/1 1/1⟩
>>>> It has no last exchange.
>>>
>>> But it has only definable exchanges which
>>> leave the numbers of O's unchanged.
>>
>> Some -- but not all -- collections have
>> _Bob-conservation_
>>
>> https://www.youtube.com/watch?v=TjAg-8qqR3g
>> _Bob_ is the O which is born on the 1/2 of
>> the fraction-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 ...
>> ...
>> The exchange ⟨1/2 2/1⟩ leaves Bob in the
>> matrix, but at 2/1.
>> -- Bob-conservation holds.
>
> Yes, at least during all definable steps.

No,
you are attributing Bob-conservation and
Bob-conservation-breaking to particular
_exchanges_ That isn't so.

Some -- but not all -- _collections_ of
_the same exchanges_ have Bob-conservation.

Bob occupies 1/2 2/1 3/1 6/1 21/1 231/1 ...

Any terminating initial segment of exchanges
leaves Bob _somewhere_ in the matrix --
where the last exchange he participates in
leaves Bob.
Bob-conservation holds _for that collection_

----
The collection of _all definable_ exchanges
does not leave Bob anywhere.

Bob occupying k/1 is followed, finitely-many
exchanges later, by Bob occupying k'/1
k' = k(k+1)/2
There is nowhere in the matrix that Bob remains.

Neither does Bob leave the matrix.

| Bob occupies a _sequence_ of matrix-locations.
| | If any of the occupied locations are outside
| the matrix, then there is a _first_ outside
| the matrix which, being first, is immediately
| preceded by a matrix location inside the matrix.
| | If Bob leaves the matrix, there must be a step
| that begins inside the matrix and ends outside
| the matrix.
| | However,
| we know things about those those steps.
| A step beginning at k/1 inside the matrix
| ends at k'/1 with k' = k(k+1)/2
| The end is inside the matrix, too.

Bob leaving the matrix contradicts what we know.

Bob isn't inside. Bob isn't outside.
The collection of _all definable_ exchanges
_does not_ have Bob-conservation.

Some -- but not all -- _collections_ of
_the same exchanges_ have Bob-conservation.

>> _All_ of the definable exchanges do not
>> have Bob-conservation.
>
> Then you could name that one where Bob leaves.

_Bob is not conserved_

Consider electric charge conservation.
We designate some closed surface, and,
_because charge is conserved_
changes in the charge inside agree with
how much charge passes through that surface.

If a certain quantity of charge is no longer there,
then it has left through that surface.
That's charge _conservation_

_Bob is not conserved_

> Then you could name that one where Bob leaves.

You reason:
If Bob is not there, then he left.
When did he leave? Where did he go?

You are using an unstated assumption that
Bob is conserved.

( It is perhaps an unfortunate choice on my part
( to name that O "Bob". Things/people named
( "Bob" are almost always conserved.
( But it's funny.
( https://www.youtube.com/watch?v=TjAg-8qqR3g

>>> No exchange changes the number of O's.
>>
>> _All_ the exchanges together change
>> the number of O's.
>
> Yes, but no definable exchange does.

All the _definable_ exchanges together
change the number of O's

> And more is not claimed.

Bob is not conserved,
even without dark exchanges.

On 11/28/2022 4:36 AM, WM wrote:
> Fritz Feldhase schrieb am Montag, 28. November 2022 um 11:38:33 UTC+1:
>
>> You don't get it, do you? He's "thinking" that you agree WITH HIM (not the other way round).
>
> He obviously agrees that no O can get lost in an individual definable step.

Nothing gets lost in Cantor Pairing.

On 11/28/2022 6:13 AM, Fritz Feldhase wrote:
> On Monday, November 28, 2022
> at 11:23:42 AM UTC+1, Jim Burns wrote:

>> After _all_ exchanges in the sequence, ...
>
> Considering usual ("standard") sequences
> there is no such point/state etc.

Bob is an O born at 1/2

Consider Kevin, an X born at 2/1

Initially, Kevin is at 2/1,
until exchange ⟨1/2,2/1⟩

After exchange ⟨1/2,2/1⟩, Kevin is at 1/2

For the matrix-location 1/2,
there is an exchange ⟨1/2,2/1⟩ after which
Kevin is (without exception) at 1/2

There is no matrix-location about which
a similar claim can be made for Bob.

Initially, Bob is at 1/2,
until exchange ⟨1/2,2/1⟩

After exchange ⟨1/2,2/1⟩, Bob is at 2/1,
until exchange ⟨2/1,3/1⟩

After exchange ⟨2/1,3/1⟩, Bob is at 3/1,
until exchange ⟨3/1,6/1⟩

After exchange ⟨3/1,6/1⟩, Bob is at 6/1,
until exchange ⟨6/1,21/1⟩

After exchange ⟨6/1,21/1⟩, Bob is at 21/1,
until exchange ⟨21/1,231/1⟩

After each exchange, Bob is _somewhere_
but only until a later exchange,
at which point he'll be _somewhere else_

There is no matrix-location for which,
after some exchange,
Bob is (without exception) at that location.

----
Let 𝐾𝑒𝑣𝑖𝑛ⱼ be the sequence of matrix locations
Kevin occupies, as a result of the first j
exchanges being exchanged.
𝐾𝑒𝑣𝑖𝑛₀ 𝐾𝑒𝑣𝑖𝑛₁ 𝐾𝑒𝑣𝑖𝑛₂ 𝐾𝑒𝑣𝑖𝑛₃ ...

There is an exchange after which
Kevin is (without exception) at the same location.
∃j, ∀k ≥ j : 𝐾𝑒𝑣𝑖𝑛ₖ = 𝐾𝑒𝑣𝑖𝑛ⱼ

I would say that,
after all exchanges, Kevin is at 𝐾𝑒𝑣𝑖𝑛ⱼ

Perhaps you do not like the way I say this.
To you, I say
∃j, ∀k ≥ j : 𝐾𝑒𝑣𝑖𝑛ₖ = 𝐾𝑒𝑣𝑖𝑛ⱼ

Let 𝐵𝑜𝑏ⱼ be the sequence of matrix locations
Bob occupies.
𝐵𝑜𝑏₀ 𝐵𝑜𝑏₁ 𝐵𝑜𝑏₂ 𝐵𝑜𝑏₃ ...

There is NO exchange after which
Bob is (without exception) at the same location.
~∃j, ∀k ≥ j : 𝐵𝑜𝑏ₖ = 𝐵𝑜𝑏ⱼ

Where is Bob?

Not at 1/1.
Bob never comes to 1/1.

Not at 1/2.
Bob starts at 1/2,
but he leaves and never returns.

Not at 2/1.
Bob comes to 2/1,
but he leaves and never returns.

Not at 1/3.
Bob never comes to 1/3.

Not at 2/2.
Bob never comes to 2/2.

For each location in the matrix,
there is a reasonable case to be made that,
after all exchanges, Bob is not there.

> But how can we DETERMINE
> the "outcome" of "all exchanges"?

We know some claims which are true of each
exchange. We can advance to further claims,
but only in ways we know keep this true-of-each
property with which we start.

If we are clever enough, this method allows us
to learn further facts about infinitely-many
exchanges, without needing to perform
infinitely-many exchanges.

We KNOW this type of result, assuming we are
clever enough to find it, despite it being about
infinitely-many and our being finite.

>> Bob has been removed from any matrix location
>
> Well, so WHERE is he now? Huh?!

Can you explain why Bob should be anywhere?

----
For some collections, it is obvious that
Bob should be somewhere -- obvious AND
there are good reasons.

But not for all collections.
It seems to me that "not for all" is
the point which Cantor wanted to make by
matching rationals to naturals.

the o's have gone dark in message below;

█n 11/28/2022 6:36 AM, WM wr█te:
> Fritz Feldhase schrieb am M█ntag, 28. November 2022 um 11:38:33 UTC+1:
>
>> You d█n't get it, d█ y█u? He's "thinking" that y█u agree WITH HIM (n█t the █ther way r█und).
>
> He █bvi█usly agrees that n█ █ can get l█st in an individual definable step.
>
> Regards, WM

On 11/28/2022 5:38 AM, Fritz Feldhase wrote:
> On Monday, November 28, 2022
> at 11:23:42 AM UTC+1, Jim Burns wrote:

>>> Hint:
>>> "Nice to see that you [Jim Burns] at least
>>> have understood what others claim to be
>>> incomprehensible for average mathematicians."
>>> (WM)
>>
>> I recommend against taking WM too seriously here.
>> He has "agreed" with me before.
>> Nothing ever came of it.
>
> You don't get it, do you?
> He's "thinking" that you agree WITH HIM
> (not the other way round).
>
> At least he's quoting you in de.sci.mathematic,
> claiming that you have understood his argument.
> :-)

If he hasn't misrepresented me, I have no objection.
"Understand" is not "agree".

I suspect that those in de.sci.mathematic who
say they don't understand his argument
are only _believing_ they don't understand
his argument, because they think
he couldn't really mean _that_
I have thought
"They couldn't really mean _that_ "
on many occasions myself, in discussions with WM
and with other posters of a certain type.

----
I would greatly prefer not being lied about
in de.sci.mathematik if that were happening.

But I am not an academic, not a mathematician
except in the broadest of senses. I don't
actually care if they think I am in cahoots
with Mückenheim. I have no reputation that needs
protecting, and someone being mistaken about
what I think doesn't change what I think.

I will let what happens in de.sci.mathematik
stay in de.sci.mathematik

On Tuesday, November 29, 2022 at 6:21:00 AM UTC+1, Jim Burns wrote:
>
> But I am not an academic, not a mathematician
> except in the broadest of senses. I don't ...

He quoted you as an example of a _mathematician_ who is able to comprehend his "argument". Together with Mr. Olcott btw. (You know Mr. Olcott? He's famous for disproving Turing's Halting Theorem and -more recently- Gödel's Incompletness Theorem(s).) He did that to prove his claim that "mathematicians are able to comprehend his argument." Well...

And now fuck off, you silly asshole.

Jim Burns schrieb am Montag, 28. November 2022 um 19:56:55 UTC+1:
> On 11/28/2022 7:22 AM, WM wrote:

> >> Bob-conservation

> > Yes, at least during all definable steps.
> No,

Name a definable or nameable step without Bob-conservation.

> Bob occupies 1/2 2/1 3/1 6/1 21/1 231/1 ...
>
> Any terminating initial segment of exchanges
> leaves Bob _somewhere_ in the matrix --

and other exchanges are undfinable and not matching Cantor's requirement to come "at an absolutely fixed position of a simple infinite sequence"

> where the last exchange he participates in
> leaves Bob.
> Bob-conservation holds _for that collection_

Only numbers wich can be last are definable numbers.
>
> ----
> The collection of _all definable_ exchanges
> does not leave Bob anywhere.

That is silly. Definable exchanges can be defined. If you cannot define the first loss of an O, then it is not lost in the definable domain.
>
> There is nowhere in the matrix that Bob remains.

There is nowhere in the visible part of the matrix.
>
> Neither does Bob leave the matrix.

This proves his presence in the darkness.
>
> | Bob occupies a _sequence_ of matrix-locations.
> |
> | If any of the occupied locations are outside
> | the matrix, then there is a _first_ outside
> | the matrix which, being first, is immediately
> | preceded by a matrix location inside the matrix.
> |
> | If Bob leaves the matrix, there must be a step
> | that begins inside the matrix and ends outside
> | the matrix.

There is a more stringent argument: A lossless exchange will never suffer from a loss.

> Bob leaving the matrix contradicts what we know.

Of course. there can be no doubt.
>
> Bob isn't inside. Bob isn't outside.

He is in the darkness.

> The collection of _all definable_ exchanges
> _does not_ have Bob-conservation.

That is nonsense. Every loss at a definable step would be definable. You claim the existence of undefinable definable steps. That is the consequence of transfinite set theory.

Regards, WM