Jim Burns schrieb am Montag, 28. November 2022 um 22:22:55 UTC+1:
> 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.

You have very nicely explained this. You are among the first who have understood it.

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

If we could not, then Cantor's enumeration would not cover all fractions.

> We know some claims which are true of each
> exchange.

The most prominent one is that never an O will leave the matrix.

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

Keep in mind this wise sentence.
>
> If we are clever enough, this method allows us
> to learn further facts about infinitely-many
> exchanges, without needing to perform
> infinitely-many exchanges.

That's it! My sequence of matrices is never completely scrutinized, but we know that they never will lose an O.

> >> Bob has been removed from any matrix location
> >
> > Well, so WHERE is he now? Huh?!
> Can you explain why Bob should be anywhere?

Yes, remember the most prominent feature.

> 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.

That is contrary to all that he said, but it is true for the result of his attempt. There are by far less natural numbers than fractions.

Regards, WM

Jim Burns schrieb am Dienstag, 29. November 2022 um 06:21:00 UTC+1:

> 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_

They are blinded by Cantor. But be assured that I did not lie about you. What I posted is this:

The left-hand-side of the exchanges ⟨1/1 1/1⟩ ⟨1/2 2/1⟩ ⟨2/1 3/1⟩ ... hold X's and O's. Their right-hand-side contains only X's. Thus, all the O's are deleted by all the exchanges. By _all_ the exchanges. Stopping at any exchange in the sequence does not delete any O's. [Jim Burns (25 Nov 2022)] {{Hinweis: Stoppen ist genau dann möglich, wenn der Austausch individuell definierbar erfolgt.}}

> 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.

And now you need to believe that exchanging the X's and O's will delete all O's in a definable way but only after all definable terms of the sequence have kept all and every O. That is almost as silly as the belief that infinite endsegments will have an empty intersection.

Regards, WM

tisdag 29 november 2022 kl. 12:51:22 UTC+1 skrev WM:
> Jim Burns schrieb am Dienstag, 29. November 2022 um 06:21:00 UTC+1:
>
> > 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_
> They are blinded by Cantor. But be assured that I did not lie about you. What I posted is this:
> The left-hand-side of the exchanges ⟨1/1 1/1⟩ ⟨1/2 2/1⟩ ⟨2/1 3/1⟩ ... hold X's and O's. Their right-hand-side contains only X's. Thus, all the O's are deleted by all the exchanges. By _all_ the exchanges. Stopping at any exchange in the sequence does not delete any O's. [Jim Burns (25 Nov 2022)] {{Hinweis: Stoppen ist genau dann möglich, wenn der Austausch individuell definierbar erfolgt.}}
> > 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.
> And now you need to believe that exchanging the X's and O's will delete all O's in a definable way but only after all definable terms of the sequence have kept all and every O. That is almost as silly as the belief that infinite endsegments will have an empty intersection.
>
> Regards, WM

No one is blinded by cantor, his arguments and stuff are understood and easy to follow. Yours is meaningless bullshit

On 11/28/2022 11:20 PM, Jim Burns wrote:
> 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".

People that know Math always looks for correctness, and that is WMs intentional weak point.
"understanding" implies "in agreement with".

> 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,

your suspicions are unfounded, resulting in conjecture.

> 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.

how do you know what someone else thinks ?

> ----
> 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.

Math is a structured language, and when someone (like WM) makes fictional stuff up, it is easily detected.
WM typically has at least one math error in every one of his posts, usually many more, one post had nine falsehoods in it.
WM is a troll.

> 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 11/29/2022 6:43 AM, WM wrote:
> Jim Burns schrieb am Montag,
> 28. November 2022 um 22:22:55 UTC+1:
>> 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:

>>>> Bob has been removed from any matrix location
>>>
>>> Well, so WHERE is he now? Huh?!
>>
>> Can you explain why Bob should be anywhere?
>
> Yes, remember the most prominent feature.

>> We know some claims which are true of each
>> exchange.
>
> The most prominent one is that
> never an O will leave the matrix.

That is not a property of _exchanges_
That is a property of some -- but not all --
_exchange-sequences_

𝐵𝑜𝑏₁ = ⟨ ⟨1/2,2/1⟩ ⟩
has Bob-conservation.

𝐵𝑜𝑏₂ = ⟨ ⟨1/2,2/1⟩ ⟨2/1,3/1⟩ ⟩
has Bob-conservation.

𝐵𝑜𝑏₃ = ⟨ ⟨1/2,2/1⟩ ⟨2/1,3/1⟩ ⟨3/1,6/1⟩ ⟩
has Bob-conservation.

Each 𝐵𝑜𝑏ⱼ is a _terminating Bob-mover_
Each terminating Bob-mover has Bob-conservation.

𝓑𝓸𝓫 is the set of all and only terminating
Bob-movers.

⋃𝓑𝓸𝓫 is the union of all and only terminating
Bob-movers.

⋃𝓑𝓸𝓫 does not have Bob-conservation.

No exchange is in ⋃𝓑𝓸𝓫
(a sequence which does not have Bob-conservation)
which is not also in some terminating Bob-mover
(a sequence which has Bob-conservation).

Bob-conservation is not a property of _exchanges_
It is a property of some -- but not all --
_exchange-sequences_

Compare to:
Life is not a property of _atoms_
It is a property of some -- but not all --
_configurations of atoms_

I had some atoms for breakfast a little while ago.
None of them were part of a living being at the time.
Some of them will be part of a living being
later today.

Your hypothesis that some
exchanges in a sequence without Bob-conservation
_must be different_ from
exchanges in a sequence with Bob-conservation
reminds me of the now-discarded theory of
_elan vital_ -- that living tissue must be
composed in part of some substance, _elan vital_
that non-living tissue does not have.

You (WM) definitely should keep beating your drum.
How do we know _for sure_ that vitalism will not
return to prominence in the science of biology?

Jim Burns schrieb am Dienstag, 29. November 2022 um 19:31:27 UTC+1:
> On 11/29/2022 6:43 AM, WM wrote:
> > Jim Burns schrieb am Montag,
> > 28. November 2022 um 22:22:55 UTC+1:
> >> 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:
>
> >>>> Bob has been removed from any matrix location
> >>>
> >>> Well, so WHERE is he now? Huh?!
> >>
> >> Can you explain why Bob should be anywhere?
> >
> > Yes, remember the most prominent feature.

Do you?

> >> We know some claims which are true of each
> >> exchange.
> >
> > The most prominent one is that
> > never an O will leave the matrix.
> That is not a property of _exchanges_
> That is a property of some -- but not all --
> _exchange-sequences_

It is a property of exchanges, namely of each and every exchange.

> 𝓑𝓸𝓫 is the set of all and only terminating
> Bob-movers.

There are no other definable moves.
>
> ⋃𝓑𝓸𝓫 is the union of all and only terminating
> Bob-movers.
>
> ⋃𝓑𝓸𝓫 does not have Bob-conservation.

Chuckle.

> Bob-conservation is not a property of _exchanges_
> It is a property of some -- but not all --
> _exchange-sequences_

Interesting, what bizarr results Cantor's theory can cause.
>
> Compare to:
> Life is not a property of _atoms_
> It is a property of some -- but not all --
> _configurations of atoms_

No, I do not compare. Lossless exchanges are lossless in every environment.
>
> I had some atoms for breakfast a little while ago.
> None of them were part of a living being at the time.
> Some of them will be part of a living being
> later today.

Hopefully they will help to clear your brain.

Regards, WM

On 11/29/2022 3:43 AM, WM wrote:
> Jim Burns schrieb am Montag, 28. November 2022 um 22:22:55 UTC+1:
>> 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.
>
> You have very nicely explained this. You are among the first who have understood it.
>
>>> But how can we DETERMINE
>>> the "outcome" of "all exchanges"?
>
> If we could not, then Cantor's enumeration would not cover all fractions.
[...]

Cantor Pairing does cover all pairs. This is includes all positive
rationals.

Chris M. Thomasson schrieb am Dienstag, 29. November 2022 um 21:07:26 UTC+1:

> Cantor Pairing does cover all pairs. This includes all positive rationals..

In every step where that can be checked we see |ℕ|*(|ℕ|-1) positive rationals are not paired. Not even a single one O is deleted.

If all are paired in the limit, this is not what is needed for a bijection, namely in Cantor's own words: "every number p/q comes at an absolutely fixed position of a simple infinite sequence" and "The infinite sequence thus defined has the peculiar property to contain the positive rational numbers completely, and each of them only once at a determined place."

Regards, WM

On 11/29/2022 2:11 PM, WM wrote:
> Jim Burns schrieb am Dienstag,
> 29. November 2022 um 19:31:27 UTC+1:
>> On 11/29/2022 6:43 AM, WM wrote:
>>> Jim Burns schrieb am Montag,
>>> 28. November 2022 um 22:22:55 UTC+1:

>>>> We know some claims which are true of each
>>>> exchange.
>>>
>>> The most prominent one is that
>>> never an O will leave the matrix.
>>
>> That is not a property of _exchanges_
>> That is a property of some -- but not all --
>> _exchange-sequences_
>
> It is a property of exchanges,
> namely of each and every exchange.
>
>> 𝓑𝓸𝓫 is the set of all and only terminating
>> Bob-movers.
>
> There are no other definable moves.
>
>> ⋃𝓑𝓸𝓫 is the union of all and only terminating
>> Bob-movers.
>>
>> ⋃𝓑𝓸𝓫 does not have Bob-conservation.
>
> Chuckle.

Each terminating Bob-mover 𝐵𝑜𝑏ⱼ leaves Bob
somewhere.

The exchanges of 𝐵𝑜𝑏ⱼ leave Bob where
𝐵𝑜𝑏ⱼ's last exchange leaves Bob: kⱼ

⋃𝓑𝓸𝓫 does not leave Bob anywhere.

For each exchange ⟨kⱼ₋₁/1,kⱼ/1⟩ in ⋃𝓑𝓸𝓫
Bob is left at kⱼ/1
_until the next exchange_ ⟨kⱼ/1,kⱼ₊₁/1⟩

None of these kⱼ/1 are last.
Each is followed by kⱼ₊₁/1
where kⱼ₊₁ = kⱼ(kⱼ+1)/2

Each matrix-place kⱼ/1 at which
⋃𝓑𝓸𝓫 _temporarily_ leaves Bob
is a matrix-place kⱼ/1 at which
𝐵𝑜𝑏ⱼ _permanently_ leaves Bob.

>> Compare to:
>> Life is not a property of _atoms_
>> It is a property of some -- but not all --
>> _configurations of atoms_
>
> No, I do not compare.
> Lossless exchanges are lossless
> in every environment.

Consider ⋃𝓑𝓸𝓫

There is nowhere that all the exchanges of
⋃𝓑𝓸𝓫 leave Bob.

However,
each exchange ⟨kⱼ₋₁/1,kⱼ/1⟩ in ⋃𝓑𝓸𝓫
when in a different environment 𝐵𝑜𝑏ⱼ
leaves Bob somewhere.

On Wednesday, November 30, 2022 at 12:56:11 AM UTC+1, Jim Burns wrote:
> ⋃𝓑𝓸𝓫 does not leave Bob anywhere..
: > There is nowhere that all the exchanges of
> ⋃𝓑𝓸𝓫 leave Bob.
:

It's getting worse with you, man.

On 11/29/2022 1:49 PM, WM wrote:
> Chris M. Thomasson schrieb am Dienstag, 29. November 2022 um 21:07:26 UTC+1:
>
>> Cantor Pairing does cover all pairs. This includes all positive rationals.
>
> In every step where that can be checked we see |ℕ|*(|ℕ|-1) positive rationals are not paired. Not even a single one O is deleted.
>
> If all are paired in the limit, this is not what is needed for a bijection, namely in Cantor's own words: "every number p/q comes at an absolutely fixed position of a simple infinite sequence" and "The infinite sequence thus defined has the peculiar property to contain the positive rational numbers completely, and each of them only once at a determined place."

Sigh. Cantor Pairing cover all pairs, period. Why are you having
difficultly understanding that? God.

On 11/29/2022 3:49 PM, WM wrote:
> Chris M. Thomasson schrieb am Dienstag, 29. November 2022 um 21:07:26 UTC+1:
>
>> Cantor Pairing does cover all pairs. This includes all positive rationals.
>
> In every step where that can be checked we see |ℕ|*(|ℕ|-1) positive rationals are not paired.

red herring. Cantor Pairing does cover all pairs.

>
> If all are paired in the limit, this is not what is needed for a bijection, namely in Cantor's own words: "every number p/q comes at an absolutely fixed position of a simple infinite sequence" and "The infinite sequence thus defined has the peculiar property to contain the positive rational numbers completely, and each of them only once at a determined place."

you are wrong.

Cantor is right, and verifies in his own words, provided by you above, that a bijection exists, a paring exists.

seems you are too dumb to understand Cantor's statements.

>
> Regards, WM

On 11/29/2022 7:08 PM, Fritz Feldhase wrote:
> On Wednesday, November 30, 2022 at 12:56:11 AM UTC+1, Jim Burns wrote:
>
>> ⋃𝓑𝓸𝓫 does not leave Bob anywhere.
> :
>> There is nowhere that all the exchanges of
>> ⋃𝓑𝓸𝓫 leave Bob.
> :
>
> It's getting worse with you, man.

Bob is one of those minions in that cartoon movie

On 11/16/2022 10:05 AM, WM wrote:
> zelos...@gmail.com schrieb am Mittwoch, 16. November 2022 um 12:52:36 UTC+1:
>> onsdag 16 november 2022 kl. 10:55:48 UTC+1 skrev WM:
>>> zelos...@gmail.com schrieb am Mittwoch, 16. November 2022 um 07:19:22 UTC+1:
>>>> tisdag 15 november 2022 kl. 18:28:09 UTC+1 skrev WM:
>>>
>>>>> "The infinite sequence thus defined has the peculiar property to contain all positive
>>>>> rational numbers and each of them only once at a determined place."
>>>>> Therefore every place can be checked step by step because it is a finite place.
>>>> You don't need to "check" it
>>>
>>> But I can do it, if it is possibke, that means if a bijection exists as Cantor describes it: "The infinite sequence thus defined has the peculiar property to contain all positive rational numbers and each of them only once at a determined place." [G. Cantor, letter to R. Lipschitz (19 Nov 1883)]
>>>
>>>> We just check that for any rational number there is a natural number assigned to it, tada, ALL has been "checked" at once
>>>
>>> As this does not explain the loss of the O's it is tad, as you say.
>
>> You can find it in this instance but that is not a requirement.
>
> Where does the first O leave the matrix? Since many are to follow, this must happen at a finite step which can be checked.

Cantor pairing for you, in the bass guitar. Now, it only pairs as much
as it can in the finite time. Trust me, is can go on forever... ;^)

https://youtu.be/XkwgJt5bxKI

On 11/29/2022 6:15 PM, Sergi o wrote:
> On 11/29/2022 3:49 PM, WM wrote:
>> Chris M. Thomasson schrieb am Dienstag, 29. November 2022 um 21:07:26
>> UTC+1:
>>
>>> Cantor Pairing does cover all pairs. This includes all positive
>>> rationals.
>>
>> In every step where that can be checked we see |ℕ|*(|ℕ|-1) positive
>> rationals are not paired.
>
> red herring.  Cantor Pairing does cover all pairs.
>
>>
>> If all are paired in the limit, this is not what is needed for a
>> bijection, namely in Cantor's own words: "every number p/q comes at an
>> absolutely fixed position of a simple infinite sequence" and "The
>> infinite sequence thus defined has the peculiar property to contain
>> the positive rational numbers completely, and each of them only once
>> at a determined place."
>
> you are wrong.
>
> Cantor is right, and verifies in his own words, provided by you above,
> that a bijection exists, a paring exists.
>
> seems you are too dumb to understand Cantor's statements.

Seems so. Or, God willing... WM is a bot and not a "teacher" at all.

Sergi o schrieb am Mittwoch, 30. November 2022 um 03:15:56 UTC+1:

> Cantor is right, and verifies in his own words, provided by you above, that a bijection exists, a paring exists.

Cantor is wrong, and claims in his own words, provided by me above, that a bijection exists, a paring exists.

Regards, WM

Jim Burns schrieb am Mittwoch, 30. November 2022 um 00:56:11 UTC+1:
> On 11/29/2022 2:11 PM, WM wrote:

> Each terminating Bob-mover 𝐵𝑜𝑏ⱼ leaves Bob
> somewhere.

There are no other definable Bob-movers. Every definable natural number is the end of a FISON.
>
> The exchanges of 𝐵𝑜𝑏ⱼ leave Bob where
> 𝐵𝑜𝑏ⱼ's last exchange leaves Bob: kⱼ
>
> ⋃𝓑𝓸𝓫 does not leave Bob anywhere..

Then it contains more than all definable Bob-movers. Otherwise there was a contradiction. The quantifier "forall" means for all and does not tolerate an exception.

> > No, I do not compare.
> > Lossless exchanges are lossless
> > in every environment.
> Consider ⋃𝓑𝓸𝓫
>
The quantifier "forall" means for all and does not tolerate an exception. All definable Bob-movers leave Bob in the matrix.

A bijection à la Cantor requires that every pair can be found, when looking for it. But we know that |ℕ|*(|ℕ|-1) O's remain in the matrix over all steps that can be found. The indices of these fractions cannot be found. The belief in Cantor's claims requires to believe that all O's will leave the matrix in individually verifiable terms of the sequence but that this is not individually verfiable for any term.

Regards, WM

On 11/29/2022 8:08 PM, Fritz Feldhase wrote:
> On Wednesday, November 30, 2022
> at 12:56:11 AM UTC+1, Jim Burns wrote:

>> ⋃𝓑𝓸𝓫 does not leave Bob anywhere.
> :
>> There is nowhere that all the exchanges of
>> ⋃𝓑𝓸𝓫 leave Bob.
> :
>
> It's getting worse with you, man.

Bob is the O born at 1/2 in the matrix.

Kevin is the X born at 2/1 in the matrix.
Stuart is the X born at 3/1 in the matrix.
Dave is the X born at 6/1 in the matrix.

There is a sequence of exchanges.
Upon each exchange, Bob, Kevin, Stuart,
and Dave are each somewhere in the matrix.
∀k, ∃!x : 𝐵𝑜𝑏ₖ = x
∀k, ∃!x : 𝐾𝑒𝑣𝑖𝑛ₖ = x
∀k, ∃!x : 𝑆𝑡𝑢𝑎𝑟𝑡ₖ = x
∀k, ∃!x : 𝐷𝑎𝑣𝑒ₖ = x

For each of them, there is a sequence of
matrix-places they are at.
For Bob
1/2 2/1 3/1 3/1 3/1 6/1 6/1 6/1 6/1 ...
For Kevin
2/1 1/2 1/2 1/2 1/2 1/2 1/2 1/2 1/2 ...
For Stuart
3/1 3/1 2/1 2/1 2/1 2/1 2/1 2/1 2/1 ...
For Dave
6/1 6/1 6/1 6/1 6/1 3/1 3/1 3/1 3/1 ...

For Dave,
there is a unique matrix-place such that,
after each exchange,
Dave is at that place --
*BUT not always _immediately_ after*

After 1 (upon 6), Dave is at 3/1
After 2 (upon 6), Dave is at 3/1
After 3 (upon 6), Dave is at 3/1
After 4 (upon 6), Dave is at 3/1
After 5 (upon 6), Dave is at 3/1
After 6 (upon 7), Dave is at 3/1
After 7 (upon 8), Dave is at 3/1
After 8 (upon 9), Dave is at 3/1
....

∃!x, ∀j, ∃k > j : 𝐷𝑎𝑣𝑒ₖ = x

For Dave, that unique place is 3/1
∀j, ∃k > j : 𝐷𝑎𝑣𝑒ₖ = 3/1
∀x: (∀j, ∃k > j : 𝐷𝑎𝑣𝑒ₖ = x) ⟹ x = 3/1

For Stuart, that unique place is 2/1
∀j, ∃k > j : 𝑆𝑡𝑢𝑎𝑟𝑡ₖ = 2/1
∀x: (∀j, ∃k > j : 𝑆𝑡𝑢𝑎𝑟𝑡ₖ = x) ⟹ x = 2/1

For Kevin, that unique place is 1/2
∀j, ∃k > j : 𝐾𝑒𝑣𝑖𝑛ₖ = 1/2
∀x: (∀j, ∃k > j : 𝐾𝑒𝑣𝑖𝑛ₖ = x) ⟹ x = 1/2

For Bob, _there is no such place_
~∃x, ∀j, ∃k > j : 𝐵𝑜𝑏ₖ = x

∀x, ∃j, ∀k > j : 𝐵𝑜𝑏ₖ ≠ x

All the exchanges leave Dave, Stuart,
and Kevin somewhere in the matrix.

However,
even though _each_ exchange leaves Bob
somewhere in the matrix,
_all_ the exchanges do not leave Bob
anywhere in the matrix.

Nor does any exchange remove Bob.

Thus,
the set of all the exchanges do not have
Bob-conservation.

onsdag 30 november 2022 kl. 12:19:31 UTC+1 skrev WM:
> Sergi o schrieb am Mittwoch, 30. November 2022 um 03:15:56 UTC+1:
>
> > Cantor is right, and verifies in his own words, provided by you above, that a bijection exists, a paring exists.
> Cantor is wrong, and claims in his own words, provided by me above, that a bijection exists, a paring exists.
>
> Regards, WM
The function fucking exists! Will you stop being a retard already!?

On 11/30/2022 6:27 AM, WM wrote:
> Jim Burns schrieb am Mittwoch,
> 30. November 2022 um 00:56:11 UTC+1:

>> Each terminating Bob-mover 𝐵𝑜𝑏ⱼ leaves Bob
>> somewhere.
>
> There are no other definable Bob-movers.
> Every definable natural number is
> the end of a FISON.

A terminating Bob-mover is a sequence
of exchanges.

It terminates.
There is a last exchange in it,
and Bob is moved by the Bob-mover
where its last exchange moves it.

>> The exchanges of 𝐵𝑜𝑏ⱼ leave Bob where
>> 𝐵𝑜𝑏ⱼ's last exchange leaves Bob: kⱼ
>>
>> ⋃𝓑𝓸𝓫 does not leave Bob anywhere.
>
> Then it contains more than all definable
> Bob-movers.

⋃𝓑𝓸𝓫 doesn't contain anything which is not in
at least one terminating Bob-mover.

However,
⋃𝓑𝓸𝓫 contains everything which IS
in at least one terminating Bob-mover.

For each matrix place to which a terminating
Bob-mover movers Bob,
there is another terminating Bob-mover
which moves Bob there and also moves him
away from there.

⋃𝓑𝓸𝓫 has _all_ terminating Bob-movers as subsets.
including both the mover-to that place and the
mover-away. Any place ⋃𝓑𝓸𝓫 moves Bob to
⋃𝓑𝓸𝓫 also moves Bob away from there.

⋃𝓑𝓸𝓫 does not leave Bob anywhere.

> Otherwise there was a contradiction.

You keep using that word.
I do not think it means
what you think it means.

> The quantifier "forall" means for all
> and does not tolerate an exception.

The quantifier "for all" means all of
_whatever we're talking about_
Other things cannot be exceptions.

A square is not an exception to
"all right triangles".
Etc.

>>> No, I do not compare.
>>> Lossless exchanges are lossless
>>> in every environment.
>>
>> Consider ⋃𝓑𝓸𝓫
>
> The quantifier "forall" means for all
> and does not tolerate an exception.
> All definable Bob-movers leave Bob in the matrix.

All terminating Bob-movers leave Bob in the
matrix.

⋃𝓑𝓸𝓫 is not a terminating Bob-mover.
⋃𝓑𝓸𝓫 does not leave Bob in the matrix.

Each exchange in ⋃𝓑𝓸𝓫 is also in a terminating
Bob-mover, which leaves Bob in the matrix.

> A bijection à la Cantor

If Cantor said what you (WM) think Cantor said,
then Cantor would be wrong.

On 11/30/2022 5:19 AM, WM wrote:
> Sergi o schrieb am Mittwoch, 30. November 2022 um 03:15:56 UTC+1:
>
>> Cantor is right, and verifies in his own words, provided by you above, that a bijection exists, a paring exists.
>
> Cantor is wrong, and claims in his own words, provided by me above, that a bijection exists, a paring exists.
>
> Regards, WM

no, you are wrong.

You have failed at every attempt, and the reason for your failure is that you do not know math.

On Wednesday, November 30, 2022 at 4:03:12 PM UTC+1, Sergi o wrote:

> You have failed at every attempt, and the reason for your failure is that you do not know math.

YES. And and now for something completely different?

Jim Burns schrieb am Mittwoch, 30. November 2022 um 12:55:06 UTC+1:
> On 11/29/2022 8:08 PM, Fritz Feldhase wrote:

> > It's getting worse with you, man.
> Bob is the O born at 1/2 in the matrix.

You can explain this in great detail in an exemplary way again and again, but Franz Fritsche does not wish to understand it, because then he would have to recognize that hitherto his attempts were in vain.
>
> However,
> even though _each_ exchange leaves Bob
> somewhere in the matrix,
> _all_ the exchanges do not leave Bob
> anywhere in the matrix.

If each exchange maintains Bob, then there is none where he could leave. If he leaves nevertheless, then it is in the darkness, i.e., such that his first absence cannot be detected.
>
> Nor does any exchange remove Bob.
>
> Thus,
> the set of all the exchanges do not have
> Bob-conservation.

What do you conclude from this? Can you find the step where Bob is missing for the first time as would be necessary when his fraction is indexed without removing this index from another indexed fraction?

Regards, WM

Jim Burns schrieb am Mittwoch, 30. November 2022 um 14:21:20 UTC+1:
> On 11/30/2022 6:27 AM, WM wrote:

> > All definable Bob-movers leave Bob in the matrix.
> All terminating Bob-movers leave Bob in the
> matrix.

There are no other definable Bob-movers. But there are undefinable Bob-movers, so-called dark Bob-movers.

> > A bijection à la Cantor
> If Cantor said what you (WM) think Cantor said,
> then Cantor would be wrong.

I told you what Cantor said. "every number p/q comes at an absolutely fixed position of a simple infinite sequence." Of course he was wrong. Dark Bob-movers cannot produce a mapping.

By the way, your argument fits Cantor's diagonal argument very nicely. Every line with finite line number does not contain the diagonal number. But all lines together contain it.

Regards, WM

On 11/21/2022 10:00 PM, Chris M. Thomasson wrote:
> On 11/21/2022 9:12 PM, FromTheRafters wrote:
>> Sergi o formulated on Friday :
>>> On 11/18/2022 3:46 PM, Chris M. Thomasson wrote:
>>>> On 11/18/2022 1:42 PM, Sergi o wrote:
>>>>> On 11/18/2022 3:33 PM, WM wrote:
>>>>>> Fritz Feldhase schrieb am Freitag, 18. November 2022 um 21:30:45
>>>>>> UTC+1:
>>>>>>> On Friday, November 18, 2022 at 7:36:20 PM UTC+1, Sergi o wrote:
>>>>>>
>>>>>>>> but you only have crap [...]
>>>>>>>
>>>>>>> Yeah, it's crap!
>>>>>>
>>>>>> Says Franz Fritsche who believes that lossless exchange can suffer
>>>>>> from infinitely many losses.
>>>>>>
>>>>>> Regards, WM
>>>>>
>>>>>
>>>>> your "lossless" exchanges creates an infinite amount of litter;
>>>>>
>>>>> 1. each time a O stickie is put onto a fraction covering it, then
>>>>> pulled off later and discarded,
>>>>> 2. each time an X pastie is put onto a fraction hiding it, or on
>>>>> top of an O stickie and then discarded,
>>>>>
>>>>> do you pick up after yourself, or just let it rot in the
>>>>> environment, like your O's ?
>>>>
>>>> I wonder if WM is familiar with lossless compression...
>>>
>>>
>>> FTX said they were a lossless exchange, too.
>>>
>>> lossless compression (the still FAT boys)=> PNG, GIF, ZIP, RAW, BMP,
>>> FLAC, ALAC, WMA
>>
>> There is lossless JPEG too, but is RAW or BMP ever even compressed?

Depends on the data stored within.

>
> I don't think so. A BMP can store lossless compressed data. Also, a PNG
> is nice because it is lossless.