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.

Regards, WM

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.
>
> Regards, WM

It doesn't because you are arguing in a step by step process WHICH IS NOT HOW FUNCTIONS FUCKING WORK!

On 11/25/2022 6:25 AM, WM wrote:
> 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.
>
> Regards, WM

LAMO => you can not fix bad math by using the term "by definition"

On 11/25/2022 4:55 AM, WM wrote:
> Jim Burns schrieb am Donnerstag,
> 24. November 2022 um 17:11:23 UTC+1:

>> It is a _sequence_ of exchanges.

>> It begins at ⟨1/1 1/1⟩

>> For each split of the sequence,
>> there is an exchange ⟨p/q k/1⟩ last before
>> and an exchange ⟨p'/q' k'/1⟩ first after
>> such that
>> k' = k⁺⁺
>> if q ≠ 1
>> then p'/q' = p⁺⁺/q⁻⁻
>> else p'/q' = 1/p⁺⁺

>> All the O's are deleted.
>
> Not in a definable way.

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.

No dark numbers are handled
in any exchange in the sequence.

It is a _sequence_ of exchanges.

Because it is a sequence,
for each split of the sequence,
there is a last-before and a first-after.

Because it is a sequence,
_any_ exchange handling dark numbers
makes necessary the existence of
some _first_ exchange to handle dark numbers,
which, as first, has an immediate predecessor
which does NOT handle dark numbers.

However,
a _first_ exchange handling dark numbers
is contradictory.
We know there is a contradiction from the
sequence-definition of last-before ⟨p/q k/1⟩
and first-after ⟨p/q k/1⟩⁺⁺
| if q ≠ 1
| then ⟨p/q k/1⟩⁺⁺ = ⟨p⁺⁺/q⁻⁻ k⁺⁺/1⟩
| else ⟨p/q k/1⟩⁺⁺ = ⟨1/p⁺⁺ k⁺⁺/1⟩

If ⟨p/q k/1⟩ doesn't handle dark numbers,
then ⟨p/q k/1⟩⁺⁺ doesn't handle dark numbers.
There can't be a first.

No _first_ exchange in the sequence
handling dark numbers
Thus,
no exchange _at all_ in the sequence
handling dark numbers.

For the exchange sequence
⟨1/1 1/1⟩ ⟨1/2 2/1⟩ ⟨2/1 3/1⟩ ...
Stopping the exchanges leaves all O's undeleted.
_All_ the exchanges delete all the O's
No dark numbers are handled in any exchange.

> That is proved by the fact, that never anybody
> will be able to find a furst O deleted.

For each exchange in the sequence,
there are infinitely-many exchanges
between that exchange and all the exchanges.

That is not how
a reallyreallyreallyreallyreallyreally large
number behaves.

Cantor's sequence is an illustration that
infinity is not
a reallyreallyreallyreallyreallyreally large
number.

On 11/23/2022 3:28 PM, Sergi o wrote:
> On 11/23/2022 1:20 PM, Chris M. Thomasson wrote:
>> On 11/23/2022 4:13 AM, FromTheRafters wrote:
>>> After serious thinking Chris M. Thomasson wrote :
>>>> On 11/22/2022 2:26 PM, Sergi o wrote:
>>>>> On 11/22/2022 3:36 PM, Chris M. Thomasson wrote:
>>>>>> On 11/22/2022 5:24 AM, FromTheRafters wrote:
>>>>>>> Chris M. Thomasson used his keyboard to write :
>>>>>>>> On 11/21/2022 9:53 PM, Chris M. Thomasson wrote:
>>>>>>>>> On 11/21/2022 8:47 PM, FromTheRafters wrote:
>>>>>>>>>> WM formulated on Thursday :
>>>>>>>>>>> FromTheRafters schrieb am Donnerstag, 17. November 2022 um
>>>>>>>>>>> 20:46:44 UTC+1:
>>>>>>>>>>>> After serious thinking WM wrote :
>>>>>>>>>>>>> The bijection is a sequence 1/1, 1/2, 2/1, 1/3, 2/2, 3/1,
>>>>>>>>>>>>> ... which
>>>>>>>>>
>>>>>>>>> [...]
>>>>>>>>>
>>>>>>>>> You are missing all a lot of Cantor parings. What about 0/1 and
>>>>>>>>> 1/0? Humm...
>>>>>>>>>
>>>>>>>>
>>>>>>>> Oops! I responded to FromTheRafters instead of WM.
>>>>>>>>
>>>>>>>> 1/1, 1/2, 2/1, 1/3, 2/2, 3/1 misses a lot of Cantor pairings.
>>>>>>>> There is 0/3, 2/0, ect...
>>>>>>>>
>>>>>>>> Why does WM seem to think that a Cantor pair must be a proper
>>>>>>>> fraction? Well, it sure seems that way.
>>>>>>>
>>>>>>> In this specific case, we are dealing with rationals whose
>>>>>>> numerator and denominator are positive, not non-negative.
>>>>>>
>>>>>> Signed zero for fun:
>>>>>>
>>>>>> 1/+0 vs 1/-0
>>>>>>
>>>>>> ;^)
>>>>>
>>>>>
>>>>>
>>>>> like a plot of  1/x^2, or 1/x^(even)
>>>>
>>>> Fwiw, I forgot one of the states of zero.
>>>>
>>>> -0, 0, +0
>>>>
>>>> We can define it as ternary. :^)
>>>
>>> Ones' Complement binary representations have a zero and a negative zero.
>>>
>>> What makes your plus zero 'different' from your zero such that it
>>> would be needed? Two ways to say nothing is somehow not enough? :)
>>
>> :^) Well, humm... perhaps... We can say a step-by-step process whose
>> limit is zero, say:
>>
>> .1, .01, .001, .0001, ...
>>
>> Gets to "positive zero", while:
>>
>> -1, -.01, -.001, -.0001, ...
>>
>> Gets to "negative zero".
>>
>> And zero, is just "zero with no sign"?
>>
>>
>> .1, .01, .001, .0001, ... = +0
>> -1, -.01, -.001, -.0001, ... = -0
>> 0 = 0
>>
>> Make any sense to you at all? Or, is that just stepping deep into the
>> heart of kookville? Humm...
>
>
> makes a difference on the approach, as with 1/x^2.

Exactly. The sign of zero can indicate how it was approached as a limit.

On Friday, November 25, 2022 at 12:07:32 PM UTC-8, Chris M. Thomasson wrote:
> On 11/23/2022 3:28 PM, Sergi o wrote:
> > On 11/23/2022 1:20 PM, Chris M. Thomasson wrote:
> >> On 11/23/2022 4:13 AM, FromTheRafters wrote:
> >>> After serious thinking Chris M. Thomasson wrote :
> >>>> On 11/22/2022 2:26 PM, Sergi o wrote:
> >>>>> On 11/22/2022 3:36 PM, Chris M. Thomasson wrote:
> >>>>>> On 11/22/2022 5:24 AM, FromTheRafters wrote:
> >>>>>>> Chris M. Thomasson used his keyboard to write :
> >>>>>>>> On 11/21/2022 9:53 PM, Chris M. Thomasson wrote:
> >>>>>>>>> On 11/21/2022 8:47 PM, FromTheRafters wrote:
> >>>>>>>>>> WM formulated on Thursday :
> >>>>>>>>>>> FromTheRafters schrieb am Donnerstag, 17. November 2022 um
> >>>>>>>>>>> 20:46:44 UTC+1:
> >>>>>>>>>>>> After serious thinking WM wrote :
> >>>>>>>>>>>>> The bijection is a sequence 1/1, 1/2, 2/1, 1/3, 2/2, 3/1,
> >>>>>>>>>>>>> ... which
> >>>>>>>>>
> >>>>>>>>> [...]
> >>>>>>>>>
> >>>>>>>>> You are missing all a lot of Cantor parings. What about 0/1 and
> >>>>>>>>> 1/0? Humm...
> >>>>>>>>>
> >>>>>>>>
> >>>>>>>> Oops! I responded to FromTheRafters instead of WM.
> >>>>>>>>
> >>>>>>>> 1/1, 1/2, 2/1, 1/3, 2/2, 3/1 misses a lot of Cantor pairings.
> >>>>>>>> There is 0/3, 2/0, ect...
> >>>>>>>>
> >>>>>>>> Why does WM seem to think that a Cantor pair must be a proper
> >>>>>>>> fraction? Well, it sure seems that way.
> >>>>>>>
> >>>>>>> In this specific case, we are dealing with rationals whose
> >>>>>>> numerator and denominator are positive, not non-negative.
> >>>>>>
> >>>>>> Signed zero for fun:
> >>>>>>
> >>>>>> 1/+0 vs 1/-0
> >>>>>>
> >>>>>> ;^)
> >>>>>
> >>>>>
> >>>>>
> >>>>> like a plot of 1/x^2, or 1/x^(even)
> >>>>
> >>>> Fwiw, I forgot one of the states of zero.
> >>>>
> >>>> -0, 0, +0
> >>>>
> >>>> We can define it as ternary. :^)
> >>>
> >>> Ones' Complement binary representations have a zero and a negative zero.
> >>>
> >>> What makes your plus zero 'different' from your zero such that it
> >>> would be needed? Two ways to say nothing is somehow not enough? :)
> >>
> >> :^) Well, humm... perhaps... We can say a step-by-step process whose
> >> limit is zero, say:
> >>
> >> .1, .01, .001, .0001, ...
> >>
> >> Gets to "positive zero", while:
> >>
> >> -1, -.01, -.001, -.0001, ...
> >>
> >> Gets to "negative zero".
> >>
> >> And zero, is just "zero with no sign"?
> >>
> >>
> >> .1, .01, .001, .0001, ... = +0
> >> -1, -.01, -.001, -.0001, ... = -0
> >> 0 = 0
> >>
> >> Make any sense to you at all? Or, is that just stepping deep into the
> >> heart of kookville? Humm...
> >
> >
> > makes a difference on the approach, as with 1/x^2.
>
> Exactly. The sign of zero can indicate how it was approached as a limit.

Zero does not have a nonzero sign. There are no negative quantities
on a quantity line. There is nothing to be taken from zero to create that.
No quantity has no signs.

On 11/25/2022 1:41 PM, mitchr...@gmail.com wrote:
> On Friday, November 25, 2022 at 12:07:32 PM UTC-8, Chris M. Thomasson wrote:
>> On 11/23/2022 3:28 PM, Sergi o wrote:
>>> On 11/23/2022 1:20 PM, Chris M. Thomasson wrote:
>>>> On 11/23/2022 4:13 AM, FromTheRafters wrote:
>>>>> After serious thinking Chris M. Thomasson wrote :
>>>>>> On 11/22/2022 2:26 PM, Sergi o wrote:
>>>>>>> On 11/22/2022 3:36 PM, Chris M. Thomasson wrote:
>>>>>>>> On 11/22/2022 5:24 AM, FromTheRafters wrote:
>>>>>>>>> Chris M. Thomasson used his keyboard to write :
>>>>>>>>>> On 11/21/2022 9:53 PM, Chris M. Thomasson wrote:
>>>>>>>>>>> On 11/21/2022 8:47 PM, FromTheRafters wrote:
>>>>>>>>>>>> WM formulated on Thursday :
>>>>>>>>>>>>> FromTheRafters schrieb am Donnerstag, 17. November 2022 um
>>>>>>>>>>>>> 20:46:44 UTC+1:
>>>>>>>>>>>>>> After serious thinking WM wrote :
>>>>>>>>>>>>>>> The bijection is a sequence 1/1, 1/2, 2/1, 1/3, 2/2, 3/1,
>>>>>>>>>>>>>>> ... which
>>>>>>>>>>>
>>>>>>>>>>> [...]
>>>>>>>>>>>
>>>>>>>>>>> You are missing all a lot of Cantor parings. What about 0/1 and
>>>>>>>>>>> 1/0? Humm...
>>>>>>>>>>>
>>>>>>>>>>
>>>>>>>>>> Oops! I responded to FromTheRafters instead of WM.
>>>>>>>>>>
>>>>>>>>>> 1/1, 1/2, 2/1, 1/3, 2/2, 3/1 misses a lot of Cantor pairings.
>>>>>>>>>> There is 0/3, 2/0, ect...
>>>>>>>>>>
>>>>>>>>>> Why does WM seem to think that a Cantor pair must be a proper
>>>>>>>>>> fraction? Well, it sure seems that way.
>>>>>>>>>
>>>>>>>>> In this specific case, we are dealing with rationals whose
>>>>>>>>> numerator and denominator are positive, not non-negative.
>>>>>>>>
>>>>>>>> Signed zero for fun:
>>>>>>>>
>>>>>>>> 1/+0 vs 1/-0
>>>>>>>>
>>>>>>>> ;^)
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>> like a plot of 1/x^2, or 1/x^(even)
>>>>>>
>>>>>> Fwiw, I forgot one of the states of zero.
>>>>>>
>>>>>> -0, 0, +0
>>>>>>
>>>>>> We can define it as ternary. :^)
>>>>>
>>>>> Ones' Complement binary representations have a zero and a negative zero.
>>>>>
>>>>> What makes your plus zero 'different' from your zero such that it
>>>>> would be needed? Two ways to say nothing is somehow not enough? :)
>>>>
>>>> :^) Well, humm... perhaps... We can say a step-by-step process whose
>>>> limit is zero, say:
>>>>
>>>> .1, .01, .001, .0001, ...
>>>>
>>>> Gets to "positive zero", while:
>>>>
>>>> -1, -.01, -.001, -.0001, ...
>>>>
>>>> Gets to "negative zero".
>>>>
>>>> And zero, is just "zero with no sign"?
>>>>
>>>>
>>>> .1, .01, .001, .0001, ... = +0
>>>> -1, -.01, -.001, -.0001, ... = -0
>>>> 0 = 0
>>>>
>>>> Make any sense to you at all? Or, is that just stepping deep into the
>>>> heart of kookville? Humm...
>>>
>>>
>>> makes a difference on the approach, as with 1/x^2.
>>
>> Exactly. The sign of zero can indicate how it was approached as a limit.
>
> Zero does not have a nonzero sign. There are no negative quantities
> on a quantity line. There is nothing to be taken from zero to create that.
> No quantity has no signs.

I take it that you are not familiar with signed zeros.

On Friday, November 25, 2022 at 1:43:36 PM UTC-8, Chris M. Thomasson wrote:
> On 11/25/2022 1:41 PM, mitchr...@gmail.com wrote:
> > On Friday, November 25, 2022 at 12:07:32 PM UTC-8, Chris M. Thomasson wrote:
> >> On 11/23/2022 3:28 PM, Sergi o wrote:
> >>> On 11/23/2022 1:20 PM, Chris M. Thomasson wrote:
> >>>> On 11/23/2022 4:13 AM, FromTheRafters wrote:
> >>>>> After serious thinking Chris M. Thomasson wrote :
> >>>>>> On 11/22/2022 2:26 PM, Sergi o wrote:
> >>>>>>> On 11/22/2022 3:36 PM, Chris M. Thomasson wrote:
> >>>>>>>> On 11/22/2022 5:24 AM, FromTheRafters wrote:
> >>>>>>>>> Chris M. Thomasson used his keyboard to write :
> >>>>>>>>>> On 11/21/2022 9:53 PM, Chris M. Thomasson wrote:
> >>>>>>>>>>> On 11/21/2022 8:47 PM, FromTheRafters wrote:
> >>>>>>>>>>>> WM formulated on Thursday :
> >>>>>>>>>>>>> FromTheRafters schrieb am Donnerstag, 17. November 2022 um
> >>>>>>>>>>>>> 20:46:44 UTC+1:
> >>>>>>>>>>>>>> After serious thinking WM wrote :
> >>>>>>>>>>>>>>> The bijection is a sequence 1/1, 1/2, 2/1, 1/3, 2/2, 3/1,
> >>>>>>>>>>>>>>> ... which
> >>>>>>>>>>>
> >>>>>>>>>>> [...]
> >>>>>>>>>>>
> >>>>>>>>>>> You are missing all a lot of Cantor parings. What about 0/1 and
> >>>>>>>>>>> 1/0? Humm...
> >>>>>>>>>>>
> >>>>>>>>>>
> >>>>>>>>>> Oops! I responded to FromTheRafters instead of WM.
> >>>>>>>>>>
> >>>>>>>>>> 1/1, 1/2, 2/1, 1/3, 2/2, 3/1 misses a lot of Cantor pairings.
> >>>>>>>>>> There is 0/3, 2/0, ect...
> >>>>>>>>>>
> >>>>>>>>>> Why does WM seem to think that a Cantor pair must be a proper
> >>>>>>>>>> fraction? Well, it sure seems that way.
> >>>>>>>>>
> >>>>>>>>> In this specific case, we are dealing with rationals whose
> >>>>>>>>> numerator and denominator are positive, not non-negative.
> >>>>>>>>
> >>>>>>>> Signed zero for fun:
> >>>>>>>>
> >>>>>>>> 1/+0 vs 1/-0
> >>>>>>>>
> >>>>>>>> ;^)
> >>>>>>>
> >>>>>>>
> >>>>>>>
> >>>>>>> like a plot of 1/x^2, or 1/x^(even)
> >>>>>>
> >>>>>> Fwiw, I forgot one of the states of zero.
> >>>>>>
> >>>>>> -0, 0, +0
> >>>>>>
> >>>>>> We can define it as ternary. :^)
> >>>>>
> >>>>> Ones' Complement binary representations have a zero and a negative zero.
> >>>>>
> >>>>> What makes your plus zero 'different' from your zero such that it
> >>>>> would be needed? Two ways to say nothing is somehow not enough? :)
> >>>>
> >>>> :^) Well, humm... perhaps... We can say a step-by-step process whose
> >>>> limit is zero, say:
> >>>>
> >>>> .1, .01, .001, .0001, ...
> >>>>
> >>>> Gets to "positive zero", while:
> >>>>
> >>>> -1, -.01, -.001, -.0001, ...
> >>>>
> >>>> Gets to "negative zero".
> >>>>
> >>>> And zero, is just "zero with no sign"?
> >>>>
> >>>>
> >>>> .1, .01, .001, .0001, ... = +0
> >>>> -1, -.01, -.001, -.0001, ... = -0
> >>>> 0 = 0
> >>>>
> >>>> Make any sense to you at all? Or, is that just stepping deep into the
> >>>> heart of kookville? Humm...
> >>>
> >>>
> >>> makes a difference on the approach, as with 1/x^2.
> >>
> >> Exactly. The sign of zero can indicate how it was approached as a limit.
> >
> > Zero does not have a nonzero sign. There are no negative quantities
> > on a quantity line. There is nothing to be taken from zero to create that.
> > No quantity has no signs.
> I take it that you are not familiar with signed zeros.

Putting a sign does not change anything about a no quantity.

Gus Gassmann schrieb am Freitag, 25. November 2022 um 18:16:49 UTC+1:
> On Friday, 25 November 2022 at 08:25:39 UTC-4, WM wrote:
> > 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.
> After finitely many steps only finitely many fractions have been "indexed", and no 'O' has been removed from the matrix.

That is the point!!! Definable steps will not succeed.

> That tells you exactly *NOTHING* about the limit of the process.

I am not interested in the limit process because it does not index any fractions in a definable way.

Regards, 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

Gus Gassmann schrieb am Freitag, 25. November 2022 um 18:14:26 UTC+1:
> On Friday, 25 November 2022 at 08:09:58 UTC-4, WM wrote:
> > zelos...@gmail.com schrieb am Freitag, 25. November 2022 um 12:38:39 UTC+1:
> > > fredag 25 november 2022 kl. 10:55:25 UTC+1 skrev WM:
> > > > Jim Burns schrieb am Donnerstag, 24. November 2022 um 17:11:23 UTC+1:
> >
> > > > > For example,
> > > > > by all of this sequence of exchanges:
> > > > > ⟨1/1 1/1⟩
> > > > > ⟨1/2 2/1⟩
> > > > > ⟨2/1 3/1⟩
> > > > > ⟨1/3 4/1⟩
> > > > > ⟨2/2 5/1⟩
> > > > > ⟨3/1 6/1⟩
> > > > > ...
> > > > >
> > > > Nice to see that you at least have understood what others claim to be incomprehensible for average mathematicians.
> > > We can get an injection from Q to N, and from N to Q, this shows the cardinality is once again, THE SAME!
> > Find the first O that is deleted.
> There is no "first O".

Correct. But if all remains O's could be abolished in a definable way, then there wass a first one.

You have no clue about limits. Try to find the first n for which 1/n = 0.. There is none. And yet lim{n --> oo} 1/n = 0. Imagine that!

Of course there is a limit. Between every definable term 1/n and the limit 0 there are almost all terms of the sequence, almost all are dark. But in case of indexing fractions, it is claimed that all fractions can be indexed individually - not only in the limit.

Regards, WM

Jim Burns schrieb am Freitag, 25. November 2022 um 18:57:33 UTC+1:

> 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.

But not in a definable way! All definable steps can be discerned.
>
> By _all_ the exchanges.
> Stopping at any exchange in the sequence
> does not delete any O's.

So it is. All visible natnumbers are ends of FISONs where you can stop. However, my proof does not stop. Potential infinity.
>
>
> No dark numbers are handled
> in any exchange in the sequence.
> It is a _sequence_ of exchanges.
> Because it is a sequence,
> for each split of the sequence,
> there is a last-before and a first-after.

Potential infinity.
>
> Because it is a sequence,
> _any_ exchange handling dark numbers
> makes necessary the existence of
> some _first_ exchange to handle dark numbers,
> which, as first, has an immediate predecessor
> which does NOT handle dark numbers.

Every individually handled number is visible and has almost all numbers as successors.
>
> However,
> a _first_ exchange handling dark numbers
> is contradictory.

Yes. With n also n^n^n is visible. That however doesno hinder that almost all numbers are succeeding n^n^n^n^n^n^n^n^n.

> We know there is a contradiction from the
> sequence-definition of last-before ⟨p/q k/1⟩
> and first-after ⟨p/q k/1⟩⁺⁺

Of course. Therefore we need darkness if we acceept completeness.

> No _first_ exchange in the sequence
> handling dark numbers
> Thus,
> no exchange _at all_ in the sequence
> handling dark numbers.

That is an error.
>
> For the exchange sequence
> ⟨1/1 1/1⟩ ⟨1/2 2/1⟩ ⟨2/1 3/1⟩ ...
> Stopping the exchanges leaves all O's undeleted.
> _All_ the exchanges delete all the O's

But only those exchanges which cannot be stooped at.

> > That is proved by the fact, that never anybody
> > will be able to find a furst O deleted.
> For each exchange in the sequence,
> there are infinitely-many exchanges
> between that exchange and all the exchanges.

Correct. And most of them cannot be stopped at.
>
> That is not how
> a reallyreallyreallyreallyreallyreally large
> number behaves.
>
> Cantor's sequence is an illustration that
> infinity is not
> a reallyreallyreallyreallyreallyreally large
> number.

But he claims that it can be covered completely. And he claims that omega is a number, even an integer.

Regards, WM

On 11/26/2022 4:18 AM, WM wrote:
> Gus Gassmann schrieb am Freitag, 25. November 2022 um 18:16:49 UTC+1:
>> On Friday, 25 November 2022 at 08:25:39 UTC-4, WM wrote:
>>> 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.
>> After finitely many steps only finitely many fractions have been "indexed", and no 'O' has been removed from the matrix.
>
> That is the point!!! Definable steps will not succeed.

so why do you persist in failure ?

>
>> That tells you exactly *NOTHING* about the limit of the process.
>
> I am not interested in the limit process

you have demonstrated you cannot understand a limit process.

>
> Regards, WM

On 11/26/2022 4:20 AM, WM wrote:
> 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 *FINITE* indices are issued step by step.

.....by clerk WM at the front door.

>
>> WHICH IS NOT HOW FUNCTIONS FUCKING WORK!
>
> That means they it work in undefinable *FINITE* ways, i.e., in the darkness.

that is your math, lost in darkness.

>
> Regards, WM

On 11/26/2022 4:23 AM, WM wrote:
> Gus Gassmann schrieb am Freitag, 25. November 2022 um 18:14:26 UTC+1:
>> On Friday, 25 November 2022 at 08:09:58 UTC-4, WM wrote:
>>> zelos...@gmail.com schrieb am Freitag, 25. November 2022 um 12:38:39 UTC+1:
>>>> fredag 25 november 2022 kl. 10:55:25 UTC+1 skrev WM:
>>>>> Jim Burns schrieb am Donnerstag, 24. November 2022 um 17:11:23 UTC+1:
>>>

>>>>>>
>>>>> Nice to see that you at least have understood what others claim to be incomprehensible for average mathematicians.
>>>> We can get an injection from Q to N, and from N to Q, this shows the cardinality is once again, THE SAME!
>> There is no "first O".
>

<snip crap>

> Regards, WM

On 11/26/2022 4:37 AM, WM wrote:
> Jim Burns schrieb am Freitag, 25. November 2022 um 18:57:33 UTC+1:
>
>> 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.
>
> But not in a definable *FINITE* way! All definable *FINITE* steps can be discerned.
>>
>> By _all_ the exchanges.
>> Stopping at any exchange in the sequence
>> does not delete any O's.
>
> So it is. All visible *FINITE* natnumbers are ends of FISONs where you can stop. However, my proof does not stop. Potential *FINITE* infinity.
>>
>>
>> No dark numbers are handled
>> in any exchange in the sequence.
>> It is a _sequence_ of exchanges.
>> Because it is a sequence,
>> for each split of the sequence,
>> there is a last-before and a first-after.
>
> Potential *FINITE* infinity.
>>
>> Because it is a sequence,
>> _any_ exchange handling dark numbers
>> makes necessary the existence of
>> some _first_ exchange to handle dark numbers,
>> which, as first, has an immediate predecessor
>> which does NOT handle dark numbers.
>
> Every individually handled *FINITE* number is visible and has almost all numbers as successors.
>>
>> However,
>> a _first_ exchange handling dark numbers
>> is contradictory.
>
> Yes. With n also n^n^n is *FINITE* visible.

>> We know there is a contradiction from the
>> sequence-definition of last-before ⟨p/q k/1⟩
>> and first-after ⟨p/q k/1⟩⁺⁺
>

>
>> No _first_ exchange in the sequence
>> handling dark numbers
>> Thus,
>> no exchange _at all_ in the sequence
>> handling dark numbers.
>
> That is an error.
>>
>> For the exchange sequence
>> ⟨1/1 1/1⟩ ⟨1/2 2/1⟩ ⟨2/1 3/1⟩ ...
>> Stopping the exchanges leaves all O's undeleted.
>> _All_ the exchanges delete all the O's
>
> But only those *FINITE* exchanges which cannot be stooped at.
>
>>> That is proved by the fact, that never anybody
>>> will be able to find a furst O deleted.
>> For each exchange in the sequence,
>> there are infinitely-many exchanges
>> between that exchange and all the exchanges.
>

>>
>> That is not how
>> a reallyreallyreallyreallyreallyreally large
>> number behaves.
>>
>> Cantor's sequence is an illustration that
>> infinity is not
>> a reallyreallyreallyreallyreallyreally large
>> number.
>

> Regards, WM

On 11/26/2022 5:37 AM, WM wrote:
> Jim Burns schrieb am Freitag,
> 25. November 2022 um 18:57:33 UTC+1:

>> 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.
>
> But not in a definable way!
> 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.

For each split of the sequence of exchanges,
there is a last-before exchange ⟨p/q k/1⟩
and a first-after exchange ⟨p/q k/1⟩⁺⁺
such that
| if q ≠ 1
| then ⟨p/q k/1⟩⁺⁺ = ⟨p⁺⁺/q⁻⁻ k⁺⁺/1⟩
| else ⟨p/q k/1⟩⁺⁺ = ⟨1/p⁺⁺ k⁺⁺/1⟩

The beginning, endlessness, last-before-split,
first-after-split are what makes it a sequence.

Describing how it begins and how it crosses
splits is what makes it _that_ sequence.

_That_ sequence,
not any finite initial segment but
_that_ sequence
matches all the fractions p/q to
all the unit fractions k/1
and
_that sequence_
never handles p q or k _other than_
those in the sequence of naturals.

The sequence of naturals
begins at 1 and has no last natural
and,
for each split of the sequence of naturals,
there is a last-before natural j
and a first-after natural j⁺⁺
such that
| j⁺⁺ ≠ 0
| if j ≠ k then j⁺⁺ ≠ k⁺⁺

The beginning, endlessness, last-before-split,
first-after-split are what makes it a sequence.

It begins and crosses splits differently from
the sequence of exchanges. It is a different
sequence, but a specific different sequence,
of naturals instead of exchanges.

>> We know there is a contradiction from the
>> sequence-definition of last-before ⟨p/q k/1⟩
>> and first-after ⟨p/q k/1⟩⁺⁺
>
> Of course.
> Therefore we need darkness
> if we accept completeness.

No.
We need to accept that
we are talking about what we are talking about,
not talking about more,
not talking about less.

Describe a sequence of exchanges:
Describe what it is to be a sequence.
Describe how it begins.
Describe how it crosses splits.
Then reason from the description.

None of that involves infinitely-many actions
which we would need to perform.

>> No _first_ exchange in the sequence
>> handling dark numbers
>> Thus,
>> no exchange _at all_ in the sequence
>> handling dark numbers.
>
> That is an error.

| Assume it is an error.
| Assume some ⟨pₓ/qₓ kₓ/1⟩ in the sequence
| handles dark numbers.
| Dark ⟨pₓ/qₓ kₓ/1⟩
| Dark θₓ = ⟨pₓ/qₓ kₓ/1⟩
| | It is a sequence of exchanges θ
| There is a split of the sequence:
| { θ' | ~∃θ" ⪳ θ' : Dark θ" } = Before
| { θ' | ∃θ" ⪳ θ' : Dark θ" } = After
| | _After_ contains any exchange which is
| dark or which follows a dark exchange
| in the sequence.
| | _Before_ contains any exchange which is
| NOT dark and which does NOT follow a
| dark exchange in the sequence.
| | It is a sequence of exchanges.
| For that split,
| there is a last-before θ₀ in Before
| and a first-after θ₁ in After
| | First-after θ₁ is dark or follows a dark
| exchange.
| But first-after θ₁ only follows exchanges in
| Before, each of which are not-dark.
| Thus, first-after θ₁ is dark.
| | Last-before θ₀ is not-dark and does not
| follow any dark exchange.
| Thus, last-before θ₀ is not-dark.
| | For that split,
| ~Dark θ₀ and Dark θ₁
| | However,
| if ~Dark θ₀ then ~Dark θ₀⁺⁺
| [1]
| The sequence of exchanges is defined
| so that θ₀⁺⁺ = θ₁
| and ~Dark θ₁
| Contradiction.

Therefore,
it is not an error.
No exchange _at all_ in the sequence
handles dark numbers.

[1]
if ~Dark θ₀ then ~Dark θ₀⁺⁺

Assume ~Dark θ₀
Let θ₀ = ⟨p₀/q₀ k₀/1⟩
~Dark p₀ q₀ k₀
~Dark p₀⁺⁺ q₀⁻⁻ k₀⁺⁺

if q₀ ≠ 1
then
⟨p₀/q₀ k₀/1⟩⁺⁺ = ⟨p₀⁺⁺/q₀⁻⁻ k₀⁺⁺/1⟩
~Dark ⟨p₀⁺⁺/q₀⁻⁻ k₀⁺⁺/1⟩
~Dark θ₀⁺⁺
else
⟨p₀/q₀ k₀/1⟩⁺⁺ = ⟨1/p₀⁺⁺ k₀⁺⁺/1⟩
~Dark ⟨1/p₀⁺⁺ k₀⁺⁺/1⟩
~Dark θ₀⁺⁺

Therefore,
if ~Dark θ₀ then ~Dark θ₀⁺⁺

On 11/26/2022 2:37 AM, WM wrote:
> Jim Burns schrieb am Freitag, 25. November 2022 um 18:57:33 UTC+1:
[...]
> But he claims that it can be covered completely. And he claims that omega is a number, even an integer.

Huh? Infinity is not a finite number. Is that what you meant?

on 11/26/2022, Chris M. Thomasson supposed :
> On 11/26/2022 2:37 AM, WM wrote:
>> Jim Burns schrieb am Freitag, 25. November 2022 um 18:57:33 UTC+1:
> [...]
>> But he claims that it can be covered completely. And he claims that omega
>> is a number, even an integer.
>
> Huh? Infinity is not a finite number. Is that what you meant?

He's confused. Omega is a number, in fact it is an ordinal number. It
doesn't represent, but rather is 'represented by' the ordered set of
all of its predecessors. That set being the ordered set of natural
numbers. It is the first limit ordinal, because it is not itself in the
set of natural numbers.

On Sunday, November 27, 2022 at 9:35:42 AM UTC+1, FromTheRafters wrote:
> on 11/26/2022, Chris M. Thomasson supposed :
> > On 11/26/2022 2:37 AM, WM wrote:
> >> Jim Burns schrieb am Freitag, 25. November 2022 um 18:57:33 UTC+1:
> > [...]
> >> But he claims that it can be covered completely. And he claims that omega
> >> is a number, even an integer.
> >
> > Huh?

Cantor considered omega, omega+1, ... to be "integers" (in a certain sense), at least he sometimes called these numbers "integers".

> He's confused.

Of course, since Mückenheim does not understand that notions may "develop". [...]

> Omega is a number, in fact it is an ordinal number. It
> doesn't represent, but rather is 'represented by' the ordered set of
> all of its predecessors. That set being the ordered set of natural
> numbers. It is the first limit ordinal, because it is not itself in the
> set of natural numbers.

In ZFC (due to von Neumann) omega usually *is* the set of all predecessors, i. e. the set of all finite ordinals, IN.

Hint: Halmos uses the symbol "omega" for denoting the set of natural numbers in his "Naive Set Theory" (which isn's naive at all).

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, not actully infinity.

> because you don't like to look at the limit, which will contradict the last twenty years of your work.

Not at all. The limit contradicts Cantor's claim "such that every element of the set stands at a definite position of this sequence" [E. Zermelo: "Georg Cantor – Gesammelte Abhandlungen mathematischen und philosophischen Inhalts", Springer, Berlin (1932) p. 152] and every definition of a bijection.

As long as we can follow the steps that "every element of the set stands at a definite position of this sequence" the number of O's is constant.

Regards, WM

Gus Gassmann schrieb am Samstag, 26. November 2022 um 15:51:19 UTC+1:
> On Saturday, 26 November 2022 at 06:23:52 UTC-4, WM wrote:

> > Correct. But if all remains O's could be abolished in a definable way, then there was a first one.
> That is not how infinity works. There is no "first one", just as there is no largest natural number.

Every definable step is in the order. If the O's were reduced in a definable way, then there was a first one. "such that every element of the set stands at a definite position of this sequence" [E. Zermelo: "Georg Cantor – Gesammelte Abhandlungen mathematischen und philosophischen Inhalts", Springer, Berlin (1932) p. 152]

> Of course every fraction can be indexed.

Every indexed fraction has all O's succeeding it. Not a single one has been deleted.

Regards, WM

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

No exchange changes the number of O's.

Regards, WM

On 11/27/2022 5:57 AM, 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.

you are not taking a limit with your matrix, you are stopping at k, only a finite case.

>
>> 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 *FINITE* mappings leave the number of O's unchanged. That and only that is the result of my *Spoof*.
>
>> 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, not actully infinity.

so you are addressing only the finite case.

>
>> because you don't like to look at the limit, which will contradict the last twenty years of your work.
>
> Not at all. The limit contradicts Cantor's claim

you fail on 3 points;

1. it is Cantors Proof, not a claim. You cannot un prove it.
1. there is no "limit". It is an infinite set.
2. you address only finite set. it is an infinite set.

> Regards, WM

On 11/27/2022 6:02 AM, WM wrote:
> Gus Gassmann schrieb am Samstag, 26. November 2022 um 15:51:19 UTC+1:
>> On Saturday, 26 November 2022 at 06:23:52 UTC-4, WM wrote:
>
>>> Correct. But if all remains O's could be abolished in a definable way, then there was a first one.
>> That is not how infinity works. There is no "first one", just as there is no largest natural number.
>
> Every definable *FINITE* step is in the order. If the O's were reduced in a definable *FINITE* way, then there was a first one.

you have lost all your Os. they were replaced by Xs remember ? or did you STOP somewhere ?

the first O is on 1/1 or did you forget again ?

"such that every element of the set stands at a definite position of this sequence" [E. Zermelo: "Georg Cantor – Gesammelte Abhandlungen mathematischen
und philosophischen Inhalts", Springer, Berlin (1932) p. 152]

which has nothing at all to do with your math.

>
>> Of course every fraction can be indexed.
>

>
> Regards, WM
>