Ben Bacarisse schrieb am Donnerstag, 8. September 2022 um 02:25:29 UTC+2:
> WM <askas...@gmail.com> writes:
>
> > Ben Bacarisse schrieb am Montag, 5. September 2022 um 17:48:48 UTC+2:
> >> WM <askas...@gmail.com> writes:
> >>
> >> > Fritz Feldhase schrieb am Montag, 5. September 2022 um 10:34:43 UTC+2:
> >> >> On Sunday, September 4, 2022 at 11:25:27 PM UTC+2, Ben Bacarisse wrote:
> >> >>
> >> >> > in WMaths (specifically for some potentially infinite sets)
> >> >
> >> > only for such!
> >> >
> >> >> it is possible to have both e ∈ S and S \ {e} = S.
> >> >
> >> > Dark elements can become visible and vice versa.
> >> I've not been following all the interminable threads, but I thought
> >> WMaths did not have dark numbers.
> >
> > It doesn't.
>
> So you justified a surprising bit of WMaths with remark about something
> not in WMaths. Are you not able to keep up?

No. Potential infinity remains the same notwithstanding whether there is a dark complement or nothing. The collection of FISONs can increase and decrease.

Regards, WM

FromTheRafters schrieb am Donnerstag, 8. September 2022 um 04:12:33 UTC+2:
> Ben Bacarisse expressed precisely :

> >> I look at it this way, Cantor proved the rationals have a one to one
> >> mapping with the naturals, it is extremely simple proof, and WM cannot
> >> un-prove it (if he could do proofs).
> >
> > Yes. And he's agreed (for the first time I can remember) that k is a
> > bijection between NxN and N.
> He must have been tired.

No, I had in mind my students and the potentially infinite collections I taught them. There is a bijection possible.

Regards, WM

zelos...@gmail.com schrieb am Donnerstag, 8. September 2022 um 07:56:09 UTC+2:
> onsdag 7 september 2022 kl. 15:46:57 UTC+2 skrev WM:

> > The reason is that shuffling the X will never remove an O.

> It is not a step by step process

Here you see the first steps: 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3, 4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ... Where would the steps cease?

Regrads, WM

On 9/8/2022 3:30 PM, WM wrote:
> Ben Bacarisse schrieb am Donnerstag, 8. September 2022 um 02:25:29 UTC+2:
>> WM <askas...@gmail.com> writes:
>>
>>> Ben Bacarisse schrieb am Montag, 5. September 2022 um 17:48:48 UTC+2:
>>>> WM <askas...@gmail.com> writes:
>>>>
>>>>> Fritz Feldhase schrieb am Montag, 5. September 2022 um 10:34:43 UTC+2:
>>>>>> On Sunday, September 4, 2022 at 11:25:27 PM UTC+2, Ben Bacarisse wrote:
>>>>>>
>>>>>>> in WMaths (specifically for some potentially infinite sets)
>>>>>
>>>>> only for such!
>>>>>
>>>>>> it is possible to have both e ∈ S and S \ {e} = S.
>>>>>
>>>>> Dark elements can become visible and vice versa.
>>>> I've not been following all the interminable threads, but I thought
>>>> WMaths did not have dark numbers.
>>>
>>> It doesn't.
>>
>> So you justified a surprising bit of WMaths with remark about something
>> not in WMaths. Are you not able to keep up?
>
> No. Potential infinity remains the same notwithstanding whether there is a dark complement or nothing. The collection of FISONs can increase and decrease.
>

Come on Man (thanks Joe), spice it up! this pseudo-math is a flop!

FISONs are like sets of counted sheeps, you can increase or decrease the number of sets of sheeps, or FISONs, using the Increase/Decrease Dial.

sheeps sets will be realocated as required to respective intersection fields, then recounted, old numbers removed, new numbers painted on.

Potential infinity is packed chock-full of billions of potential sheeps, that just wanting to get out.

All dark sheeps are repainted to black sheeps IAW specifications, no numbers, and sold to collections, such as sci.math, and they still refuse to be
counted.

> Regards, WM

On 9/8/2022 2:57 PM, WM wrote:
> Jim Burns schrieb am Donnerstag, 8. September 2022 um 00:40:48 UTC+2:
>> On 9/7/2022 4:48 PM, WM wrote:
>>> Jim Burns schrieb am Mittwoch,
>>> 7. September 2022 um 20:21:02 UTC+2:
>>
>>>> We can say about all FISONs that
>>>> they are collections with counting-orders
>>>> which begin at 0 and end somewhere.
>>>
>>> All FISONs are followed by infinite
>>> endsegments.
>> Yes.
>> That arises from FISONs being FISONs.
>
> There is an infinite set following upon all FISONs.
> To deny it by the analogy that
> ∀k ∈ ℕ ∃m ∈ ℕ: k < m ==> ∃m ∈ ℕ ∀k ∈ ℕ: k < m
> is an invalid conclusion, is an invalid conclusion and shows naivety.
>
> To claim that all FISONs are followed by infinite endsegments, but that no infinite set follows upon all FISONs is the summit of stupidity.

all FISONS is what ? ℕ ? do you mean the last FISON ?

what endsegment follows ℕ ? The Last Endsegment ?

>
> Regards, WM

still lost

On 9/7/2022 3:48 PM, WM wrote:
> Jim Burns schrieb am Mittwoch, 7. September 2022 um 20:21:02 UTC+2:
>
>> We can say about all FISONs that
>> they are collections with counting-orders
>> which begin at 0 and end somewhere.
>
> All FISONs are followed by infinite endsegments. Therefore infinitely many natural numbers are not definable and cannot be used for indexing

no. you have to show they cannot be defined, and you have to show they cannot be used for indexing.

your statement is false.

>
> Regards, WM

After serious thinking WM wrote :
> FromTheRafters schrieb am Donnerstag, 8. September 2022 um 04:12:33 UTC+2:
>> Ben Bacarisse expressed precisely :
>
>>>> I look at it this way, Cantor proved the rationals have a one to one
>>>> mapping with the naturals, it is extremely simple proof, and WM cannot
>>>> un-prove it (if he could do proofs).
>>>
>>> Yes. And he's agreed (for the first time I can remember) that k is a
>>> bijection between NxN and N.
>> He must have been tired.
>
> No, I had in mind my students and the potentially infinite collections I
> taught them. There is a bijection possible.

Ah, the potential bijection.

On Thursday, September 8, 2022 at 10:22:15 PM UTC+2, WM wrote:
> Ben Bacarisse schrieb am Donnerstag, 8. September 2022 um 02:17:23 UTC+2:
> > WM <askas...@gmail.com> writes:
>
> > We /could/ try to count number of Xs and Os but your textbook does not
> > even say how that can be done
> This logic is too simple to be covered in a book for universities. It simply states that exchanging X and O will never change the number of X's and O's. Therefore these numbers need not be counted.
> > and it would be a change of topic since
> > you started by referring to what /positions/ have Xs and Os,
> I used Cantor's prescription. The clou of the story is clearly the impossibility to change the nunbers of X's and O's.
> > and your
> > textbook defines a limit that /can/ answer this question: in the limit,
> > no /position/ has an X (given the right pattern of swaps).
> There are two mistake commited by you:
> First limits are approached, enumerations must be realized up to the last element.
> But more important is that you by dreaming of limits agree that it is not possible to index all fractions in a definable way, namely as members of the sequence 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3, 4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ... . That however is Cantor's claim:
>
> "Every well-defined set has a definite cardinality; two sets are ascribed the same cardinality if they mutually uniquely, element by element, can be mapped onto each other."
>
> "The equivalence of sets is the necessary and unmistakable criterion for the equality of their cardinal numbers. [...] If now M ~ N, then this is based on a law of assigning, by which M and N are mutually uniquely related to each other; here let the element m of M be related to the element n of N." [Cantor, p. 283f]
> >
> > You've chosen to try not to understand the argument, but that does not
> > make it any less of an argument.
> I knew your argument in advance because it is (nearly) the only possible escape of matheologians. But it is incompatible with Cantor's claim. (The other argument is that it is forbidden, first to enumerate the integer fractions.)
> >
> > > But according to Cantor's theory this should be done step by step...
> >
> > But I've shown you how to do it without Cantor! It all works just with
> > the mathematics in your textbook
> which is not suitable for proving Cantor's bijections.
> > > Cantor does not make them disappear in the limit. He claims that they
> > > disappear step by step.
> > But you could not give a citation for this.
> I can, and I did in the "pile of junk" which you obviously did not read or not understand:
>
> "Two sets are called 'equivalent' if they mutually uniquely, element by element, can be mapped onto each other."
> "Zwei Mengen werden hierbei 'äquivalent' genannt, wenn sie sich gegenseitig eindeutig, Element für Element, einander zuordnen lassen." [Cantor, pp. 380 & 441]
> >"Step by step" is a metaphor.
> It means that all natural numbers are applied in their natural order. Here you see it: 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3, 4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ... Every fraction appearing here at some step has been indexed in that step.
>
> Regards, WM

What a mess - pudding dripping down the wall and ruining the carpet.

Theorema Egregium of Muckmeatics:
You cannot nail a pudding to the wall.

Proof: Two decades of drippy pseudo-mathematical polemics by
Professor Doktor habil.(equiv.) Mückenheim, resident sage on the
Far Side of the Mountain.

On 9/8/2022 3:22 PM, WM wrote:
> Ben Bacarisse schrieb am Donnerstag, 8. September 2022 um 02:17:23 UTC+2:
>> WM <askas...@gmail.com> writes:
>
>> We /could/ try to count number of Xs and Os but your textbook does not
>> even say how that can be done
>
> This logic is too simple to be covered in a book for universities. It simply states that exchanging X and O will never change the number of X's and O's. Therefore these numbers need not be counted.

you still have not said specifically how your swaparoos take place, take (1,5) is it swapped with (17,8) ? give us your algorithum, else it remains
scattered about in non sense.

just post in text and I will convert it to math for you.

>
>> and it would be a change of topic since
>> you started by referring to what /positions/ have Xs and Os,
>
> I used Cantor's prescription. The clou of the story is clearly the impossibility to change the nunbers of X's and O's.

which means your Xs and Os are a failure.

>
>> and your
>> textbook defines a limit that /can/ answer this question: in the limit,
>> no /position/ has an X (given the right pattern of swaps).
>
> There are two mistake commited by you:
> First limits are approached, enumerations must be realized up to the last element.

He made no mistakes. there are no limits approached, and there are no "last" elements.

> But more important is that you by dreaming of limits agree that it is not possible to index all fractions in a definable way,

definable = wrong.

>namely as members of the sequence 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3, 4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ... . That however is Cantor's claim:
>
> "Every well-defined set has a definite cardinality; two sets are ascribed the same cardinality if they mutually uniquely, element by element, can be mapped onto each other."
>
> "The equivalence of sets is the necessary and unmistakable criterion for the equality of their cardinal numbers. [...] If now M ~ N, then this is based on a law of assigning, by which M and N are mutually uniquely related to each other; here let the element m of M be related to the element n of N." [Cantor, p. 283f]
>>
>> You've chosen to try not to understand the argument, but that does not
>> make it any less of an argument.
>
> I knew your argument in advance because it is (nearly) the only possible escape of matheologians. But it is incompatible with Cantor's claim.

Wrong.

>>> But according to Cantor's theory this should be done step by step...
>>
>> But I've shown you how to do it without Cantor! It all works just with
>> the mathematics in your textbook
>
> which is not suitable for proving Cantor's bijections.

so, is it WM or Cantor ?

>
>>> Cantor does not make them disappear in the limit. He claims that they
>>> disappear step by step.
>> But you could not give a citation for this.
>
> I can, and I did in the "pile of junk" which you obviously did not read or not understand:

it is very confused and not based in Math, but conjecture.
it is a pile of junk as you pull most of it from sci.math postings.

>
> "Two sets are called 'equivalent' if they mutually uniquely, element by element, can be mapped onto each other."
> "Zwei Mengen werden hierbei 'äquivalent' genannt, wenn sie sich gegenseitig eindeutig, Element für Element, einander zuordnen lassen." [Cantor, pp. 380 & 441]
>
>> "Step by step" is a metaphor
>
> It means that all natural numbers are applied in their natural order. Here you see it: 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3, 4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ... Every fraction appearing here at some step has been indexed in that step.

no. that is a sequence.

>
> Regards, WM

On 9/8/2022 4:36 PM, JVR wrote:
> On Thursday, September 8, 2022 at 10:22:15 PM UTC+2, WM wrote:
>> Ben Bacarisse schrieb am Donnerstag, 8. September 2022 um 02:17:23 UTC+2:
>>> WM <askas...@gmail.com> writes:
>>
>>> We /could/ try to count number of Xs and Os but your textbook does not
>>> even say how that can be done
>> This logic is too simple to be covered in a book for universities. It simply states that exchanging X and O will never change the number of X's and O's. Therefore these numbers need not be counted.
>>> and it would be a change of topic since
>>> you started by referring to what /positions/ have Xs and Os,
>> I used Cantor's prescription. The clou of the story is clearly the impossibility to change the nunbers of X's and O's.
>>> and your
>>> textbook defines a limit that /can/ answer this question: in the limit,
>>> no /position/ has an X (given the right pattern of swaps).
>> There are two mistake commited by you:
>> First limits are approached, enumerations must be realized up to the last element.
>> But more important is that you by dreaming of limits agree that it is not possible to index all fractions in a definable way, namely as members of the sequence 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3, 4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ... . That however is Cantor's claim:
>>
>> "Every well-defined set has a definite cardinality; two sets are ascribed the same cardinality if they mutually uniquely, element by element, can be mapped onto each other."
>>
>> "The equivalence of sets is the necessary and unmistakable criterion for the equality of their cardinal numbers. [...] If now M ~ N, then this is based on a law of assigning, by which M and N are mutually uniquely related to each other; here let the element m of M be related to the element n of N." [Cantor, p. 283f]
>>>
>>> You've chosen to try not to understand the argument, but that does not
>>> make it any less of an argument.
>> I knew your argument in advance because it is (nearly) the only possible escape of matheologians. But it is incompatible with Cantor's claim. (The other argument is that it is forbidden, first to enumerate the integer fractions.)
>>>
>>>> But according to Cantor's theory this should be done step by step...
>>>
>>> But I've shown you how to do it without Cantor! It all works just with
>>> the mathematics in your textbook
>> which is not suitable for proving Cantor's bijections.
>>>> Cantor does not make them disappear in the limit. He claims that they
>>>> disappear step by step.
>>> But you could not give a citation for this.
>> I can, and I did in the "pile of junk" which you obviously did not read or not understand:
>>
>> "Two sets are called 'equivalent' if they mutually uniquely, element by element, can be mapped onto each other."
>> "Zwei Mengen werden hierbei 'äquivalent' genannt, wenn sie sich gegenseitig eindeutig, Element für Element, einander zuordnen lassen." [Cantor, pp. 380 & 441]
>>> "Step by step" is a metaphor.
>> It means that all natural numbers are applied in their natural order. Here you see it: 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3, 4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ... Every fraction appearing here at some step has been indexed in that step.
>>
>> Regards, WM
>
> What a mess - pudding dripping down the wall and ruining the carpet.
>
> Theorema Egregium of Muckmeatics:
> You cannot nail a pudding to the wall.
>
> Proof: Two decades of drippy pseudo-mathematical polemics by
> Professor Doktor habil.(equiv.) Mückenheim, resident sage on the
> Far Side of the Mountain.

corolary: freeze it with CO2 first, then nail it, it will melt in an hour

On Thursday, September 8, 2022 at 9:43:52 PM UTC+2, WM wrote:

> Fact, not a conclusion, is that never an O disappears during the definable steps. If all the O's are lost in the limit, then they are not gone in definable steps. That proves dark steps.

That proves that you are a psychotic crank.

On 9/8/2022 3:57 PM, WM wrote:
> Jim Burns schrieb am Donnerstag,
> 8. September 2022 um 00:40:48 UTC+2:
>> On 9/7/2022 4:48 PM, WM wrote:
>>> Jim Burns schrieb am Mittwoch,
>>> 7. September 2022 um 20:21:02 UTC+2:

>>>> We can say about all FISONs that
>>>> they are collections with counting-orders
>>>> which begin at 0 and end somewhere.
>>>
>>> All FISONs are followed by infinite
>>> endsegments.

Yes,
for each FISON 𝐹,
the set 𝐸 of things-in-FISONs-following-𝐹
is infinite.
𝐸 = ⋃𝓕\𝐹

>> Yes.
>> That arises from FISONs being FISONs.
>
> There is an infinite set following upon
> all FISONs.

No.
The empty set ∅ is next[2] after
all 𝐸 = ⋃𝓕\𝐹 and ⋃𝓕

No set 𝐵 exists such that,
for each 𝐸 = ⋃𝓕\𝐹 or ⋃𝓕
𝐸 ⊇ 𝐵 ⊃ ∅

Only elements of some FISON can be in 𝐵
⋃𝓕 ⊇ 𝐵 ⊃ ∅

But,
if i ∈ 𝐵 and i ∈ 𝐹
then i ∉ 𝐸 = ⋃𝓕\𝐹
and 𝐸 ⊉ 𝐵 ⊃ ∅

But, but, if nothing is in 𝐵
𝐵 ⊅ ∅

Therefore, 𝐵 does not exist.

∅ follows upon all FISONs.

FromTheRafters <FTR@nomail.afraid.org> writes:

> After serious thinking WM wrote :
>> FromTheRafters schrieb am Donnerstag, 8. September 2022 um 04:12:33 UTC+2:
>>> Ben Bacarisse expressed precisely :
>>
>>>>> I look at it this way, Cantor proved the rationals have a one to one mapping with the naturals, it is extremely simple proof, and WM cannot
>>>>> un-prove it (if he could do proofs).
>>>> Yes. And he's agreed (for the first time I can remember) that k is a
>>>> bijection between NxN and N.
>>> He must have been tired.
>>
>> No, I had in mind my students and the potentially infinite
>> collections I taught them. There is a bijection possible.
>
> Ah, the potential bijection.

If you look at the definition in the book, it's the same as anyone
else's definition. He can call it a purple bijection if he likes, but
if the same theorems result, it's a bijection.

--
Ben.

WM <askasker48@gmail.com> writes:

> Ben Bacarisse schrieb am Donnerstag, 8. September 2022 um 01:41:11 UTC+2:
>
>> My argument uses k so that the matrix can be represented as a function
>> of one argument. The convergence of sequences of functions of two
>> arguments is much more fiddly and not in WM's textbook so he could claim
>> I am just making stuff up.
>
> No, you may use the limits you like.

Thank you. Interesting that you won't look at the argument though. Too
close to the mark, or maybe you just didn't understand?

> What do you personally conclude concerning the vast amount of O's,
> i.e., not indexed fractions, whoch are only treated "in the limit"?

Unfortunately you gave up after the very first exercise. What positions
have Os (you asked about positions originally) "in the limit" depends on
the pattern of swaps and you ducked out before giving it.

> Indexing all fractions means reaching this aim.

Not in WMaths. That's why I showed you how to get the result using just
WMaths.

>> Some patterns of swaps leave some Os "in the limit" and others
>> don't.
>
> What swaps = exchangesof X and O, delete any O's?

No swaps delete anything. Some sequences of indicator functions have
limit I(i) = 1 (i.e. no Os), others don't.

--
Ben.

WM <askasker48@gmail.com> writes:
(AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
Unendlichen" at Hochschule Augsburg.)

> Ben Bacarisse schrieb am Mittwoch, 7. September 2022 um 21:33:29 UTC+2:
>>
>> Eh? What conclusion is that? The limit of M_n is M(i) = 1. That's
>> just bog standard maths. You don't have to accept that this is "the
>> result of applying an infinite sequence of swaps" but that's not a
>> matter of conclusions. The conclusion is the limit. Interpret it how
>> you like.
>
> Fact, not a conclusion, is that never an O disappears during the
> definable steps. If all the O's are lost in the limit, then they are
> not gone in definable steps. That proves dark steps.

There are no dark anything in WMaths. You refuse to engage with the
WMath argument, either because you fear you won't understand it or
because you fear you will. But either way, ignoring it is indeed the
wisest choice.

--
Ben.

WM <monteuffl@t-online.de> writes:

> Ben Bacarisse schrieb am Donnerstag, 8. September 2022 um 02:25:29 UTC+2:
>> WM <askas...@gmail.com> writes:
>>
>> > Ben Bacarisse schrieb am Montag, 5. September 2022 um 17:48:48 UTC+2:
>> >> WM <askas...@gmail.com> writes:
>> >>
>> >> > Fritz Feldhase schrieb am Montag, 5. September 2022 um 10:34:43 UTC+2:
>> >> >> On Sunday, September 4, 2022 at 11:25:27 PM UTC+2, Ben Bacarisse wrote:
>> >> >>
>> >> >> > in WMaths (specifically for some potentially infinite sets)
>> >> >
>> >> > only for such!
>> >> >
>> >> >> it is possible to have both e ∈ S and S \ {e} = S.
>> >> >
>> >> > Dark elements can become visible and vice versa.
>> >> I've not been following all the interminable threads, but I thought
>> >> WMaths did not have dark numbers.
>> >
>> > It doesn't.
>>
>> So you justified a surprising bit of WMaths with remark about something
>> not in WMaths. Are you not able to keep up?
>
> No.

Yes you did. It's right there in the quoted text. You explain the
silly result that "it is possible to have both e ∈ S and S \ {e} = S" by
saying "Dark elements can become visible and vice versa". But the silly
result is yours from WMaths where there are no dark numbers.

Why would you flat-out lie like this? You made mistake. Whatever
explanation you have for "e ∈ S and S \ {e} = S" in WMaths, it can't
involve dark numbers.

--
Ben.

WM <monteuffl@t-online.de> writes:
(AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
Unendlichen" at Hochschule Augsburg.)

> Ben Bacarisse schrieb am Donnerstag, 8. September 2022 um 02:19:26 UTC+2:
>> WM <askas...@gmail.com> writes:
>
>> > When in every finite step not a single O is deleted. When for every X
>> > there are ℵo O's in every state that can be checked, then this is the
>> > result.
>> For every n in N, there is an infinity or Xs in the matrix, an infinity
>> of Os in the matrix, and an infinity of swaps not yet performed. It's
>> not reasonable to define "the result" when there are an infinity of
>> unperformed swaps remaining.
>
> Right.

So you agree that it is unreasonable to talk about "the result" when
there are an infinity of swaps remaining? You do see that you said
exactly what you now seem to agree is unreasonable? Flip-flop,
flip-flop.

> But this state does never change. Therefore it is not reasonable to
> claim that all fractions could be enumerated. But if this is done
> nevertheless, then we can be sure that never an O will disappear. That
> is a fact that cannot change.

Can you write that in mathematics?

The limit of a sequence does not always share all properties with the
members of the sequence. For example, can you define a sequence of
functions, f_n, from, say [0,1) to R where the integral of the f_n is 1
(for any n) but the (point-wise) limit function is the zero function?
You should be able to.

Where did all the "area" go? What step made it disappear? At every
step the area is the same and there are infinitely many non-zero
function values. All true, but as you should know, the limit can none
the less be f(x) = 0. Magic, eh? No, WMaths.

The only reasonable meaning for "the result" of an endless sequence of
swaps is a limit. And since you ask what /positions/ have Xs and Os,
the limit should be an indicator function.

> It appears in the brain of every sober mathematician that by
> exchanging X and O, even when done infinitely often, no O will
> disappear.

You'd have to say what you mean by "done infinitely often". In WMaths,
the limit of the indicator functions showing where Xs are can be
everywhere 1 (or we can switch to indicating Os if you prefer and get
the zero function). But you won't look at the argument.

--
Ben.

WM <monteuffl@t-online.de> writes:
(AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
Unendlichen" at Hochschule Augsburg.)

> Ben Bacarisse schrieb am Donnerstag, 8. September 2022 um 02:17:23 UTC+2:
>> WM <askas...@gmail.com> writes:
>
>> We /could/ try to count number of Xs and Os but your textbook does not
>> even say how that can be done
>
> This logic is too simple to be covered in a book for universities. It
> simply states that exchanging X and O will never change the number of
> X's and O's. Therefore these numbers need not be counted.

Interesting! So in WMaths the "number of Xs" means something? What is
that number and how is it defined? Are the "number of Xs" and the
"number of Os" equal, and how do you define equality for such numbers?

>> and it would be a change of topic since
>> you started by referring to what /positions/ have Xs and Os,
>
> I used Cantor's prescription. The clou of the story is clearly the
> impossibility to change the nunbers of X's and O's.

But I used WMaths to show which positions have Os at each step. You
didn't understand, but I that's what I did. All without any help from
Cantor.

>> and your
>> textbook defines a limit that /can/ answer this question: in the limit,
>> no /position/ has an X (given the right pattern of swaps).
>
> There are two mistake commited by you:

You could not even give a formula for M_1. I don't think you understood
even the start of the argument, much less the conclusion.

> First limits are approached, enumerations must be realized up to the
> last element.

I used a limit from your textbook, so presumably the limit is being
"approached".

> But more important is that you by dreaming of limits agree that it is
> not possible to index all fractions in a definable way, namely as
> members of the sequence 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2,
> 4/1, 1/5, 2/4, 3/3, 4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ...

I wrote out the formula for a WMaths function showing where the Xs and
Os were, and I showed you how to do that, in WMaths, for every step in
the sequence. You could not follow it, but that's your problem.

The limit is then simply the function sequence limit as defined in your
book.

>> You've chosen to try not to understand the argument, but that does not
>> make it any less of an argument.
>
> I knew your argument in advance because it is (nearly) the only
> possible escape of matheologians.

No, you haven't got a clue. You could not even do exercise 1, and then
you bottled out before exercise 2. If you had understood the argument
you would have pointed out what bit is not right from your textbook!
But maybe you saw it was all there. No need for completed anything.

>> But I've shown you how to do it without Cantor! It all works just with
>> the mathematics in your textbook
>
> which is not suitable for proving Cantor's bijections.

But we don't need Cantor's bijections. We need WMaths bijections. You
really were not following what I said at all, were you? Cantor never
came up -- it's WMaths all the way down.

--
Ben.

torsdag 8 september 2022 kl. 21:36:20 UTC+2 skrev WM:
> Gus Gassmann schrieb am Mittwoch, 7. September 2022 um 16:46:51 UTC+2:
> > On Wednesday, 7 September 2022 at 10:46:57 UTC-3, WM wrote:
>
> > > > > Infinitely many X's are not enough to cover the whole matrix.
> > Of course you can create a matrix with infinitely many rows and columns that contains only Xs.
>
> My statement concerned the matrix
>
> XOOO...
> XOOO...
> XOOO...
> XOOO...
> ...
>
> and all possible configurations of the symbols contained therein.
>
> Regards, WM
Your concern is the matrix and you want ot turn it into not a matrix in finite steps. Brilliant!

Jim Burns schrieb am Freitag, 9. September 2022 um 00:58:30 UTC+2:
> On 9/8/2022 3:57 PM, WM wrote:

> >>> All FISONs are followed by infinite
> >>> endsegments.
> Yes,
> for each FISON 𝐹,
> the set 𝐸 of things-in-FISONs-following-𝐹
> is infinite.

Not things-in-FISONs. FISONs are finite, therefore things in FISONs cannot be infinite.

> > There is an infinite set following upon
> > all FISONs.
> No.
> The empty set ∅ is next[2] after
> all 𝐸 = ⋃𝓕\𝐹 and ⋃𝓕

So you deny
∀k ∈ ℕ : ∩{E(1), E(2), ..., E(k+1)} = ∩{E(1), E(2), ..., E(k)} \ {k}
According to it, the empty set ∅ cannot follow directlyupon an infinite set.

Regards, WM

On Thursday, 8 September 2022 at 16:36:20 UTC-3, WM wrote:
> Gus Gassmann schrieb am Mittwoch, 7. September 2022 um 16:46:51 UTC+2:
> > On Wednesday, 7 September 2022 at 10:46:57 UTC-3, WM wrote:
>
> > > > > Infinitely many X's are not enough to cover the whole matrix.
> > Of course you can create a matrix with infinitely many rows and columns that contains only Xs.
>
> My statement concerned the matrix
>
> XOOO...
> XOOO...
> XOOO...
> XOOO...
> ...
>
> and all possible configurations of the symbols contained therein.

Whatever you call a "configuration"...

I showed you repeatedly how you can swap your Xs and Os to get

XXX...
XXX...
XXX...
....

as a limit. What you do with that information is your business. It does not alter the fact that you are a lying imbecile.

On Friday, 9 September 2022 at 05:14:34 UTC-3, WM wrote:
> So you deny
> ∀k ∈ ℕ : ∩{E(1), E(2), ..., E(k+1)} = ∩{E(1), E(2), ..., E(k)} \ {k}
> According to it, the empty set ∅ cannot follow directlyupon an infinite set.

Whatever "follow directly upon" is supposed to mean...

1/(n+1) = 1/n - 1/(n*n+1). According to you, you psychotic idiot, this means that the non-positive number 0 cannot "follow directly upon" a positive number, either. It doesn't matter, because 0 is the limit of the sequence {1/1, 1/2, 1/3, 1/4, ...}, and the limit of the sequence {E(1), E(2), E(3), E(4), ...} is the empty set.

What all of us here deny is that you have a clue, or ever had a clue. Sorry, but your lack of understanding is painfully obvious.

Ben Bacarisse schrieb am Freitag, 9. September 2022 um 02:06:41 UTC+2:
> WM <askas...@gmail.com> writes:

> > No, you may use the limits you like.
> Thank you. Interesting that you won't look at the argument though.

I told you why: When thefractions are enumerated according to Cantor by exchanging X and O, then no O will disappear. That means that most fractions will not be enumerated. If they, infinitely many, disappear "in the limit" then infinitely many cannot be put in Cantor's sequence: 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3, 4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ... That proves dark nunbers.

> > What do you personally conclude concerning the vast amount of O's,
> > i.e., not indexed fractions, whoch are only treated "in the limit"?
> What positions
> have Os (you asked about positions originally) "in the limit" depends on
> the pattern of swaps

The pattern is given by Cantor.

> >> Some patterns of swaps leave some Os "in the limit" and others
> >> don't.
> >
> > What swaps = exchangesof X and O, delete any O's?
> No swaps delete anything.

So it is. Therefore most fractions are not enumerated and are not members of Cantors above sequence.

> Some sequences of indicator functions have
> limit I(i) = 1 (i.e. no Os), others don't.

Limits are irrelevant when we ask for indexing with natural numbers, i.e. in finite steps before the limit.

Regards, WM

Gus Gassmann schrieb am Freitag, 9. September 2022 um 14:21:19 UTC+2:
> On Friday, 9 September 2022 at 05:14:34 UTC-3, WM wrote:
> > So you deny
> > ∀k ∈ ℕ : ∩{E(1), E(2), ..., E(k+1)} = ∩{E(1), E(2), ..., E(k)} \ {k}
> > According to it, the empty set ∅ cannot follow directly upon an infinite set.
> Whatever "follow directly upon" is supposed to mean...

When every intersection of endsegments is infinite, but the empty intersection follows in the limit without intermediate finite intersections, then
∀k ∈ ℕ : ∩{E(1), E(2), ..., E(k+1)} = ∩{E(1), E(2), ..., E(k)} \ {k}
is violatd.
>
> this means that the non-positive number 0 cannot "follow directly upon" a positive number, either.

It cannot and does not. Between every positive number and 0 there are infinitely many dark numbers.

> It doesn't matter, because 0 is the limit of the sequence {1/1, 1/2, 1/3, 1/4, ...}, and the limit of the sequence {E(1), E(2), E(3), E(4), ...} is the empty set.

Both is correct. In both cases there are infinitely many dark terms before the limit.

Regards, WM

Gus Gassmann schrieb am Freitag, 9. September 2022 um 14:09:41 UTC+2:
> On Thursday, 8 September 2022 at 16:36:20 UTC-3, WM wrote:

> I showed you repeatedly how you can swap your Xs and Os to get
>
> XXX...
> XXX...
> XXX...
> ...
>
> as a limit.

But I am not refuting this or any other limit! I am refuting that the O's can be removed in the course of the definable part of Cantor's sequence: 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 2/4, 3/3, 4/2, 5/1, 1/6, 2/5, 3/4, 4/3, 5/2, 6/1, ... . Enumerating individuals is a step-after-step process where every step an be checked, in principle, by its index. It is clear that by exchanging X and O never an O will disappear. If they all disappear in the limit, then their disappearance cannot be checked individually. Then we have proved dark numbers.

Regards, WM