Ben Bacarisse schrieb am Freitag, 9. September 2022 um 02:09:30 UTC+2:
> WM <askas...@gmail.com> writes:

> > 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.

But there is a potentially infinite collection of visible numbers. IF WE ASSUME an actually infinite set of numbers, then most of them must be dark.

Regards, WM

Ben Bacarisse schrieb am Freitag, 9. September 2022 um 02:13:14 UTC+2:
> WM <mont...@t-online.de> writes:

> >> 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.

I said: Potential infinity remains the same notwithstanding whether there is a dark complement or nothing. The collection of FISONs can increase and decrease.

> 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.

This obviously undesirable result is caused by potential infinity alone, whether or not dark numbers are existing.
>
> 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.

I said and you must have read it: Potential infinity remains the same notwithstanding whether there is a dark complement or nothing. The collection of FISONs can increase and decrease.

Regards, WM

Ben Bacarisse schrieb am Freitag, 9. September 2022 um 02:31:23 UTC+2:
> WM <mont...@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?

It is what we have been discussing in extensio. It is nonsense to claim the indexing of all fractions if most will be remaining not indexed forever.

> 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.

I prefer to stay with endsegments. There the direct drop from infinity to zero is impossible by the very definition:

∀k ∈ ℕ : ∩{E(1), E(2), ..., E(k+1)} = ∩{E(1), E(2), ..., E(k)} \ {k}

I assume that
∀k ∈ ℕ: E(k+1) = E(k) \ {k}
holds for all natural numbers.
Maybe you are not agreeing?
>
> 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.

Maybe. But I am interested in indexing (= disappearing O's) during the definable parts 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, ...

> > 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".

That is when all X's have reached their final positions. (There are not "all" and they will never reach, but that's the definition by Cantor: "gegenseitig eindeutige und vollständige Korrespondenz" [Cantor, p. 238]

> 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.

The argument proves dark numbers, not individually definable but only complete in whatever limit. Why should I look at the limit if each one handles infinitely many undefinable elements?

REgards, WM

Ben Bacarisse schrieb am Freitag, 9. September 2022 um 02:51:25 UTC+2:
> WM <mont...@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?

In classical mathematics, there is no alephs. But we can use classical maths to estimate, according to an argument frequently used in number theory, that in every definable step more than 1000 O's remain. For unsure and doubting readers, we could also say more than 10^10^1000 O's are remaining whenever an index settles at its finite position.

> But I used WMaths to show which positions have Os at each step.

You agreed that no O is deleted during the whole process. That's what is important! Do you wish to withdraw it?

Regards, WM

On 9/9/2022 7:28 AM, WM wrote:
> 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,

the intersection of NOT ALL endsegments is infinite.

> but the empty intersection follows

the intersection of ALL endsegments is empty.

> in the limit without intermediate finite intersections,

that is nonsense with out context.

>then
> ∀k ∈ ℕ : ∩{E(1), E(2), ..., E(k+1)} = ∩{E(1), E(2), ..., E(k)} \ {k}
> is violatd.

wrong. THat is only the relationship of adjacent endsegments. Fail.

>>
>> 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.

Wrong, you have yet to prove that. Saying it and proving it are two different things.

>
>> 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.

now you disagree with what you state above.

The limit of the sequence {E(1), E(2), E(3), E(4), ...} is E(oo), which is not the *intersection* of all endsegments, which has been proved empty.

what is E(oo) ? is there a number in there ? call it k, is k in E(oo) ? no, as k is finite. So there are no numbers in E(oo), it is empty.

notice this is a different case than intersection.

> In both cases there are infinitely many dark terms before the limit.

nope, that is dark pixie dust stuck in your eyes.

>
> Regards, WM

On 9/9/2022 8:08 AM, WM wrote:
> Ben Bacarisse schrieb am Freitag, 9. September 2022 um 02:51:25 UTC+2:
>> WM <mont...@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?
>
> In classical mathematics, there is no alephs. But we can use classical maths to estimate, according to an argument frequently used in number theory, that in every definable step more than 1000 O's remain. For unsure and doubting readers, we could also say more than 10^10^1000 O's are remaining whenever an index settles at its finite position.

wrong approach. stating a number to bound the infinite is like nailing pudding on a wall.

Your statement regarding Number Theory is a lie.

>
>> But I used WMaths to show which positions have Os at each step.
>
> You agreed that no O is deleted during the whole process. That's what is important! Do you wish to withdraw it?

that is red herring, you avoided his question.

>
> Regards, WM

On 9/9/2022 7:32 AM, WM wrote:
> 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
>
>

Your argument fails, you have not proved dark numbers anywhere. Your X and Os are obfuscation.

On 9/9/2022 7:37 AM, WM wrote:
> Ben Bacarisse schrieb am Freitag, 9. September 2022 um 02:09:30 UTC+2:
>> WM <askas...@gmail.com> writes:
>
>
>>> 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.
>
> But there is a potentially infinite collection of visible numbers.

>IF WE ASSUME an actually infinite set of numbers, then most of them must be dark.

no, You assume there are dark numbers first, then try to justify them somehow, so far you have failed to do so, but that is not a surprise because there
are no dark numbers at all.

Keep trying though. If you do prove them real, I will sell dark T Shirts with dark numbers printed on them on line.

>
> Regards, WM

On Friday, 9 September 2022 at 09:32:58 UTC-3, WM wrote:
> 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!

Of course you are not refuting it. That would require actual talent on your part and at least two functioning synapses.. But you are disputing it till kingdom come, and you pretend that limits do not matter for infinite processes. That makes you both a fucking imbecile and a fucking liar.

On 9/9/2022 4:14 AM, WM wrote:
> 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.

Yes the set of things-in-FISONs

It is incorrect to say
"A set is its elements".

You might have seen me and others write
phrases like "ℕ the natural numbers".

"ℕ the natural numbers" is technically
incorrect, but we write it anyway with the
expectation that those for whom we intend
our remarks will know that we know it's
incorrect as written and will read it as
a shortened version of
"the set ℕ of the natural numbers".

Technically, it's only
"the set ℕ of the natural numbers"
which is correct.

> FISONs are finite,

Yes.
A FISON ends somewhere.

Not all elements in FISON 𝐹
have successors in FISON 𝐹

> therefore
> things in FISONs cannot be infinite.

No.
The set of things-in-FISONs does not end
anywhere.
(Also, it does not change.)

Most of the elements in FISON 𝐹ₙ
have successors in FISON 𝐹ₙ
but not all elements.

FISON 𝐹ₙ ends somewhere.
For n in 𝐹ₙ, n+1 is not in 𝐹ₙ

However,
n+1 is still a thing-in-a-FISON
although not a thing in 𝐹ₙ

because
n+1 is a thing in 𝐹ₙ⊕⟨n+1⟩
and 𝐹ₙ⊕⟨n+1⟩ is a FISON too.

> FISONs are finite,
> therefore
> things in FISONs cannot be infinite.

The same elements can be in different sets.
Different sets can have different properties,
being finite or being infinite, for example.
Even without any element changing or
any set changing.

A FISON ends somewhere.
Each element in a FISON
is in a finite set.

The set of things-in-FISONs doesn't end
anywhere.
Each element in the set of things-in-FISONs
is in an infinite set.

They are the same things in different sets.

>>> 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 :
∀k ∈ ⋃𝓕 :

> ∩{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.

No.
"next[1]" and "next[2]" have different
meanings.

There is no end segment 𝐸
which ∅ is next[1] after.
~∃𝐸 ∈ 𝓔[⋃𝓕,ω],
~∃𝐵, 𝐸 ⊃ 𝐵 ⊃ ∅

∅ is next[2] after all end segments
~∃𝐵, ∀𝐸 ∈ 𝓔[⋃𝓕,ω], 𝐸 ⊃ 𝐵 ⊃ ∅

𝓔[⋃𝓕,ω] is the set of
all infinite end segments of ⋃𝓕

Gus Gassmann schrieb am Freitag, 9. September 2022 um 17:55:30 UTC+2:
> On Friday, 9 September 2022 at 09:32:58 UTC-3, WM wrote:
> > 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!
> But you are disputing it

No.

> you pretend that limits do not matter for infinite processes.

Limits do not matter for the question whether it is possible to enumerate all fractions by natural numbers, i.e., before any limit.

Regards, WM

On 9/9/2022 9:41 AM, Sergio wrote:
> On 9/9/2022 7:28 AM, WM wrote:

>> When every intersection of endsegments
>> is infinite,
>
> the intersection of NOT ALL endsegments
> is infinite.

A quibble.
It doesn't affect the point you're making,
but
any ⋃𝓕-end-segment collection with no last
end segment is empty. All the even end segments,
for example.

Of course, our attention is on
⋃𝓕-end-segment collections with no gaps,
no two end segments with a missing end segment
between them. Of those collections,
only the collection of ALL has no last
end segment, so, for those, your point is
100% true.

As I said, just a quibble.

On 9/9/2022 12:06 PM, Jim Burns wrote:
> On 9/9/2022 4:14 AM, WM wrote:
>> 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.
>
> Yes the set of things-in-FISONs
>
> It is incorrect to say
> "A set is its elements".

set = {element1, element2,...}

"A set is its elements" is sloppy and wrong math.

>
> You might have seen me and others write
> phrases like "ℕ the natural numbers".
>
> "ℕ the natural numbers" is technically
> incorrect, but we write it anyway with the
> expectation that those for whom we intend
> our remarks will know that we know it's
> incorrect as written and will read it as
> a shortened version of
> "the set ℕ of the natural numbers".
>
> Technically, it's only
> "the set ℕ of the natural numbers"
> which is correct.
>
>> FISONs are finite,
>
> Yes.
> A FISON ends somewhere.
>
> Not all elements in FISON 𝐹
> have successors in FISON 𝐹
>
>> therefore
>> things in FISONs cannot be infinite.
>
> No.
> The set of things-in-FISONs does not end
> anywhere.
> (Also, it does not change.)
>
> Most of the elements in FISON 𝐹ₙ
> have successors in FISON 𝐹ₙ
> but not all elements.
>
> FISON 𝐹ₙ ends somewhere.
> For n in 𝐹ₙ, n+1 is not in 𝐹ₙ
>
> However,
> n+1 is still a thing-in-a-FISON
> although not a thing in 𝐹ₙ
>
> because
> n+1 is a thing in 𝐹ₙ⊕⟨n+1⟩
> and  𝐹ₙ⊕⟨n+1⟩ is a FISON too.
>
>> FISONs are finite,
>> therefore
>> things in FISONs cannot be infinite.
>
> The same elements can be in different sets.
> Different sets can have different properties,
> being finite or being infinite, for example.
> Even without any element changing or
> any set changing.
>
> A FISON ends somewhere.
> Each element in a FISON
> is in a finite set.
>
> The set of things-in-FISONs doesn't end
> anywhere.
> Each element in the set of things-in-FISONs
> is in an infinite set.
>
> They are the same things in different sets.
>
>>>> 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 :
> ∀k ∈ ⋃𝓕 :
>
>> ∩{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.
>
> No.
> "next[1]" and "next[2]" have different
> meanings.
>
> There is no end segment 𝐸
> which ∅ is next[1] after.
> ~∃𝐸 ∈ 𝓔[⋃𝓕,ω],
> ~∃𝐵, 𝐸 ⊃ 𝐵 ⊃ ∅
>
> ∅ is next[2] after all end segments
> ~∃𝐵, ∀𝐸 ∈ 𝓔[⋃𝓕,ω], 𝐸 ⊃ 𝐵 ⊃ ∅
>
> 𝓔[⋃𝓕,ω] is the set of
> all infinite end segments of ⋃𝓕
>

On Friday, 9 September 2022 at 14:33:42 UTC-3, WM wrote:
[...]
> Limits do not matter for the question whether it is possible to enumerate all fractions by natural numbers, i.e., before any limit.

If you insist on a stepwise process, then there is a limit to consider. If you allow for a completed process, then there is no question of how to transform

XOOO...
XOOO...
XOOO...
XOOO...
....

into

XXXX...
XXXX...
XXXX...
XXXX...
....

since both matrices exist simultaneously. (Obviously.) What you want is to have both a stepwise process, and no limit. That is not possible, and I will box it about your ears every single time. I repeat: You are either a fucking idiot, or a fucking two-faced liar.

On Friday, September 9, 2022 at 7:06:37 PM UTC+2, Jim Burns wrote:
>
> It is incorrect to say "A set is its elements".

Since it isn't (and can't be) in ZFC.

Especially {a} =/= a (for any a) in ZFC, but "a set is its elements" would imply {a} = a, I'd say.

Moreover what's with the empty set? It hardly can be "its elements", I'd say (since there are no such entities).

On Friday, September 9, 2022 at 7:06:37 PM UTC+2, Jim Burns wrote:

> You might have seen me and others write phrases like "ℕ the natural numbers".

What a shame. You shouldn't write that since

> "ℕ the natural numbers" is technically incorrect,

Right.

> but we write it anyway with the
> expectation that those for whom we intend
> our remarks will know that we know it's
> incorrect as written and will read it as
> a shortened version of
> "the set ℕ of the natural numbers".

If you say so.

Actually, I like to be able to talk about the /natural numbers/ (i. e. the objects in IN), in contrast to the /set of natural numbers/, IN. :-P

When "talking" to a crank such things may matter.

On 9/9/2022 12:33 PM, WM wrote:
> Gus Gassmann schrieb am Freitag, 9. September 2022 um 17:55:30 UTC+2:
>> On Friday, 9 September 2022 at 09:32:58 UTC-3, WM wrote:
>>> 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!
>> But you are disputing it
>
> No.
>
>> you pretend that limits do not matter for infinite processes.
>
> Limits do not matter QUACK!! for the question whether it is possible to enumerate all fractions by natural numbers,

the answer is YES, it is possible to enumerate all fractions by natural numbers, IN FACT this was proven BY Cantor, and you cannot un-prove it.

>
> Regards, WM

So, where and why did you bury all the Os ?

You have a dark secret, the Graveyard of the Os.

That is what your matrix is, with X crossing out each O for good!

Swaparoofest, OR Kill the Os Fest. You decide.....

Matrix of Headstones, Graveyard of the O's

Jim Burns schrieb am Freitag, 9. September 2022 um 19:06:37 UTC+2:
> On 9/9/2022 4:14 AM, WM wrote:

> A FISON ends somewhere.
> Each element in a FISON
> is in a finite set.
>
> The set of things-in-FISONs doesn't end
> anywhere.

We know that no FISON is actually infinite, i.e. larger than every FISON. More is not required to know that everything in FISONs is finite.

> Each element in the set of things-in-FISONs
> is in an infinite set.

Potentially infinite but not really infinite.

> ∅ is next[2] after all end segments

This is my last answer to you, if your stupidity is not aible to recognize your mistake.

After all infinite endsegments with infinite intersections
∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀
the empty intersection cannot immediately follow because of
∀k ∈ ℕ : ∩{E(1), E(2), ..., E(k+1)} = ∩{E(1), E(2), ..., E(k)} \ {k}.
Only one element can be lost per step. The empty set cannot follow immediately upon infinite sets.

Regards, WM

Gus Gassmann schrieb am Freitag, 9. September 2022 um 20:10:30 UTC+2:
> On Friday, 9 September 2022 at 14:33:42 UTC-3, WM wrote:
> [...]
> > Limits do not matter for the question whether it is possible to enumerate all fractions by natural numbers, i.e., before any limit.
> If you insist on a stepwise process,

I do not insist. Cantorhas claimed that he could enumerate all fractions. The indices are a stepwise well-ordered set 1, 2, 3, ... .

> then there is a limit to consider.

What happens in the limit is not what Cantor has claimed.

"Wenn zwei wohldefinierte Mannigfaltigkeiten M und N sich eindeutig und vollständig, Element für Element, einander zuordnen lassen (was, wenn es auf eine Art möglich ist, immer auch noch auf viele andere Weisen geschehen kann), so möge für das Folgende die Ausdrucksweise gestattet sein, daß diese Mannigfaltigkeiten gleiche Mächtigkeit haben, oder auch, daß sie äquivalent sind." [Cantor, p. 119]

No limit.

"Zwei wohlgeordnete Mengen M und N heissen von gleichem Typus oder auch von gleicher Anzahl, wenn sie sich gegenseitig eindeutig und vollständig unter beidseitiger Wahrung der Rangfolge ihrer Elemente auf einander beziehen, abbilden lassen;" [G. Cantor, letter to R. Lipschitz (19 Nov 1883)]

No limit.

> If you allow for a completed process, then there is no question of how to transform
> XOOO...
> XOOO...
> XOOO...
> XOOO...
> ...
> into
>
> XXXX...
> XXXX...
> XXXX...
> XXXX...
> ...
>
> since both matrices exist simultaneously. (Obviously.) What you want is to have both a stepwise process, and no limit.

That is what Cantor claimed.

> That is not possible,

Of course not. Nevertheless it is what Cantor claimed and what his disciples usually believe. His claim cannot be realized. Dark numbers prevent it! Nice that you have grasped it after all.

Regards, WM

On 9/9/2022 2:57 PM, Fritz Feldhase wrote:
> On Friday, September 9, 2022
> at 7:06:37 PM UTC+2, Jim Burns wrote:

>> It is incorrect to say
>> "A set is its elements".
>
> Since it isn't (and can't be) in ZFC.
>
> Especially {a} =/= a (for any a) in ZFC,
> but "a set is its elements" would imply
> {a} = a, I'd say.
>
>
> Moreover what's with the empty set?
> It hardly can be "its elements", I'd say
> (since there are no such entities).

I don't know if this is current WM-dogma,
but, at one time, WM did not accept
that the empty set ∅ existed.

He used his own ∅-free version of a
well-order in a proof that dark numbers
exist.

In a WM-well-order, each subset of ℕ
contains a first element. Generate a
BS description of a natural number.
No _definable_ natural number is in 𝔹𝕊
Therefore, the first element of 𝔹𝕊 is dark.

( Note: I am not taking credit for this
( brilliance.

----
> Especially {a} =/= a (for any a) in ZFC,

Yes.
ZFC has the Axiom of Foundation.
And other, similar set theories do too.

Even in theories in which q = {q}
exists, it would be incorrect to say
"a set is its elements"
That's generally false, even there,
though, there, there are some exceptions.

What I like about _pluralities_
(which, here, I often call "collections" to
distinguish them from sets)
is that "sets" (but not sets, pluralities)
are in a totally separate domain from
elements. With pluralities,
there is no "a set is its elements".

We get to develop our ideas about
things in things without the
confusion lurking in the same object
sometimes taking the role of container
and sometimes taking the role of
contained.

ZFC and the von Neumann hierarchy of sets
are elegant and powerful, but maybe
pluralities are a better place to start.
Then, with "container" and "contained"
securely in place, we can upgrade to ZFC.

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

> Ben Bacarisse schrieb am Freitag, 9. September 2022 um 02:51:25 UTC+2:
>> But I used WMaths to show which positions have Os at each step.
>
> You agreed that no O is deleted during the whole process. That's what
> is important!

Surprisingly, more than one thing can be important.

> Do you wish to withdraw it?

Since it's obvious and true, that would be absurd. Do you want to go
back to your original question which was about /positions/ or would you
rather switch to the safe but tedious fact about "numbers" of Os? We
could then all agree!

If not, go read my argument about the positions of the Ox. Better yet,
be brave and do the exercises!

--
Ben.

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

> It is what we have been discussing in extensio. It is nonsense to
> claim the indexing of all fractions if most will be remaining not
> indexed forever.

My argument was entirely within WMaths. It used a potentially infinite
bijection between potentially infinite sets to define a potentially
infinite sequence of potentially infinite functions. It then used the
appropriate limit as defined in your book to obtain the result. Since
everything is potential, the only reasonable meaning for "the result" is
the limit produce by the potentially infinite sequence of swaps.

>> 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.
>
> I prefer to stay with endsegments.

Yet you agree that infinite non-zero points "vanish" in the limit,
something that seems to bother you in other cases like:

> There the direct drop from infinity
> to zero is impossible by the very definition:
>
> ∀k ∈ ℕ : ∩{E(1), E(2), ..., E(k+1)} = ∩{E(1), E(2), ..., E(k)} \ {k}
>
> I assume that
> ∀k ∈ ℕ: E(k+1) = E(k) \ {k}
> holds for all natural numbers.

(Aside: This is true in proper maths, but in WMaths you have not yet
been able to define set equality and difference, so I just have to take
your word for it. Any progress on that, by the way?)

> Maybe you are not agreeing?

You've been saying this for years with no one disagreeing. It's classic
crank: say something wrong and then pretend that anyone who disagrees
must not agree with something correct like this.

But, since you prefer to re-phrase my limit example in terms of
end segments, here's what you didn't know how to write:

Given e_n(i) = [i >= n], then lim_{n->oo} e_n = z where z(i) = 0.

([...] is the Iverson bracket.)

>> But you won't look at the argument.
>
> The argument proves dark numbers, not individually definable but only
> complete in whatever limit.

If you were intellectually curious (and courageous), you'd try to
understand my argument -- the one you won't look -- and you'd see that
it is entirely in WMaths and there are no dark numbers in WMaths. You
would still be free to reject it, but you should understand it first.

> Why should I look at the limit if each one
> handles infinitely many undefinable elements?

The limits in WMaths don't do that. That is exactly why you should look
at the limit.

But I am starting to think you simply don't know most of what's in your
book. The last time this can up you flatly denied that you defined
limits for any non-constant sequences of sets, but there are explicit
examples right there in the text! And then you flatly denied that your
book defines lim_{n->oo} e_n.

--
Ben.

Jim Burns <james.g.burns@att.net> writes:

> I don't know if this is current WM-dogma,
> but, at one time, WM did not accept
> that the empty set ∅ existed.

Well, it's right there in his textbook.

--
Ben.

On 9/9/2022 3:16 PM, WM wrote:
> Gus Gassmann schrieb am Freitag, 9. September 2022 um 20:10:30 UTC+2:
>> On Friday, 9 September 2022 at 14:33:42 UTC-3, WM wrote:
>> [...]
>>> Limits do not matter for the question whether it is possible to enumerate all fractions by natural numbers, i.e., before any limit.
>> If you insist on a stepwise process,
>
> I do not insist. Cantorhas claimed that he could enumerate all fractions.

Liar, Cantor has proven he can enumerate all the fractions using math. You cannot un-prove it.

>
>> then there is a limit to consider.
>

>
> That is what Cantor claimed.

nope.

>
>> That is not possible,
>

<snip bullshit>

>
> Regards, WM

On 9/9/2022 1:06 PM, Jim Burns wrote:
> On 9/9/2022 9:41 AM, Sergio wrote:
>> On 9/9/2022 7:28 AM, WM wrote:
>
>>> When every intersection of endsegments
>>> is infinite,
>>
>> the intersection of NOT ALL endsegments
>> is infinite.
>
> A quibble.
> It doesn't affect the point you're making,
> but
> any ⋃𝓕-end-segment collection with no last
> end segment is empty. All the even end segments,
> for example.
>
> Of course, our attention is on
> ⋃𝓕-end-segment collections with no gaps,
> no two end segments with a missing end segment
> between them. Of those collections,
> only the collection of ALL has no last
> end segment, so, for those, your point is
> 100% true.
>
> As I said, just a quibble.
>

ok, how about this: the intersection of any (number of) endsegments (except all endsegments) is the highest number endsegment.

intersection of E(1), E(584), E(1000) is E(1000)