WM has brought this to us :
> 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.

Endsegments are infinite, no need to add another 'infinite' to the mix
except to confuse. You confuse yourself this way.

FISONs are 'followed' by endsegments. Each natural number is a FISON
and has an associated endsegment. The set of all FISONs is the set of
natural numbers.

> Therefore infinitely many natural numbers are not definable and cannot be
> used for indexing

Non sequitur.

WM expressed precisely :
> Fritz Feldhase schrieb am Mittwoch, 7. September 2022 um 20:52:02 UTC+2:
>> On Wednesday, September 7, 2022 at 7:48:16 PM UTC+2, WM wrote:
>>
>>> Cantor [...] claims that they disappear step by step.
>>
>> Where does he claim that?!
>
> I showed you often enough. You seem to have a short memory.
> But perhaps you may understand the simple argument: Had Cantor not proved
> that the rationals were indexed step by step by 1, 2, 3, ... (that are steps)
> but by abracadabra, then no-one would have given a dime for his theory.

What seems like step by step is acually not, but the simplified
demonstration of how it works is. It shows that no element or cell of
the N X N table remains unaddressed. Nothing is "missed" in the
demonstration. The demonstration is not the function, but it uses the
function.

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.

| m is in ⋃𝓕-end-segment E(k)
is equivalent to
| k is in 𝐹ₘ the FISON ending at m

k is in E(k)

if m is in E(k), then m+1 is in E(k)
| | because 𝐹ₘ⊕⟨m+1⟩ is a FISON too,

Therefore, E(k) is infinite.

> Therefore
> infinitely many natural numbers
> are not definable and
> cannot be used for indexing

No.

| m is in ⋃𝓕-end-segment E(k)
is equivalent to
| k is in 𝐹ₘ the FISON ending at m

If
m is in all end segments E(k)
then
all FISON-ends k are in 𝐹ₘ

We know that
NOT all FISON-ends k are in 𝐹ₘ
In particular, we know
k = m+1 is NOT in 𝐹ₘ
and m+1 ends 𝐹ₘ⊕⟨m+1⟩

Sergio <invalid@invalid.com> writes:

> On 9/7/2022 2:33 PM, Ben Bacarisse wrote:
>> Sergio <invalid@invalid.com> writes:
>>
>>> On 9/7/2022 10:51 AM, Ben Bacarisse wrote:
>>>> Sergio <invalid@invalid.com> writes:
>>>>
>>>>> On 9/7/2022 8:46 AM, WM wrote:
>>>>>> Fritz Feldhase schrieb am Dienstag, 6. September 2022 um 19:14:19 UTC+2:
>>>>>>> On Tuesday, September 6, 2022 at 6:42:20 PM UTC+2, WM wrote:
>>>>>>>
>>>>>>>> Infinitely many X's are not enough to cover the whole matrix.
>>>>>>> Oh really?! Which element in the following matrix is not covered by an X?
>>>>>>>
>>>>>>> X X X X...
>>>>>>> X X X X...
>>>>>>> X X X X...
>>>>>>> X X X X...
>>>>>>> ...
>>>>>> Almost all (if you start from
>>>>>> XOO...
>>>>>> XOO...
>>>>>> XOO...
>>>>>> ...
>>>>>> and do not add further X's).
>>>>>> The reason is that shuffling the X will never remove an O.
>>>>>> Regards, WM
>>>>>
>>>>> It is wrong as it does not agree with Cantor, who proved the mapping
>>>>> using Math.
>>>>>
>>>>> your swaparoofest is simply a 'step by step' process applied to
>>>>> infinite sets and will always fail.
>>>> I think there is a reasonable definition of the result of an endless
>>>> sequence of swaps -- the limit of the sequence of indicator functions.
>>>
>>> WM doeent use limits.
>> Yes he does. His textbook is full of limits. (I see below that you
>> think I am talking about just junk collection of notes.)
>
> which book ? Guess he forgot a lot, like most of that book.

Mathematik für die ersten Semester

It is reported to have a non-trivial number of errors, but the basics
are all covered.

>>>> And the result is M(i) = 1. M(i) indicates an X at the position
>>>> numbered k(n,m) = (m + n - 1)(m + n - 2) /2 + m.
>>>
>>> no, that is the mapping of naturals to rationals, no M(i) or X needed
>>> as they are diversion.
>> Sorry, I don't understand. Did you read the argument I presented? I
>> can't see what your objection is.
>
> sure I read it, your intent may be to bridge the gap between "k(n,m) =
> (m + n - 1)(m + n - 2) /2 + m" and WM's X and O's
> But I see it as unnecessary as k(n,m) = (m + n - 1)(m + n - 2) /2 + m
> is very clear and is in no need for further stuff.

How is k enough to show that the Ox "disappear"? What's your reasoning?

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.

>>> so it must be rejected if it supports wrong conclusions.
>> 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.
>
> 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.

> So WMs conclusions using his matrix disproving Cantor is false from
> the get go.

Of course. That the Os "disappear" has nothing to do with Cantor being
wrong. The argument works in ordinary mathematics and also using only
the mathematics in WM's textbook. There is no contradiction with
anything.

> If he is serious, and knew math, he would define his "swaps" in terms
> of a specific indexing algorithm used on the rational matrix. But no,
> no math.

Yes. That was the purpose of the exercises -- to get WM to pin down the
swaps. Some patterns of swaps leave some Os "in the limit" and others
don't.

--
Ben.

On Thursday, September 8, 2022 at 1:41:11 AM UTC+2, Ben Bacarisse wrote:

> And he's agreed (for the first time I can remember) that k is a
> bijection between NxN and N.

Sorry, Ben, but you are dreaming, I'd say.

WM <askasker48@gmail.com> writes:

> Ben Bacarisse schrieb am Mittwoch, 7. September 2022 um 18:29:02 UTC+2:
>> WM <askas...@gmail.com> writes:
>
>> > They cannot and do not. Your "proof" would work with the matrix as
>> > well as with the serpent sequence, but it won't. Therefore your only
>> > attempt is to distract from the topic.
>> Not so. I would work with functions MX_i(n,m) rather than the simpler
>> M_i(n), but your book does not show how to calculate the limit of such a
>> sequence of functions, whereas it does for the simpler functions of one
>> variable.
>
> Here is no limit but the constance of the number of X's and O's.

Not so. There are lots of limits, as your book explains. There are
limits that are number and there are limits that are functions (page 201
in my PDF).

We /could/ try to count number of Xs and Os but your textbook does not
even say how that can be done and it would be a change of topic since
you started by referring to what /positions/ have Xs and Os, 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).

You've chosen to try not to understand the argument, but that does not
make it any less of an argument.

> Cantor's theory ...
....
> 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 and you don't want to put the work in
to understand how.

> 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. When, as you do, you do
mathematics by magic words, we need the actual text to know what it is
that was really written. "Step by step" is a metaphor. Proving that
lim_{n->oo} M_m = I (where I(i) = 1) is mathematics.

--
Ben.

WM <askasker48@gmail.com> writes:

> Ben Bacarisse schrieb am Mittwoch, 7. September 2022 um 17:51:38 UTC+2:
>
>> I think there is a reasonable definition of the result of an endless
>> sequence of swaps -- the limit of the sequence of indicator functions.
>
> It is hard to believe how an intelligent person can think such an
> utter nonsense. If there is a limit, then it is the limit of the
> number of O's or the ratio of O's and X's.

The limit I used is as given in your textbook, but you have ditched the
exercises soon enough that you don't need to confront that fact.

> 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.

(ℵo does not appear in you textbook so I will stick with "an infinity
of...")

--
Ben.

WM <askasker48@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?

> Nevertheless potential infinity has a growing character. Therefore
> larger finite sets are created from smaller sets.

Any progress on defining set membership, equality and difference so that
someone other than you can work with these sets? Have you even tried?

--
Ben.

Fritz Feldhase <franz.fritschee.ff@gmail.com> writes:

> On Thursday, September 8, 2022 at 1:41:11 AM UTC+2, Ben Bacarisse wrote:
>
>> And he's agreed (for the first time I can remember) that k is a
>> bijection between NxN and N.
>
> Sorry, Ben, but you are dreaming, I'd say.

Date: Thu, 1 Sep 2022 06:21:17 -0700 (PDT)
Message-ID: <e6367a34-d7e7-4d2d-ae75-43349ed2d930n@googlegroups.com>
WM: k = (m + n - 1)(m + n - 2)/2 + m. (*)
ME: I.e. k(n,m) is a bijection from NxN to N, a fact provable by any student
who has read your textbook. Do you agree that such a student could
prove this fact?
WM: Of course.

--
Ben.

On 9/7/2022 6:41 PM, Ben Bacarisse wrote:
> Sergio <invalid@invalid.com> writes:
>
>> On 9/7/2022 2:33 PM, Ben Bacarisse wrote:
>>> Sergio <invalid@invalid.com> writes:
>>>
>>>> On 9/7/2022 10:51 AM, Ben Bacarisse wrote:
>>>>> Sergio <invalid@invalid.com> writes:
>>>>>
>>>>>> On 9/7/2022 8:46 AM, WM wrote:
>>>>>>> Fritz Feldhase schrieb am Dienstag, 6. September 2022 um 19:14:19 UTC+2:
>>>>>>>> On Tuesday, September 6, 2022 at 6:42:20 PM UTC+2, WM wrote:
>>>>>>>>
>>>>>>>>> Infinitely many X's are not enough to cover the whole matrix.
>>>>>>>> Oh really?! Which element in the following matrix is not covered by an X?
>>>>>>>>
>>>>>>>> X X X X...
>>>>>>>> X X X X...
>>>>>>>> X X X X...
>>>>>>>> X X X X...
>>>>>>>> ...
>>>>>>> Almost all (if you start from
>>>>>>> XOO...
>>>>>>> XOO...
>>>>>>> XOO...
>>>>>>> ...
>>>>>>> and do not add further X's).
>>>>>>> The reason is that shuffling the X will never remove an O.
>>>>>>> Regards, WM
>>>>>>
>>>>>> It is wrong as it does not agree with Cantor, who proved the mapping
>>>>>> using Math.
>>>>>>
>>>>>> your swaparoofest is simply a 'step by step' process applied to
>>>>>> infinite sets and will always fail.
>>>>> I think there is a reasonable definition of the result of an endless
>>>>> sequence of swaps -- the limit of the sequence of indicator functions.
>>>>
>>>> WM doeent use limits.
>>> Yes he does. His textbook is full of limits. (I see below that you
>>> think I am talking about just junk collection of notes.)
>>
>> which book ? Guess he forgot a lot, like most of that book.
>
> Mathematik für die ersten Semester
>
> It is reported to have a non-trivial number of errors, but the basics
> are all covered.
>
>>>>> And the result is M(i) = 1. M(i) indicates an X at the position
>>>>> numbered k(n,m) = (m + n - 1)(m + n - 2) /2 + m.
>>>>
>>>> no, that is the mapping of naturals to rationals, no M(i) or X needed
>>>> as they are diversion.
>>> Sorry, I don't understand. Did you read the argument I presented? I
>>> can't see what your objection is.
>>
>> sure I read it, your intent may be to bridge the gap between "k(n,m) =
>> (m + n - 1)(m + n - 2) /2 + m" and WM's X and O's
>> But I see it as unnecessary as k(n,m) = (m + n - 1)(m + n - 2) /2 + m
>> is very clear and is in no need for further stuff.
>
> How is k enough to show that the Ox "disappear"? What's your reasoning?

well, I was trying to ditch the X and Os entirely,
I think all WM was doing was taking the matrix of rationals and using the first column as an index (which is crazy) then he switches to X and O, and
swaparoos... (where a swap is a step, so you have a step by step process on infinite sets, which leads to wrong conclusions)

>
> 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.

I dont think he understands that either, or that the equation k(n,m) = (m + n - 1)(m + n - 2) /2 + m is simply a formula for the indexes.
Perhaps he could at one time, or he is intentionally not understanding math...

>
>>>> so it must be rejected if it supports wrong conclusions.
>>> 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.
>>
>> 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 did, but WMsMath(t) where t is time, and he changes a lot.

>
>> So WMs conclusions using his matrix disproving Cantor is false from
>> the get go.
>
> Of course. That the Os "disappear" has nothing to do with Cantor being
> wrong. The argument works in ordinary mathematics and also using only
> the mathematics in WM's textbook. There is no contradiction with
> anything.
>
>> If he is serious, and knew math, he would define his "swaps" in terms
>> of a specific indexing algorithm used on the rational matrix. But no,
>> no math.
>
> Yes. That was the purpose of the exercises -- to get WM to pin down the
> swaps. Some patterns of swaps leave some Os "in the limit" and others
> don't.

WM does need to do that, his hand waving only warms the air

Ben Bacarisse expressed precisely :
> Sergio <invalid@invalid.com> writes:
>
>> On 9/7/2022 2:33 PM, Ben Bacarisse wrote:
>>> Sergio <invalid@invalid.com> writes:
>>>
>>>> On 9/7/2022 10:51 AM, Ben Bacarisse wrote:
>>>>> Sergio <invalid@invalid.com> writes:
>>>>>
>>>>>> On 9/7/2022 8:46 AM, WM wrote:
>>>>>>> Fritz Feldhase schrieb am Dienstag, 6. September 2022 um 19:14:19
>>>>>>> UTC+2:
>>>>>>>> On Tuesday, September 6, 2022 at 6:42:20 PM UTC+2, WM wrote:
>>>>>>>>
>>>>>>>>> Infinitely many X's are not enough to cover the whole matrix.
>>>>>>>> Oh really?! Which element in the following matrix is not covered by an
>>>>>>>> X?
>>>>>>>>
>>>>>>>> X X X X...
>>>>>>>> X X X X...
>>>>>>>> X X X X...
>>>>>>>> X X X X...
>>>>>>>> ...
>>>>>>> Almost all (if you start from
>>>>>>> XOO...
>>>>>>> XOO...
>>>>>>> XOO...
>>>>>>> ...
>>>>>>> and do not add further X's).
>>>>>>> The reason is that shuffling the X will never remove an O.
>>>>>>> Regards, WM
>>>>>>
>>>>>> It is wrong as it does not agree with Cantor, who proved the mapping
>>>>>> using Math.
>>>>>>
>>>>>> your swaparoofest is simply a 'step by step' process applied to
>>>>>> infinite sets and will always fail.
>>>>> I think there is a reasonable definition of the result of an endless
>>>>> sequence of swaps -- the limit of the sequence of indicator functions.
>>>>
>>>> WM doeent use limits.
>>> Yes he does. His textbook is full of limits. (I see below that you
>>> think I am talking about just junk collection of notes.)
>>
>> which book ? Guess he forgot a lot, like most of that book.
>
> Mathematik für die ersten Semester
>
> It is reported to have a non-trivial number of errors, but the basics
> are all covered.
>
>>>>> And the result is M(i) = 1. M(i) indicates an X at the position
>>>>> numbered k(n,m) = (m + n - 1)(m + n - 2) /2 + m.
>>>>
>>>> no, that is the mapping of naturals to rationals, no M(i) or X needed
>>>> as they are diversion.
>>> Sorry, I don't understand. Did you read the argument I presented? I
>>> can't see what your objection is.
>>
>> sure I read it, your intent may be to bridge the gap between "k(n,m) =
>> (m + n - 1)(m + n - 2) /2 + m" and WM's X and O's
>> But I see it as unnecessary as k(n,m) = (m + n - 1)(m + n - 2) /2 + m
>> is very clear and is in no need for further stuff.
>
> How is k enough to show that the Ox "disappear"? What's your reasoning?
>
> 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.
>
>>>> so it must be rejected if it supports wrong conclusions.
>>> 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.
>>
>> 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.

onsdag 7 september 2022 kl. 15:46:57 UTC+2 skrev WM:
> Fritz Feldhase schrieb am Dienstag, 6. September 2022 um 19:14:19 UTC+2:
> > On Tuesday, September 6, 2022 at 6:42:20 PM UTC+2, WM wrote:
> >
> > > Infinitely many X's are not enough to cover the whole matrix.
> > Oh really?! Which element in the following matrix is not covered by an X?
> >
> > X X X X...
> > X X X X...
> > X X X X...
> > X X X X...
> > ...
> Almost all (if you start from
> XOO...
> XOO...
> XOO...
> ...
> and do not add further X's).
>
> The reason is that shuffling the X will never remove an O.
>
> Regards, WM

It is not a step by step process

Please open paint.exe in Windows and click invert colors. Then tell me, what you have just seen.
🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑🤑

WM kirjutas Kolmapäev, 31. august 2022 kl 04:36:22 UTC-7:
> (1) Cantor has proved that all positive fractions m/n can be enumerated by all natural numbers k:
>
> k = (m + n - 1)(m + n - 2)/2 + m. (*)
>
> This is tantamount to enumerating the positive fractions by the integer fractions of the first column of the matrix
>
> 1/1, 1/2, 1/3, 1/4, ...
> 2/1, 2/2, 2/3, 2/4, ...
> 3/1, 3/2, 3/3, 3/4, ...
> 4/1, 4/2, 4/3, 4/4, ...
> ...
>
> Of course also the integer fractions belong to the fractions to be enumerated. Therefore his approach is tantamount to exchanging X's and O's in the matrix until all O's have disappeared:
>
> X, O, O, O, ...
> X, O, O, O, ...
> X, O, O, O, ...
> X, O, O, O, ...
> ...
>
> In fact by application of (*) all O's are removed from all visible or definable matrix positions. However it is clear that, by simple exchanging O's with X's, never an O will be removed from the matrix. This shows that the O's move to invisible, i.e., undefinable matrix positions. These are called dark positions.
>
> (2) The intersection of non-empty inclusion-monotonic sets like infinite endsegments E(k) = {k, k+1, k+2, ...} is not empty. Every non-empty endsegment shares at least one natural number with all non-empty endsegments. In fact every infinite endsegment shares infinitely many natural numbers with all infinite endsegments. Otherwise there would be a first endsegment sharing less natural numbers with its predecessors. This cannot happen, if all endsegments are infinite.
>
> But according to ZFC, the intersection of all endsegments is empty.
> Since all definable endsegments satisfy
>
> ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀
>
> the empty intersection cannot be accomplished by merely definable endsegments
>
> ∩{E(k) : k ∈ ℕ_def} =/= { }.
>
> Only by the presence of undefinable endsegments
>
> ∩{E(k) : k ∈ ℕ} = { }
>
> can be accomplished.
>
> (3) The simplest proof of dark natural numbers is this:
>
> Every definable natural number k is finite and belongs to a finite set
>
> {1, 2, 3, ..., k}.
>
> If there are ℵo, i.e., more than any finite number, then ℕ can can only be filled and completed by dark natural numbers. This is obvious from the simple fact
>
> ∀k ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., k}| = ℵo .
>
> Regards, WM

Sergio <invalid@invalid.com> writes:

> On 9/7/2022 6:41 PM, Ben Bacarisse wrote:
>> Sergio <invalid@invalid.com> writes:
>>
>>> On 9/7/2022 2:33 PM, Ben Bacarisse wrote:
>>>> Sergio <invalid@invalid.com> writes:
>>>>
>>>>> On 9/7/2022 10:51 AM, Ben Bacarisse wrote:
>>>>>> Sergio <invalid@invalid.com> writes:
>>>>>>
>>>>>>> On 9/7/2022 8:46 AM, WM wrote:
>>>>>>>> Fritz Feldhase schrieb am Dienstag, 6. September 2022 um 19:14:19 UTC+2:
>>>>>>>>> On Tuesday, September 6, 2022 at 6:42:20 PM UTC+2, WM wrote:
>>>>>>>>>
>>>>>>>>>> Infinitely many X's are not enough to cover the whole matrix.
>>>>>>>>> Oh really?! Which element in the following matrix is not covered by an X?
>>>>>>>>>
>>>>>>>>> X X X X...
>>>>>>>>> X X X X...
>>>>>>>>> X X X X...
>>>>>>>>> X X X X...
>>>>>>>>> ...
>>>>>>>> Almost all (if you start from
>>>>>>>> XOO...
>>>>>>>> XOO...
>>>>>>>> XOO...
>>>>>>>> ...
>>>>>>>> and do not add further X's).
>>>>>>>> The reason is that shuffling the X will never remove an O.
>>>>>>>> Regards, WM
>>>>>>>
>>>>>>> It is wrong as it does not agree with Cantor, who proved the mapping
>>>>>>> using Math.
>>>>>>>
>>>>>>> your swaparoofest is simply a 'step by step' process applied to
>>>>>>> infinite sets and will always fail.
>>>>>> I think there is a reasonable definition of the result of an endless
>>>>>> sequence of swaps -- the limit of the sequence of indicator functions.
>>>>>
>>>>> WM doeent use limits.
>>>> Yes he does. His textbook is full of limits. (I see below that you
>>>> think I am talking about just junk collection of notes.)
>>>
>>> which book ? Guess he forgot a lot, like most of that book.
>> Mathematik für die ersten Semester
>> It is reported to have a non-trivial number of errors, but the basics
>> are all covered.
>>
>>>>>> And the result is M(i) = 1. M(i) indicates an X at the position
>>>>>> numbered k(n,m) = (m + n - 1)(m + n - 2) /2 + m.
>>>>>
>>>>> no, that is the mapping of naturals to rationals, no M(i) or X needed
>>>>> as they are diversion.
>>>> Sorry, I don't understand. Did you read the argument I presented? I
>>>> can't see what your objection is.
>>>
>>> sure I read it, your intent may be to bridge the gap between "k(n,m) =
>>> (m + n - 1)(m + n - 2) /2 + m" and WM's X and O's
>>> But I see it as unnecessary as k(n,m) = (m + n - 1)(m + n - 2) /2 + m
>>> is very clear and is in no need for further stuff.
>> How is k enough to show that the Ox "disappear"? What's your reasoning?
>
> well, I was trying to ditch the X and Os entirely,

OK. Then we were talking about entirely different things.

> I think all WM was doing was taking the matrix of rationals and using
> the first column as an index (which is crazy) then he switches to X
> and O, and swaparoos... (where a swap is a step, so you have a step
> by step process on infinite sets, which leads to wrong conclusions)

I think they lead to interesting conclusions.

Consider his old tale of Scrooge McDuck who (in my version) gets $2
bills every day and burns one of them immediately. On day n, let D_n be
the set of bank notes in his possession.

There are several limits we can consider. There is the rather boring
lim_{n->oo} |D_n| which is either undefined or infinite depending on how
you define such things.

But there are also the set-sequence limits. The note numbered i is in
the limit if there is an m such that i in D_n for n > m, and note i is
/not/ in the limit if there is some m such that i not in D_n for n > m.
The limit is undefined if there is an i for which neither is the case.

It is possible, by giving different rules for which numbered notes are
burned, to have lim_{n->oo} D_n be a set of any desired size from empty
to infinite.

WM finds it baffling (and thus a "contradiction") that there is a set
sequence S_n such that |lim S_n| = 0 but lim |S_n| = oo.

>> 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.
>
> I dont think he understands that either, or that the equation k(n,m) =
> (m + n - 1)(m + n - 2) /2 + m is simply a formula for the indexes.
> Perhaps he could at one time, or he is intentionally not understanding
> math...

I am never sure if he does not understand or if he is simply putting his
fingers in his ears and going "la la la la la, I can't hear you".

For end segments of N, the point-wise limit of the sequence of indicator
functions

E_n(i) = [i >= n] (where []s is the "Iverson bracket")

is the zero function E(i) = 0. The reason I use indicator functions is
that the point-wise limit is defined in his textbook so he can't (or at
least shouldn't) deny that this limit (or the one for Xs in the matrix)
exists.

When I brought this (up for end segments) he simply flatly denied that
lim_{n->oo} E_n = E, despite my copying the definition from his book and
working through the proof! It is possible that he does not understand
what's in his book. He did keep confusing lim_{n->oo} E_n with
lim_{i->oo} E_n(i).

>> Yes. And he's agreed (for the first time I can remember) that k is a
>> bijection between NxN and N.
>
> He did, but WMsMath(t) where t is time, and he changes a lot.

That's why the book is useful.

--
Ben.

On 9/8/2022 5:41 AM, Ben Bacarisse wrote:
> Sergio <invalid@invalid.com> writes:
>
>> On 9/7/2022 6:41 PM, Ben Bacarisse wrote:
>>> Sergio <invalid@invalid.com> writes:
>>>
>>>> On 9/7/2022 2:33 PM, Ben Bacarisse wrote:
>>>>> Sergio <invalid@invalid.com> writes:
>>>>>
>>>>>> On 9/7/2022 10:51 AM, Ben Bacarisse wrote:
>>>>>>> Sergio <invalid@invalid.com> writes:
>>>>>>>
>>>>>>>> On 9/7/2022 8:46 AM, WM wrote:
>>>>>>>>> Fritz Feldhase schrieb am Dienstag, 6. September 2022 um 19:14:19 UTC+2:
>>>>>>>>>> On Tuesday, September 6, 2022 at 6:42:20 PM UTC+2, WM wrote:
>>>>>>>>>>
>>>>>>>>>>> Infinitely many X's are not enough to cover the whole matrix.
>>>>>>>>>> Oh really?! Which element in the following matrix is not covered by an X?
>>>>>>>>>>
>>>>>>>>>> X X X X...
>>>>>>>>>> X X X X...
>>>>>>>>>> X X X X...
>>>>>>>>>> X X X X...
>>>>>>>>>> ...
>>>>>>>>> Almost all (if you start from
>>>>>>>>> XOO...
>>>>>>>>> XOO...
>>>>>>>>> XOO...
>>>>>>>>> ...
>>>>>>>>> and do not add further X's).
>>>>>>>>> The reason is that shuffling the X will never remove an O.
>>>>>>>>> Regards, WM
>>>>>>>>
>>>>>>>> It is wrong as it does not agree with Cantor, who proved the mapping
>>>>>>>> using Math.
>>>>>>>>
>>>>>>>> your swaparoofest is simply a 'step by step' process applied to
>>>>>>>> infinite sets and will always fail.
>>>>>>> I think there is a reasonable definition of the result of an endless
>>>>>>> sequence of swaps -- the limit of the sequence of indicator functions.
>>>>>>
>>>>>> WM doeent use limits.
>>>>> Yes he does. His textbook is full of limits. (I see below that you
>>>>> think I am talking about just junk collection of notes.)
>>>>
>>>> which book ? Guess he forgot a lot, like most of that book.
>>> Mathematik für die ersten Semester
>>> It is reported to have a non-trivial number of errors, but the basics
>>> are all covered.
>>>
>>>>>>> And the result is M(i) = 1. M(i) indicates an X at the position
>>>>>>> numbered k(n,m) = (m + n - 1)(m + n - 2) /2 + m.
>>>>>>
>>>>>> no, that is the mapping of naturals to rationals, no M(i) or X needed
>>>>>> as they are diversion.
>>>>> Sorry, I don't understand. Did you read the argument I presented? I
>>>>> can't see what your objection is.
>>>>
>>>> sure I read it, your intent may be to bridge the gap between "k(n,m) =
>>>> (m + n - 1)(m + n - 2) /2 + m" and WM's X and O's
>>>> But I see it as unnecessary as k(n,m) = (m + n - 1)(m + n - 2) /2 + m
>>>> is very clear and is in no need for further stuff.
>>> How is k enough to show that the Ox "disappear"? What's your reasoning?
>>
>> well, I was trying to ditch the X and Os entirely,
>
> OK. Then we were talking about entirely different things.
>
>> I think all WM was doing was taking the matrix of rationals and using
>> the first column as an index (which is crazy) then he switches to X
>> and O, and swaparoos... (where a swap is a step, so you have a step
>> by step process on infinite sets, which leads to wrong conclusions)
>
> I think they lead to interesting conclusions.
>
> Consider his old tale of Scrooge McDuck who (in my version) gets $2
> bills every day and burns one of them immediately. On day n, let D_n be
> the set of bank notes in his possession.
>
> There are several limits we can consider. There is the rather boring
> lim_{n->oo} |D_n| which is either undefined or infinite depending on how
> you define such things.
>
> But there are also the set-sequence limits. The note numbered i is in
> the limit if there is an m such that i in D_n for n > m, and note i is
> /not/ in the limit if there is some m such that i not in D_n for n > m.
> The limit is undefined if there is an i for which neither is the case.
>
> It is possible, by giving different rules for which numbered notes are
> burned, to have lim_{n->oo} D_n be a set of any desired size from empty
> to infinite.

so in math the recursive formula is D_(n+1) = D_(n) + a , (n 0 to oo)
where a (net$) is changeable depending upon the structure of the problem (McDucks burning habits).
D_0 would be initial pile of bills, D_oo is the final answer, and D_k is where one stops at k
if a is constant, then answer is just D_k = D_0 + a*k

not complex enough to be interesting.

if a is not constant, but changes after a time, one can split the problem into 2 (or more) parts and sum them
One can also treat this as a classic TVM (time value of money) problem from the finance field too, use an on line calculator.
(McDuck is scrooged if he does not invest his nest egg, US Gov iBonds are paying almost 9.62% now)

One can add a probability density distribution that McDuck burns 0 to 2 bills a day, and the result will be a probability distribution as well. Now that
is an interesting problem.
OR add a time randomness to if McDuck gets paid or not on the front end.
Or add both, random payments to McDuck, and random McDuck burnings, and you get an output that is a probability distribution.
If you know the input and burning distributions, you can solve the problem using Maths.
(random payments to McDuck most likely is Poisson like, his burning is dependent upon hangover)

>
> WM finds it baffling (and thus a "contradiction") that there is a set
> sequence S_n such that |lim S_n| = 0 but lim |S_n| = oo.

on/off like a switch. EZ proof that it is that way. All WMs math in the world wont smooth that one out.

There are many areas of Math that "run out" or have a "fringe edge", or have a hole in a surface, too many dimensions, ... etc.
math works on the simpler side but not much past that. WM should know that.

>
>>> 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.
>>
>> I dont think he understands that either, or that the equation k(n,m) =
>> (m + n - 1)(m + n - 2) /2 + m is simply a formula for the indexes.
>> Perhaps he could at one time, or he is intentionally not understanding
>> math...
>
> I am never sure if he does not understand or if he is simply putting his
> fingers in his ears and going "la la la la la, I can't hear you".

He wants to keep the conversation going, while never admiting he is wrong.

>
> For end segments of N, the point-wise limit of the sequence of indicator
> functions
>
> E_n(i) = [i >= n] (where []s is the "Iverson bracket")
>
> is the zero function E(i) = 0. The reason I use indicator functions is
> that the point-wise limit is defined in his textbook so he can't (or at
> least shouldn't) deny that this limit (or the one for Xs in the matrix)
> exists.
>
> When I brought this (up for end segments) he simply flatly denied that
> lim_{n->oo} E_n = E, despite my copying the definition from his book and
> working through the proof! It is possible that he does not understand
> what's in his book. He did keep confusing lim_{n->oo} E_n with
> lim_{i->oo} E_n(i).

that all sounds kosher to me

>
>>> Yes. And he's agreed (for the first time I can remember) that k is a
>>> bijection between NxN and N.
>>
>> He did, but WMsMath(t) where t is time, and he changes a lot.
>
> That's why the book is useful.
>

On 9/7/2022 3:12 PM, Gus Gassmann wrote:
> On Wednesday, 7 September 2022 at 16:33:29 UTC-3, Ben Bacarisse wrote:
> [...]
>> Yes, that's just a pile of junk. I am talking about his published
>> textbook on basic maths.
>
> That's not much better. It is both original and correct. Unfortunately, what is correct is not original, and what is original is not correct. He cribbed liberally from the masters, but due to his incompetence he shortened and simplified things to the point where they are simply no longer true. If you read German, you can try your hands at this review by Franz Lemmermeyer: https://zbmath.org/?q=an%3A1204.00016.

good review, shows WM back then was poor at math, and injected his own Bent WM ideas into it.

"the "definition" of natural numbers by the axioms on p. 25 is outrageous; the existence of the successor is confused with the axiom of complete
induction (which the author may not care about, since the set of natural numbers in his worldview is firstly finite and secondly has large gaps,
consequently both the existence of a successor and the axiom of complete induction "exceptions suffers"); Division, one learns on p. 30, is an
abbreviation for repeated subtraction; Square roots are by definition positive on page 37, while..."

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

Sergio <invalid@invalid.com> writes:

> On 9/8/2022 5:41 AM, Ben Bacarisse wrote:
>> Sergio <invalid@invalid.com> writes:
>>

>>> I think all WM was doing was taking the matrix of rationals and using
>>> the first column as an index (which is crazy) then he switches to X
>>> and O, and swaparoos... (where a swap is a step, so you have a step
>>> by step process on infinite sets, which leads to wrong conclusions)
>>
>> I think they lead to interesting conclusions.
>> Consider his old tale of Scrooge McDuck who (in my version) gets $2
>> bills every day and burns one of them immediately. On day n, let D_n be
>> the set of bank notes in his possession.
>> There are several limits we can consider. There is the rather boring
>> lim_{n->oo} |D_n| which is either undefined or infinite depending on how
>> you define such things.
>> But there are also the set-sequence limits. The note numbered i is in
>> the limit if there is an m such that i in D_n for n > m, and note i is
>> /not/ in the limit if there is some m such that i not in D_n for n > m.
>> The limit is undefined if there is an i for which neither is the case.
>> It is possible, by giving different rules for which numbered notes are
>> burned, to have lim_{n->oo} D_n be a set of any desired size from empty
>> to infinite.
>
> so in math the recursive formula is D_(n+1) = D_(n) + a , (n 0 to oo)
> where a (net$) is changeable depending upon the structure of the
> problem (McDucks burning habits). D_0 would be initial pile of bills,
> D_oo is the final answer, and D_k is where one stops at k if a is
> constant, then answer is just D_k = D_0 + a*k

This looks numerical, I see no set sequence limits here. D_0 = {}. He
starts with no bills.

For simplicity, let's assume the bills are consecutively numbered from 1
(rather than by large non-consecutive numbers). He is given bills
numbered 1 and 2. Let's say he always burns the lowest numbered bill he
owns. Then D_1 = {2}. Now he gets bills numbered 3 and 4 and he burns
number 2. D_2 = {3, 4}...

With this pattern, lim_{n->oo} D_n = {}. No bill is survives the
flames. Of course the numerical limit of his wealth (number of dollars
owned) lim_{n->oo} |D_n| is unbounded.

And just for another example, if he always burns the highest numbered
note we get D_1 = {}, D_2 = {1}, D_3 = {1, 3}, D_3 = {1, 3, 5} and so
on. Now lim_{n->oo} D_n is the set of odd numbers and /both/
|lim_{n->oo} D_n| and lim_{n->oo} |D_n| are unbounded.

> not complex enough to be interesting.

Not sure interesting was the right word, but it's non-trivial.

--
Ben.

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.

Regards, WM

FromTheRafters schrieb am Mittwoch, 7. September 2022 um 23:51:32 UTC+2:
> WM expressed precisely :

> > I showed you often enough. You seem to have a short memory.
> > But perhaps you may understand the simple argument: Had Cantor not proved
> > that the rationals were indexed step by step by 1, 2, 3, ... (that are steps)
> > but by abracadabra, then no-one would have given a dime for his theory.
> What seems like step by step is acually not,

You are wrong. Every definable indexing of a fraction can be found as a step, finitely many steps from the start. Which one could not?

> It shows that no element or cell of
> the N X N table remains unaddressed.

No, it shows that after every index infinitely many are waiting. And my proof shows that during the whole process of indexing the number of O's remains constant and the nunmber of X's remains constant. What may happen after all definable steps, in the limit or afterwards, is individually undefinable and therefore dark.

Regards, WM

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.

Regards, WM

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.
I prove that no definable step of the formula k = (m + n - 1)(m + n - 2)/2 + m or the sequence of indexed fractions 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, ... has deleted any O.

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

> Of course. That the Os "disappear" has nothing to do with Cantor being
> wrong.

That the O's do not disappear in a definable way has to do with fractions not indexed in a definable way.

> The argument works in ordinary mathematics and also using only
> the mathematics in WM's textbook. There is no contradiction with
> anything.

My textbook concerns limits which are approached not limits which are reached. Indexing all fractions means reaching this aim.

> Some patterns of swaps leave some Os "in the limit" and others
> don't.

What swaps = exchangesof X and O, delete any O's?

Regards, WM

WM was thinking very hard :
> FromTheRafters schrieb am Mittwoch, 7. September 2022 um 23:51:32 UTC+2:
>> WM expressed precisely :
>
>>> I showed you often enough. You seem to have a short memory.
>>> But perhaps you may understand the simple argument: Had Cantor not proved
>>> that the rationals were indexed step by step by 1, 2, 3, ... (that are
>>> steps) but by abracadabra, then no-one would have given a dime for his
>>> theory.
>> What seems like step by step is acually not,
>
> You are wrong. Every definable

Ya lost me there. :)

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

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. 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.
>
> (ℵo does not appear in you textbook so I will stick with "an infinity
> of...")

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

Regards, WM