Three proofs of dark numbers

[WM]

(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

[FromTheRafters]

WM submitted this idea :

...and they all lived happily ever after.

[Ben Bacarisse]

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

> (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. (*)

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?

> 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, ...
> ...

It's equivalent to drawing arrows between the Xs and every cell
(including the Xs) so that every cell is at the beginning and end of
just one arrow. You can see how this can be done, yes?

--
Ben.

[Archimedes Plutonium]

Take this bullshit failed ignorance to sci.logic, you do not deserve to post in sci.math.
On Wednesday, August 31, 2022 at 6:36:22 AM UTC-5, WM wrote:

Take this bullshit over to sci.logic, for you are a math failure-- Your AND connector is subtraction with 2 OR 1 = 3. You do not even know a geometry proof of Fundamental Theorem of Calculus. You are a failure in geometry for your slant cut in cone is a ellipse when actually that is a oval. But worst of all-- your so stupid in science you cannot even ask the question which is the atom's true electron-- 0.5MeV particle or the muon stuck inside a 840 MeV proton torus.

You do not deserve to post in sci.math with your failed ignorance of math.

3rd published book

AP's Proof-Ellipse was never a Conic Section // Math proof series, book 1 Kindle Edition
by Archimedes Plutonium (Author)

Ever since Ancient Greek Times it was thought the slant cut into a cone is the ellipse. That was false. For the slant cut in every cone is a Oval, never an Ellipse. This book is a proof that the slant cut is a oval, never the ellipse. A slant cut into the Cylinder is in fact a ellipse, but never in a cone.

Product details
• ASIN ‏ : ‎ B07PLSDQWC
• Publication date ‏ : ‎ March 11, 2019
• Language ‏ : ‎ English
• File size ‏ : ‎ 1621 KB
• Text-to-Speech ‏ : ‎ Enabled
• Enhanced typesetting ‏ : ‎ Enabled
• X-Ray ‏ : ‎ Not Enabled
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• Lending ‏ : ‎ Enabled
•
•

Proofs Ellipse is never a Conic section, always a Cylinder section and a Well Defined Oval definition//Student teaches professor series, book 5 Kindle Edition
by Archimedes Plutonium (Author)

Last revision was 14May2022. This is AP's 68th published book of science.

Preface: A similar book on single cone cut is a oval, never a ellipse was published in 11Mar2019 as AP's 3rd published book, but Amazon Kindle converted it to pdf file, and since then, I was never able to edit this pdf file, and decided rather than struggle and waste time, decided to leave it frozen as is in pdf format. Any new news or edition of ellipse is never a conic in single cone is now done in this book. The last thing a scientist wants to do is wade and waddle through format, when all a scientist ever wants to do is science itself. So all my new news and thoughts of Conic Sections is carried out in this 68th book of AP. And believe you me, I have plenty of new news.

In the course of 2019 through 2022, I have had to explain this proof often on Usenet, sci.math and sci.physics. And one thing that constant explaining does for a mind of science, is reduce the proof to its stripped down minimum format, to bare bones skeleton proof. I can prove the slant cut in single cone is a Oval, never the ellipse in just a one sentence proof. Proof-- A single cone and oval have just one axis of symmetry, while a ellipse requires 2 axes of symmetry, hence slant cut is always a oval, never the ellipse.

Product details
• ASIN ‏ : ‎ B081TWQ1G6
• Publication date ‏ : ‎ November 21, 2019
• Language ‏ : ‎ English
• File size ‏ : ‎ 827 KB
• Simultaneous device usage ‏ : ‎ Unlimited
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#12-2, 11th published book

World's First Geometry Proof of Fundamental Theorem of Calculus// Math proof series, book 2 Kindle Edition
by Archimedes Plutonium (Author)

Last revision was 15Dec2021. This is AP's 11th published book of science.
Preface:
Actually my title is too modest, for the proof that lies within this book makes it the World's First Valid Proof of Fundamental Theorem of Calculus, for in my modesty, I just wanted to emphasis that calculus was geometry and needed a geometry proof. Not being modest, there has never been a valid proof of FTC until AP's 2015 proof. This also implies that only a geometry proof of FTC constitutes a valid proof of FTC.

Calculus needs a geometry proof of Fundamental Theorem of Calculus. But none could ever be obtained in Old Math so long as they had a huge mass of mistakes, errors, fakes and con-artist trickery such as the "limit analysis". And very surprising that most math professors cannot tell the difference between a "proving something" and that of "analyzing something". As if an analysis is the same as a proof. We often analyze various things each and every day, but few if none of us consider a analysis as a proof. Yet that is what happened in the science of mathematics where they took an analysis and elevated it to the stature of being a proof, when it was never a proof.

To give a Geometry Proof of Fundamental Theorem of Calculus requires math be cleaned-up and cleaned-out of most of math's mistakes and errors. So in a sense, a Geometry FTC proof is a exercise in Consistency of all of Mathematics. In order to prove a FTC geometry proof, requires throwing out the error filled mess of Old Math. Can the Reals be the true numbers of mathematics if the Reals cannot deliver a Geometry proof of FTC? Can the functions that are not polynomial functions allow us to give a Geometry proof of FTC? Can a Coordinate System in 2D have 4 quadrants and still give a Geometry proof of FTC? Can a equation of mathematics with a number that is _not a positive decimal Grid Number_ all alone on the right side of the equation, at all times, allow us to give a Geometry proof of the FTC?

Cover Picture: Is my hand written, one page geometry proof of the Fundamental Theorem of Calculus, the world's first geometry proof of FTC, 2013-2015, by AP.


Product details
ASIN ‏ : ‎ B07PQTNHMY
Publication date ‏ : ‎ March 14, 2019
Language ‏ : ‎ English
File size ‏ : ‎ 1309 KB
Text-to-Speech ‏ : ‎ Enabled
Screen Reader ‏ : ‎ Supported
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Lending ‏ : ‎ Enabled
Amazon Best Sellers Rank: #128,729 Paid in Kindle Store (See Top 100 Paid in Kindle Store)
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[Fritz Feldhase]

On Wednesday, August 31, 2022 at 1:36:22 PM UTC+2, WM wrote:

> the empty intersection cannot be accomplished by [only finitely many] endsegments

Actually, if M c IN, M =/= { } is finite:

> ∩{E(k) : k ∈ M} =/= { }.

If on the other hand, M c IN is infnite:

> ∩{E(k) : k ∈ M} = { }

Right, Mückenheim!

> (3) The simplest proof

that you are a crank

> is this:
>
> Every [...] natural number k is finite and belongs to a finite set

>
> {1, 2, 3, ..., k}.
>

> If there are ℵo [natural numbers], i.e., more than any finite number, then ℕ can can only be filled and completed by dark natural numbers.

Oh, really? Why no by the infinitely many "standard" natural numbers?

> This is obvious from the simple fact
>

> ∀k ∈ ℕ: |ℕ \ {1, 2, 3, ..., k}| = ℵo .

Huh?!

Hint: oo - n = oo for any "finite" number n.

You see: IN is *infinite*. {1, 2, 3, ..., k} is _finite_, for each and every k e IN. Hence IN \ {1, 2, 3, ..., k} is (still) *infinite*, for each and every k e IN.

Too hard for you to comprehend?

[Gus Gassmann]

(4) "Because I (WM) say so."

That's the only one you need, and the only one that cannot be refuted trivially.

[zelos...@gmail.com]

onsdag 31 augusti 2022 kl. 13:36:22 UTC+2 skrev WM:
> (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

Nope, very different things. One is a function from N to Q+, the other is trying to violate the definition of a 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.

Nope, because again you are trying to violate definitions and of course that is not gonna work.

>
> (2) The intersection of non-empty inclusion-monotonic sets like infinite endsegments E(k) = {k, k+1, k+2, ...} is not empty.

False, it is only non-emtpy for FINITE collections of end-segments. For INFINITE ones, it is empty.

>Every non-empty endsegment shares at least one natural number with all non-empty endsegments.

False. no matter which k you claim exist in all, E(k+1) lacks it.

>In fact every infinite endsegment shares infinitely many natural numbers with all infinite endsegments.

False, it shares NONE with all infinite endsegments. That is why it is empty.

>Otherwise there would be a first endsegment sharing less natural numbers with its predecessors.

Non-sequiter.

>This cannot happen, if all endsegments are infinite.

Or your conclusion is fucking wrong.

>
> But according to ZFC, the intersection of all endsegments is empty.

Because it is.

> Since all definable endsegments satisfy

"definable" has no meaning

>
> ∀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} =/= { }.

Made up shit for no reason.

>
> Only by the presence of undefinable endsegments
>
> ∩{E(k) : k ∈ ℕ} = { }
>
> can be accomplished.
>

Both of your attempts have failed so far.

> (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

And that was all horseshit.

[WM]

Ben Bacarisse schrieb am Mittwoch, 31. August 2022 um 17:09:18 UTC+2:
> WM <askas...@gmail.com> writes:
> (AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
> Unendlichen" at Hochschule Augsburg.)
> > (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. (*)
> 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?

Of course.

> > 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, ...
> > ...
> It's equivalent to drawing arrows between the Xs and every cell
> (including the Xs) so that every cell is at the beginning and end of
> just one arrow. You can see how this can be done, yes?

Of course. But why do you stop here? I continued: In fact by application of (*) all O's are removed from all visible or definable matrix positions.

This is what every student can prove.

> However it is clear that, by simple exchanging O's with X's, never an O will be removed from the matrix.

This is what mathematicians refuse to or cannot understand. Nevertheless it is true.

Regards, WM

[WM]

Gus Gassmann schrieb am Mittwoch, 31. August 2022 um 19:35:11 UTC+2:

> (4) "Because I (WM) say so."
>

Yes. This is so because I say so (and because every sober mind will agree): The intersection of non-empty inclusion-monotonic sets is not empty.

Regards, WM

[WM]

Fritz Feldhase schrieb am Mittwoch, 31. August 2022 um 19:15:55 UTC+2:
> On Wednesday, August 31, 2022 at 1:36:22 PM UTC+2, WM wrote:

> If on the other hand, M c IN is infnite:
>
> > ∩{E(k) : k ∈ M} = { }

The intersection of infinitely many sets need not be empty. Compare the intervals [0, 100 + 1/n]

> > (3) The simplest proof

> > Every [...] natural number k is finite and belongs to a finite set
> >
> > {1, 2, 3, ..., k}.
> >
> > If there are ℵo [natural numbers], i.e., more than any finite number, then ℕ can can only be filled and completed by dark natural numbers.
>
> Oh, really? Why no by the infinitely many "standard" natural numbers?

Because all of them satisfy

> > ∀k ∈ ℕ: |ℕ \ {1, 2, 3, ..., k}| = ℵo

and therefore cannot get rid of dark successors.

> Hint: oo - n = oo for any "finite" number n.

oo is not a number. Das Zeichen ∞, welches ich in Nr. 2 dieses Aufsatzes gebraucht habe, ersetze ich von nun an durch ω, weil das Zeichen ∞ schon vielfach zur Bezeichnung von unbestimmten [d. h. potentiellen] Unendlichkeiten verwandt wird. [Cantor]

>
> You see: IN is *infinite*. {1, 2, 3, ..., k} is _finite_, for each and every k e IN. Hence IN \ {1, 2, 3, ..., k} is (still) *infinite*, for each and every k e IN.

That shows the existence of dark numbers blowing up the set to infinity.

Regards, WM

[Gus Gassmann]

On Thursday, 1 September 2022 at 10:25:36 UTC-3, WM wrote:
> Gus Gassmann schrieb am Mittwoch, 31. August 2022 um 19:35:11 UTC+2:
>
> > (4) "Because I (WM) say so."
> >
> Yes. This is so because I say so

Good to see you made my point for me so readily.

> (and because every sober mind will agree):

This is again just (4). You really have not even any instinctual brainpower left.

> The intersection of non-empty inclusion-monotonic sets is not empty.

Not necessarily. You write SHIT as usual.

[Ben Bacarisse]

Because in the past you have denied that there is a bijection between N
and NxN. This is what NxN (or Q for that matter) being enumerated means
and I am glad that you agree. I stopped there because I am not
interested in the non-mathematical parts where things get "removed" from
matrices. That's a word game, and no fun for me.

At least we agree that NxN and Q are denumerable sets. That's a big
step for you.

--
Ben.

[Sergio]

On 9/1/2022 8:34 AM, WM wrote:
> Fritz Feldhase schrieb am Mittwoch, 31. August 2022 um 19:15:55 UTC+2:
>> On Wednesday, August 31, 2022 at 1:36:22 PM UTC+2, WM wrote:
>
>> If on the other hand, M c IN is infnite:
>>
>>> ∩{E(k) : k ∈ M} = { }
>
> The intersection of infinitely many sets need not be empty. Compare the intervals [0, 100 + 1/n]

do you understand why that does not make sense ? use Math instead, equations

what sets are you talking about ?

is it a sequence of sets ?

is your set what you call an "interval" above ?


intersection is common elements of sets under consideration

what elements are common to

[0, 100 + 1/n]

is that a set with 2 elements ?

0 is common to all, so the intersection is {0}


what if it was

[0, Cow, n + 1/n] ? ( for each n a natural number, from n= 1,2,...)

intersection = 0, Cow

what if I did not specify n ?

there would only be one set, right ?


>
> Regards, WM

What is the point of asking vague questions on simplistic matters?

you need a few years of refresher courses in sets.

[WM]

Ben Bacarisse schrieb am Donnerstag, 1. September 2022 um 16:24:05 UTC+2:
> WM <askas...@gmail.com> writes:
>
> > Ben Bacarisse schrieb am Mittwoch, 31. August 2022 um 17:09:18 UTC+2:
> >> WM <askas...@gmail.com> writes:
> >> (AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
> >> Unendlichen" at Hochschule Augsburg.)
> >> > (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. (*)

> >> > X, O, O, O, ...
> >> > X, O, O, O, ...
> >> > X, O, O, O, ...
> >> > X, O, O, O, ...
> >> > ...
> >> It's equivalent to drawing arrows between the Xs and every cell
> >> (including the Xs) so that every cell is at the beginning and end of
> >> just one arrow. You can see how this can be done, yes?
> >
> > Of course. But why do you stop here?
> Because in the past you have denied that there is a bijection between N
> and NxN.

The bijection as well as your arrows are restricted to only the definable numbers. That is the minority: |ℕ| = |{1, 2, 3, ..., n}| + ℵo . You can see that when you continue and ask for the lost O's.

> This is what NxN (or Q for that matter) being enumerated means
> and I am glad that you agree.

What I agreed to is what can be accomplished. But it is not what Cantor claimed.

> I stopped there because I am not
> interested in the non-mathematical parts where things get "removed" from
> matrices.

Not at all! Things don't get removed. Obviously no O' will be removed from the matrix. But no O can be found in the matrix after application of Cantor'sformula.

> At least we agree that NxN and Q are denumerable sets.

Collections to be precise. They can be put in bijection like all potentially infinte collections.

Regards, WM

[Ben Bacarisse]

Ah, I thought you had agreed that k(n,m) = (m + n - 1)(m + n - 2)/2 + m
is provably a bijection between NxN and N. Maybe you did not read what
you were agreeing with.

So it turns out k is /not/ a bijection between NxN and N but between
N_def x N_def and N_def (as I think you call it). How is that proved?
Is there a special form of induction?

Incidentally, how would a student know that it can't be proved for N?
If you show me the form of the proof for N_def, maybe you can point out
the step that fails for N? I'm always happy to learn some more WMaths.

--
Ben.

[Sergio]

On 9/1/2022 3:23 PM, WM wrote:
> Ben Bacarisse schrieb am Donnerstag, 1. September 2022 um 16:24:05 UTC+2:
>> WM <askas...@gmail.com> writes:
>>
>>> Ben Bacarisse schrieb am Mittwoch, 31. August 2022 um 17:09:18 UTC+2:
>>>> WM <askas...@gmail.com> writes:
>>>> (AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
>>>> Unendlichen" at Hochschule Augsburg.)
>>>>> (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. (*)
>
>>>>> X, O, O, O, ...
>>>>> X, O, O, O, ...
>>>>> X, O, O, O, ...
>>>>> X, O, O, O, ...
>>>>> ...
>>>> It's equivalent to drawing arrows between the Xs and every cell
>>>> (including the Xs) so that every cell is at the beginning and end of
>>>> just one arrow. You can see how this can be done, yes?
>>>
>>> Of course. But why do you stop here?
>> Because in the past you have denied that there is a bijection between N
>> and NxN.
>
> The bijection as well as your arrows are restricted to only the definable numbers.

Wrong. WM uses one of his magic words, "definable" which is meaningless, to miss direct.

>
>> This is what NxN (or Q for that matter) being enumerated means
>> and I am glad that you agree.
>
> What I agreed to is what can be accomplished.

which is not math.

> But it is not what Cantor claimed.

(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. (*)

and you cannot un-prove it.

>
>> I stopped there because I am not
>> interested in the non-mathematical parts where things get "removed" from
>> matrices.
>
> Not at all! Things don't get removed. Obviously no O' will be removed from the matrix.

obviously, by WM they are swaparooed, another shell game to prove nothing exists.


> Regards, WM

[zelos...@gmail.com]

That is what a crazy person that doesn't understand mathematics will say because in sensible mathematics where logical people work, there is no such implication

[zelos...@gmail.com]

It shows nothing of the sort you moron! Why do you keep thinking it does like a retard!?

[WM]

That is provable, but only for the definable part.

> Maybe you did not read what
> you were agreeing with.

Maybe you did not read what I wrote later on.

> So it turns out k is /not/ a bijection between NxN and N but between
> N_def x N_def and N_def (as I think you call it). How is that proved?
> Is there a special form of induction?

It turns out that all visible positions of the matrix are covered by X's. Cantor has proved this for ll fractions he could see:
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, ...
and many, many more. That means all visible fractions are indexed. That means there is a bijection between visible natnumbers and visible fractions.

But never an O disappears from the matrix. It disappears from the visible part only. This proves fractions without index. (It is an extremely silly approach to assume limits here where the O's suddenly disappear. In the same way the disturbing digits of Cantor's diagonal number could disappear. I will not accept such a silly explanation.)

>
> Incidentally, how would a student know that it can't be proved for N?

First he would think that all indices (X) can be applied to fractions. But he would see that never an O disappears from the matrix. In all columns there are more O's then ever will gather in the first column (in exchange for the X'sinitially residing there).

That shows him that fractions without index remain, but they cannot be identified. They are dark. This proves the existence of dark fractions. Now it is only a little step to understand that also the integer fractions and the natural numbers are mainly dark, and that all actually infinite sets contain mainly dark elements.

∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo

> If you show me the form of the proof for N_def, maybe you can point out
> the step that fails for N? I'm always happy to learn some more WMaths.

It is a pleasure for me.

Regards, WM

[WM]

zelos...@gmail.com schrieb am Freitag, 2. September 2022 um 11:26:23 UTC+2:
> torsdag 1 september 2022 kl. 15:25:36 UTC+2 skrev WM:
> > Gus Gassmann schrieb am Mittwoch, 31. August 2022 um 19:35:11 UTC+2:
> >
> > > (4) "Because I (WM) say so."
> > >
> > Yes. This is so because I say so (and because every sober mind will agree): The intersection of non-empty inclusion-monotonic sets is not empty.
> >

> That is what a crazy person that doesn't understand mathematics will say because in sensible mathematics where logical people work, there is no such implication

The states of water flowing out of a bathub will never have an empty intersection before the bathtub is empty. "Mathematics" contradicting this basic principle is rubbish.

Regards, WM

[Sergio]

here comes the misleading statement;

>but only for the definable part.


>
>> Maybe you did not read what
>> you were agreeing with.
>
> Maybe you did not read what I wrote later on.

irrelevant.

>
>> So it turns out k is /not/ a bijection between NxN and N but between
>> N_def x N_def and N_def (as I think you call it). How is that proved?
>> Is there a special form of induction?
>
> It turns out that all visible positions of the matrix are covered by X's. Cantor has proved this for ll fractions he could see:

liar.

<snip crap>

[Sergio]

do you realize this has *nothing at all* to do with the intersection of all endsegments being empty ?

Of course you do! *Diversion and deception is what you do*.


instead of bathtubs, use oil drums, or fish boats, or red herrings...

>> "Mathematics" contradicting this basic principle is rubbish.

perfect word to describe WM maths, it's rubbish. It isnt Math, it is Pretend Math at best.

>
> Regards, WM

[Ben Bacarisse]

I.e. it's /not/ provable for NxN to N. That's surprising, given what
your textbook says, but you are the only person know really knows
WMaths.

>> So it turns out k is /not/ a bijection between NxN and N but between
>> N_def x N_def and N_def (as I think you call it). How is that proved?
>> Is there a special form of induction?

You don't want to show me how it's proved? OK, I have no way to
encourage you, but I thought you might like to demonstrate how WMaths
works for actual proofs of bijections.

> It turns out that all visible positions of the matrix are covered by X's. Cantor has proved this for ll fractions he could see:
> 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, ...
> and many, many more. That means all visible fractions are
> indexed. That means there is a bijection between visible natnumbers
> and visible fractions.

Your textbook does not explain what a visible number is so I don't know
what you mean. It does talk about N and functions that are (or are not)
bijections so I thought that someone with your book could prove that
k(n,m) is a bijection between NxN and N. Which condition fails

(a) is k not provably injective?
(b) is k not provably surjective?

--
Ben.

[WM]

Ben Bacarisse schrieb am Freitag, 2. September 2022 um 18:00:32 UTC+2:

> WM <askas...@gmail.com> writes:
>
> > That means there is a bijection between visible natnumbers
> > and visible fractions.
> Your textbook does not explain what a visible number is

My textbook, like claisscal maths deals only with visible numbers. No reason to mention this. Further in 2015 when it was published I did not know about dark numbers.

> so I don't know
> what you mean.

But you know and agree that all visible places of the matrix are occupied by X's and nevertheless no O disappears from the matrix?

Regards, WM

[Ben Bacarisse]

WM <askas...@gmail.com> writes:

> Ben Bacarisse schrieb am Freitag, 2. September 2022 um 18:00:32 UTC+2:
>> WM <askas...@gmail.com> writes:
>>
>> > That means there is a bijection between visible natnumbers
>> > and visible fractions.
>> Your textbook does not explain what a visible number is
>
> My textbook, like claisscal maths deals only with visible numbers. No
> reason to mention this. Further in 2015 when it was published I did
> not know about dark numbers.

I have always been talking about the maths in your book. Did I not make
that clear enough? Sorry.

If your book is no longer adequate for this task, please say so, but I
hope it is because you are not good at answering direct questions, but
the book is there for all to see. What I wanted to know is:

(a) Is k (now cut) a function from NxN to N as the set N and the term
function are defined in your book?

(b) Is k surjective (as defined in your book)?

(c) Is k injective (as defined in your book)?

(d) Will I have to keep wrting "as define in your book" or can we that
that is implicit in everything you and I write in this thread?

--
Ben.

[Eram semper recta]

On Friday, 2 September 2022 at 09:23:22 UTC-7, WM wrote:
> Ben Bacarisse schrieb am Freitag, 2. September 2022 um 18:00:32 UTC+2:
> > WM <askas...@gmail.com> writes:
> >
> > > That means there is a bijection between visible natnumbers
> > > and visible fractions.
> > Your textbook does not explain what a visible number is
> My textbook, like claisscal maths deals only with visible numbers.

I think you need to cease introducing new names.

Numbers are not visible because a number is not something that can be seen.

Realisable and non-realisable numbers are better terms than "visible" and "dark".

You make a mockery of yourself every time you use such terms and reduce the credibility of your otherwise good arguments.

Oh, and the junk concept of "infinity" cannot be interpreted as "direction" because direction is a well-defined word:

A path or course along which something moves.

Please, we already have enough charlatans and cranks in the mainstream Church of Mathematics academia. Use words that explain what you mean.

[WM]

Ben Bacarisse schrieb am Freitag, 2. September 2022 um 18:42:47 UTC+2:
> WM <askas...@gmail.com> writes:
>
> > Ben Bacarisse schrieb am Freitag, 2. September 2022 um 18:00:32 UTC+2:
> >> WM <askas...@gmail.com> writes:
> >>
> >> > That means there is a bijection between visible natnumbers
> >> > and visible fractions.
> >> Your textbook does not explain what a visible number is
> >
> > My textbook, like claisscal maths deals only with visible numbers. No
> > reason to mention this. Further in 2015 when it was published I did
> > not know about dark numbers.
> I have always been talking about the maths in your book. Did I not make
> that clear enough? Sorry.
>
> If your book is no longer adequate for this task, please say so, but I
> hope it is because you are not good at answering direct questions, but
> the book is there for all to see. What I wanted to know is:
>
> (a) Is k (now cut) a function from NxN to N as the set N and the term
> function are defined in your book?

In my book there are, as I clearly stated in the introduction, only potentially infinite sets. That are the visible numbers which are sufficient to do classical mathematics.

> (b) Is k surjective (as defined in your book)?
>
> (c) Is k injective (as defined in your book)?
>

Of course.

But Cantor talked about actual infinity. I notice that you don't wish to consider Cantor's mapping. There we have the complete matrix of positive fractions and the complete column of integer fractions:

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, ...

5/1, 5/2, 5/3, 5/4, ...
...

Cantor's approach is modelled by exchanging X's and O's in

XOO...
XOO...
XOO...
...

until all O's have disappeared. Do you agree? Or did I misinterpret Cantor? Or do you prefer not to get involved in that question because it could be too dangerous to become a heretic matheologian? I could understand your hesitation. Even bold mathematicians like Doron Zeilberger could not resist the strong forces of the imperialists of matheology.

Regards, WM

[Sergio]

On 9/2/2022 3:31 PM, WM wrote:
> Ben Bacarisse schrieb am Freitag, 2. September 2022 um 18:42:47 UTC+2:
>> WM <askas...@gmail.com> writes:
>>
>>> Ben Bacarisse schrieb am Freitag, 2. September 2022 um 18:00:32 UTC+2:
>>>> WM <askas...@gmail.com> writes:
>>>>
>>>>> That means there is a bijection between visible natnumbers
>>>>> and visible fractions.
>>>> Your textbook does not explain what a visible number is
>>>
>>> My textbook, like claisscal maths deals only with visible numbers. No
>>> reason to mention this. Further in 2015 when it was published I did
>>> not know about dark numbers.
>> I have always been talking about the maths in your book. Did I not make
>> that clear enough? Sorry.
>>
>> If your book is no longer adequate for this task, please say so, but I
>> hope it is because you are not good at answering direct questions, but
>> the book is there for all to see. What I wanted to know is:
>>
>> (a) Is k (now cut) a function from NxN to N as the set N and the term
>> function are defined in your book?
>
> In my book there are, as I clearly stated in the introduction, only potentially infinite sets. That are the visible numbers which are sufficient to do classical mathematics.

so your book does not apply to "modern" math, but only "visible" Ants.

> Or did I misinterpret Cantor?

yes, it is the one thing you do consistantly

> Regards, WM

[Ben Bacarisse]

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

> Ben Bacarisse schrieb am Freitag, 2. September 2022 um 18:42:47 UTC+2:
>> WM <askas...@gmail.com> writes:
>>
>> > Ben Bacarisse schrieb am Freitag, 2. September 2022 um 18:00:32 UTC+2:
>> >> WM <askas...@gmail.com> writes:
>> >>
>> >> > That means there is a bijection between visible natnumbers
>> >> > and visible fractions.
>> >> Your textbook does not explain what a visible number is
>> >
>> > My textbook, like claisscal maths deals only with visible numbers. No
>> > reason to mention this. Further in 2015 when it was published I did
>> > not know about dark numbers.
>> I have always been talking about the maths in your book. Did I not make
>> that clear enough? Sorry.
>>
>> If your book is no longer adequate for this task, please say so, but I
>> hope it is because you are not good at answering direct questions, but
>> the book is there for all to see. What I wanted to know is:
>>
>> (a) Is k (now cut) a function from NxN to N as the set N and the term
>> function are defined in your book?
>
> In my book there are, as I clearly stated in the introduction, only
> potentially infinite sets. That are the visible numbers which are
> sufficient to do classical mathematics.
>
>> (b) Is k surjective (as defined in your book)?
>>
>> (c) Is k injective (as defined in your book)?
>>
> Of course.

Great. So k /is/ a bijection (with everything -- functions, bijections,
N, NxN -- as defined as in your book).

> But Cantor talked about actual infinity. I notice that you don't wish
> to consider Cantor's mapping.

Let's see how far we can get just with WMaths. The bijection you wrote

k(n,m) = (m + n - 1)(m + n - 2)/2 + m

will let us write the effect of a sequence of swaps in terms of a
sequence of WMaths functions. We may hit the rocks and find that WMaths
can not take us where we want to go, but let's try. Are you up for
that?

> There we have the complete matrix of positive fractions and the
> complete column of integer fractions:
>
> 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, ...
> 5/1, 5/2, 5/3, 5/4, ...
> ...
>
> Cantor's approach is modelled by exchanging X's and O's in
>
> XOO...
> XOO...
> XOO...
> ...
>
> until all O's have disappeared. Do you agree?

Here are the cells as numbered by your bijective mapping:

1, 3, 6, 10, 15, ...
2, 5, 9, 14, 20, ...
4, 8, 13, 19, 26, ...
7, 12, 18, 25, 33, ...
11, 17, 24, 32, 41, ...
...

With Os at 3, 5, 6, 8, 9 and so on. Let's use 1 and 0 for Xs and Os so
the matrix can be written as a function from N to {0,1} (as defined in
your book). Since these are functions of N, let's just write the first
few values rather that trying to find the ever-so messy formulas for
where there are 1s and 0s:

M_0(n) = 1, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, ...

Now what are the first few swaps? The logical choice would be to swap
the first 0 with the first following 1 so that

M_1(n) = 1, 1, 1, 0, 0, 0, 1, 0, 0, 0, 1, ...
M_2(n) = 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 1, ...
M_2(n) = 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, ...
...

I think this sequence of functions makes the pattern nice ans clear and
all the functions are just the sort of functions you manipulate in your
book. Would you like these turned back into X and O matrices? I don't
think it helps, so I haven't bothered.

Now the only meaning for the endless swapping that makes sense to me is
a limit, and fortunately you tell your students how to prove that

lim_{n->oo} M_n = M_1

where M_1 is the constant function M_1(k) = 1. You give a nice
definition of convergence for sequences of functions in your textbook,
and the function sequence M_n does indeed converge to that limit.

> Or did I misinterpret Cantor?

I don't know. Do you have a citation for his discussion of this matrix?
I does not have the neat structure of most of his arguments so I don't
think you've got it right.

> Or do you prefer not to get involved in that question because it could
> be too dangerous to become a heretic matheologian?

I'm not sure what you think is dangerous. The WMaths limit of the
WMaths function sequence M_n has no 0s in it, and that seems like the
only reasonable meaning for the effect of an endless sequence of swaps.
It seems that WMaths is up to the task. If that makes me a heretic, I
really don't care.

> I could understand your hesitation.

My only hesitancy was due to your flip-flopping about what is and is not
provable. If k (as you defined it) is not a bijection (as you define
the term in your book) between NxN (as you define it in your book) and N
(as defined in your book) then I could not even start formalising, in
WMaths, what this endless sequence of swaps might look like. Even now,
a feel sure a flip or a flop coming on... You find it hard to stick to
talking about WMaths for some reason.

But I note that /you/ still hesitate to show how a function is proved to
be bijective in WMaths. Well, hesitate is a rather weak term because I
have been asking, on and off, for /years/. It's almost as if you don't
know or are, for some reason, afraid to reveal how simple it is in
WMaths. I'm not expecting your hesitation to vanish, but I live in
hope so I am asking again.

--
Ben.

[WM]

Ben Bacarisse schrieb am Freitag, 2. September 2022 um 18:42:47 UTC+2:
> WM <askas...@gmail.com> writes:
>
> > Ben Bacarisse schrieb am Freitag, 2. September 2022 um 18:00:32 UTC+2:
> >> WM <askas...@gmail.com> writes:
> >>
> >> > That means there is a bijection between visible natnumbers
> >> > and visible fractions.
> >> Your textbook does not explain what a visible number is
> >
> > My textbook, like claisscal maths deals only with visible numbers. No
> > reason to mention this. Further in 2015 when it was published I did
> > not know about dark numbers.
> I have always been talking about the maths in your book. Did I not make
> that clear enough? Sorry.

I have been talking about Cantor's enumeration of fractions. My book does nt mention that with good reason.

>
> If your book is no longer adequate for this task,

My book, like all reasonable mathematics, is a good basis to discuss Cantor's claim. That is done here. But the basic assumption that Cantor has made must be accepted in order to discuss it.

Regards, WM

[WM]

Ben Bacarisse schrieb am Samstag, 3. September 2022 um 00:08:10 UTC+2:

> WM <askas...@gmail.com> writes:

> > There we have the complete matrix of positive fractions and the
> > complete column of integer fractions:
> >
> > 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, ...
> > 5/1, 5/2, 5/3, 5/4, ...
> > ...
> >
> > Cantor's approach is modelled by exchanging X's and O's in
> >
> > XOO...
> > XOO...
> > XOO...
> > ...
> >
> > until all O's have disappeared. Do you agree?
> Here are the cells as numbered by your bijective mapping:
>
> 1, 3, 6, 10, 15, ...
> 2, 5, 9, 14, 20, ...
> 4, 8, 13, 19, 26, ...
> 7, 12, 18, 25, 33, ...
> 11, 17, 24, 32, 41, ...
> ...

Yes, all visible cells are filled with X's as Cantor has taught us.

>
> With Os at 3, 5, 6, 8, 9 and so on.

No, the above matrix does not contain O's any longer. You have already indexed all visible cells. That means they contain X's. But you have not applied Cantor's formula correctly.

Here is it again: k = (m + n - 1)(m + n - 2)/2 + m yields 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, ...

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, ...
5/1, 5/2, 5/3, 5/4, ...
...

Integer fractions are denoted by X's:

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

In the first 3 steps we get the following matrices

XXOO...
OOOO...
XOOO...
XOOO...
...

XXOO...
XOOO...
OOOO...
XOOO...
...

XXXO...
XOOO...
OOOO...
OOOO...
...

> Now the only meaning for the endless swapping that makes sense to me is
> a limit,

Cantor does not use limits. Nobody would have believed him, if he claimed completeness of enumeration "in the limit".

and fortunately you tell your students how to prove that
>
> lim_{n->oo} M_n = M_1

Here we have no limit but a never disappearing set of O's. That is the question: How can the O's disappear without leaving the matrix!
By the way, every X comes to a position where it has infinitely many O's right to it and below it. If there is a limit, then this limit contains infinitely many O's.

> > Or did I misinterpret Cantor?
> I don't know. Do you have a citation for his discussion of this matrix?

No, the citation is written above. The matrix is but another language, expressing precisely what Cantor said.

> I does not have the neat structure of most of his arguments so I don't
> think you've got it right.

Then think over it again. I apply his formula k = (m + n - 1)(m + n - 2)/2 + m with his result in all visible cells: 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, ... But never an O disappears.

Regards, WM

[Sergio]

you use Cantors name over and over again, to try to make your math seem important.

However, your math is too riddled with obvious errors and bad math to ever be serious.

you put your PHD Ego on a subject you know nothing about. [I would replace "nothing" with "little" if you knew algebra, but you do not.]

[Sergio]

distraction => the first column is not your indexes in your matrix, you made mistakes

>
>>> Or did I misinterpret Cantor?
>> I don't know. Do you have a citation for his discussion of this matrix?
>
> No, the citation is written above. The matrix is but another language, expressing precisely what Cantor said.

*Liar*
1. you have no citation to a matrix by Cantor.
2. Your matrix does not agree with Cantor, so you made mistakes.

>
>> I does not have the neat structure of most of his arguments so I don't
>> think you've got it right.
>
> Then think over it again. I apply his formula k = (m + n - 1)(m + n - 2)/2 + m with his result in all visible cells: 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, ... But never an O disappears.

Fix your mistakes

>
> Regards, WM
>

[Ben Bacarisse]

No. I've just written some numbers down to illustrate what you
bijective mapping k looks like.

>> With Os at 3, 5, 6, 8, 9 and so on.
>
> No, the above matrix does not contain O's any longer.

I think you'd got lost at the first step. I'm using your mapping to
give write the matrix as a function from N to {0,1}. The initial matrix
has both Xs and Os. In fact it's simply this function:

M_0(i) = 1 if m = 1 and 0 otherwise where (n,m) = k'(i)

Since this is the first step, you really need to be able to see how M_0
represents the initial matrix a WMaths function from N to {0,1}. No
completed anything, everything potential.

>> Now the only meaning for the endless swapping that makes sense to me is
>> a limit,
>
> Cantor does not use limits.

I know. Do you just skim posts for keywords rather than reading them?
I said we can get rid of the Os in WMaths without using anything from
Cantor, just the maths in your book. You seemed to have stopped talking
about WMaths at the first sentence above.

>> lim_{n->oo} M_n = M_1
>
> Here we have no limit but a never disappearing set of O's. That is the
> question: How can the O's disappear without leaving the matrix!

If you want to grasp how it's done, you'd have to stick with nothing but
the maths in your book and you'll have to ask intelligent questions when
you don't follow a step. You seemed to get lost at the very start by
thinking I was talking about something other than your WMaths.

You are not good a being a student, but I could set a sequence of
exercises, based on your book, that would take you through the steps to
see how the Os "disappear" without anything from Cantor -- just WMaths.
Would you like me to do that?

>> > Or did I misinterpret Cantor?
>> I don't know. Do you have a citation for his discussion of this matrix?
>
> No, the citation is written above.

What does "no but it's above mean"? You know what a citation is? Just
say if you made it up..

--
Ben.

[WM]

Ben Bacarisse schrieb am Sonntag, 4. September 2022 um 03:33:25 UTC+2:

> I said we can get rid of the Os in WMaths without using anything from
> Cantor, just the maths in your book.

And that is clearly wrong because, according to every kind of mathematics, never an O leaves the matrix. The only change of the initial matrix

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

consists of exchanging an O and an X. Exchanging two symbols does not delete any of them.

> I could set a sequence of
> exercises, based on your book, that would take you through the steps to
> see how the Os "disappear" without anything from Cantor -- just WMaths.
> Would you like me to do that?

Please try it.

> >> > Or did I misinterpret Cantor?
> >> I don't know. Do you have a citation for his discussion of this matrix?
> >
> > No, the citation is written above.
> What does "no but it's above mean"?

Cantor said that all fractions can be indexed. That means translated to my model that all O's disappear. Can you follow that thought?

Regards, WM

[Gus Gassmann]

On Sunday, 4 September 2022 at 07:13:35 UTC-3, WM wrote:
> Ben Bacarisse schrieb am Sonntag, 4. September 2022 um 03:33:25 UTC+2:
>
> > I said we can get rid of the Os in WMaths without using anything from
> > Cantor, just the maths in your book.
> And that is clearly wrong because, according to every kind of mathematics, never an O leaves the matrix. The only change of the initial matrix
> XOOO...
> XOOO...
> XOOO...
> XOOO...
> ...
> consists of exchanging an O and an X. Exchanging two symbols does not delete any of them.

Nothing but your usual bullshit, in other words. Aside from "Because I (WM) say so", you got nothing. In particular, you have no clue about basic logic. In order to prove an inconsistency in ZFC (or any other axiom system) it is simply not enough to assume the negation of one of the axioms to derive an inconsistency. In your case, assuming that infinity (ω, or any of the alephs) does not exist is not enough to show that ZFC is inconsistent. If you had two firing synapses left, you'd understand that that. So it seems best that you just shut the fuck up and piss off.

[Sergio]

On 9/4/2022 5:13 AM, WM wrote:
> Ben Bacarisse schrieb am Sonntag, 4. September 2022 um 03:33:25 UTC+2:
>
>> I said we can get rid of the Os in WMaths without using anything from
>> Cantor, just the maths in your book.
>
> And that is clearly wrong because, according to every kind of mathematics, never an O leaves the matrix. The only change of the initial matrix
>
> XOOO...
> XOOO...
> XOOO...
> XOOO...
> ...
>
> consists of exchanging an O and an X.

>

> Cantor said that all fractions can be indexed. That means translated to my model that all O's disappear. Can you follow that thought?

Wrong, *Cantor Proved all fractions can be indexed using math*.

You cannot un-prove that.

So YOU have made mistakes in your model, and *your model is not based in Math at all*. Fail.


>
> Regards, WM

[WM]

Gus Gassmann schrieb am Sonntag, 4. September 2022 um 14:39:51 UTC+2:
> On Sunday, 4 September 2022 at 07:13:35 UTC-3, WM wrote:
> > Ben Bacarisse schrieb am Sonntag, 4. September 2022 um 03:33:25 UTC+2:
> >
> > > I said we can get rid of the Os in WMaths without using anything from
> > > Cantor, just the maths in your book.
> > And that is clearly wrong because, according to every kind of mathematics, never an O leaves the matrix. The only change of the initial matrix
> > XOOO...
> > XOOO...
> > XOOO...
> > XOOO...
> > ...
> > consists of exchanging an O and an X. Exchanging two symbols does not delete any of them.

> In particular, you have no clue about basic logic.

I have this understanding of logic: If two symbols are exchanged, then none of them is deleted.

> A In order to prove an inconsistency in ZFC

it issufficient to prove that no O can leave the matrix.

Regards, WM

[Gus Gassmann]

On Sunday, 4 September 2022 at 10:55:28 UTC-3, WM wrote:
[...]

> I have this understanding of logic: If two symbols are exchanged, then none of them is deleted.

This is correct, but insufficient.

> > A In order to prove an inconsistency in ZFC
>
> it issufficient to prove that no O can leave the matrix.

Ah, no. The fact that the reciprocal 1/n is positive for every natural number n is insufficient to conclude that the limit of the sequence {1, 1/2, 1/3, ..} cannot be 0.

[Sergio]

On 9/4/2022 8:55 AM, WM wrote:
> Gus Gassmann schrieb am Sonntag, 4. September 2022 um 14:39:51 UTC+2:
>> On Sunday, 4 September 2022 at 07:13:35 UTC-3, WM wrote:
>>> Ben Bacarisse schrieb am Sonntag, 4. September 2022 um 03:33:25 UTC+2:
>>>
>>>> I said we can get rid of the Os in WMaths without using anything from
>>>> Cantor, just the maths in your book.
>>> And that is clearly wrong because, according to every kind of mathematics, never an O leaves the matrix. The only change of the initial matrix
>>> XOOO...
>>> XOOO...
>>> XOOO...
>>> XOOO...
>>> ...
>>> consists of exchanging an O and an X. Exchanging two symbols does not delete any of them.
>> In particular, you have no clue about basic logic.
>
> I have this understanding of logic: If two symbols are exchanged, then none of them is deleted.

your exchange, or swapparoo, is a step by step process, which you have admitted before does not work on infinite sets.

>
>> A In order to prove an inconsistency in ZFC
>

> it is sufficient to prove that no O can leave the matrix.

no, it is wrong for infinite sets.

>
> Regards, WM

now correct your mistakes and stop trying to mislead people.

You still get an F, for Fail.

[Ben Bacarisse]

WM <askas...@gmail.com> writes:

> Ben Bacarisse schrieb am Sonntag, 4. September 2022 um 03:33:25 UTC+2:
>
>> I said we can get rid of the Os in WMaths without using anything from
>> Cantor, just the maths in your book.
>
> And that is clearly wrong because,

Then you will need to correct your book (again). But you won't see what
needs correcting until you follow the argument I presented based solely
on the maths in your textbook.

> according to every kind of
> mathematics, never an O leaves the matrix. The only change of the
> initial matrix
>
> XOOO...
> XOOO...
> XOOO...
> XOOO...
> ...
>
> consists of exchanging an O and an X. Exchanging two symbols does not
> delete any of them.

And yet, in WMaths, the result of an endless sequence of swaps is a matrix,
conveniently written as a function from N to {0,1}, has no 0s in the
image. Either you need to re-write something in your book or you should
define, more precisely what you mean by the result of an endless
sequence of swaps.

>> I could set a sequence of
>> exercises, based on your book, that would take you through the steps to
>> see how the Os "disappear" without anything from Cantor -- just WMaths.
>> Would you like me to do that?
>
> Please try it.

OK. (1) Use your k to write the initial matrix you show a bit of above
as a function from N to {0,1}. If there are terms you are not sure
about, please ask. Don't use any maths not in your book.

I can break this down a bit if it's too big an initial step.

>> >> > Or did I misinterpret Cantor?
>> >> I don't know. Do you have a citation for his discussion of this matrix?
>> >
>> > No, the citation is written above.
>> What does "no but it's above mean"?
>
> Cantor said that all fractions can be indexed. That means translated
> to my model that all O's disappear. Can you follow that thought?

OK, you don't have a citation. I didn't really think you did.

--
Ben.

[WM]

Gus Gassmann schrieb am Sonntag, 4. September 2022 um 16:20:32 UTC+2:
> On Sunday, 4 September 2022 at 10:55:28 UTC-3, WM wrote:
> [...]
> > I have this understanding of logic: If two symbols are exchanged, then none of them is deleted.
> This is correct, but insufficient.

It is all that is needed.

> > > A In order to prove an inconsistency in ZFC
> >
> > it issufficient to prove that no O can leave the matrix.
> Ah, no. The fact that the reciprocal 1/n is positive for every natural number n is insufficient to conclude that the limit of the sequence {1, 1/2, 1/3, ..} cannot be 0.

The limit of a sequence is a number which is approached. The absence of all O's is not a state which is approached. The sequence ℵo, ℵo, ℵo, ... has limit ℵo, not 0.

Regards, WM

[WM]

Ben Bacarisse schrieb am Sonntag, 4. September 2022 um 18:14:07 UTC+2:

> WM <askas...@gmail.com> writes:

> > according to every kind of
> > mathematics, never an O leaves the matrix. The only change of the
> > initial matrix
> >
> > XOOO...
> > XOOO...
> > XOOO...
> > XOOO...
> > ...
> >
> > consists of exchanging an O and an X. Exchanging two symbols does not
> > delete any of them.
> And yet, in WMaths, the result of an endless sequence of swaps is a matrix,
> conveniently written as a function from N to {0,1}, has no 0s in the
> image.

You are lying.

> Either you need to re-write something in your book or you should
> define, more precisely what you mean by the result of an endless
> sequence of swaps.

The result of an endless sequence ℵo, ℵo, ℵo, ... is ℵo.

> OK. (1) Use your k to write the initial matrix you show a bit of above
> as a function from N to {0,1}. If there are terms you are not sure

> about, please ask. Don't use any maths not in your bo000ok.

I prefer to write X and O in the way I did. Nothing in my book contradicts that.

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

XXOO...
OOOO...
XOOO...
XOOO...
...

XXOO...
XOOO...
OOOO...
XOOO...
...

XXXO...
XOOO...
OOOO...
OOOO...
...

...

> > Cantor said that all fractions can be indexed. That means translated
> > to my model that all O's disappear. Can you follow that thought?
> OK, you don't have a citation. I didn't really think you did.

Here is the citation. But you need a little bit of brain to derive the above sequence of matrices.

So the set of all positive rational numbers is countable as was shown by Cantor [Cantor, p. 126. G. Cantor, letter to R. Lipschitz (19 Nov 1883)] by this 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, ...

where the sum s of numerator and denominator grows stepwise by 1, and for fixed sum the numerator grows stepwise by 1 from minimum 1 to maximum s - 1. If repeating values of fractions are eliminated, every positive rational number appears exactly once, if not, we get the sequence of all positive fractions where k = (m + n - 1)(m + n - 2)/2 + m is the index or position of fraction m/n. [Cantor, p. 132]

Regards, WM

[Ben Bacarisse]

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

> Ben Bacarisse schrieb am Sonntag, 4. September 2022 um 18:14:07 UTC+2:
>> WM <askas...@gmail.com> writes:
>
>> > according to every kind of
>> > mathematics, never an O leaves the matrix. The only change of the
>> > initial matrix
>> >
>> > XOOO...
>> > XOOO...
>> > XOOO...
>> > XOOO...
>> > ...
>> >
>> > consists of exchanging an O and an X. Exchanging two symbols does not
>> > delete any of them.
>> And yet, in WMaths, the result of an endless sequence of swaps is a matrix,
>> conveniently written as a function from N to {0,1}, has no 0s in the
>> image.
>
> You are lying.

I showed you how. You flipped immediately back into whatever you are
calling maths that isn't yours these day. You appeared not to have
grasped even the first stage of my argument. Or maybe you did grasp it
and you decided you had to do something to avoid considering it.

>> Either you need to re-write something in your book or you should
>> define, more precisely what you mean by the result of an endless
>> sequence of swaps.
>
> The result of an endless sequence ℵo, ℵo, ℵo, ... is ℵo.

We're doing this with the maths in your book. I don't think ℵo is
defined in you textbook, is it? Anyway, that's not the result given by
WMaths as my argument showed.

>> OK. (1) Use your k to write the initial matrix you show a bit of above
>> as a function from N to {0,1}. If there are terms you are not sure

>> about, please ask. Don't use any maths not in your book.

>
> I prefer to write X and O in the way I did. Nothing in my book
> contradicts that.

Indeed. It all works in WMaths. If do decide to take me up on the
offer, and do the exercises, you'll see how it's done.

When you said "please try it" in reply to may offer of taking you
through some exercises to explain the argument, I thought you were
agreeing. But "please try it" could also be a challenge: "just try it
and see how for you get, mate"!

The offer still stands and, as I said, I can break it down (into two) if
you think that would help.

>> OK, you don't have a citation. I didn't really think you did.
>
> Here is the citation.

<rudeness cut>

> So the set of all positive rational numbers is countable as was shown
> by Cantor [Cantor, p. 126. G. Cantor, letter to R. Lipschitz (19 Nov
> 1883)] by this 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, ...
>
> where the sum s of numerator and denominator grows stepwise by 1, and
> for fixed sum the numerator grows stepwise by 1 from minimum 1 to
> maximum s - 1. If repeating values of fractions are eliminated, every
> positive rational number appears exactly once, if not, we get the
> sequence of all positive fractions where k = (m + n - 1)(m + n - 2)/2
> + m is the index or position of fraction m/n. [Cantor, p. 132]

Right, so you are citing something other than Cantor discussing swapping
Xs and Os in a grid. I'd really have like to see how Cantor defined the
result of and endless sequence of swaps. Maybe he does in some other
work. Do you know of any?

--
Ben.

[Tom Bola]

Ben Bacarisse schrieb:

WM does not want our math where infinity exists and he teaches this
credo to all people who are "believing" that infinity is "allowable"
which he calls dumb folks. This is WM's principal fight for many
decades now and it is the only matter in his life which he really
"posesses" as a human creature and which he is engaged in round clock.
He will never give up this "devotion" because that would also mean
to erase nearly all content in his low-IQ brain. But be sure that
he will use your great work, Ben, in a deformed shape in his further
publications. Just as an aside...

[JVR]

There is only one immutable rule in what you call WMaths and I call Muckmeatics:
"You cannot nail a pudding to the wall."
Once you know that you know everything worth knowing about this subject,
which is best described as pseudo-mathematical polemics; or, of you prefer,
polemical pseudo-mathematics.

[WM]

Ben Bacarisse schrieb am Sonntag, 4. September 2022 um 21:26:47 UTC+2:
> WM <askas...@gmail.com> writes:
> (AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
> Unendlichen" at Hochschule Augsburg.)
> > Ben Bacarisse schrieb am Sonntag, 4. September 2022 um 18:14:07 UTC+2:
> >> WM <askas...@gmail.com> writes:
> >
> >> > according to every kind of
> >> > mathematics, never an O leaves the matrix. The only change of the
> >> > initial matrix
> >> >
> >> > XOOO...
> >> > XOOO...
> >> > XOOO...
> >> > XOOO...
> >> > ...
> >> >
> >> > consists of exchanging an O and an X. Exchanging two symbols does not
> >> > delete any of them.
> >> And yet, in WMaths, the result of an endless sequence of swaps is a matrix,
> >> conveniently written as a function from N to {0,1}, has no 0s in the
> >> image.
> >
> > You are lying.
> I showed you how.

You cannot show it because it is wrong. It is nonsense to believe that by exchanging two items, one of them disappears.

> You flipped immediately back into whatever you are
> calling maths that isn't yours these day. You appeared not to have
> grasped even the first stage of my argument.

Your argument started with a wrong image of Cantor's matrix which I corrected.

> > The result of an endless sequence ℵo, ℵo, ℵo, ... is ℵo.
> We're doing this with the maths in your book. I don't think ℵo is
> defined in you textbook, is it?

There is no actual infinity in my book. There is no 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, ...
5/1, 5/2, 5/3, 5/4, ...
...

But we have to assume it in order to discuss it by means of correct mathematics from hatever books.

> Anyway, that's not the result given by
> WMaths as my argument showed.

Your argument failed to produce exchanges.

> > I prefer to write X and O in the way I did. Nothing in my book
> > contradicts that.
> Indeed. It all works in WMaths. If do decide to take me up on the
> offer, and do the exercises, you'll see how it's done.
>
> When you said "please try it" in reply to may offer of taking you
> through some exercises

I am looking foreward to your yexplaination of the argument. It deals with X and O starting from the matrix

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

not with a linear sequence.

> The offer still stands and, as I said, I can break it down (into two) if
> you think that would help.

Show what you have, but adhere to the topic and don't try to change it..

> >> OK, you don't have a citation. I didn't really think you did.
> >
> > Here is the citation.
> <rudeness cut>

No rudeness. Only a necessary hint that it requires some thinking to understand the language of matrices I use. Many are unable.

> > So the set of all positive rational numbers is countable as was shown
> > by Cantor [Cantor, p. 126. G. Cantor, letter to R. Lipschitz (19 Nov
> > 1883)] by this 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, ...
> >
> > where the sum s of numerator and denominator grows stepwise by 1, and
> > for fixed sum the numerator grows stepwise by 1 from minimum 1 to
> > maximum s - 1. If repeating values of fractions are eliminated, every
> > positive rational number appears exactly once, if not, we get the
> > sequence of all positive fractions where k = (m + n - 1)(m + n - 2)/2
> > + m is the index or position of fraction m/n. [Cantor, p. 132]

> Right, so you are citing something other than Cantor discussing swapping
> Xs and Os in a grid.

I warned you. Mathematics is interweaved with thinking.

> I'd really have like to see how Cantor defined the
> result of and endless sequence of swaps. Maybe he does in some other
> work. Do you know of any?

If he had tried it, he would have seen the result.

Cantor used the natural numbers for indexing, I use the integer fractions. That is the only new idea. Since the integer fractions also have to be indexed, we get the swaps.

Regards, WM

[WM]

Tom Bola schrieb am Sonntag, 4. September 2022 um 22:14:09 UTC+2:

> WM does not want our math where infinity exists and he teaches this
> credo to all people who are "believing" that infinity is "allowable"
> which he calls dumb folks

Wrong. I accept and assume the actually infinite 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, ...
5/1, 5/2, 5/3, 5/4, ...
...

Regards, WM

[WM]

JVR schrieb am Sonntag, 4. September 2022 um 22:14:39 UTC+2:
>
> "You cannot nail a pudding to the wall."

You are the pudding! You claimed you could explain how the O's were removed from the matrix

XOO...
XOO...
XOO...
...

by simple exchanges with the X's.

When you are challenged to show it, either you escape to insults or poems.

Regards, WM

[Tom Bola]

WM drivels:

>> WM does not want our math where infinity exists and he teaches this
>> credo to all people who are "believing" that infinity is "allowable"
>> which he calls dumb folks
>
> Wrong. I accept and assume the actually infinite 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, ...
> 5/1, 5/2, 5/3, 5/4, ...
> ...

Which is endless and: |IN| = |IN x IN|
You idiotic asshole do not WANT our infinity.

Until you accept Hilbert's Hotel it makes no sense to reply your drivel,
and you psychotic never will accept that.

Piss off, clown.

[Sergio]

On 9/4/2022 1:10 PM, WM wrote:
> Gus Gassmann schrieb am Sonntag, 4. September 2022 um 16:20:32 UTC+2:
>> On Sunday, 4 September 2022 at 10:55:28 UTC-3, WM wrote:
>> [...]
>>> I have this understanding of logic: If two symbols are exchanged, then none of them is deleted.
>> This is correct, but insufficient.
>
> It is all that is needed.

wrong, you are incompetent.

>
>>>> A In order to prove an inconsistency in ZFC
>>>
>>> it issufficient to prove that no O can leave the matrix.
>> Ah, no. The fact that the reciprocal 1/n is positive for every natural number n is insufficient to conclude that the limit of the sequence {1, 1/2, 1/3, ..} cannot be 0.
>
> The limit of a sequence is a number which is approached.

insufficient. what do YOU mean by "approached" ?


>
> Regards, WM

[Sergio]

On 9/4/2022 3:29 PM, WM wrote:
> Ben Bacarisse schrieb am Sonntag, 4. September 2022 um 21:26:47 UTC+2:
>> WM <askas...@gmail.com> writes:
>> (AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
>> Unendlichen" at Hochschule Augsburg.)
>>> Ben Bacarisse schrieb am Sonntag, 4. September 2022 um 18:14:07 UTC+2:
>>>> WM <askas...@gmail.com> writes:
>>>

<snip crap>

>
> Cantor used the natural numbers for indexing, I use the integer fractions.

that is just one of your mistakes. The other is using a step by step process (swapparoos) in an infinite matrix...

No wonder you could only Fail.

Cantor is correct and he proved he could one to one map the fractions.

You cannot.

Fail.

>
> Regards, WM

[Ben Bacarisse]

JVR <jrenne...@googlemail.com> writes:

> There is only one immutable rule in what you call WMaths and I call
> Muckmeatics: "You cannot nail a pudding to the wall."

That's my feeling, most of the time. But while you can't nail up a
whole one, I've managed, over the years, to nail up a few tiny scraps of
pudding.

For example, there is the theorem that states that in WMaths, certain
bounded monotonic sequences of rationals don't converge to a real
number. The theorem hit the rails, though, when WM could not pin down
which ones, but not before the existence of such sequences was agreed.

Sometimes the pudding scraps drop off because WM just pulls the nail out
by flat-out denial of overwhelming evidence. He has asserted that (in
WMaths) no non-constant sequences of sets converge despite his textbook
making it plain that (a) functions are just sets, and (b) some
non-constant sequences of functions converge.

But the most beautiful scrap of nailed-on pudding is the clear statement
that, in WMaths (specifically for some potentially infinite sets) it is
possible to have both e ∈ S and S \ {e} = S. WM himself described this
as a "surprise". No shit!

Pressing for an explanation resulted in WM saying that he could not
define set membership, equality and difference in a way that would
enable a student of WMaths to get this result for themselves. Quite an
admission.

--
Ben.

[FromTheRafters]

Sergio explained :

When an ant becomes a roach, it has been approached.

[Ben Bacarisse]

I do not believe that, and a curious intellectual might want to examine
the argument to see how this apparent paradox is resolved. But you
don't want (or can't) to do that, despite the argument using only the
maths in your textbook. No actual infinities. Everything potential.

>> You flipped immediately back into whatever you are
>> calling maths that isn't yours these day. You appeared not to have
>> grasped even the first stage of my argument.
>
> Your argument started with a wrong image of Cantor's matrix which I
> corrected.

Doing the exercises will avoid any such problems because /you/ will be
providing everything right from the start! Do you want to continue with
them?

>> > The result of an endless sequence ℵo, ℵo, ℵo, ... is ℵo.
>> We're doing this with the maths in your book. I don't think ℵo is
>> defined in you textbook, is it?
>
> There is no actual infinity in my book. There is no 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, ...
> 5/1, 5/2, 5/3, 5/4, ...
> ...

Indeed, but (as you often do in your book) one can write 20 typical
elements to help the reader without implying an actual infinity. This
is all WMaths after all. But you'll see when you do the exercises that
there is no need for these diagrams in WMaths, so there can be no
suspicion that an "actual infinity" has crept in.

> But we have to assume it in order to discuss it by means of correct
> mathematics from hatever books.

No! That's the great thing I am offering to help you see. Your WMaths
can do it to! No actual infinity needed.

>> Anyway, that's not the result given by
>> WMaths as my argument showed.
>
> Your argument failed to produce exchanges.

Nope. There were right there. You did cut them all from your reply. I
wonder why?

>> > I prefer to write X and O in the way I did. Nothing in my book
>> > contradicts that.
>> Indeed. It all works in WMaths. If do decide to take me up on the
>> offer, and do the exercises, you'll see how it's done.
>>
>> When you said "please try it" in reply to may offer of taking you
>> through some exercises
>
> I am looking foreward to your yexplaination of the argument.

I wrote out the argument but you misinterpreted the very first step and
then cut most of the rest of it without comment. I assume you will do
the same again, no matter how often I explain it.

No, my offer, which you seemed to accept, was to take you through it by
having /you/ produce the argument, as the answers to a series of
exercises. That, as I am sure you know, is a good way to get students
to really understand things.

>> The offer still stands and, as I said, I can break it down (into two) if
>> you think that would help.
>
> Show what you have, but adhere to the topic and don't try to change
> it.

I have the first exercise for you. I'm going to break it down since you
didn't tackle it last time when it was all-in-one.

1a. Express the initial matrix as a function, F_1, from NxN to {0,1} so
that Xs are indicated by 1 and Os by O.

1b. Can you express this initial matrix as a function M_1 from N to
{0,1}? (Hint, use F_1 and your mapping k).

Note: while you wrote a part of an actually infinite matrix, everything
from now on is going to be potentially infinite so it would be a good
idea not to write any grids in case you confuse yourself and think that
you've flipped back into actual infinity. Remember, nothing but WMaths
permitted!

--
Ben.

[zelos...@gmail.com]

fredag 2 september 2022 kl. 15:54:00 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Freitag, 2. September 2022 um 11:26:23 UTC+2:
> > torsdag 1 september 2022 kl. 15:25:36 UTC+2 skrev WM:
> > > Gus Gassmann schrieb am Mittwoch, 31. August 2022 um 19:35:11 UTC+2:
> > >
> > > > (4) "Because I (WM) say so."
> > > >
> > > Yes. This is so because I say so (and because every sober mind will agree): The intersection of non-empty inclusion-monotonic sets is not empty.
> > >
> > That is what a crazy person that doesn't understand mathematics will say because in sensible mathematics where logical people work, there is no such implication
> The states of water flowing out of a bathub will never have an empty intersection before the bathtub is empty. "Mathematics" contradicting this basic principle is rubbish.
>
> Regards, WM

Bathtub is a physical object, therefore your analogy is invalid.

"intersection" is not applicable to it either.

Everything here shows how retarded you are.

Mathematics contradicts "common sense", but guess what? Common sense has no place in mathematics.

[JVR]

You might think that it should be possible to focus on a single simple clear-cut error
and get him to understand that. But it isn't possible.

For example, in the present context Mücke keeps stating as fact that the intersection of
an infinite monotone sequence of non-empty sets is necessarily non-empty.
There is no way to disabuse him of this notion.

So the conclusion is inevitable - it doesn't matter whether his problem is
stupidity or a psychological abnormality - the effect is the same.

[Fritz Feldhase]

On Sunday, September 4, 2022 at 11:25:27 PM UTC+2, Ben Bacarisse wrote:

> in WMaths (specifically for some potentially infinite sets) it is possible to have both e ∈ S and S \ {e} = S.

From S \ {e} = S we get that ~(e ∈ S), since ~(e ∈ S \ {e}) by definition of "\". So in WMaths we have (for some e and some S) e ∈ S and ~(e ∈ S). A contradiction.

What a surprise!

Now a REASONABLE person would (at least) question some of his or her "existence assumptions" concerning "potentially infinite sets".

But WM is no REASONABLE person. Actually, even this devastating result will not have any effect on him.

[Ben Bacarisse]

Fritz Feldhase <franz.fri...@gmail.com> writes:

> On Sunday, September 4, 2022 at 11:25:27 PM UTC+2, Ben Bacarisse wrote:
>
>> in WMaths (specifically for some potentially infinite sets) it is possible to have both e ∈ S and S \ {e} = S.
>
> From S \ {e} = S we get that ~(e ∈ S), since ~(e ∈ S \ {e}) by
> definition of "\". So in WMaths we have (for some e and some S) e ∈ S
> and ~(e ∈ S). A contradiction.

I pointed out something similar, of course, based on the entirely
conventional definitions of set membership, equality and difference from
his textbook. His line is the these were "simplified" (i.e. wrong) to
help students with the basics. That's what led to my pressing for the
"real" definitions, and his admission that he could not give them.

> What a surprise!
>
> Now a REASONABLE person would (at least) question some of his or her
> "existence assumptions" concerning "potentially infinite sets".

For a long time I though WM was just playing a game. A sort of
mathematical keepy-uppie where he knew he was talking nonsense but was
having fun seeing how long he could keep any particular line of nonsense
going. But now, I think he may actually believe what he says.

> But WM is no REASONABLE person. Actually, even this devastating result
> will not have any effect on him.

If WM believes himself, there's no contradiction, just a massive hole
where the WMaths definitions of the basic set operations should be.
Perhaps the latter is easier to reconcile internally?

I often wonder what it would be like to meet a Usenet maths crank in
real life. Would it be obvious? How would a mathematical discussion go
without the opportunities for deflection and ignoring or awkward points
that Usenet facilitates?

--
Ben.

[JVR]

In the German language math group I have recently, and also a few years ago,
posted lists of errors in his book. Here is a particularly blatant one:
https://drive.google.com/file/d/1cDDjfnWyz4DU53cMv47oxoDKQVl5fuO5/view?usp=sharing

In the ensuing discussion it became clear that he copied this from somewhere else and
doesn't understand why it is nonsense. And - this is the really weird part - he insists that his
source is authoritative.

Other crass examples of his incompetence are his redefinition of continuity and his improvement of the
Peano axioms. The resulting discussion showed that he simply doesn't understand
the issues involved.

[JVR]

No, Mücke, I asked you how you constructed the infinite matrix containing the
silly X's and O's.

The point being that, since you visualize your silly X's and O's as
coming from somewhere, we can surely invert the process and simply send them
back where they came from. (Maybe you can even get a refund!)

And every time we send one back home we number the now empty seat (m,n) according to
the formula k = (m + n - 1)(m + n - 2)/2 + m.

And everybody will live happily ever after.

[Jim Burns]

Maybe like this?

| So what mental artillery have we picked up
| over the last 100 years? Alexander Luria
| studied neuropsychology in the early half of
| the century, and he found that people were
| resistent to classification, to deducing the
| hypothetical. His subjects simply couldn’t
| think about anything abstract. Consider
| this exchange:
|
| Luria: What do crows and fish have
| in common?
| Subject: Absolutely nothing. A fish swims,
| and a crow flies.
| Luria: Are they not both animals?
| Subject: Of course not, a fish is a fish,
| and a crow is a bird.
|
| The man could only think of the objects as
| how he might use them, not as abstract
| objects part of a classification system.
|
| Luria told another subject: “There are no
| camels in Germany. Hamburg is in Germany.
| Are there camels in Hamburg?” The subject
| replied, “If it’s big enough, perhaps it
| has camels.” Luria prompted him again to
| listen to the conditions, and again he
| replied that perhaps Hamburg had camels.
| He was used to camels, and he was unable to
| imagine that there weren’t any in Hamburg.

https://blog.ted.com/are-we-getting-more-intelligent-jim-flynn-at-ted2013/

[WM]

Ben Bacarisse schrieb am Sonntag, 4. September 2022 um 03:33:25 UTC+2:
> WM <askas...@gmail.com> writes:

> >> > Cantor's approach is modelled by exchanging X's and O's in
> >> >
> >> > XOO...
> >> > XOO...
> >> > XOO...
> >> > ...
> >> >
> >> > until all O's have disappeared. Do you agree?
> >> Here are the cells as numbered by your bijective mapping:
> >>
> >> 1, 3, 6, 10, 15, ...
> >> 2, 5, 9, 14, 20, ...
> >> 4, 8, 13, 19, 26, ...
> >> 7, 12, 18, 25, 33, ...
> >> 11, 17, 24, 32, 41, ...
> >> ...
> >
> > Yes, all visible cells are filled with X's as Cantor has taught us.
> No. I've just written some numbers down to illustrate what you
> bijective mapping k looks like.

> >> With Os at 3, 5, 6, 8, 9 and so on.
> >
> > No, the above matrix does not contain O's any longer.

Sorry, I misunderstood. You have described the original matrix
XOO...
XOO...
XOO...
...
But as you enumerated every position by a natural number, it looked like the final matrix to me:
XXX...
XXX...
XXX...
...

Regards, WM

[Jim Burns]

On 9/5/2022 8:14 AM, Jim Burns wrote:
> On 9/5/2022 6:24 AM, Ben Bacarisse wrote:

>> I often wonder what it would be like to
>> meet a Usenet maths crank in real life.
>> Would it be obvious?  How would a
>> mathematical discussion go without the
>> opportunities for deflection and ignoring
>> or awkward points that Usenet facilitates?

> Maybe like this?

> | Luria told another subject: “There are no
> | camels in Germany. Hamburg is in Germany.
> | Are there camels in Hamburg?” The subject
> | replied, “If it’s big enough, perhaps it
> | has camels.” Luria prompted him again to
> | listen to the conditions, and again he
> | replied that perhaps Hamburg had camels.
> | He was used to camels, and he was unable to
> | imagine that there weren’t any in Hamburg.

<WM<JB>>
>>
>> This description does immediately not rule out
>> your dark numbers and dark end segments.
>> Dark numbers could conceivably be in 𝐸
>> which are > each i in ⋃𝓕\𝐸
>>
>> However,
>> this description requires 𝐸 subset ⋃𝓕
>
> That is a wrong desription.
>
</WM<JB>>

Date: Fri, 2 Sep 2022 06:32:56 -0700 (PDT)
Subject: Re: Natural numbers and vases III

>
> https://blog.ted.com/are-we-getting-more-intelligent-jim-flynn-at-ted2013/
>

[WM]

Ben Bacarisse schrieb am Montag, 5. September 2022 um 00:06:08 UTC+2:

> WM <askas...@gmail.com> writes:

> >> I showed you how.
> >
> > You cannot show it because it is wrong. It is nonsense to believe that
> > by exchanging two items, one of them disappears.
> I do not believe that,

That is worth to be nailed to the wall.

> and a curious intellectual might want to examine
> the argument to see how this apparent paradox is resolved.

However you may try it, not by disappearing O's.

> But you
> don't want (or can't) to do that,

Then do it yourself.

> despite the argument using only the
> maths in your textbook. No actual infinities. Everything potential.

You seem to have missed the point. Without actual infinities there are no completed enumerations of infinite sets and there are no dark numbers.

> Doing the exercises will avoid any such problems because /you/ will be
> providing everything right from the start! Do you want to continue with
> them?

I have seen your description of the initial matrix. No problem. But if you think to hanlde it by potential infinity, you cannot follow Cantor.

> >> > The result of an endless sequence ℵo, ℵo, ℵo, ... is ℵo.
> >> We're doing this with the maths in your book. I don't think ℵo is
> >> defined in you textbook, is it?
> >
> > There is no actual infinity in my book. There is no 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, ...
> > 5/1, 5/2, 5/3, 5/4, ...
> > ...
> Indeed, but (as you often do in your book) one can write 20 typical
> elements to help the reader without implying an actual infinity. This
> is all WMaths after all. But you'll see when you do the exercises that
> there is no need for these diagrams in WMaths, so there can be no
> suspicion that an "actual infinity" has crept in.

The we can close the case. Dark nubers and complete enumerations are a matter of actual infinity.

> > But we have to assume it in order to discuss it by means of correct
> > mathematics from hatever books.
> No! That's the great thing I am offering to help you see. Your WMaths
> can do it to! No actual infinity needed.

Try to understand Cantor's set theory:
"In spite of significant difference between the notions of the potential and actual infinite, where the former is a variable finite magnitude, growing above all limits, the latter a constant quantity fixed in itself but beyond all finite magnitudes, it happens deplorably often that the one is confused with the other." [Cantor, p. 374]
"By the actual infinite we have to understand a quantity that on the one hand is not variable but fixed and definite in all its parts, a real constant, but at the same time, on the other hand, exceeds every finite size of the same kind by size. As an example I mention the totality, the embodiment of all finite positive integers; this set is a self-contained thing and forms, apart from the natural sequence of its numbers, a fixed, definite quantity, an , which we obviously have to call larger than every finite number." [G. Cantor, letter to A. Eulenburg (28 Feb 1886)]

Potential infinity is variable:
"Should we briefly characterize the new view of the infinite introduced by Cantor, we could certainly say: In analysis we have to deal only with the infinitely small and the infinitely large as a limit-notion, as something becoming, emerging, produced, i.e., as we put it, with the potential infinite. But this is not the proper infinite. That we have for instance when we consider the entirety of the numbers 1, 2, 3, 4, ... itself as a completed unit, or the points of a line as an entirety of things which is completely available. That sort of infinity is named actual infinite." [D. Hilbert: "Über das Unendliche", Mathematische Annalen 95 (1925) p. 167]

> > Your argument failed to produce exchanges.
> Nope. There were right there. You did cut them all from your reply. I
> wonder why?

Sorry, I did not see exchanges. What do you exchange?

> > I am looking foreward to your explaination of the argument.

> I wrote out the argument but you misinterpreted the very first step

Yes, you wrote integers which I interpreted as indices. Now I got it that you made a sequence.

> Os at 3, 5, 6, 8, 9 and so on.

> Since this is the first step, you really need to be able to see how M_0
represents the initial matrix a WMaths function from N to {0,1}.

I see. But that does not change anything except that you will never cross the diagonal / of the matrix.

>No completed anything, everything potential

and visible. No dark numbers. You need to interpret your sequence as completed.

> 1a. Express the initial matrix as a function, F_1, from NxN to {0,1} so
> that Xs are indicated by 1 and Os by O.

You did it.

>
> 1b. Can you express this initial matrix as a function M_1 from N to
> {0,1}? (Hint, use F_1 and your mapping k).

1, 1, 0, 1, 0, 0, 1, ... with 0 at 3, 5, 6, 8, 9 and so on.

>
> Note: while you wrote a part of an actually infinite matrix, everything
> from now on is going to be potentially infinite

You may stop immediately if you can't understand the difference between potential infinity (of mathematics including my book) and Cantor's actual infinity.

> Remember, nothing but WMaths
> permitted!

For analyzing the facts yes, but not for the outset.

Regards, WM

[WM]

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.

> From S \ {e} = S we get that ~(e ∈ S), since ~(e ∈ S \ {e}) by definition of "\". So in WMaths we have (for some e and some S) e ∈ S and ~(e ∈ S). A contradiction.

No, to be an element and to be not an element is not simultaneously possible.

Regards, WM

[WM]

JVR schrieb am Montag, 5. September 2022 um 13:35:29 UTC+2:
> On Sunday, September 4, 2022 at 10:34:35 PM UTC+2, WM wrote:

> I asked you how you constructed the infinite matrix containing the

> X's and O's.touk Cantor's finished infinity. A finished column of integer fractions and a finihed infinity of finihed columns of other fractions.

> The point being that, since you visualize your silly X's and O's as
> coming from somewhere, we can surely invert the process and simply send them
> back where they came from. (Maybe you can even get a refund!)

Forget Cantor and his disciples. The they all have disappeared.

>
> And every time we send one back home we number the now empty seat (m,n) according to
> the formula k = (m + n - 1)(m + n - 2)/2 + m.

I am analyzing Cantor's version. There all fractions O are remaining in the matrix and are becoming enumerated by integer fractions X

Regards, WM

[Sergio]

On 9/5/2022 8:21 AM, WM wrote:
> JVR schrieb am Montag, 5. September 2022 um 13:35:29 UTC+2:
>> On Sunday, September 4, 2022 at 10:34:35 PM UTC+2, WM wrote:
>
>> I asked you how you constructed the infinite matrix containing the
>> X's and O's.touk Cantor's finished infinity. A finished column of integer fractions and a finihed infinity of finihed columns of other fractions.
>
>> The point being that, since you visualize your silly X's and O's as
>> coming from somewhere, we can surely invert the process and simply send them
>> back where they came from. (Maybe you can even get a refund!)
>
> Forget Cantor and his disciples. The they all have disappeared.

But Cantors Proof lives on forever, and it cannot be disproved.

>>
>> And every time we send one back home we number the now empty seat (m,n) according to
>> the formula k = (m + n - 1)(m + n - 2)/2 + m.
>
> I am analyzing Cantor's version.

come on dude!

that is a 3 min task at most, and you have spent decades ?


>
> Regards, WM

[JVR]

Well - I'm definitely not going to try to make sense of this gobble-dee-gook.

[Sergio]

On 9/5/2022 8:13 AM, WM wrote:
> 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.

that is your bullshit.

[Sergio]

On 9/5/2022 8:06 AM, WM wrote:
>
> Ben Bacarisse schrieb am Montag, 5. September 2022 um 00:06:08 UTC+2:
>> WM <askas...@gmail.com> writes:
>
>>>> I showed you how.
>>>
>>> You cannot show it because it is wrong. It is nonsense to believe that
>>> by exchanging two items, one of them disappears.
>> I do not believe that,
>

<snip spoof math>


>
> Regards, WM

[Sergio]

I reviewed that book, and to me it was obvious it was a cut and paste, as the span was too great, and the depth of each level too deep, making the book
unusable, especially to students.

[Fritz Feldhase]

On Monday, September 5, 2022 at 12:24:46 PM UTC+2, Ben Bacarisse wrote:

> Fritz Feldhase writes:
>
> > On Sunday, September 4, 2022 at 11:25:27 PM UTC+2, Ben Bacarisse wrote:
> > >
> > > in WMaths (specifically for some potentially infinite sets) it is possible to have both e ∈ S and S \ {e} = S.
> > >
> > From S \ {e} = S we get that ~(e ∈ S), since ~(e ∈ S \ {e}) by
> > definition of "\". So in WMaths we have (for some e and some S) e ∈ S
> > and ~(e ∈ S). A contradiction.
> >
> I pointed out something similar, of course, based on the entirely
> conventional definitions of set membership, equality and difference from
> his textbook. His line is the these were "simplified" (i.e. wrong) to
> help students with the basics. That's what led to my pressing for the
> "real" definitions, and his admission that he could not give them.

Yes. It's clear that he only has some "notions" like "potentially infinite sets" and some rather vague ideas concerning these "notions", but no concrete system (i. e. an "alternative" set theory) in WMaths. WM is all talk.

> > Now a REASONABLE person would (at least) question some of his or her
> > "existence assumptions" concerning "potentially infinite sets".
> >
> For a long time I though WM was just playing a game. A sort of
> mathematical keepy-uppie where he knew he was talking nonsense but was
> having fun seeing how long he could keep any particular line of nonsense
> going. But now, I think he may actually believe what he says.

I'd tend to agree. Imho he's actually a so called crank.

> > But WM is no REASONABLE person. Actually, even this devastating result
> > will not have any effect on him.
> >
> If WM believes himself, there's no contradiction, just a massive hole
> where the WMaths definitions of the basic set operations should be.
> Perhaps the latter is easier to reconcile internally?
>
> I often wonder what it would be like to meet a Usenet maths crank in
> real life. Would it be obvious? How would a mathematical discussion go
> without the opportunities for deflection and ignoring or awkward points
> that Usenet facilitates?

Hard to tell, I've never met a (math) crank in real live!

Maybe he would just quit the discussion (prematurely) at some point.

[FromTheRafters]

WM laid this down on his screen :

But he has just shown you it *is* possible in your 'system' using your
WMaths. Refuse to see it if you must, but it is not a good look for a
professed professor.

Then again, it seems you *do* have a different notion of the definition
of "\" probably something to do with subtraction, bathtubs and Xs and
Os. Maybe you could share that with us?

[Python]

FromTheRafters wrote:
> WM laid this down on his screen :
>> 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.
>>
>>> From S \ {e} = S we get that ~(e ∈ S), since ~(e ∈ S \ {e}) by
>>> definition of "\". So in WMaths we have (for some e and some S) e ∈ S
>>> and ~(e ∈ S). A contradiction.
>>
>> No, to be an element and to be not an element is not simultaneously
>> possible.
>
> But he has just shown you it *is* possible in your 'system' using your
> WMaths. Refuse to see it if you must, but it is not a good look for a
> professed professor.

There is 0% chances he will ever admit being 100% wrong (and he is), but
there is a solution for that issue: ensure that he's not allow to be a
professor ever in the future, including this academic year.

As he should be for ages, and would be in almost all places where math
is taught in the entire world.

What's THAT wrong with Hochschule Augsburg?

[Fritz Feldhase]

On Monday, September 5, 2022 at 4:43:57 PM UTC+2, Python wrote:

> What's THAT wrong with Hochschule Augsburg?

Well... you see... from 2003 till 2007 he was the dean of the /faculty of general sciences/ of that "Hochschule".

(Source: https://de.wikipedia.org/wiki/Wolfgang_M%C3%BCckenheim)

It's a shame!

[Fritz Feldhase]

On Monday, September 5, 2022 at 2:15:10 PM UTC+2, Jim Burns wrote:
> On 9/5/2022 6:24 AM, Ben Bacarisse wrote:

BB: In WMaths [for a potentially infinite set S] it is possible to have both e ∈ S and S \ {e} = S.

WM: Yes.

FF From S \ {e} = S we get that ~(e ∈ S), since ~(e ∈ S \ {e}) by definition of '\'. So in WMaths we have (for some e and some S) e ∈ S and ~(e ∈ S). A contradiction.

WM: No, to be an element and to be not an element is not simultaneously possible.

Well...

[Sergio]

it is interesting on the decline of professors, one proff over here, I took a course from, I met him 30 years later at a technical meeting, and he was
pushing a idea on signal shapes that was obviously wrong on random networks, he too old and stale but still had the ego, still thought he had the best
ideas.

the news groups are one of the last refuges to get a audience for (failed ideas), so that is why WM is here.

I did download his 2015 book, and it does look cut and paste to me.

But I wont let him get away with lying about Math.

[Ben Bacarisse]

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.

Any progress on defining set membership, equality and difference so that
a student could derive this surprising result of WMaths?

--
Ben.

[Ben Bacarisse]

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

> Ben Bacarisse schrieb am Montag, 5. September 2022 um 00:06:08 UTC+2:

>> 1a. Express the initial matrix as a function, F_1, from NxN to {0,1} so
>> that Xs are indicated by 1 and Os by O.
>
> You did it.

Actually I didn't. Like you, I just skipped on to M_1.

>> 1b. Can you express this initial matrix as a function M_1 from N to
>> {0,1}? (Hint, use F_1 and your mapping k).
>
> 1, 1, 0, 1, 0, 0, 1, ... with 0 at 3, 5, 6, 8, 9 and so on.

But this is just a few values of the function. What's the formula? In
WMaths you must have formulas or things are not defined, yes?

I'll give the next exercise, but please show a formula for M_1. I think
it needs to be 100% clear that it's a well-defined WMaths function.

2a. What is the first swap, and what are the first few values of M_2,
the function from N to {0,1}, showing where the Xs and Os are after the
first swap?

2b. Does M_2 have a finite formula? I.e. is it a properly defined
WMaths function from N to {0,1}?

2c. What is the /rule/ for swapping Xs and Os that you want to follow?
Some will get us where we want to go and some won't so you have pick a
good one. Also, make it as simple as you can. The rule should enable
anyone to easily write the first dozen or so values of M_2, M_3 and M_4.

--
Ben.

[Jim Burns]

On 9/5/2022 9:13 AM, WM wrote:
> 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.

Visible elements are able to be seen.

Dark elements are not able to be seen
but are able to be able to be seen.

That seems unhelpful.

----
It would be nice if we could replace
unhelpful claims of visibility with
clear claims of FISON-membership.

A FISON is
a collection with a counting-order
which begins at 0 and ends _somewhere_

What is your latest word on whether
any elements of any FISON are dark?

----
Let 𝓕 be the collection of FISONs
∀𝐹, 𝐹 ∈ 𝓕 <-> 𝐹 is a FISON

Each FISON 𝐹 ends somewhere.
Not all things in 𝐹 in have successors in 𝐹
∀𝐹 ∈ 𝓕, ~(∀n ∈ 𝐹, n+1 ∈ 𝐹)

_All_ the FISONs _do not_ end anywhere.
∀𝐹 ∈ 𝓕, ∃𝐹' ∈ 𝓕, 𝐹 ⊂ 𝐹'

| That arises from
| 𝐹ₙ∪⟨n+1⟩ also being a FISON.

Also,
all things n in a FISON (∃𝐹∈𝓕, n∈𝐹)
do not end anywhere (∃𝐹'∈𝓕, n+1∈𝐹')
∀n, (∃𝐹∈𝓕, n∈𝐹) -> (∃𝐹'∈𝓕, n+1∈𝐹')

| That arises from
| 𝐹ₙ∪⟨n+1⟩ also being a FISON.

The collection ⋃𝓕 of things in a FISON
does not end.
∀n, (n∈⋃𝓕) -> (n+1∈⋃𝓕)

Your (WM's) claim seems to be that,
because
each FISON 𝐹 ends
but
the collection ⋃𝓕 of things in a FISON
does not end,
there must be things in ⋃𝓕 which
are not in any FISON 𝐹: your dark numbers.

The problem with your claim
(as I understand it to be)
is that ⋃𝓕 is exactly the things in FISONs
no more and no less.

The origin of your problem seems to be
in the way you attribute properties of
a set to the elements of that set.
You like to say "A set is its elements".

That doesn't work.
n in a FISON 𝐹 is in a set that ends.
n in ⋃𝓕 is in a set that does not end.
The same n is in both sets.

> Dark elements can become visible
> and vice versa.

Elements of set which do not end
can be in sets which do end

and vice versa.

>> From S \ {e} = S we get that ~(e ∈ S),
>> since ~(e ∈ S \ {e}) by definition of "\".
>> So in WMaths we have (for some e and some S)
>> e ∈ S and ~(e ∈ S). A contradiction.
>
> No,
> to be an element and to be not an element
> is not simultaneously possible.

To be an element of one set and
to be not an element of a different set
is simultaneously possible.

[zelos...@gmail.com]

måndag 5 september 2022 kl. 15:13:47 UTC+2 skrev WM:
> 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.

what the FUCK DOES THAT EVEN MEAN!?

> > From S \ {e} = S we get that ~(e ∈ S), since ~(e ∈ S \ {e}) by definition of "\". So in WMaths we have (for some e and some S) e ∈ S and ~(e ∈ S). A contradiction.
> No, to be an element and to be not an element is not simultaneously possible.

Yet you claim it

>
> Regards, WM

[WM]

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

In actual infinity the newly created visible numbers are taken from the dark part of the set, in potential infinity they are created from nothing. In both cases only the potentially infinite sets of visible numbers are part of mathematics.

Regards, WM

[WM]

Ben Bacarisse schrieb am Montag, 5. September 2022 um 18:23:55 UTC+2:
> WM <askas...@gmail.com> writes:
> (AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
> Unendlichen" at Hochschule Augsburg.)
> > Ben Bacarisse schrieb am Montag, 5. September 2022 um 00:06:08 UTC+2:
>
> >> 1a. Express the initial matrix as a function, F_1, from NxN to {0,1} so
> >> that Xs are indicated by 1 and Os by O.
> >
> > You did it.
> Actually I didn't. Like you, I just skipped on to M_1.
> >> 1b. Can you express this initial matrix as a function M_1 from N to
> >> {0,1}? (Hint, use F_1 and your mapping k).
> >
> > 1, 1, 0, 1, 0, 0, 1, ... with 0 at 3, 5, 6, 8, 9 and so on.
> But this is just a few values of the function. What's the formula? In
> WMaths you must have formulas or things are not defined, yes?

This 1, 1, 0, 1, 0, 0, 1, ... is a formula like 0.999....
>
If you can't read it, I will extend it a bit 1, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, ... .

I will not fall prey to your tedious attempt to distract from the topic.
Note that the only meaningful argument recently uttered is this: The O's will not disappear in any finite step but behind all finite steps. That is the point. You cannot accomplish more. I will not accept it.

Regards, WM

[WM]

Jim Burns schrieb am Montag, 5. September 2022 um 22:09:36 UTC+2:
> On 9/5/2022 9:13 AM, WM wrote:

> It would be nice if we could replace
> unhelpful claims of visibility with
> clear claims of FISON-membership.
>
> A FISON is
> a collection with a counting-order
> which begins at 0 and ends _somewhere_
>
> What is your latest word on whether
> any elements of any FISON are dark?

FISONs can become dark if the ressources of the systen decrease.

> > to be an element and to be not an element
> > is not simultaneously possible.
> To be an element of one set and
> to be not an element of a different set
> is simultaneously possible.

Potentially infinite collections are not sets in that sense.

Regards, WM

[zelos...@gmail.com]

WHAT FUCKING SYSTEM!? WHAT RESOURCES!? THOSE DO NOT EXIST IN MATHEMATICS!

[Sergio]

On 9/6/2022 5:19 AM, WM wrote:
> 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. Nevertheless potential infinity has a growing character. Therefore larger finite sets are created from smaller sets.

and smaller sets are created from larger sets.

.
>
> In actual infinity the newly created visible numbers are taken from the dark part of the set, in potential infinity they are created from nothing. In both cases only the potentially infinite sets of visible numbers are part of mathematics.

So you now conclude that dark numbers are not part of mathematics.


>
> Regards, WM
>

[Sergio]

On 9/6/2022 5:34 AM, WM wrote:
> Jim Burns schrieb am Montag, 5. September 2022 um 22:09:36 UTC+2:
>> On 9/5/2022 9:13 AM, WM wrote:
>
>> It would be nice if we could replace
>> unhelpful claims of visibility with
>> clear claims of FISON-membership.
>>
>> A FISON is
>> a collection with a counting-order
>> which begins at 0 and ends _somewhere_
>>
>> What is your latest word on whether
>> any elements of any FISON are dark?
>
> FISONs can become dark if the ressources of the systen decrease.

take a FISON(1) = {1} now how does that become dark using Windows 7 ?

>
>>> to be an element and to be not an element
>>> is not simultaneously possible.
>> To be an element of one set and
>> to be not an element of a different set
>> is simultaneously possible.
>
> Potentially infinite collections are not sets in that sense.

you are calling it a collection, he was posting about sets, did you miss that, Deceiver?

>
> Regards, WM

[Sergio]

On 9/6/2022 5:27 AM, WM wrote:
> Ben Bacarisse schrieb am Montag, 5. September 2022 um 18:23:55 UTC+2:
>> WM <askas...@gmail.com> writes:
>> (AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
>> Unendlichen" at Hochschule Augsburg.)
>>> Ben Bacarisse schrieb am Montag, 5. September 2022 um 00:06:08 UTC+2:
>>
>>>> 1a. Express the initial matrix as a function, F_1, from NxN to {0,1} so
>>>> that Xs are indicated by 1 and Os by O.
>>>
>>> You did it.
>> Actually I didn't. Like you, I just skipped on to M_1.
>>>> 1b. Can you express this initial matrix as a function M_1 from N to
>>>> {0,1}? (Hint, use F_1 and your mapping k).
>>>
>>> 1, 1, 0, 1, 0, 0, 1, ... with 0 at 3, 5, 6, 8, 9 and so on.
>> But this is just a few values of the function. What's the formula? In
>> WMaths you must have formulas or things are not defined, yes?
>
> This 1, 1, 0, 1, 0, 0, 1, ... is a formula like 0.999....

wrong. the first is a SEQUENCE, the second is a NUMBER.

>>
> If you can't read it, I will extend it a bit 1, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, ... .
>
> I will not fall prey to your tedious attempt to distract from the topic.

That is what WM does BIG TIME.

> Note that the only meaningful argument recently uttered is this: The O's will not disappear in any finite step but behind all finite steps. That is the point. You cannot accomplish more. I will not accept it.

you Lie with your Matrix. It is a step by step process applied to infinite sequence.
ANd you cant even tell the difference between a sequence and a number. Troll.

>
> Regards, WM

[Gus Gassmann]

On Tuesday, 6 September 2022 at 10:29:24 UTC-3, Sergio wrote:
> On 9/6/2022 5:34 AM, WM wrote:
> > FISONs can become dark if the ressources of the systen decrease.
> take a FISON(1) = {1} now how does that become dark using Windows 7 ?

Well, that's easy: Open a command window, then type “format c: /fs:ntfs”

[Sergio]

been there done that, I couldn't say Oh Shi* fast enough...
very clearing, makes Win7 dark for sure

[WM]

zelos...@gmail.com schrieb am Dienstag, 6. September 2022 um 14:39:22 UTC+2:
> tisdag 6 september 2022 kl. 12:35:03 UTC+2 skrev WM:

> > Potentially infinite collections are not sets in that sense.
> >

> WHAT FUCKING SYSTEM!? WHAT RESOURCES!? THOSE DO NOT EXIST IN MATHEMATICS!

They do. Infinitely many X's are not enough to cover the whole matrix.

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

In the first 3 steps we get the following matrices

XXOO...
OOOO...
XOOO...
XOOO...
...

XXOO...
XOOO...
OOOO...
XOOO...
...

XXXO...
XOOO...
OOOO...
OOOO...
...

and so on.

Regards, WM

[Fritz Feldhase]

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...
...

[JVR]

On Tuesday, September 6, 2022 at 6:42:20 PM UTC+2, WM wrote:

Mücke, you are a tedious, repetitive, humorless dullard.
You are boring.

[Jim Burns]

On 9/6/2022 6:34 AM, WM wrote:
> Jim Burns schrieb am Montag,
> 5. September 2022 um 22:09:36 UTC+2:
>> On 9/5/2022 9:13 AM, WM wrote:

>>> Dark elements can become visible and
>>> vice versa.

>> It would be nice if we could replace
>> unhelpful claims of visibility with
>> clear claims of FISON-membership.
>>
>> A FISON is
>> a collection with a counting-order
>> which begins at 0 and ends _somewhere_
>>
>> What is your latest word on whether
>> any elements of any FISON are dark?
>
> FISONs can become dark if
> the ressources of the systen decrease.

But your "darkness" claims aren't that
FISONs become non-FISONs, or that
non-FISONs become FISONs.
Correct?

Let us remind ourselves that a FISON is

a collection with a counting-order
which begins at 0 and ends _somewhere_

>>> to be an element and to be not an element
>>> is not simultaneously possible.
>>
>> To be an element of one set and
>> to be not an element of a different set
>> is simultaneously possible.
>
> Potentially infinite collections
> are not sets in that sense.

However,
the set of all FISONs is a set in that sense.
No FISONs become non-FISONs.
No non-FISONs become FISONs.
Correct?

The next[2] set after all
infinite end segments of ⋃𝓕
is ∅

where ⋃𝓕 us the union of the set of FISONs.

[Jim Burns]

So, have you ever worked tech support?
No reason to ask, just curious.

https://www.youtube.com/watch?v=qjCfWs_AjTc
Doctor Who:
The Woman Who Fell To Earth
- Reformatting

[Gus Gassmann]

On Tuesday, 6 September 2022 at 17:23:09 UTC-3, Jim Burns wrote:
[...]

> So, have you ever worked tech support?

I have not. More often than I care to admit, I am the guy on the right...
https://www.youtube.com/watch?v=pQHX-SjgQvQ

[Ben Bacarisse]

WM <askas...@gmail.com> writes:

> Ben Bacarisse schrieb am Montag, 5. September 2022 um 18:23:55 UTC+2:
>> WM <askas...@gmail.com> writes:
>> (AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
>> Unendlichen" at Hochschule Augsburg.)
>> > Ben Bacarisse schrieb am Montag, 5. September 2022 um 00:06:08 UTC+2:
>>
>> >> 1a. Express the initial matrix as a function, F_1, from NxN to {0,1} so
>> >> that Xs are indicated by 1 and Os by O.
>> >
>> > You did it.
>> Actually I didn't. Like you, I just skipped on to M_1.
>> >> 1b. Can you express this initial matrix as a function M_1 from N to
>> >> {0,1}? (Hint, use F_1 and your mapping k).
>> >
>> > 1, 1, 0, 1, 0, 0, 1, ... with 0 at 3, 5, 6, 8, 9 and so on.
>> But this is just a few values of the function. What's the formula? In
>> WMaths you must have formulas or things are not defined, yes?
>
> This 1, 1, 0, 1, 0, 0, 1, ... is a formula like 0.999....
>>
> If you can't read it, I will extend it a bit 1, 1, 0, 1, 0, 0, 1, 0,
> 0, 0, 1, ... .

Too hard? Sorry.

> I will not fall prey to your tedious attempt to distract from the
> topic.

The only topic I care about is helping you see how all the Os "disappear"
in WMaths as easily as in any other kind of mathematics.

I'm not sure if you /can't/ write the formula for M_1 or you'd rather
not since you know here that leads. Anyway, we'll never find out...

> Note that the only meaningful argument recently uttered is this: The
> O's will not disappear in any finite step but behind all finite steps.

I thought you'd like to see how that is done in WMaths, but you did not
follow the argument when I wrote it out for you and you've decided ditch
the exercises that will take you through it step by step.

--
Ben.

[zelos...@gmail.com]

tisdag 6 september 2022 kl. 18:42:20 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Dienstag, 6. September 2022 um 14:39:22 UTC+2:
> > tisdag 6 september 2022 kl. 12:35:03 UTC+2 skrev WM:
>
> > > Potentially infinite collections are not sets in that sense.
> > >
> > WHAT FUCKING SYSTEM!? WHAT RESOURCES!? THOSE DO NOT EXIST IN MATHEMATICS!
> They do.

No they don't!

>Infinitely many X's are not enough to cover the whole matrix.

I can make a matrix with only X in it.

>
> XOOO...
> XOOO...
> XOOO...
> XOOO...
> ...
>
> In the first 3 steps we get the following matrices
>
> XXOO...
> OOOO...
> XOOO...
> XOOO...
> ...
>
> XXOO...
> XOOO...
> OOOO...
> XOOO...
> ...
>
> XXXO...
> XOOO...
> OOOO...
> OOOO...
> ...
>
> and so on.
>
> Regards, WM

YOu are talking about a step by step process, IRRELEVANT IN MATHEMATICS!

[WM]

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