(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

WM submitted this idea :
> (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

....and they all lived happily ever after.

WM <askasker48@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.

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
• Word Wise ‏ : ‎ Not Enabled
• Print length ‏ : ‎ 20 pages
• 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
• Text-to-Speech ‏ : ‎ Enabled
• Screen Reader ‏ : ‎ Supported
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• X-Ray ‏ : ‎ Not Enabled
• Word Wise ‏ : ‎ Not Enabled
• Print length ‏ : ‎ 51 pages
• Lending ‏ : ‎ Enabled

#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
Enhanced typesetting ‏ : ‎ Enabled
X-Ray ‏ : ‎ Not Enabled
Word Wise ‏ : ‎ Not Enabled
Print length ‏ : ‎ 154 pages
Lending ‏ : ‎ Enabled
Amazon Best Sellers Rank: #128,729 Paid in Kindle Store (See Top 100 Paid in Kindle Store)
#2 in 45-Minute Science & Math Short Reads
#134 in Calculus (Books)
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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?

On Wednesday, 31 August 2022 at 08:36:22 UTC-3, WM wrote:
> (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 .

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

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

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.

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

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

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

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.

WM <askasker48@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. (*)
>> 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?

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.

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.

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

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

> 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.

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.

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

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.
>
> Regards, WM

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

torsdag 1 september 2022 kl. 15:34:54 UTC+2 skrev 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

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

Ben Bacarisse schrieb am Freitag, 2. September 2022 um 05:10:58 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 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.
> 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.

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

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

On 9/2/2022 8:51 AM, WM wrote:
> Ben Bacarisse schrieb am Freitag, 2. September 2022 um 05:10:58 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 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.
>> 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.
>
> That is provable,

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>
>
>> 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

On 9/2/2022 8:53 AM, WM wrote:
> 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.

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

WM <askasker48@gmail.com> writes:

> Ben Bacarisse schrieb am Freitag, 2. September 2022 um 05:10:58 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 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.
>> 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.
>
> That is provable, but only for the definable part.

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.

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

WM <askasker48@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.