On 9/10/2022 3:45 PM, WM wrote:
> 11 Matrizen
> Eine m × n-Matrix ist ein Raster aus m × n Koeffizienten, die in m Zeilen und n Spalten
> angeordnet sind:
> (aij)1≤i≤m,1≤j≤n (11.1)
> Zur Abkürzung werden wir auch kurz (aij) schreiben,
>
> Regards, WM

Nullmatrix oder Nullmatrix ist auch eine Matrix, die alle ihre Elemente enthält, sind Null.
Die Nullmatrix hat verschiedene Zeilen und Spalten.
Die Nullmatrizen können auch eine quadratische Matrix sein.
Wir können die Nullmatrix durch den Buchstaben "O" darstellen.
Es fungiert auch als additive Identitätsmatrix, die zu derselben Matrix führt, wenn sie hinzugefügt wird, wobei die Matrix eine Ordnung von Matrix hat,
die m x n ist.

Der Schnittpunkt aller Endsegmente ist eine leere Matrix

Die Schnittmenge aller FISONs ist {1}

On 9/10/2022 3:57 PM, WM wrote:
> Jim Burns schrieb am Samstag, 10. September 2022 um 22:07:47 UTC+2:
>
>> each 'delete-least' step
>> which starts at an infinite end segment
>> ends at an infinite end segment.
>
> Each deletion concerns only one elements. If all endsegments were infinite but the deletion leads to the empty set, then the last step (the step that produces the empty set) would be larger than one element.
>
> Regards, WM

There is no "last step" silly, this is an infinite sequence!

Es gibt keinen "letzten Schritt" albern, das ist eine unendliche Sequenz!

On 9/10/2022 3:52 PM, WM wrote:
> Gus Gassmann schrieb am Samstag, 10. September 2022 um 17:54:43 UTC+2:
>> It is important that no natural number is element of the intersection over all end segments.
>
> What do we conclude from this important fact?

no need to conclude, it is proved, and you cannot un prove it.

> The general construction formula for endsegments
> ∀k ∈ ℕ: E(k+1) = E(k) \ {k}

wrong, that is only the relation between two adjacent endsegments, you have failed.

> shows that no loss of elements can be larger than one element.

which has nothing to do with ALL ENDSEGMENTS

> This proves the existence of dark endsegments because all visible endsegments are infinite.

Du wünschst, es beweist, dass du Narr bist.

>
> Regards, WM
>

On 9/10/2022 3:40 PM, WM wrote:
> Fritz Feldhase schrieb am Samstag, 10. September 2022 um 06:17:39 UTC+2:
>> On Saturday, September 10, 2022 at 12:40:53 AM UTC+2, FromTheRafters wrote:
>>
>>> All natural numbers less than or equal to n cannot be anything else but
>>> finite.
>> Should read: "The set of all natural numbers less than or equal to n cannot be anything else but finite."
>
> Should read: For all n ∈ ℕ: The set of all natural numbers less than or equal to n cannot be anything else but finite. The set of all definable natural numbers cannot be anything else but finite.
>
> Regards, WM

definable = QUACK!!

On 9/10/2022 3:33 PM, WM wrote:
> Fritz Feldhase schrieb am Samstag, 10. September 2022 um 06:07:40 UTC+2:
>> On Friday, September 9, 2022 at 10:10:14 PM UTC+2, WM wrote:
>>>
>>> The empty set cannot follow immediately upon infinite sets.
>>
>> The set {E_n : n e IN} u {{}} is certainly ordered the following way (by the subset relation):
>>
>> E_1 > E_2 > E_3 > ... > {}.
>
> If all endsegments are infinite, then there is a discontinuity, a jump from more than one element to 0 elements. This is wrong because a jump is prohobited by
> ∀k ∈ ℕ: E(k+1) = E(k) \ {k} .
>
> Regards, WM

Sie stecken bei diesem einfachen Problem fest.
Du kennst nicht genug Mathematik, um es zu erkennen oder dich selbst zu korrigieren.

On Saturday, September 10, 2022 at 10:45:47 PM UTC+2, WM wrote:
> JVR schrieb am Samstag, 10. September 2022 um 09:40:17 UTC+2:
>
> > In fact, matrices are not really defined.
>
> 11 Matrizen
> Eine m × n-Matrix ist ein Raster aus m × n Koeffizienten, die in m Zeilen und n Spalten
> angeordnet sind:
> (aij)1≤i≤m,1≤j≤n (11.1)
> Zur Abkürzung werden wir auch kurz (aij) schreiben,
>
> Regards, WM

OK - in the Bestseller there is a 30-page chapter on 'Linear Systems of Equations'.
The first paragraph contains a rather drastic error (can you find it?), but here we
don't need to worry about that. Here we are looking for a usable definition of an
infinite matrix, by means of which we will prove Cantor wrong.

In English the definition of a matrix in the Bestseller is, as Professor Doktor habil. (approx.) Mückenheim
states above:
"An m x n matrix is a grid of m x n coefficients, arranged in m rows and n columns."

To make a useful definition out of this for the present purpose, we need m and n to
be infinite and we need something more distinct than 'coefficients' as the entries
in the 'grid'. The word 'raster' (grid) occurs nowhere else in the Bestseller.

Don't you agree? So let's get to work and define this object.

Fritz Feldhase schrieb am Samstag, 10. September 2022 um 22:59:55 UTC+2:
> On Saturday, September 10, 2022 at 10:33:50 PM UTC+2, WM wrote:

> > If all endsegments are infinite, then there is a discontinuity, a jump from more than one element to 0 elements.
> Nope, there is no jump.
>
> Hint: For each and every ednsegment E there is an endsegment E' such that E > E' > {}.
>
Irrelevant, since they are infinite. As long as all intersections of definable endsegments are infinite and the intersection of all endsegments is zero, there is either a dark realm of descending from infinity to zero or a jump.

Regards, WM

Gus Gassmann schrieb am Samstag, 10. September 2022 um 23:06:09 UTC+2:
>
> All natural numbers are not just definable, but defined.

The set is defined, not its indivduals.

> Therefore, all end segments are also defined (as well as definable).

No. But the dark is easier to see by the matrix of X's and O's.

> Since there are infinitely many of them, there intersection is empty.

No sufficient reason.

Regards, WM

Gus Gassmann schrieb am Samstag, 10. September 2022 um 23:08:12 UTC+2:
> On Saturday, 10 September 2022 at 17:52:23 UTC-3, WM wrote:
> > Gus Gassmann schrieb am Samstag, 10. September 2022 um 17:54:43 UTC+2:
> > > It is important that no natural number is element of the intersection over all end segments.
> > What do we conclude from this important fact?
The general construction formula for endsegments
∀k ∈ ℕ: E(k+1) = E(k) \ {k}
shows that no loss of elements can be larger than one element.
This proves the existence of dark endsegments because all visible endsegments are infinite.

> That the intersection over all end segment is empty,

Of course.

Regards, WM

On 9/11/2022 5:20 AM, JVR wrote:
> On Saturday, September 10, 2022 at 10:45:47 PM UTC+2, WM wrote:
>> JVR schrieb am Samstag, 10. September 2022 um 09:40:17 UTC+2:
>>
>>> In fact, matrices are not really defined.
>>
>> 11 Matrizen
>> Eine m × n-Matrix ist ein Raster aus m × n Koeffizienten, die in m Zeilen und n Spalten
>> angeordnet sind:
>> (aij)1≤i≤m,1≤j≤n (11.1)
>> Zur Abkürzung werden wir auch kurz (aij) schreiben,
>>
>> Regards, WM
>
> OK - in the Bestseller there is a 30-page chapter on 'Linear Systems of Equations'.
> The first paragraph contains a rather drastic error (can you find it?), but here we
> don't need to worry about that. Here we are looking for a usable definition of an
> infinite matrix, by means of which we will prove Cantor wrong.
>
> In English the definition of a matrix in the Bestseller is, as Professor Doktor habil. (approx.) Mückenheim
> states above:
> "An m x n matrix is a grid of m x n coefficients, arranged in m rows and n columns."
>
> To make a useful definition out of this for the present purpose, we need m and n to
> be infinite and we need something more distinct than 'coefficients' as the entries
> in the 'grid'. The word 'raster' (grid) occurs nowhere else in the Bestseller.
>
> Don't you agree? So let's get to work and define this object.

(REM: a matrix has brackets around it)

Ben Bacarisse schrieb am Sonntag, 11. September 2022 um 02:13:16 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 Freitag, 9. September 2022 um 23:06:26 UTC+2:
> >> WM <askas...@gmail.com> writes:
> >> (AKA Dr. Wolfgang Mückenheim or Mueckenheim who teaches "Geschichte des
> >> Unendlichen" at Hochschule Augsburg.)
> >> > It is what we have been discussing in extensio. It is nonsense to
> >> > claim the indexing of all fractions if most will be remaining not
> >> > indexed forever.
> >> My argument was entirely within WMaths. It used a potentially infinite
> >> bijection between potentially infinite sets to define a potentially
> >> infinite sequence of potentially infinite functions.
> >
> > There is no problem.
> You accept my argument that, in WMaths, "the result" of the swaps
> (suitably defined) is no Os in any finite position?

At the end there is in every case no definable O at any position. In my Mathematics, there are no undefinable O's. Here we discuss Cantor's theory.

> > But I have been analyzing Cantor's completed infinity with a complete
> > matrix of fractions.
> But you don't want to analyse the consequences of the swaps in potential
> infinity.

Why not? IN EVERY CASE there are no O visible at the end of the process. I never refused this. It is the proof of dark numbers.

> > Cantor has explicitly denied the potential infinity.
> So much more interesting, then, that the Xs "disappear" in potential
> infinity as well! You won't ever see how, but they do.

You mean the O's?
>
> His name is Burns. He does not disagree with what you write above. No
> one does.

As long as all intersections of definable endegments are infinite and the intersection of all endsegments is zero, there is either a dark realm of descending from infinity to zero or a jump.
>
> And you keep not answering this question:
>
> The limit is as per your book. Are there dark objects in WMaths?

Look, the limit of the sequence 1/n is zero. Between the definable terms 1/n and 0, there is either a gap or dark points.

Regards, WM

On Sunday, 11 September 2022 at 10:33:12 UTC-3, WM wrote:
> Gus Gassmann schrieb am Samstag, 10. September 2022 um 23:06:09 UTC+2:
> >
> > All natural numbers are not just definable, but defined.
> The set is defined, not its indivduals.
> > Therefore, all end segments are also defined (as well as definable).
> No. But the dark is easier to see by the matrix of X's and O's.

You have no clue about infinite matrices. But why should you? You have no clue about anything else, either.

> > Since there are infinitely many of them, there intersection is empty.
> No sufficient reason.

Go fuck yourself.

Ben Bacarisse schrieb am Sonntag, 11. September 2022 um 02:13:16 UTC+2:
> WM <askas...@gmail.com> writes:

> The limit is as per your book. Are there dark objects in WMaths?

I do not talk about the possible contents of the gaps around every definable point on the real axis. There is no next point to any definable point. Therefore either there is nothing or something dark.

> > There are no set limits in my book.
>
> Wrong. Functions are sets (see page 18).

Yes, but there is a difference between analysis where the terms can be interpreted as sets and manipulations like indexing. It seems important to fix this difference. I have not pondered about it because I was discussing always only the latter.

Regards, WM

Ben Bacarisse schrieb am Sonntag, 11. September 2022 um 02:22:53 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 Freitag, 9. September 2022 um 02:06:41 UTC+2:
>
> >> What positions
> >> have Os (you asked about positions originally) "in the limit" depends on
> >> the pattern of swaps
> >
> > The pattern is given by Cantor.
>
> You neither cited nor quoted Cantor saying anything about swaps. I
> think you made that bit up yourself, but I would love to read how Cantor
> defines the swaps and the result of an endless sequence of swaps.

Assigning an index to a fraction is expressed by assigning an X to an O in my matrix. That is simple. The only difference to Cantor's procedure is that I first assign every X to a unit-fraction. You must forbid this in order to save Cantor.

> > Limits are irrelevant when we ask for indexing with natural numbers,
> > i.e. in finite steps before the limit.
>
> You asked about "the result" of an endless sequence of swaps. That can
> not be revealed by the indicator function I_n for any n in N. Something
> else is needed to make sense of the phrase "the result".

Cantor's statemet that it is possible is taken as the basis.

> I proposed the
> limit of the sequence of indicator function. And, behold, with some
> patters of swaps, the limit is the function I(i) = 1, meaning no Os in
> any position, just as you claim is the case in set theory!

The result is in fact that no O's can be found in any position. But fact is also that they cannot have disappeared from the matrix by exchanges with X..

Regards, WM

On Sunday, September 11, 2022 at 3:29:52 PM UTC+2, WM wrote:
> Fritz Feldhase schrieb am Samstag, 10. September 2022 um 22:59:55 UTC+2:
> > On Saturday, September 10, 2022 at 10:33:50 PM UTC+2, WM wrote:
> > >
> > > If all endsegments are infinite, then there is a discontinuity, a jump from more than one element to 0 elements.
> > Nope, there is no jump.
> >
> > Hint: For each and every ednsegment E there is an endsegment E' such that E > E' > {}.
> >
> Irrelevant, since they are infinite.

Non sequitur.

> [If] the intersection of all endsegments is zero, there is either a dark realm of descending from infinity to zero or a jump.

See: https://en.wikipedia.org/wiki/False_dilemma

WM wrote on 9/11/2022 :
> Ben Bacarisse schrieb am Sonntag, 11. September 2022 um 02:22:53 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 Freitag, 9. September 2022 um 02:06:41 UTC+2:
>>>> What positions
>>>> have Os (you asked about positions originally) "in the limit" depends on
>>>> the pattern of swaps
>>>
>>> The pattern is given by Cantor.
>>
>> You neither cited nor quoted Cantor saying anything about swaps. I
>> think you made that bit up yourself, but I would love to read how Cantor
>> defines the swaps and the result of an endless sequence of swaps.
>
> Assigning an index to a fraction is expressed by assigning an X to an O in my
> matrix. That is simple. The only difference to Cantor's procedure is that I
> first assign every X to a unit-fraction. You must forbid this in order to
> save Cantor.

Or, put another way, you must not do this this way if you want to
understand Cantor.

on 9/11/2022, WM supposed :
> Gus Gassmann schrieb am Samstag, 10. September 2022 um 23:06:09 UTC+2:
>>
>> All natural numbers are not just definable, but defined.
>
> The set is defined, not its indivduals.

Then how do you prove that the 'individuals' qualify for inclusion?

Jim Burns schrieb am Sonntag, 11. September 2022 um 00:12:46 UTC+2:
> On 9/10/2022 4:57 PM, WM wrote:

> > Each deletion concerns only one element.
> And that determines which sets we are
> talking about.

But infinitely many elements remain in all endsegments we are talking about..
You believe that there is no finite endsegment. Therefore the empty set is not reached in a definable way.

> > If all endsegments were infinite
> > but the deletion leads to the empty set,
> The deletions do not lead to the intersection
> of all end segments (which = ∅)

No? Do you prefer matheologial abracadabra? I don't!
>
> There is no sequence of 'delete-least' steps
> which ends at ∅

Then you deny the universal quantifier in ∀k ∈ ℕ: E(k+1) = E(k) \ {k} ?

Whatever your opinion is: Infinitely many elements disappear in an undefinable way.

Regards, WM

On Sunday, September 11, 2022 at 6:15:58 PM UTC+2, WM wrote:

> But infinitely many elements [are] in [each and every] endsegment[.]

Indeed!

> You believe that there is no finite endsegment.

No, we do not "believe" that. We KNOW that, since the definition of the notion /endsegment/ IMPLIES that.

Hint: Each and every enegments has -by definition- a smallest element (which is a natural number). Moreover if min(E) is the smallest natural number of the endsegment E, then E contains (just) ALL natural numbers >= min(E) _by definition_.

In other words, for each and every endsegment E

E = {n e IN : n >= min(E)}

holds BY DEFINITION.

Theorem: There is no finite endsegment.

Proof:
Assume that there is a finite endsegment, say, WM. Then WM has a maximal element max(WM) e IN. So we would have: Ax e WM: x e IN & x <= max(WM). Since max(WM) e IN, max(WM) +1 e IN. Moreover max(WM)+1 > max(WM). Now, since WM is an endsegment, WM = {n e IN : n >= min(WM)}. Hence, since max(WM)+1 > max(WM) [and max(WM)+1 e IN] and (clearly) max(WM) >= min(WM), we have max(WM)+1 e WM. Contradiction! (Since max(WM)+1 > max(WM).) qed

> Therefore the empty set is <bla bla>

Yeah whatever, Mückenheim.

Note though:

> > The "deletions" do not lead to the intersection of all end segments (which = ∅)
> >
> No?

No.

> > There is no sequence of 'delete-least' steps which ends at ∅

Indeed!

> Then you deny [...] ∀k ∈ ℕ: E(k+1) = E(k) \ {k} ?

No, we don't.

Hint: There is no k e IN such that E(k) = ∅. [And ∀k ∈ ℕ: E(k+1) = E(k) \ {k} does NOT impy it.]

Proof: For all k e IN: k e E(k) = {n e IN : n >= k}. (Since k >= k, you know.) Hence for all k e IN: E(k) =/= ∅.

> Whatever your opinion is <bla bla>

Was auch immer, Mückenheim.

Fritz Feldhase schrieb am Sonntag, 11. September 2022 um 16:25:22 UTC+2:
> On Sunday, September 11, 2022 at 3:29:52 PM UTC+2, WM wrote:
> > Fritz Feldhase schrieb am Samstag, 10. September 2022 um 22:59:55 UTC+2:
> > > On Saturday, September 10, 2022 at 10:33:50 PM UTC+2, WM wrote:
> > > >
> > > > If all endsegments are infinite, then there is a discontinuity, a jump from more than one element to 0 elements.
> > > Nope, there is no jump.
> > >
> > > Hint: For each and every ednsegment E there is an endsegment E' such that E > E' > {}.
> > >
> > Irrelevant, since they are infinite.
> Non sequitur.
>
> > [If] the intersection of all endsegments is zero, there is either a dark realm of descending from infinity to zero or a jump.
>
> See: https://en.wikipedia.org/wiki/False_dilemma

No. Stupid matheologians cannot understand between all intersections of endsegments covered by the universal quantifier
∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀
and the intersection of all endsegments
∩{E(k) : k ∈ ℕ} = { }
there is a jump violating
∀k ∈ ℕ: |E(k+1)| = |E(k)| - 1
resulting from
∀k ∈ ℕ: E(k+1) = E(k) \ {k
or a dark realm.

Regards, WM

FromTheRafters schrieb am Sonntag, 11. September 2022 um 17:40:16 UTC+2:
> WM wrote on 9/11/2022 :
> > Ben Bacarisse schrieb am Sonntag, 11. September 2022 um 02:22:53 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 Freitag, 9. September 2022 um 02:06:41 UTC+2:
> >>>> What positions
> >>>> have Os (you asked about positions originally) "in the limit" depends on
> >>>> the pattern of swaps
> >>>
> >>> The pattern is given by Cantor.
> >>
> >> You neither cited nor quoted Cantor saying anything about swaps. I
> >> think you made that bit up yourself, but I would love to read how Cantor
> >> defines the swaps and the result of an endless sequence of swaps.
> >
> > Assigning an index to a fraction is expressed by assigning an X to an O in my
> > matrix. That is simple. The only difference to Cantor's procedure is that I
> > first assign every X to a unit-fraction. You must forbid this in order to
> > save Cantor.
> Or, put another way, you must not do this this way if you want to
> understand Cantor.

Or, put it yet another way, you must not check his method if you want to believe in it.

Regards, WM

FromTheRafters schrieb am Sonntag, 11. September 2022 um 17:43:02 UTC+2:
> on 9/11/2022, WM supposed :
> > Gus Gassmann schrieb am Samstag, 10. September 2022 um 23:06:09 UTC+2:
> >>
> >> All natural numbers are not just definable, but defined.
> >
> > The set is defined, not its indivduals.
> Then how do you prove that the 'individuals' qualify for inclusion?

I assume ∀k ∈ ℕ: E(k+1) = E(k) \ {k} to be correct by the definition of endsegment.

Regards, WM

WM submitted this idea :
> FromTheRafters schrieb am Sonntag, 11. September 2022 um 17:40:16 UTC+2:
>> WM wrote on 9/11/2022 :
>>> Ben Bacarisse schrieb am Sonntag, 11. September 2022 um 02:22:53 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 Freitag, 9. September 2022 um 02:06:41 UTC+2:
>>>>>> What positions
>>>>>> have Os (you asked about positions originally) "in the limit" depends on
>>>>>> the pattern of swaps
>>>>>
>>>>> The pattern is given by Cantor.
>>>>
>>>> You neither cited nor quoted Cantor saying anything about swaps. I
>>>> think you made that bit up yourself, but I would love to read how Cantor
>>>> defines the swaps and the result of an endless sequence of swaps.
>>>
>>> Assigning an index to a fraction is expressed by assigning an X to an O in
>>> my matrix. That is simple. The only difference to Cantor's procedure is
>>> that I first assign every X to a unit-fraction. You must forbid this in
>>> order to save Cantor.
>> Or, put another way, you must not do this this way if you want to
>> understand Cantor.
>
> Or, put it yet another way, you must not check his method if you want to
> believe in it.

I just checked again, and it is still working fine. BTW, I didn't check
step by step.

Fritz Feldhase schrieb am Sonntag, 11. September 2022 um 18:37:32 UTC+2:
> On Sunday, September 11, 2022 at 6:15:58 PM UTC+2, WM wrote:

> > > The "deletions" do not lead to the intersection of all end segments (which = ∅)
> > >
> > No?
> No.

Then we differ in significant precondition.
My opinion about all deletions is this
∀k ∈ ℕ: E(k+1) = E(k) \ {k}
∀k ∈ ℕ: |E(k+1)| = |E(k)| - 1.
Your deletions are brought about by matheologial abracadabra. I cannot join you.

> > > There is no sequence of 'delete-least' steps which ends at ∅
> Indeed!
>
> > Then you deny [...] ∀k ∈ ℕ: E(k+1) = E(k) \ {k} ?
>
> No, we don't.
>
Then the empty set requires a last deletion of one element.

Regards, WM

FromTheRafters schrieb am Sonntag, 11. September 2022 um 20:58:25 UTC+2:
> WM submitted this idea :

> > Or, put it yet another way, you must not check his method if you want to
> > believe in it.
> I just checked again,

You failed. The set of O's cannot be removed.

> BTW, I didn't check
> step by step.

Not necessary. We know that never, in no step, any O will vanish.

Regards, WM