On 7/11/2022 7:46 AM, WM wrote:
> Eram semper recta schrieb am Montag, 11. Juli 2022 um 10:43:22 UTC+2:
>> On Sunday, 10 July 2022 at 20:05:56 UTC+3, WM wrote:
>>> Eram semper recta schrieb am Sonntag, 10. Juli 2022 um 18:19:17 UTC+2:
>
>> I know that Cantor said "ALL" elements, but it's pretty obvious he was on the way out mentally. I personally don't think he meant "ALL".
>
> He meant all. He emphasized it frequently.
>
> He believed that the matrix
>
> XOOO...
> XOOO...
> XOOO...
> XOOO...
> ...
>
> could be covered with X's completely by moving them around according to his precription.
>
> Regards, WM

you can learn something from Mr Rectum.

On Monday, 11 July 2022 at 09:29:22 UTC-3, WM wrote:
[...]
> The sequence 10 + 1/n has no minimum. We can replace it by 10.

You really are a *FUCKING* idiot. The sequence -1, -2, -3, ... has no minimum, either. What finite number would you like to replace it by?

On Monday, 11 July 2022 at 09:51:51 UTC-3, WM wrote:
> The sequence has only infinite terms. Therefore we can replace every term by a finite term for an estimation of the lower bound of the intersection.

That is still fucking bullshit. What would a *lower* bound on the intersection look like? The only persuasive lower bound is the empty set. Congratulations!

Gus Gassmann schrieb am Montag, 11. Juli 2022 um 16:03:17 UTC+2:
> On Monday, 11 July 2022 at 09:51:51 UTC-3, WM wrote:
> > The sequence has only infinite terms. Therefore we can replace every term by a finite term for an estimation of the lower bound of the intersection.
> What would a *lower* bound on the intersection look like?

It would be dark. But it would contain, say, 100 elements which allegedly are contained in every infinite endsegment (and allegedly be deleted in the intersection of infinitely many infinite endsegments).

Regards, WM

On Monday, July 11, 2022 at 2:51:51 PM UTC+2, WM wrote <bla>

YOUR CLAIM WAS:

WM> The intersection of [the terms of] an inclusion monotonic sequence of sets is the least [term]

FF> IF THERE IS A LEAST TERM (!)

WM> There is always a least term.

No, there isn't. ESPECIALLY there's no least set in {E(n) : n e IN} and hence no least term in the sequence (E(n))_(n e IN).

Hint: The sequence (E(n))_(n e IN) is STRICTLY inclusion-monotonic. This EXCLUDES a minimal term.

You know: E(1) > E(2) > E(3) > ... > E(n) > E(n+1) > E(n+2) > ...

Got it this time?

On 7/11/2022 9:11 AM, WM wrote:
> Gus Gassmann schrieb am Montag, 11. Juli 2022 um 16:03:17 UTC+2:
>> On Monday, 11 July 2022 at 09:51:51 UTC-3, WM wrote:
>>> The sequence has only infinite terms. Therefore we can replace every term by a finite term for an estimation of the lower bound of the intersection.
>> What would a *lower* bound on the intersection look like?
>
> It would be dark.

Bullshit.

> But it would contain, say, 100 elements

Bullshit

> which allegedly are contained in every infinite endsegment

Bullshit

> (and allegedly be deleted in the intersection of infinitely many infinite endsegments).

Bullshit, "allegedly"

>
> Regards, WM

4/4 on the bullshit meter! 100% BS

On Monday, July 11, 2022 at 4:11:08 PM UTC+2, WM wrote:
> Gus Gassmann schrieb am Montag, 11. Juli 2022 um 16:03:17 UTC+2:

> > What would a *lower* bound on the intersection look like?
> >
> It would be dark. But it would contain, say, 100 elements which allegedly are contained in every infinite endsegment (and allegedly be deleted in the intersection of infinitely many infinite endsegments).

Du redest wirres Zeug, Mann!

It's getting worse with you, Mückenheim!

On Monday, 11 July 2022 at 11:11:08 UTC-3, WM wrote:
> Gus Gassmann schrieb am Montag, 11. Juli 2022 um 16:03:17 UTC+2:
> > On Monday, 11 July 2022 at 09:51:51 UTC-3, WM wrote:
> > > The sequence has only infinite terms. Therefore we can replace every term by a finite term for an estimation of the lower bound of the intersection.
> > What would a *lower* bound on the intersection look like?
> It would be dark. But it would contain, say, 100 elements which allegedly are contained in every infinite endsegment (and allegedly be deleted in the intersection of infinitely many infinite endsegments).

Nobody except you, perfosser, thinks that the same 100 elements are contained in *every* end segment. (And if it is *not* the same 100 elements, then no thinking person should expect that a fixed number of 100 elements should be contained in any infinite intersection.) Your mind has slipped to the point where you are no longer capable of rational thought. The sooner you accept this fact and stop teaching your lunatic bullshit to your students, the better for *ALL* concerned, _including_you_.

On Monday, 11 July 2022 at 15:47:06 UTC+3, WM wrote:
> Eram semper recta schrieb am Montag, 11. Juli 2022 um 10:43:22 UTC+2:
> > On Sunday, 10 July 2022 at 20:05:56 UTC+3, WM wrote:
> > > Eram semper recta schrieb am Sonntag, 10. Juli 2022 um 18:19:17 UTC+2:
>
> > I know that Cantor said "ALL" elements, but it's pretty obvious he was on the way out mentally. I personally don't think he meant "ALL".
> He meant all. He emphasized it frequently.

He may well have emphasized it frequently but that does not mean he meant "ALL". He was saying ALL in a manner of speaking which of course is not correct, but whatever... He earned the title I bestowed on him - "Father of all mainstream mathematical cranks".

In any case, the concept of countability is a pile of junk from which nothing useful can be learned.

>
> He believed that the matrix
>
> XOOO...
> XOOO...
> XOOO...
> XOOO...
> ...
>
> could be covered with X's completely by moving them around according to his precription.

Clearly delusional as any sane thinker today knows so well. If you're going to nitpick Cantor's drivel, all you will find is more drivel.
Cantor was wrong from his first line of argument wrt his diagonal bowel movement, that is, he assumed that all real numbers can be represented by an infinite decimal or radix representation. The lunatic was correct about the imaginary set of "real numbers" being uncountable - they don't exist!

Bashing Cantor on this pile of dung called sci.math is like calling the prophet Mohammed a child of the devil whilst visiting Mecca. Cantor cranks will never accept that their master was wrong at just about everything. However, one can always hope that unpolluted minds will learn.

>
> Regards, WM

Gus Gassmann schrieb am Montag, 11. Juli 2022 um 17:53:14 UTC+2:
> On Monday, 11 July 2022 at 11:11:08 UTC-3, WM wrote:
> > Gus Gassmann schrieb am Montag, 11. Juli 2022 um 16:03:17 UTC+2:
> > > On Monday, 11 July 2022 at 09:51:51 UTC-3, WM wrote:
> > > > The sequence has only infinite terms. Therefore we can replace every term by a finite term for an estimation of the lower bound of the intersection.
> > > What would a *lower* bound on the intersection look like?
> > It would be dark. But it would contain, say, 100 elements which allegedly are contained in every infinite endsegment (and allegedly be deleted in the intersection of infinitely many infinite endsegments).
> Nobody except you, thinks that the same 100 elements are contained in *every* end segment.

I can't believe that all are so stupid, but those who deny it must be very stupid.

We have a decreasing sequence of sets. As long as all sets contain more than X, they contain the same X in common. If all sets are are infinite, then infinitely many elements are common to all sets, among them are100 elements common to all sets.

Regards, WM

Gus Gassmann schrieb am Montag, 11. Juli 2022 um 16:01:06 UTC+2:
> On Monday, 11 July 2022 at 09:29:22 UTC-3, WM wrote:
> [...]
> > The sequence 10 + 1/n has no minimum. We can replace it by 10.
> The sequence -1, -2, -3, ... has no minimum, either. What finite number would you like to replace it by?

A decreasing sequence of positive numbers has a minimum or a limit.

Regards, WM

Fritz Feldhase schrieb am Montag, 11. Juli 2022 um 16:25:48 UTC+2:
> On Monday, July 11, 2022 at 2:51:51 PM UTC+2, WM wrote <bla>
> YOUR CLAIM WAS:
>
> WM> The intersection of [the terms of] an inclusion monotonic sequence of sets is the least [term]
>
> FF> IF THERE IS A LEAST TERM (!)
>
> WM> There is always a least term.
>
> No, there isn't.

In potential infinity there is always a largest term, although it changes. Same with the decreasing sequence. The least term exists but it is not constant. Nevertheless, if you dislike potential infinity, you can assume the infimum. It is infinite. Since the sequence of endsegments is decreasing, all infinite endsegments have the same infinite subset in common.

Regards, WM

On 7/11/2022 12:02 PM, WM wrote:
> Gus Gassmann schrieb am Montag, 11. Juli 2022 um 17:53:14 UTC+2:
>> On Monday, 11 July 2022 at 11:11:08 UTC-3, WM wrote:
>>> Gus Gassmann schrieb am Montag, 11. Juli 2022 um 16:03:17 UTC+2:
>>>> On Monday, 11 July 2022 at 09:51:51 UTC-3, WM wrote:
>>>>> The sequence has only infinite terms. Therefore we can replace every term by a finite term for an estimation of the lower bound of the intersection.
>>>> What would a *lower* bound on the intersection look like?
>>> It would be dark. But it would contain, say, 100 elements which allegedly are contained in every infinite endsegment (and allegedly be deleted in the intersection of infinitely many infinite endsegments).
>> Nobody except you, thinks that the same 100 elements are contained in *every* end segment.
>
> I can't believe that all are so stupid, but those who deny it must be very stupid.
>
> We have a decreasing sequence of sets. As long as all sets contain more than X, they contain the same X in common. If all sets are are infinite, then infinitely many elements are common to all sets, among them are100 elements common to all sets.
>
> Regards, WM

What part of this do you not understand WM ?

pick any element of an endsegment, call it k

but k is not in E(k+1)

therefore k is not in the intersection of all endsegments,

since we picked any k, it applies to all natural numbers

therefore the intersection of all endsegments is *empty*

On 7/11/2022 12:11 PM, WM wrote:
> Fritz Feldhase schrieb am Montag, 11. Juli 2022 um 16:25:48 UTC+2:
>> On Monday, July 11, 2022 at 2:51:51 PM UTC+2, WM wrote <bla>
>> YOUR CLAIM WAS:
>>
>> WM> The intersection of [the terms of] an inclusion monotonic sequence of sets is the least [term]
>>
>> FF> IF THERE IS A LEAST TERM (!)
>>
>> WM> There is always a least term.
>>
>> No, there isn't.
>
> In CLOWN SHOES infinity there is always a largest term, although it changes. Same with the decreasing sequence.

finite sequences

> The least term exists but it is not constant.

so it is not the least, or is it, make up your mind,

> Nevertheless, if you dislike CLOWN SHOES infinity, you can assume the infimum. It is infinite. Since the sequence of endsegments is decreasing,

Wrong.

> all infinite endsegments have the same infinite subset in common.

Wrong. Try to prove it. You cannot.

You are out of math again.

>
> Regards, WM

On 7/11/2022 12:07 PM, WM wrote:
> Gus Gassmann schrieb am Montag, 11. Juli 2022 um 16:01:06 UTC+2:
>> On Monday, 11 July 2022 at 09:29:22 UTC-3, WM wrote:
>> [...]
>>> The sequence 10 + 1/n has no minimum. We can replace it by 10.
>> The sequence -1, -2, -3, ... has no minimum, either. What finite number would you like to replace it by?
>
> A decreasing sequence of positive numbers has a minimum or a limit.

prove it

>
> Regards, WM

On Monday, July 11, 2022 at 9:00:39 AM UTC-3, WM wrote:
ith endsegments.
> (1) Every endsegment has all its elements in common with all its predecessors.
> (2) All infinite endsegments have infinitely many elements in common with all their predecessors.
> (3) A sequence of infinite endsegments consists only of infinite predecessors because every endsegment has an infinite successor.

However, each endsegment has a different infinite successor.

> (4) Therefore all infinite endsegments have an infinite set of elements in common.

Does not follow and is not true.

--
William Hughes

Eram semper recta schrieb am Montag, 11. Juli 2022 um 17:55:49 UTC+2:

> He may well have emphasized it frequently but that does not mean he meant "ALL".

If he not meant all, then his bijections were no bijections and without any value. Here is what he said to the topic:

1.1.1 Vollständig

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

"gegenseitig eindeutige und vollständige Korrespondenz" [Cantor, p. 238]

"Die sämtlichen Punkte l unsrer Menge L sind also in gegenseitig eindeutige und vollständige Beziehung zu sämtlichen Punkten f der Menge F gebracht," [Cantor, p. 241]

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

"Zwei bestimmte Mengen M und M1 nennen wir äquivalent (in Zeichen: M ~ M1), wenn es möglich ist, dieselben gesetzmäßig, gegenseitig eindeutig und vollständig, Element für Element, einander zuzuordnen." [Cantor, p. 412]

"doch gibt es immer viele, im allgemeinen sogar unzählig viele Zuordnungsgesetze, durch welche zwei äquivalente Mengen in gegenseitig eindeutige und vollständige Beziehung zueinander gebracht werden können.." [Cantor, p. 413]

"eine solche gegenseitig eindeutige und vollständige Korrespondenz hergestellt [...] irgendeine gegenseitig eindeutige und vollständige Zuordnung der beiden Mengen [...] auch eine gegenseitig eindeutige und vollständige Korrespondenz" [Cantor, p. 415]

"Zwei n-fach geordnete Mengen M und N werden 'ähnlich' genannt, wenn es möglich ist, sie gegenseitig eindeutig und vollständig, Element für Element, einander so zuzuordnen," [Cantor, p. 424]

But nowadays his followers try also to change the meaning of his words.

Regards, WM

On Monday, July 11, 2022 at 9:35:37 AM UTC-3, WM wrote:
> William schrieb am Sonntag, 10. Juli 2022 um 19:41:09 UTC+2:

> > There is no one subset that works for each endsegment.
>
> Of course it is.

Nope. It is trivial to show that given a non empty subset of natural numbers, there is an endsegment that does not contain it

--
William Hughes

On Monday, July 11, 2022 at 11:11:08 AM UTC-3, WM wrote:

> It would be dark.

|N, a Peano set, does not contain dark elements.

On Monday, July 11, 2022 at 7:11:18 PM UTC+2, WM wrote:
> Fritz Feldhase schrieb am Montag, 11. Juli 2022 um 16:25:48 UTC+2:
> > On Monday, July 11, 2022 at 2:51:51 PM UTC+2, WM wrote <bla>
> > YOUR CLAIM WAS:
> >
> > WM> The intersection of [the terms of] an inclusion monotonic sequence of sets is the least [term]
> >
> > FF> IF THERE IS A LEAST TERM (!)
> >
> > WM> There is always a least term.
> >
> > No, there isn't.
> >
> In potential infinity <bla bl bla>

We DON'T TALK ABOUT "potential infinity" here, dumbo, but about (MODERN/AXIOMATIC) SET THEORY.

Remember: "In the philosophy of mathematics, the abstraction of actual infinity involves the acceptance (if the axiom of infinity is included) of infinite entities as given, actual and completed objects. These might include the set of natural numbers, extended real numbers, transfinite numbers, or even an infinite sequence of rational numbers. Actual infinity is to be contrasted with potential infinity [...]" (Wikipedia)

What's the matter with you? Are you demented?

REMEMBER your claim was: "The intersection of [the terms of] an inclusion monotonic sequence of sets is the least [term]."

THE CONTEXT OF YOUR CLAIM IS SET THEORY, of course!

HOLY SHIT!!!

It's getting worse with you, Mückenheim!

William schrieb am Montag, 11. Juli 2022 um 19:33:16 UTC+2:
> On Monday, July 11, 2022 at 9:00:39 AM UTC-3, WM wrote:
> ith endsegments.
> > (1) Every endsegment has all its elements in common with all its predecessors.
> > (2) All infinite endsegments have infinitely many elements in common with all their predecessors.
> > (3) A sequence of infinite endsegments consists only of infinite predecessors because every endsegment has an infinite successor.
> However, each endsegment has a different infinite successor.

Highly irrelevant. Infinite sets differing by finitely many elements have an infinite intersection.

> > (4) Therefore all infinite endsegments have an infinite set of elements in common.
> Does not follow and is not true.

Is obvious because it is a decreasing sequence with only infinite endsegments. Every element of every endsegment is common to all predecessors. Never a decrease to less than infinitely many elements happens. Hence infinitely many remain in all infinite endsegments.

Regards, WM

On 7/11/2022 12:42 PM, WM wrote:
> Eram semper recta schrieb am Montag, 11. Juli 2022 um 17:55:49 UTC+2:
>
>> He may well have emphasized it frequently but that does not mean he meant "ALL".
>
> If he not meant all, then his bijections were no bijections and without any value.

Where does Cantor say "ALL" ? Also, WM you try to disprove Cantor, now you try to rely on his teachings...

Here is what he said to the topic:
>
> 1.1.1 Vollständig
>
> "Wenn zwei wohldefinierte Mannigfaltigkeiten M und N sich eindeutig und vollständig, Element für Element, einander zuordnen lassen (was, wenn es auf eine Art möglich ist, immer auch noch auf viele andere Weisen geschehen kann), so möge für das Folgende die Ausdrucksweise gestattet sein, daß diese Mannigfaltigkeiten gleiche Mächtigkeit haben, oder auch, daß sie äquivalent sind." [Cantor, p. 119]

1.1.1 Complete "If two well-defined manifolds M and N can be clearly and completely assigned to each other, element by element (which, if it is possible
in one way, can always be done in many other ways), then for the following the expression may be permitted that these manifolds have the same power, or
that they are equivalent." [Cantor, p. 119]

>
> "gegenseitig eindeutige und vollständige Korrespondenz" [Cantor, p. 238]

"mutually unambiguous and complete correspondence" [Cantor, p. 238]

>
> "Die sämtlichen Punkte l unsrer Menge L sind also in gegenseitig eindeutige und vollständige Beziehung zu sämtlichen Punkten f der Menge F gebracht," [Cantor, p. 241]

"All points l of our set L are thus brought into mutually unambiguous and complete relation to all points f of set F," [Cantor, p. 241]

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

"Two well-ordered sets M and N are said to be of the same type or number if they can be mapped to each other uniquely and completely while preserving
the order of their elements on both sides;" [G. Cantor, letter to R. Lipschitz (19 Nov 1883)]

>
> "Zwei bestimmte Mengen M und M1 nennen wir äquivalent (in Zeichen: M ~ M1), wenn es möglich ist, dieselben gesetzmäßig, gegenseitig eindeutig und vollständig, Element für Element, einander zuzuordnen." [Cantor, p. 412]

> "We call two definite sets M and M1 equivalent (in symbols: M ~ M1) if it is possible to associate them with each other lawfully, mutually uniquely
and completely, element by element." [Cantor, p. 412]

>
> "doch gibt es immer viele, im allgemeinen sogar unzählig viele Zuordnungsgesetze, durch welche zwei äquivalente Mengen in gegenseitig eindeutige und vollständige Beziehung zueinander gebracht werden können." [Cantor, p. 413]

"But there are always many, in general even innumerable laws of association, by means of which two equivalent sets can be brought into mutually
unambiguous and complete relation to one another." [Cantor, p. 413]

>
> "eine solche gegenseitig eindeutige und vollständige Korrespondenz hergestellt [...] irgendeine gegenseitig eindeutige und vollständige Zuordnung der beiden Mengen [...] auch eine gegenseitig eindeutige und vollständige Korrespondenz" [Cantor, p. 415]
>

"such a mutually unambiguous and complete correspondence established [...] any mutually unambiguous and complete association of the two sets [...] also
a mutually unambiguous and complete correspondence" [Cantor, p. 415]

> "Zwei n-fach geordnete Mengen M und N werden 'ähnlich' genannt, wenn es möglich ist, sie gegenseitig eindeutig und vollständig, Element für Element, einander so zuzuordnen," [Cantor, p. 424]

"Two n-fold ordered sets M and N are said to be 'similar' if it is possible to relate them to each other uniquely and completely, element by element,"
[Cantor, p. 424]

>

> Regards, WM

William schrieb am Montag, 11. Juli 2022 um 19:49:00 UTC+2:
> On Monday, July 11, 2022 at 11:11:08 AM UTC-3, WM wrote:
>
> > It would be dark.
>
> |N, a Peano set, does not contain dark elements.

Try to move the cursor until there are infinitely many numbers before it (the set of endsegments) and infinitely many behind (the minimum contents of infinite endsegments).

1,2,3,4,5,6,7,8,9, ..., n, | n+1, ...

Fail. As long as any element is behind the cursor, there are only finitely many before it.
"Infinitely many infinite endegments" is a self-contradiction.

Regards, WM

On Monday, July 11, 2022 at 7:45:39 PM UTC+2, William wrote:
> On Monday, July 11, 2022 at 9:35:37 AM UTC-3, WM wrote:
> > William schrieb am Sonntag, 10. Juli 2022 um 19:41:09 UTC+2:
> > >
> > > There is no one subset that works for each [and every] endsegment.
> > >
> > Of course it is.
> >
> Nope. It is trivial to show that given a non empty subset of natural numbers, there is an endsegment that does not contain it

Proof: Let S be a non empty subset of natural numbers. Then there's a minimal element min(S) in S (since IN is well-ordered).
But S !c E(n0 + 1) = {n0 + 1, n0 + 2, n0 + 3, ...}, since min(S) e S, but S !e min(S) E(n0 + 1). Hence there is an endsegment that
does not contain S (as a subset). qed

It seems that WM is not able to understand any text that starts with the word "Proof" and ends with the phrase "qed". Seems to be some curious mental defect. (Probably induced by a psychosis.)

On Monday, July 11, 2022 at 7:58:53 PM UTC+2, WM wrote:

> "Infinitely many infinite endegments" is a self-contradiction.

Hear, hear!

Time to call the doctor!