zelos...@gmail.com schrieb am Montag, 27. September 2021 um 07:02:41 UTC+2:
> fredag 24 september 2021 kl. 18:58:07 UTC+2 skrev Transfinity:
> > Serg io schrieb am Freitag, 24. September 2021 um 01:48:30 UTC+2:
> > >
> > > There is no element common to all endsegments.
> > But all have infinitely many elements in common with E(1). So they must be disjoint. Do you claim that?

> Why would they need to be disjoint?

Otherwise the intersection is not empty. Infinite and not disjoint gives an infinite intersection.

Regards, WM

On Monday, September 27, 2021 at 10:08:58 PM UTC+2, WM wrote:
> Greg Cunt schrieb am Montag, 27. September 2021 um 22:02:57 UTC+2:

Let n e IN.

> > Can you prove your idiotic claim by a quote?
> >
> Of course:

> > > > So n is in IN

Huh?! What's your problem, fool?

Completely gone nuts now, or what?

> > Please answer my question. If n is a placeholder and n is an element in IN, doesn't that mean that there are placeholders in IN?

> n is not an element of |N.

YOU ARE CRAZY, MAN. COMPLETELY NUTS. LOCO LOCO!

Jim Burns schrieb am Sonntag, 26. September 2021 um 19:14:45 UTC+2:
> On 9/24/2021 12:42 PM, WM wrote:

> "That" is a pronoun.
> That is a natural number, when I refer to a natural number.

Yes, if you refer to a certain number.
>
> "n" is variable name, or a placeholder, or call it what you like.
> n is a natural number, if I say it is.

Yes, if you can say what is its prime decomposition.

Regards, WM

On Monday, 27 September 2021 at 17:12:55 UTC-3, WM wrote:
> Gus Gassmann schrieb am Sonntag, 26. September 2021 um 03:16:34 UTC+2:
> > On Friday, 24 September 2021 at 16:57:04 UTC-3, Transfinity wrote:
> > > Gus Gassmann schrieb am Freitag, 24. September 2021 um 21:05:15 UTC+2:
> > >
> > > > The intersection over infinitely many end segments is empty.
> > > Infinitely many endsegments are more than every finite set of endsegments? What is in the infinite set that is not in at least one of all finite sets making the intersection infinite?
> > I tell you what is not in it: Elements. The intersection over an infinite number of end segments is *EMPTY*.
> The intersection over infinite endsegments is infinite.

Often stated, never proved, still false.

Intersections over infinitely many end segments are empty:

> > In particular no natural number n is an element of its successor's end segment E(n+1).

> What are the elements of the infinite endsegments? Are they also in |N?

You are repeating yourself, you are forgetting things, you are clearly dribbling at the mouth, and it seems you are losing control over your other bodily functions, too. Every end segments contains *ONLY* natural numbers. That is by definition, you moron. Every end segment has infinite cardinality. Also by definition, specifically the Peano axioms. (You have given ample evidence in the last few days that you do not remember and understand what those are, so I mention them here.) In fact, your contributions have gotten so weak-minded, so rambling, so ludicrous, that I would recommend a padded cell for you. At least there you will be safe from yourself.

On Monday, 27 September 2021 at 17:14:52 UTC-3, WM wrote:
> zelos...@gmail.com schrieb am Montag, 27. September 2021 um 07:02:41 UTC+2:
> > fredag 24 september 2021 kl. 18:58:07 UTC+2 skrev Transfinity:
> > > Serg io schrieb am Freitag, 24. September 2021 um 01:48:30 UTC+2:
> > > >
> > > > There is no element common to all endsegments.
> > > But all have infinitely many elements in common with E(1). So they must be disjoint. Do you claim that?
> > Why would they need to be disjoint?
> Otherwise the intersection is not empty. Infinite and not disjoint gives an infinite intersection.

Hallucinations and psychoses do not a proof make. Stating something over and over with not a shred of evidence does not make it true.

Every end segment contains only natural numbers (infinitely many of them), so the intersection over end segments cannot contain anything else than natural numbers. No natural number n is an element of its successor's end segment E(n+1). So no natural number n can be a member of an intersection of end segments that contains E(n+1) or any other end segment E(m) with m > n. Ergo, the infinite intersection is empty.

WM explained on 9/27/2021 :
> FromTheRafters schrieb am Sonntag, 26. September 2021 um 00:03:30 UTC+2:
>> WM brought next idea :
>>> FromTheRafters schrieb am Freitag, 24. September 2021 um 21:12:12 UTC+2:
>>>> WM used his keyboard to write :
>>>>> Consider all natural numbers which can be last of a FISON F(n). They are
>>>>> by definition the definable natnumbers.
>>>> Since each and every natural number has that property,
>>>
>>> Wrong. Provably wrong:
>>> ∀n ∈ ℕ: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
>>> But it is possible that you can't understand what follows.
>> Each FISON is finite by
>> definition and has a least and a greatest element. Subtracting finite
>> proper subsets from the naturals does not change the superset's
>> cardinality.
> But subtracting all finite proper subsets does change the superset's
> cardinality. That is just the point, the consequences of which you can't
> understand.

Wrong, it is you who sees a dichotomy which isn't there.

You conveniently ignore the fact that even your ill fated stepwise
process essentially 'removes' each and every element from the
endsegments one-by-one leaving no initial element to feed into the
successor function and thus collapsing the inductive structure.

You should be asking how it can possibly *be* infinite without an
initial element like zero or one in this case.

http://math.ucsd.edu/~csorense/teaching/math31ah/induction.pdf

måndag 27 september 2021 kl. 22:04:04 UTC+2 skrev WM:
> FromTheRafters schrieb am Sonntag, 26. September 2021 um 00:03:30 UTC+2:
> > WM brought next idea :
> > > FromTheRafters schrieb am Freitag, 24. September 2021 um 21:12:12 UTC+2:
> > >> WM used his keyboard to write :
> > >
> > >>> Consider all natural numbers which can be last of a FISON F(n). They are by
> > >>> definition the definable natnumbers.
> > >> Since each and every natural number has that property,
> > >
> > > Wrong. Provably wrong:
> > > ∀n ∈ ℕ: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
> > > But it is possible that you can't understand what follows.
> > Each FISON is finite by
> > definition and has a least and a greatest element. Subtracting finite
> > proper subsets from the naturals does not change the superset's
> > cardinality.
> But subtracting all finite proper subsets does change the superset's cardinality. That is just the point, the consequences of which you can't understand.
>
> Regards, WM

The union of all finite subsets of N is N itself.

måndag 27 september 2021 kl. 22:14:52 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Montag, 27. September 2021 um 07:02:41 UTC+2:
> > fredag 24 september 2021 kl. 18:58:07 UTC+2 skrev Transfinity:
> > > Serg io schrieb am Freitag, 24. September 2021 um 01:48:30 UTC+2:
> > > >
> > > > There is no element common to all endsegments.
> > > But all have infinitely many elements in common with E(1). So they must be disjoint. Do you claim that?
> > Why would they need to be disjoint?
> Otherwise the intersection is not empty. Infinite and not disjoint gives an infinite intersection.
>
> Regards, WM

Only if there are elements present in all of the sets. If all sets have different elements the intersection can be empty.

{a,b} {b,c} {a,c}
Intersection of all is empty, but intersetion of any 2 is not empty, yet none is disjoint.

måndag 27 september 2021 kl. 21:34:57 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Montag, 27. September 2021 um 06:48:17 UTC+2:
> > fredag 24 september 2021 kl. 15:48:09 UTC+2 skrev WM:
> > > zelos...@gmail.com schrieb am Freitag, 24. September 2021 um 07:08:11 UTC+2:
> > > > torsdag 23 september 2021 kl. 15:23:55 UTC+2 skrev WM:
> > >
> > > > > My definition: Any natnumbers must have its UPF known in principle or knowable.
> > > > >
> > > > if I say n is a natural number, it is a natural number.
> > > It is not difficult to prove you wrong. You claim that every natural number has a decimal representation. Name the decimal representation of n. Fail.
> > >
> > n has a decimal representation, doesn't mean I have to give it.
> If you claim that n has a decimal representation and this is doubted, then you have to prove it by giving it.
>
> Regards, WM
Nope, I only have to prove all natural numbers have one and that is done already.

zelos...@gmail.com schrieb am Dienstag, 28. September 2021 um 06:56:41 UTC+2:
> måndag 27 september 2021 kl. 21:34:57 UTC+2 skrev WM:

> > If you claim that n has a decimal representation and this is doubted, then you have to prove it by giving it.
> >
> Nope, I only have to prove all natural numbers have one and that is done already.

Irrelevant since it is just doubted that n is a natural number. If a honest mathematician is asked to show n's decimal representation, then he will do so, if he can.

Regards, WM

On Thursday, September 30, 2021 at 8:12:13 PM UTC+2, WM wrote:

Let n be a natural number. [Or: Let n e IN.]

> it is just doubted that n is a natural number.

BY WHOM, you silly idiot?

Was soll n denn in diesem Kontext sonst sein? Ein GUMMIBÄRCHEN?

Hint: BY DEFINITION (see above) n is an element in the set of all natural numbers. Does this set contain anything else except natural numbers?

zelos...@gmail.com schrieb am Dienstag, 28. September 2021 um 06:54:42 UTC+2:
> måndag 27 september 2021 kl. 22:04:04 UTC+2 skrev WM:

> The union of all finite subsets of N is N itself.

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

In order omega, two consecutive infinite sets are impossible. ℵo subsequent natural numbers are always remaining. They cannot be addressed. They are dark.

Regards, WM

FromTheRafters schrieb am Dienstag, 28. September 2021 um 00:18:24 UTC+2:

> You conveniently ignore the fact that even your ill fated stepwise
> process essentially 'removes' each and every element from the
> endsegments one-by-one leaving no initial element to feed into the
> successor function and thus collapsing the inductive structure.

No.
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
∀k ∈ ℕ_def: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo
See the universal quantifier?
>
Regards, WM

Greg Cunt schrieb am Donnerstag, 30. September 2021 um 20:19:56 UTC+2:
> On Thursday, September 30, 2021 at 8:12:13 PM UTC+2, WM wrote:
>
> Let n be a natural number. [Or: Let n e IN.]
> > it is just doubted that n is a natural number.
> BY WHOM,

Irrelevant.
If n is a natural number, then its decimal representation can be given.
Give it!

> Hint: BY DEFINITION (see above) n is an element in the set of all natural numbers.

No, you misunderstand. "n ∈ ℕ" is simply an abbreviation for the sentence: "Consider any natural number". By the way, you can also use k ∈ ℕ or m ∈ ℕ. Are they in trichotomy? k < m < n? Or m < k, n?

> Does this set contain anything else except natural numbers?

No, therefore it does not contain k, m, n.

Regards, WM

Gus Gassmann schrieb am Montag, 27. September 2021 um 22:45:39 UTC+2:
> On Monday, 27 September 2021 at 17:12:55 UTC-3, WM wrote:

> > The intersection over infinite endsegments is infinite.
> Often stated, never proved

Proved by inclusion monotony.
>
> Intersections over infinitely many end segments are empty:

Why are the endsegments not empty?
Why does their infinite contents differ from contents of their predecessors?

> > What are the elements of the infinite endsegments? Are they also in |N?
> You are repeating yourself, you are forgetting things, you are clearly dribbling at the mouth, and it seems you are losing control over your other bodily functions, too. Every end segments contains *ONLY* natural numbers. That is by definition

Those are in all its predecessors. And infinitely many are in all infinite successors. It is useless to continue this simple argument. Unless you show an attempt to discuss these facts, EOD

Regards, WM

Gus Gassmann schrieb am Montag, 27. September 2021 um 22:52:52 UTC+2:
> On Monday, 27 September 2021 at 17:14:52 UTC-3, WM wrote:

> Every end segment contains only natural numbers (infinitely many of them), so the intersection over end segments cannot contain anything else than natural numbers. No natural number n is an element of its successor's end segment E(n+1). So no natural number n can be a member of an intersection of end segments that contains E(n+1) or any other end segment E(m) with m > n.

Nevertheless infinitely many natural numbers remain in every endsegment. Un less they are all excorcized the intersection will not become empty.

> Ergo, the infinite intersection is empty.

Empty of definable natnumbers, filled of dark natnumbers.

Regards, WM

zelos...@gmail.com schrieb am Dienstag, 28. September 2021 um 06:56:14 UTC+2:
> måndag 27 september 2021 kl. 22:14:52 UTC+2 skrev WM:

> > > Why would they need to be disjoint?
> > Otherwise the intersection is not empty. Infinite and not disjoint gives an infinite intersection.
> >
> Only if there are elements present in all of the sets. If all sets have different elements the intersection can be empty.
>
> {a,b} {b,c} {a,c}
> Intersection of all is empty, but intersetion of any 2 is not empty, yet none is disjoint.

Just this is impossible for endsegments. Inclusion monotony!

Regards, WM

On Thursday, 30 September 2021 at 15:27:33 UTC-3, WM wrote:
> FromTheRafters schrieb am Dienstag, 28. September 2021 um 00:18:24 UTC+2:
>
> > You conveniently ignore the fact that even your ill fated stepwise
> > process essentially 'removes' each and every element from the
> > endsegments one-by-one leaving no initial element to feed into the
> > successor function and thus collapsing the inductive structure.
> No.
> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
> ∀k ∈ ℕ_def: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo
> See the universal quantifier?

Yes. So?

WM expressed precisely :
> FromTheRafters schrieb am Dienstag, 28. September 2021 um 00:18:24 UTC+2:
>
>> You conveniently ignore the fact that even your ill fated stepwise
>> process essentially 'removes' each and every element from the
>> endsegments one-by-one leaving no initial element to feed into the
>> successor function and thus collapsing the inductive structure.
>
> No.
> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
> ∀k ∈ ℕ_def: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo
> See the universal quantifier?

Your post doesn't address my statement and what you wrote makes no
sense anyway. Review
https://en.wikipedia.org/wiki/Complement_(set_theory) and recall that B
\ A, is the set of elements in B that are not in A. So, you are saying
*all* elements of {1,2,3,...} are not natural numbers.

WM explained :
> Gus Gassmann schrieb am Montag, 27. September 2021 um 22:45:39 UTC+2:
>> On Monday, 27 September 2021 at 17:12:55 UTC-3, WM wrote:
>
>>> The intersection over infinite endsegments is infinite.
>> Often stated, never proved
>
> Proved by inclusion monotony.
>>
>> Intersections over infinitely many end segments are empty:
>
> Why are the endsegments not empty?
> Why does their infinite contents differ from contents of their predecessors?
>
>>> What are the elements of the infinite endsegments? Are they also in |N?
>> You are repeating yourself, you are forgetting things, you are clearly
>> dribbling at the mouth, and it seems you are losing control over your other
>> bodily functions, too. Every end segments contains *ONLY* natural numbers.
>> That is by definition
>
> Those are in all its predecessors. And infinitely many are in all infinite
> successors. It is useless to continue this simple argument. Unless you show
> an attempt to discuss these facts, EOD

Promise?

On Thursday, 30 September 2021 at 15:12:13 UTC-3, WM wrote:
[...]
> Irrelevant since it is just doubted that n is a natural number. If a honest mathematician is asked to show n's decimal representation, then he will do so, if he can.

*That* is irrelevant. I presume you actually consider yourself a "honest mathematician" (ROFL). I also presume that you can't give the decimal representation of Graham's number. Nonetheless, it is trivial to prove that by induction it has one. (Not that that makes any difference!) You continue to display your utter ignorance of anything mathematical. In particular you have no clue about infinity, never have, and with the two or three remaining marbles there is no hope you ever will.

On Thursday, 30 September 2021 at 15:50:44 UTC-3, WM wrote:
> Gus Gassmann schrieb am Montag, 27. September 2021 um 22:45:39 UTC+2:
> > On Monday, 27 September 2021 at 17:12:55 UTC-3, WM wrote:
>
> > > The intersection over infinite endsegments is infinite.
> > Often stated, never proved
>
> Proved by inclusion monotony.

The only thing you prove by your insistent invocation of inclusion monotony is that you have no clue how it works with infinite intersections.

> > Intersections over infinitely many end segments are empty:
>
> Why are the endsegments not empty?

Because *PROVABLY* for every n in |N there is at least one set over which you intersect that does not contain it. This is really elementary stuff.

> Why does their infinite contents differ from contents of their predecessors?

Because *NO* natural number n can be contained in the end segment E(n+1).

> > > What are the elements of the infinite endsegments? Are they also in |N?
> > You are repeating yourself, you are forgetting things, you are clearly dribbling at the mouth, and it seems you are losing control over your other bodily functions, too. Every end segments contains *ONLY* natural numbers. That is by definition
>
> Those are in all its predecessors. And infinitely many are in all infinite successors. It is useless to continue this simple argument.

This is not an argument. Those aren't even coherent mathematical statements.

> Unless you show an attempt to discuss these facts, EOD

Thank god for that, and good riddance!

On Thursday, September 30, 2021 at 8:50:44 PM UTC+2, WM wrote:
> Gus Gassmann schrieb am Montag, 27. September 2021 um 22:45:39 UTC+2:
> > On Monday, 27 September 2021 at 17:12:55 UTC-3, WM wrote:
> > >
> > > The intersection over infinite endsegments is infinite.
> > >
> > Often stated, never proved
> >
> Proved by inclusion monotony.

Halts Maul, Mückenheim!

> EOD

Ufff...

torsdag 30 september 2021 kl. 20:50:44 UTC+2 skrev WM:
> Gus Gassmann schrieb am Montag, 27. September 2021 um 22:45:39 UTC+2:
> > On Monday, 27 September 2021 at 17:12:55 UTC-3, WM wrote:
>
> > > The intersection over infinite endsegments is infinite.
> > Often stated, never proved
>
> Proved by inclusion monotony.
> >
> > Intersections over infinitely many end segments are empty:
>
> Why are the endsegments not empty?
> Why does their infinite contents differ from contents of their predecessors?
>
> > > What are the elements of the infinite endsegments? Are they also in |N?
> > You are repeating yourself, you are forgetting things, you are clearly dribbling at the mouth, and it seems you are losing control over your other bodily functions, too. Every end segments contains *ONLY* natural numbers. That is by definition
>
> Those are in all its predecessors. And infinitely many are in all infinite successors. It is useless to continue this simple argument. Unless you show an attempt to discuss these facts, EOD
>
> Regards, WM

>Proved by inclusion monotony.

Does not prove that an infinite intersection is non-empty.

>Why does their infinite contents differ from contents of their predecessors?

We have told you this you stupid crank

it is empty because for any given natural number n, E(n+1) does not have it, so it is not in the intersection, and as n is arbitrary, it is true for all of them, meaning NONE is in the intersection, ergo it is EMPTY.

torsdag 30 september 2021 kl. 20:54:04 UTC+2 skrev WM:
> Gus Gassmann schrieb am Montag, 27. September 2021 um 22:52:52 UTC+2:
> > On Monday, 27 September 2021 at 17:14:52 UTC-3, WM wrote:
>
> > Every end segment contains only natural numbers (infinitely many of them), so the intersection over end segments cannot contain anything else than natural numbers. No natural number n is an element of its successor's end segment E(n+1). So no natural number n can be a member of an intersection of end segments that contains E(n+1) or any other end segment E(m) with m > n.
> Nevertheless infinitely many natural numbers remain in every endsegment. Un less they are all excorcized the intersection will not become empty.
> > Ergo, the infinite intersection is empty.
> Empty of definable natnumbers, filled of dark natnumbers.
>
> Regards, WM

Everyone is not in the intersection because there is always at least 1 intersection not containing the element. So that is why it is empty. Why is this so fucking difficult for you?