zelos...@gmail.com schrieb am Mittwoch, 6. Oktober 2021 um 07:12:15 UTC+2:
> >Everyone who says that all natural numbers are used up as indices of endsegments but infinitely many are remaining as contents of all endsegments (which used up all already) say that there is an infinite set |N (of indices) which is followed by an infinite set |N (of contents).
> This is pure nonsense, no one says anything like this

They claim that there is a bijection between the set |N and the set of endsegments. That means that all natural numbers n are used, they are mapped on endsegments E(n). None is remaining.

> >Every endsegment has infinite contents. But all endsegments exhaust |N.
> Doesn't change the fact that the intersection is empty because no element is in all sets.

Try a bit of logic!
(1) If every endsegment has infinite contents, then infinitely many natural numbers cannot be mapped on endsegments. There is no bijection between the set |N and the set of endsegments.
(2) Every infinite endsegment has an infinite intersection with all other infinite endsegments. This is proved by inclusion monotony.

Regards, WM

Greg Cunt schrieb am Mittwoch, 6. Oktober 2021 um 01:59:40 UTC+2:
> On Tuesday, October 5, 2021 at 10:16:43 AM UTC+2, WM wrote:

> > I can place n at 1, at 2, at 3, at any place I like.
> No, you can't.

Of course. Let n be a natural number allows to put n = 1.
>
> If we introduce "n" with the phrase "Let n e IN", then WE DON'T KNOW the referent of "n", but WE DO KNOW that "n" refers to a CERTAIN natural number

You cannot refute any given value, like 1 or 15 or 4711. Therefore n is not a number.

> > This is not possible with a natural number like 7 for instance, and it is not possible with a hitherto unknown number.
> Exactly. Hence it is not possible with n either.

It is, because n is the placeholder for every natnumber.

Regards, WM

Serg io schrieb am Dienstag, 5. Oktober 2021 um 20:44:03 UTC+2:
> On 10/5/2021 6:31 AM, WM wrote:

> I have a 17, can that be inserted at the position of n ?

Of course at the place where n is written. But don't confuse the positions! n has no position on the ordinal line because it is not a number. 17 however has a position there.

Regards, WM

Jim Burns schrieb am Dienstag, 5. Oktober 2021 um 20:57:40 UTC+2:
> On 10/5/2021 4:35 AM, WM wrote:

> I think that you (WM) claim that potentially infinite
> collections _change_

It is the definition: "potentially infinite ... a finite in the process of change having in each of its current states a finite size; like, for instance, the temporal duration since the beginning of the world, which, when measured in some time-unit, for instance a year, is finite in every moment, but always growing beyond all finite limits, without ever becoming really infinitely large." [G. Cantor, letter to I. Jeiler (13 Oct 1895)]" [Cantor]

The actual infinite however becomes infinitely large. Therefore the FISONs are only potentially infinite - and so are their endnumbers, the definable natural numbers.
>
> I have no complaint with you calling the collection of
> all and only FISON-enders "potentially infinite".

Obviously. FISONs are finite. Hence only finitely many symbols can be used to distinguish them. That means only finitely many can be distinguished. This figure containing FISONs

o
oo
ooo
oooo
ooooo
....

cannot be higher than broad by symmetry. I have never understood how intelligent persons can be foolish enough to claim the existnece of an actual infinity of FISONs.

> However, the collection of all and only FISON-enders
> _does not change_ A FISON exists which k ends, or
> a FISON does NOT exist which k ends, and which it is
> does not change.

That is impossible. All FISONs are finite. None is infinite. Would they not change, then there was a largest finite FISON.

Regards, WM

Gus Gassmann schrieb am Dienstag, 5. Oktober 2021 um 18:38:41 UTC+2:
> "Every end segment is not empty" is decidedly not the same thing as "there is a nonempty intersection"

If you deny mathematics, you may believe that.

> , your moronic application of inclusion monotony notwithstanding.

A sequence of non-empty sets which never gain new elements but only lose elements, cannot have an empty intersection. That is basic mathematics.

> Since there is no set of infinite cardinality common to all end segments

All infinite endsegments have an infinite intersection. Try to find a counter example, that is an infinite endsegment without an infinite intersection with all other infinite endsegments.

Regards, WM

Serg io schrieb am Dienstag, 5. Oktober 2021 um 21:24:58 UTC+2:

> ∩{E(1), E(2), ...} = { } is empty, not infinite.

Yes. But presently only infinite endsegments are concerned. For them we have
∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo
Regards, WM

Greg Cunt schrieb am Mittwoch, 6. Oktober 2021 um 02:36:27 UTC+2:
> On Tuesday, October 5, 2021 at 5:44:32 PM UTC+2, WM wrote:
>
> > I know that
> > ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo
> > in connection with
> > ∀k ∈ ℕ: E(k+1) = E(k) \ {k}
> > excludes
> > Lim E(n) = { }
> Great. Can you PROVE what you "know" (i.e. believe)?
>
The proof stands above. Further we have inclusion monotony. Try to learn and understand these things.

All infinite endsegments have an infinite intersection together. The counter-example would be an infinite endsegment which has not an infinite intersection with all other infinite endsegments. That can be reduced to two infinite endsegments without an infinite common intersection.

Regards, WM

zelos...@gmail.com schrieb am Mittwoch, 6. Oktober 2021 um 07:05:20 UTC+2:

> False, the intersection of ALL endsegments is EMPTY

Yes. But all infinite endsegments have an infinite intersection together. The counter-example would be an infinite endsegment which has not an infinite intersection with all other infinite endsegments. That can be reduced to two infinite endsegments without an infinite common intersection.

Regards, WM

On Thursday, October 7, 2021 at 12:23:19 PM UTC+2, WM wrote:
> Greg Cunt schrieb am Mittwoch, 6. Oktober 2021 um 02:36:27 UTC+2:
> > On Tuesday, October 5, 2021 at 5:44:32 PM UTC+2, WM wrote:
> > >
> > > I know that
> > > ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo
> > > in connection with
> > > ∀k ∈ ℕ: E(k+1) = E(k) \ {k}
> > > excludes
> > > Lim E(n) = { }
> > >
> > Great. Can you PROVE what you "know" (i.e. believe)?
> >
> The proof stands above.

There is no proof, you silly asshole! That's just A CLAIM.

I asked for a PROOF for this claim.

Geht's NOCH dümmer, Mückenheim?

On Thursday, October 7, 2021 at 12:25:51 PM UTC+2, WM wrote:
> zelos...@gmail.com schrieb am Mittwoch, 6. Oktober 2021 um 07:05:20 UTC+2:
> >
> > False, the intersection of ALL endsegments is EMPTY
> >
> Yes. But <bla>

FUCK YOU, Mückenheim!

On Thursday, October 7, 2021 at 11:27:46 AM UTC+2, WM wrote:
> Greg Cunt schrieb am Mittwoch, 6. Oktober 2021 um 01:59:40 UTC+2:
> > On Tuesday, October 5, 2021 at 10:16:43 AM UTC+2, WM wrote:
> > >
> > > I can place n at 1, at 2, at 3, at any place I like.

No, you can't.

If we introduce "n" with the phrase "Let n e IN", then WE DON'T KNOW the referent of "n", but WE DO KNOW that "n" refers to a CERTAIN natural number - i.e. a FIXED denotation. So you can't just stipulate (afterwards) that n = 1, or stipulate that n = 2, etc.

If you wan't "n" to denote a SPECIFIC number _of your choice_, you have to introduce "n" the following way: "Let n = 1", or "Let n = 2", etc.

On Thursday, October 7, 2021 at 11:33:45 AM UTC+2, WM wrote:

Let n e IN.

Now you claim:

> n has no position on the ordinal line because

you are a crank.

Hint: n has exactly the position n "on the ordinal line".

Greg Cunt schrieb am Donnerstag, 7. Oktober 2021 um 12:38:07 UTC+2:
> On Thursday, October 7, 2021 at 12:23:19 PM UTC+2, WM wrote:
> > Greg Cunt schrieb am Mittwoch, 6. Oktober 2021 um 02:36:27 UTC+2:
> > > On Tuesday, October 5, 2021 at 5:44:32 PM UTC+2, WM wrote:
> > > >
> > > > I know that
> > > > ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo
> > > > in connection with
> > > > ∀k ∈ ℕ: E(k+1) = E(k) \ {k}
> > > > excludes
> > > > Lim E(n) = { }
> > > >
> > > Great. Can you PROVE what you "know" (i.e. believe)?
> > >
> > The proof stands above.
> There is no proof,

Try to learn to read.
>
> I asked for a PROOF for this claim.

Since you cannot understand my proof, try this:
(1) During the steps of the sequence (E(n)), every natural number n is lost from the intersection ∩{E(1), E(2), ..., E(n)}. Therefore the intersections become empty.
(2) During the steps of the sequence, every natural number is lost from the endsegments. Therefore the endsegments become empty.

Matheologians accept (1) but don't accept (2) without any substantial reason.

Regards, WM

On Thursday, 7 October 2021 at 07:10:15 UTC-3, WM wrote:
> Gus Gassmann schrieb am Dienstag, 5. Oktober 2021 um 18:38:41 UTC+2:
> > "Every end segment is not empty" is decidedly not the same thing as "there is a nonempty intersection"
> If you deny mathematics, you may believe that.
> > , your moronic application of inclusion monotony notwithstanding.
> A sequence of non-empty sets which never gain new elements but only lose elements, cannot have an empty intersection. That is basic mathematics.

It is basic Muckenheim delusions.

> > Since there is no set of infinite cardinality common to all end segments
> All infinite endsegments have an infinite intersection. Try to find a counter example, that is an infinite endsegment without an infinite intersection with all other infinite endsegments.

So is this.

On Thursday, October 7, 2021 at 11:19:55 AM UTC+2, WM wrote:

Let's assume that

> there is a bijection between the set |N and the set of endsegments. That means that all natural numbers n are used, they are mapped on endsegments E(n). None is remaining.

Right. Hence now the set IN would be empty in mückenmath?

> There is no bijection between the set |N and the set of endsegments [in mückenmath].

No? Well in math there is.

Set_of_Endsegments := {{m e IN : m >= n} : n e IN}

Consider the function f: IN ---> Set_of_Endsegments with f(n) = {m e IN : m >= n} for all n e IN.

f is a bijection from the set IN onto the set of endsegments.

You see:

1 <-> {1, 2, 3, ...}
2 <-> {2, 3, 4, ...}
3 <-> {3, 4, 5, ...}
etc.

On Thursday, October 7, 2021 at 12:48:51 PM UTC+2, WM wrote:
> Greg Cunt schrieb am Donnerstag, 7. Oktober 2021 um 12:38:07 UTC+2:
> > On Thursday, October 7, 2021 at 12:23:19 PM UTC+2, WM wrote:
> > > Greg Cunt schrieb am Mittwoch, 6. Oktober 2021 um 02:36:27 UTC+2:
> > > > On Tuesday, October 5, 2021 at 5:44:32 PM UTC+2, WM wrote:
> > > > >
> > > > > I know that
> > > > > ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo
> > > > > in connection with
> > > > > ∀k ∈ ℕ: E(k+1) = E(k) \ {k}
> > > > > excludes
> > > > > Lim E(n) = { }
> > > > >
> > > > Great. Can you PROVE what you "know" (i.e. believe)?
> > > >
> > > The proof stands above.
> > >
> > There is no proof,
> >
> > I asked for a PROOF for this claim.
> >
> Since <bla>

I ASKED FPOOF A PROOF FPR YOUR CLAIM

| ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo
| in connection with
| ∀k ∈ ℕ: E(k+1) = E(k) \ {k}
| excludes
| Lim E(n) = { }

YOU FUCKING ASSHOLE FULL OF SHIT!

So where's your PROOF, asshole?

Hint: It seems that you know shit!

On Thursday, October 7, 2021 at 12:48:51 PM UTC+2, WM wrote:
> Greg Cunt schrieb am Donnerstag, 7. Oktober 2021 um 12:38:07 UTC+2:
> > On Thursday, October 7, 2021 at 12:23:19 PM UTC+2, WM wrote:
> > > Greg Cunt schrieb am Mittwoch, 6. Oktober 2021 um 02:36:27 UTC+2:
> > > > On Tuesday, October 5, 2021 at 5:44:32 PM UTC+2, WM wrote:
> > > > >
> > > > > I know that
> > > > > ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo
> > > > > in connection with
> > > > > ∀k ∈ ℕ: E(k+1) = E(k) \ {k}
> > > > > excludes
> > > > > Lim E(n) = { }
> > > > >
> > > > Great. Can you PROVE what you "know" (i.e. believe)?
> > > >
> > > The proof stands above.
> > >
> > There is no proof,
> >
> > I asked for a PROOF for this claim.
> >
> Since <bla>

I ASKED FOR A PROOF FOR YOUR CLAIM

| ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo
| in connection with
| ∀k ∈ ℕ: E(k+1) = E(k) \ {k}
| excludes
| Lim E(n) = { }

YOU FUCKING ASSHOLE FULL OF SHIT!

So where's your PROOF, asshole?

Hint: It seems that you know shit!

WM brought next idea :
> Greg Cunt schrieb am Mittwoch, 6. Oktober 2021 um 02:36:27 UTC+2:
>> On Tuesday, October 5, 2021 at 5:44:32 PM UTC+2, WM wrote:
>>
>>> I know that
>>> ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo
>>> in connection with
>>> ∀k ∈ ℕ: E(k+1) = E(k) \ {k}
>>> excludes
>>> Lim E(n) = { }
>> Great. Can you PROVE what you "know" (i.e. believe)?
>>
> The proof stands above. Further we have inclusion monotony. Try to learn and
> understand these things.

The only thing you have proven is that *you* don't understand these
things.

On Thursday, 7 October 2021 at 06:19:55 UTC-3, WM wrote:
> [...] There is no bijection between the set |N and the set of endsegments.

Bullshit. By your "symmetry" argument you claimed/proved a few days ago that there is. However, you also claimed that there is no bijection between the set |N and the set of FISONs. Your Alzheimer's is acting up today; take the rest of the day off.

> (2) Every infinite endsegment has an infinite intersection with all other infinite endsegments. This is proved by inclusion monotony.

Ambiguous natural language nonsense again.

WM formulated on Thursday :
> zelos...@gmail.com schrieb am Mittwoch, 6. Oktober 2021 um 07:12:15 UTC+2:
>>> Everyone who says that all natural numbers are used up as indices of
>>> endsegments but infinitely many are remaining as contents of all
>>> endsegments (which used up all already) say that there is an infinite set
>>> |N (of indices) which is followed by an infinite set |N (of contents).
>> This is pure nonsense, no one says anything like this
>
> They claim that there is a bijection between the set |N and the set of
> endsegments. That means that all natural numbers n are used, they are mapped
> on endsegments E(n). None is remaining.
>
>>> Every endsegment has infinite contents. But all endsegments exhaust |N.
>> Doesn't change the fact that the intersection is empty because no element is
>> in all sets.
>
> Try a bit of logic!
> (1) If every endsegment has infinite contents, then infinitely many natural
> numbers cannot be mapped on endsegments. There is no bijection between the
> set |N and the set of endsegments. (2) Every infinite endsegment has an
> infinite intersection with all other infinite endsegments. This is proved by
> inclusion monotony.

Show us what inclusion monotony has to do with your fantasy.

Greg Cunt used his keyboard to write :
> On Thursday, October 7, 2021 at 11:19:55 AM UTC+2, WM wrote:
>
> Let's assume that
>
>> there is a bijection between the set |N and the set of endsegments. That
>> means that all natural numbers n are used, they are mapped on endsegments
>> E(n). None is remaining.
>
> Right. Hence now the set IN would be empty in mückenmath?

No, they are just exhausted and need to rest up a little. In Muckymath
they grow back since they were only potentially infinite in the first
place.

>> There is no bijection between the set |N and the set of endsegments [in
>> mückenmath].
>
> No? Well in math there is.
>
> Set_of_Endsegments := {{m e IN : m >= n} : n e IN}
>
> Consider the function f: IN ---> Set_of_Endsegments with f(n) = {m e IN : m
> >= n} for all n e IN.
>
> f is a bijection from the set IN onto the set of endsegments.
>
> You see:
>
> 1 <-> {1, 2, 3, ...}
> 2 <-> {2, 3, 4, ...}
> 3 <-> {3, 4, 5, ...}
> etc.

Looks like exclusion to me.

The second endsegment has one reason to negate the infinite
intersection -- that is, the number one symbol is missing, that is
enough. The next has two reasons as it is missing the one and two
symbols. The number of missing symbols is strictly monotonic increasing
but WM ignores this because it is not confusing enough to hide his
subterfuge.

Greg Cunt schrieb am Donnerstag, 7. Oktober 2021 um 12:42:06 UTC+2:
> On Thursday, October 7, 2021 at 11:27:46 AM UTC+2, WM wrote:

> > > > I can place n at 1, at 2, at 3, at any place I like.
>
> No, you can't.

1, 2, 3, 4, 5, ...
n, 2, 3, 4, 5, ...
1, n, 3, 4, 5, ...
1, 2, n, 4, 5, ...
1, 2, 3, 4, 5, ..., n, ...

Regards, WM

Greg Cunt schrieb am Donnerstag, 7. Oktober 2021 um 12:44:51 UTC+2:
> On Thursday, October 7, 2021 at 11:33:45 AM UTC+2, WM wrote:
>
> Let n e IN.
>
> Now you claim:
> > n has no position on the ordinal line because
> you are a crank.
>
> Hint: n has exactly the position n "on the ordinal line".

Alas there is no position n: 1, 2, 3, 4, 5, ...

Regards, WM

Greg Cunt schrieb am Donnerstag, 7. Oktober 2021 um 12:53:29 UTC+2:
> On Thursday, October 7, 2021 at 11:19:55 AM UTC+2, WM wrote:
>
> Let's assume that
> > there is a bijection between the set |N and the set of endsegments. That means that all natural numbers n are used, they are mapped on endsegments E(n). None is remaining.
> Right. Hence now the set IN would be

completely used. No natnumber remains after all used ones. Otherwise it would no be a mapping from |N to the set of endsegments. Links total rechts eindeutig.

Regards, WM

Gus Gassmann schrieb am Donnerstag, 7. Oktober 2021 um 14:05:12 UTC+2:
> On Thursday, 7 October 2021 at 06:19:55 UTC-3, WM wrote:
> > If every endsegment has infinite contents, then infinitely many natural numbers cannot be mapped on endsegments. There is no bijection between the set |N and the set of endsegments.

> By your "symmetry" argument you claimed/proved a few days ago that there is.

Of course there is a bijection. But no natural numbers are remaining such that all endsegments in this mapping could have infinite contents.

Regards, WM