On 10/10/2021 12:14 PM, WM wrote:
> Serg io schrieb am Samstag, 9. Oktober 2021 um 18:30:26 UTC+2:
>
>> PROOF;
>> for any k, where k is a natural number
>> k is not in the endsegments E(k+1), E(k+2), ...
>> therefore k is not in the intersection of all endsegments.
>> since we chose ANY k,

I put back in your snippage;

" and proved there are no such k's in the intersection of all endsegments, then there are no natural numbers in the intersection."

>
> No! Every endsegment is infinite. For infinitely many k your proof fails.

Wrong. My proof holds for all, every, and each natural number in N.

This has been Proven above.

Try to name a natural number that is in the intersection of all endsegments. Fail. There are none.

> > Regards, WM
>

On 10/10/2021 12:44 PM, WM wrote:
> zelos...@gmail.com schrieb am Sonntag, 10. Oktober 2021 um 19:21:59 UTC+2:
>> söndag 10 oktober 2021 kl. 19:15:01 UTC+2 skrev WM:
>>> Serg io schrieb am Samstag, 9. Oktober 2021 um 18:30:26 UTC+2:
>>>
>>>> PROOF;
>>>> for any k, where k is a natural number
>>>> k is not in the endsegments E(k+1), E(k+2), ...
>>>> therefore k is not in the intersection of all endsegments.
>>>> since we chose ANY k,
>>> No! Every endsegment is infinite. For infinitely many k your proof fails.
>>>
>> It is ANY k that means it applies to ALL of them, NONE does it fail for.
>
> What are the infinitely many elements which remain in every infinite endsegment?

each Endsegment is a infinite fixed set of natural numbers.

there is no "removal", "remain", "acquire" with Endsegments as they are fixed sets, there are no changes ever.

>Does the proof prove that they disappear such that none remains.

the Proof assumes some natural number k, is in the intersection of all endsegments, however k is not in E(K+1) so K cannot be in the intersection of all
endsegments. So the assumption is false. Which proves there are no natural numbers in the intersection of all endsegments.

>How can all the endsegments remain infinite?

no endsegments changed at all, they are fixed infinite sets.

>
> Regards, WM
>

Serg io schrieb am Sonntag, 10. Oktober 2021 um 20:00:32 UTC+2:
> On 10/10/2021 12:44 PM, WM wrote:

> >How can all the endsegments remain infinite?
> no endsegments changed at all, they are fixed infinite sets.

"Remain infinite" means that infinitely many elements are contained in all terms of the sequence of endsegments (E(n)).
The infinite contents that is in all endsegments E(n) from the first one on cannot be used to index endsegments.

Regards, WM

Serg io schrieb am Sonntag, 10. Oktober 2021 um 19:54:47 UTC+2:
> On 10/10/2021 12:14 PM, WM wrote:

> > No! Every endsegment is infinite. For infinitely many k your proof fails.
> Wrong. My proof holds for all, every, and each natural number in N.

But you said infinitely many natnumbers exist in all endsegments because all endsegments are infinite. So you cannot prove that they all are omitted.
>
> Try to name a natural number that is in the intersection of all endsegments. Fail. There are none.

What is the contents of the infinite endsegments?

Regards, WM

On 10/10/2021 12:46 PM, WM wrote:
> zelos...@gmail.com schrieb am Sonntag, 10. Oktober 2021 um 19:32:27 UTC+2:
>> fredag 8 oktober 2021 kl. 18:58:43 UTC+2 skrev WM:
>
>>> The contents of endsegments is what is not used to index its predecessors.
>> Irrelevant.
>
> Chuckle. Your only way to pretend ignorance?
>
>>> If all endsegments have infinite contents, then infinitely many numbers are not used to index all predecessors, i.e., all endsegments. The complement, used to index them, is finite.
>> Meaningless empty assertion.
>
> Meaningful and very important for mathematicians who can think.
>
> Regards, WM
>

nope, it is silly. Try expressing that thought in equations instead...

we are waiting....

the clock is ticking....

On 10/10/2021 1:13 PM, Transfinity wrote:
> Serg io schrieb am Sonntag, 10. Oktober 2021 um 19:54:47 UTC+2:
>> On 10/10/2021 12:14 PM, WM wrote:
>
>>> No! Every endsegment is infinite. For infinitely many k your proof fails.
>> Wrong. My proof holds for all, every, and each natural number in N.
>
> But you said infinitely many natnumbers exist in all endsegments because all endsegments are infinite. So you cannot prove that they all are omitted.

I proved using correct Math, that the intersection of all endsegments is empty.

if you do not believe it, then just name a natural number that is in the intersection of all endsegments. Fail

>>
>> Try to name a natural number that is in the intersection of all endsegments. Fail. There are none.
>
> What is the contents of the infinite endsegments?

diversion.

>
> Regards, WM
>

On Sunday, 10 October 2021 at 14:32:30 UTC-3, WM wrote:
> Jim Burns schrieb am Freitag, 8. Oktober 2021 um 21:57:31 UTC+2:
> > On 10/8/2021 9:35 AM, WM wrote:
[...]

> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
>
> This shows all FISONs are finite. But without changing, there would be a largest one. There is none. Therefore the set is potentially infinite.

Sixteen+ years of debating in newsgroups, preceded by doctoral studies in physics, and *ZERO* comprehension of mathematics. Wow! Muckenheim, your string of failures is impressive!

Serg io schrieb am Sonntag, 10. Oktober 2021 um 20:29:54 UTC+2:
> On 10/10/2021 1:13 PM, Transfinity wrote:
> > Serg io schrieb am Sonntag, 10. Oktober 2021 um 19:54:47 UTC+2:
> >> On 10/10/2021 12:14 PM, WM wrote:

> > But you said infinitely many natnumbers exist in all endsegments because all endsegments are infinite. So you cannot prove that they all are omitted.
> I proved using correct Math, that the intersection of all endsegments is empty.

I am not interested in the intersection. You said that all endsegments are infinite. So they all contain natnumbers, don't they?
>
> > What is the contents of the infinite endsegments?
> diversion.

Diversion is not in any endsegment. Every endsegment contains infinitely many natural numbers. They cannot be indices of endsegments.

Regards, WM

On Sunday, 10 October 2021 at 14:25:09 UTC-3, WM wrote:
[...]
> The contents of an endsegment cannot be used to index its predecessors:
>
> E(1) = {1, 2, 3, 4, 5, 6, ...} has no predecessor.
> E(1), E(2) = {2, 3, 4, 5, 6, ...}
> E(1), E(2), E(3) = {3, 4, 5, 6, ...}
> E(1), E(2), E(3), E(4) = {4, 5, 6, ...}
> The contents of E(4) cannot be an index of the predecessors of E(4) which are E(1), E(2), E(3).
>
> If now every endsegment has infinite contents {n, n+1, ...} , then infinitely many natural numbers cannot be used to index the predecessors of any endsegment. Therefore only finitely many endsegments can be indexed and can exist.

What a silly old fool you are. Of course, *NO* end segment has more than finitely many predecessors. I presume even you accept and understand that. Your last sentence is again your damned quantifier dyslexia. It does not follow from the preceding and is *WRONG*. Give it up! You have no clue of mathematics, never had a clue, and never will have a clue.

Serg io schrieb am Sonntag, 10. Oktober 2021 um 20:27:02 UTC+2:
> On 10/10/2021 12:46 PM, WM wrote:
> > zelos...@gmail.com schrieb am Sonntag, 10. Oktober 2021 um 19:32:27 UTC+2:
> >> fredag 8 oktober 2021 kl. 18:58:43 UTC+2 skrev WM:
> >
> >>> The contents of endsegments is what is not used to index its predecessors.
> >> Irrelevant.
> >
> > Chuckle. Your only way to pretend ignorance?
> >
> >>> If all endsegments have infinite contents, then infinitely many numbers are not used to index all predecessors, i.e., all endsegments. The complement, used to index them, is finite.
> >> Meaningless empty assertion.
> >
> > Meaningful and very important for mathematicians who can think.
> >
> Try expressing that thought in equations instead...
>
Why? Every endsegment has the cardinality ℵo. For consecutive sets ℵo + ℵo is larger than ℵo. n + ℵo = ℵo is possible. Hence there are only finitely many endsegments.

Regards, WM

Gus Gassmann schrieb am Sonntag, 10. Oktober 2021 um 22:34:03 UTC+2:
> On Sunday, 10 October 2021 at 14:25:09 UTC-3, WM wrote:
> [...]
> > The contents of an endsegment cannot be used to index its predecessors:
> >
> > E(1) = {1, 2, 3, 4, 5, 6, ...} has no predecessor.
> > E(1), E(2) = {2, 3, 4, 5, 6, ...}
> > E(1), E(2), E(3) = {3, 4, 5, 6, ...}
> > E(1), E(2), E(3), E(4) = {4, 5, 6, ...}
> > The contents of E(4) cannot be an index of the predecessors of E(4) which are E(1), E(2), E(3).
> >
> > If now every endsegment has infinite contents {n, n+1, ...} , then infinitely many natural numbers cannot be used to index the predecessors of any endsegment. Therefore only finitely many endsegments can be indexed and can exist.
> *NO* end segment has more than finitely many predecessors.

Correct. The set of infinite endsegments is not actually infinite. Only finitely many indices are available.

Regards, WM

WM brought next idea :
> Serg io schrieb am Sonntag, 10. Oktober 2021 um 20:29:54 UTC+2:
>> On 10/10/2021 1:13 PM, Transfinity wrote:
>>> Serg io schrieb am Sonntag, 10. Oktober 2021 um 19:54:47 UTC+2:
>>>> On 10/10/2021 12:14 PM, WM wrote:
>
>>> But you said infinitely many natnumbers exist in all endsegments because
>>> all endsegments are infinite. So you cannot prove that they all are
>>> omitted.
>> I proved using correct Math, that the intersection of all endsegments is
>> empty.
>
> I am not interested in the intersection. You said that all endsegments are
> infinite. So they all contain natnumbers, don't they?

No, they are all proper subsets of the natural numbers.

On Sunday, October 10, 2021 at 10:37:09 PM UTC+2, WM wrote:

> ℵo + ℵo is larger than ℵo.

No. ℵo + ℵo = ℵo in set theory (say ZFC).

See: https://ncatlab.org/nlab/show/cardinal+arithmetic#properties

On Sunday, October 10, 2021 at 10:39:47 PM UTC+2, WM wrote nonsense.

Actually:

> The set of [...] endsegments is [...] infinite

Proof: The function n |-> E(n) is a bijection from IN onto {E(n) : n e IN} (i.e. the set of all endsegments). qed

Hint: E(n) = {m e IN : m >= n} by definition.

On 10/10/2021 3:37 PM, WM wrote:
> Serg io schrieb am Sonntag, 10. Oktober 2021 um 20:27:02 UTC+2:
>> On 10/10/2021 12:46 PM, WM wrote:
>>> zelos...@gmail.com schrieb am Sonntag, 10. Oktober 2021 um 19:32:27 UTC+2:
>>>> fredag 8 oktober 2021 kl. 18:58:43 UTC+2 skrev WM:
>>>
>>>>> The contents of endsegments is what is not used to index its predecessors.
>>>> Irrelevant.
>>>
>>> Chuckle. Your only way to pretend ignorance?
>>>
>>>>> If all endsegments have infinite contents, then infinitely many numbers are not used to index all predecessors, i.e., all endsegments. The complement, used to index them, is finite.
>>>> Meaningless empty assertion.
>>>
>>> Meaningful and very important for mathematicians who can think.
>>>
>> Try expressing that thought in equations instead...
>>
> Why? Every endsegment has the cardinality ℵo.

agree

> For consecutive sets ℵo + ℵo is larger than ℵo.

Wrong.

ℵ0 + ℵ0 = ℵ0

and ℵ0 * ℵ0 = ℵ0

Google for it.

> n + ℵo = ℵo is possible.

what is n ?

> Hence there are only finitely many endsegments.

wrong. the number of endsegments is infinite.

>
> Regards, WM
>

Why are you posting such obviously wrong math ?

On 10/10/2021 3:33 PM, WM wrote:
> Serg io schrieb am Sonntag, 10. Oktober 2021 um 20:29:54 UTC+2:
>> On 10/10/2021 1:13 PM, Transfinity wrote:
>>> Serg io schrieb am Sonntag, 10. Oktober 2021 um 19:54:47 UTC+2:
>>>> On 10/10/2021 12:14 PM, WM wrote:
>
>>> But you said infinitely many natnumbers exist in all endsegments because all endsegments are infinite. So you cannot prove that they all are omitted.
>> I proved using correct Math, that the intersection of all endsegments is empty.
>
> I am not interested in the intersection. You said that all endsegments are infinite. So they all contain natnumbers, don't they?
>>
>>> What is the contents of the infinite endsegments?
>> diversion.
>
> Diversion is not in any endsegment. Every endsegment contains infinitely many natural numbers. They cannot be indices of endsegments.
>
> Regards, WM
>

your statement "What is the contents of the infinite endsegments?" I say is diversion, you tried to change the topic, it is a red herring, you failed.

The topic is "the intersection of all endsegments is empty" you said it was infinite, then you agreed it was empty, then you said it was infinite, make
up your mind or prove it one way or the other. OR change the subject.

söndag 10 oktober 2021 kl. 19:44:26 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Sonntag, 10. Oktober 2021 um 19:21:59 UTC+2:
> > söndag 10 oktober 2021 kl. 19:15:01 UTC+2 skrev WM:
> > > Serg io schrieb am Samstag, 9. Oktober 2021 um 18:30:26 UTC+2:
> > >
> > > > PROOF;
> > > > for any k, where k is a natural number
> > > > k is not in the endsegments E(k+1), E(k+2), ...
> > > > therefore k is not in the intersection of all endsegments.
> > > > since we chose ANY k,
> > > No! Every endsegment is infinite. For infinitely many k your proof fails.
> > >
> > It is ANY k that means it applies to ALL of them, NONE does it fail for..
> What are the infinitely many elements which remain in every infinite endsegment? Does the proof prove that they disappear such that none remains. How can all the endsegments remain infinite?
>
> Regards, WM
It proves that ALL natural numbers are missing in some of the endsegments, that is all it needs to do to show the intersection is emtpy!

The rest is you being an idiot

söndag 10 oktober 2021 kl. 19:46:46 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Sonntag, 10. Oktober 2021 um 19:32:27 UTC+2:
> > fredag 8 oktober 2021 kl. 18:58:43 UTC+2 skrev WM:
>
> > >The contents of endsegments is what is not used to index its predecessors.
> > Irrelevant.
> Chuckle. Your only way to pretend ignorance?
> > >If all endsegments have infinite contents, then infinitely many numbers are not used to index all predecessors, i.e., all endsegments. The complement, used to index them, is finite.
> > Meaningless empty assertion.
> Meaningful and very important for mathematicians who can think.
>
> Regards, WM

>Chuckle. Your only way to pretend ignorance?

Going like Gerbil now? What I am saying is that it is IRRELEVANT like so much you pull up! I am sick of you pulling shit that is of no relevance so I will call you out on it.

>Meaningful and very important for mathematicians who can think.

No it isn't! It is only meanignful to YOU and important to YOU but to mathematics it is meaningless empty assertions of no importance!

You cannot express things in mathematical terms because you are not a mathematician and you do not understand mathematics!

On Monday, October 11, 2021 at 6:56:10 AM UTC+2, zelos...@gmail.com wrote:

> You [WM] cannot express things in mathematical terms because you are not a mathematician and you do not understand mathematics!

See: https://mathvault.ca/math-glossary/#pseudomath
and: https://en.wikipedia.org/wiki/Crank_(person)

On Sunday, October 10, 2021 at 10:33:13 PM UTC+2, WM wrote:

> Every endsegment contains infinitely many natural numbers [and ONLY natural numbers].

Right.

> They cannot be indices of endsegments.

Huh?! Since each and every natural number is an index of an (one) endsegment, each and every element in an endsegement is an index of an endsegment.

CPU kaputt, oder was, Mückenheim?

On Monday, October 11, 2021 at 4:07:13 AM UTC+2, Serg io wrote:
> On 10/10/2021 3:33 PM, WM wrote bullshit.
>
> The topic is "the intersection of all endsegments is empty" you said it was infinite,
> then you agreed it was empty, then you said it was infinite, make up your mind or [...]

Well. it depends on the day of week and/or time of day.

On Sunday, 10 October 2021 at 17:39:47 UTC-3, WM wrote:
> [...] The set of infinite endsegments is not actually infinite. Only finitely many indices are available.

Arguing the existence of potentially infinite sets is a clear sign of progressing insanity, my friend. It has nothing whatsoever to do with ZFC, or with reality.

On Sunday, 10 October 2021 at 17:37:09 UTC-3, WM wrote:
> [...] Every endsegment has the cardinality ℵo. For consecutive sets ℵo + ℵo is larger than ℵo.

Bullshit. {1, 3, 5, 7, ..., 2, 4, 6, 8,... } is well-ordering of the natural numbers of order type ℵo + ℵo. You clearly have no idea about infinity, well-orders, the natural numbers, set, or really *ANYTHING* that has to do with mathematics. You are a demented fool.

On Monday, October 11, 2021 at 4:10:15 PM UTC+2, Gus Gassmann wrote:
> On Sunday, 10 October 2021 at 17:37:09 UTC-3, WM wrote:
> >
> > [...] Every endsegment has the cardinality ℵo. For consecutive sets ℵo + ℵo is larger than ℵo.
> >
> Bullshit. {1, 3, 5, 7, ..., 2, 4, 6, 8,... } is well-ordering of the natural numbers of order type ℵo + ℵo.

I guess you meant omega + omega here.

Anyway, WM's claim "ℵo + ℵo is larger than ℵo" is clearly bullshit, since ℵo + ℵo = ℵo-

Without any doubt:

> You clearly have no idea about infinity, well-orders, the natural numbers, set, or really *ANYTHING* that has to do with mathematics. You are a demented fool.

On 10/11/2021 9:06 AM, Gus Gassmann wrote:
> On Sunday, 10 October 2021 at 17:39:47 UTC-3, WM wrote:
>> [...] The set of infinite endsegments is not actually infinite. Only finitely many indices are available.
>
> Arguing the existence of potentially infinite sets is a clear sign of progressing insanity, my friend. It has nothing whatsoever to do with ZFC, or with reality.
>

yep, that WM is way down the path to

potential insanity => progressing insanity => actual insanity (completed insanity)