FromTheRafters schrieb am Sonntag, 12. September 2021 um 20:36:13 UTC+2:
> WM wrote on 9/12/2021 :
> > FromTheRafters schrieb am Sonntag, 12. September 2021 um 01:15:47 UTC+2:
> >
> >>> All endsegments are infinite. What do they contain?
> >> Natural numbers, naturally.
> >
> > Natural numbers are remaining in endsegments as long as endsegments are
> > infinite.
> Which they are, so no need to mention it as if it were a choice.

Your claim is wrong if there are infinitely many endsegments:

Here is the proof
E(1), E(2), E(3), ..., E(n) = (n, n+1, n+2, n+3, ...)
aleph_0 natnumbers before the equals sign and after it is not possible.

Regards, WM

Serg io schrieb am Sonntag, 12. September 2021 um 20:36:46 UTC+2:

> > Therefore aleph_0 endsegments with aleph_0 elements each can only be assumed by peopel who cannot count 1 + 1.
> there are aleph_0 endsegments each endsegment has aleph_0 elements, deal with it.

Why should I deal with foolish ideas? It is enough that I deal with fools.

E(1), E(2), E(3), ..., E(n) = (n, n+1, n+2, n+3, ...)

Regards, WM

On Sunday, 12 September 2021 at 13:42:34 UTC-3, Transfinity wrote:
> Gus Gassmann schrieb am Sonntag, 12. September 2021 um 02:37:38 UTC+2:
> > On Saturday, 11 September 2021 at 17:22:50 UTC-3, WM wrote:
> > > Gus Gassmann schrieb am Samstag, 11. September 2021 um 17:16:18 UTC+2:
> > > > On Saturday, 11 September 2021 at 10:20:38 UTC-3, WM wrote:
> > > > > William schrieb am Samstag, 11. September 2021 um 02:18:54 UTC+2:
>
> > > > And why should it? For any n in N, the *FINITE* (got that?) intersection of end segments {k, k+1, k+2, ...} over k <= n is just E(n).
> > > And there is no finite n leading to an infinite intersection. But there are only finite n. Hence there is no empty intersection.
> > For the record, there are infinitely many natural numbers. each of them finite.
> But infinitely many must be spared for the contents of the endsegments. Therefore only finitely many n can be used to index the endsegments E(n). Note that in order omega two consecutive infinite sets of card aleph_0 are not possible. Therefore aleph_0 endsegments with aleph_0 elements each can only be assumed by peopel who cannot count 1 + 1.

What utter bullshit you are blowing out your arse today! Note: Every natural number n can *OF COURSE* serve two purposes: To be an element of an end segment and to be the starting element (what you call the "index") of an end segment. How the *FUCK* do you think that E(1) and E(2) can exist simultaneously? Note: 2 is both an element of E(1) and the "index" of E(2). You are not able to think clearly any more. Get oot of the business while you might still have a *SHRED* of credibility (with some people).

On Sunday, 12 September 2021 at 15:45:47 UTC-3, WM wrote:
[...]
> Here is the proof
> E(1), E(2), E(3), ..., E(n) = (n, n+1, n+2, n+3, ...)
> aleph_0 natnumbers before the equals sign and after it is not possible.

This is even more clearly nonsense than your usual drivel. There are only finitely many natural numbers (*n* of them) on the left of the equal sign, and aleph_0 on the right. You even counted one of them twice, you stupid cunt.

On 9/12/2021 12:28 PM, WM wrote:
> Jim Burns schrieb
> am Sonntag, 12. September 2021 um 00:42:52 UTC+2:
>> On 9/11/2021 4:27 PM, WM wrote:

>>> All endsegments are infinite.
>>> What do they contain?
>> Infinitely-many not-forever-numbers.
>
> If endsegments are forever, then numbers must be forever.

No end segment is the last end segment.
No number is the last number.

For each 0,...,k, there is k+1 not in it.

Also, and without contradicting the previous:
Each end segment is finitely-preceded.
Each number is finitely-preceded.

For each k, there is 0,...,k, such that
0,...,k begins at 0, ends at k,
for each _split_ B,A of 0,...,k,
_adjacent_ i,j exist, i in B, j in A, j=i+1.

>> ( Note that this is by design.
>> ( Forever-numbers can't be counted with.
>
> Then endsegments can't contain them.

A natural number that cannot be counted to, even in principle,
is not a natural number.

Forever-numbers not-in end segments is not an objection.
In the same way, flying rainbow sparkle ponies not-in
end segments is not an objection. They _shouldn't_ be.

>> ( an intersection of end segments is the last end segment
>> ( or the empty set.
>
> Forever non-empty endsegments contain more than nothing.
> This is contained in all endsegments
> and hence in their intersection.

"More than nothing".
What is more than nothing? Anything can be more.
You're treating it like binary more/not-more.
The answer you asked for:
there is one of the places your argument is wrong.

----
Each number is finitely-preceded.
If a finitely-preceded number m is in the intersection,
every number after m is each end segment.
*BECAUSE THAT'S WHAT AN INTERSECTION IS*

If a finitely-preceded number m is in the intersection,
there are only finitely-many numbers k < m to which
two end segments E,E' can answer "Is k in you?" differently.
End segments that give all the same answers are
the same end segment.

If a finitely-preceded number m is in the intersection,
there are only finitely-many different end segments,
and they each start with some finitely-preceded k =< m.

Therefore,
if a finitely-preceded number m is in the intersection,
end segments are not forever.

On Sunday, September 12, 2021 at 1:21:17 PM UTC-3, WM wrote:
> William schrieb am Samstag, 11. September 2021 um 23:25:41 UTC+2:
> > On Saturday, September 11, 2021 at 5:22:50 PM UTC-3, WM wrote:
> >
> > > And there is no finite n leading to a [finite] intersection. But there are onkly finite n.
> >
> > So what? We are talking about a property of the sequence. There are only finite n. But the sequence is made from
> > and infinite number of finite n.
> Impossible for definable n:

So what? I did not restrict n to things you can write down.

--
William Hughes

On 9/12/2021 1:47 PM, WM wrote:
> Serg io schrieb am Sonntag, 12. September 2021 um 20:36:46 UTC+2:
>
>>> Therefore aleph_0 endsegments with aleph_0 elements each can only be assumed by peopel who cannot count 1 + 1.
>> there are aleph_0 endsegments each endsegment has aleph_0 elements, deal with it.
>
> Why should I deal with foolish ideas? It is enough that I deal with fools.
>
> E(1), E(2), E(3), ..., E(n) = (n, n+1, n+2, n+3, ...)

that is meaningless in math.

A,B,C,D = something IS MEANINGLESS

go take algebra!

>
> Regards, WM
>

On Sunday, September 12, 2021 at 8:45:47 PM UTC+2, WM wrote:

> ... E(n) = (n, n+1, n+2, n+3, ...)

Hint: E(n) = {n, n+1, n+2, n+3, ...}, you silly crank!

On Sunday, September 12, 2021 at 9:39:21 PM UTC+2, Gus Gassmann wrote:

Mückenheim wrote:

> > E(1), E(2), E(3), ..., E(n) = (n, n+1, n+2, n+3, ...)

What the hell IS this?

Any ideas?

Note that E(n) = {n, n+1, n+2, n+3, ...} (i.e. (E(n) is a _set_ of natural numbers, not a SEQUENCE with natural terms).

On Sunday, 12 September 2021 at 18:58:05 UTC-3, Greg Cunt wrote:
> On Sunday, September 12, 2021 at 9:39:21 PM UTC+2, Gus Gassmann wrote:
> Mückenheim wrote:
>
> > > E(1), E(2), E(3), ..., E(n) = (n, n+1, n+2, n+3, ...)
> What the hell IS this?
>
> Any ideas?
>
> Note that E(n) = {n, n+1, n+2, n+3, ...} (i.e. (E(n) is a _set_ of natural numbers, not a SEQUENCE with natural terms).

Your guess is as good as mine. In fact, I misread the parentheses on the right as curly braces. It didn't make sense even then. I have no clue what the equation sign is supposed to do there. It clearly does not equate E(1), E(2), E(3), ..., E(n) with the expression on the right. He clearly has no clue how to write mathematics. But a while back he was saying something to the effect that you need aleph_0 natural numbers to index the end segments, and then _another_ aleph_0 to describe an end segment's elements. My hunch is that he thinks the line you quoted somehow makes this obvious. The only thing it makes painfully obvious is that he has no idea what the issues are or how to express himself in a way that one could even start to have a debate.

Bottom line: That stuff is not even wrong.

On 9/12/2021 5:15 PM, Gus Gassmann wrote:
> On Sunday, 12 September 2021 at 18:58:05 UTC-3, Greg Cunt wrote:
>> On Sunday, September 12, 2021 at 9:39:21 PM UTC+2, Gus Gassmann wrote:
>> Mückenheim wrote:
>>
>>>> E(1), E(2), E(3), ..., E(n) = (n, n+1, n+2, n+3, ...)
>> What the hell IS this?
>>
>> Any ideas?
>>
>> Note that E(n) = {n, n+1, n+2, n+3, ...} (i.e. (E(n) is a _set_ of natural numbers, not a SEQUENCE with natural terms).
>
> Your guess is as good as mine. In fact, I misread the parentheses on the right as curly braces. It didn't make sense even then. I have no clue what the equation sign is supposed to do there. It clearly does not equate E(1), E(2), E(3), ..., E(n) with the expression on the right. He clearly has no clue how to write mathematics. But a while back he was saying something to the effect that you need aleph_0 natural numbers to index the end segments, and then _another_ aleph_0 to describe an end segment's elements. My hunch is that he thinks the line you quoted somehow makes this obvious. The only thing it makes painfully obvious is that he has no idea what the issues are or how to express himself in a way that one could even start to have a debate.
>
> Bottom line: That stuff is not even wrong.
>

he's lost in Palookaville again, suggest he go back to that winning turtle with Framed Achilles the loser...

WM submitted this idea :
> FromTheRafters schrieb am Sonntag, 12. September 2021 um 20:36:13 UTC+2:
>> WM wrote on 9/12/2021 :
>>> FromTheRafters schrieb am Sonntag, 12. September 2021 um 01:15:47 UTC+2:
>>>
>>>>> All endsegments are infinite. What do they contain?
>>>> Natural numbers, naturally.
>>>
>>> Natural numbers are remaining in endsegments as long as endsegments are
>>> infinite.
>> Which they are, so no need to mention it as if it were a choice.
>
> Your claim is wrong if there are infinitely many endsegments:

Your usual horseshit.

> Here is the proof
> E(1), E(2), E(3), ..., E(n) = (n, n+1, n+2, n+3, ...)
> aleph_0 natnumbers before the equals sign and after it is not possible.

The LHS is a finite collection of infinite objects and the RHS is an
infinite collection of finite objects. So what? Is that special in your
matheology?

Gus Gassmann schrieb am Sonntag, 12. September 2021 um 21:36:24 UTC+2:
> On Sunday, 12 September 2021 at 13:42:34 UTC-3, Transfinity wrote:
> > Gus Gassmann schrieb am Sonntag, 12. September 2021 um 02:37:38 UTC+2:
> > > On Saturday, 11 September 2021 at 17:22:50 UTC-3, WM wrote:
> > > > Gus Gassmann schrieb am Samstag, 11. September 2021 um 17:16:18 UTC+2:
> > > > > On Saturday, 11 September 2021 at 10:20:38 UTC-3, WM wrote:
> > > > > > William schrieb am Samstag, 11. September 2021 um 02:18:54 UTC+2:
> >
> > > > > And why should it? For any n in N, the *FINITE* (got that?) intersection of end segments {k, k+1, k+2, ...} over k <= n is just E(n).
> > > > And there is no finite n leading to an infinite intersection. But there are only finite n. Hence there is no empty intersection.
> > > For the record, there are infinitely many natural numbers. each of them finite.
> > But infinitely many must be spared for the contents of the endsegments. Therefore only finitely many n can be used to index the endsegments E(n). Note that in order omega two consecutive infinite sets of card aleph_0 are not possible.
Every natural number n can *OF COURSE* serve two purposes: To be an element of an end segment and to be the starting element (what you call the "index") of an end segment.

But if there are infinitely many starting elements, then there are no successors. You cannot count beyond all infinitely many starting elements.

How the *FUCK* do you think that E(1) and E(2) can exist simultaneously? Note: 2 is both an element of E(1) and the "index" of E(2).

No problem. Only if there were infinitely many endsegments, no consecutive infinite set would be possible, not even a single finite natural number.

Regards, WM

Gus Gassmann schrieb am Sonntag, 12. September 2021 um 21:39:21 UTC+2:
> On Sunday, 12 September 2021 at 15:45:47 UTC-3, WM wrote:
> [...]
> > Here is the proof
> > E(1), E(2), E(3), ..., E(n) = (n, n+1, n+2, n+3, ...)
> > aleph_0 natnumbers before the equals sign and after it is not possible.
> There are only finitely many natural numbers (*n* of them) on the left of the equal sign,

Yes, but there would be infinitely many if infinitely many endsegments existed.

> and aleph_0 on the right.

That is only possible for finitely many endsegments. Infinitely many infinite endsegments cannot exist.

> You even counted one of them twice.

No problem.

Regards, WM

On Monday, September 13, 2021 at 9:32:54 AM UTC-3, WM wrote:

> But if there are infinitely many starting elements, then there are no successors. You cannot count beyond all infinitely many starting elements.

Piffle. There are infinitely many starting elements but each starting element is finite. There is no starting element that does not have a successor, no starting element that you cannot "count beyond".

--
William Hughes

torsdag 9 september 2021 kl. 15:01:10 UTC+2 skrev WM:
> zelos...@gmail.com schrieb am Donnerstag, 9. September 2021 um 07:03:00 UTC+2:
> > onsdag 8 september 2021 kl. 21:38:49 UTC+2 skrev WM:
> > > Jim Burns schrieb am Mittwoch, 8. September 2021 um 00:33:30 UTC+2:
> > >
> > > > If there is no last end segment of all end segments,
> > > > then the intersection of all end segments is
> > > Then there is no "all endsegments". "All" can only be proved by showing the last.
> > >
> > Where does it say that "all" must mean a last?
> In linear order nobody should believe the claim of all without being shown the last element.
>
> Regards, WM

that is only true for FINITE sets, not INFINITE that means NO END.

måndag 13 september 2021 kl. 14:36:20 UTC+2 skrev WM:
> Gus Gassmann schrieb am Sonntag, 12. September 2021 um 21:39:21 UTC+2:
> > On Sunday, 12 September 2021 at 15:45:47 UTC-3, WM wrote:
> > [...]
> > > Here is the proof
> > > E(1), E(2), E(3), ..., E(n) = (n, n+1, n+2, n+3, ...)
> > > aleph_0 natnumbers before the equals sign and after it is not possible.
> > There are only finitely many natural numbers (*n* of them) on the left of the equal sign,
> Yes, but there would be infinitely many if infinitely many endsegments existed.
> > and aleph_0 on the right.
> That is only possible for finitely many endsegments. Infinitely many infinite endsegments cannot exist.
>
> > You even counted one of them twice.
>
> No problem.
>
> Regards, WM

There is nothing in mathematics that says infinitely many endsegments cannot exist. It is an empty assertion of yours.

On Monday, 13 September 2021 at 09:36:20 UTC-3, WM wrote:
> Gus Gassmann schrieb am Sonntag, 12. September 2021 um 21:39:21 UTC+2:
> > On Sunday, 12 September 2021 at 15:45:47 UTC-3, WM wrote:
> > [...]
> > > Here is the proof
> > > E(1), E(2), E(3), ..., E(n) = (n, n+1, n+2, n+3, ...)
> > > aleph_0 natnumbers before the equals sign and after it is not possible.
> > There are only finitely many natural numbers (*n* of them) on the left of the equal sign,
> Yes, but there would be infinitely many if infinitely many endsegments existed.
> > and aleph_0 on the right.
> That is only possible for finitely many endsegments. Infinitely many infinite endsegments cannot exist.
>
> > You even counted one of them twice.
>
> No problem.

Problem, you blithering idiot! You write down a bunch of symbols, concatenate them in a completely asinine way that makes absolutely no sense and think you have a poof of something. I repeat: You have no clue about mathematics, never had a clue, and never will have. All your recent posts show is a frightening descent into a mental miasma. You are no longer compos mentis. The sooner you get professional help, the better.

On 9/13/2021 7:36 AM, WM wrote:
> Gus Gassmann schrieb am Sonntag, 12. September 2021 um 21:39:21 UTC+2:
>> On Sunday, 12 September 2021 at 15:45:47 UTC-3, WM wrote:
>> [...]
>>> Here is the proof
>>> E(1), E(2), E(3), ..., E(n) = (n, n+1, n+2, n+3, ...)
>>> aleph_0 natnumbers before the equals sign and after it is not possible.
>> There are only finitely many natural numbers (*n* of them) on the left of the equal sign,
>
> Yes, but there would be infinitely many if infinitely many endsegments existed.

wrong. E(n) is infinite for all n (n is called a variable WM)

>
>> and aleph_0 on the right.
>
> That is only possible for finitely many endsegments.

wrong. The cursor function says there are infinitely many endsegments,
It is written E(n) as n goes to infinity.
So Shall it be done.

> Infinitely many infinite endsegments cannot exist.

Wrong, BS on its face.

>
>> You even counted one of them twice.
>
> No problem.

why are you counting twice?

>
> Regards, WM
>

mush in => mush out

(do not look at below mis-wroten and foul non mathematical "equation"!)

mush(1),mush(2),mush(3),mush(n) = mushnik(n,(n, n+1, n+2, n+3, ...)

On 9/13/2021 7:32 AM, WM wrote:
> Gus Gassmann schrieb am Sonntag, 12. September 2021 um 21:36:24 UTC+2:
>> On Sunday, 12 September 2021 at 13:42:34 UTC-3, Transfinity wrote:
>>> Gus Gassmann schrieb am Sonntag, 12. September 2021 um 02:37:38 UTC+2:
>>>> On Saturday, 11 September 2021 at 17:22:50 UTC-3, WM wrote:
>>>>> Gus Gassmann schrieb am Samstag, 11. September 2021 um 17:16:18 UTC+2:
>>>>>> On Saturday, 11 September 2021 at 10:20:38 UTC-3, WM wrote:
>>>>>>> William schrieb am Samstag, 11. September 2021 um 02:18:54 UTC+2:
>>>
>>>>>> And why should it? For any n in N, the *FINITE* (got that?) intersection of end segments {k, k+1, k+2, ...} over k <= n is just E(n).
>>>>> And there is no finite n leading to an infinite intersection. But there are only finite n. Hence there is no empty intersection.
>>>> For the record, there are infinitely many natural numbers. each of them finite.
>>> But infinitely many must be spared for the contents of the endsegments. Therefore only finitely many n can be used to index the endsegments E(n). Note that in order omega two consecutive infinite sets of card aleph_0 are not possible.
> Every natural number n can *OF COURSE* serve two purposes: To be an element of an end segment and to be the starting element (what you call the "index") of an end segment.

what else can a natural number n serve ?

>
> But if there are infinitely many starting elements, then there are no successors. You cannot count beyond all infinitely many starting elements.

Wrong, you state that infinity is used up.

>
> How the *FUCK* do you think that E(1) and E(2) can exist simultaneously? Note: 2 is both an element of E(1) and the "index" of E(2).

All endsegments exist simultaneously. Those with small brains cannot think of such things, and must stick to counting rocks.

>
> No problem. Only if there were infinitely many endsegments, no consecutive infinite set would be possible, not even a single finite natural number.

obviously thousands of brain cells were liquidated during the generation of that sentence

>
> Regards, WM
>

Starting Ants
Simultaneously Ants

On Monday, September 13, 2021 at 3:56:35 PM UTC+2, Serg io wrote:

> wrong. E(n) is infinite for all n (n is called a variable WM)

No, Sergio, "n" is called a variable. n is called a natural number (since it is one), if n e IN.

WM is constantly mixing that up, so please...

On Monday, September 13, 2021 at 4:04:40 PM UTC+2, Serg io wrote:

> what else can a natural number n serve ?

See? You are calling it (n) a natural number, here. :-)

On 9/13/2021 9:36 AM, Greg Cunt wrote:
> On Monday, September 13, 2021 at 3:56:35 PM UTC+2, Serg io wrote:
>
>> wrong. E(n) is infinite for all n (n is called a variable WM)
>
> No, Sergio, "n" is called a variable. n is called a natural number (since it is one), if n e IN.
>
> WM is constantly mixing that up, so please...
>

agree, n "n" ,

"n"o misdirectio"n" i"n"te"n"ded

gad! its used in words too!!

FromTheRafters schrieb am Montag, 13. September 2021 um 03:54:55 UTC+2:
> WM submitted this idea :
> > FromTheRafters schrieb am Sonntag, 12. September 2021 um 20:36:13 UTC+2:
> >> WM wrote on 9/12/2021 :
> >>> FromTheRafters schrieb am Sonntag, 12. September 2021 um 01:15:47 UTC+2:
> >>>
> >>>>> All endsegments are infinite. What do they contain?
> >>>> Natural numbers, naturally.
> >>>
> >>> Natural numbers are remaining in endsegments as long as endsegments are
> >>> infinite.
> >> Which they are, so no need to mention it as if it were a choice.
> >
> > Your claim is wrong if there are infinitely many endsegments:
> Your usual horseshit.
> > Here is the proof
> > E(1), E(2), E(3), ..., E(n) = (n, n+1, n+2, n+3, ...)
> > aleph_0 natnumbers before the equals sign and after it is not possible.
> The LHS is a finite collection of infinite objects and the RHS is an
> infinite collection of finite objects.

The RHS is E(n). The natural numbers are used in E(1), E(2), and so on. Only what remains after the "and so on" can be the contents of the next endsegment. But if there is no last endsegment, then nothing remains for the contents.

Regards, WM

William schrieb am Sonntag, 12. September 2021 um 23:10:48 UTC+2:
> On Sunday, September 12, 2021 at 1:21:17 PM UTC-3, WM wrote:
> > William schrieb am Samstag, 11. September 2021 um 23:25:41 UTC+2:
> > > On Saturday, September 11, 2021 at 5:22:50 PM UTC-3, WM wrote:
> > >
> > > > And there is no finite n leading to a [finite] intersection. But there are only finite n.
> > >
> > > So what? We are talking about a property of the sequence. There are only finite n. But the sequence is made from
> > > an infinite number of finite n.
> > Impossible for definable n:
> So what? I did not restrict n to things you can write down.

So you accept also numbers n which have no finite representation? As I said, it is impossible to have aleph_0 natnumbers with finite representation:

Representation of the natural numbers: o, oo, ooo,...
Representation of the number of terms: o, oo, ooo, ...
Both sequences will never deviate.

Regards, WM