On Wednesday, 13 October 2021 at 08:11:34 UTC-3, WM wrote:
[...]
> All infinite endsegments share infinitely many with all infinite endsegments. Inclusion monotony.

Garbled nonsense.

On 10/13/2021 6:11 AM, WM wrote:
> zelos...@gmail.com schrieb am Dienstag, 12. Oktober 2021 um 06:59:24 UTC+2:
>> måndag 11 oktober 2021 kl. 20:38:15 UTC+2 skrev WM:
>
>> No endsegment share all with the rest
>
> All infinite endsegments share infinitely many with all infinite endsegments. All infinite endsegments share infinitely many with all infinite endsegments
>
> Regards, WM
>

True => No endsegment share all with the rest

False => All infinite endsegments share infinitely many with all infinite endsegments

Red Herring => Inclusion monotony

On 10/13/2021 6:15 AM, WM wrote:
> zelos...@gmail.com schrieb am Dienstag, 12. Oktober 2021 um 07:00:25 UTC+2:
>> måndag 11 oktober 2021 kl. 21:06:55 UTC+2 skrev WM:
>
>>>> What I am saying is that it is IRRELEVANT
>>> Two consecutive sets in ω cannot have cardinality ℵo both.
>>> If all endsegments have card ℵo then the set of their indices is finite.
>
>> No one says that you dishonest shit.
>
> But it is fact. omega cannot be divided in two consecutive infinite sets.

Bullshit. You do not know math, nor infinity operations. You novice.

No wonder you make so many mistakes.

Gus Gassmann formulated the question :
> On Wednesday, 13 October 2021 at 08:11:34 UTC-3, WM wrote:
> [...]
>> All infinite endsegments share infinitely many with all infinite
>> endsegments. Inclusion monotony.
>
> Garbled nonsense.

Garbled nonsense with inclusion monotony on top.

Gus Gassmann schrieb am Mittwoch, 13. Oktober 2021 um 14:25:50 UTC+2:
> On Wednesday, 13 October 2021 at 08:11:34 UTC-3, WM wrote:
> [...]
> > All infinite endsegments share infinitely many with all infinite endsegments. Inclusion monotony.
> Garbled nonsense.

What is the first infinite endsegment that shares less?

Regards, WM

zelos...@gmail.com schrieb am Mittwoch, 13. Oktober 2021 um 13:19:40 UTC+2:
> onsdag 13 oktober 2021 kl. 13:11:34 UTC+2 skrev WM:
> > zelos...@gmail.com schrieb am Dienstag, 12. Oktober 2021 um 06:59:24 UTC+2:
> > > måndag 11 oktober 2021 kl. 20:38:15 UTC+2 skrev WM:
> >
> > > No endsegment share all with the rest
> > All infinite endsegments share infinitely many with all infinite endsegments. Inclusion monotony.
> >
> incorrect, they share NOTHING with all of them. No matter what element you say is in all of them, I can find an endsegment missing it.

You can only name natnumbers which have infinitely many successors. They belong to infinite endsegments. Infinite endsegments cannot have lost all natnumbers.
>
> That is why the intersection is empty

But not the intersection of all infinite endsegments. If you can't understand how it works, try to find the first infinite endsegment that shares less than infinitely many natnumbers with all predecessors.

Regards, WM

zelos...@gmail.com schrieb am Mittwoch, 13. Oktober 2021 um 13:18:50 UTC+2:
> onsdag 13 oktober 2021 kl. 13:15:14 UTC+2 skrev WM:

> > But it is fact. omega cannot be divided in two consecutive infinite sets. If it is divided, then the stop is at a finite position.
> >
> no one says that again,

For infinite endsegments you need ℵo natnumbers as indexes. If they all are infinite you need ℵo natnumbers as contents in each one. Too much.

Regards

Crank Wolfgang Mueckenheim, aka WM wrote:
> zelos...@gmail.com schrieb am Mittwoch, 13. Oktober 2021 um 13:19:40 UTC+2:
>> onsdag 13 oktober 2021 kl. 13:11:34 UTC+2 skrev WM:
>>> zelos...@gmail.com schrieb am Dienstag, 12. Oktober 2021 um 06:59:24 UTC+2:
>>>> måndag 11 oktober 2021 kl. 20:38:15 UTC+2 skrev WM:
>>>
>>>> No endsegment share all with the rest
>>> All infinite endsegments share infinitely many with all infinite endsegments. Inclusion monotony.
>>>
>> incorrect, they share NOTHING with all of them. No matter what element you say is in all of them, I can find an endsegment missing it.
>
> You can only name natnumbers which have infinitely many successors.

All natnumbers have infinitely many successors.

> They belong to infinite endsegments.

All natnumbers belong to some infinite endsegment, so what?

> Infinite endsegments cannot have lost all natnumbers.

They don't have to "lost" all natnumbers for the intersection to
be empty.

>> That is why the intersection is empty

> But not the intersection of all infinite endsegments.

It is empty. Trivial fact.

On Wednesday, October 13, 2021 at 5:04:40 PM UTC+2, WM wrote:

> For infinite[ly man] endsegments you need ℵo natnumbers as indexes. If they all are infinite you need ℵo natnumbers as contents in each one. Too much.

Holy shit!

So in your psychotic kingdom there is no sequence

(N, N, N, N, ...) ,

right? After all, "for infinitely many Ns you need ℵo natnumbers as indexes. If they all are infinite you need ℵo natnumbers as contents in each one. Too much."

Greg Cunt schrieb am Mittwoch, 13. Oktober 2021 um 17:17:19 UTC+2:
> On Wednesday, October 13, 2021 at 5:04:40 PM UTC+2, WM wrote:
>
> > For infinitely many endsegments you need ℵo natnumbers as indexes. If they all are infinite you need ℵo natnumbers as contents in each one. Too much.
>
> So there is no sequence
>
> (N, N, N, N, ...) ,
>
> right? After all, "for infinitely many Ns you need ℵo natnumbers as indexes.

Here we have parallel sets as contents and indexes.

For endsegments one and the same set is used to index the endsegments and to fill them. No index of its predecessors E(1), E(2), ..., E(n-1) can be in the contents of any endsegment E(n).

Example: E(1), E(2), E(3), E(4) = {4, 5, 6, ...}.

Regards, WM

Python schrieb am Mittwoch, 13. Oktober 2021 um 17:17:08 UTC+2:

> > You can only name natnumbers which have infinitely many successors.
> All natnumbers have infinitely many successors.

No, then the successors would not be natnumbers.
All natnumbers are in |N, and |N has no successor.

Regards, WM

On Wednesday, 13 October 2021 at 11:52:53 UTC-3, WM wrote:
> Gus Gassmann schrieb am Mittwoch, 13. Oktober 2021 um 14:25:50 UTC+2:
> > On Wednesday, 13 October 2021 at 08:11:34 UTC-3, WM wrote:
> > [...]
> > > All infinite endsegments share infinitely many with all infinite endsegments. Inclusion monotony.
> > Garbled nonsense.
> What is the first infinite endsegment that shares less?

Still garbled nonsense. Shares less what with what? Fucking learn to express yourself clearly, or shut the fuck up.

On Wednesday, 13 October 2021 at 12:31:19 UTC-3, WM wrote:
> Python schrieb am Mittwoch, 13. Oktober 2021 um 17:17:08 UTC+2:
>
> > > You can only name natnumbers which have infinitely many successors.
> > All natnumbers have infinitely many successors.
>
> Regards, WM
> No, then the successors would not be natnumbers.
> All natnumbers are in |N, and |N has no successor.

Cute two-step around natural language ambiguity. As a mathematical contribution, it is *SHIT*, of course.

On Wednesday, 13 October 2021 at 12:04:40 UTC-3, WM wrote:
> For infinite endsegments you need ℵo natnumbers as indexes. If they all are infinite you need ℵo natnumbers as contents in each one. Too much.

Crap, as usual. You seem to be claiming that E(2) cannot be defined, because 2 is already an element of E(1). I don't believe you are thinking before you write, *EVER*, and that leads me to think that you *CANNOT* think any longer. Your brain is fried, you have wasted your last brain cell, and from here on it, you can't even *SHIT* properly any longer. You should reside in a padded cell, with 24 hour supervision.

Gus Gassmann schrieb am Mittwoch, 13. Oktober 2021 um 17:34:28 UTC+2:
> On Wednesday, 13 October 2021 at 12:31:19 UTC-3, WM wrote:
> > Python schrieb am Mittwoch, 13. Oktober 2021 um 17:17:08 UTC+2:
> >
> > > > You can only name natnumbers which have infinitely many successors.
> > > All natnumbers have infinitely many successors.
> >
> > Regards, WM
> > No, then the successors would not be natnumbers.
> > All natnumbers are in |N, and |N has no successor.
> Cute two-step around natural language ambiguity.

No ambiguity. All natural numbers are in |N. None is missing.
All natnumbers which can be defined have aleph_0 successors and hence are only a minority. Collecting them into a set |N_def would leave almost all natnumbers outside.

Regards, WM

Gus Gassmann schrieb am Mittwoch, 13. Oktober 2021 um 17:32:27 UTC+2:
> On Wednesday, 13 October 2021 at 11:52:53 UTC-3, WM wrote:
> > Gus Gassmann schrieb am Mittwoch, 13. Oktober 2021 um 14:25:50 UTC+2:
> > > On Wednesday, 13 October 2021 at 08:11:34 UTC-3, WM wrote:
> > > [...]
> > > > All infinite endsegments share infinitely many with all infinite endsegments. Inclusion monotony.
> > > Garbled nonsense.
> > What is the first infinite endsegment that shares less?
> Shares less what with what?
Are you really too stupid to interpret this simple sentence?
What is the first infinite endsegment that shares less than infinitely many elements with one or two or many or all infinite endsegments?

Regards, WM

Gus Gassmann schrieb am Mittwoch, 13. Oktober 2021 um 17:38:17 UTC+2:
> On Wednesday, 13 October 2021 at 12:04:40 UTC-3, WM wrote:
> > For infinite endsegments you need ℵo natnumbers as indexes. If they all are infinite you need ℵo natnumbers as contents in each one. Too much.
> You seem to be claiming that E(2) cannot be defined, because 2 is already an element of E(1).

Sorry, you confuse everything. All elements of E(1) will later become indexes of endsegments. Number n will appear as index of E(n) and as its first element. Then it disappeares from the sequence. No further endegment carries it as an index nor as contents.

But if infinitely many natnumbers will remain as contents, then they cannot become indexes. Therefore only the complement, which is finite, can be used as indexes.

Regards, WM

On Wednesday, 13 October 2021 at 12:41:42 UTC-3, WM wrote:
> Gus Gassmann schrieb am Mittwoch, 13. Oktober 2021 um 17:32:27 UTC+2:
> > On Wednesday, 13 October 2021 at 11:52:53 UTC-3, WM wrote:
> > > Gus Gassmann schrieb am Mittwoch, 13. Oktober 2021 um 14:25:50 UTC+2:
> > > > On Wednesday, 13 October 2021 at 08:11:34 UTC-3, WM wrote:
> > > > [...]
> > > > > All infinite endsegments share infinitely many with all infinite endsegments. Inclusion monotony.
> > > > Garbled nonsense.
> > > What is the first infinite endsegment that shares less?
> > Shares less what with what?
> Are you really too stupid to interpret this simple sentence?

If you are too stupid or lazy to bother writing down what you mean, you are welcome to fuck off. The question is ambiguous and has two possible answers, as you perfectly well know.

What is the first infinite endsegment that shares less than infinitely many elements with one or two or many or all infinite endsegments?

The intersection of finitely many intersections E(k_1), E(k_2), ..., E(k_n) has cardinality aleph_0 and equals the end segment with the largest of the "indices" k_i, i=1...n. Since there is no largest natural number in an infinite set of indices, the intersection of end segments whose index set is infinite, has cardinality 0.

On Wednesday, 13 October 2021 at 13:45:36 UTC-3, WM wrote:
> Gus Gassmann schrieb am Mittwoch, 13. Oktober 2021 um 17:38:17 UTC+2:
> > On Wednesday, 13 October 2021 at 12:04:40 UTC-3, WM wrote:
> > > For infinite endsegments you need ℵo natnumbers as indexes. If they all are infinite you need ℵo natnumbers as contents in each one. Too much.
> > You seem to be claiming that E(2) cannot be defined, because 2 is already an element of E(1).
> Sorry, you confuse everything. All elements of E(1) will later become indexes of endsegments. Number n will appear as index of E(n) and as its first element. Then it disappeares from the sequence. No further endegment carries it as an index nor as contents.
>
> But if infinitely many natnumbers will remain as contents, then they cannot become indexes. Therefore only the complement, which is finite, can be used as indexes.

Your usual bullshit. There is no largest natural number, hence no last index, and hence end segment. It is clear that you cannot grasp this simple fact, which is why I have continually said that you are too fucking stupid to understand infinity, or any mathematics, for that matter.

On 10/13/2021 9:52 AM, WM wrote:
> Gus Gassmann schrieb am Mittwoch, 13. Oktober 2021 um 14:25:50 UTC+2:
>> On Wednesday, 13 October 2021 at 08:11:34 UTC-3, WM wrote:
>> [...]
>>> All infinite endsegments share infinitely many with all infinite endsegments. Inclusion monotony.
>> Garbled nonsense.
>
> What is the first infinite endsegment that shares less?
>
> Regards, WM
>

Endsegments are fixed infinite series, they never "share", nor "acquire" nor "lose" elements.

try using an equation for that

On 10/13/2021 10:41 AM, WM wrote:
> Gus Gassmann schrieb am Mittwoch, 13. Oktober 2021 um 17:32:27 UTC+2:
>> On Wednesday, 13 October 2021 at 11:52:53 UTC-3, WM wrote:
>>> Gus Gassmann schrieb am Mittwoch, 13. Oktober 2021 um 14:25:50 UTC+2:
>>>> On Wednesday, 13 October 2021 at 08:11:34 UTC-3, WM wrote:
>>>> [...]
>>>>> All infinite endsegments share infinitely many with all infinite endsegments. Inclusion monotony.
>>>> Garbled nonsense.
>>> What is the first infinite endsegment that shares less?
>> Shares less what with what?
> Are you really too stupid to interpret this simple sentence?
> What is the first infinite endsegment that shares less than infinitely many elements with one or two or many or all infinite endsegments?

Each endsegment is a separate fixed infinite series of natural numbers, there is no sharing.

if you want to share, use AN EQUATION

Are you really too stupid to use an EQUATION ?

>
> Regards, WM
>

On 10/13/2021 10:39 AM, WM wrote:
> Gus Gassmann schrieb am Mittwoch, 13. Oktober 2021 um 17:34:28 UTC+2:
>> On Wednesday, 13 October 2021 at 12:31:19 UTC-3, WM wrote:
>>> Python schrieb am Mittwoch, 13. Oktober 2021 um 17:17:08 UTC+2:
>>>
>>>>> You can only name natnumbers which have infinitely many successors.
>>>> All natnumbers have infinitely many successors.
>>>
>>> Regards, WM
>>> No, then the successors would not be natnumbers.
>>> All natnumbers are in |N, and |N has no successor.
>> Cute two-step around natural language ambiguity.
>
> No ambiguity. All natural numbers are in |N. None is missing.

why would natural numbers be "missing" from |N ??

> All natnumbers which can be defined have aleph_0 successors and hence are only a minority. Collecting them into a set |N_def would leave almost all natnumbers outside.

wrong. "minority" and "almost all" are subjective meaningless terms are therefore are rejected.

Use equations, or go post in sci.fake.math

>
> Regards, WM
>

On 10/13/2021 10:28 AM, WM wrote:
> Greg Cunt schrieb am Mittwoch, 13. Oktober 2021 um 17:17:19 UTC+2:
>> On Wednesday, October 13, 2021 at 5:04:40 PM UTC+2, WM wrote:
>>
>>> For infinitely many endsegments you need ℵo natnumbers as indexes. If they all are infinite you need ℵo natnumbers as contents in each one. Too much.
>>
>> So there is no sequence
>>
>> (N, N, N, N, ...) ,
>>
>> right? After all, "for infinitely many Ns you need ℵo natnumbers as indexes.
>
> Here we have parallel sets as contents and indexes.
>
> For endsegments one and the same set is used to index the endsegments and to fill them. No index of its predecessors E(1), E(2), ..., E(n-1) can be in the contents of any endsegment E(n).
>
> Example: E(1), E(2), E(3), E(4) = {4, 5, 6, ...}.

that is malignant and disordered thinking

>
> Regards, WM
>
>

On 10/13/2021 12:11 PM, Gus Gassmann wrote:
> On Wednesday, 13 October 2021 at 13:45:36 UTC-3, WM wrote:
>> Gus Gassmann schrieb am Mittwoch, 13. Oktober 2021 um 17:38:17 UTC+2:
>>> On Wednesday, 13 October 2021 at 12:04:40 UTC-3, WM wrote:
>>>> For infinite endsegments you need ℵo natnumbers as indexes. If they all are infinite you need ℵo natnumbers as contents in each one. Too much.
>>> You seem to be claiming that E(2) cannot be defined, because 2 is already an element of E(1).
>> Sorry, you confuse everything. All elements of E(1) will later become indexes of endsegments. Number n will appear as index of E(n) and as its first element. Then it disappeares from the sequence. No further endegment carries it as an index nor as contents.
>>
>> But if infinitely many natnumbers will remain as contents, then they cannot become indexes. Therefore only the complement, which is finite, can be used as indexes.
>
> Your usual bullshit. There is no largest natural number, hence no last index, and hence end segment. It is clear that you cannot grasp this simple fact, which is why I have continually said that you are too fucking stupid to understand infinity, or any mathematics, for that matter.
>

he understands most words, not all, separately, but seems unable to hook them up in a meaningful sentence

Gus Gassmann schrieb am Mittwoch, 13. Oktober 2021 um 19:11:57 UTC+2:
> There is no largest natural number, hence no last index, and hence end segment.

Hence no empty intersection. Every removed number has infinitely many not removed successors.

Regards, WM