It happens that Serg io formulated :
> 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 ??

They got called upon to serve as 'indexes' and never returned to base.

>> 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.

We do have 'almost all' and 'almost homomorphisms'.

On Wednesday, 13 October 2021 at 15:43:22 UTC-3, Transfinity wrote:
> 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.

Oh bullshit. You have no clue about how to determine infinite intersections, either. Hint: It is not a stepwise process.

On Wednesday, October 13, 2021 at 8:53:27 PM UTC+2, FromTheRafters wrote:
> It happens that Serg io formulated :
> >
> > why would natural numbers be "missing" from |N ??
> >
> They got called upon to serve as 'indexes' and never returned to base.

Or they've gone AWOL!

On 10/13/2021 11:41 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?
>
> one

There is no first.

> or two

There is no first.

> or many

Finitely-many? There is no first.

> or all infinite endsegments?

N.

Shares less what with what?

After serious thinking Greg Cunt wrote :
> On Wednesday, October 13, 2021 at 8:53:27 PM UTC+2, FromTheRafters wrote:
>> It happens that Serg io formulated :
>>>
>>> why would natural numbers be "missing" from |N ??
>>>
>> They got called upon to serve as 'indexes' and never returned to base.
>
> Or they've gone AWOL!

A whole detachment of FISON endsegments has been potentially activated
and infinitry units stand ready. The high capacity natural sponge
inclusion monotony cannons should absorb *all* of the spilled numbers
allowing a clear path to the hotel parking lot -- which is full of
buses waiting for the new wing to be built by Mucky Construction Inc.

No worries, it is over halfway complete already.

On 10/13/2021 4:30 PM, FromTheRafters wrote:
> After serious thinking Greg Cunt wrote :
>> On Wednesday, October 13, 2021 at 8:53:27 PM UTC+2, FromTheRafters wrote:
>>> It happens that Serg io formulated :
>>>>
>>>> why would natural numbers be "missing" from |N ??
>>>>
>>> They got called upon to serve as 'indexes' and never returned to base.
>>
>> Or they've gone AWOL!
>
> A whole detachment of FISON endsegments has been potentially activated and infinitry units stand ready. The high capacity natural sponge inclusion
> monotony cannons should absorb *all* of the spilled numbers allowing a clear path to the hotel parking lot -- which is full of buses waiting for the new
> wing to be built by Mucky Construction Inc.
>
> No worries, it is over halfway complete already.

good, almost potentially completed housing for sponge and bus contents

Crank Wolfgang Mueckenheim, aka 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.
>
> No,

Yes

> then the successors would not be natnumbers.
> All natnumbers are in |N,

Every members of N (i.e. natnumbers) have infinitely many successors
(all in N also).

> and |N has no successor.

Irrelevant.

onsdag 13 oktober 2021 kl. 16:58:26 UTC+2 skrev 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

> You can only name natnumbers which have infinitely many successors. They belong to infinite endsegments. Infinite endsegments cannot have lost all natnumbers.

Which is all natural numbers, yes there is no empty endsegment but there is no element in common to all endsegments.

> 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.

Non-sequitor. What they share with predecessors is not relevant, what is relevant is what they share with ALL endsegments which is NO elements.

onsdag 13 oktober 2021 kl. 17:04:40 UTC+2 skrev 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

Meaningless shite

zelos...@gmail.com schrieb am Donnerstag, 14. Oktober 2021 um 06:47:31 UTC+2:
> onsdag 13 oktober 2021 kl. 16:58:26 UTC+2 skrev WM:

> > You can only name natnumbers which have infinitely many successors. They belong to infinite endsegments. Infinite endsegments cannot have lost all natnumbers.
> Which is all natural numbers, yes there is no empty endsegment but there is no element in common to all endsegments.

Contradiction with mathematics: Inclusion monotony.

> > 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.
> Non-sequitor. What they share with predecessors is not relevant, what is relevant is what they share with ALL endsegments which is NO elements.

(1) ALL endsegments are predecessors, because there is no last one.
(2) Try to find the first infinite endsegment that shares less than infinitely many natnumbers with all other infinite endsegments.

Regards, WM

Python schrieb am Donnerstag, 14. Oktober 2021 um 01:13:31 UTC+2:
> Wolfgang Mueckenheim, aka 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.
> >
> > No, then the successors would not be natnumbers.
> > All natnumbers are in |N,
> Every members of N (i.e. natnumbers) have infinitely many successors
> (all in N also).

All can be subtracted, then none remains. Hence also all successors can be subtracted. No successors remain.

> > and |N has no successor.
> Irrelevant.

Very relevant, because |N contains only natural numbers. If there was not a last one, no last one could be removed and always successors would remain.

Regaeds, WM

Gus Gassmann schrieb am Mittwoch, 13. Oktober 2021 um 19:08:15 UTC+2:

> 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,

Nevertheless there are only natural numbers in the set of indices. As long as only endsegments containing natural numbers and having natural indices are intersected, the intersection is infinite.

> the intersection of end segments whose index set is infinite, has cardinality 0.

Why that? Magic of matheology? In mathematics the intersection of infinite endegments is infinite.

Look, if you take the infinite set of infinite endsegments and try to find any infinite endsegment that has less than infinitely many natural numbers in common with any infinite endsegment or any set of infinite endsegments or all predecessors or all successors or all infinite endsegments, then you will fail. Therefore your claim is absolutely foolish and contrary to mathematics and logic.

Regards, WM

Jim Burns schrieb am Mittwoch, 13. Oktober 2021 um 22:46:57 UTC+2:
> On 10/13/2021 11:41 AM, WM wrote:

> > What is the first infinite endsegment that shares
> > less than infinitely many elements with
> > one
> > or two
> > or many
> > or all infinite endsegments?
> >
> There is no first.
>
> > or two
>
> There is no first.
>
> > or many
>
> Finitely-many? There is no first.
>
> > or all infinite endsegments?
>
> Shares less what with what?

Look, if you take the infinite set of infinite endsegments and try to find any infinite endsegment that has less than infinitely many natural numbers in common with any infinite endsegment or any set of infinite endsegments or all predecessors or all successors or all infinite endsegments, then you will fail. Therefore it is contrary to mathematics and logic to claim that *infinite* endsegments can have an empty intersection.

Regards, WM

On Thursday, 14 October 2021 at 06:49:59 UTC-3, WM wrote:
> Gus Gassmann schrieb am Mittwoch, 13. Oktober 2021 um 19:08:15 UTC+2:
>
> > 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,
> Nevertheless there are only natural numbers in the set of indices. As long as only endsegments containing natural numbers and having natural indices are intersected, the intersection is infinite.

No, it's not. Intersections over end segments can only contain natural numbers at least as large as the largest "index". In an infinite collection there is no largest index, and hence there are no natural numbers in the intersection. Ergo, the intersection is empty.

> > the intersection of end segments whose index set is infinite, has cardinality 0.

> Why that? Magic of matheology? In mathematics the intersection of infinite endegments is infinite.
>
> Look, if you take the infinite set of infinite endsegments and try to find any infinite endsegment that has less than infinitely many natural numbers in common with any infinite endsegment

finite intersection, hence the cardinality is aleph_0

> or any set of infinite endsegments

crap statement. It depends on the cardinality of the set you intersect over..

> or all predecessors

finite set, hence the cardinality of the intersection is aleph_0

> or all successors or all infinite endsegments,

Infinite sets both, so the cardinality of the intersection is 0.

> then you will fail.

The only one who fails to apply logic in a consistent manner is you. You throw words around in arbitrary order, you invent ever new terms to paper over your insecurities, and you have never seen two quantifiers you didn't interchange in heartbeat. You practice Muckemyth, not mathematics. And your dementia and mental defects run so deep that you can't even see that any more..

After serious thinking WM wrote :
> zelos...@gmail.com schrieb am Donnerstag, 14. Oktober 2021 um 06:47:31 UTC+2:
>> onsdag 13 oktober 2021 kl. 16:58:26 UTC+2 skrev WM:
>
>>> You can only name natnumbers which have infinitely many successors. They
>>> belong to infinite endsegments. Infinite endsegments cannot have lost all
>>> natnumbers.
>> Which is all natural numbers, yes there is no empty endsegment but there is
>> no element in common to all endsegments.
>
> Contradiction with mathematics: Inclusion monotony.

Stop lying. Successive FISON sets are inclusive, their associated
successive endsegment sets are exclusive. A single element's exclusion
from an endsegment set is sufficient to make the intersection of all
endsegment sets empty.

WM presented the following explanation :
> Gus Gassmann schrieb am Mittwoch, 13. Oktober 2021 um 19:08:15 UTC+2:
>
>> 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,
>
> Nevertheless there are only natural numbers in the set of indices. As long as
> only endsegments containing natural numbers and having natural indices are
> intersected, the intersection is infinite.
>
>> the intersection of end segments whose index set is infinite, has
>> cardinality 0.
>
> Why that? Magic of matheology? In mathematics the intersection of infinite
> endegments is infinite.
>
> Look, if you take the infinite set of infinite endsegments and try to find
> any infinite endsegment that has less than infinitely many natural numbers in
> common with any infinite endsegment or any set of infinite endsegments or all
> predecessors or all successors or all infinite endsegments, then you will
> fail.

Paraphrased:

Look, if you take the -infinite- set of -infinite- endsegments and try
to find any -infinite- endsegment that has less than infinitely many
natural numbers in common with any -infinite- endsegment or any set of
-infinite- endsegments or all predecessors or all successors or all
-infinite- endsegments, then you will fail.

Even cleared of obsessively superfluous infinities you still make no
sense. Of course, you will see it as if we don't get it, but in
reality, you need medical assistance.

On 10/14/2021 4:49 AM, WM wrote:
> Gus Gassmann schrieb am Mittwoch, 13. Oktober 2021 um 19:08:15 UTC+2:
>
>> 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,
>
> Nevertheless there are only natural numbers in the set of indices. As long as only endsegments containing natural numbers and having natural indices are intersected, the intersection is infinite.
>
>> the intersection of end segments whose index set is infinite, has cardinality 0.
>
> Why that? Magic of matheology? In mathematics the intersection of infinite endegments is infinite.
>
> Look, if you take the infinite set of infinite endsegments and try to find any infinite endsegment that has less than infinitely many natural numbers in common with any infinite endsegment or any set of infinite endsegments or all predecessors or all successors or all infinite endsegments, then you will fail. Therefore your claim is absolutely foolish and contrary to mathematics and logic.
>
> Regards, WM
>
>

You are not using truth or logic.
You are failing extremely basic logic AND you are being intenitonally dishonest.
You pull statements and "conclusions" out of your arse with NOTHING supporting them.

On 10/14/2021 5:00 AM, WM wrote:
> Jim Burns schrieb am Mittwoch, 13. Oktober 2021 um 22:46:57 UTC+2:
>> On 10/13/2021 11:41 AM, WM wrote:
>
>>> What is the first infinite endsegment that shares
>>> less than infinitely many elements with
>>> one
>>> or two
>>> or many
>>> or all infinite endsegments?
>>>
>> There is no first.
>>
>>> or two
>>
>> There is no first.
>>
>>> or many
>>
>> Finitely-many? There is no first.
>>
>>> or all infinite endsegments?
>>
>> Shares less what with what?
>
> Look, if you take the infinite set of infinite endsegments and try to find any infinite endsegment that has less than infinitely many natural numbers in common with any infinite endsegment or any set of infinite endsegments or all predecessors or all successors or all infinite endsegments, then you will fail. Therefore it is contrary to mathematics and logic to claim that *infinite* endsegments can have an empty intersection.
>
> Regards, WM
>

You are not using truth or logic.
You are failing extremely basic logic AND you are being intenitonally dishonest.
You pull statements and "conclusions" out of your arse with NOTHING supporting them.

On 10/14/2021 4:36 AM, WM wrote:
> zelos...@gmail.com schrieb am Donnerstag, 14. Oktober 2021 um 06:47:31 UTC+2:
>> onsdag 13 oktober 2021 kl. 16:58:26 UTC+2 skrev WM:
>
>>> You can only name natnumbers which have infinitely many successors. They belong to infinite endsegments. Infinite endsegments cannot have lost all natnumbers.
>> Which is all natural numbers, yes there is no empty endsegment but there is no element in common to all endsegments.
>
> Contradiction with mathematics: Inclusion monotony.
>
>>> 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.
>> Non-sequitor. What they share with predecessors is not relevant, what is relevant is what they share with ALL endsegments which is NO elements.
>
> (1) ALL endsegments are predecessors, because there is no last one.
> (2) Try to find the first infinite endsegment that shares less than infinitely many natnumbers with all other infinite endsegments.
>
> Regards, WM
>

You are not using truth or logic.
You are failing extremely basic logic AND you are being intenitonally dishonest.
You pull statements and "conclusions" out of your arse with NOTHING supporting them.

On Thursday, 14 October 2021 at 06:36:40 UTC-3, WM wrote:
[...]
> (1) ALL endsegments are predecessors, because there is no last one.
> (2) Try to find the first infinite endsegment that shares less than infinitely many natnumbers with all other infinite endsegments.

Try to reformulate (2) without the ambiguity you are so fond of, or fuck off.

On 10/14/2021 6:00 AM, WM wrote:
> Jim Burns schrieb
> am Mittwoch, 13. Oktober 2021 um 22:46:57 UTC+2:
>> On 10/13/2021 11:41 AM, 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

>>> What is the first infinite endsegment that shares
>>> less than infinitely many elements with
>>> one
>>> or two
>>> or many
>>> or all infinite endsegments?
>>>
>> There is no first.
>>
>>> or two
>>
>> There is no first.
>>
>>> or many
>>
>> Finitely-many? There is no first.
>>
>>> or all infinite endsegments?
>>
>> Shares less what with what?
>
> Look,

Shares less what with what?

----
> Look, if you take the infinite set of infinite endsegments

For example, FISON-enders.

After each FISON, there is another FISON.
After each FISON-ender, there is another FISON-ender.
There is no last FISON-ender.
For each FISON-ender, there is an end segment.
There is no last end segment.
For each end segment, there is no last FISON-ender in it.

> and try to find any infinite endsegment that has less than
> infinitely many natural numbers in common with any infinite
> endsegment or any set of infinite endsegments or all
> predecessors or all successors or all infinite endsegments,
> then you will fail. Therefore

"Therefore" is what you write when you've closed your eyes
and started randomly punching buttons.

> then you will fail. Therefore it is contrary to mathematics
> and logic to claim that *infinite* endsegments can have an
> empty intersection.

A FISON-ender in every end segment is in no FISON.
No FISON-ender is in no FISONs.
Their intersection is empty.

Gus Gassmann schrieb am Donnerstag, 14. Oktober 2021 um 12:17:53 UTC+2:
> On Thursday, 14 October 2021 at 06:49:59 UTC-3, WM wrote:
> > Gus Gassmann schrieb am Mittwoch, 13. Oktober 2021 um 19:08:15 UTC+2:
> >
> > > 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,
> > Nevertheless there are only natural numbers in the set of indices. As long as only endsegments containing natural numbers and having natural indices are intersected, the intersection is infinite.
> No, it's not. Intersections over end segments can only contain natural numbers at least as large as the largest "index". In an infinite collection there is no largest index,

Wrong. It is not definable. But since there are infinitely many natural numbers in every infinite endsegment, each one has a finite natnumber as its index. It is simply not a fixed largest index recognizable.

Regards, WM

FromTheRafters schrieb am Donnerstag, 14. Oktober 2021 um 14:14:56 UTC+2:
> WM presented the following explanation :

> > Why that? Magic of matheology? In mathematics the intersection of infinite
> > endegments is infinite.
> >
> > Look, if you take the infinite set of infinite endsegments and try to find
> > any infinite endsegment that has less than infinitely many natural numbers in
> > common with any infinite endsegment or any set of infinite endsegments or all
> > predecessors or all successors or all infinite endsegments, then you will
> > fail.
> Paraphrased:
>
> Look, if you take the -infinite- set of -infinite- endsegments and try
> to find any -infinite- endsegment that has less than infinitely many
> natural numbers in common with any -infinite- endsegment or any set of
> -infinite- endsegments or all predecessors or all successors or all
> -infinite- endsegments, then you will fail.
>
> Even cleared of obsessively superfluous infinities

It is not superfluous. Every infinite endsegment contains natural numbers. The smallest one is the index of the endsegment. Endsegments with finite index and infinite contents cannot be empty. Such endegments cannot fail to have a last index defining the intersection. But this last index is not definable. Nevertheless it exists. Have you ever seen an infinite endsegment without an index?

Regards, WM

On Friday, 15 October 2021 at 15:29:18 UTC-3, WM wrote:
> Gus Gassmann schrieb am Donnerstag, 14. Oktober 2021 um 12:17:53 UTC+2:
> > On Thursday, 14 October 2021 at 06:49:59 UTC-3, WM wrote:
> > > Gus Gassmann schrieb am Mittwoch, 13. Oktober 2021 um 19:08:15 UTC+2:
> > >
> > > > 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,
> > > Nevertheless there are only natural numbers in the set of indices. As long as only endsegments containing natural numbers and having natural indices are intersected, the intersection is infinite.
> > No, it's not. Intersections over end segments can only contain natural numbers at least as large as the largest "index". In an infinite collection there is no largest index,
> Wrong. It is not definable. But since there are infinitely many natural numbers in every infinite endsegment, each one has a finite natnumber as its index. It is simply not a fixed largest index recognizable.

Meaningless gibberish. *EVERY* natural number is definable, since *EVERY* natural number other than 0 (or 1 if you prefer) is the (unique and direct) successor of another natural number, and so on, all the way back to 0.

After serious thinking WM wrote :
> FromTheRafters schrieb am Donnerstag, 14. Oktober 2021 um 14:14:56 UTC+2:
>> WM presented the following explanation :
>
>>> Why that? Magic of matheology? In mathematics the intersection of infinite
>>> endegments is infinite.
>>>
>>> Look, if you take the infinite set of infinite endsegments and try to find
>>> any infinite endsegment that has less than infinitely many natural numbers
>>> in common with any infinite endsegment or any set of infinite endsegments
>>> or all predecessors or all successors or all infinite endsegments, then
>>> you will fail.
>> Paraphrased:
>>
>> Look, if you take the -infinite- set of -infinite- endsegments and try
>> to find any -infinite- endsegment that has less than infinitely many
>> natural numbers in common with any -infinite- endsegment or any set of
>> -infinite- endsegments or all predecessors or all successors or all
>> -infinite- endsegments, then you will fail.
>>
>> Even cleared of obsessively superfluous infinities
>
> It is not superfluous. Every infinite endsegment

There you go again. Your endsegments are defined as infinite, so the
word "infinite" before the word "endsegment" is superfluous.

You do it in the hopes that it will confuse others about the difference
between the infinite collection of endsegments as a set, and the set
size of the individual endsegments.

Stop lying.