On Wednesday, December 15, 2021 at 6:55:18 AM UTC-4, WM wrote:
> William schrieb am Dienstag, 14. Dezember 2021 um 15:49:09 UTC+1:
> > On Tuesday, December 14, 2021 at 5:28:43 AM UTC-4, WM wrote:
> > > William schrieb am Montag, 13. Dezember 2021 um 16:27:06 UTC+1:
> >
> > > > Each endsegment has infinite content There is no one set that works for
> > > > *all*l endsegments
> > > Of course the [elements of any single set are] dark.
> >
> > And thus not natural numbers. The Peano set of natural numbers has no "dark" elements. Whatever the elements of this putative set are they are not natural numbers.
> Infinite sets can be exhausted

Whatever you mean by the term "exhausted", it still remains true that the elements of the putative set common to all endsegments must be "dark" and the set of natural numbers does not not contain "dark" elements. (The set of natural numbers is a Peano set and a Peano set does not contain "dark" elements. It does contain elements that cannot be written down.)

--
William Hughes

On 12/15/2021 5:00 AM, WM wrote:
> zelos...@gmail.com schrieb am Mittwoch, 15. Dezember 2021 um 06:11:58 UTC+1:
>> tisdag 14 december 2021 kl. 11:00:04 UTC+1 skrev WM:
>
>>
>>> Infinite sets like FISON and endsegments cannot differ by more than finite subsets.
>> FISONs are finite sets
>>
>> infinite sets can differ by infinite subsets and have nothing in comon.
>
> That is not possible for segments in the natural order of ℕ. As long as ℵo elements are following, the preceding set is finite.
>
>
>>> The intersection of endsegments is ℕ\L where L ist the set of numbers lost by n ∉ E(n+1).
>> L=N
>
> AS long as the endsegments are infinite, they have not lost all natnumbers.

Endsegments do not lose elements! They are fixed sets.

>
> Your claim that infinite endsegments have lost all natural numbers shows that you are a fool of matheology.
>
>>> C is the collection of all endsegments which have infinite intersection
>> Intersection WITH WHAT!?
>
> The intersection of all infinite endsegments is infinite

Wrong, because n ∉ E(n+1), therefore there is no n in the intersection of all endsegments.

your L is diversion, red herring.

> because L is finite as long as only infinite endsegments are concerned.
>
> Regards, WM

William schrieb am Mittwoch, 15. Dezember 2021 um 15:47:19 UTC+1:
> On Wednesday, December 15, 2021 at 6:55:18 AM UTC-4, WM wrote:
> > William schrieb am Dienstag, 14. Dezember 2021 um 15:49:09 UTC+1:
> > > On Tuesday, December 14, 2021 at 5:28:43 AM UTC-4, WM wrote:
> > > > William schrieb am Montag, 13. Dezember 2021 um 16:27:06 UTC+1:
> > >
> > > > > Each endsegment has infinite content There is no one set that works for
> > > > > *all*l endsegments
> > > > Of course the [elements of any single set are] dark.
> > >
> > > And thus not natural numbers. The Peano set of natural numbers has no "dark" elements. Whatever the elements of this putative set are they are not natural numbers.
> > Infinite sets can be exhausted
> Whatever you mean by the term "exhausted"

For instance, infinite sets can be completely in a bijection with infinite sets.

> , it still remains true that the elements of the putative set common to all endsegments must be "dark" and the set of natural numbers does not not contain "dark" elements.

But what are the elements of infinite endsegments?

> (The set of natural numbers is a Peano set and a Peano set does not contain "dark" elements. It does contain elements that cannot be written down.)

Do you claim that all endsegments are infinite? Do you claim that infinite endegments do not contain natural numbers? Can you agree that the intersection of endsegments is ℕ\L where L ist the set of numbers lost by n ∉ E(n+1).

Regards, WM

On 12/15/2021 12:00 PM, WM wrote:
> William schrieb am Mittwoch, 15. Dezember 2021 um 15:47:19 UTC+1:
>> On Wednesday, December 15, 2021 at 6:55:18 AM UTC-4, WM wrote:
>>> William schrieb am Dienstag, 14. Dezember 2021 um 15:49:09 UTC+1:
>>>> On Tuesday, December 14, 2021 at 5:28:43 AM UTC-4, WM wrote:
>>>>> William schrieb am Montag, 13. Dezember 2021 um 16:27:06 UTC+1:
>>>>
>>>>>> Each endsegment has infinite content There is no one set that works for
>>>>>> *all*l endsegments
>>>>> Of course the [elements of any single set are] dark.
>>>>
>>>> And thus not natural numbers. The Peano set of natural numbers has no "dark" elements. Whatever the elements of this putative set are they are not natural numbers.
>>> Infinite sets can be exhausted
>> Whatever you mean by the term "exhausted"
>
> For instance, infinite sets can be completely in a bijection with infinite sets.
>
>> , it still remains true that the elements of the putative set common to all endsegments must be "dark" and the set of natural numbers does not not contain "dark" elements.
>
> But what are the elements of infinite endsegments?
>
>> (The set of natural numbers is a Peano set and a Peano set does not contain "dark" elements. It does contain elements that cannot be written down.)
>
> Do you claim that all endsegments are infinite? Do you claim that infinite endegments do not contain natural numbers? Can you agree that the intersection of endsegments is ℕ\L where L ist the set of numbers lost by n ∉ E(n+1).
>
> Regards, WM

you can't wiggle out of this one, MrWMormie

1. endsegments are sets

2. Sets do not change

3. endsegments are infinite

4. endegments elements are natural numbers

5. the intersection of all endsegments is empty

6. the following are red herrings strewn about: dark numbers, L ist, lost, exhausted, none remain

On Wednesday, December 15, 2021 at 2:00:51 PM UTC-4, WM wrote:
> William schrieb am Mittwoch, 15. Dezember 2021 um 15:47:19 UTC+1:
>> ... it still remains true that the elements of the putative set common to all endsegments must be "dark" and the set of natural numbers does not not contain "dark" elements.
> But what are the elements of infinite endsegments?

All endsegments are infinite. Each consists of a different subset of elements of the natural numbers |N_P and nothing else.

> > (The set of natural numbers is a Peano set and a Peano set does not contain "dark" elements. It does contain elements that cannot be written down.)
> Do you claim that all endsegments are infinite? Do you claim that infinite endegments do not contain natural numbers? Can you agree that the intersection of endsegments is ℕ\L where L ist the set of numbers lost by n ∉ E(n+1).

L, the set of numbers lost by some natural number (not the same natural number for every element of L) is |N_P, the set of natural numbers. The intersection of endsegments is |N_P\L the empty set,

--
William Hughes

William schrieb am Mittwoch, 15. Dezember 2021 um 21:56:02 UTC+1:
> On Wednesday, December 15, 2021 at 2:00:51 PM UTC-4, WM wrote:
> > William schrieb am Mittwoch, 15. Dezember 2021 um 15:47:19 UTC+1:
> >> ... it still remains true that the elements of the putative set common to all endsegments must be "dark" and the set of natural numbers does not not contain "dark" elements.
> > But what are the elements of infinite endsegments?
> All endsegments are infinite. Each consists of a different

Is each subset remaining there from the first endsegment E(1)? Or have some natnumbers been newly acquired in the course of the sequence?

> subset of elements of the natural numbers |N_P and nothing else.

Each endsegment E(n) has not lost all natnumbers. Therefore up to each E(n) the intersection is infinite. If there are no others, then the intersection over all is infinite.

But each natnumber of |N_P has been lost.

> > > (The set of natural numbers is a Peano set and a Peano set does not contain "dark" elements. It does contain elements that cannot be written down.)
> > Do you claim that all endsegments are infinite? Do you claim that infinite endegments do not contain natural numbers? Can you agree that the intersection of endsegments is ℕ\L where L ist the set of numbers lost by n ∉ E(n+1).
> L, the set of numbers lost by some natural number (not the same natural number for every element of L) is |N_P, the set of natural numbers.

You claim that all endsegments are infinite. How can any infinite endsegment have lost all its contents?

Regards, WM

On Wednesday, December 15, 2021 at 6:17:36 PM UTC-4, WM wrote:

> How can any infinite endsegment have lost all its contents?

The endsegment has to "lose" an infinite number of elements.

The set of natural numbers does not contain "dark" elements. Hence the endsegments which are subsets of the natural numbers do not contain "dark" elements. Thus there is no element common to every endsegment.

--
William Hughes

On 12/15/2021 4:17 PM, WM wrote:
> William schrieb am Mittwoch, 15. Dezember 2021 um 21:56:02 UTC+1:
>> On Wednesday, December 15, 2021 at 2:00:51 PM UTC-4, WM wrote:
>>> William schrieb am Mittwoch, 15. Dezember 2021 um 15:47:19 UTC+1:
>>>> ... it still remains true that the elements of the putative set common to all endsegments must be "dark" and the set of natural numbers does not not contain "dark" elements.
>>> But what are the elements of infinite endsegments?
>> All endsegments are infinite. Each consists of a different
>
> Is each subset remaining there from the first endsegment E(1)? Or have some natnumbers been newly acquired in the course of the sequence?
>
>> subset of elements of the natural numbers |N_P and nothing else.
>
> Each endsegment E(n) has not lost all natnumbers. Therefore up to each E(n) the intersection is infinite. If there are no others, then the intersection over all is infinite.
>
> But each natnumber of |N_P has been lost.
>
>
>>>> (The set of natural numbers is a Peano set and a Peano set does not contain "dark" elements. It does contain elements that cannot be written down.)
>>> Do you claim that all endsegments are infinite? Do you claim that infinite endegments do not contain natural numbers? Can you agree that the intersection of endsegments is ℕ\L where L ist the set of numbers lost by n ∉ E(n+1).
>> L, the set of numbers lost by some natural number (not the same natural number for every element of L) is |N_P, the set of natural numbers.
>
> You claim that all endsegments are infinite. How can any infinite endsegment have lost all its contents?

the intersection of all endsegments is not an endsegment, and it is empty.

sets do not lose elements.

>
> Regards, WM

On 12/15/2021 3:06 AM, WM wrote:
> Jim Burns schrieb am Mittwoch, 15. Dezember 2021 um 00:41:43 UTC+1:
>> On 12/14/2021 4:53 AM, WM wrote:
>
>> Generalizing,
>> there really is no end of ∪{FISON'}
>
> But there is an end of the set of all natnumbers.
[...]

Wtf are you talking about! There is no end to the natural numbers!
Fucking moron!

onsdag 15 december 2021 kl. 12:01:01 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Mittwoch, 15. Dezember 2021 um 06:11:58 UTC+1:
> > tisdag 14 december 2021 kl. 11:00:04 UTC+1 skrev WM:
>
> >
> > >Infinite sets like FISON and endsegments cannot differ by more than finite subsets.
> > FISONs are finite sets
> >
> > infinite sets can differ by infinite subsets and have nothing in comon.
> That is not possible for segments in the natural order of ℕ. As long as ℵo elements are following, the preceding set is finite.
> > >The intersection of endsegments is ℕ\L where L ist the set of numbers lost by n ∉ E(n+1).
> > L=N
> AS long as the endsegments are infinite, they have not lost all natnumbers.
>
> Your claim that infinite endsegments have lost all natural numbers shows that you are a fool of matheology.
> > >C is the collection of all endsegments which have infinite intersection
> > Intersection WITH WHAT!?
> The intersection of all infinite endsegments is infinite because L is finite as long as only infinite endsegments are concerned.
>
> Regards, WM

>That is not possible for segments in the natural order of ℕ. As long as ℵo elements are following, the preceding set is finite.

You said sets, generic, not segments of N, specific, :) Different kind of things!

>AS long as the endsegments are infinite, they have not lost all natnumbers..

There is no "lost", but for any all n, there is always an endsegment not containing n, ergo L=N

>Your claim that infinite endsegments have lost all natural numbers shows that you are a fool of matheology.

I have made no such claim, I do basic mathematics which you are too stupid to understand.

>The intersection of all infinite endsegments is infinite because L is finite as long as only infinite endsegments are concerned.

False, L=N, so not finite.

On Wednesday, December 15, 2021 at 9:20:35 PM UTC-8, zelos...@gmail.com wrote:
> onsdag 15 december 2021 kl. 12:01:01 UTC+1 skrev WM:
> > zelos...@gmail.com schrieb am Mittwoch, 15. Dezember 2021 um 06:11:58 UTC+1:
> > > tisdag 14 december 2021 kl. 11:00:04 UTC+1 skrev WM:
> >
> > >
> > > >Infinite sets like FISON and endsegments cannot differ by more than finite subsets.
> > > FISONs are finite sets
> > >
> > > infinite sets can differ by infinite subsets and have nothing in comon.
> > That is not possible for segments in the natural order of ℕ. As long as ℵo elements are following, the preceding set is finite.
> > > >The intersection of endsegments is ℕ\L where L ist the set of numbers lost by n ∉ E(n+1).
> > > L=N
> > AS long as the endsegments are infinite, they have not lost all natnumbers.
> >
> > Your claim that infinite endsegments have lost all natural numbers shows that you are a fool of matheology.
> > > >C is the collection of all endsegments which have infinite intersection
> > > Intersection WITH WHAT!?
> > The intersection of all infinite endsegments is infinite because L is finite as long as only infinite endsegments are concerned.
> >
> > Regards, WM
>
> >That is not possible for segments in the natural order of ℕ. As long as ℵo elements are following, the preceding set is finite.
> You said sets, generic, not segments of N, specific, :) Different kind of things!
> >AS long as the endsegments are infinite, they have not lost all natnumbers.
> There is no "lost", but for any all n, there is always an endsegment not containing n, ergo L=N
> >Your claim that infinite endsegments have lost all natural numbers shows that you are a fool of matheology.
> I have made no such claim, I do basic mathematics which you are too stupid to understand.
> >The intersection of all infinite endsegments is infinite because L is finite as long as only infinite endsegments are concerned.
> False, L=N, so not finite.

BABBALANJA--My lord Media, there are things infinite in the finite;
and dualities in unities. Our eyes are pleased with the redness of the
rose, but another sense lives upon its fragrance. Its redness you must
approach, to view: its invisible fragrance pervades the field. So,
with the Koztanza. Its mere beauty is restricted to its form: its
expanding soul, past Mardi does embalm." -- Melville

William schrieb am Donnerstag, 16. Dezember 2021 um 00:08:04 UTC+1:
> On Wednesday, December 15, 2021 at 6:17:36 PM UTC-4, WM wrote:
>
> > How can any infinite endsegment have lost all its contents?
> The endsegment has to "lose" an infinite number of elements.

But it can't as long as it remains infinite. The natnumbers are 1, 2, 3, ...., n, n+1, ... . As long as there is an infinite rest n+1, ..., the lost part is finite.
>
> The set of natural numbers does not contain "dark" elements. Hence the endsegments which are subsets of the natural numbers do not contain "dark" elements. Thus there is no element common to every endsegment.

But there are infinitely many natnumbers common to all infinite endsegments.. Otherwise there must be a first infinite endsegment that has less than infinitely many numbers kept from all its predecessors. It must have acquired them in the course of the sequence.

This idea is so silly that it is hard to formulate it, let alone to believe it.

Regards, WM

zelos...@gmail.com schrieb am Donnerstag, 16. Dezember 2021 um 06:20:35 UTC+1:
> onsdag 15 december 2021 kl. 12:01:01 UTC+1 skrev WM:

> >The intersection of all infinite endsegments is infinite because L is finite as long as only infinite endsegments are concerned.
> False, L=N, so not finite.

For all infinite endsegments L =/= ℕ. Contradiction.

Regards, WM

On 12/16/2021 9:43 AM, WM wrote:
> William schrieb am Donnerstag, 16. Dezember 2021 um 00:08:04 UTC+1:
>> On Wednesday, December 15, 2021 at 6:17:36 PM UTC-4, WM wrote:
>>
>>> How can any infinite endsegment have lost all its contents?
>> The endsegment has to "lose" an infinite number of elements.
>
> But it can't as long as it remains infinite. The natnumbers are 1, 2, 3, ..., n, n+1, ... . As long as there is an infinite rest n+1, ..., the lost part is finite.

no. It remains infinite. Sets are not variables, they do not change elements. that is what operators do.

>>
>> The set of natural numbers does not contain "dark" elements. Hence the endsegments which are subsets of the natural numbers do not contain "dark" elements. Thus there is no element common to every endsegment.
>
> But there are infinitely many natnumbers common to all infinite endsegments.

so what ? they are defined that way. your "But" is misleading.

> Otherwise there must be a first infinite endsegment

called E(1)

> that has less than infinitely many numbers kept from all its predecessors.

nope.

> It must have acquired them in the course of the sequence.

wrong, sets are not variables, they are fixed.

>
> This idea is so silly that it is hard to formulate it, let alone to believe it.

its called WMs Math

>
> Regards, WM

On 12/16/2021 9:50 AM, WM wrote:
> zelos...@gmail.com schrieb am Donnerstag, 16. Dezember 2021 um 06:20:35 UTC+1:
>> onsdag 15 december 2021 kl. 12:01:01 UTC+1 skrev WM:

(below got snipped by WM)

You said sets, generic, not segments of N, specific, :) Different kind of things!

>AS long as the endsegments are infinite, they have not lost all natnumbers.

There is no "lost", but for any all n, there is always an endsegment not containing n, ergo L=N

>Your claim that infinite endsegments have lost all natural numbers shows that you are a fool of matheology.

I have made no such claim, I do basic mathematics which you are too stupid to understand.

>The intersection of all infinite endsegments is infinite because L is finite as long as only infinite endsegments are concerned.

False, L=N, so not finite

>
>>> The intersection of all infinite endsegments is infinite because L is finite as long as only infinite endsegments are concerned.
>> False, L=N, so not finite.
>
> For all infinite endsegments L =/= ℕ. Contradiction.
>
> Regards, WM

On Thursday, December 16, 2021 at 11:43:39 AM UTC-4, WM wrote:
> William schrieb am Donnerstag, 16. Dezember 2021 um 00:08:04 UTC+1:
> > On Wednesday, December 15, 2021 at 6:17:36 PM UTC-4, WM wrote:
> >
> > > How can any infinite endsegment have lost all its contents?
> > The endsegment has to "lose" an infinite number of elements.
> But it can't as long

There is no "as long". There is just one "step", an operator applied to the set
(E(1), E(2). E(3) ...) . You cannot analyse this problem by looking at the state of a process at finite steps (even if you consider an infinite number of finite steps.) You can show the operator results in the empty set by noting that any element common to all endsegments must be "dark" and no endsegment contains a "dark" element.
--
William Hughes

torsdag 16 december 2021 kl. 16:50:16 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Donnerstag, 16. Dezember 2021 um 06:20:35 UTC+1:
> > onsdag 15 december 2021 kl. 12:01:01 UTC+1 skrev WM:
>
> > >The intersection of all infinite endsegments is infinite because L is finite as long as only infinite endsegments are concerned.
> > False, L=N, so not finite.
> For all infinite endsegments L =/= ℕ. Contradiction.
>
> Regards, WM

False, you said L = {n e N: n+1 /e E(n)}

then L=N

zelos...@gmail.com schrieb am Freitag, 17. Dezember 2021 um 06:17:24 UTC+1:
> torsdag 16 december 2021 kl. 16:50:16 UTC+1 skrev WM:
> > zelos...@gmail.com schrieb am Donnerstag, 16. Dezember 2021 um 06:20:35 UTC+1:
> > > onsdag 15 december 2021 kl. 12:01:01 UTC+1 skrev WM:
> >
> > > >The intersection of all infinite endsegments is infinite because L is finite as long as only infinite endsegments are concerned.
> > > False, L=N, so not finite.
> > For all infinite endsegments L =/= ℕ. Contradiction.
> >
> False, you said L = {n e N: n+1 /e E(n)}
>
> then L=N

But you said that all endsegments are infinite. What is the contents of the infinite endsegments which have lost all natural numbers in ℕ?

Regards, WM

William schrieb am Donnerstag, 16. Dezember 2021 um 18:25:21 UTC+1:
> On Thursday, December 16, 2021 at 11:43:39 AM UTC-4, WM wrote:
> > William schrieb am Donnerstag, 16. Dezember 2021 um 00:08:04 UTC+1:
> > > On Wednesday, December 15, 2021 at 6:17:36 PM UTC-4, WM wrote:
> > >
> > > > How can any infinite endsegment have lost all its contents?
> > > The endsegment has to "lose" an infinite number of elements.
> > But it can't as long
> There is no "as long".

For all definable numbers of the sequence ℕ there is an "as long".

> There is just one "step", an operator applied to the set
> (E(1), E(2). E(3) ...) . You cannot analyse this problem by looking at the state of a process at finite steps (even if you consider an infinite number of finite steps.)

That is possible however for all endsegments E(n) with definable indexes n. Your claim shows that not all natural nunmbers are definable.

> You can show the operator results in the empty set by noting that any element common to all endsegments must be "dark" and no endsegment contains a "dark" element.

What is the contents of all infinite endsegments which do not contain natural numbers?

Regards, WM

zelos...@gmail.com schrieb am Freitag, 17. Dezember 2021 um 06:17:24 UTC+1:
> torsdag 16 december 2021 kl. 16:50:16 UTC+1 skrev WM:
> > zelos...@gmail.com schrieb am Donnerstag, 16. Dezember 2021 um 06:20:35 UTC+1:
> > > onsdag 15 december 2021 kl. 12:01:01 UTC+1 skrev WM:
> >
> > > >The intersection of all infinite endsegments is infinite because L is finite as long as only infinite endsegments are concerned.
> > > False, L=N, so not finite.
> > For all infinite endsegments L =/= ℕ. Contradiction.
> >
> > Regards, WM
> False, you said L = {n e N: n+1 /e E(n)}

No, I said L = {n ∈ ℕ | "n ∉ E(n+1)}
>
> then L=N

But you said that all endsegments are infinite. What is the contents of the infinite endsegments which have lost all natural numbers in ℕ?

Regards, WM

WM formulated on Friday :
> William schrieb am Donnerstag, 16. Dezember 2021 um 18:25:21 UTC+1:
>> On Thursday, December 16, 2021 at 11:43:39 AM UTC-4, WM wrote:
>>> William schrieb am Donnerstag, 16. Dezember 2021 um 00:08:04 UTC+1:
>>>> On Wednesday, December 15, 2021 at 6:17:36 PM UTC-4, WM wrote:
>>>>
>>>>> How can any infinite endsegment have lost all its contents?
>>>> The endsegment has to "lose" an infinite number of elements.
>>> But it can't as long
>> There is no "as long".
>
> For all definable numbers of the sequence ℕ there is an "as long".
>
>> There is just one "step", an operator applied to the set
>> (E(1), E(2). E(3) ...) . You cannot analyse this problem by looking at the
>> state of a process at finite steps (even if you consider an infinite number
>> of finite steps.)
>
> That is possible however for all endsegments E(n) with definable indexes n.
> Your claim shows that not all natural nunmbers are definable.
>
>> You can show the operator results in the empty set by noting that any
>> element common to all endsegments must be "dark" and no endsegment contains
>> a "dark" element.
>
> What is the contents of all infinite endsegments which do not contain natural
> numbers?

You have not yet defined "all -infinite- endsegments which do not
contain natural numbers" in Muckymath so go ahead and do so. It appears
you will also need to explain what "all empty sets" might mean in your
system.

fredag 17 december 2021 kl. 11:06:25 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Freitag, 17. Dezember 2021 um 06:17:24 UTC+1:
> > torsdag 16 december 2021 kl. 16:50:16 UTC+1 skrev WM:
> > > zelos...@gmail.com schrieb am Donnerstag, 16. Dezember 2021 um 06:20:35 UTC+1:
> > > > onsdag 15 december 2021 kl. 12:01:01 UTC+1 skrev WM:
> > >
> > > > >The intersection of all infinite endsegments is infinite because L is finite as long as only infinite endsegments are concerned.
> > > > False, L=N, so not finite.
> > > For all infinite endsegments L =/= ℕ. Contradiction.
> > >
> > > Regards, WM
> > False, you said L = {n e N: n+1 /e E(n)}
> No, I said L = {n ∈ ℕ | "n ∉ E(n+1)}
> >
> > then L=N
>
> But you said that all endsegments are infinite. What is the contents of the infinite endsegments which have lost all natural numbers in ℕ?
>
> Regards, WM

>No, I said L = {n ∈ ℕ | "n ∉ E(n+1)}

My apologies, I typoed severely

But with that, congrats, you have L=N :)

because n ∉ E(n+1) is true for ALL n

>But you said that all endsegments are infinite. What is the contents of the infinite endsegments which have lost all natural numbers in ℕ?

Who says any endsegment is empty?

On 12/17/2021 4:02 AM, WM wrote:
> William schrieb am Donnerstag, 16. Dezember 2021 um 18:25:21 UTC+1:
>> On Thursday, December 16, 2021 at 11:43:39 AM UTC-4, WM wrote:
>>> William schrieb am Donnerstag, 16. Dezember 2021 um 00:08:04 UTC+1:
>>>> On Wednesday, December 15, 2021 at 6:17:36 PM UTC-4, WM wrote:
>>>>
>>>>> How can any infinite endsegment have lost all its contents?
>>>> The endsegment has to "lose" an infinite number of elements.
>>> But it can't as long
>> There is no "as long".
>
> For all definable numbers of the sequence ℕ there is an "as long".

"as long" is a conditional operator, should be "as long as what"

so, where/what is the WHAT ?

>
>> There is just one "step", an operator applied to the set
>> (E(1), E(2). E(3) ...) . You cannot analyse this problem by looking at the state of a process at finite steps (even if you consider an infinite number of finite steps.)
>
> That is possible however for all endsegments E(n) with definable indexes n. Your claim shows that not all natural nunmbers are definable.
>
>> You can show the operator results in the empty set by noting that any element common to all endsegments must be "dark" and no endsegment contains a "dark" element.
>
> What is the contents of all infinite endsegments which do not contain natural numbers?

really?

[ What is the contents of a container that does not have contents ?]

>
> Regards, WM

On 12/17/2021 3:57 AM, WM wrote:
> zelos...@gmail.com schrieb am Freitag, 17. Dezember 2021 um 06:17:24 UTC+1:
>> torsdag 16 december 2021 kl. 16:50:16 UTC+1 skrev WM:
>>> zelos...@gmail.com schrieb am Donnerstag, 16. Dezember 2021 um 06:20:35 UTC+1:
>>>> onsdag 15 december 2021 kl. 12:01:01 UTC+1 skrev WM:
>>>
>>>>> The intersection of all infinite endsegments is infinite because L is finite as long as only infinite endsegments are concerned.
>>>> False, L=N, so not finite.
>>> For all infinite endsegments L =/= ℕ. Contradiction.
>>>
>> False, you said L = {n e N: n+1 /e E(n)}
>>
>> then L=N
>
> But you said that all endsegments are infinite. What is the contents of the infinite endsegments which have lost all natural numbers in ℕ?
>
> Regards, WM

know what a "definition" is ?

endsegments are defined as infinite fixed sets.

they are not variable sets.

They do not lose elements.

your statement is invalid, wrong.

On Friday, December 17, 2021 at 6:02:54 AM UTC-4, WM wrote:
> William schrieb am Donnerstag, 16. Dezember 2021 um 18:25:21 UTC+1:

> > There is just one "step", an operator applied to the set
> > (E(1), E(2). E(3) ...) . You cannot analyse this problem by looking at the state of a process at finite steps (even if you consider an infinite number of finite steps.)
> That is possible

No it is not possible, The state "at infinity" may not have the same properties as the state at finite steps.

William Hughes