On Wednesday, November 24, 2021 at 10:36:11 AM UTC+1, WM wrote:

> [...] every set of [...] endsegments has a last element

Oh boy!

WM laid this down on his screen :
> FromTheRafters schrieb am Mittwoch, 24. November 2021 um 10:28:11 UTC+1:
>> WM was thinking very hard :
>>> zelos...@gmail.com schrieb am Dienstag, 23. November 2021 um 10:55:20
>>> UTC+1:
>>>> tisdag 23 november 2021 kl. 10:32:18 UTC+1 skrev WM:
>>>>> Yes, but all infinite endsegments have infinitely many numbers in common.
>>>> All infinitely many endsegments have NO elements in common with ALL
>>>> endsegments, that is why it is EMPTY
>>>
>>> The set of infinite endsegments has an infinite intersection. Inclusion
>>> monotony.
>> "The set of [Edit: -infinite-] endsegments has an infinite
>> intersection. Inclusion monotony."
>>
>> this only works if this set has a last element.
>
> Why should that be a condition?

Seriously? In your FISON removing stepwise process each and every
previously 'removed' FISON 'removes' another 'not last' element from
consideration in an intersection. This 'process' does not stop.

[...]

> Of course every set of infinite endsegments has a last element,

Every *finite* set of -infinite- endsegments has a greatest or last
element in this order. Intersections require matching elements not just
counting cardinality of sets. Just because a set has aleph_zero
cardinality doesn't mean it has a "37" element in it.

onsdag 24 november 2021 kl. 10:25:33 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Mittwoch, 24. November 2021 um 09:59:25 UTC+1:
> > onsdag 24 november 2021 kl. 09:50:59 UTC+1 skrev WM:
> > > zelos...@gmail.com schrieb am Dienstag, 23. November 2021 um 10:55:20 UTC+1:
> > > > tisdag 23 november 2021 kl. 10:32:18 UTC+1 skrev WM:
> > >
> > > > >Yes, but all infinite endsegments have infinitely many numbers in common.
> > > > All infinitely many endsegments have NO elements in common with ALL endsegments, that is why it is EMPTY
> > > The set of infinite endsegments has an infinite intersection. Inclusion monotony.
> > > >
> > > > EACH endsegment has it with ANY OTHER endsegment but for all, it is not true.
> > > Of course it is true then for all. How could you contradict this? You argue that every natnumbers n is not contained in E(n+1). But als long as E(n+1) contains infinitely many elements, this argument is invalid for infinitely many elements. It is invalid for many, many more elements than it is valid for.
> > >
> > > Regards, WM
> >
> >
> > >The set of infinite endsegments has an infinite intersection. Inclusion monotony.
> > False and an empty assertion.
> >
> > The intersection is EMPTY
> > >Of course it is true then for all. How could you contradict this?
> > It is NOT true and I have PROVEN it.
> > >You argue that every natnumbers n is not contained in E(n+1).
> > Which is a definitional truth
> > >But als long as E(n+1) contains infinitely many elements
> > It has no effect on the fact that n is NOT in E(n+1)
> > >this argument is invalid for infinitely many elements. It is invalid for many, many more elements than it is valid for.
> > False, it is valid for ALL of them and the cardinality of E(m) is irrelevant.
> As long as the endsegments are infinite, infinitely many numbers are not lost. If all endsegments are infinite, infinitely many numbers are never lost.
>
> Regards, WM
none is ever lost. and no element is in ALL endsegments which is why the intersection is empty!

That is ALL THAT MATTERS!

onsdag 24 november 2021 kl. 10:36:11 UTC+1 skrev WM:
> FromTheRafters schrieb am Mittwoch, 24. November 2021 um 10:28:11 UTC+1:
> > WM was thinking very hard :
> > > zelos...@gmail.com schrieb am Dienstag, 23. November 2021 um 10:55:20 UTC+1:
> > >> tisdag 23 november 2021 kl. 10:32:18 UTC+1 skrev WM:
> > >
> > >>> Yes, but all infinite endsegments have infinitely many numbers in common.
> > >> All infinitely many endsegments have NO elements in common with ALL
> > >> endsegments, that is why it is EMPTY
> > >
> > > The set of infinite endsegments has an infinite intersection. Inclusion
> > > monotony.
> > "The set of [Edit: -infinite-] endsegments has an infinite
> > intersection. Inclusion monotony."
> >
> > this only works if this set has a last element.
> Why should that be a condition?
> >
> Such a last
> > element would be in every endsegment, and there would be infinitely
> > many of them.
> Of course every set of infinite endsegments has a last element, namely an index which infinitely many endsegments are following upon. This index belongs to a finite initial segment of natural numbers because it is followed by infinitely many natnumbers.
>
> It works as long as all endsegments are infinite. By separation we can collect all endsegments with
> ∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀
> into a set. Then no additional endsegment is in the set. All endsegments have infinite itersection.
> |∩{E(k) : k ∈ ℕ}| = ℵ₀ .
> Note that the intersection does not depend on the order of the endsegments.
>
> Regards, WM

False and a complete non-sequitor

On 11/24/2021 2:46 AM, WM wrote:
> William schrieb am Dienstag, 23. November 2021 um 18:18:46 UTC+1:
>> On Tuesday, November 23, 2021 at 4:29:51 AM UTC-5, WM wrote:
>> "Between each element, n, of the Peano set of natural numbers there is a set M(n) of cardinality aleph_o between n and omega (note M(n) can change).

<snip crap>

On 11/24/2021 2:50 AM, WM wrote:
> zelos...@gmail.com schrieb am Dienstag, 23. November 2021 um 10:55:20 UTC+1:
>> tisdag 23 november 2021 kl. 10:32:18 UTC+1 skrev WM:
>
>>> Yes, but all infinite endsegments have infinitely many numbers in common.
>> All infinitely many endsegments have NO elements in common with ALL endsegments, that is why it is EMPTY
>
> The set of infinite endsegments has an infinite intersection. Inclusion monotony.

Fail. there are no elements in that set.

>>
>> EACH endsegment has it with ANY OTHER endsegment but for all, it is not true.
>
> Of course it is true then for all.

wrong, intersection is not an endsegment.

> How could you contradict this?

using simple math. You do not understand simple math.

> You argue that every natnumbers n is not contained in E(n+1). But als long as E(n+1) contains infinitely many elements, this argument is invalid for infinitely many elements.

wrong again. E(n+1) contains infinitely many elements, but not any below n+1

> It is invalid for many, many more elements than it is valid for.

wrong.

>
> Regards, WM
>

you fail to understand the math of infinity.

On Wednesday, November 24, 2021 at 3:46:36 AM UTC-5, WM wrote:
it has no largest element. But every element has almost all ℵo natnumbers beyond itself. Rhis distance cannot be bridged by Peano elements.
> > The Peano set of natural numbers does not contain "dark" elements.
> It is not actually infinite.

Piffle, A Peano set has cardinality aleph_0

--
William Hughes

William schrieb am Mittwoch, 24. November 2021 um 17:47:22 UTC+1:
> On Wednesday, November 24, 2021 at 3:46:36 AM UTC-5, WM wrote:
> it has no largest element. But every element has almost all ℵo natnumbers beyond itself. This distance cannot be bridged by Peano elements.
> > > The Peano set of natural numbers does not contain "dark" elements.
> > It is not actually infinite.
> Piffle, A Peano set has cardinality aleph_0
>
Here you use ℵo to denote a potentially infinite set. Cantor has desinated ℵo to denote actually infinite sets where every element has ℵo successors which can be exhausted collectively in mappings or Cantor lists but cannot be exhausted individually because all individuals defined by Peano cannot and do not exhaust the set.

You understand that there is a difference between exhausting and not exhausting?

Regards, WM

Fritz Feldhase schrieb am Mittwoch, 24. November 2021 um 10:48:32 UTC+1:
> On Wednesday, November 24, 2021 at 9:50:59 AM UTC+1, WM wrote:
> > zelos...@gmail.com schrieb am Dienstag, 23. November 2021 um 10:55:20 UTC+1:
> > > EACH and EVERY endsegment has [an infinite intersection] with ANY OTHER endsegment but for all, it is not true.
> > >
> > Of course it is true then for all.
> No, it isn't.
> > How could you contradict this?
> By a so called MATHEMATICAL PROOF, you know.

Mathematical proof shows by inclusion monotony: All infinite endsegments have an infinite contents in common. If your "proof" shows the contrary, it is based upon a wrong theory. The contents of any endsegment is remaining there from the first one E(1) = |N. As long as there is contents, the set of non-empty endsegments has a non-empty intersection. You know? Or are you a matheologian? They believe that there is no infinite set dark numbers between every natnumber and ω. But they cannot find any addressable number which is closer to ω. Not to be taken seriously.

Regards, WM

On Thursday, November 25, 2021 at 10:03:58 AM UTC+1, WM wrote:
> William schrieb am Mittwoch, 24. November 2021 um 17:47:22 UTC+1:
> >
> > A Peano set has cardinality aleph_0
> >
> Here you use ℵo to denote a potentially infinite set.

HEILIGE SCHEISSE, Mückenheim!

ES GIBT KEINE "potentiell unendlichen Mengen"!

THERE ARE NO "potentially infinite sets".

==> IN is INFINITE and has cardinality aleph.

FromTheRafters schrieb am Mittwoch, 24. November 2021 um 11:01:57 UTC+1:
> WM laid this down on his screen :
> > FromTheRafters schrieb am Mittwoch, 24. November 2021 um 10:28:11 UTC+1:
> >> WM was thinking very hard :
> >>> zelos...@gmail.com schrieb am Dienstag, 23. November 2021 um 10:55:20
> >>> UTC+1:
> >>>> tisdag 23 november 2021 kl. 10:32:18 UTC+1 skrev WM:

> >>> The set of infinite endsegments has an infinite intersection. Inclusion
> >>> monotony.
> >> this only works if this set has a last element.
> >
> > Why should that be a condition?
> Seriously? In your FISON removing stepwise process each and every
> previously 'removed' FISON 'removes' another 'not last' element from
> consideration in an intersection. This 'process' does not stop.

If it does not stop, then there is no result. But you claim that there is a result. All elements have to be removed. But not only from the intersection but also from the endsegments. As long as there are infinite endsegments the process has not a result.

> > Of course every set of infinite endsegments has a last element,
> Every *finite* set of -infinite- endsegments has a greatest or last
> element in this order.

As long as the contents is actually infinite, the number of endsegments cannot be actually infinite. There are no two actually infinite consecutive sets in |N. There is no chance to use natnumbers for both applications, indices and contents.

> Intersections require matching elements not just
> counting cardinality of sets.

The contents of any endsegment is remaining there from the first one E(1) = |N. As long as there is contents, the set of non-empty endsegments has a non-empty intersection. You know? Or are you a matheologian? They believe that there is no infinite set of dark numbers between every natnumber and ω. But they cannot find any addressable number which is closer to ω. Not to be taken seriously.

Regards, WM

On Thursday, November 25, 2021 at 10:39:25 AM UTC+1, WM wrote:
> Fritz Feldhase schrieb am Mittwoch, 24. November 2021 um 10:48:32 UTC+1:
> > >
> > > How could you contradict this?
> > >
> > By a so called MATHEMATICAL PROOF, you know.
> >
> Mathematical proof shows <bla bla bla>

Show _the proof_, you psychotic asshole full of shit - or shut up!

zelos...@gmail.com schrieb am Mittwoch, 24. November 2021 um 13:16:13 UTC+1:
> onsdag 24 november 2021 kl. 10:25:33 UTC+1 skrev WM:

> > As long as the endsegments are infinite, infinitely many numbers are not lost. If all endsegments are infinite, infinitely many numbers are never lost.
> >
> none is ever lost.

n is lost in E(n+1)

> and no element is in ALL endsegments

But infinitely many elements are in all infinite endsegments.

> which is why the intersection is empty!

The contents of any endsegment is remaining there from the first one E(1) = |N. As long as there is contents, the set of non-empty endsegments has a non-empty intersection. You know? Or are you a matheologian? They believe that there is no infinite set of dark numbers between every natnumber and ω. But they cannot find any addressable number which is closer to ω. Not to be taken seriously.

Regards, WM

Fritz Feldhase schrieb am Donnerstag, 25. November 2021 um 10:46:44 UTC+1:
> On Thursday, November 25, 2021 at 10:03:58 AM UTC+1, WM wrote:
> > William schrieb am Mittwoch, 24. November 2021 um 17:47:22 UTC+1:
> > >
> > > A Peano set has cardinality aleph_0
> > >
> > Here you use ℵo to denote a potentially infinite set.

> THERE ARE NO "potentially infinite sets".
>
The definable natural numbers are a potentially infinite set. Only dark numbers can complete the actually infinite set.

You believe that there is no infinite set of dark numbers between every natnumber and ω. But you cannot find any number which is closer to ω.. That is matheology, inconsistent nonsense.

Regards, WM

On Thursday, November 25, 2021 at 10:49:55 AM UTC+1, WM wrote:
> zelos...@gmail.com schrieb am Mittwoch, 24. November 2021 um 13:16:13 UTC+1:
> >
> > no element is in ALL endsegments
> >
> But infinitely many elements are in all [...] endsegments.

https://en.wikipedia.org/wiki/Quantifier_shift

On Thursday, November 25, 2021 at 10:54:10 AM UTC+1, WM wrote:
> Fritz Feldhase schrieb am Donnerstag, 25. November 2021 um 10:46:44 UTC+1:
> > On Thursday, November 25, 2021 at 10:03:58 AM UTC+1, WM wrote:
> > > William schrieb am Mittwoch, 24. November 2021 um 17:47:22 UTC+1:
> > > >
> > > > A Peano set has cardinality aleph_0
> > > >
> > > Here you use ℵo to denote a potentially infinite set.
> > >
> > THERE ARE NO "potentially infinite sets".
> >
> The definable natural numbers <bla bla bla>

NOCHMAL MÜCKENHEIM: ES GIBT KEINE "potentiell unendlichen Mengen"!

THERE ARE NO "potentially infinite sets".

==> IN is INFINITE and has cardinality aleph.

torsdag 25 november 2021 kl. 10:39:25 UTC+1 skrev WM:
> Fritz Feldhase schrieb am Mittwoch, 24. November 2021 um 10:48:32 UTC+1:
> > On Wednesday, November 24, 2021 at 9:50:59 AM UTC+1, WM wrote:
> > > zelos...@gmail.com schrieb am Dienstag, 23. November 2021 um 10:55:20 UTC+1:
> > > > EACH and EVERY endsegment has [an infinite intersection] with ANY OTHER endsegment but for all, it is not true.
> > > >
> > > Of course it is true then for all.
> > No, it isn't.
> > > How could you contradict this?
> > By a so called MATHEMATICAL PROOF, you know.
> Mathematical proof shows by inclusion monotony: All infinite endsegments have an infinite contents in common. If your "proof" shows the contrary, it is based upon a wrong theory. The contents of any endsegment is remaining there from the first one E(1) = |N. As long as there is contents, the set of non-empty endsegments has a non-empty intersection. You know? Or are you a matheologian? They believe that there is no infinite set dark numbers between every natnumber and ω. But they cannot find any addressable number which is closer to ω. Not to be taken seriously.
>
> Regards, WM

>Mathematical proof shows by inclusion monotony:

Shows not what you want

>All infinite endsegments have an infinite contents in common

They don't and is a non-sequitor.

>If your "proof" shows the contrary, it is based upon a wrong theory.

wrong in YOUR OPINION, in MATHEMATICS it is shown EASY you are wrong because you are commiting logical FALLACIES!

>The contents of any endsegment is remaining there from the first one E(1) = |N. As long as there is contents, the set of non-empty endsegments has a non-empty intersection.

False

>You know?

I do not know false things, only an idiot accept those.

>Or are you a matheologian?

I am a mathematician doing mathematics, unlike you. THere is no "Matheologian", there is however cranks and you're one!

>They believe that there is no infinite set dark numbers between every natnumber and ω.

You cannot define "dark" in a meaningful manner that doesn't render the set empty.

>But they cannot find any addressable number which is closer to ω. Not to be taken seriously.

"addressable" is not defined in mathematics.

torsdag 25 november 2021 kl. 10:49:55 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Mittwoch, 24. November 2021 um 13:16:13 UTC+1:
> > onsdag 24 november 2021 kl. 10:25:33 UTC+1 skrev WM:
>
> > > As long as the endsegments are infinite, infinitely many numbers are not lost. If all endsegments are infinite, infinitely many numbers are never lost.
> > >
> > none is ever lost.
>
> n is lost in E(n+1)
> > and no element is in ALL endsegments
> But infinitely many elements are in all infinite endsegments.
> > which is why the intersection is empty!
> The contents of any endsegment is remaining there from the first one E(1) = |N. As long as there is contents, the set of non-empty endsegments has a non-empty intersection. You know? Or are you a matheologian? They believe that there is no infinite set of dark numbers between every natnumber and ω. But they cannot find any addressable number which is closer to ω. Not to be taken seriously.
>
> Regards, WM

n is NOT IN E(n+1)

Nothing was lost.

>But infinitely many elements are in all infinite endsegments.

not that they all share!

WM brought next idea :
> FromTheRafters schrieb am Mittwoch, 24. November 2021 um 11:01:57 UTC+1:
>> WM laid this down on his screen :
>>> FromTheRafters schrieb am Mittwoch, 24. November 2021 um 10:28:11 UTC+1:
>>>> WM was thinking very hard :
>>>>> zelos...@gmail.com schrieb am Dienstag, 23. November 2021 um 10:55:20
>>>>> UTC+1:
>>>>>> tisdag 23 november 2021 kl. 10:32:18 UTC+1 skrev WM:
>
>>>>> The set of infinite endsegments has an infinite intersection. Inclusion
>>>>> monotony.
>>>> this only works if this set has a last element.
>>>
>>> Why should that be a condition?
>> Seriously? In your FISON removing stepwise process each and every
>> previously 'removed' FISON 'removes' another 'not last' element from
>> consideration in an intersection. This 'process' does not stop.
>
> If it does not stop, then there is no result. But you claim that there is a
> result.

Silly. I was attempting to speak your language while showing you your
error. That is why I said *your* step-by-step process. There is no
actual step-by-step process involved in the intersection.

WM explained on 11/25/2021 :
> FromTheRafters schrieb am Mittwoch, 24. November 2021 um 11:01:57 UTC+1:
>> WM laid this down on his screen :
>>> FromTheRafters schrieb am Mittwoch, 24. November 2021 um 10:28:11 UTC+1:
>>>> WM was thinking very hard :
>>>>> zelos...@gmail.com schrieb am Dienstag, 23. November 2021 um 10:55:20
>>>>> UTC+1:
>>>>>> tisdag 23 november 2021 kl. 10:32:18 UTC+1 skrev WM:
>
>>>>> The set of infinite endsegments has an infinite intersection. Inclusion
>>>>> monotony.
>>>> this only works if this set has a last element.
>>>
>>> Why should that be a condition?
>> Seriously? In your FISON removing stepwise process each and every
>> previously 'removed' FISON 'removes' another 'not last' element from
>> consideration in an intersection. This 'process' does not stop.
>
> If it does not stop, then there is no result. But you claim that there is a
> result. All elements have to be removed. But not only from the intersection
> but also from the endsegments. As long as there are infinite endsegments the
> process has not a result.
>
>>> Of course every set of infinite endsegments has a last element,
>> Every *finite* set of -infinite- endsegments has a greatest or last
>> element in this order.
>
> As long as the contents is actually infinite, the number of endsegments
> cannot be actually infinite. There are no two actually infinite consecutive
> sets in |N. There is no chance to use natnumbers for both applications,
> indices and contents.
>
>> Intersections require matching elements not just
>> counting cardinality of sets.
>
> The contents of any endsegment is remaining there from the first one E(1) =
> |N.

If E(1) is to be a first endsegment in your naturals then your E(n+1)
definition for endsegemnts suggests that your 'n' is zero and missing
from the endsegment when compared to the naturals with zero.

You keep switching from the naturals with zero to them without zero.
Can't you do *anything* right?

On Thursday, November 25, 2021 at 11:31:52 AM UTC+1, FromTheRafters asked WM:

> Can't you do *anything* right?

No, he can't.

On Thursday, 25 November 2021 at 05:49:55 UTC-4, WM wrote:
[...]
> But infinitely many elements are in all infinite endsegments.

Ambiguous nonsense, as usual. Shut the fuck up!

On Thursday, November 25, 2021 at 4:03:58 AM UTC-5, WM wrote:
> ...potentially infinite set.
Nonsense.

--
William Hughes

On Thursday, November 25, 2021 at 4:54:10 AM UTC-5, WM wrote:
> Fritz Feldhase schrieb am Donnerstag, 25. November 2021 um 10:46:44 UTC+1:

> > THERE ARE NO "potentially infinite sets".
> >
> The definable natural numbers are

the numbers you can write down and are a finite set. "Potentially infinite set"
nonsense.

--
William Hughes

On Thursday, November 25, 2021 at 1:54:10 AM UTC-8, WM wrote:
> Fritz Feldhase schrieb am Donnerstag, 25. November 2021 um 10:46:44 UTC+1:
> > On Thursday, November 25, 2021 at 10:03:58 AM UTC+1, WM wrote:
> > > William schrieb am Mittwoch, 24. November 2021 um 17:47:22 UTC+1:
> > > >
> > > > A Peano set has cardinality aleph_0
> > > >
> > > Here you use ℵo to denote a potentially infinite set.
> > THERE ARE NO "potentially infinite sets".
> >
> The definable natural numbers are a potentially infinite set.

Your 'definable natural numbers' are/is not a set, as you have elsewhere admitted.