FromTheRafters schrieb am Samstag, 20. November 2021 um 20:33:51 UTC+1:
> After serious thinking WM wrote :
> > Jim Burns schrieb am Freitag, 19. November 2021 um 22:56:40 UTC+1:
> >> On 11/19/2021 7:52 AM, WM wrote:
> >>
> >>> How can infinitely many endsegments contain infinitely
> >>> many natural numbers each?
> >> Something you (WM) haven't done yet is ask yourself
> >> what "infinite" or "finite" mean.
> >
> > Infinite here means ℵo, more than any natnumber n.
> 2^ℵo is also 'larger' than any natural number. Are you saying that ℵo
> and 2^ℵo are equal?

2^ℵo has not a place in mathematics. Of course there are more fractions than natnumbers. But I need not consider larger infinities than ℵo in case of endsegments.

Regards, WM

Fritz Feldhase schrieb am Samstag, 20. November 2021 um 21:51:52 UTC+1:
> On Friday, November 19, 2021 at 9:25:16 AM UTC+1, WM wrote:

> > By the axiom of separation [natural numbers satifying this predicate] can be collected in a set ℕ_def
>
> Indeed, we may define the set
>
> ℕ_def := {k e IN : ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀} .
>
> Now it's easy to show that

In order to get rid of the nonsense postulatd by set theory, and to see that it is nonsense, we define the set

|∩{E(k) : k ∈ ℕ_def}| = ℵ₀ .

Regards, WM

WM wrote :
> FromTheRafters schrieb am Samstag, 20. November 2021 um 20:33:51 UTC+1:
>> After serious thinking WM wrote :
>>> Jim Burns schrieb am Freitag, 19. November 2021 um 22:56:40 UTC+1:
>>>> On 11/19/2021 7:52 AM, WM wrote:
>>>>
>>>>> How can infinitely many endsegments contain infinitely
>>>>> many natural numbers each?
>>>> Something you (WM) haven't done yet is ask yourself
>>>> what "infinite" or "finite" mean.
>>>
>>> Infinite here means ℵo, more than any natnumber n.
>> 2^ℵo is also 'larger' than any natural number. Are you saying that ℵo
>> and 2^ℵo are equal?
>
> 2^ℵo has not a place in mathematics. Of course there are more fractions than
> natnumbers. But I need not consider larger infinities than ℵo in case of
> endsegments.

You deny the continuum and yet use its features in attempts to disprove
the the fact that the cardinality of the rational numbers is the same
as the cardinality of the naturals.

On Sunday, November 21, 2021 at 5:17:26 AM UTC-5, WM wrote:
> William schrieb am Samstag, 20. November 2021 um 20:16:37 UTC+1:
> > On Saturday, November 20, 2021 at 1:01:09 PM UTC-5, WM wrote:
>
> > > > > But if it exists on the ordinal line, then all natnumbers must be positioned before omega.
> > > > Correct
> > > That is an end.
> > Nope, omega is not an element of the Peano set of natural numbers. There is no last natural number.
> If there is no last natural number, then there is an empty space before omega.

Nope, 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).
There is no ordinal between the Peano set of natural numbers and omega.
--
William Hughes

On 11/21/2021 4:17 AM, WM wrote:
> William schrieb am Samstag, 20. November 2021 um 20:16:37 UTC+1:
>> On Saturday, November 20, 2021 at 1:01:09 PM UTC-5, WM wrote:
>
>>>>> But if it exists on the ordinal line, then all natnumbers must be positioned before omega.
>>>> Correct
>>> That is an end.
>> Nope, omega is not an element of the Peano set of natural numbers. There is no last natural number.
>
> If there is no last natural number, then there is an empty space before omega. Why?

Why should their be a "last" empty space before omega ?

> If there are all natural numbers, then either there is a last one or infinitely many cannot be distinguished; they are dark.

This is your personal misunderstanding of infinity, you keep repeating it, it is wrong.
you have yet to provide a proof of dark numbers, only your imagination. There is no "last" natural number.

> Regards, WM
>

William schrieb am Sonntag, 21. November 2021 um 14:55:03 UTC+1:
> On Sunday, November 21, 2021 at 5:17:26 AM UTC-5, WM wrote:
> > William schrieb am Samstag, 20. November 2021 um 20:16:37 UTC+1:
> > > On Saturday, November 20, 2021 at 1:01:09 PM UTC-5, WM wrote:
> >
> > > > > > But if it exists on the ordinal line, then all natnumbers must be positioned before omega.
> > > > > Correct
> > > > That is an end.
> > > Nope, omega is not an element of the Peano set of natural numbers. There is no last natural number.
> > If there is no last natural number, then there is an empty space before omega.
> Nope, 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).
> There is no ordinal between the Peano set of natural numbers and omega.
>
This proves that the set has more elements than those which have infinite distance M(n), that can change but remains infinite, from omega.

Regards, WM

On Monday, November 22, 2021 at 9:44:30 AM UTC-5, WM wrote:
> William schrieb am Sonntag, 21. November 2021 um 14:55:03 UTC+1:
> > On Sunday, November 21, 2021 at 5:17:26 AM UTC-5, WM wrote:
> > > William schrieb am Samstag, 20. November 2021 um 20:16:37 UTC+1:
> > > > On Saturday, November 20, 2021 at 1:01:09 PM UTC-5, WM wrote:
> > >
> > > > > > > But if it exists on the ordinal line, then all natnumbers must be positioned before omega.
> > > > > > Correct
> > > > > That is an end.
> > > > Nope, omega is not an element of the Peano set of natural numbers. There is no last natural number.
> > > If there is no last natural number, then there is an empty space before omega.
> > Nope, 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).
> > There is no ordinal between the Peano set of natural numbers and omega.
> >
> This proves that the set has more elements

Nope. It is sufficient that the Peano set of natural numbers has no largest element..

--
William Hughes

fredag 19 november 2021 kl. 11:01:59 UTC+1 skrev WM:
>Important to remember that it holds for all infinitely many endsegments.

Not the way you want it.

>n is removed by E(n+1). Infinitely many successors of n remain in all endsegments.

n is not IN E(n+1), nothign was removed, it is not a process!

>Infinitely many elements are in all sets.

Not relevant, the fact no element is in all sets means the empty set is the intersection.

>Infinitely many elements are in the intersection of all infinite endsegments.

False, 0 elements is in the intersection of all infinitely many endsegments.

> Regards, WM

fredag 19 november 2021 kl. 13:47:34 UTC+1 skrev WM:
> Fritz Feldhase schrieb am Freitag, 19. November 2021 um 13:06:14 UTC+1:
> > On Friday, November 19, 2021 at 11:01:59 AM UTC+1, WM wrote:
> >
> > > Infinitely many elements are in the intersection of all [...] endsegments.
> >
> > Nonsense. The intersection of all endsegments [each of which is infinite] is empty.
>
> You are an enemy of mathematics. In mathemathics the intersection of infinite endsegments is infinite.
> >
> > 1. The intersection of [the set of] all endsegments is empty.
>
> Yes, but the infinite endsegments maintain infinitely many natural numbers from the first to every infinite endsegment.
>
> > 2. Each and every endsegment is empty.
>
> No. Every definable endsegment is infinite. Bur infinitely many endsegments are exactly aqs many as infinitely many natural numbers. How could more natural numbers exist than endsegments. This would be necessary if all infinitely many endsegments had not lost all natural numbers.
>
> Regards, WM

The enemy of mathematics is you whom fail to do basic logic

William schrieb am Montag, 22. November 2021 um 19:09:21 UTC+1:
> On Monday, November 22, 2021 at 9:44:30 AM UTC-5, WM wrote:
> > William schrieb am Sonntag, 21. November 2021 um 14:55:03 UTC+1:
> > > On Sunday, November 21, 2021 at 5:17:26 AM UTC-5, WM wrote:
> > > > William schrieb am Samstag, 20. November 2021 um 20:16:37 UTC+1:
> > > > > On Saturday, November 20, 2021 at 1:01:09 PM UTC-5, WM wrote:
> > > >
> > > > > > > > But if it exists on the ordinal line, then all natnumbers must be positioned before omega.
> > > > > > > Correct
> > > > > > That is an end.
> > > > > Nope, omega is not an element of the Peano set of natural numbers.. There is no last natural number.
> > > > If there is no last natural number, then there is an empty space before omega.
> > > Nope, 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).
> > > There is no ordinal between the Peano set of natural numbers and omega.
> > >
> > This proves that the set has more elements
> Nope. It is sufficient that the Peano set of natural numbers has no largest element..
>
Not at all. Why should this influence the facts acknowledged by yourself? "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)." Yes, M(n) can change but it cannot decrease below aleph_o for every definable n.

Regards, WM

zelos...@gmail.com schrieb am Dienstag, 23. November 2021 um 06:00:30 UTC+1:
> fredag 19 november 2021 kl. 11:01:59 UTC+1 skrev WM:

> > Infinitely many elements are in all sets.
>
> Not relevant, the fact no element is in all sets means the empty set is the intersection.

It contradicts: Infinitely many elements are in every endsegment.
>
> >Infinitely many elements are in the intersection of all infinite endsegments.
>
> False, 0 elements is in the intersection of all infinitely many endsegments.

Yes, but all infinite endsegments have infinitely many numbers in common.

Regards, WM

zelos...@gmail.com schrieb am Dienstag, 23. November 2021 um 06:00:57 UTC+1:

> The enemy of mathematics is

who does not understand this simple truth: How could more natural numbers exist than endsegments? This would be necessary if all infinitely many endsegments had not lost all natural numbers.

Regards, WM

tisdag 23 november 2021 kl. 10:35:17 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Dienstag, 23. November 2021 um 06:00:57 UTC+1:
>
> > The enemy of mathematics is
> who does not understand this simple truth: How could more natural numbers exist than endsegments? This would be necessary if all infinitely many endsegments had not lost all natural numbers.
>
> Regards, WM

Nope, it is still you.

The cardinality of all endsegmetns and natural numbers are the same.

The rest is just you going stupid

tisdag 23 november 2021 kl. 10:32:18 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Dienstag, 23. November 2021 um 06:00:30 UTC+1:
> > fredag 19 november 2021 kl. 11:01:59 UTC+1 skrev WM:
>
> > > Infinitely many elements are in all sets.
> >
> > Not relevant, the fact no element is in all sets means the empty set is the intersection.
> It contradicts: Infinitely many elements are in every endsegment.
> >
> > >Infinitely many elements are in the intersection of all infinite endsegments.
> >
> > False, 0 elements is in the intersection of all infinitely many endsegments.
> Yes, but all infinite endsegments have infinitely many numbers in common.
>
> Regards, WM

>It contradicts: Infinitely many elements are in every endsegment.

It contradicts nothing. You claim it does but logically there is none.

>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

EACH endsegment has it with ANY OTHER endsegment but for all, it is not true.

On 11/23/2021 3:35 AM, WM wrote:
> zelos...@gmail.com schrieb am Dienstag, 23. November 2021 um 06:00:57 UTC+1:
>
>> The enemy of mathematics is
>
> who does not understand this simple truth: How could more natural numbers exist than endsegments? This would be necessary if all infinitely many endsegments had not lost all natural numbers.
>
> Regards, WM
>

Errors;

1 your premise is wrong, it is not a truth.

2 endsegments are fixed sets, they do not "lose" numbers.

3 natural numbers came first, then sets, then endsegments, you have it backwards.

4 there is no "necessary"

5 what You call a simple truth, is a question.

On 11/23/2021 3:32 AM, WM wrote:
> zelos...@gmail.com schrieb am Dienstag, 23. November 2021 um 06:00:30 UTC+1:
>> fredag 19 november 2021 kl. 11:01:59 UTC+1 skrev WM:
>
>>> Infinitely many elements are in all sets.
>>
>> Not relevant, the fact no element is in all sets means the empty set is the intersection.
>
> It contradicts: Infinitely many elements are in every endsegment.
>>
>>> Infinitely many elements are in the intersection of all infinite endsegments.
>>
>> False, 0 elements is in the intersection of all infinitely many endsegments.
>
> Yes, but all infinite endsegments have infinitely many numbers in common.
>
> Regards, WM
>

errors,

1 the intersection is not an endsegment.

2. you confuse Union with Intersection

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).

Correct, every element of the Peano set of natural numbers is finite. So what? The Peano set of natural numbers is infinite. The Peano set of natural numbers does not contain "dark" elements.

> ...every definable n.

The finite set of elements that can be written down is irrelevant.

--
William Hughes

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).

But it has ℵo elements.

> Correct, every element of the Peano set of natural numbers is finite. So what?

It is not only finite but belongs to a FISON which is vanishing compared to the remaining endsgement.

> The Peano set of natural numbers is infinite.

Yes, 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. It does not contain ℵo elements.
> > ...every definable n.
>
> The finite set of elements that can be written down is irrelevant.

It is all that we can apply in mathematics. Every definable natnumber has infinite distance from ω. But the set of all natnumbers has no distance from ω. It is even called ω in modern set theory. So we have a big difference between definable natnumbers and undefinable nat numbers.

Regards, 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

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.

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

On Wednesday, November 24, 2021 at 9:46:36 AM UTC+1, WM wrote:
> William schrieb am Dienstag, 23. November 2021 um 18:18:46 UTC+1:
> >
> > The [...] set of natural numbers does not contain "dark" elements.
> >
> It is not actually infinite. It does not contain ℵo elements.

Holy shit!!!

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."

Liar, this only works if this set has a last element. Such a last
element would be in every endsegment, and there would be infinitely
many of them.

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

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.

> You argue that [for each and every] natnumbers [n:] n is not contained in E(n+1) [and hence is not contained in INTERSECTION {E(k) : k e IN}; with other words, THERE IS NO natural number contained in INTERSECTION {E(k) : k e IN}]. But als long as [for each and every natural number n] E(n+1) contains infinitely many elements, this argument <bla>

The argument does NOT depend on the number of elements in the sets E(n) (with n e IN).

Hint: It works for ARBITRARY sets A(n) c IN (with n e IN): If for all n e IN: there is an m e IN such that n !e A(m), then INTERSECTION {A(n) : n e IN} = {}.

An e IN: n !e E(n+1) is just a "special case".