William schrieb am Mittwoch, 17. November 2021 um 16:51:46 UTC+1:
> On Wednesday, November 17, 2021 at 5:59:09 AM UTC-5, WM wrote:
> > William schrieb am Montag, 15. November 2021 um 16:21:51 UTC+1:
> > > On Monday, November 15, 2021 at 6:25:17 AM UTC-5, WM wrote:
> >
> > > > What makes it differ from the Peano set
> > > The axiom of induction does not hold.
> > There is no way to reach ω by induction.
> So What? omega is not a natural number. The natural numbers are a Peano set. Your |N is not a Peano set.

Set theory is not Peano theory. Zermelo's axiom of infinity is not Peano's theory.

Regards, WM

zelos...@gmail.com schrieb am Mittwoch, 17. November 2021 um 13:56:52 UTC+1:
> onsdag 17 november 2021 kl. 11:59:09 UTC+1 skrev WM:

> > ∀ n∈ℕ ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo
> Means that for any given n, we have that there is an infinite proper subset of E(n)
> > ~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| = ℵo .
> Means that there exist no proper subset of N that is infinite such that such that it is ALL endsegments.

That implies smaller endsegments.

In ZF even ~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| > 0 is true. That means that there is nothing beyond all natnumbers n. Hence not all endsegments can be infinite.

Regards, WM

On 11/17/2021 6:50 AM, WM wrote:

> 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 .

∀j ∈ ℕ : E(j) ⊇ ∩{E(k) : k ∈ ℕ}

~∃i ∈ ℕ, ∀j ∈ ℕ : E(j) ⊇ E(i)

~∃i ∈ ℕ : E(i) = ∩{E(k) : k ∈ ℕ}

∩{E(k) : k ∈ ℕ} ∉ {E(k) : k ∈ ℕ}

On 11/17/2021 12:34 PM, WM wrote:
> zelos...@gmail.com schrieb am Mittwoch, 17. November 2021 um 13:56:52 UTC+1:
>> onsdag 17 november 2021 kl. 11:59:09 UTC+1 skrev WM:
>
>>> ∀ n∈ℕ ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo
>> Means that for any given n, we have that there is an infinite proper subset of E(n)
>>> ~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| = ℵo .
>> Means that there exist no proper subset of N that is infinite such that such that it is ALL endsegments.
>
> That implies smaller endsegments.

all endsegments are infinite.

>
> In ZF even ~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| > 0 is true. That means that there is nothing beyond all natnumbers n.

>Hence not all endsegments can be infinite.

all endsegments are infinite.

>
> Regards, WM
>

2 wrongs in one post.

On Wednesday, November 17, 2021 at 1:30:47 PM UTC-5, WM wrote:

> Set theory

says that the natural numbers are a Peano set. Your |N is not a Peano set.

--
William Hughes

onsdag 17 november 2021 kl. 19:34:45 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Mittwoch, 17. November 2021 um 13:56:52 UTC+1:
> > onsdag 17 november 2021 kl. 11:59:09 UTC+1 skrev WM:
>
> > > ∀ n∈ℕ ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo
> > Means that for any given n, we have that there is an infinite proper subset of E(n)
> > > ~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| = ℵo .
> > Means that there exist no proper subset of N that is infinite such that such that it is ALL endsegments.
>
> That implies smaller endsegments.
>
> In ZF even ~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| > 0 is true. That means that there is nothing beyond all natnumbers n. Hence not all endsegments can be infinite.
>
> Regards, WM

>In ZF even ~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| > 0 is true.

Becuase the empty set is the only subset of all endsegments, which is why it is the intersection.

>That means that there is nothing beyond all natnumbers n. Hence not all endsegments can be infinite.

Means nothing of the sort.

William schrieb am Mittwoch, 17. November 2021 um 21:49:06 UTC+1:
> On Wednesday, November 17, 2021 at 1:30:47 PM UTC-5, WM wrote:
>
> > Set theory
>
> says that the natural numbers are a Peano set.

You have to distinguish the natural numbers limited by omega and the natural numbers without limits. I show by means of two very simple arguments that the natural numbers limited by omega contain dark numbers:

Every infinite endsegment receives and carries its contents from the first one
∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀
By the axiom of separation these endsegments can be collected in a set ℕ_def such that
|∩{E(k) : k ∈ ℕ_def}| = ℵ₀ .
ZFC says
|∩{E(k) : k ∈ ℕ}| = 0
which would be a contradiction in case ℕ_def = ℕ.

Start with ZFC:
~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| > 0
and ask yourself: Why does not even a set of card 1 exist for all n∈ℕ?
Answer: Because
~∀ n∈ℕ ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo
Contradicting ZFC here
∀ n∈ℕ ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo
unless we distinguish ℕ and ℕ_def.
∀ n∈ℕ_def ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo

Regards, WM

zelos...@gmail.com schrieb am Donnerstag, 18. November 2021 um 06:22:56 UTC+1:
> onsdag 17 november 2021 kl. 19:34:45 UTC+1 skrev WM:

> >In ZF even ~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| > 0 is true.
> Becuase the empty set is the only subset of all endsegments, which is why it is the intersection.
> >That means that there is nothing beyond all natnumbers n. Hence not all endsegments can be infinite.
> Means nothing of the sort.

Ask yourself or better ask a person with clear brain and common sense: Why does not even a set of card 1 exist for all infinite endsegments?

Every infinite endsegment receives and carries its contents from the first one
∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀
By the axiom of separation these endsegments can be collected in a set ℕ_def such that
|∩{E(k) : k ∈ ℕ_def}| = ℵ₀ .
ZFC says
|∩{E(k) : k ∈ ℕ}| = 0
which would be a contradiction in case ℕ_def = ℕ.

Start with ZFC:
~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| > 0
and ask yourself: Why does not even a set of card 1 exist for all n∈ℕ?
Answer: Because
~∀ n∈ℕ ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo
Contradicting ZFC here
∀ n∈ℕ ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo
unless we distinguish ℕ and ℕ_def.
∀ n∈ℕ_def ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo

Regards, WM

Jim Burns schrieb am Mittwoch, 17. November 2021 um 20:23:48 UTC+1:
> On 11/17/2021 6:50 AM, WM wrote:
>
> > 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 . (*)

> ∀j ∈ ℕ : E(j) ⊇ ∩{E(k) : k ∈ ℕ}

Of curse. and E(j) is infinite.
>
> ~∃i ∈ ℕ, ∀j ∈ ℕ : E(j) ⊇ E(i)

Why not? There does not exist a *definable* i such that E(i) is the empty set.
>
> ~∃i ∈ ℕ : E(i) = ∩{E(k) : k ∈ ℕ}

Why not? There does not exist a *definable* i such that E(i) is the empty set.
>
> ∩{E(k) : k ∈ ℕ} ∉ {E(k) : k ∈ ℕ}

Hence we obtain a contradiction with (*). Or do you refuse the axiom of separation?

Regards, WM

>Ask yourself or better ask a person with clear brain and common sense:

I have a brain and I way smarter than you :) I can do logic, you can't!

>Why does not even a set of card 1 exist for all infinite endsegments?

Because they are all in bijection with a proper subset of themselves.

>which would be a contradiction in case ℕ_def = ℕ.

it is not a contradiction because you ASSUME WITHOUT PROOF that |∩{E(k) : k ∈ ℕ_def}| = ℵ₀ .

And that is a false assumption you do. Demonstrating you lack a fucking brain and cannot do logic.

>∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀
>By the axiom of separation these endsegments can be collected in a set ℕ_def such that

which means N_def=N

>Start with ZFC:
>~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| > 0

Correct

>and ask yourself: Why does not even a set of card 1 exist for all n∈ℕ?

because the element would not be in a later endsegment.

>~∀ n∈ℕ ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo

is true if the former is true which it is.

>Contradicting ZFC here

No contradiction.

>∀ n∈ℕ ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo

does not do what you want it to.

>unless we distinguish ℕ and ℕ_def.

There is nothign to distinguish, because every time you try to define N_def, you get N

>∀ n∈ℕ_def ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo

False

torsdag 18 november 2021 kl. 11:17:17 UTC+1 skrev WM:
> Jim Burns schrieb am Mittwoch, 17. November 2021 um 20:23:48 UTC+1:
> > On 11/17/2021 6:50 AM, WM wrote:
> >
> > > 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 . (*)
> > ∀j ∈ ℕ : E(j) ⊇ ∩{E(k) : k ∈ ℕ}
> Of curse. and E(j) is infinite.
> >
> > ~∃i ∈ ℕ, ∀j ∈ ℕ : E(j) ⊇ E(i)
> Why not? There does not exist a *definable* i such that E(i) is the empty set.
> >
> > ~∃i ∈ ℕ : E(i) = ∩{E(k) : k ∈ ℕ}
> Why not? There does not exist a *definable* i such that E(i) is the empty set.
> >
> > ∩{E(k) : k ∈ ℕ} ∉ {E(k) : k ∈ ℕ}
> Hence we obtain a contradiction with (*). Or do you refuse the axiom of separation?
>
> Regards, WM
your "definable" is meaningless here.

zelos...@gmail.com schrieb am Donnerstag, 18. November 2021 um 11:20:55 UTC+1:

> you ASSUME WITHOUT PROOF that |∩{E(k) : k ∈ ℕ_def}| = ℵ₀ .
>
> >∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀
> >By the axiom of separation these endsegments can be collected in a set ℕ_def such that
> which means N_def=N

The intersection of endsegments which yield an infinite intersection ist infinite in every case.

> >Start with ZFC:
> >~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| > 0
> Correct
> >and ask yourself: Why does not even a set of card 1 exist for all n∈ℕ?
> because the element would not be in a later endsegment.

All infinite endsegments carry infinitely many such elements.

> There is nothign to distinguish, because every time you try to define N_def, you get N

Then ZFC is false.
> >∀ n∈ℕ_def ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo
> False
ZFC says even ∀ n∈ℕ ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo

Regards, WM

>The intersection of endsegments which yield an infinite intersection ist infinite in every case.

Every FINITE case

>All infinite endsegments carry infinitely many such elements.

Yes but they all are excluded by infinitely many endsegments so touch shit.

>ZFC says even ∀ n∈ℕ ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo

I misread that one, my bad

zelos...@gmail.com schrieb am Donnerstag, 18. November 2021 um 11:35:24 UTC+1:
> >The intersection of endsegments which yield an infinite intersection ist infinite in every case.
> Every FINITE case

All infinite endsegments belong to finite cases. Please find an infinite endsegment which does not belong to a finite case.

> >All infinite endsegments carry infinitely many such elements.
> Yes but they all are excluded by infinitely many endsegments

Not before all have gone from the endsegments!

How could the intersection, which is the smallest endsegment, be empty when the smallest endsegment is infinite?

You claim that all n are lost but do not consider that all n have infinitely many succesors which are never lost in infinite endsegments. Therefore never an empty intersection can be resulting.

Regards, WM

torsdag 18 november 2021 kl. 11:53:19 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Donnerstag, 18. November 2021 um 11:35:24 UTC+1:
> > >The intersection of endsegments which yield an infinite intersection ist infinite in every case.
> > Every FINITE case
> All infinite endsegments belong to finite cases. Please find an infinite endsegment which does not belong to a finite case.
> > >All infinite endsegments carry infinitely many such elements.
> > Yes but they all are excluded by infinitely many endsegments
> Not before all have gone from the endsegments!
>
> How could the intersection, which is the smallest endsegment, be empty when the smallest endsegment is infinite?
>
> You claim that all n are lost but do not consider that all n have infinitely many succesors which are never lost in infinite endsegments. Therefore never an empty intersection can be resulting.
>
> Regards, WM

>ll infinite endsegments belong to finite cases. Please find an infinite endsegment which does not belong to a finite case.

All irrelevant to when you do an infinite intersection.

>Not before all have gone from the endsegments!

for a given number n, it is excluded in all m>n, E(m) endsegments.

>How could the intersection, which is the smallest endsegment, be empty when the smallest endsegment is infinite?

False, the intersection is not the smallest endsegment. it is the largest subset IN COMMON with all endsegments. which is the empty set!

>You claim that all n are lost but do not consider that all n have infinitely many succesors which are never lost in infinite endsegments.

Except they are ALL excluded by AT LEAST 1 endsegment!

for a given number n, it is excluded in all m>n, E(m) endsegments.

On Thursday, November 18, 2021 at 5:04:49 AM UTC-5, WM wrote:
> William schrieb am Mittwoch, 17. November 2021 um 21:49:06 UTC+1:
> > On Wednesday, November 17, 2021 at 1:30:47 PM UTC-5, WM wrote:
> >
> > > Set theory
> >
> > says that the natural numbers are a Peano set.
> You have to distinguish the natural numbers limited by omega and the natural numbers without limits. I

Nope. There is only one set of natural numbers, a Peano set. The ordinal omega is greater than any natural number.

--
William Hughes

On 11/18/2021 4:53 AM, WM wrote:
> zelos...@gmail.com schrieb am Donnerstag, 18. November 2021 um 11:35:24 UTC+1:
>>> The intersection of endsegments which yield an infinite intersection ist infinite in every case.
>> Every FINITE case
>
> All infinite endsegments belong to finite cases.

there is no finite cases.

> Please find an infinite endsegment which does not belong to a finite case.

any endsegment.

>
>>> All infinite endsegments carry infinitely many such elements.
>> Yes but they all are excluded by infinitely many endsegments
>
> Not before all have gone from the endsegments!
>
> How could the intersection, which is the smallest endsegment, be empty when the smallest endsegment is infinite?

there is no smallest endsegment, they all are infinite.

>
> You claim that all n are lost but do not consider that all n have infinitely many succesors which are never lost in infinite endsegments. Therefore never an empty intersection can be resulting.

still wrong.

>
> Regards, WM
>

On Thursday, November 18, 2021 at 11:04:49 AM UTC+1, WM wrote:

> ~∀n∈ℕ ∃M⊂ℕ: M⊂E(n) ∧ |M| = ℵo

No, you silly asshole.

For each and every n ∈ ℕ there's a set M ⊂ E(n), namely E(n), such that M ⊂ E(n) and |M| = ℵo (since for each and every n ∈ ℕ: E(n) ⊂ E(n) and |E(n)| = ℵo). In other words: ∀n∈ℕ ∃M⊂ℕ: M⊂E(n) ∧ |M| = ℵo

> Contradicting <bla>

Halt doch einfach mal Dein Maul, Mensch! Du redest nur Scheiße.

On Thursday, November 18, 2021 at 11:08:19 AM UTC+1, WM wrote:

> ~∃M⊂ℕ ∀n∈ℕ : M⊂E(n) ∧ |M| > 0
>
> Why does not even a set [M⊂ℕ] of card 1 exist [such that] for all n ∈ ℕ [M⊂E(n)]?
>
> Answer:

|M| > 0 mans that M is not empty. Now any nonempty set of natural numbers (condition M⊂ℕ) has a least element. Assume that n_0 is the least element of M. But since n_0 !e E(n_0+1), M isn't a subset of E(n_0+1).. Hence M isn't a subset of ALL endsegments (i. e. ∀n∈ℕ : M⊂E(n) doesn't hold).

On Wednesday, November 17, 2021 at 12:50:07 PM UTC+1, WM wrote:

> |∩{E(k) : k ∈ ℕ}| = ℵ₀ .

Nope. Your claim is false.

Actually, |∩{E(k) : k ∈ ℕ}| = 0 since ∩{E(k) : k ∈ ℕ} = {}.

Fritz Feldhase schrieb am Donnerstag, 18. November 2021 um 19:28:38 UTC+1:
> On Wednesday, November 17, 2021 at 12:50:07 PM UTC+1, WM wrote:
>
> > |∩{E(k) : k ∈ ℕ}| = ℵ₀ .
> Nope. Your claim is false.
>
> Actually, |∩{E(k) : k ∈ ℕ}| = 0 since ∩{E(k) : k ∈ ℕ} = {}.

This is a contradiction.
Every infinite endsegment receives its contents from the first one
∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀
By the axiom of separation these endsegments can be collected in a set ℕ_def such that
|∩{E(k) : k ∈ ℕ_def}| = ℵ₀ .

∩{E(k) : k ∈ ℕ} = {} holds only for k which have ℵ₀ successors.

Regards, WM

zelos...@gmail.com schrieb am Donnerstag, 18. November 2021 um 13:47:15 UTC+1:

> >ll infinite endsegments belong to finite cases. Please find an infinite endsegment which does not belong to a finite case.
> All irrelevant to when you do an infinite intersection.

That is not possible since all definable endsegments are less than infinitely many.

> >Not before all have gone from the endsegments!
> for a given number n, it is excluded in all m>n, E(m) endsegments.

For a given number n, the endsegment is infinite and belongs to a finite set.

> >How could the intersection, which is the smallest endsegment, be empty when the smallest endsegment is infinite?
> False, the intersection is not the smallest endsegment.

Inclusion monotony shows that the intersection is the smallest endsegment.

> >You claim that all n are lost but do not consider that all n have infinitely many succesors which are never lost in infinite endsegments.
> Except they are ALL excluded by AT LEAST 1 endsegment!
> .
> for a given number n, it is excluded in all m>n, E(m) endsegments.

For every given number n the endsegment E(n) is infinite. This does not not change during the whole set of endsegments. Therefore not all numbers are removed. Infinitely many remain in all endsegments.

Never thought about this simple contradiction?

Regards, WM

fredag 19 november 2021 kl. 09:31:09 UTC+1 skrev WM:
>That is not possible since all definable endsegments are less than infinitely many.

"definable" is meaningless

And there are infinitely many endsegments

>For a given number n, the endsegment is infinite and belongs to a finite set

Belongs also to the infinite set of all endsegments, irrelevant.

>Inclusion monotony shows that the intersection is the smallest endsegment.

It shows no such thing you retarded fuck.

>For every given number n the endsegment E(n) is infinite.

Correct

>This does not not change during the whole set of endsegments.

It applies to all by the statement itself, no need to reiterate it.

>Therefore not all numbers are removed. Infinitely many remain in all endsegments.

No elements are removed, but no element is in all sets, ergo no element is in the intersection

>Never thought about this simple contradiction?

You provided no contradiction. You stated a truth, then rubbish.

zelos...@gmail.com schrieb am Freitag, 19. November 2021 um 10:07:37 UTC+1:
> fredag 19 november 2021 kl. 09:31:09 UTC+1 skrev WM:

> >For every given number n the endsegment E(n) is infinite.
> Correct
> >This does not not change during the whole set of endsegments.
> It applies to all by the statement itself, no need to reiterate it.

Important to remember that it holds for all infinitely many endsegments.

> >Therefore not all numbers are removed. Infinitely many remain in all endsegments.
> No elements are removed,

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

> but no element is in all sets,

Infinitely many elements are in all sets.
∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵo

> ergo no element is in the intersection

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

Regards, WM

WM expressed precisely :
> zelos...@gmail.com schrieb am Donnerstag, 18. November 2021 um 13:47:15
> UTC+1:
>
>>> ll infinite endsegments belong to finite cases. Please find an infinite
>>> endsegment which does not belong to a finite case.
>> All irrelevant to when you do an infinite intersection.
>
> That is not possible since all definable endsegments are less than infinitely
> many.
>
>>> Not before all have gone from the endsegments!
>> for a given number n, it is excluded in all m>n, E(m) endsegments.
>
> For a given number n, the endsegment is infinite and belongs to a finite set.
>
>>> How could the intersection, which is the smallest endsegment, be empty when
>>> the smallest endsegment is infinite?
>> False, the intersection is not the smallest endsegment.
>
> Inclusion monotony shows that the intersection is the smallest endsegment.
>
>>> You claim that all n are lost but do not consider that all n have
>>> infinitely many succesors which are never lost in infinite endsegments.
>> Except they are ALL excluded by AT LEAST 1 endsegment!
>> .
>> for a given number n, it is excluded in all m>n, E(m) endsegments.
>
> For every given number n the endsegment E(n) is infinite. This does not not
> change during the whole set of endsegments. Therefore not all numbers are
> removed. Infinitely many remain in all endsegments.
>
> Never thought about this simple contradiction?

It is not a contradiction, it is just you failing to grasp the ideas
behind infinite sets.