William schrieb am Samstag, 13. November 2021 um 20:32:41 UTC+1:
> On Saturday, November 13, 2021 at 2:21:25 PM UTC-5, WM wrote:
> > William schrieb am Samstag, 13. November 2021 um 20:10:41 UTC+1:
> > > On Saturday, November 13, 2021 at 2:02:03 PM UTC-5, WM wrote:
> > > > ℕ\M which is a finite set
> > > |N is not the natural numbers and M does not exist.
> > ℕ ist the genuine set of natural numbers
> Nope. the set of natural numbers is a Peano set. Your |N is not a Peano set.

The ℕ used here is the official set of natural numbers:

∀ n∈ℕ ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo (*)
~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| = ℵo .

> Any element of M has to be an element of E(n) for every n in |N_F. There are no such elements

There are no such elements? Every n has ℵo successors. There are infinitely many for every n according to (*). Do you deny the first line? Do you deny that ℕ\M with |M| = ℵo is a finite set? If so, then you refuse to see the logical consequences. Small wonder that set theory is free of contradictions when its advocates refuse to see them.

Regards, WM

Python schrieb am Samstag, 13. November 2021 um 21:07:06 UTC+1:

> > In ∀ n∈ℕ_def ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo (*)
> M cannot be fixed (i.e. independant of n)

Why should that matter? Any idea? Do you wish to claim that M could disappear? Do you claim that the set ℕ\M could become ℕ in the infinite? Do you try to cheat and deceive?

In every single case the set M is infinite. That is enough to see that the set ℕ\M is finite. All n that can be chosen are from ℕ\M. Do you deny that (*) is correct? Do you claim the existence of two consecutive actually infinite sets in the natural order of ℕ? Where does the first one end and the second one begin?

Please declare these delusions in clear words that every youg student can recognize what a mess your matheology is.

Regards, WM

> The ℕ used here is the official set of natural numbers:

The set of natural numbers is a Peano set. Your |N is not a Peano set.

>... ℕ\M. ... |M| = ℵo

Nonsense. A set that is not the natural numbers and s set that does not exist.

--
William Hughes

William schrieb am Samstag, 13. November 2021 um 22:40:56 UTC+1:
> > The ℕ used here is the official set of natural numbers:
> The set of natural numbers is a Peano set. Your |N is not a Peano set.
>
> >... ℕ\M. ... |M| = ℵo
>
> Nonsense. A set that is not the natural numbers and s set that does not exist.

Okay, you distrust logig or are incapable of understanding it. Nobody can help you. Console yourself with the thought that you are not the only one and stay with your delusions. Facts are here: With the endsegments E(n) = {n, n+1, n+2, ...} we get
∀ n∈ℕ_def ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo
~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| = ℵo .
For ℕ_def = ℕ there is a contradiction. Hence for every sober mind: Either there are dark numbers or actual infinity is blatant nonsense.

Regards, WM

crank Wolfgang Mückenheim, aka WM wrote:
> Python schrieb am Samstag, 13. November 2021 um 21:07:06 UTC+1:
>
>>> In ∀ n∈ℕ_def ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo (*)
>> M cannot be fixed (i.e. independant of n)
>
> Why should that matter? Any idea?

because this explains why this is also false:

∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| = ℵo .

> Do you wish to claim that M could disappear?

No. I claim that there is no single M indepedent of n. Basic
simple logic almost everyone can get but you.

This does not even (as usual) have anything specific about
dealing with infinite sets, even for finite sets quantifier
exchange is (in general) invalid.

> Do you try to cheat and deceive?

No. But you are. As usual, crank Wolfgang Mückenheim, from Hochschule
Augsburg. This is the very basis of your hole infamous, shameful
"career" ; Once you'll pay for your CRIMES, Mückenheim.

On Saturday, November 13, 2021 at 4:46:57 PM UTC-5, WM wrote:
> William schrieb am Samstag, 13. November 2021 um 22:40:56 UTC+1:
> > > The ℕ used here is the official set of natural numbers:
> > The set of natural numbers is a Peano set. Your |N is not a Peano set.
> >
> > >... ℕ\M. ... |M| = ℵo
> >
> > Nonsense. A set that is not the natural numbers and s set that does not exist.
> Okay, you distrust

Does not change the fact that |N is not a Peano set and M does not exist.

--
William Hughes

WM formulated on Saturday :
> FromTheRafters schrieb am Samstag, 13. November 2021 um 20:30:28 UTC+1:
>> WM laid this down on his screen :
>>> William schrieb am Samstag, 13. November 2021 um 20:10:41 UTC+1:
>>>> On Saturday, November 13, 2021 at 2:02:03 PM UTC-5, WM wrote:
>>>>> ℕ\M which is a finite set
>>>>> N is not the natural numbers and M does not exist.
>>>
>>> ℕ ist the genuine set of natural numbers defined by Zermelo or v. Neumann
>>> or Peano.
>> Finally, you admit that the natural numbers are defined.
>
> The natural numbers are defined collectively.

Ah, so they were all 'definable' after all. Thanks for clearing that
up.

WM laid this down on his screen :
> William schrieb am Samstag, 13. November 2021 um 22:40:56 UTC+1:
>>> The ℕ used here is the official set of natural numbers:
>> The set of natural numbers is a Peano set. Your |N is not a Peano set.
>>
>>> ... ℕ\M. ... |M| = ℵo
>>
>> Nonsense. A set that is not the natural numbers and s set that does not
>> exist.
>
> Okay, you distrust logig or are incapable of understanding it. Nobody can
> help you. Console yourself with the thought that you are not the only one and
> stay with your delusions. Facts are here: With the endsegments E(n) = {n,
> n+1, n+2, ...} we get ∀ n∈ℕ_def ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo ~∃ M⊂ℕ ∀ n∈ℕ :
> M⊂E(n) ∧ |M| = ℵo . For ℕ_def = ℕ there is a contradiction. Hence for every
> sober mind: Either there are dark numbers or actual infinity is blatant
> nonsense.

Stop lying and trying to insult your betters.

Python schrieb am Samstag, 13. November 2021 um 23:09:37 UTC+1:
> cWolfgang Mückenheim, aka WM wrote:
> > Python schrieb am Samstag, 13. November 2021 um 21:07:06 UTC+1:
> >
> >>> In ∀ n∈ℕ_def ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo (*)
> >> M cannot be fixed (i.e. independant of n)
> >
> > Why should that matter? Any idea?
> because this explains why this is also false:
>
> ∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| = ℵo .

No, it does not explain why the above is wrong and why this is correct:
~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| = ℵo is correct.

> > Do you wish to claim that M could disappear?
> No. I claim that there is no single M indepedent of n.

You are wrong. The intersection of gapless infinite subsets of ℕ, like the infinite endsegments, is infinite. This is the infinite set M.

> Basic simple logic almost everyone can get but you.

Basic simple logic is this: Every natnumber that you can apply in your argument (n not in E(n+1)) has infinitely many successors. Therefore your argument about the emptiness of the intersection of all infinite endsegments is simply wrong. The intersection is empty of the potentially infinite sets of natnumbers which have infinitely many successors; but this contains an actually infinite set.

Regards, WM

William schrieb am Samstag, 13. November 2021 um 23:24:47 UTC+1:

> Does not change the fact that |N is not a Peano set and M does not exist.

No, no, no!!!

This is a correct statement
~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| = ℵo
with ℕ being the natural numbers of ZF and M being an actually infinite set.

The intersection of gapless infinite subsets of ℕ, like the infinite endsegments, is infinite. This is the infinite set M.

Why did you believe the contrary before you could understand this? Here comes the explanation: Every natnumber that you can apply in your argument (n not in E(n+1)) has infinitely many successors. Therefore your argument about the emptiness of the intersection of all infinite endsegments concerns only these numbers but it does noprove anything about the actually infinite set M.

Regards, WM

In ZFC the finite ordinal n has much more ordinals alpha
with n < alpha, they are at least uncountable many, actually

the ordinals alpha with n < alpha is a proper class, so
you cannot assign it a cardinal. This class here:

{ alpha | n < alpha }

Is not only dark. It is ultra dark, the darkest darks
of the darks, its more dark than vantablack.

WM schrieb am Sonntag, 14. November 2021 um 10:57:15 UTC+1:
> William schrieb am Samstag, 13. November 2021 um 23:24:47 UTC+1:
>
> > Does not change the fact that |N is not a Peano set and M does not exist.
> No, no, no!!!
>
> This is a correct statement
> ~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| = ℵo
> with ℕ being the natural numbers of ZF and M being an actually infinite set.
> The intersection of gapless infinite subsets of ℕ, like the infinite endsegments, is infinite. This is the infinite set M.
> Why did you believe the contrary before you could understand this? Here comes the explanation: Every natnumber that you can apply in your argument (n not in E(n+1)) has infinitely many successors. Therefore your argument about the emptiness of the intersection of all infinite endsegments concerns only these numbers but it does noprove anything about the actually infinite set M.
>
> Regards, WM

Well maybe you can assign it a cardinal with suitable
large cardinal axioms, and what ever. But since you
can view cardinals as selected ordinals, and ordinals

as selected sets, assigning a cardinal to it would say
putting { alpha | n < alpha } in bijection with a set.
Which is impossible for a proper class.

Its similar to Dans NOT PURPLE, i.e. V \ p, only
here the class difference is On \ n.

Mostowski Collapse schrieb am Sonntag, 14. November 2021 um 12:50:42 UTC+1:
> In ZFC the finite ordinal n has much more ordinals alpha
> with n < alpha, they are at least uncountable many, actually
>
> the ordinals alpha with n < alpha is a proper class, so
> you cannot assign it a cardinal. This class here:
>
> { alpha | n < alpha }
>
> Is not only dark. It is ultra dark, the darkest darks
> of the darks, its more dark than vantablack.
> WM schrieb am Sonntag, 14. November 2021 um 10:57:15 UTC+1:
> > William schrieb am Samstag, 13. November 2021 um 23:24:47 UTC+1:
> >
> > > Does not change the fact that |N is not a Peano set and M does not exist.
> > No, no, no!!!
> >
> > This is a correct statement
> > ~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| = ℵo
> > with ℕ being the natural numbers of ZF and M being an actually infinite set.
> > The intersection of gapless infinite subsets of ℕ, like the infinite endsegments, is infinite. This is the infinite set M.
> > Why did you believe the contrary before you could understand this? Here comes the explanation: Every natnumber that you can apply in your argument (n not in E(n+1)) has infinitely many successors. Therefore your argument about the emptiness of the intersection of all infinite endsegments concerns only these numbers but it does noprove anything about the actually infinite set M.
> >
> > Regards, WM

The purple paradox for On \ n is now, instead if a finite
ordinal n, you can also take a finite ot infinite ordinal
beta, this here will never have a cardinality:

On \ beta

Can DC Proof, prove that?

Mostowski Collapse schrieb am Sonntag, 14. November 2021 um 12:51:41 UTC+1:
> Well maybe you can assign it a cardinal with suitable
> large cardinal axioms, and what ever. But since you
> can view cardinals as selected ordinals, and ordinals
>
> as selected sets, assigning a cardinal to it would say
> putting { alpha | n < alpha } in bijection with a set.
> Which is impossible for a proper class.
>
> Its similar to Dans NOT PURPLE, i.e. V \ p, only
> here the class difference is On \ n.
> Mostowski Collapse schrieb am Sonntag, 14. November 2021 um 12:50:42 UTC+1:
> > In ZFC the finite ordinal n has much more ordinals alpha
> > with n < alpha, they are at least uncountable many, actually
> >
> > the ordinals alpha with n < alpha is a proper class, so
> > you cannot assign it a cardinal. This class here:
> >
> > { alpha | n < alpha }
> >
> > Is not only dark. It is ultra dark, the darkest darks
> > of the darks, its more dark than vantablack.
> > WM schrieb am Sonntag, 14. November 2021 um 10:57:15 UTC+1:
> > > William schrieb am Samstag, 13. November 2021 um 23:24:47 UTC+1:
> > >
> > > > Does not change the fact that |N is not a Peano set and M does not exist.
> > > No, no, no!!!
> > >
> > > This is a correct statement
> > > ~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| = ℵo
> > > with ℕ being the natural numbers of ZF and M being an actually infinite set.
> > > The intersection of gapless infinite subsets of ℕ, like the infinite endsegments, is infinite. This is the infinite set M.
> > > Why did you believe the contrary before you could understand this? Here comes the explanation: Every natnumber that you can apply in your argument (n not in E(n+1)) has infinitely many successors. Therefore your argument about the emptiness of the intersection of all infinite endsegments concerns only these numbers but it does noprove anything about the actually infinite set M.
> > >
> > > Regards, WM

A small correction On \ beta gives { alpha | beta =< alpha }, for von
Neuman ordinals where < is the same as in, needs some thinking.

Mostowski Collapse schrieb:
> The purple paradox for On \ n is now, instead if a finite
> ordinal n, you can also take a finite ot infinite ordinal
> beta, this here will never have a cardinality:
>
> On \ beta
>
> Can DC Proof, prove that?
>
> Mostowski Collapse schrieb am Sonntag, 14. November 2021 um 12:51:41 UTC+1:
>> Well maybe you can assign it a cardinal with suitable
>> large cardinal axioms, and what ever. But since you
>> can view cardinals as selected ordinals, and ordinals
>>
>> as selected sets, assigning a cardinal to it would say
>> putting { alpha | n < alpha } in bijection with a set.
>> Which is impossible for a proper class.
>>
>> Its similar to Dans NOT PURPLE, i.e. V \ p, only
>> here the class difference is On \ n.
>> Mostowski Collapse schrieb am Sonntag, 14. November 2021 um 12:50:42 UTC+1:
>>> In ZFC the finite ordinal n has much more ordinals alpha
>>> with n < alpha, they are at least uncountable many, actually
>>>
>>> the ordinals alpha with n < alpha is a proper class, so
>>> you cannot assign it a cardinal. This class here:
>>>
>>> { alpha | n < alpha }
>>>
>>> Is not only dark. It is ultra dark, the darkest darks
>>> of the darks, its more dark than vantablack.
>>> WM schrieb am Sonntag, 14. November 2021 um 10:57:15 UTC+1:
>>>> William schrieb am Samstag, 13. November 2021 um 23:24:47 UTC+1:
>>>>
>>>>> Does not change the fact that |N is not a Peano set and M does not exist.
>>>> No, no, no!!!
>>>>
>>>> This is a correct statement
>>>> ~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| = ℵo
>>>> with ℕ being the natural numbers of ZF and M being an actually infinite set.
>>>> The intersection of gapless infinite subsets of ℕ, like the infinite endsegments, is infinite. This is the infinite set M.
>>>> Why did you believe the contrary before you could understand this? Here comes the explanation: Every natnumber that you can apply in your argument (n not in E(n+1)) has infinitely many successors. Therefore your argument about the emptiness of the intersection of all infinite endsegments concerns only these numbers but it does noprove anything about the actually infinite set M.
>>>>
>>>> Regards, WM

On Sunday, November 14, 2021 at 4:57:15 AM UTC-5, WM wrote:
> William schrieb am Samstag, 13. November 2021 um 23:24:47 UTC+1:
>
> > Does not change the fact that |N is not a Peano set and M does not exist.
> No, no, no!!!
>

Does not change the fact that your |N is not a Peano set.

> This is a correct statement
> ~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| = ℵo
> with ℕ being the natural numbers

Nope, the set of natural numbers is a Peano set. Your |N is not a Peano set.

--
William Hughes

On 11/14/2021 1:40 AM, FromTheRafters wrote:
> WM formulated on Saturday :
>> FromTheRafters schrieb am Samstag, 13. November 2021 um 20:30:28 UTC+1:
>>> WM laid this down on his screen :
>>>> William schrieb am Samstag, 13. November 2021 um 20:10:41 UTC+1:
>>>>> On Saturday, November 13, 2021 at 2:02:03 PM UTC-5, WM wrote:
>>>>>> ℕ\M which is a finite set N is not the natural numbers and M does not exist.
>>>>
>>>> ℕ ist the genuine set of natural numbers defined by Zermelo or v. Neumann or  Peano.
>>> Finally, you admit that the natural numbers are defined.
>>
>> The natural numbers are defined collectively.
>
> Ah, so they were all 'definable' after all. Thanks for clearing that up.

I have seen flocks of sheeps do that, they will collect into an area (like a set), then define themselves all at once, bleating, then one by one they
remove themselves.

William schrieb am Sonntag, 14. November 2021 um 14:43:44 UTC+1:
> On Sunday, November 14, 2021 at 4:57:15 AM UTC-5, WM wrote:
> > William schrieb am Samstag, 13. November 2021 um 23:24:47 UTC+1:
> >
> > > Does not change the fact that |N is not a Peano set and M does not exist.
> > No, no, no!!!
> >
> Does not change the fact that your |N is not a Peano set.
> > This is a correct statement
> > ~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| = ℵo
> > with ℕ being the natural numbers
> Nope, the set of natural numbers is a Peano set. Your |N is not a Peano set.
>
What makes it differ from the Peano set in your opinion?

Regards, WM

lördag 13 november 2021 kl. 19:05:59 UTC+1 skrev WM:
> William schrieb am Samstag, 13. November 2021 um 13:50:24 UTC+1:
> > On Saturday, November 13, 2021 at 4:22:51 AM UTC-5, WM wrote:
> >
> ∀ n∈ℕ_def ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo
> ~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| = ℵo .
> > > No infinite set
> > that does not change exists according to the second statement. There is an infinite set that changes
> Irrelevant. It remains infinite.
> > (i.e. a different infinite set for each element of |N_F) according to the first statement.
> E(n) differs. M is the fixed subset ℵo. That means natural numbers can only be chosen from ℕ\M which has not cardinality ℵo, because there are not two sets of ℵo in the natural order of ℕ.
>
> > No contradiction.
>
> The second statement covers all naturakl numbers, more than the first statement.
>
> Regards, WM

Nope, still wrong

lördag 13 november 2021 kl. 20:05:52 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Samstag, 13. November 2021 um 14:26:22 UTC+1:
>
> > > ∀ n∈ℕ ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo
> > > ~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| = ℵo .
> > Those are not contradictions
>
> A simply unfounded and wrong claim.
> Between all n∈ℕ and ω there are infinitely many natnumbers according to the first line. There is nothing between ℕ and ω according to
> ~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| > 0.
> A contradiction.
>
> Regards, WM

>A simply unfounded and wrong claim.

Nope, it is correct unlike yours.

>Between all n∈ℕ and ω there are infinitely many natnumbers according to the first line.

Nope, it just says there is a subset in the endsegment in question that is infinite.

>There is nothing between ℕ and ω according to
>~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| > 0.

False, that say there is no subset of all endsegments with a cardinality greater than zero.

>A contradiction.

Nope

"For each endsegment there is a subset with infinite cardinality"
and
"There is no set that is subset of all endsegments with a cardinality greater than zero"

are not contradictory.

On Monday, November 15, 2021 at 6:25:17 AM UTC-5, WM wrote:
> William schrieb am Sonntag, 14. November 2021 um 14:43:44 UTC+1:

>> ... the set of natural numbers is a Peano set. Your |N is not a Peano set.
> >
> What makes it differ from the Peano set

The axiom of induction does not hold.

--
Wiliam Hughes

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. But ω exists in set theory. I use the ℕ of set theory. ZFC says

∀ n∈ℕ ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo
Every n has infinite distance from ω.
~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| = ℵo .
There is no infinite distance of every n from ω.

The Peano set has no ω and therefore no distance to ω.

Regards, WM

zelos...@gmail.com schrieb am Montag, 15. November 2021 um 13:25:06 UTC+1:
> lördag 13 november 2021 kl. 20:05:52 UTC+1 skrev WM:

> "For each endsegment there is a subset with infinite cardinality"
> and
> "There is no set that is subset of all endsegments with a cardinality greater than zero"
> are not contradictory.

They are clear contradictions because of inclusion monotony. I cannot forbid your way of "arguing", I can only try to drag this nonsense into the public eye.

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 .

Regards, WM

onsdag 17 november 2021 kl. 11:59:09 UTC+1 skrev WM:
> 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. But ω exists in set theory. I use the ℕ of set theory. ZFC says
>
> ∀ n∈ℕ ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo
> Every n has infinite distance from ω.
> ~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| = ℵo .
> There is no infinite distance of every n from ω.
>
> The Peano set has no ω and therefore no distance to ω.
>
> Regards, WM

That is NOT what those means!

> ∀ 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.

None of your fucking shit.

onsdag 17 november 2021 kl. 12:50:07 UTC+1 skrev WM:
> zelos...@gmail.com schrieb am Montag, 15. November 2021 um 13:25:06 UTC+1:
> > lördag 13 november 2021 kl. 20:05:52 UTC+1 skrev WM:
>
> > "For each endsegment there is a subset with infinite cardinality"
> > and
> > "There is no set that is subset of all endsegments with a cardinality greater than zero"
> > are not contradictory.
> They are clear contradictions because of inclusion monotony. I cannot forbid your way of "arguing", I can only try to drag this nonsense into the public eye.
>
> 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 .
>
> Regards, WM

>They are clear contradictions because of inclusion monotony.

False, they are not and a computer can demonstrate it.

>I cannot forbid your way of "arguing", I can only try to drag this nonsense into the public eye.

It is not non-sense to point out where you are being stupid.

>By separation we can collect all endsegments with
>∀k ∈ ℕ: ∩{E(1), E(2), ..., E(k)} = E(k) /\ |E(k)| = ℵ₀

That is a true statement but does not help you.

>into a set. Then no additional endsegment is in the set. All endsegments have infinite itersection.
>|∩{E(k) : k ∈ ℕ}| = ℵ₀

False, the first one does NOT lead to the second.

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.

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William Hughes