|N_F

[William]

The following statements of about the set |N_F are all true

(i) Every element of |N_F is finite

(ii) |N_F, like any Peano set, has cardinality aleph_0

(iii) Between every element of |N_F and omega there are infinitely many natural numbers.

(iv) There is no natural number between the set |N_F and omega.

(iv) is true, not because there is an element of |N_F which is not separated from omega by an infinite set of natural numbers, but because there is no largest natural number.

--
William Hughes


*i

[WM]

William schrieb am Samstag, 6. November 2021 um 22:42:36 UTC+1:
> The following statements of about the set |N_F are all true
>
> (i) Every element of |N_F is finite
>
> (ii) |N_F, like any Peano set, has cardinality aleph_0
>
> (iii) Between every element of |N_F and omega there are infinitely many natural numbers.
>
> (iv) There is no natural number between the set |N_F and omega.

What is left-hand side next to omega on the ordinal line?

>
> (iv) is true, not because there is an element of |N_F which is not separated from omega by an infinite set of natural numbers, but because there is no largest natural number.
>

Please explain with that model what elements are in all infinitely many infinite endsegments, the intersection of which is empty.

Regards, WM

[Gus Gassmann]

On Sunday, 7 November 2021 at 07:56:59 UTC-4, WM wrote:
> William schrieb am Samstag, 6. November 2021 um 22:42:36 UTC+1:
> > The following statements of about the set |N_F are all true
> >
> > (i) Every element of |N_F is finite
> >
> > (ii) |N_F, like any Peano set, has cardinality aleph_0
> >
> > (iii) Between every element of |N_F and omega there are infinitely many natural numbers.
> >
> > (iv) There is no natural number between the set |N_F and omega.
> What is left-hand side next to omega on the ordinal line?

This is totally unclear. "Left-hand side" refers to driving rules, or perhaps to equations, but not to ordinal lines. If you are asking what is the immediate predecessor of omega, the answer is: There is none. Omega does not have a predecessor, it is a "limit ordinal".

> > (iv) is true, not because there is an element of |N_F which is not separated from omega by an infinite set of natural numbers, but because there is no largest natural number.
> >
> Please explain with that model what elements are in all infinitely many infinite endsegments, the intersection of which is empty.

This is easy: Every end segment E(n) contains infinitely many natural numbers, but never n - 1.

[William]

On Sunday, November 7, 2021 at 7:56:59 AM UTC-4, WM wrote:
> William schrieb am Samstag, 6. November 2021 um 22:42:36 UTC+1:
> > The following statements of about the set |N_F are all true
> >
> > (i) Every element of |N_F is finite
> >
> > (ii) |N_F, like any Peano set, has cardinality aleph_0
> >
> > (iii) Between every element of |N_F and omega there are infinitely many natural numbers.
> >
> > (iv) There is no natural number between the set |N_F and omega.
> What is left-hand side next to omega on the ordinal line?

"next to omega on the left-hand side" is meaningless. (N_F has no largest element).

--
William Hughes

[WM]

Gus Gassmann schrieb am Sonntag, 7. November 2021 um 14:04:27 UTC+1:
> On Sunday, 7 November 2021 at 07:56:59 UTC-4, WM wrote:
> > William schrieb am Samstag, 6. November 2021 um 22:42:36 UTC+1:
> > > The following statements of about the set |N_F are all true
> > >
> > > (i) Every element of |N_F is finite
> > >
> > > (ii) |N_F, like any Peano set, has cardinality aleph_0
> > >
> > > (iii) Between every element of |N_F and omega there are infinitely many natural numbers.
> > >
> > > (iv) There is no natural number between the set |N_F and omega.
> > What is left-hand side next to omega on the ordinal line?
> This is totally unclear. "Left-hand side" refers to driving rules, or perhaps to equations, but not to ordinal lines.

The ordinal line like the real line shows increasing values from left to right.

> If you are asking what is the immediate predecessor of omega, the answer is: There is none. Omega does not have a predecessor, it is a "limit ordinal".

Limits destroy set theory. If you cannot count all natural numbers then countability is nonsense. Therefore you have to count until only omega remains.

> > > (iv) is true, not because there is an element of |N_F which is not separated from omega by an infinite set of natural numbers, but because there is no largest natural number.
> > >
> > Please explain with that model what elements are in all infinitely many infinite endsegments, the intersection of which is empty.
> This is easy: Every end segment E(n) contains infinitely many natural numbers, but never n - 1.

If you collect only endsegments which do not have an empty intersection together (collect as long as the intersection is infinite, then stop) then you get
|∩{E(k) : k ∈ ℕ_def}| = ℵ₀ .

What gives
|∩{E(k) : k ∈ ℕ}| = 0 ?

Regards, WM

[WM]

It is not meaningless if omega exists. Then there is this point somewhere on the ordinal line. Then there is the question for next sensible.

> (N_F has no largest element).

N_F is followed by omega. Every definable element is not followed by omega.

Regards, WM

[Serg io]

On 11/7/2021 1:24 PM, WM wrote:
> Gus Gassmann schrieb am Sonntag, 7. November 2021 um 14:04:27 UTC+1:
>> On Sunday, 7 November 2021 at 07:56:59 UTC-4, WM wrote:
>>> William schrieb am Samstag, 6. November 2021 um 22:42:36 UTC+1:
>>>> The following statements of about the set |N_F are all true
>>>>
>>>> (i) Every element of |N_F is finite
>>>>
>>>> (ii) |N_F, like any Peano set, has cardinality aleph_0
>>>>
>>>> (iii) Between every element of |N_F and omega there are infinitely many natural numbers.
>>>>
>>>> (iv) There is no natural number between the set |N_F and omega.
>>> What is left-hand side next to omega on the ordinal line?
>> This is totally unclear. "Left-hand side" refers to driving rules, or perhaps to equations, but not to ordinal lines.
>
> The ordinal line like the real line shows increasing values from left to right.
>
>> If you are asking what is the immediate predecessor of omega, the answer is: There is none. Omega does not have a predecessor, it is a "limit ordinal".
>
> Limits destroy set theory. If you cannot count all natural numbers then countability is nonsense. Therefore you have to count until only omega remains.

Your convoluted thinking has got you in a very small box. "Therefore you have to count until only omega remains."

>
>>>> (iv) is true, not because there is an element of |N_F which is not separated from omega by an infinite set of natural numbers, but because there is no largest natural number.
>>>>
>>> Please explain with that model what elements are in all infinitely many infinite endsegments, the intersection of which is empty.
>> This is easy: Every end segment E(n) contains infinitely many natural numbers, but never n - 1.
>
> If you collect only endsegments which do not have an empty intersection together (collect as long as the intersection is infinite, then stop) then you get
> |∩{E(k) : k ∈ ℕ_def}| = ℵ₀ .

"collect" is not a math term. Define what your "collect" means in terms of Math.

your equation has 2 errors, please correct them.

>
> What gives
> |∩{E(k) : k ∈ ℕ}| = 0 ?

what number is in every endsegments ? None.

>
> Regards, WM
>

[Gus Gassmann]

On Sunday, 7 November 2021 at 15:24:20 UTC-4, WM wrote:
> Gus Gassmann schrieb am Sonntag, 7. November 2021 um 14:04:27 UTC+1:
> > On Sunday, 7 November 2021 at 07:56:59 UTC-4, WM wrote:
> > > William schrieb am Samstag, 6. November 2021 um 22:42:36 UTC+1:
> > > > The following statements of about the set |N_F are all true
> > > >
> > > > (i) Every element of |N_F is finite
> > > >
> > > > (ii) |N_F, like any Peano set, has cardinality aleph_0
> > > >
> > > > (iii) Between every element of |N_F and omega there are infinitely many natural numbers.
> > > >
> > > > (iv) There is no natural number between the set |N_F and omega.
> > > What is left-hand side next to omega on the ordinal line?
> > This is totally unclear. "Left-hand side" refers to driving rules, or perhaps to equations, but not to ordinal lines.
> The ordinal line like the real line shows increasing values from left to right.

So what, sirra, is the left-hand side of the ordinal line? I'd say, there is no better description than 1 (or 0 if you prefer).

> > If you are asking what is the immediate predecessor of omega, the answer is: There is none. Omega does not have a predecessor, it is a "limit ordinal".
> Limits destroy set theory. If you cannot count all natural numbers then countability is nonsense. Therefore you have to count until only omega remains.

As usual you try to shoehorn a natural language interpretation onto the term "countable". Cantor made very clear that a. this does not work and b. what needs to be used instead. I am not going to trot all of this out again here.

> > > > (iv) is true, not because there is an element of |N_F which is not separated from omega by an infinite set of natural numbers, but because there is no largest natural number.
> > > >
> > > Please explain with that model what elements are in all infinitely many infinite endsegments, the intersection of which is empty.
> > This is easy: Every end segment E(n) contains infinitely many natural numbers, but never n - 1.
> If you collect only endsegments which do not have an empty intersection together (collect as long as the intersection is infinite, then stop) then you get
> |∩{E(k) : k ∈ ℕ_def}| = ℵ₀ .

You do not. You require a proof that ℕ_def is finite. You can't even supply that. (In fact, ℕ_def = ℕ and thus has cardinality ℵ₀.) In addition, you studiously avoid talking about the nub of the problem:

>> Every end segment E(n) contains infinitely many natural numbers, but never n - 1.

The intersection of all end segments *CAN* be (and (*IS*) empty, even though each end segment has cardinality ℵ₀, inclusion monotony and all other smokescreens you have thrown up over the years notwithstanding.

> What gives
> |∩{E(k) : k ∈ ℕ}| = 0 ?

This is a trivial consequence of the fact that ℕ has cardinality ℵ₀.

[WM]

Gus Gassmann schrieb am Sonntag, 7. November 2021 um 22:21:25 UTC+1:
> On Sunday, 7 November 2021 at 15:24:20 UTC-4, WM wrote:
> > Gus Gassmann schrieb am Sonntag, 7. November 2021 um 14:04:27 UTC+1:
> > > On Sunday, 7 November 2021 at 07:56:59 UTC-4, WM wrote:
> > > > William schrieb am Samstag, 6. November 2021 um 22:42:36 UTC+1:
> > > > > The following statements of about the set |N_F are all true
> > > > >
> > > > > (i) Every element of |N_F is finite
> > > > >
> > > > > (ii) |N_F, like any Peano set, has cardinality aleph_0
> > > > >
> > > > > (iii) Between every element of |N_F and omega there are infinitely many natural numbers.
> > > > >
> > > > > (iv) There is no natural number between the set |N_F and omega.
> > > > What is left-hand side next to omega on the ordinal line?
> > > This is totally unclear. "Left-hand side" refers to driving rules, or perhaps to equations, but not to ordinal lines.
> > The ordinal line like the real line shows increasing values from left to right.
> So what, sirra, is the left-hand side of the ordinal line? I'd say, there is no better description than 1 (or 0 if you prefer).

The question is different from what you pretended to have understood: What is left-hand side next to omega on the ordinal line?

> > > If you are asking what is the immediate predecessor of omega, the answer is: There is none. Omega does not have a predecessor, it is a "limit ordinal".
> > Limits destroy set theory. If you cannot count all natural numbers then countability is nonsense. Therefore you have to count until only omega remains.
> As usual you try to shoehorn a natural language interpretation onto the term "countable".

No. ∀n ∈ ℕ is a clear and formal statement.

Cantor made very clear that a. this does not work and b. what needs to be used instead.

Yes, he used natural language because he had no formal yet: "eine solche gegenseitig eindeutige und vollständige Korrespondenz hergestellt [...] irgendeine gegenseitig eindeutige und vollständige Zuordnung der beiden Mengen [...] auch eine gegenseitig eindeutige und vollständige Korrespondenz" [Cantor, p. 415] In case of enumerating this means all natnural numbers are required until omega.

> I am not going to trot all of this out again here.

You know that you are wrong.

> > If you collect only endsegments which do not have an empty intersection together (collect as long as the intersection is infinite, then stop) then you get
> > |∩{E(k) : k ∈ ℕ_def}| = ℵ₀ .
> You do not. You require a proof that ℕ_def is finite.

No, I use what there is. ℕ_def is not finite but |ℕ_def| < ℵ₀ .

> You can't even supply that. (In fact, ℕ_def = ℕ and thus has cardinality ℵ₀.)

Look, if you collect only those endsegments which give an infinite intersection, then the cardinality cannot be ℵ₀.

> >> Every end segment E(n) contains infinitely many natural numbers, but never n - 1.
> The intersection of all end segments *CAN* be (and (*IS*) empty, even though each end segment has cardinality ℵ₀, inclusion monotony and all other smokescreens

> Inclusion monotony is so straight and simple that it cannot be denunciated as wrong by fools of matheology.

> > What gives
> > |∩{E(k) : k ∈ ℕ}| = 0 ?
> This is a trivial consequence of the fact that ℕ has cardinality ℵ₀.

You should try to understand! All definable numbers are lost but all endsegments with empty intersection contain infinitely many natnumbers. Which numbers remain when the intersection is empty?

Regards, WM

[WM]

If you don't know it, try to learn it without bothering me. But it is not so important because all endsegments have an empty intersection, that is, they do not contain any natural number in common, but nevertheless all endsegments are infinite, that is, they all contain infinitely many natural numbers each. What numbers are that?

Regards, WM

[WM]

William schrieb am Samstag, 6. November 2021 um 22:42:36 UTC+1:
>

> (iii) Between every element of |N_F and omega there are infinitely many natural numbers.
>
> (iv) There is no natural number between the set |N_F and omega.

How can that be? Every set of definable natnumbers reaches exactly as far as its elements: Yes, between every definable element and omega there are infinitely many natural numbers. There is a reason however, why the properties of infinite sets differ from the properties of their elements and why sets reach farther, until omega. Guess why.

Regards, WM

[Gus Gassmann]

On Sunday, 7 November 2021 at 17:45:28 UTC-4, WM wrote:
> Gus Gassmann schrieb am Sonntag, 7. November 2021 um 22:21:25 UTC+1:
> > On Sunday, 7 November 2021 at 15:24:20 UTC-4, WM wrote:
> > > Gus Gassmann schrieb am Sonntag, 7. November 2021 um 14:04:27 UTC+1:
> > > > On Sunday, 7 November 2021 at 07:56:59 UTC-4, WM wrote:

[...]

> > > > > What is left-hand side next to omega on the ordinal line?
> > > > This is totally unclear. "Left-hand side" refers to driving rules, or perhaps to equations, but not to ordinal lines.
> > > The ordinal line like the real line shows increasing values from left to right.
> > So what, sirra, is the left-hand side of the ordinal line? I'd say, there is no better description than 1 (or 0 if you prefer).
> The question is different from what you pretended to have understood: What is left-hand side next to omega on the ordinal line?

As I said: You do not know how to express yourself properly. However, the answer is equally easy: three dots ("...") or perhaps a squiggly or zigzagging line ("/\/\/\").

> > > > If you are asking what is the immediate predecessor of omega, the answer is: There is none. Omega does not have a predecessor, it is a "limit ordinal".
> > > Limits destroy set theory. If you cannot count all natural numbers then countability is nonsense. Therefore you have to count until only omega remains.
> > As usual you try to shoehorn a natural language interpretation onto the term "countable".
> No. ∀n ∈ ℕ is a clear and formal statement.

That is decidedly not what you wrote before. You nonsensically opined that "[i]f you cannot count all natural numbers then countability is nonsense".

> Cantor made very clear that a. this does not work and b. what needs to be used instead.
> Yes, he used natural language because he had no formal yet: "eine solche gegenseitig eindeutige und vollständige Korrespondenz hergestellt [...] irgendeine gegenseitig eindeutige und vollständige Zuordnung der beiden Mengen [...] auch eine gegenseitig eindeutige und vollständige Korrespondenz" [Cantor, p. 415] In case of enumerating this means all natnural numbers are required until omega.
> > I am not going to trot all of this out again here.
> You know that you are wrong.

I truly know no such thing. What I *DO* know is that you have not presented *ONE* coherent argument in this forum. Trying to simply deny this without actually presenting a coherent argument strikes me as self-defeating.

> > > If you collect only endsegments which do not have an empty intersection together (collect as long as the intersection is infinite, then stop) then you get
> > > |∩{E(k) : k ∈ ℕ_def}| = ℵ₀ .
> > You do not. You require a proof that ℕ_def is finite.
> No, I use what there is. ℕ_def is not finite but |ℕ_def| < ℵ₀ .

This is clearly about as daft as can be. There is no concept of "slightly infinite" or " almost infinite but not quite". (At least in ZFC there is not, but if you try to show ZFC inconsistent, you will have to prove two contradictory statements *WITHIN* ZFC. Since nothing you have done has come even close to showing even the most rudimentary understanding of the issues, I am not holding my breath that there will be anything substantive from you in the next, oh, omega years.)

> > You can't even supply that. (In fact, ℕ_def = ℕ and thus has cardinality ℵ₀.)
> Look, if you collect only those endsegments which give an infinite intersection, then the cardinality cannot be ℵ₀.

Granted, but that is not using the set ℕ_def. How hard can that be to comprehend?

> > >> Every end segment E(n) contains infinitely many natural numbers, but never n - 1.
> > The intersection of all end segments *CAN* be (and (*IS*) empty, even though each end segment has cardinality ℵ₀, inclusion monotony and all other smokescreens

> > Inclusion monotony is so straight and simple that it cannot be denunciated as wrong by fools of matheology.

The level of indentation is again wrong. This line is your crap. I can assure you that I would never use the word "denunciated". Cute portmanteau, but nonetheless ridiculous.

> > > What gives
> > > |∩{E(k) : k ∈ ℕ}| = 0 ?
> > This is a trivial consequence of the fact that ℕ has cardinality ℵ₀.
> You should try to understand! All definable numbers are lost but all endsegments with empty intersection contain infinitely many natnumbers. Which numbers remain when the intersection is empty?

Several things about this mishmash of ideas: First, nothing is ever "lost" from an intersection. Second, the intersection contains exactly those numbers that are contained in every end segment. Since there are infinitely many end segments E(n), and no end segment E(n+1) contains n, the intersection is empty. In other words, it does not contain anything, and in particular it doesn't contain numbers.

[Gus Gassmann]

On Sunday, 7 November 2021 at 18:17:06 UTC-4, WM wrote:
> [...] Yes, between every definable element and omega there are infinitely many natural numbers. There is a reason however, why the properties of infinite sets differ from the properties of their elements and why sets reach farther, until omega. Guess why.

Aside from the fact that sets do not "reach", I'd say a simple reason is that omega is a limit ordinal.

[FromTheRafters]

WM presented the following explanation :

Proper subsets. Can you prove that they are natural numbers as you
claim they are? Can you show how the predecessor function connects each
element to the initial element, zero or one, your choice?

[zelos...@gmail.com]

so just N

[William]

On Sunday, November 7, 2021 at 3:27:31 PM UTC-4, WM wrote:
> William schrieb am Sonntag, 7. November 2021 um 14:08:04 UTC+1:
> > On Sunday, November 7, 2021 at 7:56:59 AM UTC-4, WM wrote:
> > > William schrieb am Samstag, 6. November 2021 um 22:42:36 UTC+1:
> > > > The following statements of about the set |N_F are all true
> > > >
> > > > (i) Every element of |N_F is finite
> > > >
> > > > (ii) |N_F, like any Peano set, has cardinality aleph_0
> > > >
> > > > (iii) Between every element of |N_F and omega there are infinitely many natural numbers.
> > > >
> > > > (iv) There is no natural number between the set |N_F and omega.
> > > What is left-hand side next to omega on the ordinal line?
> > "next to omega on the left-hand side" is meaningless.
> It is not meaningless if omega exists.

Nope, An ordinal "next to omega on the left hand side" would have to be the largest element of |N_F. |N_F, like any Peano set,
does not have a last element.

Recall our starting assumption, "|N_F is a Peano set". You need to show this assumption leads to a contradiction. not to results (e.g. there is no natural number "next to omega on the left hand side") you do not like.

--
Wiliam Hughes

[William]

On Sunday, November 7, 2021 at 6:17:06 PM UTC-4, WM wrote:
> William schrieb am Samstag, 6. November 2021 um 22:42:36 UTC+1:
> >
> > (iii) Between every element of |N_F and omega there are infinitely many natural numbers.
> >
> > (iv) There is no natural number between the set |N_F and omega.
> How can that be?

The set |N_F does not have a last element.

--
William Hughes

[Serg io]

On 11/7/2021 3:51 PM, WM wrote:
> Serg io schrieb am Sonntag, 7. November 2021 um 20:33:31 UTC+1:
>> On 11/7/2021 1:24 PM, WM wrote:
>
>>> If you collect only endsegments which do not have an empty intersection together (collect as long as the intersection is infinite, then stop) then you get
>>> |∩{E(k) : k ∈ ℕ_def}| = ℵ₀ .
>> "collect" is not a math term. Define what your "collect" means in terms of Math.
>
> If you don't know it, try to learn it without bothering me.

So you admit you cannot define it in terms of math nor equations.

>But it is not so important because all endsegments have an empty intersection, that is, they do not contain any natural number in common,

you finally accept the proofs, and truth, good.

> but nevertheless all endsegments are infinite, that is, they all contain infinitely many natural numbers each. What numbers are that?

cant you answer your own question ?



>
> Regards, WM
>

[William]

On Sunday, November 7, 2021 at 5:51:35 PM UTC-4, WM wrote:
>[the set of]l endsegments [has] an empty intersection, that is, they do not contain any natural number in common,

correct.

>but nevertheless all endsegments are infinite,

Correct. This follows from an element of the set |N_F is finite, the set |N_F is infinite.

>that is, they all contain infinitely many natural numbers each. What numbers are that?

A different set for each .

--
William Hughes

[WM]

Name some elements which are not removed by the intersection:
∀k ∈ ℕ: E(k+1) = E(k) \ {k}
You know: There remains only the empty set. But nevertheless all endsegments are infinite. There remains an infinite set. May it be different for each endsegment. Name only some natnumbers which are not removed.

Regards, WM

[WM]

William schrieb am Montag, 8. November 2021 um 14:50:02 UTC+1:
> On Sunday, November 7, 2021 at 6:17:06 PM UTC-4, WM wrote:
> > William schrieb am Samstag, 6. November 2021 um 22:42:36 UTC+1:
> > >
> > > (iii) Between every element of |N_F and omega there are infinitely many natural numbers.
> > >
> > > (iv) There is no natural number between the set |N_F and omega.
> > How can that be?

It is not. There are many natural numbers between "all definable natnumbers" ans omega.

> The set |N_F does not have a last element.

Neither does the set ℕ, but its elements touch omega. That is not a contradiction because most natnumbers are dark and are not in a discernible order. The simplest picture is the set of geometrical points of unit fractions. They are all there on the real line but there is no last before zero discernible.

Regards, WM

[WM]

William schrieb am Montag, 8. November 2021 um 14:44:38 UTC+1:
> On Sunday, November 7, 2021 at 3:27:31 PM UTC-4, WM wrote:

> > > > > (iv) There is no natural number between the set |N_F and omega.
> > > > What is left-hand side next to omega on the ordinal line?
> > > "next to omega on the left-hand side" is meaningless.
> > It is not meaningless if omega exists.
> Nope, An ordinal "next to omega on the left hand side" would have to be the largest element of |N_F.

If it were definable.

> |N_F, like any Peano set,
> does not have a last element.

But omega is not arbitrarily far from the natnumbers. It is following next upon all natnumbers, the "der Größe nach zunächst folgende Zahl" (Cantor). So there is no gap. And if it were, then we could ask where it starts.

>
> Recall our starting assumption, "|N_F is a Peano set". You need to show this assumption leads to a contradiction. not to results (e.g. there is no natural number "next to omega on the left hand side") you do not like.

The Peano set is potentially infinite. Between it and omega there are almost all aleph_0 natural numbers - thise which make all endsegments infinite although all Peano numbers are lost in the sequence of endsegments

Regards, WM

[William]

On Tuesday, November 9, 2021 at 12:47:19 PM UTC-4, WM wrote:
> William schrieb am Montag, 8. November 2021 um 14:44:38 UTC+1:
> > On Sunday, November 7, 2021 at 3:27:31 PM UTC-4, WM wrote:
>
> > > > > > (iv) There is no natural number between the set |N_F and omega.
> > > > > What is left-hand side next to omega on the ordinal line?
> > > > "next to omega on the left-hand side" is meaningless.
> > > It is not meaningless if omega exists.
> > Nope, An ordinal "next to omega on the left hand side" would have to be the largest element of |N_F.
> If it were definable.

Nope, whether or not you can write it down, It would still have to be the largest element of |N_F.

>... set is potentially infinite.

Nonsense.


--
William Hughes

[WM]

FromTheRafters schrieb am Montag, 8. November 2021 um 00:59:37 UTC+1:
> WM presented the following explanation :

> > all endsegments have an empty intersection, that is, they
> > do not contain any natural number in common, but nevertheless all endsegments
> > are infinite, that is, they all contain infinitely many natural numbers each.
> > What numbers are that?
> Proper subsets.

Of course. But no element can be defined.

> Can you prove that they are natural numbers as you
> claim they are?

Endsegments contain only natural numbers.

> Can you show how the predecessor function connects each
> element to the initial element, zero or one, your choice?

It does not connect the dark natural numbers. All natural numbers connected by the successor function (which is tantamount with "being the last number of a FISON") get lost endsegment by endsegment:
∀k ∈ ℕ_def: E(k+1) = E(k) \ {k}
Nevertheless infinitely many natural numbers remain in *every* endsegment.

Regards, WM

[William]

On Tuesday, November 9, 2021 at 12:35:47 PM UTC-4, WM wrote:
>... natnumbers are dark and are not in a discernible order.

Nope, something that is not in the order is not a natural number. The fact that you cannot write a natural number down does not mean in is not in the order. "dark natural numbers" are not natural numbers. A natural number is an element of the Peano set |N_F.

--
William Hughes

[William]

On Tuesday, November 9, 2021 at 12:28:29 PM UTC-4, WM wrote:
> William schrieb am Dienstag, 9. November 2021 um 15:51:32 UTC+1:
> > On Sunday, November 7, 2021 at 5:51:35 PM UTC-4, WM wrote:
> > >[the set of]l endsegments [has] an empty intersection, that is, they do not contain any natural number in common,
> >
> > correct.
> > >but nevertheless all endsegments are infinite,
> > Correct. This follows from an element of the set |N_F is finite, the set |N_F is infinite.
> > >that is, they all contain infinitely many natural numbers each. What numbers are that?
> > A different set for each .
> Name some elements which are not removed by the intersection:

There are no elements of |N_F that are not "removed by the intersection". There is no element that in in every one of the different sets.

--
William Hughes

[William]

On Tuesday, November 9, 2021 at 12:55:26 PM UTC-4, WM wrote:
>... infinitely many natural numbers remain in *every* endsegment.

But not the sane natural numbers.

[WM]

Gus Gassmann schrieb am Sonntag, 7. November 2021 um 23:29:33 UTC+1:
> On Sunday, 7 November 2021 at 18:17:06 UTC-4, WM wrote:
> > [...] Yes, between every definable element and omega there are infinitely many natural numbers. There is a reason however, why the properties of infinite sets differ from the properties of their elements and why sets reach farther, until omega. Guess why.
>
> Aside from the fact that sets do not "reach",

Sets of ordinal numbers have an extension on the ordinal line.

> I'd say a simple reason is that omega is a limit ordinal.

Cantor does not know that the set of natnumbers must reach until omega because he only knows definable numbers. But it seems that he has recognized that there is no gap: omega follows next upon the natural numbers.

Oder aber die Zahlen β enthalten keine größte, dann besitzen sie (nach dem zweiten Erzeugungsprinzip) eine "Grenze" β', welche auf alle β zunächst folgt, (Cantor, p. 208f).

Regards, WM

[WM]

Gus Gassmann schrieb am Sonntag, 7. November 2021 um 23:26:49 UTC+1:
> On Sunday, 7 November 2021 at 17:45:28 UTC-4, WM wrote:

> > The question is different from what you pretended to have understood: What is left-hand side next to omega on the ordinal line?
> As I said: You do not know how to express yourself properly. However, the answer is equally easy: three dots ("...") or perhaps a squiggly or zigzagging line ("/\/\/\").

What comes before the three dots? Note that we are in actual infinity.

> > > As usual you try to shoehorn a natural language interpretation onto the term "countable".
> > No. ∀n ∈ ℕ is a clear and formal statement.
> That is decidedly not what you wrote before. You nonsensically opined that "[i]f you cannot count all natural numbers then countability is nonsense".

So it is.

> if you try to show ZFC inconsistent, you will have to prove two contradictory statements *WITHIN* ZFC.

It is sufficient to show that dark numbers are implied by actual infinity.

> > Look, if you collect only those endsegments which give an infinite intersection, then the cardinality cannot be ℵ₀.
> Granted, but that is not using the set ℕ_def.

There is no set but only a potentially infinite collection.

> > > >> Every end segment E(n) contains infinitely many natural numbers, but never n - 1.
> > > The intersection of all end segments *CAN* be (and (*IS*) empty, even though each end segment has cardinality ℵ₀

So it is . This proves that no numbers are staying in all endsegments and infinitely many numbers are staying in all endsegments (because there are none entering the scene). This is a contradiction if all are definable. But obviously they are not.

> > You should try to understand! All definable numbers are lost but all endsegments with empty intersection contain infinitely many natnumbers. Which numbers remain when the intersection is empty?
> Several things about this mishmash of ideas: First, nothing is ever "lost" from an intersection.

Step by step the sequece of intersections decreases:

∀k ∈ ℕ: E(k+1) = E(k) \ {k}

element by element.

> Second, the intersection contains exactly those numbers that are contained in every end segment. Since there are infinitely many end segments E(n), and no end segment E(n+1) contains n, the intersection is empty. In other words, it does not contain anything, and in particular it doesn't contain numbers.

What do all infinite endsegments contain? What natnumbers are in every endsegment but do not contribute to the intersection of all endsegments (which otherwise would be infinite).

Regards, WM

[WM]

William schrieb am Dienstag, 9. November 2021 um 18:02:51 UTC+1:
> On Tuesday, November 9, 2021 at 12:55:26 PM UTC-4, WM wrote:
> >... infinitely many natural numbers remain in *every* endsegment.
>

> But not the same natural numbers.

Name only some of your choice in any endsegment of your choice which are not lost by the mechanism proving that all are lost from the intersection:
∀k ∈ ℕ: E(k+1) = E(k) \ {k} .
Since not all are lost from the endsegments, there must infinitely many others remain in every infinite endsegment.

Regards, WM

[WM]

Yes. But infinitely many natural numbers remain in every endsegment, in all endsegments of the infinite sequence. I claim that these natural numbers are dark. Can you identify one of them?

Regards, WM

[WM]

William schrieb am Dienstag, 9. November 2021 um 17:59:48 UTC+1:
> On Tuesday, November 9, 2021 at 12:35:47 PM UTC-4, WM wrote:
> >... natnumbers are dark and are not in a discernible order.
>
> Nope, something that is not in the order is not a natural number.

It is not a definable natural number, and it is certainly not a familiar object. But without them there is no actually infinite set ℕ.

> The fact that you cannot write a natural number down does not mean in is not in the order.

There is a complete set ℕ filling the ordinal line between 0 and ω or, if you insist, some point before omega.

> "dark natural numbers" are not natural numbers.

They are not definable natural numbers, mostly.

> A natural number is an element of the Peano set |N_F.

For all these numbers we get

∀n ∈ ℕ_Peano: |ℕ \ {1, 2, 3, ..., n}| = ℵo .

It is strictly excluded by logic that they are actually infinite.

Regards, WM

[William]

On Tuesday, November 9, 2021 at 1:35:59 PM UTC-4, WM wrote:

> ...Since not all are lost from the endsegments,

Nope. One element of |N_F is "lost" from every endsegment (a different element of each endsegment). There are an infinite number of endsegments. All are "lost from the endsegments".

--
William Hughes

[WM]

But all endsegments remain infinite. What blows them up if all definable elements are lost?

Regards, WM

[William]

On Tuesday, November 9, 2021 at 1:38:48 PM UTC-4, WM wrote:
> William schrieb am Dienstag, 9. November 2021 um 18:00:01 UTC+1:
> > On Tuesday, November 9, 2021 at 12:28:29 PM UTC-4, WM wrote:
>
> > > Name some elements which are not removed by the intersection:
> > There are no elements of |N_F that are not "removed by the intersection".

> Yes. But infinitely many natural numbers remain in every endsegment

Indeed,
(iii) Between every element of |N_F and omega there is an infinite set of natural numbers (a different set for each element).

However, there is no element that in in every one of the different sets.

(iv) There is no natural number between the set |N_F and omega.

--
William Hughes

[William]

On Tuesday, November 9, 2021 at 1:45:41 PM UTC-4, WM wrote:
> William schrieb am Dienstag, 9. November 2021 um 17:59:48 UTC+1:
> > On Tuesday, November 9, 2021 at 12:35:47 PM UTC-4, WM wrote:
> > >... natnumbers are dark and are not in a discernible order.
> >
> > Nope, something that is not in the order is not a natural number.
> It is not a definable natural number,

Indeed, it is not any type of natural number.

> and it is certainly not a familiar object. But without them there is no actually infinite set ℕ.
> > The fact that you cannot write a natural number down does not mean in is not in the order.
> There is a complete set ℕ filling the ordinal line between 0 and ω or, if you insist, some point before omega.
> > "dark natural numbers" are not natural numbers.
> They are not definable natural numbers

Indeed, they are not any type of natural number.

--
William Hughes

[William]

On Tuesday, November 9, 2021 at 1:56:25 PM UTC-4, WM wrote:
> all
Ah the ambiguous "all".

--
William Hughes

[WM]

William schrieb am Dienstag, 9. November 2021 um 18:58:51 UTC+1:
> On Tuesday, November 9, 2021 at 1:38:48 PM UTC-4, WM wrote:
> > William schrieb am Dienstag, 9. November 2021 um 18:00:01 UTC+1:
> > > On Tuesday, November 9, 2021 at 12:28:29 PM UTC-4, WM wrote:
> >
> > > > Name some elements which are not removed by the intersection:
> > > There are no elements of |N_F that are not "removed by the intersection".
> > Yes. But infinitely many natural numbers remain in every endsegment
> Indeed,
> (iii) Between every element of |N_F and omega there is an infinite set of natural numbers (a different set for each element).

It is this one: ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.

The sets are not very different however, because almost all elements, namely ℵo, remain the same.

>
> However, there is no element that in in every one of the different sets.

There are almost all elements the same in these sets, because finite is much, much smaller than infinite. How could an infinite set be reduced to zero/empty in finite steps? Steps which according to
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
never touch the infinite majority?
Your claim therefore is blatantly wrong.

> (iv) There is no natural number between the set |N_F and omega.

Why does the set differ so much from its elements? Your claim unwittingly establishes dark numbers. They are what is required to make the properties of the set differ from the properties of all finite subsets of its definable elements.

Regards, WM

[William]

On Tuesday, November 9, 2021 at 2:23:54 PM UTC-4, WM wrote:

> William schrieb am Dienstag, 9. November 2021 um 18:58:51 UTC+1:

> > (iv) There is no natural number between the set |N_F and omega.
> Why does the set differ so much from its elements?

It does not "differ so much", However no element of |N_F is infinite. The set |N_F is infinite, not because it contains
an infinite element, but because, like any Peano set, it has no last element (Note, |N_F does not contain any "dark" elements but it is infinte)

--
William Hughes

[WM]

It is much smaller however than every set of aleph_0 elements. Please observe logic:
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵ₀.
That proves that *all* elements of ℕ_def of ℕ_F do not change the infinity of the remaining set of aleph_0 dark numbers. Your impression that ℕ_def could be actually infinite is simply wrong and contradicts the above statement, i.e., the simplest logic. If you consciously violate logic there is no reason to continue. If you did unconsciously, then it's time to recognize the truth.

Regards, WM

[FromTheRafters]

WM pretended :

> FromTheRafters schrieb am Montag, 8. November 2021 um 00:59:37 UTC+1:
>> WM presented the following explanation :
>>> all endsegments have an empty intersection, that is, they
>>> do not contain any natural number in common, but nevertheless all
>>> endsegments are infinite, that is, they all contain infinitely many
>>> natural numbers each. What numbers are that?
>> Proper subsets.
>
> Of course. But no element can be defined.
>
>> Can you prove that they are natural numbers as you
>> claim they are?
>
> Endsegments contain only natural numbers.

That is not a proof outside of muckymath.

[FromTheRafters]

WM brought next idea :

Until you have defined finite endsegments, you should leave out the
'infinite' part of 'infinite endsegments'. It only shows that you like
to muddy the waters.

[William]

On Tuesday, November 9, 2021 at 3:32:00 PM UTC-4, WM wrote:
> William schrieb am Dienstag, 9. November 2021 um 19:51:47 UTC+1:
> > On Tuesday, November 9, 2021 at 2:23:54 PM UTC-4, WM wrote:
> > > William schrieb am Dienstag, 9. November 2021 um 18:58:51 UTC+1:
> >
> > > > (iv) There is no natural number between the set |N_F and omega.
> > > Why does the set differ so much from its elements?
> > It does not "differ so much", However no element of |N_F is infinite. The set |N_F is infinite, not because it contains
> > an infinite element, but because, like any Peano set, it has no last element (Note, |N_F does not contain any "dark" elements but it is infinte)
> It is much smaller however than every set of aleph_0 elements.

Piffle. Every element of the set N_F heads a set that "much smaller" than a set with cardinality aleph_0. The set |N_F, like any Peano set, has cardinality aleph_0

> Please observe logic:
> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵ₀.
> That proves that *all* elements of ℕ_def of ℕ_F do not change the infinity of the remaining set of aleph_0 dark numbers.

|N_F does not contain "dark numbers".

--
William Hughes

[FromTheRafters]

on 11/9/2021, WM supposed :

> William schrieb am Montag, 8. November 2021 um 14:44:38 UTC+1:
>> On Sunday, November 7, 2021 at 3:27:31 PM UTC-4, WM wrote:
>
>>>>>> (iv) There is no natural number between the set |N_F and omega.
>>>>> What is left-hand side next to omega on the ordinal line?
>>>> "next to omega on the left-hand side" is meaningless.
>>> It is not meaningless if omega exists.
>> Nope, An ordinal "next to omega on the left hand side" would have to be the
>> largest element of |N_F.
>
> If it were definable.
>
>>> N_F, like any Peano set,
>> does not have a last element.
>
> But omega is not arbitrarily far from the natnumbers. It is following next
> upon all natnumbers, the "der Größe nach zunächst folgende Zahl" (Cantor). So
> there is no gap. And if it were, then we could ask where it starts.
>>
>> Recall our starting assumption, "|N_F is a Peano set". You need to show this
>> assumption leads to a contradiction. not to results (e.g. there is no
>> natural number "next to omega on the left hand side") you do not like.
>
> The Peano set is potentially infinite.

Stop lying.

[Gus Gassmann]

On Tuesday, 9 November 2021 at 13:19:36 UTC-4, WM wrote:
> Gus Gassmann schrieb am Sonntag, 7. November 2021 um 23:26:49 UTC+1:
> > On Sunday, 7 November 2021 at 17:45:28 UTC-4, WM wrote:
>
> > > The question is different from what you pretended to have understood: What is left-hand side next to omega on the ordinal line?
> > As I said: You do not know how to express yourself properly. However, the answer is equally easy: three dots ("...") or perhaps a squiggly or zigzagging line ("/\/\/\").
> What comes before the three dots?

A finite number of integers. Any finite set will do.

[...]

> > if you try to show ZFC inconsistent, you will have to prove two contradictory statements *WITHIN* ZFC.
> It is sufficient to show that dark numbers are implied by actual infinity.

Yeah, well, good luck with that.

[...]

> > > > The intersection of all end segments *CAN* be (and (*IS*) empty, even though each end segment has cardinality ℵ₀
> So it is . This proves that no numbers are staying in all endsegments and infinitely many numbers are staying in all endsegments (because there are none entering the scene).

I will comment on this natural language word salad. It is so bad, it is "not even wrong".

> > > You should try to understand! All definable numbers are lost but all endsegments with empty intersection contain infinitely many natnumbers. Which numbers remain when the intersection is empty?
> > Several things about this mishmash of ideas: First, nothing is ever "lost" from an intersection.
> Step by step the sequece of intersections decreases:
> ∀k ∈ ℕ: E(k+1) = E(k) \ {k}
> element by element.

Forming the intersection is not a sequential process, no matter how hard that is for you to comprehend.

> > Second, the intersection contains exactly those numbers that are contained in every end segment. Since there are infinitely many end segments E(n), and no end segment E(n+1) contains n, the intersection is empty. In other words, it does not contain anything, and in particular it doesn't contain numbers.
> What do all infinite endsegments contain? What natnumbers are in every endsegment but do not contribute to the intersection of all endsegments (which otherwise would be infinite).

Oh *SHUT* *UP*, you blithering idiot. No natural number n is an element of the end segment E(n+1). We had this, and it is tiresome to have to correct your moronic mistakes every single time. Muckenheim: Not a single coherent thought since 2005.

[zelos...@gmail.com]

tisdag 9 november 2021 kl. 17:28:29 UTC+1 skrev WM:
> William schrieb am Dienstag, 9. November 2021 um 15:51:32 UTC+1:
> > On Sunday, November 7, 2021 at 5:51:35 PM UTC-4, WM wrote:

> > >[the set of]l endsegments [has] an empty intersection, that is, they do not contain any natural number in common,
> >
> > correct.

> > >but nevertheless all endsegments are infinite,

> > Correct. This follows from an element of the set |N_F is finite, the set |N_F is infinite.

> > >that is, they all contain infinitely many natural numbers each. What numbers are that?

> > A different set for each .

> Name some elements which are not removed by the intersection:

> ∀k ∈ ℕ: E(k+1) = E(k) \ {k}

> You know: There remains only the empty set. But nevertheless all endsegments are infinite. There remains an infinite set. May it be different for each endsegment. Name only some natnumbers which are not removed.
>
> Regards, WM
why is this the hill you die on? This is so stupid that you go on and on about this

[WM]

William schrieb am Dienstag, 9. November 2021 um 20:46:07 UTC+1:
> On Tuesday, November 9, 2021 at 3:32:00 PM UTC-4, WM wrote:
> Every element of the set N_F heads a set that "much smaller" than a set with cardinality aleph_0. The set |N_F, like any Peano set, has cardinality aleph_0

Impossible as long as |ℕ \ ℕ_def| = ℵ₀.
Two consecutive infinite sets of ℵ₀ elements are impossible here.
The instantaneous vanishing of |ℕ \ ℕ_def| is not mathematics but credo in absurdum.
Therefore your claim is silly.

> > Please observe logic:
> > ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵ₀.
> > That proves that *all* elements of ℕ_def of ℕ_F do not change the infinity of the remaining set of aleph_0 dark numbers.
> |N_F does not contain "dark numbers".

No, but ℕ does. All visible numbers are restricted to ℕ_def in

∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵ₀.

Regards, WM

[WM]

Gus Gassmann schrieb am Dienstag, 9. November 2021 um 20:55:16 UTC+1:
> On Tuesday, 9 November 2021 at 13:19:36 UTC-4, WM wrote:
> > Gus Gassmann schrieb am Sonntag, 7. November 2021 um 23:26:49 UTC+1:
> > > On Sunday, 7 November 2021 at 17:45:28 UTC-4, WM wrote:
> >
> > > > The question is different from what you pretended to have understood: What is left-hand side next to omega on the ordinal line?
> > > As I said: You do not know how to express yourself properly. However, the answer is equally easy: three dots ("...") or perhaps a squiggly or zigzagging line ("/\/\/\").
> > What comes before the three dots?
> A finite number of integers. Any finite set will do.

No, the ordinal line does not change upon your order.

> > Step by step the sequece of intersections decreases:
> > ∀k ∈ ℕ: E(k+1) = E(k) \ {k}
> > element by element.
> Forming the intersection is not a sequential process, no matter how hard that is for you to comprehend.

That implies the existence of dark parts of the sequence.

> > What do all infinite endsegments contain? What natnumbers are in every endsegment but do not contribute to the intersection of all endsegments (which otherwise would be infinite).

> No natural number n is an element of the end segment E(n+1).

That is well known. You need not repeat it over and over again. The question is: What is remaining to make all endsegments infinite? And what is passing them? (Every infinite endsegment, even every non-empty endsegment is passed by some natnumbers.)

Regards, WM

[Gus Gassmann]

On Wednesday, 10 November 2021 at 12:49:08 UTC-4, WM wrote:
> Gus Gassmann schrieb am Dienstag, 9. November 2021 um 20:55:16 UTC+1:
> > On Tuesday, 9 November 2021 at 13:19:36 UTC-4, WM wrote:
> > > Gus Gassmann schrieb am Sonntag, 7. November 2021 um 23:26:49 UTC+1:
> > > > On Sunday, 7 November 2021 at 17:45:28 UTC-4, WM wrote:
> > >
> > > > > The question is different from what you pretended to have understood: What is left-hand side next to omega on the ordinal line?
> > > > As I said: You do not know how to express yourself properly. However, the answer is equally easy: three dots ("...") or perhaps a squiggly or zigzagging line ("/\/\/\").
> > > What comes before the three dots?
> > A finite number of integers. Any finite set will do.
> No, the ordinal line does not change upon your order.

Oh please. There are infinitely many natural numbers, and there is no largest. Why the hell do you think you need the three dots in the first place?

[William]

On Wednesday, November 10, 2021 at 12:42:42 PM UTC-4, WM wrote:
> William schrieb am Dienstag, 9. November 2021 um 20:46:07 UTC+1:
> > On Tuesday, November 9, 2021 at 3:32:00 PM UTC-4, WM wrote:
> > Every element of the set N_F heads a set that "much smaller" than a set with cardinality aleph_0. The set |N_F, like any Peano set, has cardinality aleph_0
> Impossible as long as |ℕ \ ℕ_def| = ℵ₀.

The finite set of natural numbers that can be written down, |N_def, is irrelevant as is your set |N which is not the set of natrural numbers.
(Your set |N is not a Peano set, the set of natural numbers is a Peano set).

--
William Hughes

[zelos...@gmail.com]

Is false, given all definitions you have given for N_def, N_def=N

[WM]

They represent some definable and all undefinable finite ordinals.

Set theorists confess that every natural number n has an infinite distance to ω
∀n ∈ ℕ: |ℕ \ {1, 2, 3, ..., n}| = ℵo
but that the set ℕ has no distance to ω. So either the distance is removed by some magic action when forming the set, or undefinable numbers are remaining and are simply included as individually unspecified elements in "the infinite set ℕ". These are the alternatives known to me. Are there more?

Regards, WM

[WM]

zelos...@gmail.com schrieb am Donnerstag, 11. November 2021 um 07:09:19 UTC+1:
> onsdag 10 november 2021 kl. 17:42:42 UTC+1 skrev WM:

> > All visible numbers are restricted to ℕ_def in
> > ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵ₀.
>

> >Impossible as long as |ℕ \ ℕ_def| = ℵ₀.

> Is false, given all definitions you have given for N_def, N_def=N

[WM]

But all Peano natural numbers n have an infinite distance to ω
∀n ∈ ℕ: |ℕ \ {1, 2, 3, ..., n}| = ℵo
The Peano set ℕ has no distance to ω. So either the distance is removed by some magic action when forming the set, or undefinable numbers are remaining and are simply included as individually unspecified elements in "the infinite Peano set ℕ". These are the alternatives known to me. Are there more?

Regards, WM

[Gus Gassmann]

On Thursday, 11 November 2021 at 07:28:59 UTC-4, WM wrote:
> Gus Gassmann schrieb am Mittwoch, 10. November 2021 um 18:46:04 UTC+1:
> > On Wednesday, 10 November 2021 at 12:49:08 UTC-4, WM wrote:
>
> > > No, the ordinal line does not change upon your order.
> > Oh please. There are infinitely many natural numbers, and there is no largest. Why the hell do you think you need the three dots in the first place?
> They represent some definable and all undefinable finite ordinals.

All natural numbers have been defined, hence are "definable". Your objection is risible.

> Set theorists confess that every natural number n has an infinite distance to ω
> ∀n ∈ ℕ: |ℕ \ {1, 2, 3, ..., n}| = ℵo
> but that the set ℕ has no distance to ω. So either the distance is removed by some magic action when forming the set, or undefinable numbers are remaining and are simply included as individually unspecified elements in "the infinite set ℕ". These are the alternatives known to me. Are there more?

Given your utter ignorance of mathematics, the answer is, of course there are more. You don't even for a moment consider that the "magic action" you invoke might be the simple taking of a limit, do you? How stupid does one have to be before Hochschule Augsburg considers one suitable for a professorship?

[William]

On Thursday, November 11, 2021 at 7:32:04 AM UTC-4, WM wrote:
> William schrieb am Mittwoch, 10. November 2021 um 22:00:53 UTC+1:
> > On Wednesday, November 10, 2021 at 12:42:42 PM UTC-4, WM wrote:
> > > William schrieb am Dienstag, 9. November 2021 um 20:46:07 UTC+1:
> > > > On Tuesday, November 9, 2021 at 3:32:00 PM UTC-4, WM wrote:
> > > > Every element of the set N_F heads a set that "much smaller" than a set with cardinality aleph_0. The set |N_F, like any Peano set, has cardinality aleph_0
> > > Impossible as long as |ℕ \ ℕ_def| = ℵ₀.
> > The finite set of natural numbers that can be written down, |N_def, is irrelevant as is your set |N which is not the set of natrural numbers.
> > (Your set |N is not a Peano set, the set of natural numbers is a Peano set).

> But all [elements of a Peano set] n have an infinite distance to ω

> ∀n ∈ ℕ: |ℕ \ {1, 2, 3, ..., n}| = ℵo

Nope, you have used your set |N which is not a Peano set. |N_F us a Peano set. What you want is

∀n ∈ |N_F: ||N_F\ {1, 2, 3, ..., n}| = ℵo

Indeed,. Note that ∀n ∈ |N_F: {1, 2, 3, ..., n} has a last element, this is a statement that is true for every element of the set |N_F. The set |N_F does not have a lase element.

> The Peano set ℕ has no distance to ω. So either the distance is removed by some magic action when forming the set, or undefinable numbers are remaining and are simply included as individually unspecified elements in "the infinite Peano set ℕ"

Your ℕ is not a Peano set. A Peano set does not have "unspecified" (your latest way of saying dark) elements.

--
William Hughes

[Serg io]

On 11/9/2021 11:08 AM, WM wrote:
> Gus Gassmann schrieb am Sonntag, 7. November 2021 um 23:29:33 UTC+1:
>> On Sunday, 7 November 2021 at 18:17:06 UTC-4, WM wrote:
>>> [...] Yes, between every definable element and omega there are infinitely many natural numbers. There is a reason however, why the properties of infinite sets differ from the properties of their elements and why sets reach farther, until omega. Guess why.
>>
>> Aside from the fact that sets do not "reach",
>
> Sets of ordinal numbers have an extension on the ordinal line.
>
>> I'd say a simple reason is that omega is a limit ordinal.
>
> Cantor does not know that the set of natnumbers must reach until omega because he only knows definable numbers.

wrong Cantor is dead, long time.

> But it seems that he has recognized that there is no gap: omega follows next upon the natural numbers.

your conjecture.

[zelos...@gmail.com]

Yes, the alternative which is the accurate is that you are a crank, an idiot, and you do not understand mathematics!

[WM]

Gus Gassmann schrieb am Donnerstag, 11. November 2021 um 13:35:05 UTC+1:

> All natural numbers have been defined, hence are "definable".

Then this is a contradiction because of ∀k ∈ ℕ: E(k+1) = E(k) \ {k} .

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

Regards, WM

[Transfinity]

William schrieb am Donnerstag, 11. November 2021 um 16:19:09 UTC+1:

> Your ℕ is not a Peano set. A Peano set does not have "unspecified" (your latest way of saying dark) elements.

Then your Peano set causes a contradiction:

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

Note that ∀k ∈ ℕ: E(k+1) = E(k) \ {k} .

Regards, WM

[William]

On Friday, November 12, 2021 at 4:42:23 PM UTC-5, Transfinity wrote:
> William schrieb am Donnerstag, 11. November 2021 um 16:19:09 UTC+1:
>
> > Your ℕ is not a Peano set. A Peano set does not have "unspecified" (your latest way of saying dark) elements.
> Then your Peano set causes a contradiction:

You say something irrelevant about your set |N which is not a Peano set and is not the set of natural numbers. What you want is

∀ n∈|N_F ∃ M(n)⊂|N_F: M(n)⊂E(n) ∧ |M(n)| = ℵo

Note that the set M(n) may change when the element of |N_F changes.

~∃ Q⊂|N_F ∀ n∈|N_F : Q⊂E(n) ∧ |Q| = ℵo .

No contradiction. For every element, n, of the set |N_F, there exits the set M(n) that depends on the element of |N_F.
There is no set, Q, that does not depend on the element of |N_F.


--
William Hughes

[WM]

William schrieb am Freitag, 12. November 2021 um 23:34:32 UTC+1:
> On Friday, November 12, 2021 at 4:42:23 PM UTC-5, Transfinity wrote:
> > William schrieb am Donnerstag, 11. November 2021 um 16:19:09 UTC+1:
> >
> > > Your ℕ is not a Peano set. A Peano set does not have "unspecified" (your latest way of saying dark) elements.
> > Then your Peano set causes a contradiction:
> You say something irrelevant about your set |N which is not a Peano set and is not the set of natural numbers. What you want is
>
> ∀ n∈|N_F ∃ M(n)⊂|N_F: M(n)⊂E(n) ∧ |M(n)| = ℵo
>
> Note that the set M(n) may change when the element of |N_F changes.

It may but need not change. But importantly every set M(n) is infinite |M(n)| = ℵo.

>
> ~∃ Q⊂|N_F ∀ n∈|N_F : Q⊂E(n) ∧ |Q| = ℵo .
>

> No contradiction. For every element, n, of the set |N_F, there exists the set M(n) that depends on the element of |N_F.

That is said in the first statement already, not in the second. The second statement says that no infinite set M (you changed its name to Q but that does not change the contents of the statement) existst outside of |N_F - in fact not even one single element.

> There is no set, Q, that does not depend on the element of |N_F.

No infinite set outside of |N_F can be existing according to the second statement, but there is an infinite set outside of |N_F according to the first statement. This is a contradiction if |N_F is the same in both cases.

Regards, WM

[William]

On Saturday, November 13, 2021 at 4:22:51 AM UTC-5, WM wrote:

> No infinite set
that does not change exists according to the second statement. There is an infinite set that changes
(i.e. a different infinite set for each element of |N_F) according to the first statement. No contradiction.

--
William Hughes

[zelos...@gmail.com]

That causes no contradition.

>∀ n∈ℕ ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo

NEITHER DOES THIS

>~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| = ℵo .

You can just say ~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n)

[zelos...@gmail.com]

Those are not contradictions

[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

[William]

On Saturday, November 13, 2021 at 1:05:59 PM UTC-5, WM wrote:
> (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

There is no fixed subset.

> That means natural numbers can only be chosen from ℕ\M

Nope you are using your set |N which is not a Peano set and M which does not exist. This is nonsense.

--
William Hughes

[WM]

William schrieb am Samstag, 13. November 2021 um 19:19:12 UTC+1:
> On Saturday, November 13, 2021 at 1:05:59 PM UTC-5, WM wrote:
> > (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
> There is no fixed subset.

That is irrelevant, as long all sets are infinite. Why should it be of any relevance? Can you tell me? In every case natural numbers can only be chosen from ℕ\M which is a finite set. Or do you believe in two actually infinite concecutive sets in ℕ? Where does the second one start?

> Nope you are using your set |N which is not a Peano set and M which does not exist.

I am not using it but ℕ which exists like M according to ZF
∀ n∈ℕ ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo .

It turns out however that ℕ causes a contradiction: Between all n∈ℕ and ω there are infinitely many natnumbers. The second statement contradicts this one:

~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| = ℵo .

There is nothimg between ℕ and ω. A contradiction.

Note that the set of positive fractions smaller than (n+1)/n is infinite. It has a fixed an infinite minimum. Same with M.

Regards, WM

[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

[William]

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. Nonsense.

--
William Hughes

[WM]

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.

Again: Why do you deny that every M, even if not constant but never smaller than ℵo, when subtracted from this genuine ℕ restricts the set ℕ\M to be finite?

Note that every set of positive fractions smaller than (n+1)/n is infinite. It has a fixed an infinite minimum. Same with M. But M contains ℵo natural numbers. Hence |ℕ\M| < ℵo for every possible M.

Regards, WM

[FromTheRafters]

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.

[William]

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


--
William Hughes

[Python]

crank Wolfgang Mückenheim, aka WM wrote:
> 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.

ℵo is not a subset of N, crank Wolfgang Mückenheim, from Hochschule
Augsburg. In ∀ n∈ℕ_def ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo, M cannot be fixed
(i.e. independant of n)

[WM]

The natural numbers are defined collectively. But most are not defined individually. The difference is clearly visible here:

∀ n∈ℕ_def ∃ M⊂ℕ: M⊂E(n) ∧ |M| = ℵo

~∃ M⊂ℕ ∀ n∈ℕ : M⊂E(n) ∧ |M| = ℵo .

Regards, WM

[WM]

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

[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

[William]

> 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

[WM]

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

[Python]

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.

[William]

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

[FromTheRafters]

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.

[FromTheRafters]

WM laid this down on his screen :

Stop lying and trying to insult your betters.

[WM]

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

[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

[Mostowski Collapse]

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.

[Mostowski Collapse]

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]

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]

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:

[William]

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

[Serg io]

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.

[WM]

What makes it differ from the Peano set in your opinion?

Regards, WM

[zelos...@gmail.com]

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

[zelos...@gmail.com]

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.

[William]

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

[WM]

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

[WM]

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

[zelos...@gmail.com]

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.

[zelos...@gmail.com]

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.

[William]

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