On 12/26/2021 9:05 AM, WM wrote:
> William schrieb am Samstag, 25. Dezember 2021 um 17:23:11 UTC+1:
>> On Saturday, December 25, 2021 at 6:21:04 AM UTC-4, WM wrote:
>
>>> All FISONs are an infinite sequence
>> The set of FISONs is ordered and infinite, so the set o FISONs is also an infinite sequence.
>> As every element of the set of natural numbers is the largest element of a FISON (trivial induction), when you subtract the union of all elements of the infinite -sequence of FISONs from the natural numbes you get the empty set.
>
> Maybe. But every FISON, when individually subtracted from ℕ, leaves ℵ₀ natnumbers.

no. It leaves an endsegment. You use sets and elements interchangeable, which is error.

> Therefore the union or sequence or set of all FISONs, when subtracted from ℕ, leaves less - nothing as you say.

so you agree the union of all FISONs is ℕ.

> This proves: By collectively subtracting, more is subtracted than by individually subtracting.

no. Nothing was proved.

you are using your weasel words again, collective => infinite individual => finite

>
>>> and all FISONs, when subtracted from ℕ, leave ℵ₀ natnumbers.
>> Nope you are confusing "all FISONs", the infinite sequence of FISONs, with the elements of the infinite sequence of FISONs.
>
> Only the elements which can be treated individually will leave ℵo numbers:

"treated" is not a math term. do you mean removed from a set ?

> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo
> The infinite sequence as collection or set does not leave anything.
>
> Now we have fixed the clear distinction between natnumbers which can be handled individually and the infinite dark remainder.

which is....

>
> Regards, WM

William schrieb am Sonntag, 26. Dezember 2021 um 17:52:56 UTC+1:
> On Sunday, December 26, 2021 at 11:05:31 AM UTC-4, WM wrote:
> >... By collectively subtracting, more is subtracted than by individually subtracting.
>
> Collectively subtracting,: subtracting an infinite set.
> Individually subtracting,: subtracting a finite set.
> Indeed, the two are different,

Fine.
> However the Peano set of natural numbers does not contain dark elements.
>
But there's this disturbing little difference.

ℕ contains numbers almost all of which cannot be handled individually. Proof: Never anyone has or will have individually treated more than the elements of a finite intial segment. And what is more important: Never could this be done because the Peano set of natural numbers contains elements which never could be shown to belong to the second half of the complete well-ordered set. (Note: If all elements exist well-ordered, then also two halves exist, unless almost are dark.)

Regards, WM

On 12/27/2021 6:24 AM, WM wrote:
> William schrieb am Sonntag, 26. Dezember 2021 um 17:52:56 UTC+1:
>> On Sunday, December 26, 2021 at 11:05:31 AM UTC-4, WM wrote:
>>> ... By collectively subtracting, more is subtracted than by individually subtracting.
>>
>> Collectively subtracting,: subtracting an infinite set.
>> Individually subtracting,: subtracting a finite set.
>> Indeed, the two are different,
>
> Fine.
>
>> However the Peano set of natural numbers does not contain dark elements.
>>
> But there's this disturbing little difference.
>
> ℕ contains numbers almost all of which cannot be handled individually.

Wrong. Any number in ℕ can be "handled individually", each has their own individual name/number.

>Proof: Never anyone has or will have individually treated more than the elements of a finite intial segment.

Wrong. FISON(1) was "individually treated" when the very first sheep was counted.

> And what is more important: Never could this be done because the Peano set of natural numbers contains elements which never could be shown to belong to the second half of the complete well-ordered set.

You need to publish, expose and report such "second half of the complete well-ordered set."

> (Note: If all elements exist well-ordered, then also two halves exist, unless almost are dark.)

What is the meaning of "almost are dark", is that gray or...

>
> Regards, WM

On Monday, December 27, 2021 at 8:24:20 AM UTC-4, WM wrote:
> William schrieb am Sonntag, 26. Dezember 2021 um 17:52:56 UTC+1:
> > On Sunday, December 26, 2021 at 11:05:31 AM UTC-4, WM wrote:
> > >... By collectively subtracting, more is subtracted than by individually subtracting.
> >
> > Collectively subtracting,: subtracting an infinite set.
> > Individually subtracting,: subtracting a finite set.
> > Indeed, the two are different,
> Fine.
> > However the Peano set of natural numbers does not contain dark elements /
> >
> But there's this disturbing little difference.
>
> ℕ contains numbers almost all of which cannot be handled individually.

Nope, "handled individually" means finite, and all natural numbers. elements of the Peano set N_P, are finite (trivial induction).

> <skip> ...second half

"half" is meaningless when applied to Peano sets. A Peano set does not consist of a first and a second half. (trivial induction.
Assume there is a "first half". Then 1 is in the first half. If n is in the first half then n+1 is in the first half, By induction all elements of the Peano set are in the first half, Contradiction. Thus "first half" is meaningless.

--
William Hughes

William schrieb am Dienstag, 28. Dezember 2021 um 04:00:07 UTC+1:
> On Monday, December 27, 2021 at 8:24:20 AM UTC-4, WM wrote:
> > William schrieb am Sonntag, 26. Dezember 2021 um 17:52:56 UTC+1:
> > > On Sunday, December 26, 2021 at 11:05:31 AM UTC-4, WM wrote:
> > > >... By collectively subtracting, more is subtracted than by individually subtracting.
> > >
> > > Collectively subtracting,: subtracting an infinite set.
> > > Individually subtracting,: subtracting a finite set.
> > > Indeed, the two are different,
> > Fine.
> > > However the Peano set of natural numbers does not contain dark elements /
> > >
> > But there's this disturbing little difference.
> >
> > ℕ contains numbers almost all of which cannot be handled individually.
> Nope, "handled individually" means finite, and all natural numbers. elements of the Peano set N_P, are finite (trivial induction).

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

> "half" is meaningless when applied to Peano sets. A Peano set does not consist of a first and a second half. (trivial induction.

"Complete" is meaningless when applied to Peano sets. A Peano set is no set.. Assume it is complete. That means no element is missing. Double each element, then you get twice as many which before had been considered complete.
> Assume there is a "first half". Then 1 is in the first half. If n is in the first half then n+1 is in the first half, By induction all elements of the Peano set are in the first half, Contradiction. Thus "first half" is meaningless.

Assume there is ℵo. Then there is everything in the first ℵo/n for every definable n.

Regards, WM

On Tuesday, December 28, 2021 at 6:22:50 AM UTC-4, WM wrote:
> William schrieb am Dienstag, 28. Dezember 2021 um 04:00:07 UTC+1:

>
> > "half" is meaningless when applied to Peano sets. A Peano set does not consist of a first and a second half. (trivial induction.
> "Complete" is meaningless when applied to Peano sets.
Nope. A Peano set exists. Thus a "complete" set without last element exists.

> A Peano set is no set. Assume it is complete That means no element is missing.

Indeed, an element of the natural numbers is an element of N_P

>Double each element, then you get

A proper subset with cardinality aleph_0 (the even numbers). Each element of the natural numbers, when doubled, is a natural number
and thus an element of N_P (this is a consequence of the fact that N_P has no last element). Doubling each element does not produce new elements .

--
William Hughes

William schrieb am Dienstag, 28. Dezember 2021 um 17:24:33 UTC+1:
> On Tuesday, December 28, 2021 at 6:22:50 AM UTC-4, WM wrote:
> > William schrieb am Dienstag, 28. Dezember 2021 um 04:00:07 UTC+1:
>
> >
> > > "half" is meaningless when applied to Peano sets. A Peano set does not consist of a first and a second half. (trivial induction.
> > "Complete" is meaningless when applied to Peano sets.
> Nope. A Peano set exists. Thus a "complete" set without last element exists.

Sorry, you can't believe in logic?

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

Or why don't you understand that this refutes your claim?

What happens when subtracting the union of FISONs from ℕ? There are two alternatives:
(1) Either also almost all natnumbers remain. This would prove that they are dark. So dark numbers exist.
(2) Or no natnumbers remain. Then these natnumbers can be subtracted collectively (by the union of FISONs) but not individually (by any FISON). These natnumbers are dark.
Therefore both alternatives prove the existence of dark numbers. Therefore there is no full power set. Alas this prevents bijections between infinite sets at all.

> > A Peano set is no set. Assume it is complete. That means no element is missing.
>
> Indeed, an element of the natural numbers is an element of N_P
> >Double each element, then you get
> A proper subset with cardinality aleph_0 (the even numbers). Each element of the natural numbers, when doubled, is a natural number
> and thus an element of N_P (this is a consequence of the fact that N_P has no last element). Doubling each element does not produce new elements .

I meant not only to consider the doubled numbers but also those between them. Then you have twice as many as before. Have you created them? No! Before they have been existing too, but they were dark.

Regards, WM

On 12/29/2021 4:42 AM, WM wrote:
> William schrieb am Dienstag, 28. Dezember 2021 um 17:24:33 UTC+1:
>> On Tuesday, December 28, 2021 at 6:22:50 AM UTC-4, WM wrote:
>>> William schrieb am Dienstag, 28. Dezember 2021 um 04:00:07 UTC+1:
>>
>>>
>>>> "half" is meaningless when applied to Peano sets. A Peano set does not consist of a first and a second half. (trivial induction.
>>> "Complete" is meaningless when applied to Peano sets.
>> Nope. A Peano set exists. Thus a "complete" set without last element exists.
>
> Sorry, you can't believe in logic?
>
> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo .

that is only an Endsegment, E(n+1)

>
> Or why don't you understand that this refutes your claim?
>
> What happens when subtracting the union of FISONs from ℕ? There are two alternatives:
> (1) Either also almost all natnumbers remain. This would prove that they are dark. So dark numbers exist.
> (2) Or no natnumbers remain. Then these natnumbers can be subtracted collectively (by the union of FISONs) but not individually (by any FISON). These natnumbers are dark.
> Therefore both alternatives prove the existence of dark numbers. Therefore there is no full power set. Alas this prevents bijections between infinite sets at all.
>
>
>>> A Peano set is no set. Assume it is complete. That means no element is missing.
>>
>> Indeed, an element of the natural numbers is an element of N_P
>>> Double each element, then you get
>> A proper subset with cardinality aleph_0 (the even numbers). Each element of the natural numbers, when doubled, is a natural number
>> and thus an element of N_P (this is a consequence of the fact that N_P has no last element). Doubling each element does not produce new elements .
>
> I meant not only to consider the doubled numbers but also those between them. Then you have twice as many as before. Have you created them? No! Before they have been existing too, but they were dark.

totally unworkable, just because you are not directly looking at the number, does not mean it is "Dark"

>
> Regards, WM

On Wednesday, December 29, 2021 at 6:43:04 AM UTC-4, WM wrote:
> William schrieb am Dienstag, 28. Dezember 2021 um 17:24:33 UTC+1:
> > On Tuesday, December 28, 2021 at 6:22:50 AM UTC-4, WM wrote:
> > > William schrieb am Dienstag, 28. Dezember 2021 um 04:00:07 UTC+1:
> >
> > >
> > > > "half" is meaningless when applied to Peano sets. A Peano set does not consist of a first and a second half. (trivial induction.
> > > "Complete" is meaningless when applied to Peano sets.
> > Nope. A Peano set exists. Thus a "complete" set without last element exists.
> Sorry, you can't believe in logic?
>
>∀n ∈ ℕ_P: |ℕ_P \ {1, 2, 3, ..., n}| = ℵo .
>
> Or why don't you understand that this refutes your claim?

This is a (correct) statement that every element of the set of Natural numbers is finite. We also have
∀n ∈ ℕ_P: ℕ_P \ {1, 2, 3, ..., n} is a subset of N_P. After

>
>
> > ><snip> ... A Peano set is no set. Assume it is complete. That means no element is missing.
> >
> > Indeed, an element of the natural numbers is an element of N_P
> > >Double each element, then you get
> > A proper subset with cardinality aleph_0 (the even numbers). Each element of the natural numbers, when doubled, is a natural number
> > and thus an element of N_P (this is a consequence of the fact that N_P has no last element). Doubling each element does not produce new elements .
> I meant not only to consider the doubled numbers but also those between them. Then you have twice as many as before.

Nope, doubling a natural number produces a natural number. This procedure (taking the union of even integers with the set of integers). does not change the set. You do not have "twice as many as before" you have as many as before.

--
William Hughes

William schrieb am Mittwoch, 29. Dezember 2021 um 17:08:30 UTC+1:
> On Wednesday, December 29, 2021 at 6:43:04 AM UTC-4, WM wrote:
> > > Nope. A Peano set exists. Thus a "complete" set without last element exists.
> > Sorry, you can't believe in logic?
> >
> >∀n ∈ ℕ_P: |ℕ_P \ {1, 2, 3, ..., n}| = ℵo .
> >
> > Or why don't you understand that this refutes your claim?
> This is a (correct) statement that every element of the set of Natural numbers is finite.

For every element we have |{1, 2, 3, ..., n}|/|{n+1, n+2, n+3, ...}| = 0. Therefore there is no chance to get an actually infinite set by definable natnumbers. All provably fail.

> > > A proper subset with cardinality aleph_0 (the even numbers). Each element of the natural numbers, when doubled, is a natural number
> > > and thus an element of N_P (this is a consequence of the fact that N_P has no last element). Doubling each element does not produce new elements .
> > I meant not only to consider the doubled numbers but also those between them. Then you have twice as many as before.
> Nope, doubling a natural number produces a natural number. This procedure (taking the union of even integers with the set of integers). does not change the set. You do not have "twice as many as before" you have as many as before.

If all are doubled (such that none is missing) then your statement is wrong..

Regards, WM

On Thursday, 30 December 2021 at 11:41:18 UTC+1, WM wrote:
> William schrieb am Mittwoch, 29. Dezember 2021 um 17:08:30 UTC+1:

> > Nope, doubling a natural number produces a natural number. This procedure
>> (taking the union of even integers with the set of integers). does not change the
> > set. You do not have "twice as many as before" you have as many as before.
> If all are doubled (such that none is missing) then your statement is wrong.

You are a fucking disgrace, Wolfgang-my-ass, and yet another agent of the enemy:
thanks to you and almost two decades of your relentless spamming and shitting
nonsense across the channel, no proper and serious discussion of the patently
broken standard infinity has ever been even possible, you fucking nazi piece of
shit and agent of the enemy. ESAT, you and your retarded genocidal employers.

Really, fuck you man, fuck you those who still play with you, you fucking retarded
bastards, you stupid shameless sociopathic retarded cunts who can only pollute
each and every single fucking public corner and pond of this forgotten planet.

I have got to vouching for a mass extinction, you piece of disturbed retarded shit
will just never ever do the right thing ever...

*Plonk*

Julio

On 12/30/2021 4:41 AM, WM wrote:
> William schrieb am Mittwoch, 29. Dezember 2021 um 17:08:30 UTC+1:
>> On Wednesday, December 29, 2021 at 6:43:04 AM UTC-4, WM wrote:
>
>>>> Nope. A Peano set exists. Thus a "complete" set without last element exists.
>>> Sorry, you can't believe in logic?
>>>
>>> ∀n ∈ ℕ_P: |ℕ_P \ {1, 2, 3, ..., n}| = ℵo .
>>>
>>> Or why don't you understand that this refutes your claim?
>> This is a (correct) statement that every element of the set of Natural numbers is finite.
>
> For every element we have |{1, 2, 3, ..., n}|/|{n+1, n+2, n+3, ...}| = 0. Therefore there is no chance to get an actually infinite set by definable natnumbers. All provably fail.

see? you stop at n again. also he was addressing "elements", not sets.

>
>>>> A proper subset with cardinality aleph_0 (the even numbers). Each element of the natural numbers, when doubled, is a natural number
>>>> and thus an element of N_P (this is a consequence of the fact that N_P has no last element). Doubling each element does not produce new elements .
>>> I meant not only to consider the doubled numbers but also those between them. Then you have twice as many as before.
>> Nope, doubling a natural number produces a natural number. This procedure (taking the union of even integers with the set of integers). does not change the set. You do not have "twice as many as before" you have as many as before.
>
> If all are doubled (such that none is missing) then your statement is wrong.

you are wrong, why you always wrong?

>
> Regards, WM

On Thursday, December 30, 2021 at 6:41:18 AM UTC-4, WM wrote:
> William schrieb am Mittwoch, 29. Dezember 2021 um 17:08:30 UTC+1:
> > On Wednesday, December 29, 2021 at 6:43:04 AM UTC-4, WM wrote:
>
> > > > Nope. A Peano set exists. Thus a "complete" set without last element exists.
> > > Sorry, you can't believe in logic?
> > >
> > >∀n ∈ ℕ_P: |ℕ_P \ {1, 2, 3, ..., n}| = ℵo .
> > >
> > > Or why don't you understand that this refutes your claim?
> > This is a (correct) statement that every element of the set of Natural numbers is finite.
> For every element we have |{1, 2, 3, ..., n}|/|{n+1, n+2, n+3, ...}| = 0.

Correct, Each element of N_P is finite. However, N_P is infinite

> > > <snip> ...I meant not only to consider the doubled numbers but also those between them. Then you have twice as many as before.
> > Nope, doubling a natural number produces a natural number. This procedure (taking the union of even integers with the set of integers). does not change the set. You do not have "twice as many as before" you have as many as before.
> If all are doubled (such that none is missing) then your statement is wrong.

Nope. Doubling every element of the natural numbers produces a subset of the natural numbers. (Proof by induction. Doubling 1 produces a natural number. If doubling n produces a natural number, doubling (n+1) produces a natural number)

--
William Hughes

On Thursday, December 30, 2021 at 6:59:12 AM UTC-8, Serg io wrote:
> On 12/30/2021 4:41 AM, WM wrote:
> > William schrieb am Mittwoch, 29. Dezember 2021 um 17:08:30 UTC+1:
> >> On Wednesday, December 29, 2021 at 6:43:04 AM UTC-4, WM wrote:
> >
> >>>> Nope. A Peano set exists. Thus a "complete" set without last element exists.
> >>> Sorry, you can't believe in logic?
> >>>
> >>> ∀n ∈ ℕ_P: |ℕ_P \ {1, 2, 3, ..., n}| = ℵo .
> >>>
> >>> Or why don't you understand that this refutes your claim?
> >> This is a (correct) statement that every element of the set of Natural numbers is finite.
> >
> > For every element we have |{1, 2, 3, ..., n}|/|{n+1, n+2, n+3, ...}| = 0. Therefore there is no chance to get an actually infinite set by definable natnumbers. All provably fail.
> see? you stop at n again. also he was addressing "elements", not sets.
> >
> >>>> A proper subset with cardinality aleph_0 (the even numbers). Each element of the natural numbers, when doubled, is a natural number
> >>>> and thus an element of N_P (this is a consequence of the fact that N_P has no last element). Doubling each element does not produce new elements .
> >>> I meant not only to consider the doubled numbers but also those between them. Then you have twice as many as before.
> >> Nope, doubling a natural number produces a natural number. This procedure (taking the union of even integers with the set of integers). does not change the set. You do not have "twice as many as before" you have as many as before.
> >
> > If all are doubled (such that none is missing) then your statement is wrong.
> you are wrong, why you always wrong?
>
> >
> > Regards, WM

Right- "y u no right yourself...".

This meme is "why don't you..." or "y u no".

There's probably also subcultures where it's also "why don't you ...",
but in the sense of return my love or take risk.

The ants, let's hear see the ;point that at least one variety of Sergio's ants,
is equi-interpretable with the flying rainbow sparkle pony, of James Burns,
and also the Elephant, and the Relephant.

Real numbers, as a set....

I'm under the impression WM has pathetic drunk syndrome.
Being familar with pathetic drunk myself, is for dance rummy.

Let's be clear that "y u no see the point", at some point is
not just "ignorant", which is not a defense but at least pathetic,
what they call the "ingrate" or "invincible idiot".

"Invincible idiot: never defeated: on the idiot scale...."

About N_F, and univalency and the illative, and,
what for topologists makes for category theory ,
I can type this much faster in my text editor than this Internet gateway,
let's agree that most people call it a conservative extension of ZF.

On Thursday, December 30, 2021 at 7:47:57 AM UTC-8, William wrote:
> On Thursday, December 30, 2021 at 6:41:18 AM UTC-4, WM wrote:
> > William schrieb am Mittwoch, 29. Dezember 2021 um 17:08:30 UTC+1:
> > > On Wednesday, December 29, 2021 at 6:43:04 AM UTC-4, WM wrote:
> >
> > > > > Nope. A Peano set exists. Thus a "complete" set without last element exists.
> > > > Sorry, you can't believe in logic?
> > > >
> > > >∀n ∈ ℕ_P: |ℕ_P \ {1, 2, 3, ..., n}| = ℵo .
> > > >
> > > > Or why don't you understand that this refutes your claim?
> > > This is a (correct) statement that every element of the set of Natural numbers is finite.
> > For every element we have |{1, 2, 3, ..., n}|/|{n+1, n+2, n+3, ...}| = 0.
> Correct, Each element of N_P is finite. However, N_P is infinite
>
>
> > > > <snip> ...I meant not only to consider the doubled numbers but also those between them. Then you have twice as many as before.
> > > Nope, doubling a natural number produces a natural number. This procedure (taking the union of even integers with the set of integers). does not change the set. You do not have "twice as many as before" you have as many as before.
> > If all are doubled (such that none is missing) then your statement is wrong.
> Nope. Doubling every element of the natural numbers produces a subset of the natural numbers. (Proof by induction. Doubling 1 produces a natural number. If doubling n produces a natural number, doubling (n+1) produces a natural number)
>
> --
> William Hughes

Also there's an even arbitrarily _larger_ one than that.

On Thursday, December 30, 2021 at 3:11:53 AM UTC-8, ju...@diegidio.name wrote:
> On Thursday, 30 December 2021 at 11:41:18 UTC+1, WM wrote:
> > William schrieb am Mittwoch, 29. Dezember 2021 um 17:08:30 UTC+1:
>
> > > Nope, doubling a natural number produces a natural number. This procedure
> >> (taking the union of even integers with the set of integers). does not change the
> > > set. You do not have "twice as many as before" you have as many as before.
> > If all are doubled (such that none is missing) then your statement is wrong.
> You are a fucking disgrace, Wolfgang-my-ass, and yet another agent of the enemy:
> thanks to you and almost two decades of your relentless spamming and shitting
> nonsense across the channel, no proper and serious discussion of the patently
> broken standard infinity has ever been even possible, you fucking nazi piece of
> shit and agent of the enemy. ESAT, you and your retarded genocidal employers.
>
> Really, fuck you man, fuck you those who still play with you, you fucking retarded
> bastards, you stupid shameless sociopathic retarded cunts who can only pollute
> each and every single fucking public corner and pond of this forgotten planet.
>
> I have got to vouching for a mass extinction, you piece of disturbed retarded shit
> will just never ever do the right thing ever...
>
> *Plonk*
>
> Julio

Yeah, thanks, Julio, well put.

William schrieb am Donnerstag, 30. Dezember 2021 um 16:47:57 UTC+1:
> On Thursday, December 30, 2021 at 6:41:18 AM UTC-4, WM wrote:
> > William schrieb am Mittwoch, 29. Dezember 2021 um 17:08:30 UTC+1:
> > > On Wednesday, December 29, 2021 at 6:43:04 AM UTC-4, WM wrote:

> Correct, Each element of N_P is finite. However, N_P is infinite

How does it get actually infinite when all definable natural numbers fail?
∀n ∈ ℕ_def: |{1, 2, 3, ..., n}|/|{n+1, n+2, n+3, ....}| = 0.
>
> > If all are doubled (such that none is missing) then your statement is wrong.
> Nope. Doubling every element of the natural numbers produces a subset of the natural numbers. (Proof by induction. Doubling 1 produces a natural number. If doubling n produces a natural number, doubling (n+1) produces a natural number)

Doubling a natural number produces a natural number. But if there was a fixed set, then doubling would double its elements. Therefore there is no fixed set of numbers which can be doubled. (Proof by induction.)

Regards, WM

On Thursday, December 30, 2021 at 2:29:24 PM UTC-4, WM wrote:
> William schrieb am Donnerstag, 30. Dezember 2021 um 16:47:57 UTC+1:

> > Correct, Each element of N_P is finite. However, N_P is infinite
> How does it get actually infinite when all definable natural numbers fail?

It does not "get actually infinite" . N_P is not the result of a process. N_P is a Peano set and has cardinality aleph_0

N_P does not have a last element. Every element of N_P is the largest element of a set with last element.
Thus the set N_P has a property that no element of N_P has.

> Doubling a natural number produces a natural number.

Correct

> But if there was a fixed set, then doubling would double its [cardinality]

Simply false.

--
William Hughes

On Thursday, December 30, 2021 at 1:21:17 PM UTC-4, Ross A. Finlayson wrote:
> On Thursday, December 30, 2021 at 7:47:57 AM UTC-8, William wrote:

> > Nope. Doubling every element of the natural numbers produces a subset of the natural numbers. (Proof by induction. Doubling 1 produces a natural number. If doubling n produces a natural number, doubling (n+1) produces a natural number)

> Also there's an even arbitrarily _larger_ one than that.

Induction, used once, proves something about *every* element in the set N_P, even "arbitrarily larger" ones.

--
William Hughes

On Thursday, December 30, 2021 at 11:13:57 AM UTC-8, William wrote:
> On Thursday, December 30, 2021 at 1:21:17 PM UTC-4, Ross A. Finlayson wrote:
> > On Thursday, December 30, 2021 at 7:47:57 AM UTC-8, William wrote:
>
> > > Nope. Doubling every element of the natural numbers produces a subset of the natural numbers. (Proof by induction. Doubling 1 produces a natural number. If doubling n produces a natural number, doubling (n+1) produces a natural number)
> > Also there's an even arbitrarily _larger_ one than that.
> Induction, used once, proves something about *every* element in the set N_P, even "arbitrarily larger" ones.
>
> --
> William Hughes

Yes, induction is "arbitrarily larger". There is some arbitrarily large case that according to induction,
it results the same as infinite induction. Of course, induction has that that doesn't exist.

Or, "reductio ad absurdam, induction shows induction doesn't exist", i.e.,
the same case that makes for proof by induction and infinite induction
also makes sure to _bound_ induction.

Integral moduli in the numbers are _number theory_'s.

Here that's real-valued number systems and a neat subset an integers.
(Of course that reflects the inclusion of classes of those as sets in
a set theory.)

I.e. each number isn't large enough to actually say only via induction the guarantee of its case,
what results must be a deduction of the case of the set, or its access as a set, is that
guarantees of induction are sound, both from the case and the case either closing or not closing,
when here the "set" is of the same structure of each of the elements.

For example, an infinite set of ordinals, it _is_ an ordinal.

You wouldn't say that's guaranteed by induction though, here it also is, ....

Anyways numbers are number theory's and matters of their density
for example are about integral moduli, then infinite sets have that
infinite induction over them does not guarantee transfinite induction
with their "ordinal container" the cases: then some infinite sets when
for example it _is_ the ordinal container, it's guaranteed they do,
or for example don't.

That the infinite set of ordinals, is itself just another ordinal,
is called Burali-Forti's paradox, or a feature of ordinals in
ordering theory and with respect to how they're modeled
in sets where sets are regular or rather no infinite descending chains,
while, the infinite ordering has infinity at the end.

This way it's as sensibly: for ordinals: there's usually a way to
apply the transfer principle, and otherwise not. That's just like
talking about real-valued numbers, where 1 = 1.0 in value
while as sets 1 elt N =/= 1.0 elt R, to define what are still "sets" and "ordinals",
but behave like numbers and spaces.

Real-valued spaces..., ..., number theory's.

William schrieb am Donnerstag, 30. Dezember 2021 um 20:13:38 UTC+1:
> On Thursday, December 30, 2021 at 2:29:24 PM UTC-4, WM wrote:
> > William schrieb am Donnerstag, 30. Dezember 2021 um 16:47:57 UTC+1:
>
> > > Correct, Each element of N_P is finite. However, N_P is infinite
> > How does it get actually infinite when all definable natural numbers fail?
> It does not "get actually infinite" . N_P is not the result of a process. N_P is a Peano set and has cardinality aleph_0
>
> N_P does not have a last element. Every element of N_P is the largest element of a set with last element.
> Thus the set N_P has a property that no element of N_P has.

But the set ℕ has a property that the set ℕ_def = ℕ_P does not havee, namely a number of elements that is larger than *all* elements.

> > Doubling a natural number produces a natural number.
> Correct
>
> > But if there was a fixed set, then doubling would double its [cardinality]
>
> Simply false.

With cardinality yes. But cardinality is self-contradictory as we see when scrutinizing cantors "bijection"
{1, 2, 3, 4, 5, ...} --> {1/1, 1/2, 2/1, 1/3, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 5/1, 1/6, ...}.

Collect all rationals of the interval (0, 1] into a reservoir, map one of them by 1 and remove it from the reservoir. Then collect all rationals of the interval (1, 2] into the reservoir, map one of them by 2 and remove it from the reservoir. Then collect all rationals of the interval (2, 3] into the reservoir, map one of them by 3 and remove it from the reservoir. Continue until all intervals will have been worked through. Then the reservoir will be empty.

Of course this conclusion is contrary to mathematics. The reservoire will not be empty.

But what I said is: Doubling a natural number produces a natural number. But if there was a fixed set, then doubling would double its elements. And that is simply mathematics.

Regards, WM

William schrieb am Donnerstag, 30. Dezember 2021 um 20:13:57 UTC+1:

> Induction, used once, proves something about *every* element in the set N_P, even "arbitrarily larger" ones.
>
It proves that all definable FISONs will never cover ℕ. The union of the set of FISONs however cannot surpass all FISONs.

{1}
{2, 1}
{3, 2, 1}
{4, 3, 2, 1}
{5, 4, 3, 2, 1}
....

If there are ℵo natnumbers, then they are in the first column. If they are in the first column, then they are in the whole figure. There are not any two lines in the figure which contain more than one of them. No line contains ℵo natnumbers.

But the set ℕ is much, much larger that ℕ_def:

{1, 2, 3, 4, 5, ..., ω/1000} is not a FISON although it contains only numbers less than ω.

Regards, WM

On Friday, December 31, 2021 at 5:34:43 AM UTC-4, WM wrote:

> {1}
> {2, 1}
> {3, 2, 1}
> {4, 3, 2, 1}
> {5, 4, 3, 2, 1}
> ...
>
> If there are ℵo natnumbers, then they are in the first column. If they are in the first column, then they are in the whole figure.
> There are not any two lines in the figure which contain more than one of them. No line contains ℵo natnumbers.

Correct the whole figure, the set, contains aleph_0 elements. But no line, the elements of the figure, contain aleph_0 elements. The set has a property that no element does.

>
> But the set ℕ is much, much larger that ℕ_def:
>
> {1, 2, 3, 4, 5, ..., ω/1000} is not a FISON although it contains only numbers less than ω.

Nope, ω/1000= w. The only ordinals less than ω are the natural numbers.

--
William Hughes

On Friday, December 31, 2021 at 5:26:21 AM UTC-4, WM wrote:
> William schrieb am Donnerstag, 30. Dezember 2021 um 20:13:38 UTC+1:
> > On Thursday, December 30, 2021 at 2:29:24 PM UTC-4, WM wrote:
> > > William schrieb am Donnerstag, 30. Dezember 2021 um 16:47:57 UTC+1:
> >
> > > > Correct, Each element of N_P is finite. However, N_P is infinite
> > > How does it get actually infinite when all definable natural numbers fail?
> > It does not "get actually infinite" . N_P is not the result of a process. N_P is a Peano set and has cardinality aleph_0
> >
> > N_P does not have a last element. Every element of N_P is the largest element of a set with last element.
> > Thus the set N_P has a property that no element of N_P has.
> But the set ℕ has a property that the set ℕ_def = ℕ_P

nope |N_def, the set of natural numbers that can be written down, is a finite set, N_P is an infinite set. The two are not the same.

>does not have, namely a number of elements that is larger than *all* elements.

yes, your |N.which is not the natural numbers, has paradoxical elements that are greater than themselves. N_P only has finite elements,
|N def has a last element, N_P has no last element

> > > Doubling a natural number produces a natural number.
> > Correct
> >
> > > But if there was a fixed set, then doubling would double its [cardinality]
> >
> > Simply false.
> With cardinality yes.

Cardinality is the only way to talk about the "Number of elements" in a set with no largest element.

On 12/31/2021 3:26 AM, WM wrote:
> William schrieb am Donnerstag, 30. Dezember 2021 um 20:13:38 UTC+1:
>> On Thursday, December 30, 2021 at 2:29:24 PM UTC-4, WM wrote:
>>> William schrieb am Donnerstag, 30. Dezember 2021 um 16:47:57 UTC+1:
>>
>>>> Correct, Each element of N_P is finite. However, N_P is infinite
>>> How does it get actually infinite when all definable natural numbers fail?
>> It does not "get actually infinite" . N_P is not the result of a process. N_P is a Peano set and has cardinality aleph_0
>>
>> N_P does not have a last element. Every element of N_P is the largest element of a set with last element.
>> Thus the set N_P has a property that no element of N_P has.
>
> But the set ℕ has a property that the set ℕ_def = ℕ_P does not havee, namely a number of elements that is larger than *all* elements.

Prove it.

>
>>> Doubling a natural number produces a natural number.
>> Correct
>>
>>> But if there was a fixed set, then doubling would double its [cardinality]
>>
>> Simply false.
>
> With cardinality yes. But cardinality is self-contradictory as we see when scrutinizing cantors "bijection"
> {1, 2, 3, 4, 5, ...} --> {1/1, 1/2, 2/1, 1/3, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, 5/1, 1/6, ...}.

there is no self contradiction.

>
> Collect all rationals of the interval (0, 1] into a reservoir, map one of them by 1 and remove it from the reservoir. Then collect all rationals of the interval (1, 2] into the reservoir, map one of them by 2 and remove it from the reservoir. Then collect all rationals of the interval (2, 3] into the reservoir, map one of them by 3 and remove it from the reservoir. Continue until all intervals will have been worked through. Then the reservoir will be empty.

your leaky rubber sets are failed math, grasping at obtuse illogical constructs to open a crack for your dark ones.

>
> Of course this conclusion is contrary to mathematics. The reservoire will not be empty.
>
> But what I said is: Doubling a natural number produces a natural number. But if there was a fixed set, then doubling would double its elements. And that is simply mathematics.

not with infinite sets. ℵo = 2*ℵo
you still do not get it, or you like to troll with it,
ℵo bottles of beer...

>
> Regards, WM