On 4/21/2022 9:34 AM, WM wrote:
> William schrieb am Mittwoch, 20. April 2022 um 19:26:48 UTC+2:
>> On Wednesday, April 20, 2022 at 9:38:20 AM UTC-3, WM wrote:
>>> William schrieb am Dienstag, 19. April 2022 um 18:04:38 UTC+2:
>>>
>>>> ...We apply induction exactly once, and show something about each of the infinitely many elements of N_p.
>>> The most important result of induction is this: Every number covered by induction
>> I.e any element. a we know by definition that each element of a Peano set is covered by induction.
>>> belongs to a finite set which is followed by an infinite set
>> trivial. Note that, by the induction we have already done, we know that, as this infinite set contains only elements of N_p, the infinite set does not contain a "dark number".
>

<snip oppositional crap>
>

> Regards, WM

On 4/21/2022 9:16 AM, WM wrote:
> William schrieb am Mittwoch, 20. April 2022 um 19:02:38 UTC+2:
>> On Wednesday, April 20, 2022 at 9:38:20 AM UTC-3, WM wrote:
>>> William schrieb am Dienstag, 19. April 2022 um 18:04:38 UTC+2:
>>>
>>>> A "potentially infinite set" is nonsense.
>>> Alas it is the only set that we can apply in mathematics.
>> Piffle. A result you do not like is not a contradiction,
>

<snip oppositional crap>
>

> Regards, WM

On Thursday, April 21, 2022 at 11:34:06 AM UTC-3, WM wrote:
> William schrieb am Mittwoch, 20. April 2022 um 19:26:48 UTC+2:
> > On Wednesday, April 20, 2022 at 9:38:20 AM UTC-3, WM wrote:
> > > William schrieb am Dienstag, 19. April 2022 um 18:04:38 UTC+2:
> > >
> > > > ...We apply induction exactly once, and show something about each of the infinitely many elements of N_p.
> > > The most important result of induction is this: Every number covered by induction
> > I.e any element. a we know by definition that each element of a Peano set is covered by induction.
> > >belongs to a finite set which is followed by an infinite set
> > trivial. Note that, by the induction we have already done, we know that, as this infinite set contains only elements of N_p, the infinite set does not contain a "dark number".
> They are the buffer between N_p and the limit ω.
Piffle. By definition of ω there is nothing between N_p and ω.
> Proof by induction for all elements of N_p.
Piffle, the only thing you prove by induction is the trivial fact that every element of N_p is followed by an infinite set of elements of N_p. By our previous induction we know that no element of this set is "dark"

--
William Hughes

On Thursday, April 21, 2022 at 11:16:32 AM UTC-3, WM wrote:
> William schrieb am Mittwoch, 20. April 2022 um 19:02:38 UTC+2:
> > On Wednesday, April 20, 2022 at 9:38:20 AM UTC-3, WM wrote:
> > > William schrieb am Dienstag, 19. April 2022 um 18:04:38 UTC+2:
> > >
> > > > A "potentially infinite set" is nonsense.
> > > Alas it is the only set that we can apply in mathematics.
> > Piffle. A result you do not like is not a contradiction,
> By induction we prove that every definable natural number, i.e., every natural number which is subject to induction,

i.e. every natural number (every element of a Peano set is subject to induction)

> belongs to a FISON

correct,

>... has ℵo successors

none of which are dark

--
William Hughes

On Wednesday, 20 April 2022 at 15:47:13 UTC+3, WM wrote:
> horand....@gmail.com schrieb am Dienstag, 19. April 2022 um 17:56:26 UTC+2:
> > On Tuesday, 19 April 2022 at 12:29:56 UTC-3, WM wrote:
> > > horand....@gmail.com schrieb am Montag, 18. April 2022 um 18:24:04 UTC+2:
> > [...]
> > > > The stepwise process does not end. However, the stepwise process defined by the sequence 1/1, 1/2, 1/3, 1/4, ... does not end, either. And yet it does have a *LIMIT*.
> > > which is not reached.
> > Yes, the limit is not reached, in the same way that the limit 0 is not reached by the sequence 1/1, 1/2, 1/3, 1/4, ... . Nonetheless, it is the limit.
> What does this mean? The difference between the size f(n) and the limit f(ω) becomes smaller than every desired eps: |f(n) - f(ω)| < eps. But the number of terms between n and ω remains infinite: |ω - n| = ℵo.

One can say the same about the flawed mainstream definition of derivative:

f'(x) = lim_{h->0} [f(x+h)-f(x)]/h

As h decreases indefinitely, the finite differences approach a "limit":

[f(x+h_1)-f(x)]/h_1 ; [f(x+h_2)-f(x)]/h_2; ... ; [f(x+h_n)-f(x)]/h_n; ...

The difference between [f(x+h_n)-f(x)]/h_n and the "ultimate ratio" [f(x+h_ω )-f(x)]/h_ω becomes smaller than every
desired ε: |[f(x+h_n)-f(x)]/h_n - [f(x+h_ω )-f(x)]/h_ω|<ε. But the number of terms between n and ω remains "infinite" |ω-n|=א‎0.

It's the same bullshit - just a different flavour? :-)

>
> In set theory however, we do not need the sizes of terms but the distances. That is what you confuse.
>
> Regards, WM

On Thursday, 21 April 2022 at 20:19:58 UTC+3, William wrote:
> On Thursday, April 21, 2022 at 11:16:32 AM UTC-3, WM wrote:
> > William schrieb am Mittwoch, 20. April 2022 um 19:02:38 UTC+2:
> > > On Wednesday, April 20, 2022 at 9:38:20 AM UTC-3, WM wrote:
> > > > William schrieb am Dienstag, 19. April 2022 um 18:04:38 UTC+2:
> > > >
> > > > > A "potentially infinite set" is nonsense.
> > > > Alas it is the only set that we can apply in mathematics.
> > > Piffle. A result you do not like is not a contradiction,
> > By induction we prove that every definable natural number, i.e., every natural number which is subject to induction,
> i.e. every natural number (every element of a Peano set is subject to induction)
>
> > belongs to a FISON
>
> correct,
>
> >... has ℵo successors
>
> none of which are dark

Quite an ignorant statement because most successors are not known (dark?) and more importantly cannot be known.

For in order to be known, one must assume that the so-called set of natural numbers that acts as an index, is complete.

But if complete, then there is a largest natural number which is obviously false.

>
> --
> William Hughes

You fucctaard. Stop wasting bandwith with this garbage

On Thursday, April 21, 2022 at 1:27:05 PM UTC-4, Eram semper recta wrote:
> On Wednesday, 20 April 2022 at 15:47:13 UTC+3, WM wrote:
> > horand....@gmail.com schrieb am Dienstag, 19. April 2022 um 17:56:26 UTC+2:
> > > On Tuesday, 19 April 2022 at 12:29:56 UTC-3, WM wrote:
> > > > horand....@gmail.com schrieb am Montag, 18. April 2022 um 18:24:04 UTC+2:
> > > [...]
> > > > > The stepwise process does not end. However, the stepwise process defined by the sequence 1/1, 1/2, 1/3, 1/4, ... does not end, either. And yet it does have a *LIMIT*.
> > > > which is not reached.
> > > Yes, the limit is not reached, in the same way that the limit 0 is not reached by the sequence 1/1, 1/2, 1/3, 1/4, ... . Nonetheless, it is the limit.
> > What does this mean? The difference between the size f(n) and the limit f(ω) becomes smaller than every desired eps: |f(n) - f(ω)| < eps.. But the number of terms between n and ω remains infinite: |ω - n| = ℵo.
> One can say the same about the flawed mainstream definition of derivative:
>
> f'(x) = lim_{h->0} [f(x+h)-f(x)]/h
>
> As h decreases indefinitely, the finite differences approach a "limit":
>
> [f(x+h_1)-f(x)]/h_1 ; [f(x+h_2)-f(x)]/h_2; ... ; [f(x+h_n)-f(x)]/h_n; ...
>
> The difference between [f(x+h_n)-f(x)]/h_n and the "ultimate ratio" [f(x+h_ω )-f(x)]/h_ω becomes smaller than every
> desired ε: |[f(x+h_n)-f(x)]/h_n - [f(x+h_ω )-f(x)]/h_ω|<ε. But the number of terms between n and ω remains "infinite" |ω-n|=א‎0.
>
> It's the same bullshit - just a different flavour? :-)
> >
> > In set theory however, we do not need the sizes of terms but the distances. That is what you confuse.
> >
> > Regards, WM

On Friday, 22 April 2022 at 04:51:28 UTC+3, MORON in 3D driveled:
> You fucctaard. Stop wasting bandwith with this garbage
> On Thursday, April 21, 2022 at 1:27:05 PM UTC-4, Eram semper recta wrote:
> > On Wednesday, 20 April 2022 at 15:47:13 UTC+3, WM wrote:
> > > horand....@gmail.com schrieb am Dienstag, 19. April 2022 um 17:56:26 UTC+2:
> > > > On Tuesday, 19 April 2022 at 12:29:56 UTC-3, WM wrote:
> > > > > horand....@gmail.com schrieb am Montag, 18. April 2022 um 18:24:04 UTC+2:
> > > > [...]
> > > > > > The stepwise process does not end. However, the stepwise process defined by the sequence 1/1, 1/2, 1/3, 1/4, ... does not end, either. And yet it does have a *LIMIT*.
> > > > > which is not reached.
> > > > Yes, the limit is not reached, in the same way that the limit 0 is not reached by the sequence 1/1, 1/2, 1/3, 1/4, ... . Nonetheless, it is the limit.
> > > What does this mean? The difference between the size f(n) and the limit f(ω) becomes smaller than every desired eps: |f(n) - f(ω)| < eps. But the number of terms between n and ω remains infinite: |ω - n| = ℵo.
> > One can say the same about the flawed mainstream definition of derivative:
> >
> > f'(x) = lim_{h->0} [f(x+h)-f(x)]/h
> >
> > As h decreases indefinitely, the finite differences approach a "limit":
> >
> > [f(x+h_1)-f(x)]/h_1 ; [f(x+h_2)-f(x)]/h_2; ... ; [f(x+h_n)-f(x)]/h_n; ....
> >
> > The difference between [f(x+h_n)-f(x)]/h_n and the "ultimate ratio" [f(x+h_ω )-f(x)]/h_ω becomes smaller than every
> > desired ε: |[f(x+h_n)-f(x)]/h_n - [f(x+h_ω )-f(x)]/h_ω|<ε. But the number of terms between n and ω remains "infinite" |ω-n|=א‎0.
> >
> > It's the same bullshit - just a different flavour? :-)

Wolfgang:
Do you see how my statement infuriates butthurt mainstream cranks?

The statement was directed at your selective (biased) logic, but because it strikes too close to home where mainstream beliefs are concerned, it is undesirable.

The moron MythIn3D was so infuriated that he had to post his drivel.

Your argument wrt the topic is correct. However, you cannot see that your alternate belief wrt the mainstream derivative is false and can be rejected in the same way as you reject the drivel of retards who claim all successors are possible.

> > > Regards, WM

On Thursday, 21 April 2022 at 20:27:05 UTC+3, Eram semper recta wrote:
> On Wednesday, 20 April 2022 at 15:47:13 UTC+3, WM wrote:
> > horand....@gmail.com schrieb am Dienstag, 19. April 2022 um 17:56:26 UTC+2:
> > > On Tuesday, 19 April 2022 at 12:29:56 UTC-3, WM wrote:
> > > > horand....@gmail.com schrieb am Montag, 18. April 2022 um 18:24:04 UTC+2:
> > > [...]
> > > > > The stepwise process does not end. However, the stepwise process defined by the sequence 1/1, 1/2, 1/3, 1/4, ... does not end, either. And yet it does have a *LIMIT*.
> > > > which is not reached.
> > > Yes, the limit is not reached, in the same way that the limit 0 is not reached by the sequence 1/1, 1/2, 1/3, 1/4, ... . Nonetheless, it is the limit.
> > What does this mean? The difference between the size f(n) and the limit f(ω) becomes smaller than every desired eps: |f(n) - f(ω)| < eps.. But the number of terms between n and ω remains infinite: |ω - n| = ℵo.
> One can say the same about the flawed mainstream definition of derivative:
>
> f'(x) = lim_{h->0} [f(x+h)-f(x)]/h
>
> As h decreases indefinitely, the finite differences approach a "limit":
>
> [f(x+h_1)-f(x)]/h_1 ; [f(x+h_2)-f(x)]/h_2; ... ; [f(x+h_n)-f(x)]/h_n; ...
>
> The difference between [f(x+h_n)-f(x)]/h_n and the "ultimate ratio" [f(x+h_ω )-f(x)]/h_ω becomes smaller than every
> desired ε: |[f(x+h_n)-f(x)]/h_n - [f(x+h_ω )-f(x)]/h_ω|<ε. But the number of terms between n and ω remains "infinite" |ω-n|=א‎0.

So, while you are busy with your epsilonic arguments, you fail to see that there is no ratio [f(x+h_ω )-f(x)]/h_ω that can be the limit (derivative) and yet the derivative itself is a *ratio of magnitudes*, viz. rise/run of tangent line.

The epsilonic argument is nothing but a property of the limit and it proves nothing unless you already know the limit.

>
> It's the same bullshit - just a different flavour? :-)
> >
> > In set theory however, we do not need the sizes of terms but the distances. That is what you confuse.
> >
> > Regards, WM

horand....@gmail.com schrieb am Donnerstag, 21. April 2022 um 16:43:05 UTC+2:
> On Thursday, 21 April 2022 at 11:14:46 UTC-3, WM wrote:

> > f(ω) is its limit like 0.
> > > Your habit of simply plugging ω into the functional equation is what has gotten you in trouble time and again. You simply can't do it that way.
> > If ω is the limit of the sequence (n), then this can be done.
> Only if f is continuous,

A function or sequence f(n) has domain n ∈ ℕ. It cannot be discontinuous at ω because there is no value to be attached at ω. ω does not belong to ℕ. f(ω) can only be the limit calculated from the function for n --> oo.

Regards, WM

sergio schrieb am Donnerstag, 21. April 2022 um 16:57:15 UTC+2:
> On 4/21/2022 9:14 AM, WM wrote:

> > f is a function like (1/n).

> >> And what is f(ω)?
> >
> > f(ω) is its limit like 0.

> No. If so, then what is f(ω-1) ?

It is dark like the values at almost all arguments n.

Regards, WM

William schrieb am Donnerstag, 21. April 2022 um 19:12:05 UTC+2:
> On Thursday, April 21, 2022 at 11:34:06 AM UTC-3, WM wrote:

> > They are the buffer between N_p and the limit ω.
> Piffle. By definition of ω there is nothing between N_p and ω.
> > Proof by induction for all elements of N_p.
> Piffle, the only thing you prove by induction is the trivial fact that every element of N_p is followed by an infinite set of elements of N_p.

That is wrong. If all elements of N_p could be exhausted by this proof (i.e.., if it would hold for all elements of N_p), there would nothing remain before omega.

Regards, WM

William schrieb am Donnerstag, 21. April 2022 um 19:19:58 UTC+2:
> On Thursday, April 21, 2022 at 11:16:32 AM UTC-3, WM wrote:
> > William schrieb am Mittwoch, 20. April 2022 um 19:02:38 UTC+2:
> > > On Wednesday, April 20, 2022 at 9:38:20 AM UTC-3, WM wrote:
> > > > William schrieb am Dienstag, 19. April 2022 um 18:04:38 UTC+2:
> > > >
> > > > > A "potentially infinite set" is nonsense.
> > > > Alas it is the only set that we can apply in mathematics.
> > > Piffle. A result you do not like is not a contradiction,
> > By induction we prove that every definable natural number, i.e., every natural number which is subject to induction,
> i.e. every natural number (every element of a Peano set is subject to induction)
>
> > belongs to a FISON
>
> correct,
>
> >... has ℵo successors
>
> none of which are dark

You cannot divide |N in two consecutive sets of infinitely many natnumbers in FISONs which are followed by infinitely many natnumbers. Induction holds only for a potentially infinite set of definable natnumbers which are followed by infinitely many dark natnumbers.

Regards, WM

Eram semper recta schrieb am Donnerstag, 21. April 2022 um 19:31:36 UTC+2:
> On Thursday, 21 April 2022 at 20:19:58 UTC+3, William wrote:
> > On Thursday, April 21, 2022 at 11:16:32 AM UTC-3, WM wrote:
> > > William schrieb am Mittwoch, 20. April 2022 um 19:02:38 UTC+2:
> > > > On Wednesday, April 20, 2022 at 9:38:20 AM UTC-3, WM wrote:
> > > > > William schrieb am Dienstag, 19. April 2022 um 18:04:38 UTC+2:
> > > > >
> > > > > > A "potentially infinite set" is nonsense.
> > > > > Alas it is the only set that we can apply in mathematics.
> > > > Piffle. A result you do not like is not a contradiction,
> > > By induction we prove that every definable natural number, i.e., every natural number which is subject to induction,
> > i.e. every natural number (every element of a Peano set is subject to induction)
> >
> > > belongs to a FISON
> >
> > correct,
> >
> > >... has ℵo successors
> >
> > none of which are dark
> Quite an ignorant statement because most successors are not known (dark?) and more importantly cannot be known.

Correct!
>
> For in order to be known, one must assume that the so-called set of natural numbers that acts as an index, is complete.
>
> But if complete, then there is a largest natural number which is obviously false.

They believe that each of ℵo natnumbers is followed by ℵo natnumbers. That is nonsense. Between every number and the limit there are almost all numbers,

Regards, WM

On 4/22/2022 10:01 AM, WM wrote:
> Eram semper recta schrieb am Donnerstag, 21. April 2022 um 19:31:36 UTC+2:
>> On Thursday, 21 April 2022 at 20:19:58 UTC+3, William wrote:
>>> On Thursday, April 21, 2022 at 11:16:32 AM UTC-3, WM wrote:
>>>> William schrieb am Mittwoch, 20. April 2022 um 19:02:38 UTC+2:
>>>>> On Wednesday, April 20, 2022 at 9:38:20 AM UTC-3, WM wrote:
>>>>>> William schrieb am Dienstag, 19. April 2022 um 18:04:38 UTC+2:
>>>>>>
>>>>>>> A "potentially infinite set" is nonsense.
>>>>>> Alas it is the only set that we can apply in mathematics.
>>>>> Piffle. A result you do not like is not a contradiction,
>>>> By induction we prove that every definable natural number, i.e., every natural number which is subject to induction,
>>> i.e. every natural number (every element of a Peano set is subject to induction)
>>>
>>>> belongs to a FISON
>>>
>>> correct,
>>>
>>>> ... has ℵo successors
>>>
>>> none of which are dark
>> Quite an ignorant statement because most successors are not known (dark?) and more importantly cannot be known.
>
> Correct!
>>
>> For in order to be known, one must assume that the so-called set of natural numbers that acts as an index, is complete.
>>
>> But if complete, then there is a largest natural number which is obviously false.
>
> They believe that each of ℵo natnumbers is followed by ℵo natnumbers. That is nonsense. Between every number and the limit there are almost all numbers,
>
> Regards, WM

you are out of math again, spewing your imagination. No wonder you do not use equations.

On 4/22/2022 9:51 AM, WM wrote:
> William schrieb am Donnerstag, 21. April 2022 um 19:12:05 UTC+2:
>> On Thursday, April 21, 2022 at 11:34:06 AM UTC-3, WM wrote:
>
>>> They are the buffer between N_p and the limit ω.
>> Piffle. By definition of ω there is nothing between N_p and ω.
>>> Proof by induction for all elements of N_p.
>> Piffle, the only thing you prove by induction is the trivial fact that every element of N_p is followed by an infinite set of elements of N_p.
>
> That is wrong. If all elements of N_p could be exhausted by this proof (i.e., if it would hold for all elements of N_p), there would nothing remain before omega.
>
> Regards, WM

nope, you are wrong, you are wrong in EVERY POST.

On 4/22/2022 9:55 AM, WM wrote:
> William schrieb am Donnerstag, 21. April 2022 um 19:19:58 UTC+2:
>> On Thursday, April 21, 2022 at 11:16:32 AM UTC-3, WM wrote:
>>> William schrieb am Mittwoch, 20. April 2022 um 19:02:38 UTC+2:
>>>> On Wednesday, April 20, 2022 at 9:38:20 AM UTC-3, WM wrote:
>>>>> William schrieb am Dienstag, 19. April 2022 um 18:04:38 UTC+2:
>>>>>
>>>>>> A "potentially infinite set" is nonsense.
>>>>> Alas it is the only set that we can apply in mathematics.
>>>> Piffle. A result you do not like is not a contradiction,
>>> By induction we prove that every definable natural number, i.e., every natural number which is subject to induction,
>> i.e. every natural number (every element of a Peano set is subject to induction)
>>
>>> belongs to a FISON
>>
>> correct,
>>
>>> ... has ℵo successors
>>
>> none of which are dark
>
> You cannot divide |N in two consecutive sets of infinitely many natnumbers in FISONs which are followed by infinitely many natnumbers. Induction holds only for a potentially infinite set of definable natnumbers which are followed by infinitely many dark natnumbers.
>
> Regards, WM

that is nonsense on its face.

On 4/22/2022 9:45 AM, WM wrote:
> sergio schrieb am Donnerstag, 21. April 2022 um 16:57:15 UTC+2:
>> On 4/21/2022 9:14 AM, WM wrote:
>
>>> f is a function like (1/n).
>
>>>> And what is f(ω)?
>>>
>>> f(ω) is its limit like 0.
>
>> No. If so, then what is f(ω-1) ?
>
> It is dark like the values at almost all arguments n.
>
> Regards, WM

Wrong. It is your not knowing how to use common math nomenclature. I don't have the time or inclination to teach you algebra.

your knowledge of Math is dark, black hole dark, non existent dark.

On Friday, 22 April 2022 at 11:41:55 UTC-3, WM wrote:
[...]
> A function or sequence f(n) has domain n ∈ ℕ. It cannot be discontinuous at ω because there is no value to be attached at ω. ω does not belong to ℕ. f(ω) can only be the limit calculated from the function for n --> oo.

Nonsense! If f(.) is the card(.) function, then your method will go wrong. If you actually think that your preceding bullshit is proper mathematics, then it is no wonder that you run into so many seeming contradictions. Hint: *YOUR* thinking is wrong, not mathematics.

On Friday, April 22, 2022 at 11:55:22 AM UTC-3, WM wrote:
> You cannot divide |N in two consecutive sets of infinitely many natnumbers

Every *element*. F_n (there are an infinite number) of the set of FISONs is a finite set which is followed by, S(F_n) (this set may change if the FISON changes and there is no set which works for every FISON) an infinite set of elements of N_p. There are no elements of N_p which follow the *set* of FISONs. No Peano set contains a "dark" element

--
William Hughes

On Friday, April 22, 2022 at 11:51:18 AM UTC-3, WM wrote:
> > the only thing you prove by induction is the trivial fact that every element of N_p is followed by an infinite set of elements of N_p.
> That is wrong.

Piffle .

>If all elements of N_p could be exhausted by this proof
> (i.e., if it would hold for all elements of N_p), there would nothing remain before omega.

The result that "there would nothing remain before omega" is not a contradiction, it is a result you do not like. It follows trivially from the fact that a Peano set has no last element. All elements of N_p are "before omega" omega is a limit ordinal so there this no last element of N_p before omega.

--
William Hughes

horand....@gmail.com schrieb am Freitag, 22. April 2022 um 18:58:19 UTC+2:
> On Friday, 22 April 2022 at 11:41:55 UTC-3, WM wrote:
> [...]
> > A function or sequence f(n) has domain n ∈ ℕ. It cannot be discontinuous at ω because there is no value to be attached at ω. ω does not belong to ℕ. f(ω) can only be the limit calculated from the function for n --> oo.
> If f(.) is the card(.) function, then your method will go wrong.

There is no card function. If you go through all definable n you will never reach ω

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

If you remove all n collectively, only ω remains.

|ℕ \ {1, 2, 3, ...}| = 0 .

Hence most natnumbers are dark and cannot contribute to mappings and cardinality.

Regards, WM

Eram semper recta schrieb am Freitag, 22. April 2022 um 09:01:00 UTC+2:

> Your argument wrt the topic is correct. However, you cannot see that your alternate belief wrt the mainstream derivative is false and can be rejected in the same way

I do not reject the existence of a limit like 0 of the sequence (1/n). What I reject is the definability of all terms 1/n.

Regards, WM

William schrieb am Freitag, 22. April 2022 um 19:05:11 UTC+2:
> There are no elements of N_p which follow the *set* of FISONs. No Peano set contains a "dark" element

We cannot remove all natnumbers by removing them individually, for instance their FISONs, because all are not sufficient:

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

But we can remove all natnumbers collectively.

|ℕ \ {1, 2, 3, ...}| = 0 .

That proves a difference between individually and collectively treating numbers. You should be able to recognize that.

Regards, WM

William schrieb am Freitag, 22. April 2022 um 19:24:02 UTC+2:
> On Friday, April 22, 2022 at 11:51:18 AM UTC-3, WM wrote:

> >If all elements of N_p could be exhausted by this proof
> > (i.e., if it would hold for all elements of N_p), there would nothing remain before omega.
> The result that "there would nothing remain before omega" is not a contradiction, it is a result you do not like.

It is wrong. We cannot remove all natnumbers by removing them individually by induction, for instance their FISONs, because all are not sufficient:

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

> It follows trivially from the fact that a Peano set has no last element. All elements of N_p are "before omega" omega is a limit ordinal so there this no last element of N_p before omega.

No last *definable* element. Nevertheless you can remove all natnumbers collectively:

|ℕ \ {1, 2, 3, ...}| = 0 .

That proves a difference between individually and collectively treating numbers. You should be able to recognize that.

Regards, WM