On 2/28/2022 12:06 PM, Fritz Feldhase wrote:
> On Monday, February 28, 2022 at 6:54:23 PM UTC+1, Fritz Feldhase wrote:
>
> Seems that your "argument" is the following:
>
> [WM:] "In my opinion..."
>
>> | "it should also be possible to model this configuration by exchanging pairs of fractions [...]
>
> Well, opinions aren't "good enough" in mathematics, Mückenheim.
>
>> | but it is not."
>
> Big surprise!
>
> Your fellow crank: "In my opinion it should be possible to divide by 0. But it is not! Something's not right in mainstream math!"
>
> Yeah, a convincing argument!

WM has at least one mistake (in math) in every one of his posts.

I have him classed as pre-algebra with memory loss. I havent seen any real math from him at all.
I think he posts a max of 5 equations over and over. Kaput.

On Monday, February 28, 2022 at 6:54:23 PM UTC+1, Fritz Feldhase wrote:
> On Sunday, February 27, 2022 at 10:34:55 PM UTC+1, WM wrote:
> >
> > Find the first failure.
>
> The first (serious) failure in your text is your claim that "[i]t should also be possible to reach this configuration by exchanging pairs of fractions."

Correction. The first (serious) failure in your text appears in the follwoing part:

"...gather them in the first column (k, 1). Then all other columns of the matrix are empty."

Such a "matrix" does not exist in mathematics.

But you may define and then consider a matrix (m_ij) such that for all i e IN and all j e IN \ {1} m_ij = 0, and m_k1 = a_k for all k e IN (where (a_k) is the sequence of all positive fractions, defined by Cantor).

Fritz Feldhase schrieb am Montag, 28. Februar 2022 um 17:45:39 UTC+1:

> ∀n ∈ ℕ: |ℕ \ {1, 2, 3, ..., n}| = ℵo ,
>
> of course. In other words,
>
> ∀n ∈ ℕ: |{k e ℕ : k > n}| = ℵo ..
>
> What a surprise!

Collect all these elements. Then you have ℕ _def. Or do you think that by collecting them they will lose their property?

Regards, WM

On Monday, February 28, 2022 at 8:05:00 PM UTC+1, WM wrote:
> Fritz Feldhase schrieb am Montag, 28. Februar 2022 um 17:45:39 UTC+1:
> >
> > ∀n ∈ ℕ: |ℕ \ {1, 2, 3, ..., n}| = ℵo ,
> >
> > of course. In other words,
> >
> > ∀n ∈ ℕ: |{k e ℕ : k > n}| = ℵo .
> >
> > What a surprise!
> >
> Collect all these elements. Then you have ℕ_def.

Yaeh, THAT would be a definition, namely:

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

Then we get ℕ_def = ℕ, since {n e ℕ : |ℕ \ {1, 2, 3, ..., n}| = ℵo} = ℕ.

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

horand....@gmail.com schrieb am Montag, 28. Februar 2022 um 19:57:15 UTC+1:
> On Monday, 28 February 2022 at 14:20:55 UTC-4, WM wrote:
> > Fritz Feldhase schrieb am Montag, 28. Februar 2022 um 14:57:30 UTC+1:
> > > On Monday, February 28, 2022 at 11:49:25 AM UTC+1, WM wrote:
> >
> > > > > We KNOW, since we can PROVE it, that there is a bijection from {2*n : n e IN} onto {2*n + 1 : n e IN}, for example.
> > > > >
> > > > You can prove it for every definable natnumber has infinitely many successors, infinitely many of which cannot become defined:
> > ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., k}| = ℵo .
> > You cannot prove it for all natnumbers.
> Ah, the patented quantifier exchange is being set up.

If there are all natnumbers, then all can be subtracted from ℕ such that nothing remains But when subtracting all definable individuals then almost all remains.

Regards, WM

Fritz Feldhase schrieb am Montag, 28. Februar 2022 um 19:50:10 UTC+1:
> On Monday, February 28, 2022 at 11:54:07 AM UTC+1, WM wrote:

> Hint: {n e ℕ : |ℕ \ {1, 2, 3, ..., n}| = ℵo} = ℕ.

What is the reason of ℵo remainings? Why can they not be subtracted?
Hint: They can, but not individually.

Regards, WM

Jim Burns schrieb am Montag, 28. Februar 2022 um 19:24:59 UTC+1:
> On 2/28/2022 5:46 AM, WM wrote:

> >> The infinite set doesn't end,
> >> but _our reasoning_ can end.
> >
> > It does. It cannot avoid that every definable natnumber
> > has infinitely many successors, infinitely many of which
> > cannot become defined:
> > ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., k}| = ℵo . (*)
> >
> > Either they do not exist at all. Then we have
> > potential infinity, no completeness,
> For what you (WM) mean by completeness,
> you require elements that are not equal to themselves.

No. They are equal to themselves. But we can't know them.
>
> We exclude from consideration elements which are not
> equal to themselves.

All natnumbers are equal to themselves.
>
> > Or we have completenes such that
> > ℕ \ {1, 2, 3, ...} = { }
> > but then this subtraction can only be performed
> > collectively, not individually as in (*).
> We can _reason about_ ℕ \ {1, 2, 3, ...} by
> describing one of the elements of ℕ \ {1, 2, 3, ...}
> and then, taking only reliable statement-steps,
> arrive at claims in addition to the description.

No. Show the step from ℕ \ {1, 2, 3, ..., n} =/= { } to ℕ \ {1, 2, 3, ...} = { }.

> All of the elements are equal to themselves.
> Apparently, things which are equal to themselves
> are not dark

Wrong.

Regards, WM

On Monday, February 28, 2022 at 8:00:52 PM UTC+1, WM wrote:
> Fritz Feldhase schrieb am Montag, 28. Februar 2022 um 18:54:23 UTC+1:
> > On Sunday, February 27, 2022 at 10:34:55 PM UTC+1, WM wrote:
> >
> > | "It should also be possible to reach this configuration by exchanging pairs of fractions."
> >
> > But "reach" in what sense exactly?
> >
> In this sense: Every fraction of Cantor's sequence 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, ... [appears] in the first column [of one of the matrices in the sequence].

BEWARE OF THE QUNATIFIER SHIFT!

Well, no such matrix appears in the sequence of matrices you defined.

So no, it should NOT "be possible to reach this configuration by exchanging pairs of fractions."

> > So my question is just: __WHY?__
> >
> This is what Cantor prescribes.

No, Cantor did NOT "prescribe" your sequence of matrices, but a certain sequence of fractions.

> > No justification is given for YOUR claim "It should also be possible to <bla bla bla>".
> >
> It is

nonsense.

On Monday, 28 February 2022 at 15:11:48 UTC-4, WM wrote:
> horand....@gmail.com schrieb am Montag, 28. Februar 2022 um 19:57:15 UTC+1:
> > On Monday, 28 February 2022 at 14:20:55 UTC-4, WM wrote:
> > > Fritz Feldhase schrieb am Montag, 28. Februar 2022 um 14:57:30 UTC+1:
> > > > On Monday, February 28, 2022 at 11:49:25 AM UTC+1, WM wrote:
> > >
> > > > > > We KNOW, since we can PROVE it, that there is a bijection from {2*n : n e IN} onto {2*n + 1 : n e IN}, for example.
> > > > > >
> > > > > You can prove it for every definable natnumber has infinitely many successors, infinitely many of which cannot become defined:
> > > ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., k}| = ℵo .
> > > You cannot prove it for all natnumbers.
> > Ah, the patented quantifier exchange is being set up.
> If there are all natnumbers, then all can be subtracted from ℕ such that nothing remains But when subtracting all definable individuals then almost all remains.

Like I said, quantifier swap. Why would it surprise you that you can't apply that here? (No, I know that already: You are incapable of doing any mathematics. I think sergio has you pegged: "pre-algebra, with memory loss".)

On Monday, February 28, 2022 at 8:20:36 PM UTC+1, WM wrote:

> Show the step from ℕ \ {1, 2, 3, ..., n} =/= { } to ℕ \ {1, 2, 3, ...} = { }.

Since there is no such "step", he can't show you that step.

You know, there is no largest natural number (and hence no predecessor of "it").

Why are you asking for such a silly task?

On 2/28/2022 2:20 PM, WM wrote:
> Jim Burns schrieb
> am Montag, 28. Februar 2022 um 19:24:59 UTC+1:
>> On 2/28/2022 5:46 AM, WM wrote:

>>>> The infinite set doesn't end,
>>>> but _our reasoning_ can end.
>>>
>>> It does. It cannot avoid that every definable natnumber
>>> has infinitely many successors, infinitely many of which
>>> cannot become defined:
>>> ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., k}| = ℵo . (*)
>>>
>>> Either they do not exist at all. Then we have
>>> potential infinity, no completeness,
>>
>> For what you (WM) mean by completeness,
>> you require elements that are not equal to themselves.
>
> No. They are equal to themselves. But we can't know them.

A set 𝕊 is either like {hydrogen,helium,lithium} and
can be matched inside some ⟨1,...,𝑛⟩

or 𝕊 is like the set ℝ of real numbers and
𝕊 contains at least one endless, non-repeating sequence
⟨𝑥₁,𝑥₂,𝑥₃,...⟩

Suppose that 𝕊 is like ℝ

Define 𝑠𝑢𝑐𝑐 on ⟨𝑥₁,𝑥₂,𝑥₃,...⟩
𝑠𝑢𝑐𝑐(𝑥ₖ) = 𝑥ₖ₊₁

Define 𝑖𝑑 on 𝕊
𝑖𝑑(𝑥) = 𝑥

Define 𝑓 to be 𝑠𝑢𝑐𝑐 on ⟨𝑥₁,𝑥₂,𝑥₃,...⟩
Define 𝑓 to be 𝑖𝑑 on 𝕊\⟨𝑥₁,𝑥₂,𝑥₃,...⟩ everything else.

𝑓 is a bijection between 𝕊 and its
proper subset 𝕊\{𝑥₁}

There are no sets containing some ⟨𝑥₁,𝑥₂,𝑥₃,...⟩
which conserve sheep.

>> We can _reason about_ ℕ \ {1, 2, 3, ...} by
>> describing one of the elements of ℕ \ {1, 2, 3, ...}
>> and then, taking only reliable statement-steps,
>> arrive at claims in addition to the description.
>
> No.
> Show the step from
> ℕ \ {1, 2, 3, ..., n} =/= { }
> to
> ℕ \ {1, 2, 3, ...} = { }.

ℕ is like ℝ
It contains some ⟨𝑥₁,𝑥₂,𝑥₃,...⟩

ℕ is not like {hydrogen,helium,lithium}
It cannot be matched with any ⟨1,...,𝑛⟩

It happens that WM formulated :
> FromTheRafters schrieb am Montag, 28. Februar 2022 um 17:36:17 UTC+1:
>> WM presented the following explanation :
>
>>> Of the fact that every definable natnumber has infinitely many successors,
>>> infinitely many of which cannot become defined: ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3,
>>> ..., k}| = ℵo . This defines ℕ_def.
>>>
>>> These however are not all natnumbers because collectively all can be
>>> handled such that not infinitely many successors remain, not even one: ℕ \
>>> {1, 2, 3, ...} = { } . This defines ℕ.
>> It says that there is no difference between the set of natural numbers
>> and the set of natural numbers.
>
> But the set of definable natural numbers differs:

No they don't.

On 2/28/2022 12:23 PM, WM wrote:
> FromTheRafters schrieb am Montag, 28. Februar 2022 um 17:36:17 UTC+1:
>> WM presented the following explanation :
>
>>> Of the fact that every definable natnumber has infinitely many successors,
>>> infinitely many of which cannot become defined: ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3,
>>> ..., k}| = ℵo . This defines ℕ_def.
>>>
>>> These however are not all natnumbers because collectively all can be handled
>>> such that not infinitely many successors remain, not even one: ℕ \ {1, 2, 3,
>>> ...} = { } . This defines ℕ.
>> It says that there is no difference between the set of natural numbers
>> and the set of natural numbers.
>
> But the set of definable natural numbers

no such set.

>
> Regards, WM

On 2/28/2022 1:14 PM, WM wrote:
> Fritz Feldhase schrieb am Montag, 28. Februar 2022 um 19:50:10 UTC+1:
>> On Monday, February 28, 2022 at 11:54:07 AM UTC+1, WM wrote:
>
>> Hint: {n e ℕ : |ℕ \ {1, 2, 3, ..., n}| = ℵo} = ℕ.
>
> What is the reason of ℵo remainings? Why can they not be subtracted?
> Hint: They can, but not individually.
>
> Regards, WM

So we go round and round the loops in your mind that you have reasoned out, using the two or three equations you are familiar with, it seems to make
sense to you. But that is not math at all.

You are trying to play three-card monte with one card.

On Monday, February 28, 2022 at 9:10:32 PM UTC+1, Jim Burns wrot
>
> Define 𝑖𝑑 on 𝕊
> 𝑖𝑑(𝑥) = 𝑥

Remember:

"There is no function. There is no identity mapping from ℕ to ℕ."

As well as:

"there are no mappings between infinite sets possible, not even betwenn ℕ and ℕ."

Why do you ignore such clear statements?

FromTheRafters schrieb am Montag, 28. Februar 2022 um 21:38:35 UTC+1:
> It happens that WM formulated :
> > FromTheRafters schrieb am Montag, 28. Februar 2022 um 17:36:17 UTC+1:
> >> WM presented the following explanation :
> >
> >>> Of the fact that every definable natnumber has infinitely many successors,
> >>> infinitely many of which cannot become defined: ∀n ∈ ℕ_def: |ℕ \ {1, 2, 3,
> >>> ..., k}| = ℵo . This defines ℕ_def.
> >>>
> >>> These however are not all natnumbers because collectively all can be
> >>> handled such that not infinitely many successors remain, not even one: ℕ \
> >>> {1, 2, 3, ...} = { } . This defines ℕ.
> >> It says that there is no difference between the set of natural numbers
> >> and the set of natural numbers.
> >
> > But the set of definable natural numbers differs:
> No they don't.

Counterfactual caim. You can remove every definable number. Almost all will remain.

Regards, WM

On 2/28/2022 4:18 PM, Fritz Feldhase wrote:
> On Monday, February 28, 2022 at 9:10:32 PM UTC+1,
> Jim Burns wrot

>> Define 𝑖𝑑 on 𝕊
>> 𝑖𝑑(𝑥) = 𝑥
>
> Remember:
>
> "There is no function.
> There is no identity mapping from ℕ to ℕ."
>
> As well as:
>
> "there are no mappings between infinite sets possible,
> not even betwenn ℕ and ℕ."
>
> Why do you ignore such clear statements?

Because I have to either ignore that or ignore this

>>>> For what you (WM) mean by completeness,
>>>> you require elements that are not equal to themselves.
>>>
>>> No. They are equal to themselves. But we can't know them.

So, I take turns.
Why not?

Fritz Feldhase schrieb am Montag, 28. Februar 2022 um 20:22:17 UTC+1:
> On Monday, February 28, 2022 at 8:00:52 PM UTC+1, WM wrote:

> Well, no such matrix appears in the sequence of matrices you defined.

There is no last term of Cantor's sequence. There is no last term of the sequence of matrices.
>
> So no, it should NOT "be possible to reach this configuration by exchanging pairs of fractions."

If Cantor's sequence can be complete, then the sequence can be completely in the first column of a matrix.

> > > So my question is just: __WHY?__
> > >
> > This is what Cantor prescribes.
> No, Cantor did NOT "prescribe" your sequence of matrices, but a certain sequence of fractions.

This sequence can be everywhere including the first column of a matrix. No? Everywhere but not there?

Regards, WM

Jim Burns schrieb am Montag, 28. Februar 2022 um 21:10:32 UTC+1:
> On 2/28/2022 2:20 PM, WM wrote:

> >> We can _reason about_ ℕ \ {1, 2, 3, ...} by
> >> describing one of the elements of ℕ \ {1, 2, 3, ...}
> >> and then, taking only reliable statement-steps,
> >> arrive at claims in addition to the description.
> >
> > No.
> > Show the step from
> > ℕ \ {1, 2, 3, ..., n} =/= { }
> > to
> > ℕ \ {1, 2, 3, ...} = { }.
> ℕ is like ℝ
> It contains some ⟨𝑥₁,𝑥₂,𝑥₃,...⟩
>
> ℕ is not like {hydrogen,helium,lithium}

Therefore you cannot take only reliable statement-steps from ℕ_def to ℕ.

Regards, WM

On 2/28/2022 1:18 PM, Fritz Feldhase wrote:
> On Monday, February 28, 2022 at 9:10:32 PM UTC+1, Jim Burns wrot
>>
>> Define 𝑖𝑑 on 𝕊
>> 𝑖𝑑(𝑥) = 𝑥
>
> Remember:
>
> "There is no function. There is no identity mapping from ℕ to ℕ."
>
> As well as:
>
> "there are no mappings between infinite sets possible, not even betwenn ℕ and ℕ."
>
> Why do you ignore such clear statements?

Mapping the infinite natural numbers to infinite even natural numbers:

[0] = 0
[1] = 2
[2] = 4
[3] = 6
[4] = 8
[5] = 10
[6] = 12
[7] = 14
[8] = 16
[9] = 18
....

[n] = n * 2

Is that Kosher?

On 2/28/2022 1:02 AM, FromTheRafters wrote:
> Chris M. Thomasson expressed precisely :
>> On 2/27/2022 10:56 PM, Chris M. Thomasson wrote:
>>> On 2/27/2022 1:14 AM, FromTheRafters wrote:
>>>> Chris M. Thomasson submitted this idea :
>>>>> On 2/25/2022 3:49 AM, FromTheRafters wrote:
>>>>>> Chris M. Thomasson pretended :
>>>>>>> On 2/24/2022 4:54 AM, WM wrote:
>>>>>>>> Fritz Feldhase schrieb am Mittwoch, 23. Februar 2022 um 16:18:03
>>>>>>>> UTC+1:
>>>>>>>>
>>>>>>>>> Can you actually PROVE your claim? (I got the impression that
>>>>>>>>> actually, say, (M_k)_1,2 --> oo. So far I see (M_k)_1,2 = (2/1,
>>>>>>>>> 3/1, 6/1, ...) and a little consideration shows that (M_k)_1,2
>>>>>>>>> diverges, no?)
>>>>>>>>
>>>>>>>> Like the sequence 1, 2, 3, ... . Important is only that never an
>>>>>>>> empty place can occur when two fractions are exchanged.
>>>>>>>
>>>>>>> 1, 2, 3, ... can be used to index every fraction.
>>>>>>
>>>>>> I disagree. We should be talking about rational numbers, not
>>>>>> fractions.
>>>>>
>>>>> I am just thinking that 1, 2, 3... is infinite. Therefore it can be
>>>>> used to index any other infinite set?
>>>>
>>>> It is only countably so. I retract my statement about fractions
>>>> though. It is just that I was taught long ago that fractions are
>>>> part of a whole, not just simply a whole number ratio as is
>>>> evidently the case now. In fact 'whole numbers' weren't used at all
>>>> after about the third grade. The naturals are 'positive integers'
>>>> and the naturals with zero are non-negative integers.
>>>>
>>>> If you take a circle's diameter as the unit, and wrap that unit
>>>> around the circumference you get three 'whole units' as an 'integer
>>>> part' and a 'fractional part' of that whole which can be expressed
>>>> in decimal radix as <integer part>.<fractional part> which
>>>> fractional part is not even necessarily in Q.
>>>>
>>>> The Continued Fractional Expansion of the value can sometimes tell
>>>> us more about it being irrational by revealing an unending pattern
>>>> such as with the noble numbers and other clear unending patterns.
>>>> The rational fractions will terminate.
>>>
>>> The rational fractions will repeat, and then can be terminated. But,
>>> how many times do we have to iterate the digits in order find a point
>>> of repetition?
>>
>> Think of a decimal whose digits were created by a TRNG from [0...9].
>> There would be no repeat, and no pattern.
>
> No formula either. Can we say it 'represents' a number?

Humm... Theoretically there can be a sequence of infinite dice rolls
that just happen to perfectly equal pi? Is that okay to even ponder on?

> IMO we can say
> it has a place on the real number line if we can just 'nail down' which
> number it represents. Since we cannot do that with 'random' sequences I
> don't consider them.
>
>> Toss a ten sided die for each digit... When does it perfectly repeat
>> itself?

Fritz Feldhase schrieb am Montag, 28. Februar 2022 um 20:42:51 UTC+1:
> On Monday, February 28, 2022 at 8:20:36 PM UTC+1, WM wrote:
>
> > Show the step from ℕ \ {1, 2, 3, ..., n} =/= { } to ℕ \ {1, 2, 3, ...} = { }.
> Since there is no such "step", he can't show you that step.

But he claims that all ℕ can be stepped through. It is impossible. Why is it impossible? Note that all ℕ_def can be stepped through.
> You know, there is no largest natural number (and hence no predecessor of "it").

But there is nothing left when all are removed to ℕ \ {1, 2, 3, ....}.

By the way "no largest natural number" implies "no last fraction" in 1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1, 1/5, ... neither in any column nor elsewhere. Only initial segments. All possible initial segments are in my sequence of matrices. Nowhere is more. Therefore I do exactly what Cantor does --- only a bit more carefully.

Regards, WM

On 2/28/2022 5:51 AM, Fritz Feldhase wrote:
> On Monday, February 28, 2022 at 12:38:30 AM UTC+1, Chris M. Thomasson wrote:
>> On 2/27/2022 10:30 AM, Fritz Feldhase wrote:
>>> On Sunday, February 27, 2022 at 4:20:33 AM UTC+1, Chris M. Thomasson wrote:
>>>
>>>> I am just thinking that 1, 2, 3... is infinite. Therefore it can be used
>>>> to index any other infinite set?
>>>
>>> No, it can not be used to index, say, IR. (You will always miss some elements in IR.)
>>>
>>> See: https://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument
>>>
>> The the natural numbers should be able to be used to index any
>> irrational. Think of the digits of pi:
>>
>> [0] = 3
>> [1] = 1
>> [2] = 4
>> [3] = 1
>> [4] = 5
>> [5] = 9
>> [...] = [...]
>>
>> PI has infinite digits, and the natural numbers are infinite in nature, so it
>> can be used to index an irrational?
>
> THIS wasn't your question, idiot. [Now you are just "indexing" the "digits" of pi.]

Oh, shit. Sorry.

> Your question was:
>>>> I am just thinking that 1, 2, 3... is infinite. Therefore it can be used
>>>> to index any other infinite set?
>
> Answer: No, it can't.
>
> EOD.

If an infinite set is comprised of elements, then each element can be
indexed by the natural numbers, right?

On 2/28/2022 1:11 AM, FromTheRafters wrote:
> Chris M. Thomasson laid this down on his screen :
>> On 2/27/2022 10:30 AM, Fritz Feldhase wrote:
>>> On Sunday, February 27, 2022 at 4:20:33 AM UTC+1, Chris M. Thomasson
>>> wrote:
>>>
>>>> I am just thinking that 1, 2, 3... is infinite. Therefore it can be
>>>> used
>>>> to index any other infinite set?
>>>
>>> No, it can not be used to index, say, IR. (You will always miss some
>>> elements in IR.)
>>>
>>> See: https://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument
>>
>> The the natural numbers should be able to be used to index any
>> irrational. Think of the digits of pi:
>>
>> [0] = 3
>> [1] = 1
>> [2] = 4
>> [3] = 1
>> [4] = 5
>> [5] = 9
>> [...] = [...]
>>
>>
>> index 3 = digit 1 of the irrational in question. It should work. PI
>> has infinite digits, and the natural numbers are infinite in nature,
>> so it can be used to index an irrational?
>
> That's not how I see it, that is not indexing. Say, for the sake of
> argument, that pi is the twenty-first irrational or transcendental
> number. By indexing, what is the twenty-sixth transcendental number in
> that ordering?
>
> Hint: Iterate your successor function.

You would have to get a function that can create pi on a digit-by-digit
basis. You give it a position, say 26'th, and it creates that specific
digit. So, it iterates 26 times to get it. Not sure how to get it
directly without some sort of iteration...

WM has brought this to us :
> FromTheRafters schrieb am Montag, 28. Februar 2022 um 21:38:35 UTC+1:
>> It happens that WM formulated :
>>> FromTheRafters schrieb am Montag, 28. Februar 2022 um 17:36:17 UTC+1:
>>>> WM presented the following explanation :
>>>>> Of the fact that every definable natnumber has infinitely many
>>>>> successors, infinitely many of which cannot become defined: ∀n ∈ ℕ_def:
>>>>> |ℕ \ {1, 2, 3, ..., k}| = ℵo . This defines ℕ_def.
>>>>>
>>>>> These however are not all natnumbers because collectively all can be
>>>>> handled such that not infinitely many successors remain, not even one: ℕ
>>>>> \ {1, 2, 3, ...} = { } . This defines ℕ.
>>>> It says that there is no difference between the set of natural numbers
>>>> and the set of natural numbers.
>>>
>>> But the set of definable natural numbers differs:
>> No they don't.
>
> Counterfactual caim. You can remove every definable number. Almost all will
> remain.

You said 'definable natural number' before. For those, each defined one
is followed by the next (hence defined) one, so all are defined and
therefore were definable. It is just the natural numbers again, nothing
hiding in darkness.