William schrieb am Freitag, 24. Juni 2022 um 19:06:04 UTC+2:
> On Friday, June 24, 2022 at 1:24:52 PM UTC-3, WM wrote:
> > William schrieb am Freitag, 24. Juni 2022 um 18:05:45 UTC+2:
> at appears to work if not checked
> > > |N exists and is a Peano set (axiom of infinity). f(n) exists for every element of |N (induction). Hence the bijection { (n,f(n) | n in |N} exists.
> <snip>
> >That function does not exist except the first few terms.
> Nope. |N is a Peano set thus induction holds for every element of |N

Induction holds for every matrix of the infinite sequence

1/1, 1/2, 1/3, 1/4, ...
2/1, 2/2, 2/3, 2/4, ...
3/1, 3/2, 3/3, 3/4, ...
4/1, 4/2, 4/3, 4/4, ...
5/1, 5/2, 5/3, 5/4, ...
....

1/1, 2/1, 1/3, 1/4, ... 1/1, 3/1, 1/3, 1/4, ... 1/1, 3/1, 4/1, 1/4, ... 1/1, 3/1, 4/1, 1/4, ...
1/2, 2/2, 2/3, 2/4, ... 1/2, 2/2, 2/3, 2/4, ... 1/2, 2/2, 2/3, 2/4, ... 1/2, 5/1, 2/3, 2/4, ...
3/1, 3/2, 3/3, 3/4, ... 2/1, 3/2, 3/3, 3/4, ... 2/1, 3/2, 3/3, 3/4, ... 2/1, 3/2, 3/3, 3/4, ...
4/1, 4/2, 4/3, 4/4, ... 4/1, 4/2, 4/3, 4/4, ... 1/3, 4/2, 4/3, 4/4, ... 1/3, 4/2, 4/3, 4/4, ...
5/1, 5/2, 5/3, 5/4, ... 5/1, 5/2, 5/3, 5/4, ... 5/1, 5/2, 5/3, 5/4, ... 2/2, 5/2, 5/3, 5/4, ...
.... ... ... ...

and proves that never the attempted matrix

1/1, __, __, __, ...
1/2, __, __, __, ...
1/3, __, __, __, ...
2/2, __, __, __, ...
....

will be reached.

> f(n) exist for every element of |N (induction),

f(n) in Cantor's function shows that always (i.e., for every index you can check) ℵo terms are not checked.

Regards, WM

On Saturday, June 25, 2022 at 9:29:56 AM UTC-3, WM wrote:

> [Induction] proves that never the attempted matrix
>
> 1/1, __, __, __, ...
> 1/2, __, __, __, ...
> 1/3, __, __, __, ...
> 2/2, __, __, __, ...
> ...
>
> will be reached.

Indeed, you can use induction to show that your method does not work. The fact that your method does not work does not mean there is no method that can show that the set { (n,f(n) | n in |N} exists.

> > f(n) exist for every element of |N (induction),
> f(n) in Cantor's function shows that always (i.e., for every index you can check) ℵo terms are not checked.

So what? Induction can be used to prove many things. In particular, induction shows that f(n) exists for every element of |N.
One does not "check" every element of |N, one uses induction.

--
William Hughes

William schrieb am Samstag, 25. Juni 2022 um 15:25:47 UTC+2:
> On Saturday, June 25, 2022 at 9:29:56 AM UTC-3, WM wrote:
>
> > [Induction] proves that never the attempted matrix
> >
> > 1/1, __, __, __, ...
> > 1/2, __, __, __, ...
> > 1/3, __, __, __, ...
> > 2/2, __, __, __, ...
> > ...
> >
> > will be reached.
> Indeed, you can use induction to show that your method does not work.

It is Cantor's method. You can use induction to show that never all fractions are enumerated. Always almost all are missing.

> The fact that your method does not work does not mean there is no method that can show that the set { (n,f(n) | n in |N} exists.

The topic here is Cantor's method. This method does not work.

> > > f(n) exist for every element of |N (induction),
> > f(n) in Cantor's function shows that always (i.e., for every index you can check) ℵo terms are not checked.
> So what? Induction can be used to prove many things. In particular, induction shows that f(n) exists for every element of |N.

Yes. But "every element of |N" is leading astray because always infinitely many elements will follow "every element of |N" (which therefore is not every element of |N).

> One does not "check" every element of |N, one uses induction.

But in the complete sequence |N one could, if desired, check every element of |N. Likewise I can check every desired matrix. But even for those never checked I can prove that they are not completely covered by indexes or, alternatively, that not all elements can be gathered in the first column.

That disproves Cantor's claim. And only that is the topic here!

Regards, WM

On Sunday, June 26, 2022 at 10:06:59 AM UTC-3, WM wrote:

> ... "every element of |N"
is the set |N and is not followed by an element of the set |N. Hence "every element of |N" is "every element of |N" even though each element of |N is followed by an infinite number of elements

>But in the complete sequence |N one could, if desired, check every element of |N.

Nope, the complete sequence of |N (a sequence without last element) contains an infinite number of elements. You cannot "check" each element in an infinite set. To show that something is true for each element of |N you use induction.
--
William Hughes

On Sunday, June 26, 2022 at 3:06:59 PM UTC+2, WM wrote:
> William schrieb am Samstag, 25. Juni 2022 um 15:25:47 UTC+2:
>
> > So what? Induction can be used to prove many things. In particular, induction shows that f(n) exists for every element of |N.
> >
> Yes. But "every element of |N" is leading astray because always infinitely many elements will follow "every element of |N" (which therefore is not every element of |N).

Errr? are you claiming that induction does not prove that A(n) holds for all n e IN, even though it states that A(n) holds for all n e IN (if so)?

Please elaborate!

> > One does not "check" every element of |N, one uses induction.

Right.

> But in the complete sequence |N one could, if desired, check every element of |N.

Really? Then I'd like you to check n + n = 2*n for every element n in IN. Thank you in advance!

> Likewise I can check every desired matrix.

Hell, you are like Chuck Norris!

Remember: He counted to infinity - twice!

But please start with checking n + n = 2*n for every element n in IN

On Sunday, 26 June 2022 at 12:14:42 UTC-3, Fritz Feldhase wrote:
[...]
> > Likewise I can check every desired matrix.
> Hell, you are like Chuck Norris!

*Potentially* half as good....

William schrieb am Sonntag, 26. Juni 2022 um 17:02:21 UTC+2:
> On Sunday, June 26, 2022 at 10:06:59 AM UTC-3, WM wrote:
>
> > ... "every element of |N"
> is the set |N and is not followed by an element of the set |N. Hence "every element of |N" is "every element of |N" even though each element of |N is followed by an infinite number of elements
> >But in the complete sequence |N one could, if desired, check every element of |N.
> Nope, the complete sequence of |N (a sequence without last element) contains an infinite number of elements.

Then not every natural number can be checked. That proves dark = not accessible numbers!

> You cannot "check" each element in an infinite set.

Then not every natural number is accessible, That proves dark = not accessible numbers!

> To show that something is true for each element of |N you use induction.

But that does not concern all elements. Induction is only valid for elements which in principle could be checked. (Proof by induction.)

Regards, WM