Re: Incompleteness of Cantor's enumeration of the rational numbers

Am Fri, 22 Nov 2024 13:08:28 +0100 schrieb WM:

On 22.11.2024 08:49, Moebius wrote:

Am 22.11.2024 um 03:58 schrieb Chris M. Thomasson:

On 11/21/2024 1:45 PM, WM wrote:

On 21.11.2024 22:05, joes wrote:

 

Counting concerns every single number.

Every single natural can be counted to.

Nonsense.

Proof by induction:

Induction proves that every initial segment of endsegments has an
infinite intersection.

True. But not the intersection of all of them, since that is the (ugh)
infinite initial segment, and "infinite" itself is not a natural number
covered by induction - only all the naturals themselves.

1 can be counted to (obviously). If n (where n is a natural number) can
be counted to, then n+1 can be counted to (obviously). Hence for each
and every natural numbers n: n can be counted too. qed

But not all endsegments have an infinite intersection.

Yes they do (with what?).

All endsegments have an empty intersection.

No, all segments have an infinite intersection with each other,
namely the one that comes "later", with a larger index.

Since every endsegment can lose only one
number, there must be infinitely many endsegments involved in reducing
the intersection from infinite to empty.

Exactly. That is all of them, there are infinitely segments.

What one cannot be counted to?

Just the indices involved in reducing the intersection from infinite to
empty. They are dark.

They don't exist. That is why they are disregarded.

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Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.