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

Am Sat, 30 Nov 2024 22:56:00 +0100 schrieb WM:

On 30.11.2024 22:45, Jim Burns wrote:

On 11/30/2024 2:07 PM, WM wrote:

>

But there is a sequence of endsegments E(1), E(2), E(3), ...

With an empty set of common finite cardinals.

Inclusion monotony prevents an empty set of common finite cardinals
without an empty endsegment.

Wrong.

In non-empty endsegments there are common finite cardinals.

Only for finite intersections.
 

and a sequence of their intersections E(1), E(1)∩E(2), E(1)∩E(2)∩E(3),
Both are identical -
from the first endsegment on until every existing endsegment.

Yes, they are identical. And identically empty.

You said that the endsegments are never empty.

The limit is empty.

 

Each infinite collection of end segments of
finite cardinals has an empty intersection.

Because an infinite collection of endsegments requires infinitely many
indices, that is all indices. As long as some contents n, n+1, n+2, ...
is in endsegments, the indices reach only from 1 to n-1.

And as long as you only consider a finite number of segments,

you don't
have infinitely many of them.