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

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. In non-empty endsegments there are common finite cardinals.

 

and a sequence of their intersections
E(1), E(1)∩E(2), E(1)∩E(2)∩E(3), ... .

 With an empty set of common finite.cardinals.
 

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.

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.
Regards, WM