Re: Incompleteness of Cantor's enumeration of the rational numbers (extra-ordinary)

On 04.12.2024 21:36, Jim Burns wrote:

On 12/4/2024 12:29 PM, WM wrote:

The set of all endsegments
can be subdivided into two sets,
one of which is finite and the other is infinite.

 The intersection of the infinite one is empty.

The intersection of the two sets is not empty, wherever the cut is made.

E(1)∩E(2)∩...∩E(n) = E(n).

No finite.cardinal is in the intersection of
all end.segments.

No finite cardinal is in all endsegments.
The sequences of E(1)∩E(2)∩...∩E(n) and of E(n) both have an empty limit.

 No intersection of
more.than.finitely.many end.segments
of the finite.cardinals
holds a finite.cardinal,  or
is non.empty.

Small wonder. More than finitely many endsegments require infinitely many indices, i.e., all indices. No natnumbers are remaining in the contents.
Regards, WM