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

On 28.11.2024 17:45, joes wrote:

Am Thu, 28 Nov 2024 11:39:05 +0100 schrieb WM:

 

A simpler arguments is this: All endsegments are in a decreasing
sequence.

There is no decrease, they are all infinite.

Every endsegment has one number less than its predecessor.
That is called decrease.

Before the decrease has reached finite endsegments, all are infinite
and share an infinite contents from E(1) = ℕ on. They have not yet had
the chance to reduce their infinite subset below infinity.

All segments are infinite. Nothing can come "afterwards".

Then the intersection is never empty.