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

On 10.01.2025 10:15, joes wrote:

Am Thu, 09 Jan 2025 22:55:13 +0100 schrieb WM:

On 09.01.2025 21:17, joes wrote:

Am Thu, 09 Jan 2025 19:25:19 +0100 schrieb WM:

>

Losing all numbers but keeping infinitely many is impossible in
inclusion-monotonic sequences.

This case doesn't occur.

Loss of all numbers is proven by the empty intersection.
Keeping infinitely many is poved by Fritsche.

...for different cases. There is no empty segment, each is infinite.

Without empty endsegment, not all numbers become indices. Note that bijections need all the indices. There is no limit accepted.

What element of ℕ does not become an index?

Those which remain in all endsegments. If they did not remain, not all endsegments could be infinite, because of inclusion monotony.
Regards, WM
Regards, WM