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

On 29.11.2024 22:36, Jim Burns wrote:

The end.segments are infinite.
Their intersection is empty.

Contradiction in terms of inclusion monotony! The intersection is an endsegment.

Nothing is infinite and empty.
 

Up to every infinite endsegment E(n) the index n is finite and the intersection is infinite.
∀k ∈ ℕ_def: ∩{E(1), E(2), ..., E(k)} = E(k).
There is no infinite set of indices in ℕ followed by the infinite contents of endsegments. Therefore there is no infinite set of infinite endsegments possible. Either the set of indices is infinite, then the remaining contents is empty, or the remaining contents is infinite, then the set of indices is finite.
Try to show a counter example: Infinitely many indices and simultaneously infinite remainig contents.
Regards, WM