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

On 11.12.2024 20:27, Jim Burns wrote:

 ⋂{E(i):i} = {}.

Of course. But all intersections with finite contents are invisible.

Therefore,
one.element.emptier ℕ\{0}
is not.smaller.than ℕ

It is a smaller set. Cardinalities are not useful.

 That is the reason that
the end.segments of ℕ
stay the same larger.than.any.finite size
as they lose each element of ℕ.

The intersection of all is empty. The sequence of intersections cannot lose more than one element per term. The terms immediately before the empty intersection are finite.
Regards, WM