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

On 30.11.2024 18:45, joes wrote:

Am Sat, 30 Nov 2024 18:20:51 +0100 schrieb WM:

For an intersection, the "smallest" set matters, which there isn't in
this infinite sequence, only a "biggest".

If all sets are infinite, then there is no smaller set than an infinite
set.

True. All endsegments are infinite. But they form a chain of inclusion,
and there is no smallest set, because that chain is infinite.

There is an infinite sequence of endsegments E(1), E(2), E(3), ...
and an infinite sequence of their intersections
E(1), E(1)∩E(2), E(1)∩E(2)∩E(3), ... .
Both are identical - from the first endsegment on until every existing endsegment.

The intersection of the "finite initial segment" of endsegments is
∩{E(1), E(2), ..., E(k)} = E(k)
is a function which remains infinite for all infinite endsegments. If
all endsegments remain infinite forever, then this function remains
infinite forever.

It does for all finite k.

Of course. Only for finite k the endsegments are infinite.

All natural k are finite.

Then all endsegments are infinite like their intersections.
Regards, WM