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

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

On 30.11.2024 16:58, joes wrote:

Am Sat, 30 Nov 2024 13:19:46 +0100 schrieb WM:

On 30.11.2024 12:54, FromTheRafters wrote:
>

Finite sets versus infinite sets. Finite ordered sets have a last
element which can be in the intersection of all previously considered
finite sets. Infinite ordered sets have no such last element.

But they have infinitely many elements which contribute to the
intersection.

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.

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.

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Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.