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

Am Wed, 27 Nov 2024 20:59:43 +0100 schrieb WM:

On 27.11.2024 16:57, Jim Burns wrote:

On 11/27/2024 6:04 AM, WM wrote:

 

However,
one makes a quantifier shift, unreliable, to go from that to ⛔⎛ there
is an end segment such that ⛔⎜ for each number (finite cardinal) ⛔⎝

the

number isn't in the end segment.

 
If all endsegments are infinite then infinitely
many natbumbers remain in all endsegments.

Yes.

Infinite endsegments with an empty intersection are excluded by
inclusion monotony.

Not by intersecting infinitely many segments.

Because that would mean infinitely many different
numbers in infinite endsegments.

Weird way to put it.

Each end.segment is infinite.

That means it has infinitely many numbers in common with every other
infinite endsegment. If not, then there is an infinite endsegment with
infinitely many numbers but not with infinitely many numbers in common
with other infinite endsegments. Contradiction by inclusion monotony.

That is true.

Their intersection of all is empty. These claims do not conflict.

It conflicts with the fact, that the endsegments can lose elements but
never gain elements.

They are infinite, they don't need to gain elements.

In an infinite endsegment numbers are remaining.
In many infinite endsegments infinitely many numbers are the same.

And the intersection of all, which isn't any end.segment, is empty.

Wrong. Up to every endsegment the intersection is this endsegment. Up to
every infinite endsegment the intersection is infinite. This cannot
change as long as infinite endsegments exist.

I.e. never.

--
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.