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

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.

Don't blather nonsense. If all endsegments are infinite then infinitely many natbumbers remain in all endsegments.
Infinite endsegments with an empty intersection are excluded by inclusion monotony. Because that would mean infinitely many different numbers in infinite endsegments.

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.

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.

 

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.
Regards, WM