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

On 26.11.2024 20:44, Jim Burns wrote:

On 11/26/2024 2:15 PM, WM wrote:

On 26.11.2024 19:49, Jim Burns wrote:

 

There are no last end.segments of ℕᶠⁱⁿ
There are no finitely.sized end segments of ℕᶠⁱⁿ
There are no finite cardinals common to
 each end.segment of ℕᶠⁱⁿ

>
That is a contradiction.

 It contradicts ℕᶠⁱⁿ being finite, nothing else.

It contradicts inclusion monotony.

 

If there are no common numbers,
then all numbers must have been lost.
But then no numbers are remaining.

 Yes.

Then also no numbers are remaining in the endsegments.

 Each finite.cardinal k is countable.past to
  k+1 which indexes
   Eᶠⁱⁿ(k+1) which doesn't hold
    k which is not common to
     all end segments.
 Each finite.cardinal k is not.in
  the intersection of all end segments,
  the set of elements common to all end.segments,
   which is empty.
 No numbers are remaining.

That is true. But you claimed that every endsegment is infinite. In an infinite endsegment numbers are remaining. In many infinite endsegments infinitely many numbers are the same.

 

Then there are finite endsegments because
∀k ∈ ℕ: |E(k+1)| = |E(k)| - 1.

 For each cardinal.which.can.change.by.1 j

That does not contradict the fact that infinite endsegments have infinitely many numbers in common and hence an infinite intersection.
Regards, WM