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

On 04.12.2024 19:06, FromTheRafters wrote:

WM laid this down on his screen :

In two sets A and B which are non-empty both but have an empty intersection, there must be at least two elements a and b which are in one endsegment but not in the other:
a ∈ A but a ∉ B and b ∉ A but b ∈ B.

 Finite thinking.

Correct thinking is always valid.

 

Same with a set of endsegments.

 No, because they are infinite and have no last element to be in every participating endsegment.

But they have all elements, all of which are finite and obey
E(1)∩E(2)∩...∩E(n) = E(n).
If all endsegments are non-empty, then they have a non-empty intersection unless the above condition holds.
Regards, WM