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

On 04.12.2024 11:33, FromTheRafters wrote:

WM formulated the question :

On 03.12.2024 21:34, Jim Burns wrote:

On 12/3/2024 8:02 AM, WM wrote:

>

E(1)∩E(2)∩...∩E(n) = E(n).
Sequences which are identical in every term
have identical limits.

>
An empty intersection does not require
  an empty end.segment.

>
A set of non-empty endsegments has a non-empty intersection. The reason is inclusion-monotony.

 Conclusion not supported by facts.

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.
Same with a set of endsegments. It can be divided into two sets for both of which the same is required.
Regards, WM