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

On 04.12.2024 18:16, Jim Burns wrote:

On 12/4/2024 8:31 AM, WM wrote:

Below: two is finite.

No, they may be finite or infinite.

 

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.

 No.
Not all sets of end.segments
can be subdivided into two FINITE sets.

The set of all endsegments can be subdivided into two sets, one of which is finite and the other is infinite.

GREATERS is inclusion.monotonic and {}.free.

Then

⋂GREATERS = {}

is wrong.
E(1), E(2), E(3), ...
and
E(1), E(1)∩E(2), E(1)∩E(2)∩E(3), ...
are identical for every n and in the limit because
E(1)∩E(2)∩...∩E(n) = E(n).
For very naive readers I recommend the bathtub. All its states have a non-empty intersection unless one of the states is the empty state.
Regards, WM
Regards, WM