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

On 12/4/2024 12:29 PM, WM wrote:

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 your proof, it's two.

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.

The intersection of the infinite one is empty.

GREATERS is inclusion.monotonic and {}.free.

>
Then
>

⋂GREATERS = {}

>
is wrong.

⎛ Assume that you (WM) are correct.
⎜ Assume 𝔊 ∈ ⋂GREATERS
⎜
⎜ 𝔊 is in common with each greater.segment
⎜ In particular, 𝔊 ∈ G(𝔊) ∈ GREATERS
⎜
⎜ 𝔊 <ᵉᵃᶜʰ G(𝔊)
⎜ Since 𝔊 ∈ G(𝔊),  𝔊 < 𝔊
⎜
⎜ However,
⎜ ¬(𝔊 < 𝔊)
⎝ Contradiction.
Therefore,
you (WM) are incorrect.
⋂GREATERS = {}

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).

⋂GREATERS = {}
{} ∉ GREATERS
⋂ENDS = (⋂GREATERS)∩ℕ = {}∩ℕ = {}
{} ∉ ENDS

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.

No.
No finite.cardinal is in
more.than.finitely.many end.segments.
No finite.cardinal is in common with
all more.than.finitely.many end.segments.
No finite.cardinal is in the intersection of
all end.segments.
No intersection of
more.than.finitely.many end.segments
of the finite.cardinals
holds a finite.cardinal,  or
is non.empty.