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

On 12/5/2024 4:00 AM, WM wrote:

On 04.12.2024 21:36, Jim Burns wrote:

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

[...]

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

>
Small wonder.
More than finitely many endsegments
require
infinitely many indices, i.e., all indices.
No natnumbers are remaining in the contents.

⎛ That's the intersection.
⎜
⎜ The intersection of
⎜ more.than.finitely.many end.segments
⎜ of the finite.cardinals
⎝ is empty.
⎛ For each finite cardinal,
⎜ there are more.than.that.many finite.cardinals.
⎜
⎜ There are
⎝ more.than.finitely.many finite.cardinals.
⎛ For each finite cardinal,
⎜ there is an end.segment
⎜
⎜ There are
⎜ more.than.finitely.many end.segments
⎝ of the finite.cardinals
⎛ The intersection of
⎜ more.than.finitely.many end.segments
⎜ of the finite.cardinals
⎜ is empty.
⎜
⎜ The intersection of all end.segments
⎜ of the finite.cardinals
⎝ is empty.
Also...
⎛ For each finite.cardinal,
⎜ the union of
⎜ its end.segment and its fore.segment
⎜ holds all the finite.cardinals, and
⎜ is an infinite set.
⎜
⎜ For each finite.cardinal,
⎜ the union of
⎜ its fore.segment (finite) and any finite set
⎜ is a finite set.
⎜
⎜ For each finite.cardinal,
⎝ its end.segment is not a finite set..
⎛ The intersection of all end.segments
⎜ of the finite.cardinals (each infinite)
⎝ is empty.
----

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.

>
The intersection of the two sets is not empty,
wherever the cut is made.

E(1)∩E(2)∩...∩E(n) = E(n).

Our sets don't change.
E(n)∩{} = {}
E(1)∩E(2)∩...∩E(n) = E(n)
E(n+1)∩E(n+2)∩... = {}
The intersection of the two sets is empty,
wherever the cut is made.
∀k ∈ E(n+1):
⎛ k ∉ E(k+1) ∈ {E(n+1),E(n+2),...}
⎝ k ∉ ⋂{E(n+1),E(n+2),...}
⋂{E(n+1),E(n+2),...} = {}
E(n+1)∩E(n+2)∩... = {}
(E(1)∩E(2)∩...∩E(n)) ∩ (E(n+1)∩E(n+2)∩...) =
E(n)∩{} =
{}

No finite.cardinal is in the intersection of
all end.segments.

>
No finite cardinal is in all endsegments.
The sequences of E(1)∩E(2)∩...∩E(n) and
of E(n) both have an empty limit.
>

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

>
Small wonder.
More than finitely many endsegments
require
infinitely many indices, i.e., all indices.
No natnumbers are remaining in the contents.

Yes.