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

On 12/5/2024 2:30 PM, WM wrote:

On 05.12.2024 18:12, Jim Burns wrote:

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

On 04.12.2024 21:36, Jim Burns 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.

>
And it is the empty endsegment.

Depending upon how 'end.segment' is defined,
{} either is or isn't an end.segment.
Consider the options.
⎛ With {} as an end.segment,
⎜ there are more.than.finite.many end.segments,
⎜ too many for any finite.cardinal to be
⎜ in common with all end segments.
⎜ And therefore,
⎜ the intersection of all holds no finite cardinal.
⎜
⎜ With {} NOT as an end.segment,
⎜ there STILL are more.than.finite.many end.segments,
⎜ too many for any finite.cardinal to be
⎜ in common with all end segments.
⎜ And therefore,
⎝ the intersection of all STILL holds no finite cardinal.
The intersection of all non.empty.end.segments
of the finite.cardinals,
which are each infinite non.empty.end.segments,
is empty.
Because,
⎛ for each finite.cardinal,
⎜ there are fewer finite.cardinals before it
⎝ than there are finite.cardinals after it.
Not because
( an end.segment is empty.