Sujet : Re: Incompleteness of Cantor's enumeration of the rational numbers (extra-ordinary)
De : james.g.burns (at) *nospam* att.net (Jim Burns)
Groupes : sci.mathDate : 05. Dec 2024, 23:20:10
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <ae606e53-0ded-4101-9685-fa33c9a35cb9@att.net>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27
User-Agent : Mozilla Thunderbird
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.