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

Am Sun, 08 Dec 2024 11:50:08 +0100 schrieb WM:

On 08.12.2024 00:38, Jim Burns wrote:

On 12/7/2024 4:37 PM, WM wrote:

On 07.12.2024 20:59, Jim Burns wrote:

On 12/7/2024 6:09 AM, WM wrote:

On 06.12.2024 19:17, Jim Burns wrote:

On 12/6/2024 3:19 AM, WM wrote:

On 05.12.2024 23:20, Jim Burns wrote:

 

⎜ With {} NOT as an end.segment,

all endsegments hold content.

But no common.to.all finite.cardinals.

Show two endsegments which do not hold common content.

I will, after you show me a more.than.finitely.many two.

There are no more than finitely many natural numbers which can be
shown.
All which can be shown have common content.

All endsegments which can be shown (by their indices) have common
content.

Nope, not in common with all, only for finite intersections.

Each end.segment is more.than.finite and the intersection of the
end.segments is empty.

∀n ∈ ℕ: E(1)∩E(2)∩...∩E(n) = E(n)
What can't I understand here?

That all those n only produce finite intersections.

This is not gibberish but mathematics:
∀n ∈ ℕ: E(1)∩E(2)∩...∩E(n) = E(n).
Every counter argument has to violate this.

False. For example, see above.

>

Each finite.cardinal is not in common with more.than.finitely.many
end.segments.

Of course not. All non-empty endsegments belong to a finite set with an
upper bound.

There are definitely more than finitely many naturals and segments
(none are empty).

Each end.segment has, for each finite.cardinal,
a subset larger than that cardinal.

That is not true for the last dark endsegments. It changes at the dark
finite cardinal ω/2.

What changes to what? Why exactly there?

--
Am Sat, 20 Jul 2024 12:35:31 +0000 schrieb WM in sci.math:
It is not guaranteed that n+1 exists for every n.