Sujet : Re: Incompleteness of Cantor's enumeration of the rational numbers (extra-ordinary)
De : wolfgang.mueckenheim (at) *nospam* tha.de (WM)
Groupes : sci.mathDate : 08. Dec 2024, 11:50:08
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <vj3tl0$3nktg$2@dont-email.me>
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
User-Agent : Mozilla Thunderbird
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.
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 you understand here?
This is not gibberish but mathematics:
∀n ∈ ℕ: E(1)∩E(2)∩...∩E(n) = E(n).
True.
Every counter argument has to violate this.
False.
For example, see above.
Every counter argument violates this.
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.
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.
Regards, WM
…