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.logicDate : 23. Nov 2024, 22:20:25
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <96af151c-285d-4161-842a-63019cac9699@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
User-Agent : Mozilla Thunderbird
On 11/23/2024 12:23 PM, WM wrote:
On 23.11.2024 16:33, Jim Burns wrote:
For _your_ end.segments,
k ↦ k+1 : one.to.one
E(k) → E(k+1) : one.to.one
|E(k)| ≤ |E(k+1)|
>
No.
|E(k)| ≥ |E(k+1)| = |E(k)| - 1.
E(k) ⊇ E(k+1)
|E(k)| ≥ |E(k+1)|
doesn't contradict
|E(k)| ≤ |E(k+1)|
Together,
|E(k)| = |E(k+1)|
and
|E(k+1)| doesn't lose one number.
Finite cardinalities can lose one number.
After all the finite cardinalities,
there are the infinite cardinalities,
which aren't finite
and can't lose one number.
|E(k)| = |E(k+1)| is infinite.
----
Do you (WM) object to
k ↦ k+1 : one.to.one
Do you (WM) object to
E(k) → E(k+1) : one.to.one
If you object, why?
If you don't object,
one consequence is
|E(k)| ≤ |E(k+1)|
and
|E(k)| = |E(k+1)|
and
|E(k)| = |E(k+1)| is infinite.
…