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 : 24. Nov 2024, 20:26:36
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <09f8a86f-3f75-4af8-a190-0def76c1ab82@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
User-Agent : Mozilla Thunderbird
On 11/24/2024 9:05 AM, WM wrote:
On 23.11.2024 23:10, Jim Burns wrote:
On 11/23/2024 4:39 PM, WM wrote:
On 23.11.2024 22:20, Jim Burns wrote:
Do you (WM) object to
k ↦ k+1 : one.to.one
>
I don't know what that waffle should mean.
>
k ↦ k+1 means the successor operation.
>
'One.to.one' means that,
if j≠k then j+1≠k+1
different numbers have different successors.
>
I am claiming that
different numbers have different successors.
>
Ok.
"Ok, I understand you"
or
"Ok, different numbers have different successors"
?
E(k) ⊇ E(k+1)
|E(k)| ≥ |E(k+1)|
doesn't contradict
|E(k)| ≤ |E(k+1)|
>
It does.
>
Then you (WM) also don't know
what a contradiction is.
>
|E(k)| ≤ |E(k+1)| is wrong.
Different numbers have different successors.
The successor operation is one.to.one
from E(k) to E(k+1)
What we mean by
|E(k)| ≤ |E(k+1)|
is that
there is a one.to.one function
from E(k) to E(k+1)
The successor operation, for example.
So, there is.
So, |E(k)| ≤ |E(k+1)| isn't wrong.
----
Also,
because E(k) ⊇ E(k+1)
|E(k)| ≥ |E(k+1)|
Because |E(k)| ≤ |E(k+1)| and |E(k)| ≥ |E(k+1)|
|E(k)| = |E(k+1)|
Because |E(k)| = |E(k+1)|
|E(k)| = |E(k+1)| doesn't change by 1
and
|E(k)| = |E(k+1)| is an infinite cardinality.
…