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

On 1/10/25 5:12 AM, WM wrote:

On 10.01.2025 04:22, Moebius wrote:
 

Again, referring to the sucessor operation s, we have
>
            {1, 2, 3, 4, ...} = {s0, s1, s2, s3, ...} .
>
If we NOW compare
>
                    {s0, s1, s2, s3, ...}       (= {1, 2, 3, 4, ...})
with
                    { 0,  1,  2,  3, ...} ,
>
does ist STILL make sense to claim "everybody can see that they are not equal in size"?

 The set of natural numbers reaches from zero to omega exclusively: (0, ω). The set {s0, s1, s2, s3, ...} can be mapped on the set {0,  1,  2, 3, ...}. Then all pairs (sn, n) do exist. If however s0 is shifted, i.e., mapped on 1, s1 on 2, and so on, then 0 has no longer a partner and the last sn is mapped on ω.
 Note: Actual infinity is brought about by the vast realm of dark numbers.
 Regards, WM
 

Nop, Dark Numbers don't exist, but are a figment of your imagination cause by your need to try to invent numbers that don't exist in the set you are looking.
Your "logic" is based FUNDAMENTALLY on the LIE that you can imagine the "last" value of an actually infinte series, which BY DEFINITION doesn't have a last element.
This LIE causes your logic to blow itself up into smithereens, taking your reasoning power with it, by the contradictions created by your lies.