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

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