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

On 12/17/24 4:51 PM, WM wrote:

On 17.12.2024 13:34, Richard Damon wrote:
 

According to Cantor the "bijection" uses all n and nothing more.

>
Right, but no FISON uses contains ALL n.

 But all FISONs contain/are all n.
 Regards, WM
 

???? The FISONs are all finite, so NONE of them contain *ALL* n.
You can have a FISON for ANY n, but a give FISON isn't for ALL n.
You don't seem to understand the difference.
You don't seem to understand that the whole can be more than the sum of its parts.