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

On 12/29/2024 3:09 AM, WM wrote:

On 28.12.2024 20:42, Moebius wrote:
 

Hint: For each and every n e IN there is a singleton such that n is in it (namely {n}).
>
BUT there's no singleton such that each and every n e IN is in it.

 My theorem: Every union of FISONs which stay below a certain threshold stays below that threshold.

The threshold is finite.
{ 1 }, { 1, 2 }, { 1, 2, 3 }, ...
threshold = 3
{ 1 }, { 1, 2 }, { 1, 2 (3) }, three is invalid so:
{ 1 }, { 1, 2 }
// combining things and removing duplicates
{ 1, 2 }
This is finite thinking.

 Find a counterexample. Don't claim it but prove it. Fail.
 Regards, WM