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

On 17.12.2024 00:52, Richard Damon wrote:

On 12/16/24 3:30 AM, WM wrote:

On 15.12.2024 21:21, joes wrote:

 

Therefore we use all [1, n].

Those are all finite.

All n are finite.

But N isn't, so the sets [1, n] aren't what the bijection is defined
on.

Every element is the last element of a FISON [1, n]. ℕ is the set of all
FISONs.

I use all FISONs.

All intervals do it because there is no n outside of all intervals [1,
n]. My proof applies all intervals.

And all the intervals are finite, and thus not the INFINITE set N,
which is where the bijection occurs.

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