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. ∀n ∈ ℕ: f([1, n]) =< 1/10.
Ever heard of the effect of the universal quantifier?

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.
Regards, WM