Re: The set of necessary FISONs

On 14.02.2025 18:32, joes wrote:

You claim that UF = X and also U({}) = X.

No. You and others have claimed that there is a set of FISONs containing all natural numbers. I have accepted that a s premise.

If UF=N and N={}, then UF={}.

No, learn what an implication is! In W. Mückenheim: "Mathematik für die ersten Semester", 4th ed., De Gruyter, Berlin (2015) for instance it is explained for beginners.
There is no FISON that is in the claimed set. Therefore
If UF = ℕ then ℕ = {}.
This does *not* prove UF = {}. F contains sets and UF contains natural numbers, but not enough natural numbers, not ℕ.
Regards, WM