Re: The set of necessary FISONs

On 20.02.2025 03:23, Richard Damon wrote:

On 2/19/25 11:52 AM, WM wrote:

Proof: If UF = ℕ is assumed, then F(1) can be omitted without changing the union of the remainder. And if F(n) can be omitted without changing this union, then also F(n+1) can be omitted without changing this union. That makes the omitted FISONs the inductive collection of all FISONs and proves the implication: If UF = ℕ, then { } = ℕ.

But induction doesn't subtract elements.

By induction we can prove what FISONs are useless and can be subtracted, namely all FISONs satisfying |ℕ \ {1, 2, 3, ..., n}| = ℵo.

 All you have shown is that the no element in the set of all FISON is neeeded.

And no FISON is useful.

 The problem is your "set" UF isn't being defined by a proper set theory, but just by Naive Set Theory.

My set is defined by induction like the set of definable numbers or FISONs is defined by Peano, Dedekind, Cantor, Zermelo, Schmidt, v. Neumann, Lorenzen.
Regards, WM