Re: The set of necessary FISONs

On 2/20/25 4:44 AM, WM wrote:

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.

No, you can prove that none of them are NECESSARY, but they can do the job.

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

 And no FISON is useful.

No, YOU (and your logic) is not 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.

But sets aren't defined "by induction", so your set isn't defined.
Please re-read those theories and see how to actually construct a set.
Of course, if you think the axiom of infinity was proven by something, that shows you don't understand the basic meaning of the words, since axioms, BY DEFINITION, can not be proven.

 Regards, WM