Re: The set of necessary FISONs

On 2/19/25 11:57 AM, WM wrote:

Am 19.02.2025 um 13:17 schrieb Richard Damon:
 

Name a FISON that can not be put into a set that is sufficient set, one whose union is the set of Natural Numbers.

 Every FISON. There is no sufficient set. If it 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.
 Regards, WM
 

Look up the meaning of sufficient. I don't think you know the meaning of the word.
Note, your subject line uses the word you mean, "necessary", but you ignore the fact that a set of necessary elements doesn't need to exist.
The fact that you can omit something, doesn't make it not sufficient.
Also, you don't understand what "induction" does, it doesn't MAKE a set, it TESTS a set. Of course, it could be that in your Naive logic, you do things different, but Naive logic, like Naive Set Theory, is just incorrect.