Re: The set of necessary FISONs

On 23.01.2025 16:18, Jim Burns wrote:

On 1/23/2025 3:43 AM, WM wrote:

ℕ is a superset of each FISON
Each set which is a superset of each FISON
  is a superset of ℕ

Wrong.

That's a longer.winded way to say that
ℕ is the union of all FISONs

The union of all FISONs does not cover ℕ. Otherwise Cantor's theorem would require the existence of a first necessary FISON.

 Each FISON is a proper subset of ℕ
Each FISON is not ℕ

Therefore each FISON can be dropped from the set of candidates. Nothing remains.

"All" is not more than the repeated "each".

 'All' is complete,
whatever you (WM) mean by 'complete'.
  From the finite.length description of a FISON,
we know that up to a FISON
is not complete, is not all.

Up to every FISON |ℕ \ {1, 2, 3, ..., n}| = ℵo. Since every FISON is the union of all its predecessors we get
F(n): |ℕ \ UF(n)| = ℵo.
If you don't believe in the union of all F(n), find the first exception.
Regards, WM