Re: The set of necessary FISONs

On 21.01.2025 21:35, Jim Burns wrote:

On 1/21/2025 6:45 AM, WM wrote:

For each finite set, there is
a (finite) FISON larger than it,
and that FISON subsets ℕ

It is a proper subset and therefore not necessary but completely useless in the union.

 For each finite set, there is
a (finite) FISON larger than it.

That means the set is potentially infinite.

All can be dropped.

 Each is not all.

All consists of each and not more.

Each can be dropped.
All cannot be dropped.

If you were right, there would be a first FISON required according to Cantor's theorem.

 

But according to Cantor's Theorem B,
every non-empty set of different numbers of
the first and the second number class
has a smallest number, a minimum.
This proves that
the set of indices n of necessary F(n),
by not having a first element,
is empty.

 ℕ = ⋃{FISON}

Contradicted by mathematics, namely Cantor's theorem.
Regards, wM