Re: The set of necessary FISONs

Am Fri, 24 Jan 2025 10:37:51 +0100 schrieb WM:

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.

For every natural, {1, …, n} exists.
>

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.

Cantor's theorem does not force the existence of a necessary set.
>

"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.

N is not a FISON.
>

Since every FISON is the
union of all its predecessors we get F(n): |ℕ \ UF(n)| = ℵo.

Is there supposed to be a universal quantifier there?
>

If you don't believe in the union of all F(n), find the first exception.

WDYM believe in the union?