Re: The set of necessary FISONs

On 22.01.2025 18:17, Jim Burns wrote:

On 1/22/2025 5:29 AM, WM wrote:

For each (finite) FISON, there is
a (finite) FISON larger than it,
and
it is a proper subset of that larger FISON,
which is a proper subset of ℕ

Right. The sequence of FISONs is potentially infinite. For each FISON F(n) there exists F(n^n^n).

 

and therefore not necessary
but completely useless in the union.

 For each FISON,
each natural number in it
is also in a later, fuller.by.one, and larger FISON,
and
is not dropped by dropping that (earlier) FISON.

Right.

 For the union of all (finite) FISONs,
there isn't any
(finite) FISON larger than it.

There is no constant "all" in potential infinity.

 The union of all (finite) FISONs is not finite.

Anyhow there is no set of FISONs the union of which would be ℕ.

 

Each can be dropped.
All cannot be dropped.

>

"All" is not more than the repeated "each". If all could not be dropped, then there was a first FISON that could not be dropped. However, none of the following FISONs is useful or necessary:
∀n ∈ ℕ: |ℕ \ {1, 2, 3, ..., n}| = ℵo.

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