Re: how

Le 22/06/2024 à 19:34, Jim Burns a écrit :

On 6/22/2024 8:11 AM, WM wrote:

if existing [nonempty], must have a smallest element - according to Cantor.
But it has not.
Therefore there is no necessary FISON.

 Yes.
No FISON is necessary  ⇔
Each FISON is unnecessary  ⇐
For each FISON, a proper.superset.FISON exists.

The set has no first element. ==> The set is empty.
If ℕ is union of FISONs, then ℕ is empty.

 

We cannot find a necessary FISON because
each one covers only a tiny subset of ℕ.

 Yes.
 

ℵo Elements are missing.

 No.

∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo

 Each FISON is in {FISON}

Each FISON is Union of all predecessors.

FISON.union U{FISON} and
minimal.inductive ⋂{inductive} are
both inductive  and
both well.ordered.

Minimal set of neccessary FISONs is not well ordered. It has no first element.

Infinity is not finite.

But it is not matheology either.

But every FISON is a very, very proper subset:
∀n ∈ ℕ_def: |ℕ \ {1, 2, 3, ..., n}| = ℵo.
That statement covers all FISONs.

 U{FISON} = ⋂{inductive}
 ∀n ∈ U{FISON}: |U{FISON}\{1,2,3,...,n}| = ℵ₀

Each FISON is Union of all predecessors.
U{FISON} is a FISON.
No FISON has ℵ₀ elements.
No union of FISONs has ℵ₀ elements.
Regards, WM