Re: how

Le 21/06/2024 à 21:13, Jim Burns a écrit :

However,
if all FISONs are removed,
no remaining FISON is a superset.

The set of necerssary FISONs, if existing, must have a smallest element - according to Cantor. But it has not. Therefore there is no necessary FISON. We cannot find a necessary FISON because each one covers only a tiny subset of ℕ. ℵo Elements are missing. That is not expalint by quantifier magic but by mathematical facts: Dark numbers.

The convention is that ⋃{} = {}
But, either way, ⋃{} ≠ ℕ

Of course not. My proof only shows that IF ℕ is the union of FISONs, THEN ℕ = ⋃{}.

 

Among the FISONs of ℕ there is not,
in any enumeration,
a first one that is required to yield
the union ℕ.

 For each FISON, another FISON is proper superset.
∀ᶠⁱˢᵒⁿ⟨0…j⟩ ∃ᶠⁱˢᵒⁿ⟨0…k⟩≠⟨0…j⟩: ⟨0…j⟩ ⊂ ⟨0…k⟩
 Therefore,
there is NO FISON superset all FISONs.

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