Re: how

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

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,

F such that ⋃{FISON}≠⋃({FISON}\{F})

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.

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

Yes.

ℵo Elements are missing.

No.
Each FISON is in {FISON}
FISON.union U{FISON} and
minimal.inductive ⋂{inductive} are
both inductive  and
both well.ordered.
NO FIRST jₓ ∈ U{FISON}:  jₓ ∉ ⋂{inductive}
because
jₓ⁻¹ ∈ ⋂{inductive}  ⇒  jₓ ∈ ⋂{inductive}
¬(jₓ⁻¹ ∈ ⋂{inductive}  ∧  jₓ ∉ ⋂{inductive})
NO FIRST jₓ ∈ U{FISON}:  jₓ ∉ ⋂{inductive}
NO jₓ ∈ U{FISON}:  jₓ ∉ ⋂{inductive}
U{FISON} ⊆ ⋂{inductive}
NO FIRST kₓ ∈ ⋂{inductive}:  kₓ ∉ U{FISON}
because
kₓ⁻¹ ∈ U{FISON}  ⇒  kₓ ∈ U{FISON}
¬(kₓ⁻¹ ∈ U{FISON}  ∧  kₓ ∉ U{FISON})
NO FIRST kₓ ∈ ⋂{inductive}:  kₓ ∉ U{FISON}
NO kₓ ∈ ⋂{inductive}:  kₓ ∉ U{FISON}
⋂{inductive} ⊆ U{FISON}
U{FISON} = ⋂{inductive}

That is not expalint by quantifier magic
but by mathematical facts:

Infinity is not finite.

Dark numbers.

ToMAYto, toMAHto,

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.

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