Re: The set of necessary FISONs

On 2/8/2025 5:35 AM, WM wrote:

On 07.02.2025 22:45, Jim Burns wrote:

On 2/7/2025 11:39 AM, WM wrote:

On 07.02.2025 17:10, Jim Burns wrote:

On 2/6/2025 2:32 PM, WM wrote:

On 06.02.2025 19:54, Jim Burns wrote:

On 2/6/2025 11:55 AM, WM wrote:

I prefer Wikipedia:
∀P (P(1) /\ ∀k(P(k) ==> P(k+1)) ==> ∀n (P(n)).

>
That's intended to be part of the definition of ℕ₁

>
As well it is
the definition of the collection of all FISONs.

>
Which is curious, when one considers that
the collection of all FISONs appears nowhere in it.

>
The axiom of induction says:
If any property or predicate P satifies
(P(1) /\ ∀k(P(k) ==> P(k+1)),
then it describes all elements of
an inductive = infinite set.

>
All inductive sets are infinite.
Not all infinite sets are inductive.

>
Irrelevant.

inductive ≠ infinite
inductive ≠ valid.for.induction
(despite the name)
minimal.inductive = valid.for.induction

That is satisfied by
the set M of all FISONs which
are useless in  U(A(n)) = ℕ.

>
Only not.followed FISONs are not (your term) uselessᵂᴹ.
Each FISON is followed.

>
Therefore all are useless.
>
All not needed elements belong to an inductive set.

...which is also
a  minimal.inductive = valid.for.induction  set.

Inductive sets have no last element.

Yes.
inductive 𝕎  ⇒  ∀ᵂj ∃ᵂk j<k

Therefore
the set of FISONs which
could have the union ℕ

FISON F(k) = {i:i≤k}
Set of FISONs {{i:i≤k}:k}
Set of sets 𝒜 of FISONs
{𝒜⊆{{i:i≤k}:k}}
Set of sets 𝒜 of FISONs with union ℕ
{𝒜⊆{{i:i≤k}:k}:⋃𝒜=ℕ}
Set of FISONs common to sets with union ℕ
⋂{𝒜⊆{{i:i≤k}:k}:⋃𝒜=ℕ}
Is
⋂{𝒜⊆{{i:i≤k}:k}:⋃𝒜=ℕ}
a set of FISONs with union ℕ
?
Is
⋂{𝒜⊆{{i:i≤k}:k}:⋃𝒜=ℕ} ∈ {𝒜⊆{{i:i≤k}:k}:⋃𝒜=ℕ}
?
If you say yes, how do you know?
Union of FISONs common to sets with union ℕ
⋃⋂{𝒜⊆{{i:i≤k}:k}:⋃𝒜=ℕ}

has no first element. That means it is empty.
>
Therefore U(A(n)) = ℕ

Is
⋃⋂{𝒜⊆{{i:i≤k}:k}:⋃𝒜=ℕ} = ℕ
?

Therefore U(A(n)) = ℕ ==> U{ } = { } = ℕ.
This is false.
By contraposition we get
~{ } = ℕ ==> ~ U(A(n)) = ℕ.