Re: The set of necessary FISONs

On 2/8/2025 4:54 PM, WM wrote:

On 08.02.2025 18:43, Jim Burns wrote:

On 2/8/2025 9:21 AM, Jim Burns wrote:

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

Inductive sets have no last element.

Do you use ω-1 ?

>
Not in this proof.:
>
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.

Not for all inductive sets.
For all minimal.inductive sets.
minimal.inductive ≠ inductive ≠ infinite

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

Without exception,
the union of FISONs.after is ℕ
∀ᴺj′:  ⋃{{i:i≤k}:j′<k} = ℕ
⎛ ∀ᴺj′:
⎜ ∀ᴺi′ ∃ᴺk′:  i′≤k′ ∧ j′<k′ ∧
⎜  i′ ∈ {i:i≤k′} ∈ {{i:i≤k}:j′<k}
⎜
⎜ Proof: Let k′ = max{i′,j′+1}
⎜
⎜ ∀ᴺj′:
⎜ ∀ᴺi′:  i′ ∈ ⋃{{i:i≤k}:j′<k}
⎜
⎝ ∀ᴺj′:  ⋃{{i:i≤k}:j′<k} ⊇ ℕ

Inductive sets have no last element.

⎛ ∀ᴺj′:  ⋃{{i:i≤k}:j′<k} ⊇ ℕ
⎝ without exception, the union of FISONs.after is ℕ
because
⎛ ∀ᴺj′ ∃ᴺk′: j′<k′
⎝ inductive sets have no last element.

Therefore U(A(n)) = ℕ ==>
U{ } = { } = ℕ.

Why that '==>' ?
My best guess at why you claim  U{ } = { } = ℕ
is that you (WM) are assuming that,
for some FISON (ie, F(ω-1)) such that
there are no FISONs.after.
You (WM) haven't given any other reason.