Re: The set of necessary FISONs

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.

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. Inductive sets have no last element. Therefore the set of FISONs which could have the union ℕ has no first element. That means it is empty.
Therefore U(A(n)) = ℕ ==> U{ } = { } = ℕ. This is false. By contraposition we get ~{ } = ℕ ==> ~ U(A(n)) = ℕ.
Regards, WM