Re: The set of necessary FISONs

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:

If the set M is described as
the smallest set satisfying
1 ∈ M and n ∈ M ==> n+1 ∈ M
then ℕ\M = Ø.

>
ℕ₁ = ∅ satisfies that definition.

>
No. 1 is not in ∅,

...and ∅/M = ∅
That's why we don't define ℕ₁ like that.

Better:
ℕ₁ is the emptiest set M such that
1 ∈ M and n ∈ M ⇒ n+1 ∈ M
Thus:
1 ∈ ℕ₁ and n ∈ ℕ₁ ⇒ n+1 ∈ ℕ₁
∀P:(1 ∈ P and n ∈ P ⇒ n+1 ∈ P) ⇒ ℕ₁ ⊆ P
>
Is ℕ₁ the emptiest set M such that
1 ∈ M and n ∈ M ⇒ n+1 ∈ M ?

 Relevant is the set of FISONs.

ℕ₁ is the emptiest superset M of each FISON.
∀F: FISON F ⇒ F ⊆ ℕ₁
∀M: (∀F: FISON F ⇒ F ⊆ M) ⇒ ℕ₁ ⊆ M

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.

Which is curious, when one considers that
ℕ₁ appears nowhere in it.

>
The axiom of induction holds for
all predicates P which satisfy induction.

Even better.to.know than "which predicates?" is
"which objects are being discussed?"

If the set M is described as the smallest set satisfying
F(1) ∈ M and F(n) ∈ M ==> F(n+1) ∈ M
then M contains
all FISONs which can be subtracted from U(Fn))
without changing the assumed result ℕ.

Also, any superset of (emptiest) M
contains at least what M contains, and
thus also contains all FISONs.
M and its supersets can't all be the single ℕ₁
Our sets don't change.
We define the single ℕ₁ to be
the single (emptiest) M