Re: The set of necessary FISONs

On 06.02.2025 19:54, Jim Burns wrote:

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

On 06.02.2025 15:57, Jim Burns wrote:

 

The key is that  ∀ᴺ¹n: ∃ᴺ¹j′: n<j′

>
The key is that
the set ℕ is created by induction.

 The set ℕ₁ is described as having induction valid for it.

Then it is the collection ℕ_def of definable numbers.

 Sets missing natural numbers and
sets with extra, non.inducible, un.natural numbers
are not ℕ₁

Then it is the set ℕ of all natural numbers.
You contradict yourself.

 

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 ∅,

 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.

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
ℕ₁ appears nowhere in it.

The axiom of induction holds for all predicates P which satisfy induction.
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.
Regards, WM