Re: The set of necessary FISONs

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

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.
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

There is nothing remaining.
>

By the axiom of induction
the result is the empty set.

>
You (WM) agree that  ∀ᴺ¹n: ∃ᴺ¹j′: n<j′

>
No. I agree to what I wrote.

Is ℕ₁ the emptiest set M such that
1 ∈ M and n ∈ M ⇒ n+1 ∈ M ?
If ℕ₁ is, then  ∀ᴺ¹n: ∃ᴺ¹j′: n<j′

The axiom of induction:
∀P:ℕ₁→{⊤,⊥}:

>
That appears like nonsense.

{⊤,⊥} is the set of truth.values.
I could have written {T,F} but
I thought that the polite thing to do was to
not.assume that everyone reading my post
is a native English.speaker.
P:ℕ₁→{⊤,⊥} is a function from ℕ₁ to truth.values.
Put another way, P is a predicate.

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
Which is curious, when one considers that
ℕ₁ appears nowhere in it.
That's why I prefer
∀P:ℕ₁→{⊤,⊥}:
∀P ∈ {⊤,⊥}ᴺ¹:
works, as well.