Re: The set of necessary FISONs

On 22.02.2025 21:29, Jim Burns wrote:

On 2/22/2025 5:17 AM, WM wrote:

On 21.02.2025 20:40, Jim Burns wrote:

 

However,
all sets Z which satisfy Infinityᶻᶠᶜ
HAVE A SUBSET which
is usable in a proof.by.inductionᶻᶠᶜ

>
Zermelo calls that set Z_0 or set of numbers.

 The usable SUBSET ℕⁱᵒᵒⁱˢˢ of Zermelo's asserted Z
is its.own.only.inductive.sub.set (iooiss).
Our sets do not change.
 I'm pretty sure your ℕ_def = ℕⁱᵒᵒⁱˢˢ
Our sets do not change.

Yes, except that Zermelo starts with zero or empty set.
Please stop your silly labeling. Zermelo calls it Z_0.

⎛ You (WM) say 'inductive S' means
⎜ ∀k:S∋k⇒S∋k+1 ∧ S∋1
⎜ Then
⎜ ℕ₁ⁱᵒᵒⁱˢˢ = {1,2,3,...}

Yes.

⎜
⎜ We (matheologists) say 'inductive S' means
⎜ ∀k:S∋k⇒S∋k+1 ∧ S∋0
⎜ Then
⎜ ℕ₀ⁱᵒᵒⁱˢˢ = {0,1,2,...}

That is not a significant difference.

For each two meanings 'inductiveₓ' and 'inductiveᵥ',

That is not a significant difference.

We can remove all numbers from Z_0 and produce the empty set.

 We can remove all numbers
from {137} and produce the empty set.
 {137} ≠ 137

If we remove all numbers from {137} we produce { }, if we remove all numbers from 137 nothing remains. It is a matter of taste whether the empty set is nothing.

 

Homework: Show the same for the set of FISONs.

Regards, WM