Re: The set of necessary FISONs

On 2/21/2025 12:17 PM, WM wrote:

On 21.02.2025 11:37, joes wrote:

Am Fri, 21 Feb 2025 10:39:46 +0100 schrieb WM:

On 21.02.2025 02:18, Richard Damon wrote:

In order to secure the existance of "infinite" sets,
we still need the following axioms.

>
Almost correct, but only singular:
we still need the following axiom.
>

There is no mention of "induction" in the statement.

>
The following axiom is the axiom of induction.
The set Z contains the empty set as an element and
with every element a also the element {a}.

>
Yet not the element Z.

>
Zermelo claims to secure the existence of set Z.

The ZFC axiom Infinityᶻᶠᶜ secures the existence of Zᶻᶠᶜ
∃Zᶻᶠᶜ:  Zᶻᶠᶜ ∋ {}  ∧  ∀a: Zᶻᶠᶜ ∋ a ⇒ Zᶻᶠᶜ ∋ a∪{a}
Not all sets Z which satisfy Infinityᶻᶠᶜ
are usable in a proof.by.inductionᶻᶠᶜ
Let 0 = {}, k+1 = k∪{k}
Yes,
if Z = ω = {0,1,2,...} = [0,ω)ᵛᴺ
then
Z satisfies Infinityᶻᶠᶜ
and,
for all predicates A(k),
A(0) ∧ ∀k∈Z: A(k)⇒A(k+1)  ⇒  ∀n∈Z: A(n)
However,
if Z = ω+ω = {0,1,2,...;ω,ω+1,...} = [0,ω+ω)ᵛᴺ
then
Z satisfies Infinityᶻᶠᶜ
but,
for some predicates B(k)
B(0) ∧ ∀k∈Z: B(k)⇒B(k+1)  ∧  ¬∀n∈Z: B(n)
For example,
for Z = ω+ω
and B(k) == "[0,k)ᵛᴺ is finite"
B(k) is an inductive predicate.
B(0) ∧ ∀k∈(ω+ω): B(k)⇒B(k+1)
B(k) is true in [0,ω)ᵛᴺ
B(k) is false in [ω,ω+ω)ᵛᴺ
¬∀n∈(ω+ω): B(n)
----
Not all sets Z which satisfy Infinityᶻᶠᶜ
are usable in a proof.by.inductionᶻᶠᶜ
However,
all sets Z which satisfy Infinityᶻᶠᶜ
HAVE A SUBSET which
is usable in a proof.by.inductionᶻᶠᶜ
⋂𝒫ⁱⁿᵈ(Z) = ⋂{S⊆Z:inductiveᶻᶠᶜ.S}  is
that subset of Z usable in proofs by induction.
Zᶻᶠᶜ existing follows Infinityᶻᶠᶜ
⋂𝒫ⁱⁿᵈ(Z) existing follows Zᶻᶠᶜ existing and
other axioms, such as PowerSetᶻᶠᶜ and Separationᶻᶠᶜ

This set can be handled,
for instance removed from Z.
A set containing Z is not useful in this context.
 Regards, WM