Re: The set of necessary FISONs

On 2/8/2025 9:21 AM, Jim Burns wrote:

On 2/8/2025 5:35 AM, WM wrote:

On 07.02.2025 22:45, Jim Burns wrote:

On 2/7/2025 11:39 AM, WM wrote:

On 07.02.2025 17:10, Jim Burns wrote:

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:

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.

>
The axiom of induction says:
If any property or predicate P satifies
(P(1) /\ ∀k(P(k) ==> P(k+1)),
then it describes all elements of
an inductive = infinite set.

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All inductive sets are infinite.
Not all infinite sets are inductive.

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Irrelevant.

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inductive ≠ infinite
>
inductive ≠ valid.for.induction
(despite the name)
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minimal.inductive = valid.for.induction
>

That is satisfied by
the set M of all FISONs which
are useless in  U(A(n)) = ℕ.

>
Only not.followed FISONs are not (your term) uselessᵂᴹ.
Each FISON is followed.

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Therefore all are useless.
>
All not needed elements belong to an inductive set.

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...which is also
a  minimal.inductive = valid.for.induction  set.
>

Inductive sets have no last element.

>
Yes.
inductive 𝕎  ⇒  ∀ᵂj ∃ᵂk j<k
>

Therefore
the set of FISONs which
could have the union ℕ

>
FISON F(k) = {i:i≤k}
>
Set of FISONs {{i:i≤k}:k}
>
Set of sets 𝒜 of FISONs
{𝒜⊆{{i:i≤k}:k}}
>
Set of sets 𝒜 of FISONs with union ℕ
{𝒜⊆{{i:i≤k}:k}:⋃𝒜=ℕ}
>
Set of FISONs common to sets with union ℕ
⋂{𝒜⊆{{i:i≤k}:k}:⋃𝒜=ℕ}
>
Is
⋂{𝒜⊆{{i:i≤k}:k}:⋃𝒜=ℕ}
a set of FISONs with union ℕ
?

∀ᴺk′ ∃ᴺk″: k′ ≨ k″
∀⁽ꟳ⁾F(k′) ∃⁽ꟳ⁾F(k″): F(k′) ⊊ F(k″)
  Let k′ ≨ k″
∀⁽ꟳ⁾F(k′) ∃𝒜″∈{𝒜⊆{F}:⋃𝒜=ℕ}:  F(k′) ∉ 𝒜″
⎛ Let k′ ≨ k″
⎜ F(k′) ⊊ F(k″)
⎜ Let ⋃𝒜′ = ℕ
⎜ Let 𝒜″ = (𝒜′\{F(k′)})∪{F(k″)}
⎜
⎜ ⋃𝒜″ = ℕ
⎜ F(k′) ∉ 𝒜″ ∈ {𝒜⊆{F}:⋃𝒜=ℕ}
⎝ F(k′) ∉ ⋂{𝒜⊆{F}:⋃𝒜=ℕ}
∀⁽ꟳ⁾F(k′):  F(k′) ∉ ⋂{𝒜⊆{F}:⋃𝒜=ℕ}
⋂{𝒜⊆{F}:⋃𝒜=ℕ} = {}

Is
⋂{𝒜⊆{{i:i≤k}:k}:⋃𝒜=ℕ} ∈ {𝒜⊆{{i:i≤k}:k}:⋃𝒜=ℕ}
?

No.
⋂{𝒜⊆{F}:⋃𝒜=ℕ} = {}
⋂{} ≠ ℕ

If you say yes, how do you know?

Do you use ω-1 ?

Union of FISONs common to sets with union ℕ
⋃⋂{𝒜⊆{{i:i≤k}:k}:⋃𝒜=ℕ}
>

has no first element. That means it is empty.
>
Therefore U(A(n)) = ℕ

>
Is
⋃⋂{𝒜⊆{{i:i≤k}:k}:⋃𝒜=ℕ} = ℕ
?
>

Therefore U(A(n)) = ℕ ==> U{ } = { } = ℕ.
This is false.
By contraposition we get
~{ } = ℕ ==> ~ U(A(n)) = ℕ.