Re: The set of necessary FISONs

On 01.03.2025 21:10, Jim Burns wrote:

On 3/1/2025 1:28 PM, WM wrote:

On 01.03.2025 18:47, Jim Burns wrote:

On 3/1/2025 7:51 AM, WM wrote:

 

Z_0 contains only 0, {0}, {{0}}, ...

>
Z₀ = {0,{0},{{0}},...} is
the only subset of Z₀ which
holds 0 and, for each a, holds {a}

>
Z₀ is defined by induction.

 inductive(Z)
 inductive(Z) :⇔ 0∈Z ∧ ∀a:a∈Z⇒{a}∈Z
 Z₀ is the emptiest (<EZ>"einfachste"?)
inductive set.

The simplest example.

 inductive(W) ⇒ Z₀ ⊆ W
inductive(Z₀)
 Z₀ is not constructed by supertask.

Inductive Z₀: { } ∈ Z₀, and if {{{...{{{ }}}...}}} with n curly brackets ∈ Z₀ then {{{...{{{ }}}...}}} with n+1 curly brackets ∈ Z₀.

Z₀ = ⋂𝒫ⁱⁿᵈ(Z)

Also true.

 

Likewise UF is defined by induction.
ℕ \ F(1) \ F(2) \ F(3) \ ... = ℵo
= ℕ \ (F(1) U F(2) U F(3) U ...) = ℵo.
>

In narrower Z₀,
inductivity identifies a unique set.

>
So it is.

 Thank you.
 {S⊆Z₀:inductive.S} = {Z₀} ∧
inductive{i:A(i)}  ⇒
{i:A(i)} ∈ {SsZ₀:inductive.S} = {Z₀}  ⇒
Z₀ = {i:A(i)}
 Z₀ = {i:A(i)} ∧
∀n ∈ {i:A(i)}: A(n)  ⇒
∀n ∈ Z₀: A(n)
 Our inductive conclusions are justifiedly universal
because they are in narrower Z₀
not because of any supertask.

Induction abbreviates a supertask. If 1 then 2, if 2 then 3, and so on. But supertasks will never pass through the dark numbers. They can only extend the defined numbers without end, never crossing the infinitely larger domain of dark numbers - if such exist at all!
Regards, WM