Sujet : Re: The set of necessary FISONs
De : james.g.burns (at) *nospam* att.net (Jim Burns)
Groupes : sci.mathDate : 01. Mar 2025, 21:10:21
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <cd2a1798-4145-4127-a71c-95462cffd034@att.net>
References : 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26
User-Agent : Mozilla Thunderbird
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.
inductive(W) ⇒ Z₀ ⊆ W
inductive(Z₀)
Z₀ is not constructed by supertask.
Z₀ = ⋂𝒫ⁱⁿᵈ(Z)
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.