Sujet : Re: The set of necessary FISONs
De : james.g.burns (at) *nospam* att.net (Jim Burns)
Groupes : sci.mathDate : 02. Mar 2025, 01:00:57
Autres entêtes
Organisation : A noiseless patient Spider
Message-ID : <aa1dbc81-afc7-469f-a287-d2675f3f9bca@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
User-Agent : Mozilla Thunderbird
On 3/1/2025 5:55 PM, Ross Finlayson wrote:
On 03/01/2025 12:10 PM, Jim Burns wrote:
Our inductive conclusions are justifiedly universal
because they are in narrower Z₀
not because of any supertask.
>
Like Zeno's that show motion is impossible?
Let's play "Where's Zeno?"
A property P is pre.inductive iff,
if each set in set 𝒞 of sets has P,
then its intersection ⋂𝒞 has P
∀S∈𝒞:P(𝒞) ⇒ P(⋂𝒞)
For example,
'holding 7,11,13' is pre.inductive.
If each set of a set 𝒞 of sets holds 7,11,13
then ⋂𝒞 holds 7,11,13.
Another example is 'being inductive'.
Consider a set Z such that P(Z)
for pre.inductive property P
The intersection ⋂𝒫ᴾ(Z) of subsets having P
has P.
The only subset of ⋂𝒫ᴾ(Z) which has P
is ⋂𝒫ᴾ(Z)
⎛ Assume, for {i:A(i)} ⊆ ⋂𝒫ᴾ(Z)
⎜ P{i:A(i)}
⎜
⎜ There is only one subset of ⋂𝒫ᴾ(Z) which has P
⎜ {i:A(i)} = ⋂𝒫ᴾ(Z)
⎜
⎜ By definition of {i:A(i)}
⎜ ∀n ∈ {i:A(i)}: A(n)
⎜
⎜ {i:A(i)} = ⋂𝒫ᴾ(Z)
⎜
⎝ ∀n ∈ ⋂𝒫ᴾ(Z): A(n)
Therefore,
if P{i:A(i)}
then ∀n ∈ ⋂𝒫ᴾ(Z): A(n)
For
P(S) ⇔ 0∈S ∧ ∀k∈⋂𝒫ᴾ(Z):k∈S⇒k+1∈S
and
ℕ = ⋂𝒫ᴾ(Z):
A(0) ∧ ∀k∈ℕ:A(k)⇒A(k+1) ⇒ ∀n∈ℕ:A(n)
Where's Zeno?