Re: The set of necessary FISONs

On 03/01/2025 04:00 PM, Jim Burns wrote:

On 3/1/2025 5:55 PM, Ross Finlayson wrote:

On 03/01/2025 12:10 PM, Jim Burns wrote:

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Our inductive conclusions are justifiedly universal
because they are in narrower Z₀
not because of any supertask.

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Like Zeno's that show motion is impossible?

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Let's play "Where's Zeno?"
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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)
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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.
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Another example is 'being inductive'.
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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)
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⎛ 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)
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Therefore,
if P{i:A(i)}
then ∀n ∈ ⋂𝒫ᴾ(Z): A(n)
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For
P(S) ⇔ 0∈S ∧ ∀k∈⋂𝒫ᴾ(Z):k∈S⇒k+1∈S
and
ℕ = ⋂𝒫ᴾ(Z):
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A(0) ∧ ∀k∈ℕ:A(k)⇒A(k+1)  ⇒  ∀n∈ℕ:A(n)
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Where's Zeno?
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He's standing right next to you.
Yet, you must've crossed the bridge.
He yells loud, "you are on the other side".
The geometric series you'd have is "complete", then?
I.e., attaching your cases to the geometric series.
  ⋂𝒫ᴾ(Z) = ℕ+