English
If nhds x contains s, then there exists a finite index set I and a piecewise construction x,y so that the piecewise element lies in s.
Русский
Если nhds x содержит s, существует конечный набор индексов I и разбиение piecewise x y ∈ s.
LaTeX
$$$\\displaystyle \\exists I : \\mathrm{Finset}\\ ι, I.piecewise x y \\in s$$$
Lean4
theorem pi_generateFrom_eq {A : ι → Type*} {g : ∀ a, Set (Set (A a))} :
(@Pi.topologicalSpace ι A fun a => generateFrom (g a)) =
generateFrom {t | ∃ (s : ∀ a, Set (A a)) (i : Finset ι), (∀ a ∈ i, s a ∈ g a) ∧ t = pi (↑i) s} :=
by
refine le_antisymm ?_ ?_
· apply le_generateFrom
rintro _ ⟨s, i, hi, rfl⟩
letI := fun a => generateFrom (g a)
exact isOpen_set_pi i.finite_toSet (fun a ha => GenerateOpen.basic _ (hi a ha))
·
classical
refine le_iInf fun i => coinduced_le_iff_le_induced.1 <| le_generateFrom fun s hs => ?_
refine GenerateOpen.basic _ ⟨update (fun i => univ) i s, { i }, ?_⟩
simp [hs]