English
In a Lindelöf regular space, two disjoint closed sets admit a separating cover.
Русский
В Линдельёфовом регулярном пространстве два дисjoint-т closed множества допускают разделяющий покрытие.
LaTeX
$$$HasSeparatingCover s t$ при условии $IsClosed(s)$, $IsClosed(t)$, $Disjoint(s,t)$ и [LindelöfSpace X], [RegularSpace X]$$
Lean4
/-- This technique to witness `HasSeparatingCover` in regular Lindelöf topological spaces
will be used to prove regular Lindelöf spaces are normal. -/
theorem HasSeparatingCover {s t : Set X} [LindelofSpace X] [RegularSpace X] (s_cl : IsClosed s) (t_cl : IsClosed t)
(st_dis : Disjoint s t) : HasSeparatingCover s t := by
-- `IsLindelof.indexed_countable_subcover` requires the space be Nonempty
rcases isEmpty_or_nonempty X with empty_X | nonempty_X
· rw [subset_eq_empty (t := s) (fun ⦃_⦄ _ ↦ trivial) (univ_eq_empty_iff.mpr empty_X)]
exact
hasSeparatingCovers_iff_separatedNhds.mpr (SeparatedNhds.empty_left t) |>.1
-- This is almost `HasSeparatingCover`, but is not countable. We define for all `a : X` for use
-- with `IsLindelof.indexed_countable_subcover` momentarily.
have (a : X) : ∃ n : Set X, IsOpen n ∧ Disjoint (closure n) t ∧ (a ∈ s → a ∈ n) :=
by
wlog ains : a ∈ s
· exact ⟨∅, isOpen_empty, SeparatedNhds.empty_left t |>.disjoint_closure_left, fun a ↦ ains a⟩
obtain ⟨n, nna, ncl, nsubkc⟩ :=
((regularSpace_TFAE X).out 0 3 :).mp ‹RegularSpace X› a tᶜ <| t_cl.compl_mem_nhds (disjoint_left.mp st_dis ains)
exact
⟨interior n, isOpen_interior,
disjoint_left.mpr fun ⦃_⦄ ain ↦ nsubkc <| (IsClosed.closure_subset_iff ncl).mpr interior_subset ain, fun _ ↦
mem_interior_iff_mem_nhds.mpr nna⟩
-- By Lindelöf, we may obtain a countable subcover witnessing `HasSeparatingCover`
choose u u_open u_dis u_nhds using this
obtain ⟨f, f_cov⟩ :=
s_cl.isLindelof.indexed_countable_subcover u u_open (fun a ainh ↦ mem_iUnion.mpr ⟨a, u_nhds a ainh⟩)
exact ⟨u ∘ f, f_cov, fun n ↦ ⟨u_open (f n), u_dis (f n)⟩⟩
