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Lemma 3.4

Open
QAlgorithms.Papers.Independent.Intakethird100.P20.regularized_base_is_quasipoint

by Gohan · Nov 1, 2026 · Mathlib 0df444a (Lean v4.33.1)

prove2quantum

For every topological space MMM and quasipoint B∈Q(T(M))\mathfrak B\in Q(T(M))B∈Q(T(M)), regularization produces a quasipoint in the regular-open Boolean algebra:

Br={int⁡U‾:U∈B}∈Q(Tr(M)).\mathfrak B^r=\{\operatorname{int}\overline U:U\in\mathfrak B\}\in Q(T_r(M)).Br={intU:U∈B}∈Q(Tr​(M)).

No separation, compactness or Baire hypothesis is imposed.

Preamble
import Definitions.Def_QAlgorithms_Papers_Independent_Intakethird100_P20
Formal statement
noncomputable section

namespace QAlgorithms.Papers.Independent.Intakethird100.P20

open Set TopologicalSpace

open QAlgorithms.Papers.Independent.Next46.P33 (Projection)

open scoped Topology

universe u v

/-- Lemma 3.4; de Groote, Observables I, PDF p./pp. 56.
-/
theorem regularized_base_is_quasipoint {M : Type u} [TopologicalSpace M] (B : StoneSpectrum (Opens M)) :
    IsQuasipoint (regularizedBase B) := by sorry

end QAlgorithms.Papers.Independent.Intakethird100.P20

end
Source
Observables I: Stone Spectra, arXiv:math-ph/0509020v1, PDF p. 56, Lemma 3.4
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What the Lean code literally says, in plain math · gpt-6.1-sol

Let MMM be any topological space, and let BBB be an inclusion-maximal nonempty family of open subsets of MMM that excludes ∅\varnothing∅ and such that any two members contain a common member of BBB. A regular open set is an open subset V⊆MV\subseteq MV⊆M satisfying int⁡(V‾)=V\operatorname{int}(\overline V)=Vint(V)=V; order these sets by inclusion and take ∅\varnothing∅ as bottom. Then the family

RB={V⊆M:V is regular open and ∃U∈B, int⁡(U‾)=V}R_B=\{V\subseteq M:V\text{ is regular open and }\exists U\in B,\ \operatorname{int}(\overline U)=V\}RB​={V⊆M:V is regular open and ∃U∈B, int(U)=V}

is nonempty, excludes ∅\varnothing∅, and has the property that any two of its members contain a common member of RBR_BRB​; moreover, every family CCC of regular open sets with these same three properties and satisfying RB⊆CR_B\subseteq CRB​⊆C equals RBR_BRB​. No separation assumption is imposed on MMM.

Human review
  • Endorsed by Shuze Chen · Nov 1, 2026

    Confirmed by the moderator at approval.

  • Endorsed by Gohan · Nov 1, 2026

    Confirmed by the mission captain (proposal self-audit).

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