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Proposition 3.7

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QAlgorithms.Papers.Independent.Intakethird100.P20.frame_stone_extremally_disconnected

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

prove2quantum

Let LLL be a complete lattice satisfying the frame law a∧⋁S=⋁b∈S(a∧b)a\wedge\bigvee S=\bigvee_{b\in S}(a\wedge b)a∧⋁S=⋁b∈S​(a∧b). Then its Stone spectrum is extremally disconnected, and

⋃a∈SQa(L)‾=Q⋁S(L)(S⊆L).\overline{\bigcup_{a\in S}Q_a(L)}=Q_{\bigvee S}(L)\qquad(S\subseteq L).a∈S⋃​Qa​(L)​=Q⋁S​(L)(S⊆L).

Thus the closure of every open subset of the spectrum is open. The frame law is the source's meaning of complete distributivity, not the stronger modern two-index law.

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

/-- Proposition 3.7; de Groote, Observables I, PDF p./pp. 45.
-/
theorem frame_stone_extremally_disconnected {L : Type u} [CompleteLattice L] (hframe : HasFrameLaw L) :
    ExtremallyDisconnected (StoneSpectrum L) ∧
      ∀ s : Set L, closure (⋃ a ∈ s, stoneBasic a) = stoneBasic (sSup s) := by sorry

end QAlgorithms.Papers.Independent.Intakethird100.P20

end
Source
Observables I: Stone Spectra, arXiv:math-ph/0509020v1, PDF p. 45, Proposition 3.7
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What the Lean code literally says, in plain math · gpt-6.1-sol

Let LLL be a complete lattice satisfying a∧sup⁡s=sup⁡{a∧b:b∈s}a\wedge\sup s=\sup\{a\wedge b:b\in s\}a∧sups=sup{a∧b:b∈s} for every a∈La\in La∈L and every subset s⊆Ls\subseteq Ls⊆L, including the empty subset. Let Q(L)\mathcal Q(L)Q(L) consist of the inclusion-maximal nonempty subsets B⊆LB\subseteq LB⊆L excluding ⊥\bot⊥ such that any two members have a common lower bound in BBB, and give it the topology generated by D(a)={B∈Q(L):a∈B}D(a)=\{B\in\mathcal Q(L):a\in B\}D(a)={B∈Q(L):a∈B}. Then Q(L)\mathcal Q(L)Q(L) is extremally disconnected, meaning the closure of every open subset is open, and for every subset s⊆Ls\subseteq Ls⊆L,

⋃a∈sD(a)‾=D(sup⁡s).\overline{\bigcup_{a\in s}D(a)}=D(\sup s).a∈s⋃​D(a)​=D(sups).

This includes s=∅s=\varnothings=∅, for which sup⁡s=⊥\sup s=\botsups=⊥ and both sides are empty.

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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