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38 Chapter 6. Some special sublocales Thus, for the left-hand resp. right hand side A resp. B of (∗) we have
A = A ∨ (S ∩ T ) = (A ∨ S) ∩ (A ∨ T ) = (B ∨ S) ∩ (B ∨ T ) = B ∨ (S ∩ T ) = B.
⇐: First observe that we have in Sl(L) the implication
S1 ⊂S2 ⇒ ∃T suchthatT∩S1 =O̸=T∩S2. (∗∗)
(Indeed: consider, using 6.5, S1 as i∈J Ci with complemented Ci. Then there is an i such that S2 Ci. The complement T of Ci has the required property.) Nowlet(∗)holdforS.SetS∗ = {S′ |S′∩S=O}.By(∗),S∩S∗ =O. Suppose that S ∨S∗ ̸= L. Then there is, by (∗∗), a T such that T ∩(S ∨S∗) = O (and hence T ∩ S = O = T ∩ S∗) and T = T ∩ L ̸= O. But since T ∩ S = O we have T ⊆ S∗ and hence T = T ∩ S∗ = O.
Remark. Note that, more generally, in any complete distributive lattice satisfy- ing (∗∗) the complemented elements are precisely the ones satisfying property (∗) (the proof is the same).
6.8. Spatial sublocales. For a subset A of a locale L set mA = { B | B ⊆ A}.
By Proposition 3.5.2, L is spatial iff mPt(L) = L. We easily obtain:
6.8.1. Proposition. Each complemented sublocale of a spatial locale is spatial.
Proof.LetS,T ⊆LbesublocalessuchthatS∩T =OandS∨T =L.Then for each a ∈ Pt(L), a = s ∧ t with s ∈ S and t ∈ T . Since a is meet-irreducible, a = s ∈ S or a = t ∈ T (recall 1.6). Thus Pt(L) = Pt(S) ∪ Pt(T). Now mPt(S) ⊆ S, mPt ⊆ T and
S∨T =L=mPt(L)=mPt(S)∨mPt(T).
Hence S = mPt(S) and T = mPt(T).
6.8.2. For a locale L and a ∈ L, b(a) is itself a locale so we may consider its Boolean sublocale
bb(a)(a) = {x → a | x ∈ b(a)} = {(x → a) → a | x ∈ L}.
Evidently, a ∈ bb(a)(a) since (a → a) → a = 1 → a = a and for any sublocale S
of L a ∈ S implies immediately bb(a)(a) ⊆ S.
Lemma. For any meet-irreducible p of L, bb(p)(p) = {p, 1}.