Page 79 - Textos de Matemática Vol. 41

10.4. Some characterizations of subfitness 65
Proposition. The following statements about a locale L are equivalent: (i) L is fit.
(ii)ForanysublocalesSandTofL,S′=T′ ⇒S=T. (iii) For each sublocale S of L, S = {o(x) | νS (x) = 1}. (iv) Each sublocale is an intersection of open sublocales.
(v) Each closed sublocale is an intersection of open sublocales.
Proof. (i)⇒(ii): Let S′ = T′ and let b ∈ T, b ̸= 1. Set a = νS(b). Suppose a∨c=1andb∨c≤a1 ∈S.Thena1 ≥a∨c=1sothatb∨c∈/S′ =T′.By 1.3.1(H9), however, (b∨c)∧(c→b) ≤ b and hence b∨c ≤ (c→b)→b ∈ T so that (c→b)→b = 1 and c→b = b by 1.3.1(H3) and (H4). Therefore, by (Fit), a ≤ b and hence b = νS(b) ∈ S.
(ii)⇒(iii): The inclusion ⊆ follows from the Lemma. On the other hand, if a∈T ={o(x)|νS(x)=1}
we have x→a = a whenever νS(x) = 1. Thus, if νS(a) = 1 we have a = a→a = 1.HenceT{1}⊆S′,consequentlyT′ ⊆S′ ⊆T′,andby(ii),S=T.
(iii)⇒(iv)⇒(v) is trivial.
(v)⇒(i): By 6.6.1, if ↑a is an intersection of some open sublocales then it is the
intersection of all open sublocales o(c) with ν↑a(c) = 1. As ν↑a(c)={s|a≤s, c≤s}=a∨c,
we have c(a) =↑a = {o(c) | a ∨ c = 1} and hence, if c → b = b for all c such that a ∨ c = 1 then a ≤ b.
10.4. Some characterizations of subfitness.
Proposition. The following statements about a locale L are equivalent:
(i) L is subfit.
(ii) For each sublocale S of L, S {1} is cofinal in L {1} only if S = L.
(iii) If S ̸= L for a sublocale S of L then there is a closed c(x) ̸= O such that
S ∩ c(x) = O.
(iv) For each open sublocale o(a), o(a) = {c(x) | x ∨ a = 1}.
(v) Each open sublocale is a join of closed sublocales.