Page 99 - Textos de Matemática Vol. 41

12.6.
Complete regularity 85 A locale L is completely regular if for each a ∈ L,
a = { b | b ≺≺ a } .
(A space X is completely regular iff the locale Ω(X) is completely regular. The implication ⇒ is immedi- ate; the proof of ⇐ is not quite so straightforward as in 11.6.2, it requires the construction of suitable real functions; this can be done by a procedure similar to the standard proof of the Urysohn Lemma of classical topology and we omit it.)
12.6.1. Lemma. Let E be a symmetric entourage. Then, for every a ∈ L, E ◦ (a ⊕ a) ⊆ (E2a ⊕ E2a).
Proof. Let (x,y) ∈ E, (y,z) ∈ a⊕a, with x,y,z ̸= 0. Then y∨z ≤ a. Since (y, x) ∈ E, then (x, x), (y, y) ∈ E2. Thus (x ∨ y, x ∨ y) ∈ E2. But (x ∨ y) ∧ a = (a ∧ x) ∨ y ≥ y ̸= 0. Hence x ≤ x ∨ y ≤ E2a. On the other hand,
a=a∧( x)={a∧x|(x,x)∈E,a∧x̸=0}≤Ea, (x,x)∈E
which implies that z ≤ Ea ≤ E2a. Hence (x, z) ∈ E2a ⊕ E2a.
12.6.2. Proposition. Let E be a (basis of a) uniformity on L. Then: (1) E is interpolative.
(2) aE b⇒a≺≺b.
(3) A locale L admits a uniformity iff it is completely regular.
Proof. (1) Suppose that aE b. Then there is some E ∈ E such that E◦(a⊕a) ⊆ (b ⊕ b). Consider a symmetric F ∈ E satisfying F 4 ⊆ E. By Lemma 12.6.1, a E F2a. On the other hand, F2a E b. In fact, by 12.6.1,
F ◦ (F2a ⊕ F2a) ⊆ F2(F2a) ⊕ F2(F2a)
and it is easy to check that F2(F2a) ≤ F4a ≤ Ea ≤ b. (2) is an immediate consequence of (1) and 12.3.2(3).