Page 103 - Textos de Matemática Vol. 41

12.10. Uniformities on localic groups 89 12.10. Uniformities on localic groups. For N = {x ∈ G | ε∗(x) = 1} (the
neighbourhood filter of the unit of G) one has
(1) x,y∈N⇒x∧y∈N,
(2) x ∈ N ⇒ ι∗(x) ∈ N (by the second equation in (12.9.1)), and (3) foranyx∈N thereexistsy∈N suchthaty⊕y⊆m∗(x)
(indeed, since ε∗ = (ε⊕ε)∗m∗ then 1 = ε(x) = {ε∗(a)⊕ε∗(b) | a⊕b ⊆ m∗(x)} so there is some a ⊕ b ⊆ m∗(x) with ε∗(a) = ε∗(b) = 1; take y = a ∧ b).
12.10.1.Foreachx∈N defineE(x)∈G⊕Gby
E(x) = (idG ⊕ ı)∗m∗(x) = {a ⊕ ı∗(b) | a ⊕ b ≤ m∗(x)} (recall (12.4.1)).
12.10.2. Lemma.
(1) Each E(x) is an entourage of G.
(2) E(x)−1 = E(ı∗(x)).
(3) Let x,a ∈ G and b ∈ N satisfying a ⊕ b ⊆ m∗(x). For any symmetric entourage F of G such that F2 ⊆E(b), F ◦(a⊕a)⊆x⊕x.
Proof. (1) ∆∗(E(x)) = ∆∗(idG ⊕ ı∗)m∗(x) = ε∗(x) = 1. ∗∗
(2) Put m (x) = i∈J(ai ⊕bi). Then E(x) = i∈J(ai ⊕ı (bi)). On the other hand, since m∗ı∗ = r∗(ı∗ ⊕ ı∗)m∗, we have
m∗(ı∗(x)) = (ı∗(bi) ⊕ ı∗(ai)) i∈J
and, therefore, E(ı∗(x)) = i∈J (bi ⊕ ı∗(ai)).
(3) Assume (α,β) ∈ F, (β,γ) ≤ (a,a) with α,β,γ ̸= 0. We need to show that both α and γ are below x. Clearly γ ≤ a ≤ x. Let us show that also α ≤ x. Of course, (α,β) ∈ F2 and, by the symmetry of F, (β,α) ∈ F2, which forces (α ∨ β,α ∨ β) ∈ F2 ⊆ E(b), as (α,α) and (β,β) also belong to F2. Moreover (α∨β)∧a ̸= 0. Thus ı∗(α∨β)⊕(α∨β) ⊆ m∗(b). On the other hand, a⊕b ⊆ m∗(x). Consequently, a ⊕ b ⊕ a ⊆ (m∗ ⊕ idG)(x ⊕ a) which, in turn, implies that
a⊕μ∗(b)⊕a⊆(idG ⊕m∗ ⊕idG)(m∗ ⊕idG)(x⊕a). Thus, we have
a⊕ı∗(α∨β)⊕a⊆(idG ⊕m∗ ⊕idG)(m∗ ⊕idG)(x⊕a).