Page 104 - Textos de Matemática Vol. 41

90 Chapter 12. By the associativity of m,
(idG ⊕m∗ ⊕idG)·(m∗ ⊕idG) = =
=
Entourages and (quasi-)uniformities
(idG ⊕m∗)m∗⊕idG (m∗ ⊕ idG)m∗ ⊕ idG
(m∗ ⊕ idG ⊕ idG)(m∗ ⊕ idG).
Thus,
a⊕ı∗(α∨β)⊕(α∨β)⊕a⊆(m∗ ⊕idG ⊕idG)(m∗ ⊕idG)(x⊕a).
Applying (∆∗(idG ⊕ ı∗)) ⊕ idG ⊕ idG to both sides we get
(σ∗ε∗ ⊕idG ⊕idG)(m∗ ⊕idG)(x⊕a)
= (((σ∗ ⊕ idG)(ε∗ ⊕ idG)m∗) ⊕ idG)(x ⊕ a)
(a∧(α∨β))⊕(α∨β)⊕a ⊆
= ((σ∗ε∗ ⊕ idG)m∗ ⊕ idG)(x ⊕ a)
= (σ∗ ⊕idG ⊕idG)(x⊕a).
But (σ∗ ⊕idG)(x) = 1⊕x so (a∧(α∨β))⊕(α∨β)⊕a ⊆ 1⊕x⊕a. Since a∧(α∨β) ̸= 0, we finally obtain (α∨β)⊕a ⊆ x⊕a. In conclusion, α ≤ α∨β ≤ x.
12.10.3. Theorem. B = {E(x) | x ∈ N} forms a basis for a uniformity E on G.
Proof. Obviously, E(x) ∩ E(y) = E(x ∧ y). Therefore, by 12.10(1), B is a filter basis of (Ent(G), ⊆).
The symmetry is an immediate consequence of Lemma 12.10.2(2) and 12.10(2).
Consider E(x) ∈ B and take y ∈ N given by 12.10(3). We claim that E(y) ◦ E(y) ⊆ E(x). In fact, by Lemma 2.5, E(y) ◦ E(y) is the entourage
(ı∗(a) ⊕ b) ◦ (ı∗(a) ⊕ b) =
(a,b)∈m∗(y) (a,b)∈m∗(y)
= (ı∗(a) ⊕ b) ◦
(ı∗(a) ⊕ b).
Take (ı∗(a), b) with (a, b) ∈ m∗(y) and (ı∗(c), d) with (c, d) ∈ m∗(y) such that
(a,b)∈m∗(y) (a,b)∈m∗(y)
b ∧ ı∗(c) ̸= 0. From the inclusion y ⊕ y ⊆ m∗(x) it follows that
m∗(y) ⊕ m∗(y) ⊆ (m∗ ⊕ m∗)(m∗(x)) = (idG ⊕ m∗ ⊕ idG)(m∗ ⊕ idG)(m∗(x)).
Therefore
a⊕b⊕c⊕d⊆(idG ⊕m∗ ⊕idG)(m∗ ⊕idG)(m∗(x)).