Proposition 5.1.13. Let $(X, \fU)$ be a uniform space and $A \subset X$, then:
For any symmetric entourage $V \in \fU$, $V(A) \in \cn(A)$.
Let $\fB \subset \fU$ be the family of all symmetric entourages, then $\ol{A}= \bigcap_{U \in \fB}U(A)$.
Proof. (1): For each $x \in A$, $x \in V(x) \subset V(A)$. By (F1) of $\cn(x)$, $V(A) \in \cn(x)$ for all $x \in A$. Thus $V(A) \in \cn(A)$.
(2): Let $x \in X$, then the following are equivalent:
For all $V \in \fB$, $x \in V(A)$.
For all $V \in \fB$, there exists $y \in A$ such that $x \in V(y)$.
For all $V \in \fB$, $V(x) \cap A \ne \emptyset$.
Since $\bracs{V(y): V \in \fB}$ is a fundamental system of neighbourhoods at $y$ (Lemma 5.1.9), (c) is equivalent to $x \in \overline{A}$. Therefore $\ol{A}= \bigcap_{U \in \fB}U(A)$.$\square$