Proposition 6.1.12 ([Proposition 2.1.2, Bou13]).label Let $(X, \fU)$ be a uniform space, $V \in \fU$ be a symmetric entourage, and $M \subset X \times X$, then:
- (1)
$V \circ M \circ V \in \cn(M)$.
- (2)
Let $\fB$ be the set of all symmetric entourages, then $\ol{M}= \bigcap_{V \in \fB}V \circ M \circ V$.
with respect to the product topology on $X \times X$.
Proof. (1): Let $(x, y) \in V \circ M \circ V$. By Lemma 6.1.11, there exists $(p, q) \in M$ such that $(x, y) \in V(p) \times V(q) \in \cn(p, q)$. In particular, if $(x, y) \in M$, then this implies that $V \circ M \circ V \in \cn(x, y)$. Thus $V \circ M \circ V \in \cn(M)$.
(2): Let $(x, y) \in X \times X$. By Lemma 6.1.11, the following are equivalent:
- (a)
$(x, y) \in V \circ M \circ V$ for all $V \in \fB$.
- (b)
For every $V \in \fB$, there exists $(p, q) \in M$ such that $(p, q) \in V(x) \times V(y)$.
As $\bracsn{V(x) \times V(y)| V \in \fB}$ is a fundamental system of neighbourhoods at $(x, y)$, (b) is equivalent to $(x, y) \in \ol{M}$. Therefore
$\square$