Lemma 5.1.15. Let $(X, \fU)$ be a uniform space, $U \in \fU$ be a symmetric, open entourage, and $M \subset X$, then $U(M)$ is open.
Proof. Let $U \in \fV$ and $y \in U(M)$. Since $U \subset X \times X$ is open, there exists $x \in M$, $V \in \cn(x)$, and $V' \in \cn(y)$ such that $(x, y) \subset V \times V' \subset U$. In which case, $U(M) \supset V' \in \cn(y)$. Hence $U(M)$ is open.$\square$