Proposition 5.3.8. Let $(X, \fU)$ be a uniform space and $\seqi{d}$ be a family of pseudometrics that induces the topology on $X$, then the following are equivalent:
$X$ is separated.
For any $x, y \in X$ with $x \ne y$, there exists $i \in I$ such that $d_{i}(x, y) > 0$.
For any $x, y \in X$ with $x \ne y$, there exists a uniformly continuous pseudometric $d$ on $X$ such that $d(x, y) > 0$.
Proof. (1) $\Rightarrow$ (2): By assumption, there exists $U \in \fU$ such that $(x, y) \not\in U$, so there exists $J \subset I$ finite and $r > 0$ such that
In which case, there must exist $j \in J$ such that $d_{j}(x, y) \ge r > 0$.
(3) $\Rightarrow$ (1): Let $x, y \in X$ with $x \ne y$ and $d$ be a continuous pseudometric on $X$ such that $r = d(x, y) > 0$. By Theorem 5.3.7, $E(d, r) \in \fU$. Therefore $\bigcup_{U \in \fU}U = \Delta$, and $X$ is separated by Definition 5.1.17.$\square$