Proposition 5.1.6. Let $(X, \fU)$ be a uniform space and $A \subset X$, then the subspace topology of $A$ coincides with the topology induced by the subspace uniformity on $A$.
Proof. Let $x \in A$ and $U \in \fU$, then $U(x) \cap A = [U \cap (A \times A)](x)$. Thus $V \subset X$ is a neighbourhood of $x$ with respect to the subspace topology if and only if it is a neighbourhood of $x$ with respect to the topology on $A$ induced by the subspace uniformity.$\square$