Very minor fix, checked pagination

ch-metric.tex

@@ -949,7 +949,7 @@ \subsection{Topology}
 open in $X$, such that $U = W \cap V$.  Intersection of two open sets
 is open so $U$ is open in $X$.
 
-Now suppose $U$ is open in $X$, then $U = U \cap V$. So
+Now suppose $U$ is open in $X$.  Then $U = U \cap V$. So
 $U$ is open in $V$ again by \propref{prop:topology:subspaceopen}.
 \end{proof}