Lemma 5.5.4. Let $(X, \fU)$ be a uniform space and $A \subset X$ be a dense subset, then $X$ is complete if and only if every Cauchy filter on $A$ converges to at least one point in $X$.
Proof. Let $\fF \subset X$ be a Cauchy filter. Using Proposition 5.4.8, assume without loss of generality that $\fF$ is minimal. By Proposition 5.4.9, $\fF$ admits a base consisting of open sets. Hence $E \cap A \ne \emptyset$ for all $E \in \fF$. Let $\fF_{A} = \bracs{E \cap A| E \in \fF}$, then $\fF_{A}$ is a filter base on $A$.
By assumption, there exists $x \in X$ such that $\fF_{A}$ converges to $x$. Since $\fF$ is a Cauchy filter contained in the filter generated by $\fF_{A}$ in $X$, $x$ is also a limit point of $\fF$ by Proposition 5.4.10.$\square$