Proposition 9

Let $X$ be a uniform space and let $A$ be a dense subset of $X$ such that every Cauchy filter base on $A$ converges in $X$. Then $X$ is complete.

It is enough to show that every minimal Cauchy filter $\mathfrak{F}$ on $X$ is convergent. Since $A$ is dense and since every set of $\mathfrak{F}$ has a non-empty interior (no. 2, Corollary 4 to Proposition 5), the trace $\mathfrak{F}_A$ of $\mathfrak{F}$ on $A$ is a Cauchy filter on $A$, hence converges to a point $x_0 \in X$. Since $\mathfrak{F}$ is coarser than the filter on $X$ generated by $\mathfrak{F}_A$, it follows that $\mathfrak{F}$ converges to $x_0$ (no. 2, Corollary 3 to Proposition 5).