Proposition 6.4.3.label Let $X$ be a uniform space, then the following are equivalent:
- (1)
$X$ is compact.
- (2)
$X$ is complete and totally bounded.
In particular, $X$ is totally bounded if and only if its Hausdorff completion is compact.
Proof. (1) $\Rightarrow$ (2): By (1) of Definition 5.16.1, $X$ is totally bounded. Let $\fF$ be a Cauchy filter, then by (4) of Definition 5.16.1, $\fF$ has at least one accumulation point. Since $\fF$ is Cauchy, the accumulation points and limit points of $\fF$ coincide by (2) of Proposition 6.6.12.
(2) $\Rightarrow$ (1): By (2) of Definition 6.4.1, every ultrafilter is Cauchy, and hence converges, which satisfies (4) of Definition 5.16.1.$\square$