Definition 6.4.1.label Let $(X, \fU)$ be a uniform space, then the following are equivalent:
- (1)
For every $U \in \fU$, there exists $\seqf{x_j}\subset U$ such that $X = \bigcup_{j = 1}^{n} U(x_{j})$.
- (2)
Every ultrafilter on $X$ is Cauchy.
- (3)
For every filter $\fF_{0} \subset 2^{X}$, there exists $\fF \supset \fF_{0}$ such that $\fF$ is Cauchy.
If the above holds, then $(X, \fU)$ is totally bounded.
Proof. (1) $\Rightarrow$ (2): Let $\fF \subset 2^{X}$ be an ultrafilter and $U \in \fU$, then there exists $U$-small sets $\seqf{E_j}\subset 2^{X}$ such that $X = \bigcup_{j = 1}^{n} E_{j}$. By (3) of Definition 5.2.9, there exists $1 \le j \le n$ such that $E_{j} \in \fF$, so $\fF$ is Cauchy.
(2) $\Rightarrow$ (3): By the ultrafilter lemma.
$\neg (1) \Rightarrow \neg (3)$: Let $U \in \fU$ be symmetric such that for all $X_{0} \subset X$ finite, $\bigcup_{x \in X_0}U(x) \subsetneq X$. Let $\fF_{0}$ be the filter generated by
then for any $U$-small set $E \subset X$ and $x \in E$, $E \cap [X \setminus U(x)] = \emptyset$. Therefore no filter finer than $\fF_{0}$ contains a $U$-small set.$\square$