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It would be appreciated if someone could check if these proofs are correct (in Schechter's HAF, the 'only if' direction of the theorem below is essentially left as an exercise with a hint, so I wanted to make sure it is correct); the converse direction of the theorem is based on the argument in Kelley but I include this for completeness. Also, I assume AC throughout and make no explicit mention to its use.

Theorem Let $(X, \mathcal U)$ be a uniform space. Then, $X$ is totally bounded iff every filter on $X$ has a Cauchy refinement.

Proof. For the 'only if' direction, it is enough to show every ultrafilter on $X$ is Cauchy; to this end, let $\mathcal F$ be an ultrafilter on $X$, and fix an entourage $V$. Pick symmetric $U$ such that $U \circ U \subseteq V$, and pick any $F \in \mathcal F$. Since $X$ is totally bounded, it follows $F$ is totally bounded, hence $F \subseteq \bigcup_{j=1}^n U[x_j]$ for some $x_1, \ldots, x_n \in F$. By virtue of being an ultrafilter, some $U[x_i]$ belongs to $\mathcal F$, and we have $U[x_i] \times U[x_i] \subseteq V$ by symmetry of $U$. So, $\mathcal F$ contains a $V$-small set.

For the converse, it is equivalent to consider nets, and suppose $X$ is not totally bounded; we construct a net (a sequence in fact) with no Cauchy subnet. Let $P$ be a family of pseudometrics generating the uniformity; because $X$ is not totally bounded, there is some $\varepsilon>0$ and some $p \in P$ such that $X \setminus \bigcup_{j=1}^m B_p(y_j, \varepsilon) \neq \emptyset$ for all finite sequences $(y_j)_{j=1,\ldots, m}$. Let $x_0 \in X$ be arbitrary, and suppose $x_0, \ldots, x_{n-1}$ have been chosen with the property $x_k \in X \setminus \bigcup_{j=1}^{k-1} B_p(x_j, \varepsilon)$ for $1 \leq k \leq n-1$. By the previous remarks, it is possible to choose $x_n \in X \setminus \bigcup_{j=1}^{n-1} B_p(x_j, \varepsilon)$. Suppose now $(x_{\varphi(\alpha)})_{\alpha \in A}$ is a subnet of $(x_n)_n$. Let $\alpha \in A$; because the image of $\varphi$ is cofinal in $\mathbb N$ and since $\varphi$ is increasing, there is $\beta \in A$ such that $\alpha \leq \beta$ and $\varphi(\alpha) < \varphi(\beta)$; hence, $p(x_{\varphi(\beta)}, x_{\varphi(\alpha)}) \geq \varepsilon$ by the construction above. Hence $(x_{\varphi(\alpha)})_{\alpha}$ is not Cauchy.

And a corollary:

Corollary If $(X, \mathcal U), (Y, \mathcal V)$ are uniform spaces, then every Cauchy continuous map $f:X \to Y$ sends totally bounded sets to totally bounded sets.

Remark: There is an easy adaptation of the argument here (Uniformly continuous function (onto) sends totally bounded set to totally bounded set.) using nets, though as an exercise, I rewrote the proof using ultrafilters instead.

Proof. Suppose $A \subseteq X$ is totally bounded, and let $\mathcal F$ be a filter consisting of subsets of $f(A)$. Choose an ultrafilter of subsets $\mathcal G$ of $A$ refining the filterbase $A \cap f^{-1}(\mathcal F) := \{A \cap f^{-1}(F) | F \in \mathcal F\}$. Then, $f(\mathcal G) \supseteq f(A \cap f^{-1}(\mathcal F))$ and $f(A \cap f^{-1}(\mathcal F))$ generates a filter which is finer than $\mathcal F$ (since $F \supseteq f(A \cap f^{-1}(F))$ for all $F \in \mathcal F$). Moreover, since $\mathcal G$ is Cauchy by virtue of being an ultrafilter of subsets of the totally bounded set $A$, it follows by Cauchy continuity that $f(\mathcal G)$ generates a Cauchy ultrafilter $\mathcal H$. By the preceding remarks, $\mathcal H \supseteq \mathcal F$.

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