# Proof verification: totally bounded iff every filter has a cauchy superfilter

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, $$\mathcal H \supseteq \mathcal F$$.