Three definition of total boundness (for uniform spaces)

Question

The following are three definitions of a totally bounded uniform spaces on a set $U$:

1. For every entourage $E$ there exists a finite cover $S$ of $U$ such that $\forall A\in S:A\times A\subseteq E$.
2. For every entourage $E$ there exists a finite set $B$ such that $E[B]=U$ (here $E[B]$ is the image of the set $B$ by the relation $E$).
3. Every proper filter on $U$ contains a Cauchy filter.

I know the proof that $(1)\Leftrightarrow(2)$ (see my book).

But how to prove equivalence with the third (for any uniform spaces)?

Answer 1

Point 3. should be formulated as: "every filter on $U$ has a Cauchy refinement", where a refinement of a filter is a larger filter, a superset.

Suppose $U$ is a totally bounded uniform space. Let $\mathcal{F}$ be a proper filter on $U$ and let $\mathcal{F}'$ be a refining ultrafilter. I claim $\mathcal{F}'$ is Cauchy:

let $D$ be an entourage of $U$ and let $E$ be a symmetric entourage of $U$ such that $E \circ E \subset D$. Then there exists a finite subset $F$ of $U$ such that $E[F] = U$, by total boundedness. Then each of the sets $E[x], x \in F$ is $D$-small and they form a finite cover of $U$, so one of them is in the ultrafilter $\mathcal{F}'$, as required. So $\mathcal{F}'$ is Cauchy.

On the other hand, suppose that 3. is true, and $U$ is not totally bounded. The latter means that there exists an entourage $D$ such that $X \setminus D[F]$ is non-empty, for every finite subset $F$ of $U$. All such sets $D[F]$ are closed under finite unions, so all sets $X \setminus D[F]$ form a filter base. Let $\mathcal{F}$ be the filter generated by this base, and by assumption $\mathcal{F}'$ is a Cauchy filter that contains $\mathcal{F}$.

Let $M$ be a $D$-small (for this $D$) member of $\mathcal{F}'$. Let $F$ be a finite subset of $U$. Then $X \setminus D[F] \in \mathcal{F} \subset \mathcal{F}'$ and so $M$ intersects this set, say in $x$. Then $M \subset D[x]$ (as $M \times M \subset D$, as $M$ is $D$-small) and so $M \subset D[F \cup \{x\}]$. But this means that $M \in \mathcal{F}'$ does not intersect $X \setminus D[F \cup \{x\}] \in \mathcal{F} \subset \mathcal{F}'$, which is a contradiction. This shows that $U$ is totally bounded.