Cauchy filters in metric spaces Some terminology: Let $(X,d)$ be a metric space. 


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*A filter $\mathcal F \subseteq \mathcal P (X)$ is Cauchy if $\forall \epsilon >0 \exists x\in X: B_\epsilon(x)\in \mathcal F$. 

*A filter $\mathcal F \subseteq \mathcal P (X)$ is round if $\forall F\in \mathcal F \exists \epsilon>0 \forall x\in X:B_\epsilon(x)\in \mathcal F \implies B_\epsilon(x)\subseteq F$. 

*A Cauchy filter $\mathcal F$ is said to be minimal if no filter that is properly contained in it is Cauchy. 


I found an article that remarks, as a matter of fact, that round Cauchy filters and minimal Cauchy filters coincide. Showing that a round Cauchy filter must be minimal is straightforward. But I can't immediately see how to prove the other implication. So, how does one show that a minimal Cauchy fiter must be round? Also, any references for the standard constructions of metric theory convergence by using filters (I know it's not strictly required, as sequences suffice, but still, I'm curious) are welcome. Thanks.
 A: I suppose it would be easy to solve this by referring to the sequential completion, but that might be considered cheating.
I will use $N_\epsilon(A)$ to denote the $\epsilon$-neighbourhood of a set $A$.
For a given filter $\mathcal{F}$ the family 
$\{ N_\epsilon(A) \mid \epsilon > 0,\, A \in \mathcal{F} \}$ is the base of a
filter $\mathcal{F}^\circ \subset \mathcal{F}$. Clearly if $\mathcal{F}$ is 
a Cauchy filter, then so is $\mathcal{F}^\circ$.
I claim that $\mathcal{F}^\circ$ is also round.
Proof: 
Take an arbitrary $V \in \mathcal{F}^\circ$. By construction, there are $\epsilon > 0$ and $A \in \mathcal{F}$ such that $N_\epsilon(A) \subset V$.
Put $\delta = \epsilon / 3$. Now if there is some $x$ such that 
$B_\delta(x) \in \mathcal{F}^\circ$, then $B_\delta(x)$ meets $N_\delta(A)$, which means that $x \in N_{2\delta}(A)$, and therefore $B_\delta(x) \subset N_\epsilon(A) \subset V$.
To conclude: for every Cauchy filter there is a smaller, round Cauchy filter,
therefore any minimal Cauchy filter is round.
P.S.: I have not really investigated it, but it seems this might apply more generally to uniform spaces.
