Given a filter $\mathcal F$ on a countable set $X$, consider the family $$\mathcal F^+:=\{A\subset X:\forall F\in\mathcal F\;(A\cap F\ne\emptyset)\}.$$

The following characterization is well-known.

**Theorem S.** A filter $\mathcal F$ on a countable set $X$ is not an ultrafilter if and only if the family $\mathcal F^+$ contains two disjoint sets.

A topological counterpart of this theorem is less known and holds for analytic filters. Let us recall that a filter $\mathcal F$ on a countable set $X$ is *analytic* if it is analytic (i.e., a continuous image of a Polish space) as a subspace of the power-set $\mathcal P(X)$ endowed with the natural compact metrizable topology.

**Theorem T.** An analytic filter $\mathcal F$ on a countable topological space $X$ is divergent if and only if the family $\mathcal F^+$ contains two disjoint closed subsets of $X$.

This theorem can be (more or less) easily proved using Talagrand's characterization of meager filters.

Now the **question**: *Is the following metric counterpart of Theorems S and T true?*

**Conjecture M.** An analytic filter $\mathcal F$ on a countable metric space $(X,d)$ is not Cauchy if and only if the family $\mathcal F^+$ contains two sets $A,B$ with positive distance, i.e., such that $d(A,B):=\inf\{d(a,b):a\in A,\;b\in B\}>0$.

**Remark.** Conjecture M is true for metric spaces which are uniformly homeomorphic to subspaces of $\mathbb R^\omega$ endowed with the natural uniformity. In particular, it is true for totally bounded metric spaces.

On the other hand, the unit ball of an infinite-dimensional Banach space is not uniformly homeomorphic to a subspace of $\mathbb R^\omega$.