First-countable uniform space: Is the filter definition of complete equivalent to the sequential definition?

Question

This question is motivated mainly by the topology of first-countable spaces being uniquely determined by their notions of convergence of sequences: the closure of a set $A$ is the same as the set of sequential limit points of $A$. So topological notions involving filters should be equivalent to similar definitions involving sequences in first-countable spaces.

We say a uniform space is complete if every Cauchy filter converges.

We could also define sequential completeness from the convergence of every elementary Cauchy filter, in other words the convergence of every Cauchy sequence.

Are the notions of completeness and sequential completeness equivalent in a first-countable uniform space?

While the first paragraph motivates "yes", a uniform structure is more than a topological notion.

The way I thought to approach it is to try and find a way to associate to each filter a set of elementary filters, where the question of convergence of the filter is equivalent to the convergence of one elementary filter (see here), but the way I thought to do it does not work.

Answer 1

No, they are not equivalent, even for uniform spaces that are first countable (as topological spaces). The standard example would be $X = \omega_1$ in its uniformity (which is even unique, in this special case). The space $X$ is first countable and sequentially compact (which implies it is totally bounded as a uniform space) and sequentially complete as well (as a Cauchy sequence with a convergent subsequence is convergent). It it were complete, then it would be compact (as totally bounded complete uniform spaces are), which it is not ($\omega_1 + 1$ is its unique compactification).

Answer 2

Just for the record, your statement is almost true. The issue is with first countability, which is a property of topological spaces, this notion is not "uniform enough". If $(X, \Phi)$ is the uniform space (where $\Phi$ is the set of entourages), let us say that the uniform structure $\Phi$ is countably generated, if there is a countable subset $F \subseteq \Phi$ such that every $A \in \Phi$ contains as a subset some element of $F$. Every countably generated uniform structure induces a first countable topology, but as the example of $\omega_1$ shows, the converse is false.

If the uniform structure on $X$ is countably generated, then $X$ is complete if and only if it is sequentially complete, because for every Cauchy filter we can find a suitable Cauchy sequence. This generalizes the result saying that a pseudometric space is a complete uniform space if and only it is sequentially complete.

One can also show that if $X$ is a uniform space with countably generated uniform structure, then its (Hausdorff) completion is the same as its sequential completion (just like in the case of pseudometric spaces).