Is there a nice characterization for the complete metric subspaces of $\mathbb{Q}$ (with the usual metric)?
It seems like a such a subspace must have empty interior; if it contained an open interval then it would clearly contain a non-converging Cauchy sequence. But this isn't enough, because consider the subset $\left\{\frac{1}{n} : n \in \mathbb{N} \right\} \subseteq \mathbb{Q}$. This has empty interior but is not complete because it doesn't contain $0$. In other words, a complete subspace of $\mathbb{Q}$ must be closed (which I think is true about metric spaces in general).
So a complete subspace of $\mathbb{Q}$ must be closed and have empty interior, but are these two sufficient to imply completeness? If so, how do I prove this, and if not, what am I missing? I'm not necessarily looking for a complete answer, just some pointers in the right direction.
There is a complete characterization of all countable complete metric spaces. Ordinals.
More specifically, one can show that if $\alpha$ is a countable ordinal then the order topology on $\alpha$ is completely metrizable.
Of course that there are complete subspaces of $\Bbb Q$ which are not order isomorphic to an ordinal; however they are homeomorphic to an ordinal (once we choose the metric, and we forget about the order).
[In my previous post I wrote something which was false; that successor ordinals were the complete ones, however I neglected the fact that a sequence whose distances from the first element approach $\infty$ need not have a limit. Successor ordinals are the compact metric spaces, though. So a subset of $\Bbb Q$ is compact if and only if it is homeomorphic to a successor ordinal.]