Is there a metric $d$ such that $(\mathbb{Q},d)$ is complete? Is there a metric $d$ such that $(\mathbb{Q},d)$ is complete? If so why?
A question I've been pondering about the last few days, but my maths knowledge isn't advanced enough to draw a "mathematical" conclusion
 A: Let $X=\{0\}\cup\left\{\frac1n:n\in\Bbb Z^+\right\}$; this is a compact subset of $\Bbb R$, so it’s complete in the usual metric. Let $h:\Bbb Q\to X$ be any bijection, and define a metric $d$ on $\Bbb Q$ by $d(p,q)=|h(p)-h(q)|$; then $\langle\Bbb Q,d\rangle$ is isometric to $X$ and therefore is a complete metric space.
$X$ can be replaced with any countably infinite complete metric space, and there are many of these. For instance, let $\alpha$ is any countable limit ordinal; the ordinal space $\alpha+1$ is countable and compact and can be embedded in the real line. There are uncountably many of these spaces — $\omega_1$ of them, to be precise — and no two of them are homeomorphic.
A: The discrete metric $d:\Bbb Q \times \Bbb Q \rightarrow \{0,1\}$ with $d(x,y)=0 \Leftrightarrow x=y$ turns $\Bbb Q$ into a complete metric space. It’s a bit useless though, since the underlying topology is the discrete one, so you don‘t have a distiction between open subsets, closed subsets and arbitrary subsets...
