$\mathbb Q$-metric spaces: how much is lost?

Question

I want to know how weak (literally "punctured") the theory of metric spaces becomes if we impose the condition that the distance function can assume only rational (nonnegative) values.

Answer 1

Clearly a $\Bbb Q$-metrizable space must be zero-dimensional. The space ${}^\omega\omega$ of irrationals is universal for separable zero-dimensional metrizable spaces and is $\Bbb Q$-metrizable, so every separable zero-dimensional metrizable space is $\Bbb Q$-metrizable. To define such a metric on ${}^\omega\omega$, for distinct $x=\langle x_n:n\in\omega\rangle$ and $y=\langle y_n:n\in\omega\rangle$ in ${}^\omega\omega$ let $\delta(x,y)=\min\{n\in\omega:x_n\ne y_n\}$ and $d(x,y)=2^{-\delta(x,y)}$.

More generally, let $D_\kappa$ be the discrete space of cardinality $\kappa$. Then ${}^\omega D_\kappa$ with the product topology is universal for strongly zero-dimensional metrizable spaces of weight at most $\kappa$, and a $\Bbb Q$-metric can be defined on it in the same way as for ${}^\omega\omega$. (These are in fact non-Archimedean metrics, also known as ultrametrics.)

There are (necessarily non-separable) zero-dimensional metric spaces that are not strongly zero-dimensional; I don’t know whether they admit $\Bbb Q$-metrics.