$\mathbf{Q}_{p}^{n}$ has a countable base?

Question

We can find a countable base for $\mathbf{R}^{n}$ using fact that $\mathbf{Q}$ is dense in the reals (example of proof by someone else is here: Proof that $\mathbf{R}^n$ has a countable base.).

We also have countable base for $\mathbf{Q}_{p}$ since the set of all balls is countable (Proposition 1.6 in this paper: https://www2.math.ethz.ch/education/bachelor/seminars/hs2011/p-adic/report5.pdf).

But $\mathbf{Q}$ is also dense in $\mathbf{Q}_{p}$, can we just use similar reasoning to the $\mathbf{R}^{n}$ to show that there is countable base for $\mathbf{Q}^{p}$, or is this not true? Thank you.

Answer 1

In technical terms:

For metric (or metrizable) spaces, separability is equivalent to second-countability.

That is

Every metric space with a countable dense subset has a countable base.

Indeed, let $(X,d)$ be a metric space and $Q\subseteq X$ a countable dense subset. Then $(X,d)$ has a countable base, consisting of the balls of rational radius around points in $Q$. $\square$

Also, the product of two (or finitely many) second-countable spaces is second-countable (by taking the obvious rectangles as base).

Answer 2

Yes, in any metric space $(X,d)$ (and $\mathbf{Q}_p$ has the $p$-adic metric), if we have a countable dense set $D$ we have a countable base: we can take the set of all balls $\{B_d(x,\frac{1}{m}): x \in D, m \in \mathbb{N}\}$ as such a base.