In an ultrametric space, is every open set closed?

Question

I saw the following well-known fact for ultrametric spaces

Every open ball is closed.

So this stimulates me to think whether this is true for open set or not.

By an ultramtric space, it's a metric space $(M,d)$ whose metric satisfies the following condition (stronger than triangle inequality): $$ d(x,z) \leqslant \max \{ d(x,y), d(y,z)\}, \;\; \forall \; x,y,z \in M. $$

My attempt:

After I try to prove this statement is true by contradiction argument, I realized there is always a gap. So I believe this is false now. But I can't still find a counterexample.

I also try to google some key words, but things I can find out are for open balls. I don't see any discussion for my problem.

Answer 1

No. In the $p$-adic numbers $\Bbb Q_p$, one-point subsets such as $\{0\}$ are closed, but not open. The complement of a one-point subset is open, but not closed.

Answer 2

For a simple ad hoc example, take $\mathbb R$ or $\mathbb Q$ and define $d(x,y)=\max(|x|,|y|)$ for $x\ne y$; the set $\{0\}$ is closed but not open, so its complement is open but not closed.