# In an ultrametric space, is every open set closed?

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.

• If every open set is closed, then every closed set is open. In a metric space, every one-point set is closed., So in effect you are asking whether every ultrametric space is discrete. – bof Sep 14 '19 at 6:21
• @bof I don't think this is my question, but it's close. I am asking whether every ultrametric space is "almost" discrete. This terminology is taken from topospaces.subwiki.org/wiki/Almost_discrete_space. – Yung-Hsiang Huang Sep 14 '19 at 6:33
• Yes, but in a $T_1$ space, if every closed set is open, then every one-point set is open, so every set is open. – bof Sep 14 '19 at 6:36
• @bof Got it. Thanks. – Yung-Hsiang Huang Sep 15 '19 at 0:54

## 2 Answers

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.

• Thank you so much!!! This is an example I cannot find out by myself. – Yung-Hsiang Huang Sep 14 '19 at 6:28
• Or more generally take any ultrametric space which does not induce the discrete topology; then there is a one-point set which is closed but not open, and its complement will be open but not closed. – bof Sep 14 '19 at 6:38
• The Cantor cube $\{0,1\}^\mathbb{N}$ is an ultrametric space in the metric $d(x,y)=2^{-min(n: x_n \neq y_n)}$ and every singleton is closed and not open, and its complement open and not closed. – Henno Brandsma Sep 14 '19 at 16:00

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.