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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.

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    $\begingroup$ 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. $\endgroup$ – bof Sep 14 '19 at 6:21
  • $\begingroup$ @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. $\endgroup$ – Yung-Hsiang Huang Sep 14 '19 at 6:33
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    $\begingroup$ 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. $\endgroup$ – bof Sep 14 '19 at 6:36
  • $\begingroup$ @bof Got it. Thanks. $\endgroup$ – Yung-Hsiang Huang Sep 15 '19 at 0:54
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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.

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  • $\begingroup$ Thank you so much!!! This is an example I cannot find out by myself. $\endgroup$ – Yung-Hsiang Huang Sep 14 '19 at 6:28
  • $\begingroup$ 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. $\endgroup$ – bof Sep 14 '19 at 6:38
  • $\begingroup$ 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. $\endgroup$ – Henno Brandsma Sep 14 '19 at 16:00
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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.

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