Let $(X,d)$ be a metric space. For all points $x,y \in X$ we define the metric segment between them as the following set:

$$\left [ x,y \right ] = \left \{ z \in X : d(x,z)+d(z,y)=d(x,y)\right \}$$

We then say that a set $S\subseteq X$ is convex if for all $x,y \in S$ it holds true that $\left [ x,y \right ] \subseteq S$.

It can be easily shown that arbitrary intersection of convex sets in metric spaces is a convex set. Therefore, for each subset $S \subseteq X$ of a metric space $(X,d)$ we define its convex hull as the set $\mathrm{conv}(S)=\bigcap_{}^{} \left \{ U \supseteq S : U \; \mathrm{convex} \right \}$.

We say that a set is a convex polytope if it is a convex hull of a finite set.

My question is are convex polytopes in metric spaces closed sets?

  • $\begingroup$ May we use alternative definitions of convex hull? The one I'm thinking of is $\text{conv}(S) = \{ \sum_{i=1}^n a_i s_i : \{a_i\}_{i=1}^n\subseteq [0,1], \{s_i\}_{i=1}^n\subseteq S, \sum a_i = 1\}$ $\endgroup$
    – gist076923
    Sep 28, 2022 at 15:00
  • $\begingroup$ Now that I think about it, I'm not sure if my definition works in this context $\endgroup$
    – gist076923
    Sep 28, 2022 at 15:01
  • 3
    $\begingroup$ No, it is only an equivalent definition in strictly convex normed vector spaces. We can not, of course, always define addition and scalar multiplication in metric spaces. $\endgroup$ Sep 28, 2022 at 15:07
  • $\begingroup$ so I think this is true for the discrete metric, but no clue if this would be true in general. $\endgroup$
    – Zim
    Sep 29, 2022 at 14:57
  • 3
    $\begingroup$ Of course, every subset of a discrete space is closed/open. $\endgroup$ Sep 29, 2022 at 15:07

2 Answers 2


A (perhaps) simplified version of what Eric Wofsey wrote. Define $X_1=\Bbb Q^2\cap [0,1]^2$ and $X_2=\{x\in [0,1]^2\,:\, x\text{ isn't on any line that joins two points of }\Bbb Q^2\}$

Notice that $X_2$ is dense in $[0,1]^2$ (say, because of Baire category theorem).

Call $X=X_1\cup X_2$, with the metric induced by $\Bbb R^2$. Notice that, since $[x,y]$ only depends on the distance, given $S\subseteq (M,d)$ with the subspace metric and $x,y\in S$, $[x,y]_S=[x,y]_M\cap S$. This makes it clear that $X_1$ is convex in $X$. It's also obvious that it's contained in all $X$-convex sets containing the corners of $[0,1]^2$.

Therefore $X_1=\operatorname{conv}_X\{(0,1)(0,0),(1,0),(1,1)\}$, but it isn't closed. In fact, it's dense with dense complement.

  • $\begingroup$ Good idea, removing the need for barycentric coordinates. However, the (for me) complicated part of Eric's proof was that $X_2 \neq \emptyset$. Once you think about linear (in)dependence of points from $\mathbb R^2$ over $\mathbb Q$ that becomes clear, but I don't think it is absolutely trivial. $\endgroup$
    – Ingix
    Sep 30, 2022 at 8:39
  • $\begingroup$ @Ingix Since the condition is algebraic you can have explicit ways to do it, but a quick way to see it is that lines are closed subsets with empty interior in a complete metric space and you are removing countably many of those. Therefore by Baire category theorem $X_2$ is not only non-empty, but dense as well. $\endgroup$ Sep 30, 2022 at 9:07

Take three noncollinear points $x,y,z\in\mathbb{R}^2$. Recall that every point in the triangle formed by $x,y,z$ can be written uniquely in barycentric coordinates as $ax+by+cz$ where $a,b,c\in[0,1]$ and $a+b+c=1$. Let $X$ be the set of such points whose barycentric coordinates $(a,b,c)$ are all rational. Note that for any point on a line segment between two points of $X$, the three barycentric coordinates are linearly dependent over $\mathbb{Q}$. In particular, there is a point $p$ in the triangle that is not on any line segment between points of $X$ (just take the first two barycentric coordinates to be small irrationals $a$ and $b$ such that $\{1,a,b\}$ is linearly independent).

Now consider $X$ as a subset of the metric space $Y=X\cup\{p\}$ (with the Euclidean metric). Then $X$ is the convex hull of $\{x,y,z\}$ in $Y$, but $X$ is not closed in $Y$.

  • $\begingroup$ I see how you are trying to construct a counterexample. However, I am not sure how we can add countable dense subsets of line segments such that no two points are collinear with $a$. You also don't specify what you mean by "iterate". On what set? What process? $\endgroup$ Sep 29, 2022 at 20:16
  • $\begingroup$ Actually, you can make the example completely explicit--I've edited. $\endgroup$ Sep 29, 2022 at 20:38
  • $\begingroup$ This makes it clearer. Thank you! $\endgroup$ Oct 4, 2022 at 9:16

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