# Are convex polytopes closed in arbitrary metric spaces?

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?

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

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.

• 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. Sep 30, 2022 at 8:39
• @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. 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 $$Y$$.

• 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? Sep 29, 2022 at 20:16
• Actually, you can make the example completely explicit--I've edited. Sep 29, 2022 at 20:38
• This makes it clearer. Thank you! Oct 4, 2022 at 9:16