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?
Best Answer
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.