Are convex polytopes closed in arbitrary metric spaces?

Question

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?

Answer 1

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.

Answer 2

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 $Y$.