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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 $\mathrm{conv}(S)=\bigcap_{}^{} \left \{ U \supseteq S : U \; \mathrm{convex} \right \}$.

My question is for a metric space $(X,d)$ and a open subset $S \subseteq X$, is the set $\mathrm{conv}(S)$ open?

Note: This question has been associated to my post, but it uses an inequivalent notion of convexity that cannot be used in metric spaces.

Tian Vlasic
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1 Answers1

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I think I have a counterexample.

Look at $X=\mathbb{R}^2\setminus\{(x,y)\in\mathbb{R}^2:|x|<1,y\neq 0\}$. Below is a picture of $X$ in red as a subset of $\mathbb{R}^2$ in A.

Counterexample to convex hulls of open sets in metric spaces.

The metric on $X$ comes from the Euclidean metric. $X$ consists of three parts: "the left part", "the right part" and "the middle part". In each part separately, distances and metric segments are defined as induced by the intrinsic metric from $\mathbb{R}$ - see B for some examples. For two points that lie in different parts, one has to pass through one (or two) "choke points" at $Q_1, Q_2$, but except for that the metric segments consists of two or three usual Euclidean segments.

Now take as open set $S$ the union of two open balls: $S=B((-3,0);1)\cup B((3,0);1)$ for $P_1=(-3,0),P_2=(3,0)$. The convex hull $C=\text{conv}(S)$ should be $S$ together with two "open sectorlike shapes" and the "middle part" as depicted in C. If we look at the two "choke points" $Q_1, Q_2$ we see $Q_1,Q_2\in C$, however no open subset containing any of the two choke points is contained in $C$. An open ball around $Q_1$ is depicted in green in D - note that only the green part of the black disc is part of the ball.

Now this is not a formal proof, but I hope it fits. If my intuition derailed me somewhere, please object and let me know.

Remark: In this metric on $X$ all open balls should be identical to their convex hull. If we take the metric induced by the $1$-norm we get open balls that are squares oriented as "diamonds", while their convex hull should be squares oriented as "boxes". If we replace the Euclidean metric by the $1$-metric in the construction with $X$, I think that we do not get a counterexample. If the opening angle at choke points would not be $180^{\circ}$ but larger (suitably adjusting $X$), I guess one would also get a counterexample coming from the $1$-metric. By "$180^{\circ}$" I mean the usual Euclidean angles on $\mathbb{R}^2$ - nothing which has to do with the counterexample metrics.

Tian Vlasic
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Thomas Preu
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    +1, but I think you mean "choke" points. – Cheerful Parsnip Aug 06 '22 at 22:22
  • From what I can see in picture B, I think you might be using the intrinsic metric from $\mathbb{R}^2$? – Tian Vlasic Aug 07 '22 at 17:33
  • After quickly browsing the wikipedia article on intrinsic metric, which was previously unknown to me: probably yes. – Thomas Preu Aug 07 '22 at 17:49
  • The idea of my example - at least I understand it that way - is a play with dimension: The metric space consists of different parts of different dimensions (naively or, if you want, covering dimension or something like that) $1$ and $2$. At the gluing points, where these parts meet, interesting phenomena arise. – Thomas Preu Aug 07 '22 at 17:56
  • Because I included the boundary points at $x\in{\pm 1}$ the space $X$ is even complete. Except for the two choke points it is locally a Riemannian manifold, but it is not geodesically complete. How one would go on to construct a geodesically complete Riemannian manifold $X$ such that its induced metric space has the counterexample property is beyond me - it might be true that there are no counterexamples in that case. Anyways potential counterexamples needed to be of a completely different nature, because (connected) Riemannian manifolds have the same (local) dimension everywhere - I think. – Thomas Preu Aug 07 '22 at 18:26
  • @TianVlašić Why would this not be the case? The topology on $X$ is induced as the subspace topology from Euclidean $\mathbb{R}^2$. This means that each open set $U$ of X w.r.t. the topology induced by the metric is of the form $U=V\cap X$ for some open set $V$ of $\mathbb{R}^2$ in the Euclidean topology. Why is this wrong? On another note: where did your question about semispaces go, which I answered to the best of my efforts? – Thomas Preu Aug 14 '22 at 19:35
  • Note, from picture B, you are using the intrinsic metric, so it is not the metric induced from $\mathbb{R}$. Also, about the question on semispaces, I will again post it with an additional question about the convexity of complements of semispaces. – Tian Vlasic Aug 14 '22 at 19:45
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    @TianVlašić The metric is not induced from $\mathbb{R}^2$, I agree. But the topology is. On the other question: Do you expect me to write my answer again? I put some work into that. And note: policy is one question at a time. – Thomas Preu Aug 14 '22 at 19:53