TITLE: What is the relation between convex metric spaces and convex sets?
QUESTION [5 upvotes]: Here's another question that came to mind when I was reading the article on convex metric spaces in Wikipedia:
According to the article, "a circle, with the distance between two points measured along the shortest arc connecting them, is a (complete) convex metric space. Yet, if $x$ and $y$ are two points on a circle diametrically opposite to each other, there exist two metric segments conecting them (the two arcs into which these points split the circle), and those two arcs are metrically convex, but their intersection is the set $\{x,y\}$ which is not metrically convex."

In the article, a circle is treated as closed subset of Euclidean space, and according to another part of the article, "[a]ny convex set in a Euclidean space is a convex metric space with the induced Euclidean norm. For closed sets the converse is also true: if a closed subset of a Euclidean space together with the induced distance is a convex metric space, then it is a convex set."
If this is true, that is, if a circle is a closed subset of Euclidean space with an induced norm (the length of a segment along the shortest path between any two points on the circle) and is a convex metric space, being therefore a convex set, why isn't the intersection $\{x,y\}$ also metrically convex? Are the points $\{x,y\}$ as a subset of the circle not metrically convex on their own?

REPLY [2 votes]: If this is true, that is, if a circle is a closed subset of Euclidean space with an induced norm (the length of a segment along the shortest path between any two points on the circle) and is a convex metric space, being therefore a convex set,

The induced norm is not the length of the shortest path along the circle. It is the norm induced by the metric of the ambient space, i.e. the length of the line segment in the Euclidean plane joining two points on the circle. With this norm, the circle is not a convex metric space, nor is it a convex set.

why isn't the intersection $\{x,y\}$ also metrically convex? Are the points $\{x,y\}$ as a subset of the circle not metrically convex on their own?

No. By definition, for a space to be metrically convex there must be a third point $z$ distinct from both $x$ and $y$ satisfying $d(x,z)+d(z,y)=d(x,y)$. But there is no third point in $\{x,y\}$.