# What is the relation between convex metric spaces and convex sets?

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?

• "A circle is a closed subset of Euclidean space" is not universally true. That's just one way to think about a circle. Another notion of a circle is a space (manifold) with its own path metric, which is not identical to any subset of the plane, or indeed of any Euclidean space. Which subsets are convex and which are not depends on what notion of a circle we use. – user53153 Feb 25 '13 at 22:02
• The notion seems a bit odd when considering subsets. Why would one want to require only some point in between to belong to the subset, not all of them? – Martin Feb 25 '13 at 22:08
• @5pm Edited to remove that ambiguity. – bright-star Feb 26 '13 at 2:55
• You misread that "another part of the article": it says "closed subsets of Euclidean spaces are convex metric spaces if and only if they are convex sets." – user53153 Feb 26 '13 at 2:57
• If you think that all closed subsets of a Euclidean space are convex, then indeed, there is a serious misunderstanding. – user53153 Feb 26 '13 at 3:46

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 $\{x,y\}$.