5
$\begingroup$

This question is motivated by the post Uncountable intersections of open balls in a separable metric space.

The general problem is the following: given a connected Riemannian manifold $M$, what are the sets that can be written as an (not necessarily countable) intersection of open balls?

If $M$ is a round sphere, then any subset is an intersection of open balls. If $M$ is a flat euclidean space, then my guess is that sets that are intersection of open balls are precisely the bounded convex sets. This seems to be related to the cut locus of points.

A specific question is:

What are the compact connected Riemannian manifolds on which all sets are intersection of open balls?

$\endgroup$
  • 1
    $\begingroup$ A related question might be: what are the metric spaces in which every point is the complement of an open ball? This seems to be the property enabling your examples on the circle and the sphere. $\endgroup$ – Joel David Hamkins May 10 '15 at 17:36
  • 2
    $\begingroup$ Let the hull of a set be the minimal superset which can be written in this form (it is unique because the intersection of two such supersets yields a smaller such superset). Then I believe a set is the intersection of open balls if it contains the hull of every pair of points in the set. This is analogous to the case of convex sets, where the hull of two points is a segment. $\endgroup$ – Yoav Kallus May 10 '15 at 18:29
  • $\begingroup$ Any Wiedersehen manifold has this property. These are manifolds for which the cut locus of any point is a single point. $\endgroup$ – Oliver Jones May 29 '15 at 23:59
  • $\begingroup$ Actually, it's easy to see that any manifold satisfying the condition must have the property that the complement of a point is an open ball. $\endgroup$ – Oliver Jones Jun 10 '15 at 23:27
  • 1
    $\begingroup$ It's not true that every bounded convex set in the Euclidean plane is the intersection of open balls. For example, an open square is bounded and convex, but any open disk that contains it must contain all of the points on the boundary of the square other than the corners. $\endgroup$ – Jim Belk Jun 15 '15 at 16:44
4
$\begingroup$

Any manifold that satisfies this condition must have the property that the complement of a point is an open ball. This means that the antipodal set of any point is just a point. Compact symmetric spaces satisfying this condition have been classified by Liu and Deng. See "The antipodal sets of compact symmetric spaces" in Balkan J. Geom. Appl. 19, no. 1, 73-79.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.