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Apr 13, 2017 at 12:58 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
Jun 15, 2015 at 16:44 comment added Jim Belk 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.
Jun 14, 2015 at 1:34 answer added Oliver Jones timeline score: 4
Jun 10, 2015 at 23:27 comment added Oliver Jones 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.
May 29, 2015 at 23:59 comment added Oliver Jones Any Wiedersehen manifold has this property. These are manifolds for which the cut locus of any point is a single point.
S May 12, 2015 at 5:48 history edited Marco Golla CC BY-SA 3.0
improved formatting, added riemannian-geometry tag
S May 12, 2015 at 5:48 history suggested user73589 CC BY-SA 3.0
improved formatting
May 12, 2015 at 5:33 review Suggested edits
S May 12, 2015 at 5:48
May 10, 2015 at 18:29 comment added Yoav Kallus 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.
May 10, 2015 at 18:22 history edited coudy CC BY-SA 3.0
add a word
May 10, 2015 at 17:36 comment added Joel David Hamkins 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.
May 10, 2015 at 17:10 history asked coudy CC BY-SA 3.0