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Dec 22, 2020 at 20:34 history edited user111 CC BY-SA 4.0
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Mar 4, 2020 at 22:34 comment added Dmitri Panov Dan, yes, for example, a good metric would be such that each fiber $\mathbb R^n$ is isometric to an open half of a unite sphere (it is identified with $\mathbb R^n$ by the projection from its centre). Such a metric is good because convex subsets of $\mathbb R^n$ correspond to convex subsets of the half-sphere.
Mar 4, 2020 at 20:32 comment added Dan Petersen Thanks! For completeness, I guess a natural way to ensure that this construction exhausts $U$ is to put a Riemannian metric on the total space of $E$ and then in step 1) note that we can make sure that each $B_x$ contains all vectors whose distance to the complement of $U$ is at least $1/n$.
Mar 4, 2020 at 18:24 history edited Dan Petersen CC BY-SA 4.0
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Mar 4, 2020 at 18:24 vote accept Dan Petersen
Mar 4, 2020 at 18:12 history answered Dmitri Panov CC BY-SA 4.0