Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am wondering if it is always possible to find disjoint sets on any manifold such that these sets are balls when mapped to their locally Euclidean space $such$ $that$ there are an infinite number of such sets.

The result is, for example, obvious when the manifold is itself Euclidean; not sure if this is true in general.

share|cite|improve this question
What does it mean "are balls when mapped to their locally Euclidean space"? It seems to me that this condition is not well defined. – Emanuele Paolini Jan 29 '13 at 21:12
Every smooth manifold has a countable basis of regular coordinate balls. A smooth coordinate ball is a smooth coordinate domain whose image under a smooth coordinate map is a ball in ordinary Euclidean space. – Squirtle Jan 29 '13 at 21:18
But the same manifold can have different coordinate maps, and hence a set could be mapped to a ball by a coordinate map but not by another. – Emanuele Paolini Jan 29 '13 at 21:23

You could always look at an open $\Bbb R^n$-homeomorphic subset of your manifold and say "this is basically Euclidian, so I can find an infinitude of balls within this subset." Then you're done.

Also, note that for any open subset of a manifold homeomorphic to Euclidian space, there is no unique such homeomorphism, so a set that looks like a ball under one homeomorphism can look like a cube under another.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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