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Let $M$ be a smooth manifold. Let's call $M$ quasi-seperated if $M$ has the following property: If $B,C \subseteq M$ are open balls, then $B \cap C \subseteq M$ is a finite(!) union of open balls. By an open ball I mean an open submanifold, which is diffeomorphic to some $D^n$.

Is every manifold quasi-separated? If not, are open balls quasi-separated? Is there a simple characterization of quasi-separated manifolds?

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In the plane you can find two open subsets whose intersection has infinitely components so, in particular, they do not intersect in finitely many balls. – Mariano Suárez-Alvarez Sep 2 '10 at 15:58
Could you give an example of a quasi-separated manifold? Where does this come from? It seems to me that a smooth manifold is quasi-separated if and only if $R^n$ is. – Deane Yang Sep 2 '10 at 16:00
(Foe example, let one of the sets be the open upper half plane unioned with all open balls of radius $1/4$ centered at the points $(n,0)$ with $n\in\ZZ$, and the other set be the reflection of the first one on the $x$-axis.) – Mariano Suárez-Alvarez Sep 2 '10 at 16:00
@Mariano: I already had this example in mind, but I had some doubt that this set is actually an open ball. Thus we can conclude that every quasi-separated manifold is $1$-dimensional, and vice versa? – Martin Brandenburg Sep 2 '10 at 16:54
up vote 1 down vote accepted

Using the Riemann mapping theorem you can easily show that $\mathbb{R}^2$ is not quasi separated.

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Oh, I see that Mariano already posted that as a comment. – Michael Bächtold Sep 2 '10 at 16:00

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