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Is every nonempty open set $U\subseteq\mathbb{R}^n$ diffeomorphic to $\mathbb{R}^n$?

I think this is false; perhaps I can take $U$ to be the disjoint union of two open balls. But how can I prove that this is not diffeomorphic to $\mathbb{R}^n$?

More generally, do diffeomorphisms preserve connectedness, or number of connected components?

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2 Answers 2

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Diffeomorphism saves connectedness. But here $\mathbb R^n$ is connected and union of two disjoint open sets is not connected.

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For two sets to be diffeomorphic, they must be homeomorphic. Connectedness (and number of connected compotents) is a topological property and is therefore preserved under homeomorphism. Therefore $\mathbb{R}^n$ is not homeomorphic to any disconnected open subset of itself.

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  • $\begingroup$ and if we assume $U$ connected? $\endgroup$
    – Veridian Dynamics
    Dec 6, 2017 at 19:23
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    $\begingroup$ Still not. Take the annulus $\endgroup$
    – eepperly16
    Dec 6, 2017 at 19:25
  • $\begingroup$ @VeridianDynamics The statement may hold if your set is open and convex or perhaps open and star-like. $\endgroup$
    – eepperly16
    Dec 6, 2017 at 19:29
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    $\begingroup$ @Arthur Unfortunately, no. There are rather surprising counterexamples. But even better; there are contractible open subsets of $\Bbb R^3$ which are not even homeomorphic to $\Bbb R^3$: eg, the Whitehead manifold $\endgroup$
    – Balarka Sen
    Dec 6, 2017 at 20:46
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    $\begingroup$ @eepperly16 This is indeed true for open star-like subsets, but is rather hard to prove. See, eg, this MO post $\endgroup$
    – Balarka Sen
    Dec 6, 2017 at 20:51

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