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

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?

## 2 Answers

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.

• and if we assume $U$ connected? – Veridian Dynamics Dec 6 '17 at 19:23
• Still not. Take the annulus – eepperly16 Dec 6 '17 at 19:25
• @VeridianDynamics The statement may hold if your set is open and convex or perhaps open and star-like. – eepperly16 Dec 6 '17 at 19:29
• @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 – Balarka Sen Dec 6 '17 at 20:46
• @eepperly16 This is indeed true for open star-like subsets, but is rather hard to prove. See, eg, this MO post – Balarka Sen Dec 6 '17 at 20:51

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