# Subset of $\mathbb R^n$ homeomorphic to sphere?

Let $C$ be a subset of $\mathbb R^n$ with the following properties attached to it:

• Convex
• Compact
• Non-empty interior

Is the boundary of $C$ homeomorphic to the ball of dimension $n-1$? Why?

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I think this might be hepful: math.stackexchange.com/questions/165629/… –  Tomás Feb 8 '13 at 21:26
@Tomás In general, the boundaries of homeomorphic subspaces need not be homeomorphic. –  Hagen von Eitzen Feb 8 '13 at 21:28
My apologies, it's corrected now. English isn't my first language and didn't know the correct terminology. –  Ulibniss Feb 8 '13 at 21:32
Interesting. Please, can you give me a example? –  Tomás Feb 8 '13 at 21:33
@Tomás $\mathbb R^2$, $\mathbb R\times (0,\infty)$, $\mathbb R\times (0,1)$ and the open unit disk are all homeomorphic, but their boundaries are not (empty, the real line, two disjoint lines, the unit circle). Another funny set is $\mathbb R^2\setminus\bigcup_{(n,m)\in\mathbb Z^2\setminus\{(0,0)\}} (n,m)\cdot [1,\infty)$. –  Hagen von Eitzen Feb 9 '13 at 13:09

Let $O$ be an interior point of $C$. Then the central projection $f\colon\partial C\to S^{n-1}$ along rays ending at $O$ turns out to be a homeomorphism: By convexity of $C$, $f$ is injective. Because $C$ is bounded, $f$ is also surjective. Remains to show that both $f$ and its inverse are continuous. For $f$ itself, this is clear (using that an open ball around $O$ does not intersect $\partial C$). For the inverse, the argument is also quite easy (using convexity and again an open ball $\subset C$ around $O$).
@Tomás: For $x\in \partial C$, the unique ray starting at $O$ and passing through $x$ intersects $S^{n-1}$ in exactly one point $y$. Let $f(x)=y$. –  Hagen von Eitzen Feb 9 '13 at 13:01