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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?

Thanks in advance!

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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

2 Answers 2

up vote 3 down vote accepted

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 $O$).

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(You do not really need to show contiuity of the inverse, once you know that the map is continuous and bijective, because these are compact Hausdorff spaces) –  Mariano Suárez-Alvarez Feb 8 '13 at 21:31
What is this central projection? –  Tomás Feb 8 '13 at 21:34
@Tomás, have you a least tried to google for «central projection»? –  Mariano Suárez-Alvarez Feb 8 '13 at 21:35
No I did not, le me try it. –  Tomás Feb 8 '13 at 21:38
@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

This is an old thread, but resurrecting for sake of including more detail. We begin with a lemma.

Lemma. Let $E$ be a compact subset of a metrix space $X$. Then $\partial E$ is compact.

Proof. Since $E^\circ$ is open in $X$, it is also open in $E$. Since $E$ is compact, $E$ is closed, and $E = \overline{E}$. Thus $\partial E = \overline{E}\,\backslash\, E^\circ = E \,\backslash\, E^\circ$, so $\partial E$ is closed as a subset of $E$, and therefore compact. $\square$

We want to show that if $X = \mathbb{R}^n$ and $E$ is a compact convex subset with non-empty interior, then $\partial E$ is homeomorphic to the sphere $\{(x_1, \dots, x_n) \text{ }|\text{ }\sum_{i=1}^n x_i^2 = 1\}$.

We may assume that $0 \in E^\circ$ since the problem is translation invariant. Let $u$ be a unit vector. We show that there is a unique $s_u$ such that $s_uu \in \partial E$; indeed, take $s = \sup\{t\text{ }|\text{ }tu \in E\}$. It is clear that $s_u \in \partial E$, and it is also clear that if $t > s_u$ then $tu \notin \partial E$ since $\partial E \subset E$ as seen in our proof of the lemma. For any $u$, $s_u > 0$ since $0$ is interior to $E$, and $s_u < \infty$ because $E$ is bounded. Now, if $t = (1-\lambda)s_u$ with $0 < \lambda < 1$, and $N_\epsilon(0) \subset E$, then by convexity of $E$, $N_{\lambda \epsilon}(tu) \subset E$, so $tu \notin \partial E$. Thus $F(u) = t_uu$ is a bijection from there sphere to $\partial E$. Now, $F^{-1}$ is the restriction to $\partial E$ of the map $x \mapsto x/|x|$ which is continuous on $\mathbb{R}^n \,\backslash\,0$. Thus $F^{-1}$ is a continuous bijection from $\partial E$ to the sphere, and since $\partial E$ is compact by the lemma, $(F^{-1})^{-1}$ is also continuous.

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