Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
I think this might be hepful:… – 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
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 $O$).

share|cite|improve this answer
(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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.