Boundary of Compact Convex Set with Nonempty Interior is Homeomorphic to Sphere

Theorem

Let $n \in \N_{> 0}$.

Let $C \subseteq \R^n$ be a compact convex subset of real Euclidean $n$-space.

Suppose that the interior $C^\circ$ is non-empty.

Then, the boundary $\partial C$ is homeomorphic to $\Bbb S^{n - 1}$, the unit $n - 1$-sphere.

Proof

Let $\bsx_0 \in C^\circ$ be an element of $C^\circ$.

Define $\phi : \R^n \setminus \set {\bsx_0} \to \Bbb S^{n - 1}$ as:

$\map \phi \bsx = \dfrac 1 {\norm {\bsx - \bsx_0}} \paren {\bsx - \bsx_0}$

As Normed Vector Space is Hausdorff Topological Vector Space, it follows that $\phi$ is continuous.

Define $\phi^* : \partial C \to \Bbb S^{n - 1}$ as:

$\map {\phi^*} \bsx = \map \phi \bsx$

which is well-defined as the boundary is disjoint from the interior by definition.

By Restriction of Continuous Mapping is Continuous:

$\phi^*$ is continuous.

$\Box$

It remains to show that $\phi^*$ is a bijection.

Let $\bsy \in \Bbb S^{n - 1}$ be arbitrary.

We need to prove that there is a unique $\bsx \in \partial C$ such that:

$\map \phi \bsx = \bsy$

By Ray from Interior of Compact Convex Set Meets Boundary Exactly Once, there is a unique $t \in \R_{> 0}$ such that:

$\bsx_0 + t \bsy \in \partial C$

Therefore:

\(\ds \map \phi {\bsx_0 + t \bsy}\)

\(=\)

\(\ds \frac 1 {\norm {t \bsy} } \paren {t \bsy}\)

Definition of $\phi$

\(\ds \)

\(=\)

\(\ds \bsy\)

Definition of Unit Sphere

Now, $\bsx \in \partial C$ satisfies:

$\map \phi \bsx = \bsy$.

Then:

\(\ds \bsy\)

\(=\)

\(\ds \map \phi \bsx\)

\(\ds \)

\(=\)

\(\ds \frac 1 {\norm {\bsx - \bsx_0} } \paren {\bsx - \bsx_0}\)

\(\ds \leadsto \ \ \)

\(\ds \norm {\bsx - \bsx_0} \bsy\)

\(=\)

\(\ds \bsx - \bsx_0\)

\(\ds \leadsto \ \ \)

\(\ds \bsx\)

\(=\)

\(\ds \bsx_0 + \norm {\bsx - \bsx_0} \bsy\)

Therefore:

$\bsx_0 + \norm {\bsx - \bsx_0} \bsy \in \partial C$

so:

$\norm {\bsx - \bsx_0} = t$

Hence:

$\bsx = \bsx_0 + t \bsy$

Therefore, for every $\bsy \in \Bbb S^{n - 1}$, there is a unique $\bsx \in \partial C$ such that:

$\map {\phi^*} \bsx = \map \phi \bsx = \bsy$

and so $\phi^*$ is a bijection by definition.

$\Box$

We have shown that $\phi^*$ is a continuous bijection.

By Boundary of Compact Set in Hausdorff Space is Compact, $\partial C$ is compact.

Therefore, by Continuous Bijection from Compact to Hausdorff is Homeomorphism:

$\phi^*$ is a homeomorphism from $\partial C$ to $\Bbb S^{n - 1}$

$\blacksquare$