Compact Convex Set with Nonempty Interior is Homeomorphic to Cone on Boundary
Theorem
Let $n \in \N_{> 0}$.
Let $T \subseteq \R^n$ be a compact convex subset of real Euclidean $n$-space.
Suppose the interior of $T$ is non-empty.
Then, $T$ is homeomorphic to the cone on its boundary.
Proof
Let $\bsx_0 \in T^\circ$ be an interior point of $T$.
Let $C \partial T$ denote the cone on the boundary of $T$.
Define $\phi : C \partial T \to T$ as:
- $\map \phi {\eqclass {\tuple {e, \bsx, t}} \RR} = t \bsx + \paren {1 - t} \bsx_0$
where:
- $e$ is the unique element of the trivial topological space used in the construction of the cone.
- $\RR$ is the equivalence relation used to define the join
For $\phi$ to be well-defined, it is necessary that if:
- $\tuple {\tuple {e, \bsx, t}, \tuple {e', \bsx', t'}} \in \RR$
then:
- $t \bsx + \paren {1 - t} \bsx_0 = t' \bsx' + \paren {1 - t'} \bsx_0$
By definition of join, the hypothesis implies that one of:
- $t = t' = 0$ and $e = e'$
- $t = t' \in \openint 0 1$ and $\tuple {e, \bsx, t} = \tuple {e', \bsx', t'}$
- $t = t' = 1$ and $\bsx = \bsx'$
is true.
If the first holds, then:
- $t \bsx + \paren {1 - t} \bsx_0 = \bsx_0 = t' \bsx' + \paren {1 - t'} \bsx_0$
If the second holds, then:
- $e = e'$
- $\bsx = \bsx'$
- $t = t'$
and the equality holds trivially.
If the third holds, then:
- $t \bsx + \paren {1 - t} \bsx_0 = \bsx = \bsx' = t' \bsx' + \paren {1 - t'} \bsx_0$
Thus, in every case, $\phi$ is well-defined.
Bijection
Next, we will show that $\phi$ is a bijection.
Let $\bsy \in T$ be arbitrary.
If $\bsy = \bsx_0$, then clearly:
- $\map \phi {\eqclass {\tuple {e, \bsx, t}} \RR} = \bsx_0 \iff t = 0$
which is unique by definition of $\RR$ and of trivial topological space.
Therefore, suppose $\bsy \ne \bsx_0$.
Then, by Ray from Interior of Compact Convex Set Meets Boundary Exactly Once, there is a unique $t > 0$ such that:
- $\bsx_0 + t \paren {\bsy - \bsx_0} \in \partial T$
Let $\bsx = \bsx_0 + t \paren {\bsy - \bsx_0} \in \partial T$.
Thus:
\(\ds t \bsy\) | \(=\) | \(\ds \bsx - \bsx_0 + t \bsx_0\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \bsy\) | \(=\) | \(\ds \frac 1 t \bsx + \paren {1 - \frac 1 t} \bsx_0\) |
Furthermore, if some $\bsx' \in \partial T$ and $t' > 0$ satisfies:
- $\bsy = t' \bsx' + \paren {1 - t'} \bsx_0$
we get:
\(\ds t' \bsx'\) | \(=\) | \(\ds \bsy + \paren {t' - 1} \bsx_0\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \bsx'\) | \(=\) | \(\ds \frac 1 {t'} \bsy + \paren {1 - \frac 1 {t'} } \bsx_0\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \bsx_0 + \frac 1 {t'} \paren {\bsy - \bsx_0}\) | ||||||||||||
\(\ds \) | \(\in\) | \(\ds \partial T\) | Definition of $\bsx'$ | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds t\) | \(=\) | \(\ds \frac 1 {t'}\) | Uniqueness of $t$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \bsx'\) | \(=\) | \(\ds \bsx\) |
It remains to show that $\dfrac 1 t \in \hointl 0 1$.
By Ray from Bounded Set Meets Boundary, there exists some $s \ge 0$ such that:
- $\bsy + s \paren {\bsx - \bsx_0} \in \partial T$
Therefore:
- $\dfrac 1 t \bsx + \paren {1 - \dfrac 1 t} \bsx_0 + s \paren {\bsx - \bsx_0} \in \partial T$
That is:
- $\bsx_0 + \paren {\dfrac 1 t + s} \paren {\bsx - \bsx_0} \in \partial T$
By Ray from Interior of Compact Convex Set Meets Boundary Exactly Once, there is a unique $u > 0$ such that:
- $\bsx_0 + u \paren {\bsx - \bsx_0} \in \partial T$
Since $\dfrac 1 t + s > 0$:
- $\dfrac 1 t + s = u$
As clearly $u = 1$ satisfies:
- $\bsx_0 + u \paren {\bsx - \bsx_0} = \bsx \in \partial T$
We have:
- $\dfrac 1 t + s = 1$
Therefore:
- $\dfrac 1 t \le 1$
Thus, for every $\bsy \in T$ with $\bsy \ne \bsx_0$, there is a unique $\dfrac 1 t$ and $\bsx$ such that:
- $\map \phi {\eqclass {\tuple {e, \bsx, \dfrac 1 t} } \RR} = \bsy$
Combined with the case $\bsy = \bsx_0$ above, we get that:
- $\phi$ is a bijection
$\Box$
Continuity
The last step is to prove that $\phi$ is continuous.
Let $X = \eqclass {\tuple {e, \bsx, t}} \RR$ be arbitrary, where $\bsx \in \partial T$ and $t \in \closedint 0 1$.
Let $N$ be a neighborhood of $\map \phi X$.
By definition, there is some open ball:
- $\map {B_\epsilon} {\map \phi X} \subseteq N$
We will separately examine the cases of $t = 0$ and $t > 0$.
If $t = 0$, then $\map \phi X = \bsx_0$.
By defining the continuous mapping $\psi : \partial T \to \R_{\ge 0}$ as:
- $\map \psi \bsz = \norm {\bsz - \bsx_0}$
and applying:
- Boundary of Compact Set in Hausdorff Space is Compact
- Continuous Image of Compact Space is Compact
- Compact Subspace of Metric Space is Bounded
we have that:
- $\set {\norm {\bsz - \bsz_0} : \bsz \in \partial T}$
is bounded above.
Let $\delta = \dfrac \epsilon {\sup_{\bsz \in \partial T} \norm {\bsz - \bsx_0}}$
Then, for any $\bsz \in \partial T$ and $t \in \hointr 0 \delta$:
\(\ds \norm {\map \phi {\eqclass {\tuple {e, \bsz, t} } \RR } - \bsx_0}\) | \(=\) | \(\ds \norm {t \bsz + \paren {1 - t} \bsx_0 - \bsx_0}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds t \norm {\bsz - \bsx_0}\) | ||||||||||||
\(\ds \) | \(<\) | \(\ds \delta \norm {\bsz - \bsx_0}\) | Assumption on $t$ | |||||||||||
\(\ds \) | \(\le\) | \(\ds \delta \sup_{\bsz \in T} \norm {\bsz - \bsx_0}\) | Definition of Supremum of Subset of Real Numbers | |||||||||||
\(\ds \) | \(=\) | \(\ds \epsilon\) | Definition of $\delta$ |
We have shown that the image of:
- $A_\delta = \set {\eqclass {\tuple {e, \bsz, t}} \RR : \bsz \in \partial T \land t \in \hointr 0 \delta}$
under $\phi$ is a subset of $\map {B_\epsilon} {\map \phi X}$, itself, a subset of $N$.
Thus, we only need to show that $A_\delta$ is a neighborhood of $\bsx_0$ in $C \partial T$.
We have that $\set e \times \partial T \times \hointr 0 \delta$ is open in the product space underlying the join operation.
- This is by definition, since it is exactly $\pr_3^{-1} \hointr 0 \delta$
But then $A_\delta$ is open in $C \partial T$, by definition of quotient topology.
$\Box$
Suppose $t > 0$.
Then, $\norm {\map \phi X - \bsx_0} > 0$.
By replacing $\epsilon$ with $\map \min {\epsilon, \norm {\map \phi X - \bsx_0}}$, we can ensure:
- $\map {B_\epsilon} {\map \phi X} \subseteq N \setminus \set \bszero$
Now, let:
- $\delta = \dfrac \epsilon 2$
- $\delta' = \dfrac \epsilon {2 \sup_{\bsz \in \partial T} \norm {\bsz - \bsx_0}}$
We will consider:
- $R = \set e \times \paren {\map {B_\delta} \bsx \cap \partial T} \times \paren {\openint {t - \delta'} {t + \delta'} \cap \closedint 0 1}$
which is open in:
- $\set e \times \partial T \times \closedint 0 1$
by the definitions of the subspace and product topologies.
Furthermore, as $t - \delta' > 0$, we have:
- $R = q_\RR^{-1} \sqbrk {q_\RR \sqbrk R}$
Therefore, $q_\RR \sqbrk R$ is open in $C \partial T$ by definition of quotient topology.
It remains to show that:
- $\phi \sqbrk {q_\RR \sqbrk R} \subseteq \map {B_\epsilon} X$
We have:
\(\ds \forall \eqclass {e, \bsx', t'} \RR \in q_\RR \sqbrk R: \, \) | \(\ds \norm {\map \phi {\eqclass {e, \bsx', t'} \RR} - \map \phi {\eqclass {e, \bsx, t} \RR} }\) | \(=\) | \(\ds \norm {\paren {t' \bsx' + \paren {1 - t'} \bsx_0} - \paren {t \bsx + \paren {1 - t} \bsx_0} }\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \norm {t' \paren {\bsx' - \bsx} + \paren {t' - t} \paren {\bsx - \bsx_0} }\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \size {t'} \norm {\bsx' - \bsx} + \size {t' - t} \norm {\bsx - \bsx_0}\) | ||||||||||||
\(\ds \) | \(<\) | \(\ds \size {t'} \delta + \delta' \norm {\bsx - \bsx_0}\) | ||||||||||||
\(\ds \) | \(\le\) | \(\ds \delta + \delta' \sup_{\bsz \in \partial T} \norm {\bsz - \bsx_0}\) | $t' \in \closedint 0 1$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac \epsilon 2 + \frac \epsilon 2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \epsilon\) |
Therefore:
- $\phi \sqbrk {q_\RR \sqbrk R} \subseteq \map {B_\epsilon} X$
$\Box$
Therefore, $\phi$ is continuous by definition.
$\Box$
By:
the domain of $\phi$ is compact.
By Subspace of Hausdorff Space is Hausdorff, the codomain of $\phi$ is Hausdorff.
By Continuous Bijection from Compact to Hausdorff is Homeomorphism:
- $\phi$ is a homeomorphism from $C \partial T$ to $T$.
$\blacksquare$