Compact Convex Set with Nonempty Interior is Homeomorphic to Cone on Boundary

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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:

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 $T$.

$\blacksquare$

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