Ray from Interior of Compact Convex Set Meets Boundary Exactly Once

Theorem

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

Let $\bsx_0 \in A^\circ$ be an interior point of $A$.

Then, for every $\bsy \in \R^n \setminus \bszero$, there is a unique $t \in \R_{> 0}$ such that:

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

Proof

By Compact Subspace of Metric Space is Bounded:

$A$ is bounded

Then, by Ray from Bounded Set Meets Boundary, there is some $t \in \R_{\ge 0}$ such that:

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

But, if $t = 0$, then:

$\bsx_0 \in \partial A$

contradicting the definition of boundary, since $\bsx_0 \in A^\circ$.

Thus, $t > 0$.

Now, suppose that $t' \in \R_{> 0}$ also satisfies:

$\bsx_0 + t' \bsy \in \partial A$

Without loss of generality, suppose $t' \le t$.

By definition of interior point, there is some $\epsilon > 0$ such that:

$\map {B_\epsilon} {\bsx_0} \subseteq A$

Let $\epsilon' = \dfrac {t - t'} t \epsilon$.

Aiming for a contradiction, suppose $t' < t$.

Then, $\epsilon' > 0$.

Let $\bsx' \in \map {B_{\epsilon'}} {\bsx_0 + t' \bsy}$ be arbitrary.

Define:

$\bsx = \bsx_0 + \dfrac t {t - t'} \paren {\bsx' - \paren {\bsx_0 + t' \bsy}}$

We have:

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

\(=\)

\(\ds \frac t {t - t'} \norm {\bsx' - \paren {\bsx_0 + t' \bsy} }\)

Norm Axiom $\text N 2$: Positive Homogeneity

\(\ds \)

\(<\)

\(\ds \frac t {t - t'} \epsilon'\)

\(\ds \)

\(=\)

\(\ds \frac t {t - t'} \frac {t - t'} t \epsilon\)

Definition of $\epsilon'$

\(\ds \)

\(=\)

\(\ds \epsilon\)

Therefore:

$\bsx \in \map {B_\epsilon} {\bsx_0} \subseteq A$

Finally:

\(\ds \frac {t - t'} t \bsx\)

\(=\)

\(\ds \frac {t - t'} t \bsx_0 + \bsx' - \paren {\bsx_0 + t' \bsy}\)

\(\ds \leadsto \ \ \)

\(\ds \bsx'\)

\(=\)

\(\ds \frac {t - t'} t \bsx - \frac {t - t'} t \bsx_0 + \bsx_0 + t' \bsy\)

\(\ds \)

\(=\)

\(\ds \frac {t - t'} t \bsx + \frac {t'} t \bsx_0 + t' \bsy\)

\(\ds \)

\(=\)

\(\ds \paren {1 - \frac {t'} t} \bsx + \frac {t'} t \paren {\bsx_0 + t \bsy}\)

As $0 < t' < t$, it follows that:

$0 < \dfrac {t'} t < 1$

Additionally, by Compact Subspace of Hausdorff Space is Closed:

$\partial A \subseteq A$

Therefore:

$\bsx_0 + t \bsy \in A$

So, by definition of convex:

$\bsx' \in A$

As $\bsx' \in \map {B_{\epsilon'}} {\bsx_0 + t' \bsy}$ was arbitrary, it follows that:

$\map {B_{\epsilon'}} {\bsx_0 + t' \bsy} \subseteq A$

Therefore, by definition:

$\bsx_0 + t' \bsy \in A^\circ$

But we assumed that:

$\bsx_0 + t' \bsy \in \partial A$

which is a contradiction.

Therefore, by Proof by Contradiction:

$t' = t$

$\blacksquare$