Ray from Bounded Set Meets Boundary

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Theorem

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

Let $\bsx_0 \in A$ be point of $A$.

Then, for every $\bsy \in \R^n \setminus \set \bszero$, there is some $t \in \R_{\ge 0}$ such that:

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

where $\partial A$ denotes the boundary of $A$.


Proof

For every $t \in \R$, let:

$\map \bsx t = \bsx_0 + t \bsy$

Define:

$D = \set {t \in \R_{\ge 0} : \map \bsx t \in A}$

As $\map \bsx 0 = \bsx_0 \in A$:

$0 \in D$

so:

$D$ is non-empty


By definition of bounded, there exists some $K \in \R$ such that:

$\forall \bsx \in A: \size {\bsx - \bsx_0} \le K$

Suppose for arbitrary $t \in \R_{\ge 0}$ that $\map \bsx t \in A$.

Then:

\(\ds K\) \(\ge\) \(\ds \norm {\map \bsx t - \bsx_0}\)
\(\ds \) \(=\) \(\ds \norm {t \bsy}\)
\(\ds \) \(=\) \(\ds t \norm \bsy\) Norm Axiom $\text N 2$: Positive Homogeneity
\(\ds \leadsto \ \ \) \(\ds t\) \(\le\) \(\ds \frac K {\norm \bsy}\)

Therefore, $D$ is bounded above.

By the Least Upper Bound Property, let:

$t_D = \sup D$

We will show that:

$\map \bsx {t_D} \in \partial A$


Let $N \subseteq \R^n$ be a neighborhood of $\map \bsx {t_D}$.

By definition, there exists some $\delta > 0$ such that:

$\map {B_\delta} {\map \bsx {t_D}} \subseteq N$

We want to prove that:

$A \cap \map {B_\delta} {\map \bsx {t_D}} \ne \O$
$\overline A \cap \map {B_\delta} {\map \bsx {t_D}} \ne \O$


By Characterizing Property of Supremum of Subset of Real Numbers, there exists some $t_\delta \in D$ such that:

$t_D - \dfrac \delta {\norm \bsy} < t_\delta \le t_D$

By definition of $D$:

$\map \bsx {t_\delta} \in A$

Also:

\(\ds \norm {\map \bsx {t_D} - \map \bsx {t_\delta} }\) \(=\) \(\ds \norm {t_D \bsy - t_\delta \bsy}\)
\(\ds \) \(=\) \(\ds \paren {t_D - t_\delta} \norm \bsy\) Norm Axiom $\text N 2$: Positive Homogeneity and $t_D \ge t_\delta$
\(\ds \) \(<\) \(\ds \frac \delta {\norm \bsy} \norm \bsy\)
\(\ds \) \(=\) \(\ds \delta\)

Therefore:

$\map \bsx {t_\delta} \in \map {B_\delta} {\map \bsx {t_D}}$

It immediately follows that:

$A \cap \map {B_\delta} {\map \bsx {t_D}} \ne \O$


Now, let:

$t'_\delta = t_D + \dfrac \delta {2 \norm \bsy}$

By definition of supremum, $t_D$ is an upper bound for $D$.

Therefore, since:

$t'_\delta > t_D$

it follows that:

$t'_\delta \notin D$

That is:

$\map \bsx {t'_\delta} \notin A$

or:

$\map \bsx {t'_\delta} \in \overline A$

Finally:

\(\ds \norm {\map \bsx {t'_\delta} - \map \bsx {t_D} }\) \(=\) \(\ds \norm {\paren {t_D + \frac \delta {2 \norm \bsy} } \bsy - t_D \bsy}\)
\(\ds \) \(=\) \(\ds \frac \delta {2 \norm \bsy} \norm \bsy\) Norm Axiom $\text N 2$: Positive Homogeneity
\(\ds \) \(=\) \(\ds \frac \delta 2\)
\(\ds \) \(<\) \(\ds \delta\)

That is:

$\map \bsx {t'_\delta} \in \map {B_\delta} {\map \bsx {t_D}}$

From the above, we have:

$\map \bsx {t'_\delta} \in \overline A \cap \map {B_\delta} {\map \bsx {t_D}}$

That is:

$\overline A \cap \map {B_\delta} {\map \bsx {t_D}} \ne \O$


From:

$A \cap \map {B_\delta} {\map \bsx {t_D}} \ne \O$
$\overline A \cap \map {B_\delta} {\map \bsx {t_D}} \ne \O$

we have proven that:

$\map \bsx {t_D} \in \partial A$

by definition of boundary.

Therefore:

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

$\blacksquare$

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