Theorem: Ray from Bounded Set Meets Boundary

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:
{{begin-eqn}}
{{eqn | l = K
      | o = \ge
      | r = \norm {\map \bsx t - \bsx_0}
}}
{{eqn | r = \norm {t \bsy}
}}
{{eqn | r = t \norm \bsy
      | c = {{NormAxiomVector|2}}
}}
{{eqn | ll = \leadsto
      | l = t
      | o = \le
      | r = \frac K {\norm \bsy}
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = \norm {\map \bsx {t_D} - \map \bsx {t_\delta} }
      | r = \norm {t_D \bsy - t_\delta \bsy}
}}
{{eqn | r = \paren {t_D - t_\delta} \norm \bsy
      | c = {{NormAxiomVector|2}} and $t_D \ge t_\delta$
}}
{{eqn | o = <
      | r = \frac \delta {\norm \bsy} \norm \bsy
}}
{{eqn | r = \delta
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = \norm {\map \bsx {t'_\delta} - \map \bsx {t_D} }
      | r = \norm {\paren {t_D + \frac \delta {2 \norm \bsy} } \bsy - t_D \bsy}
}}
{{eqn | r = \frac \delta {2 \norm \bsy} \norm \bsy
      | c = {{NormAxiomVector|2}}
}}
{{eqn | r = \frac \delta 2
}}
{{eqn | o = <
      | r = \delta
}}
{{end-eqn}}
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:
:$\bsx_0 + t_D \bsy \in \partial A$
{{qed}}
Category:Bounded Metric Spaces
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