Complement of Bounded Set has Exactly One Unbounded Component

== Theorem ==
Let $n \in \N_{> 1}$
Let $A \subseteq \R^n$ be a bounded subspace of real Euclidean $n$-space.
Then, $\R^n \setminus A$ has exactly one unbounded component.
== Proof ==
By definition of bounded, there exists some $\bsx_0 \in \R^n$ and $\epsilon > 0$ such that:
:$A \subseteq \map {B_\epsilon} {\bsx_0}$
=== Existence ===
First, we will show that there is some unbounded component in $\R^n \setminus A$.
Let $t \ge \epsilon$ be arbitrary.
Let $\bsx_t = \bsx_0 + \tuple {t, 0, \dotsc, 0}$.
Then:
Therefore:
:$\bsx_t \notin \map {B_\epsilon} {\bsx_0} \supseteq A$
Hence:
:$\bsx_t \in \R^n \setminus A$
for all $t \ge \epsilon$.
Let $T = \set {\bsx_t : t \ge \epsilon}$.
From the above remarks:
:$T \subseteq \R^n \setminus A$
We will show that $T$ is connected and unbounded.
It will then immediately follows that the component containing $T$ has the same properties.
 $T$ is bounded.
Then, by definition of bounded, there exists some $K \in \R$ such that:
:$\forall \bsx \in T: \map d {\bsx, \bsx_\epsilon} \le K$
As:
:$\map d {\bsx_\epsilon, \bsx_\epsilon} \le K$
we have by :
:$0 \le K$
But then, we have:
which is a contradiction.
Therefore, by Proof by Contradiction:
:$T$ is unbounded.
Let $\bsx_t, \bsx_{t'} \in T$ be arbitrary.
, suppose $t \le t'$.
Define $f : \closedint 0 1 \to T$ as:
:$\map f s = \tuple {t + \paren {t' - t} s, 0, \dotsc, 0}$
$f$ is well-defined, as for any $s \in \closedint 0 1$, we have:
:$0 \le \paren {t' - t} s \le t' - t$
and therefore:
:$t \le t + \paren {t' - t} s \le t'$
$f$ is also continuous, as:
:$\size {s - s'} < \dfrac \epsilon {\size {t' - t}} \implies \norm {\map f s - \map f {s'}} = \size {s - s'} \norm {\tuple {t' - t, 0, \dotsc, 0}} < \epsilon$
for every $\epsilon$, provided that $t \ne t'$.
If $t = t'$, then $f$ is continuous by Constant Mapping is Continuous.
Therefore:
:$\bsx_t$ and $\bsx_{t'}$ are path-connected in $T$
As $\bsx_t$ and $\bsx_{t'}$ were arbitrary:
:$T$ is path-connected
By Path-Connected Space is Connected:
:$T$ is connected
=== Uniqueness ===
We will show that $\R^n \setminus \map {B_\epsilon} {\bsx_0}$ is connected.
Then, any component $C$ other than the one containing $\R^n \setminus \map {B_\epsilon} {\bsx_0}$ satisfies:
:$C \subseteq \map {B_\epsilon} {\bsx_0}$
and is thus bounded by definition.
Let $\bsx, \bsx' \in \R^n \setminus \map {B_\epsilon} {\bsx_0}$ be arbitrary.
Define:
:$\bsy = \bsx_0 + \dfrac \epsilon {\norm {\bsx - \bsx_0}} \paren {\bsx - \bsx_0}$
:$\bsy' = \bsx_0 + \dfrac \epsilon {\norm {\bsx' - bsx_0}} \paren {\bsx' - \bsx_0}$
We have:
Therefore, both $\bsy$ and $\bsy'$ lie on the $n - 1$-sphere with center $\bsx_0$ and radius $\epsilon$.
By Sphere is Connected and Connected Set in Subspace:
:$\bsy$ and $\bsy'$ are connected in $\R^n \setminus \map {B_\epsilon} {\bsx_0}$
By Connectedness of Points is Equivalence Relation, it remains to show that:
:$\bsx$ is connected to $\bsy$
:$\bsx'$ is connected to $\bsy'$
Define $f, g : \closedint 0 1 \to \R^n \setminus \map {B_\epsilon} \bsx_0$ as:
:$\map f t = \bsx_0 + \paren {1 + t \paren {\dfrac \epsilon {\norm {\bsx - \bsx_0}} - 1}} \paren {\bsx - \bsx_0}$
:$\map g t = \bsx_0 + \paren {1 + t \paren {\dfrac \epsilon {\norm {\bsx' - \bsx_0}} - 1}} \paren {\bsx' - \bsx_0}$
Both $f$ and $g$ are continuous by Combination Theorem for Continuous Mappings on Metric Space.
We also have that:
In exactly the same manner:
:$\norm {\map g t - \bsx_0} \ge \epsilon$
This shows that, indeed:
:$f \closedint 0 1 \subseteq \R^n \setminus \map {B_\epsilon} {\bsx_0}$
:$g \closedint 0 1 \subseteq \R^n \setminus \map {B_\epsilon} {\bsx_0}$
Additionally:
:$\map f 0 = \bsx$
:$\map f 1 = \bsy$
:$\map g 0 = \bsx'$
:$\map g 1 = \bsy'$
Therefore:
:$\bsx$ is path-connected to $\bsy$ in $\R^n \setminus \map {B_\epsilon} {\bsx_0}$
:$\bsx'$ is path-connected to $\bsy'$ in $\R^n \setminus \map {B_\epsilon} {\bsx_0}$
By:
:Points are Path-Connected iff Contained in Path-Connected Set
:Path-Connected Space is Connected
:
we have that:
:$\bsx$ is connected to $\bsy$
:$\bsx'$ is connected to $\bsy'$
completing our proof by the remarks above.