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:
| \(\ds \norm {\bsx_t - \bsx_0}\) | \(=\) | \(\ds \norm {\bsx_0 + \tuple {t, 0, \dotsc, 0} - \bsx_0}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \norm {\tuple {t, 0, \dotsc, 0} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds t\) | ||||||||||||
| \(\ds \) | \(\ge\) | \(\ds \epsilon\) |
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.
Aiming for a contradiction, suppose $T$ 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 Metric Space Axiom $(\text M 1)$:
- $0 \le K$
But then, we have:
| \(\ds \map d {\bsx_{\epsilon + K + 1}, \bsx_\epsilon}\) | \(=\) | \(\ds \norm {\bsx_0 + \tuple {\epsilon + K + 1, 0, \dotsc, 0} - \bsx_0 - \tuple {\epsilon, 0, \dotsc, 0} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \norm {\tuple {K + 1, 0, \dotsc, 0} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds K + 1\) | ||||||||||||
| \(\ds \) | \(>\) | \(\ds K\) |
which is a contradiction.
Therefore, by Proof by Contradiction:
- $T$ is unbounded.
$\Box$
Let $\bsx_t, \bsx_{t'} \in T$ be arbitrary.
Without loss of generality, 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
$\Box$
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:
| \(\ds \norm {\bsy - \bsx_0}\) | \(=\) | \(\ds \norm {\dfrac \epsilon {\norm {\bsx - \bsx_0} } \paren {\bsx - \bsx_0} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac \epsilon {\norm {\bsx - \bsx_0} } \norm {\bsx - \bsx_0}\) | Norm Axiom $\text N 2$: Positive Homogeneity | |||||||||||
| \(\ds \) | \(=\) | \(\ds \epsilon\) | ||||||||||||
| \(\ds \norm {\bsy' - \bsx_0}\) | \(=\) | \(\ds \norm {\dfrac \epsilon {\norm {\bsx' - \bsx_0} } \paren {\bsx' - \bsx_0} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac \epsilon {\norm {\bsx' - \bsx_0} } \norm {\bsx' - \bsx_0}\) | Norm Axiom $\text N 2$: Positive Homogeneity | |||||||||||
| \(\ds \) | \(=\) | \(\ds \epsilon\) |
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:
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:
| \(\ds \norm {\map f t - \bsx_0}\) | \(=\) | \(\ds \norm {\paren {1 + t \paren {\dfrac \epsilon {\norm {\bsx - \bsx_0} } - 1} } \paren {\bsx - \bsx_0} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {1 + t \paren {\dfrac \epsilon {\norm {\bsx - \bsx_0} } - 1} } \norm {\bsx - \bsx_0}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \norm {\bsx - \bsx_0} + t \paren {\epsilon - \norm{\bsx - \bsx_0} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds t \epsilon + \paren {1 - t} \norm {\bsx - \bsx_0}\) | ||||||||||||
| \(\ds \) | \(\ge\) | \(\ds t \epsilon + \paren {1 - t} \epsilon\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \epsilon\) |
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
- Definition of Connected Points (Topology)
we have that:
completing our proof by the remarks above.
$\blacksquare$