Complement of Bounded Set in Complex Plane has at most One Unbounded Component

Theorem

Let $S \subseteq \C$ be bounded.

Then $\C \setminus S$ has at most one unbounded component.

Proof

If $\C \setminus S$ has no unbounded components, we are done.

Suppose that $\C \setminus S$ has at least one unbounded component.

We must show that it has at most one.

Since $S$ is bounded, there exists $R > 0$ such that:

$\cmod z \le R$ for all $z \in S$.

Let $C$ be an unbounded component of $\C \setminus S$.

Since $C$ is unbounded, there exists $z_0 \in C$ with $\cmod {z_0} > R$.

Hence:

$\ds z_0 \in C \cap \set {z \in \C : \cmod z > R}$

From Complement of Closed Disk in Complex Plane is Path-Connected, we have that:

$\set {z \in \C : \cmod z > R}$ is path-connected.

From Path-Connected Space is Connected:

$\set {z \in \C : \cmod z > R}$ is connected.

Hence from Union of Connected Sets with Common Point is Connected, we have that:

$C \cup \set {z \in \C : \cmod z > R}$ is connected.

From definition 3 of a component, we have that:

$C$ is a maximal connected set.

Since:

$C \subseteq C \cup \set {z \in \C : \cmod z > R}$

we therefore have:

$C = C \cup \set {z \in \C : \cmod z > R}$

So we have:

$\set {z \in \C : \cmod z > R} \subseteq C$

Let $C'$ be a component of $\C \setminus S$ that is distinct from $C$.

Then $C'$ is disjoint from $C$.

In particular, $C'$ is disjoint from $\set {z \in \C : \cmod z > R}$.

So, we have:

$C' \subseteq \C \setminus C \subseteq \set {z \in \C : \cmod z \le R}$

That is:

$C'$ is bounded.

Hence:

$C$ is the only unbounded component of $\C \setminus S$.

This was the demand.

$\blacksquare$