Equivalence of Definitions of Exterior Point (Complex Analysis)

Theorem
Let $S \subseteq \C$ be a subset of the complex plane.

Let $z_0 \in \C$.

Proof
Let $S \subseteq \C$.

Definition 1 implies Definition 2
Let $z_0$ be an exterior point of $S$ by definition 1.

Let $\map {N_\epsilon} {z_0}$ be an $\epsilon$-neighborhood of $z_0$ such that $\map {N_\epsilon} {z_0} \cap S = \O$.

That is, $\map {N_\epsilon} {z_0}$ has no points which are also in $S$.

By definition, it follows that $z_0$ is not a boundary point of $S$.

$z_0$ is an interior point of $S$.

Let $\map {N_{\epsilon'} } {z_0}$ be an $\epsilon'$-neighborhood such that $\map {N_{\epsilon'} } {z_0} \subseteq S$.

By Empty Intersection iff Subset of Complement:
 * $\map {N_\epsilon} {z_0} \subseteq \relcomp \C S$

Thus $z_0$ is an exterior point of $S$ by definition 2.

Definition 2 implies Definition 1
Let $z_0$ be an exterior point of $S$ by definition 2.

As $z_0$ is not an interior point of $S$ there exists no $\epsilon$-neighborhood of $z_0$ which is disjoint from $\relcomp \C S$.

That is, every $\epsilon$-neighborhood of $z_0$ contains points which are not in $S$.

As $z_0$ is not a boundary point of $S$, there exists at least one $\epsilon$-neighborhood of $z_0$ which does not contain both points in $S$ and points not in $S$.

But as every $\epsilon$-neighborhood of $z_0$ contains points which are not in $S$, it follows there must be at least one $\epsilon$-neighborhood of $z_0$ disjoint from $S$.

Thus $z_0$ is an exterior point of $S$ by definition 1.