Limit Point of Set in Complex Plane not Element is Boundary Point

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

Let $z \in \C$ be a limit point of $S$ such that $z \notin S$.

Then $z$ is a boundary point of $S$.

Proof
Let $z \in \C$ be a limit point of $S$ such that $z \notin S$.

Suppose $z$ is not a boundary point of $S$.

Then there exists an $\epsilon$-neighborhood $\map {N_\epsilon} z$ of $z$ such that either:
 * $(1): \quad$ All elements of $\map {N_\epsilon} z$ are in $S$

or
 * $(2): \quad$ All elements of $\map {N_\epsilon} z$ are not in $S$.

If $(1)$, then $z \in S$ which is a contradicts $z \notin S$.

If $(2)$, then $z$ is not a limit point of $S$, which also contradicts the conditions on $z$.