Equivalence of Definitions of Open Set (Complex Analysis)

Proof
Let $S \subseteq \C$.

Definition 1 implies Definition 2
Let $S$ be an open set in $\C$ by definition 1.

That is:
 * $\forall z_0 \in S: \exists \epsilon \in \R_{>0}: N_{\epsilon} \left({z_0}\right) \subseteq S$

Thus every point in $S$ has an $\epsilon$-neighborhood $N_{\epsilon} \left({z_0}\right)$ such that $N_{\epsilon} \left({z_0}\right) \subseteq S$.

That is, by definition, every point of $S$ is an interior point.

Thus $S$ is an open set in $\C$ by definition 2.

Definition 2 implies Definition 1
Let $S$ be an open set in $\C$ by definition 2.

That is, every point of $S$ is an interior point.

This means:
 * $\forall z_0 \in S: \exists \epsilon \in \R_{>0}: N_{\epsilon} \left({z_0}\right) \subseteq S$

Thus $S$ is an open set in $\C$ by definition 1.