Finite Union of Open Sets in Complex Plane is Open

Theorem
Let $S_1, S_2, \ldots, S_n$ be open sets of $\C$.

Then $\ds \bigcup_{k \mathop = 1}^n S_k$ is an open set of $\C$.

Proof
Let $\ds z \in \bigcup_{k \mathop = 1}^n S_k$.

Then by definition of finite union:
 * $\exists k \in \set {1, 2, \ldots, n}: z \in S_k$

By definition of open set:
 * $\exists \epsilon \in \R_{>0}: \map {N_\epsilon} z \subseteq S_k$

where $\map {N_\epsilon} z$ is the $\epsilon$-neighborhood of $z$ for $\epsilon$.

By Set is Subset of Union:
 * $S_k \subseteq S$

Thus by Subset Relation is Transitive:
 * $\map {N_\epsilon} z \subseteq S$

As $z$ is any arbitrary element of $S$, it follows that $S$ is an open set of $\C$ by definition.