# Complement of Closed Set in Complex Plane is Open

## Theorem

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

Then the complement of $S$ in $\C$ is open.

## Proof

 $\displaystyle$  $\displaystyle$ $S$ is closed $\displaystyle$ $\leadsto$ $\displaystyle$ $\forall z \in \C: z$ is a limit point of $S \implies z \in S$ Definition of Closed Set $\displaystyle$ $\leadsto$ $\displaystyle$ $\forall z \in \C: z \notin S \implies z$ is not a limit point of $S$ Rule of Transposition $\displaystyle$ $\leadsto$ $\displaystyle$ $\forall z \in \C \setminus S \implies z$ is not a limit point of $S$ Definition of Relative Complement $\displaystyle$ $\leadsto$ $\displaystyle$ $\forall z \in \C \setminus S:$ there exists a deleted $\epsilon$-neighborhood $\map {N_\epsilon} z \setminus \set z$ of $z$ entirely in $S$ Definition of Limit Point $\displaystyle$ $\leadsto$ $\displaystyle$ $\C \setminus S$ is open in $\C$ Definition of Open Set

$\blacksquare$