# Complement of Open Set in Complex Plane is Closed

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## Theorem

Let $S \subseteq \C$ be an open subset of the complex plane $\C$.

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

## Proof

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

$\blacksquare$