Complement of Closed Set in Complex Plane is Open

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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

$\blacksquare$


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