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

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

$\blacksquare$


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