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.

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$S$ is closed

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$\leadsto$

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$\forall z \in \C: z$ is a limit point of $S \implies z \in S$

$\quad$ Definition of Closed Set

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$\leadsto$

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$\forall z \in \C: z \notin S \implies z$ is not a limit point of $S$

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$\leadsto$

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$\forall z \in \C \setminus S \implies z$ is not a limit point of $S$

$\quad$ Definition of Relative Complement

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$\leadsto$

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$\forall z \in \C \setminus S:$ there exists a deleted $\epsilon$-neighborhood $\map {N_\epsilon} z \setminus \set z$ of $z$ entirely in $S$

$\quad$ Definition of Limit Point

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$\leadsto$

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$\C \setminus S$ is open in $\C$

$\quad$ Definition of Open Set

$\blacksquare$