# 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 |

$\blacksquare$

## Sources

- 1981: Murray R. Spiegel:
*Theory and Problems of Complex Variables*(SI ed.) ... (previous) ... (next): $1$: Complex Numbers: Supplementary Problems: Point Sets: $124$