Complement of Open Set in Complex Plane is Closed

From ProofWiki
Jump to navigation Jump to search

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