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