Closed Implies Complement is Open (Vice Versa)

Theorem

Let $A \subseteq R$ . Then $A$ is open iff $A^{c}$ (the Complement) is Closed (Toplogical). This is the same as saying that $A$ is closed iff $A^{c}$ is open.

Proof

( $\Rightarrow$ ): Suppose $A$ is open. Let's show $A^{c}$ is closed.

Let $x$ be a limit point of $A^{c}$ . We want to show that $x \in A^{c}$ , so assume for contradiction that $x \in A$ , so then $\exists ϵ > 0$ where $V_{ϵ} (x) \subseteq A$ . But then for any sequence of points in $A^{c}$ , no element can be within $ϵ$ distance of $A$ , so $x$ cannot be a limit point of $A$ . This is a contradiction, so $x \notin A$ .

( $\Leftarrow$ ): Suppose $A^{c}$ is closed. Let $y \in A$ . Then $y$ is not a limit point of $A^{c}$ . Then $\exists ϵ > 0$ such that:

V_{ϵ} (y) \cap A^{c} = \emptyset

But if that's the case then $V_{ϵ} (y) \subseteq A$ as desired.

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