Complex Analysis/Appendix/Proofs/Theorem 1.1

We prove that a set $𝔊\subset ℂ$ is closed if and only if it contains all of its limit points.

We assume that $𝔊$ contains all of its limit points and we show that its complement is open. Let $z_0\in ℂ-𝔊$. Then, since $z_0$ is not a limit point of $𝔊$, there is a ball $B_{\delta }(z_0)$ that contains no point of $𝔊$, that is $B_{\delta }(z_0)\subset ℂ-𝔊$. Since this is true for all $z_0\in ℂ-𝔊$, it follows that $𝔊$ is closed.

We now assume that there is a limit point of $𝔊$ in $ℂ-𝔊$ and show that $𝔊$ is not closed. Let $z_0\in ℂ-𝔊$ be a limit point of $𝔊$. Then There is no neighborhood of $z_0$ that is contained in the complement of $𝔊$, and therefore $𝔊$ is not closed.