Complex Analysis/Appendix/Proofs/Theorem 1.1

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

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

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