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 z₀ is not a limit point of $\mathfrak{G}$, there is a ball B_δ(z₀) 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 z₀ that is contained in the complement of $\mathfrak{G}$, and therefore $\mathfrak{G}$ is not closed.