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.