# Complex Analysis/Appendix/Proofs/Theorem 1.1

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

We assume that ${\displaystyle {\mathfrak {G}}}$ contains all of its limit points and we show that its complement is open. Let ${\displaystyle z_{0}\in \mathbb {C} -{\mathfrak {G}}}$. Then, since ${\displaystyle z_{0}}$ is not a limit point of ${\displaystyle {\mathfrak {G}}}$, there is a ball ${\displaystyle B_{\delta }(z_{0})}$ that contains no point of ${\displaystyle {\mathfrak {G}}}$, that is ${\displaystyle B_{\delta }(z_{0})\subset \mathbb {C} -{\mathfrak {G}}}$. Since this is true for all ${\displaystyle z_{0}\in \mathbb {C} -{\mathfrak {G}}}$, it follows that ${\displaystyle {\mathfrak {G}}}$ is closed.

We now assume that there is a limit point of ${\displaystyle {\mathfrak {G}}}$ in ${\displaystyle \mathbb {C} -{\mathfrak {G}}}$ and show that ${\displaystyle {\mathfrak {G}}}$ is not closed. Let ${\displaystyle z_{0}\in \mathbb {C} -{\mathfrak {G}}}$ be a limit point of ${\displaystyle {\mathfrak {G}}}$. Then There is no neighborhood of ${\displaystyle z_{0}}$ that is contained in the complement of ${\displaystyle {\mathfrak {G}}}$, and therefore ${\displaystyle {\mathfrak {G}}}$ is not closed.