# Closed Sets in the Complex Plane

Recall from the Open Sets in the Complex Plane page that for $z \in \mathbb{C}$ and $r > 0$ then open disk centered at $z$ with radius $r$ is defined as the following set of points:

(1)We also said that if $A \subseteq \mathbb{C}$ then $A$ is said to be open if for every $z \in A$ there exists an open disk centered at $z$ fully contained in $A$, i.e., there exists an $r > 0$ such that $D(z, r) \subseteq A$.

We are now ready to define closed sets in the complex plane.

Definition: Let $C \subseteq \mathbb{C}$. Then $C$ is said to be Closed in $\mathbb{C}$ if $C^c = \mathbb{C} \setminus C$ is open. |

Somewhat trivially (again), the emptyset $\emptyset$ and whole set $\mathbb{C}$ are closed sets. We've already noted that these sets are also open, so they're both open and closed (a rather unintuitive definition!)

We now state a similar proposition regarding unions and intersections of closed sets.

Proposition 1: The closed sets of $\mathbb{C}$ satisfy the following properties:a) $\emptyset$ and $\mathbb{C}$ are closed in $\mathbb{C}$.b) If $\{ U_i : i \in I \}$ is an arbitrary collection of closed sets in $\mathbb{C}$ then $\displaystyle{\bigcap_{i \in I} U_i}$ is open in $\mathbb{C}$.c) If $\{ U_1, U_2, ..., U_n \}$ is a finite collection of closed sets in $\mathbb{C}$ then $\displaystyle{\bigcup_{i=1}^{n} U_i}$ is open in $\mathbb{C}$. |

The following theorem gives us a nice criterion for determining whether or not a set $C \subseteq \mathbb{C}$ is closed.

Theorem 1: Let $C \subseteq \mathbb{C}$. Then $C$ is closed if and only if every convergent sequence $(z_n)_{n=1}^{\infty}$ contained in $C$ converges to some $Z \in C$. |

**Proof:**$\Rightarrow$ Suppose that $C$ is closed and let $(z_n)_{n=1}^{\infty}$ be a sequence of complex numbers in $C$ that converges to some $Z$. Let $r > 0$ and consider $D(Z, r)$. Since $(z_n)_{n=1}^{\infty}$ converges to $Z$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then $z_n \in D(Z, r)$. But then this means that $D(Z, r) \not \subseteq \mathbb{C} \setminus \mathbb{C}$ for any $r > 0$. So $Z \not \in \mathrm{int} (\mathbb{C} \setminus C) = \mathbb{C} \setminus C$ (since $\mathbb{C} \setminus C$ is open). So $Z \in C$.

- $\Leftarrow$ We will prove this by contradiction. Suppose that every sequence of complex numbers in $C$ converges in $C$.

- Suppose instead that $C$ is not closed. Then $\mathbb{C} \setminus C$ is not open. Let $Z \in \mathbb{C} \setminus C$. Then for every $r > 0$, $D(Z, r) \not \subseteq \mathbb{C} \setminus C$. So within each of these disks there exists a point in $C$.

- So for each $n \in \mathbb{N}$ choose a point $z_n \in D \left ( Z, \frac{1}{n} \right ) \cap C$. Then $(z_n)_{n=1}^{\infty}$ is a convergent sequence of complex numbers in $C$ converging to $Z$ not in $C$ - a contradiction.

- So $C$ must have been closed. $\blacksquare$