Compact Sets of Complex Numbers

# Compact Sets of Complex Numbers

 Definition: Let $A \subseteq \mathbb{C}$. A Cover of $A$ is a collection of sets $\{ U_{\alpha} \}_{\alpha \in \Gamma}$ ($\Gamma$ is some indexing set) such that $A \subseteq \bigcup_{\alpha \in \Gamma} U_{\alpha}$. An Open Cover of $A$ is a cover of $A$ consisting of only open sets, and a Closed Cover of $A$ is a cover of $A$ consisting of only closed sets.
 Definition: A set $A \subseteq \mathbb{C}$ is said to be Compact if every open cover of $A$ has a finite open subcover.

By "subcover" we mean a subset of the cover of $A$ which also covers $A$.

We will now prove some important properties of compact sets.

 Theorem 1: Let $A \subseteq \mathbb{C}$. If $A$ is compact then $A$ is bounded.
• Proof: Let $A$ be a compact set of complex numbers. Then every open cover of $A$ has a finite open subcover. Let $z_0 \in A$ and let
(1)
\begin{align} \quad \mathcal F = \{ B(z_0, r) : r > 0 \} \end{align}
• Then $\mathcal F$ is an open cover of $A$. Since $A$ is compact, $\mathcal F$ has a finite open subcover:
(2)
\begin{align} \quad \mathcal F = \{ B(z_0, r_1), B(z_0, r_2), ..., B(z_0, r_n) \} \end{align}
• Let $R = \max \{ r_1, r_2, ..., r_n \}$. Then $B(z_0, r_k) \subseteq B(z_0, R)$ for every $k \in \{ 1, 2, ..., n \}$ and furthermore:
(3)
\begin{align} \quad A \subseteq \bigcup_{k=1}^{n} B(z_0, r_k) = B(z_0, R) \end{align}
• So $A$ is bounded. $\blacksquare$
 Theorem 2: Let $A \subseteq \mathbb{C}$. If $A$ is compact then $A$ is closed.
 Theorem 3: Let $A \subseteq \mathbb{C}$. Then $A$ is compact if and only if $A$ is closed and bounded.

It is very important to remark the significance of Theorem 3. In general metric spaces, Theorem 3 is NOT true in general - that is, there exists metric spaces which contain closed and bounded sets which are not compact!

 Theorem 4: Every closed subset of a compact set of complex numbers is also compact.
• Proof: Let $A$ and $B$ be sets of complex numbers such that $A \subseteq B$ and let $A$ be closed and let $B$ be compact.
• Let $\mathcal F$ be an open cover of $A$. Since $A$ is closed, $\mathbb{C} \setminus A$ is open. Observe that since $A \subseteq B$ we have that $\mathcal F \cup \{ \mathbb{C} \setminus A \}$ is an open cover of $B$.
• Since $B$ is compact, there exists a finite open subcover of $\mathcal F \cup \{ \mathbb{C} \setminus A \}$, say:
(4)
\begin{align} \quad \{ U_1, U_2, ..., U_n, \mathbb{C} \setminus A \} \end{align}
• But then $\{ U_1, U_2, ..., U_n \}$ is a finite open subcover of $\mathcal F$, so $A$ is compact. $\blacksquare$