The Closure of a Set of Complex Numbers

# The Closure of a Set of Complex Numbers

 Definition: Let $A$ be a set of complex numbers. The Closure of $A$, denoted $\overline{A}$, is defined to be the smallest closed set containing $A$.

By "smallest closed set containing $A$" we mean that any other closed set containing $A$ also contains $\overline{A}$.

We will now prove some properties about the closure of a set of complex numbers.

 Theorem 1: Let $A$ be a set of complex numbers. Then $A$ is closed if and only if $A = \overline{A}$.
• Proof: $\Rightarrow$ Suppose that $A$ is closed. Then clearly the smallest closed set containing $A$ is $A$ itself, so $A = \overline{A}$.
• $\Leftarrow$ Suppose that $A = \overline{A}$. By definition, $\mathrm{int}(A)$ is the smallest closed set containing $A$, and so $\overline{A} = A$ is closed. $\blacksquare$
 Theorem 2: Let $A$ and $B$ be sets of complex numbers. Then: a) $\overline{A} \cup \overline{B} = \overline{A \cup B}$. b) $\overline{A} \cap \overline{B} \supseteq \overline{A \cap B}$.

Observe that $\overline{A} \cap \overline{B}$ is not contained in $\overline{A \cap B}$ in general. For example, let $A = [0, 1)$ and let $B = (1, 2]$. Then $\overline{A} = [0, 1]$ and $\overline{B} = [1, 2]$. So $\overline{A} \cap \overline{B} = \{ 1 \}$.

Now $A \cap B = \emptyset$ and $\overline{A \cap B} = \emptyset$.

Clearly $\overline{A} \cap \overline{B} \not \subseteq \overline{A \cap B}$.