Answer
See the solution.
Work Step by Step
$Proof.$
Let $A$ be a set, we know $\overline A =\{x: x\notin A\}$ and $\overline {\overline A} =\{x: x\notin \overline A\}$. To show $A=\overline {\overline A}$, we will show $A \subseteq \overline {\overline A}$ and $\overline {\overline A} \subseteq A$. So let $x\in A$, then by definition of $\overline A$, $x \notin \overline A$. Hence by definition of $\overline {\overline A}$, $x \in \overline {\overline A}$. Therefore, $A \subseteq \overline {\overline A}$. Now suppose $x \in \overline {\overline A}$. Then since $\overline {\overline A}$ is the set of elements not in $\overline A$, $x \notin \overline A$. Thus since $\overline A$ is the set of elements that are not in $A$, we must have $x \in A$. Hence $\overline {\overline A} \subseteq A$, and, therefore, $A=\overline {\overline A}$.
