 Solution

Complements are relative to the same universal set $\mathcal{U}$. Suppose $A\subseteq B$ and $x\in \overline{B}$. Then $x\in \mathcal{U}$ and $x\notin B$. If $x$ belonged to $A$, the assumed set inclusion would imply $x\in B$, a contradiction. Thus $x\notin A$ and $x\in \overline{A}$, provingThere is only one set inclusion in this conclusion; a reverse inclusion is neither needed nor generally true. The request for double inclusion applies to the equalities in (i) and (ii).