1. **State the problem:** Show using a Venn diagram and set theory that if $A \subset B$ then $B' \subset A'$, and conversely, if $B' \subset A'$ then $A \subset B$. 2. **Recall definitions:** - $A \subset B$ means every element of $A$ is also in $B$. - $A'$ (the complement of $A$) is the set of elements not in $A$ but in the universal set. 3. **If $A \subset B$, prove $B' \subset A'$:** - Take an element $x \in B'$. - By definition of complement, $x \notin B$. - Since $A \subset B$, all elements of $A$ are also in $B$, so $x$ cannot be in $A$. - Therefore, $x \notin A$, implying $x \in A'$. - Hence, $B' \subset A'$. 4. **Conversely, if $B' \subset A'$, prove $A \subset B$:** - Take an element $y \in A$. - If $y \notin B$, then $y \in B'$. - Since $B' \subset A'$, $y \in A'$, which contradicts $y \in A$. - Hence, there is no $y \in A$ such that $y \notin B$, so $A \subset B$. 5. **Summary:** - The inclusion $A \subset B$ flips to inclusion $B' \subset A'$ for complements. The Venn diagram visually confirms these inclusions by showing $A$ completely inside $B$, and complements $B'$ and $A'$ as areas outside their respective circles, illustrating $B' \subset A'$ visually.