1. **Problem statement:** (i) Prove that if $A \subseteq B$ then $P(A) \subseteq P(B)$. (ii) Determine if $P(A \cup B) = P(A) \cup P(B)$ and justify. 2. **Recall definitions:** - $P(X)$ is the power set of $X$, the set of all subsets of $X$. - $A \subseteq B$ means every element of $A$ is in $B$. 3. **Proof for (i):** - Assume $A \subseteq B$. - Take any $S \in P(A)$, so $S \subseteq A$. - Since $A \subseteq B$, by transitivity $S \subseteq B$. - Therefore, $S \in P(B)$. - Hence, every element of $P(A)$ is in $P(B)$, so $P(A) \subseteq P(B)$. 4. **Check (ii):** - Is $P(A \cup B) = P(A) \cup P(B)$? - $P(A \cup B)$ contains all subsets of $A \cup B$. - $P(A) \cup P(B)$ contains subsets of $A$ or subsets of $B$. - Consider $A = \{1\}$, $B = \{2\}$. - $A \cup B = \{1,2\}$. - $P(A \cup B) = \{\emptyset, \{1\}, \{2\}, \{1,2\}\}$. - $P(A) = \{\emptyset, \{1\}\}$, $P(B) = \{\emptyset, \{2\}\}$. - $P(A) \cup P(B) = \{\emptyset, \{1\}, \{2\}\}$. - $\{1,2\} \in P(A \cup B)$ but not in $P(A) \cup P(B)$. - So $P(A \cup B) \neq P(A) \cup P(B)$. 5. **Conclusion:** - (i) is true: $P(A) \subseteq P(B)$ if $A \subseteq B$. - (ii) is false: $P(A \cup B) \neq P(A) \cup P(B)$ in general. Final answers: (i) $P(A) \subseteq P(B)$ if $A \subseteq B$. (ii) $P(A \cup B) \neq P(A) \cup P(B)$ in general.