Question:
Is it true that for any sets $A$ and $B, P(A) \cup P(B)=P(A \cup B)$ ? Justify your answer.
Solution:
False.
Let $X \in P(A) \cup P(B)$
$\Rightarrow X \in P(A)$ or $X \in P(B)$
$\Rightarrow X \subset A$ or $X \subset B$
$\Rightarrow X \subset(A \cup B)$
$\Rightarrow X \in P(A \cap B)$
$\therefore P(A) \cup P(B) \subset P(A \cup B)$ ...(1)
Again, let $X \in P(A \cup B)$
But $X \notin P(A)$ or $x \notin P(B)$
[For example let $A=\{2,5\}$ and $B=\{1,3,4\}$ and take $X=\{1,2,3,4\}$ ]
So, $X \notin P(A) \cup P(B)$
Thus, $P(A \cup B)$ is not necessarily a subset of $P(A) \cup P(B)$.