Is it true that for any sets $A$ and $B$,$P(A) \cup P(B) = P(A \cup B)$? Justify your answer.

(N/A) The statement is $False$.
Let $A = \{0, 1\}$ and $B = \{1, 2\}$.
Then $A \cup B = \{0, 1, 2\}$.
The power set $P(A) = \{\varnothing, \{0\}, \{1\}, \{0, 1\}\}$.
The power set $P(B) = \{\varnothing, \{1\}, \{2\}, \{1, 2\}\}$.
Therefore,$P(A) \cup P(B) = \{\varnothing, \{0\}, \{1\}, \{2\}, \{0, 1\}, \{1, 2\}\}$.
The power set $P(A \cup B) = \{\varnothing, \{0\}, \{1\}, \{2\}, \{0, 1\}, \{0, 2\}, \{1, 2\}, \{0, 1, 2\}\}$.
Comparing the two sets,we observe that $\{0, 2\} \in P(A \cup B)$ but $\{0, 2\} \notin P(A) \cup P(B)$.
Thus,$P(A) \cup P(B) \neq P(A \cup B)$.