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For any sets $\mathrm{A}$ and $\mathrm{B}$, show that
$P(A \cap B)=P(A) \cap P(B).$
Solution
Let $X \in P\left( {A \cap B} \right).$ Then $X \subset A \cap B.$ So, $X \subset A$ and $X \subset B.$ Therefore, $X \in P\left( A \right)$ and $X \in P\left( B \right)$ which implies $X \in P\left( A \right) \cap P\left( B \right).$ This given $P\left( {A \cap B} \right) \subset P\left( A \right) \cap P\left( B \right).$ Let $Y \in P\left( A \right) \cap P\left( B \right).$ Then $Y \in P\left( A \right)$ and $Y \in P\left( B \right).$ So, $Y \subset A$ and $Y \subset B$ Therefore, $Y \subset A \cap B,$ Which implies $Y \in P\left( {A \cap B} \right).$ This gives
$P\left( A \right) \cap P\left( B \right) \subset P\left( {A \cap B} \right)$
Hence $P\left( {A \cap B} \right) = P\left( A \right) \cap P\left( B \right)$
