Let $A, B$ be sets. Show that $\mathcal P(A ∩ B) = \mathcal P(A) ∩ \mathcal P(B)$. [duplicate]

Solution 1:

Let $X \in P(A \cap B)$.

Then each element of $X$ is an element of $A$ and $B$, hence $X$ is also in $P(A)$ and $P(B) \implies X \in P(A)\cap P(B)$.

Now

Let $Y \in P(A)\cap P(B)$.

Then $Y \in P(A) $ and $Y \in P(B)$. Therefore each element of $Y$ is an element of $A$ and $B$. Hence each element of $Y$ is in $A\cap B \implies Y \in P(A\cap B)$.

Notice that $X$ and $Y$ are arbitrary, hence we have shown that any set in $P(A \cap B)$ is in $P(A)\cap P(B)$ and vice versa.

From this we can conclude that the two sets have identical composition and are thus equal.

Solution 2:

$\mathcal{P}(A\cap B)$ is the set of all and only the sets which are at the same time subsets of $A$ and $B$. But by definition these are exactly the elements of $\mathcal{P}(A)\cap\mathcal{P}(B)$.