For all sets $A$ and $B$, $P(A \times B) = P(A) \times P(B)$ 
Prove each statement that is true and find a counterexample for each statement that is false.
For all sets $A$ and $B$, $P(A × B) = P(A) × P(B)$.
For all sets $A$ and $B$, $P(A ∩ B) = P(A) ∩ P(B)$.

Here $P$ is the power set.
Can some show or explain to me how to do this? I'm very confused. Thanks.
 A: The first statement, $\mathcal{P}(A \times B) = \mathcal{P}(A) \times \mathcal{P}(B)$, is false. To disprove this, all you have to do is consider a special case like $A = \{1\}, B = \{1\}$. What are the two sets in this case? Or, count the sizes of the two sets.
The second statement is true.
To prove the equality of sets $X = Y$, a very common strategy is to first show $X \subseteq Y$, and then show $Y \subseteq X$. To show $X \subseteq Y$, you start with an arbitrary $x \in X$ and you prove that $x \in Y$.
So we want to show the equality $\mathcal{P}(A \cap B) = \mathcal{P}(A) \cap \mathcal{P}(B)$.
Suppose $x \in \mathcal{P}(A \cap B)$. Then $x \subseteq A \cap B$. It follows since $A \cap B \subseteq A$ and $A \cap B \subseteq B$ that $x \subseteq A$ and $x \subseteq B$. So $x \in \mathcal{P}(A)$ and $x \in \mathcal{P}(B)$. So $x \in \mathcal{P}(A) \cap \mathcal{P}(B)$.
That proves $\mathcal{P}(A \cap B) \subseteq \mathcal{P}(A) \cap \mathcal{P}(B)$. Can you do the other direction?
A: As suggested above, the first proposition is false. Remember that the size of the power set of a set $S$ is $2^{|S|}$. It is easy to find a counter example, say with $A$ being a set of size 2 and $B$ being a set of size 1. Size meaning cardinality.
Here is a proof of the second question.
Take $S \in P(A \cap B)$. Then we know $S \subseteq A \cap B$. So $S$ is a subset of both $A$ and $B$ individually. So $S$ is an element of the power sets of both $A$ and $B$ individually.
Hence $S \in P(A) \cap P(B)$. This shows that $P(A \cap B) \subseteq P(A) \cap P(B)$.
Now take $S\in P(A) \cap P(B).$ Then $S\subseteq A$ and $S\subseteq B$ individually. Hence $S \subseteq A\cap B$. So $S \in  P(A \cap B)$. 
We can conclude that $P(A \cap B) = P(A) \cap P(B)$.
