If $\mathcal{P}(A \cup B) = \mathcal{P}(A) \cap \mathcal{P}(B)$, then $A = B$ I'm confused trying proving this problem. I tried the following:
Suppose $A\neq B$, then $\mathcal{P}(A) \cap \mathcal{P}(B)=\emptyset\implies \mathcal{P}(A \cup B) =\emptyset \implies A \cup B =\emptyset \implies A=B$, but this is a contradiction. Thus, $A = B$. The thing with this though is that I'm not quite sure about $\mathcal{P}(A) \cap \mathcal{P}(B)=\emptyset$, because $A$ and $B$ can have a common element and still be different.
 A: Let's see.
$$\mathcal P(A\cup B)=\mathcal P(A)\cap\mathcal P(B).$$
In words: every subset of $A\cup B$ is a subset of $A$ and a subset of $B$ (and vice versa).
Well, $A$ is a subset of $A\cup B$ (you knew that, right?), so $A$ is a subset of $A$ (so what, we already knew that) and $A$ is a subset of $B$ (now we're getting somewhere).
We've shown that $A$ is a subset of $B$. The problem is half done, because to show that $A=B$ we just have to show that $A$ is a subset of $B$ and $B$ is a subset of $A$. I'll leave the other half to you.
A: Note that for any set $A$ we have that $A$ is the maximal element of $\mathcal P(A)$. So if the given property is true then any set in $\mathcal P(A\cup B)$ (especially $A,B$) must be in $\mathcal P(A)$ and in $\mathcal P(B)$ and thus subsets of $A$ and of $B$. Thus $A\subset B$ and $B\subset A$.
Also your prove is not correct. Even if $A,B$ are disjoint (which they do not need to by), the powersets share at least the $\emptyset$.
A: You are right to be concerned about your assumption that if $A \neq B$, then $\mathcal{P}(A) \cap \mathcal{P}(B) = \emptyset$. For example, if $A = \{1, 2\}$ and $B = \{2, 3\}$, then $A \neq B$, but $\mathcal{P}(A) \cap \mathcal{P}(B) = \{ \emptyset, \{2\}\}$. In fact $\mathcal{P}(A) \cap \mathcal{P}(B)$ always has $\emptyset$ as a member. (In general, $\mathcal{P}(A) \cap \mathcal{P}(B) = \mathcal{P}(A \cap B)$.)
To solve your problem, let's assume $\mathcal{P}(A \cup B) = \mathcal{P}(A) \cap \mathcal{P}(B)$. Then for any $a \in A$, we have: $$\{a\} \in \mathcal{P}(A \cup B) = \mathcal{P}(A) \cap \mathcal{P}(B) \subseteq \mathcal{P}(B)$$ so $\{a\} \subseteq B$, which implies that $a \in B$. So $A \subseteq B$. Likewise starting with $b \in B$, our assumption gives us that $b \in A$, so that $B \subseteq A$. Hence our assumption implies that $A = B$.
