Prove $P(A \cup B)= P(A)\cup P(B) \leftrightarrow (A \subseteq B) \vee (B \subseteq A)$ I want to prove this claim:
$$P(A \cup B)= P(A)\cup P(B) \leftrightarrow (A \subseteq B) \vee (B \subseteq A)$$
I need to prove the both sides? Or which side I need to start with?
I would like to get some hints.
thanks!

The reverse implication should be entirely clear to you.
Suppose now that $\mathcal P(A \cup B) = \mathcal P(A) \cup \mathcal P(B)$.
Then in particular $\mathcal P(A\cup B) \subseteq \mathcal P(A) \cup \mathcal P(B)$.
Can you see how to proceed from there? Think about what sets are in $\mathcal P(A) \cup \mathcal P(B)$.
