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!
 A: 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 $\mathcal P(A) \cup \mathcal P(B)$.
A: You need to prove both sides of the implication. So, assume first that either $A\subseteq B$ or $B\subseteq A$. Without loss of generality, suppose $A\subseteq B$. Now porve that $\mathcal P(A\cup B)=\mathcal P(A)\cup \mathcal P(B)$. This is most easily done by noticing that in this case $A\cup B=B$.
Next, assume that $\mathcal P(A\cup B)=\mathcal P(A)\cup \mathcal (B)$. You now wish to show that either $A\subseteq B$ or $B\subseteq A$. Suppose neither holds. Can you find a contradiction?
A: Suppose $\mathcal P(A \cup B)= \mathcal P(A)\cup \mathcal P(B)$. Since $A\cup B\subseteq A\cup B$, $A\cup B \in \mathcal P(A\cup B)$ and thus $A\cup B\in \mathcal P(A) \cup \mathcal P(B)$. So either, $A\cup B\subseteq A$ or $A\cup B\subseteq B$ and it is easy to show that the two statements imply $B \subseteq A$ and $A \subseteq B$ respectively.
Suppose $A \subseteq B$ or $B \subseteq A$. Without loss of generality assume $A \subseteq B$. Then it is easy to show that $\mathcal P(A)\subseteq \mathcal P(B)$. Now prove that $X\subseteq Y \implies X\cup Y=Y$. So $A\cup B = B$ and $\mathcal P(A)\cup\mathcal P(B)= \mathcal P(B)$. Thus $\mathcal P(A \cup B)= \mathcal P(B)= \mathcal P(A)\cup \mathcal P(B)$.
A: ($\Longleftarrow$) If $A \subseteq B$, then $A \cup B = B$, also $P(A) \subseteq P(B)$ and finally $P(A) \cup P(B) = P(B)$, hence $$P(A \cup B) = P(B) = P(A) \cup P(B).$$ Analogously for $B \subseteq A$.
($\Longrightarrow$) Suppose that $A \not\subseteq B$ and $B \not\subseteq A$, then there exist elements $a \in A$ and $b \in B$ such that $a \notin B$ and $b \notin A$. We have that 
$$\{a,b\} \subseteq A \cup B,\quad \text{ so }\quad \{a,b\} \in P(A \cup B),$$ 
but 
$$\{a,b\} \not\subseteq A \quad \text{ implies }\quad \{a,b\} \notin P(A),$$
$$\{a,b\} \not\subseteq B\quad\text{ implies }\quad\{a,b\} \notin P(B),$$ 
so $$P(A \cup B)\ \ \ni\ \ \{a,b\}\ \ \notin \ \ P(A) \cup P(B),$$
hence, $P(A \cup B) \neq P(A) \cup P(B)$.
I hope this helps ;-)
