How do you prove $P(A) \cup P(B) = P(A \cup B) \Rightarrow (A \subseteq B) \lor (B \subseteq A)$ I’m learning proofs on my own and I’m on the section about cases.  I have to prove the following but don’t how to do this.  Can someone tell me how you would prove this? (where $P()$ represents the power set):
$P(A) \cup P(B) = P(A \cup B) \Rightarrow (A \subseteq B) \lor (B \subseteq A)$
Your help is greatly appreciated! :)
 A: Try to prove this with contraposition (meaning $(A \Rightarrow B) \equiv (\neg B \Rightarrow \neg A)$):
Assume neither $A \subseteq B$ nor $B \subseteq A$ holds, hence you get elementes $a \in A \setminus B$ and $b \in B \setminus A$. Now consider the subset $\{a,b\} \in P(A \cup B)$. Is this in $P(A) \cup P(B)$ (hence in $P(A)$ or $P(B)$)?
A: A similar way without the need for the contrapositive: get an arbitrary $x$ such that $x\in A$ and an arbitrary $y$ such that $y\in B$, then the set $\{x,y\}\in\mathcal P(A\cup B)$, by definition of the power set.
But now it must hold that $\{x,y\}\in\mathcal P(A)\cup\mathcal P(B)$, this mean that $\{x,y\}\in\mathcal P(A)$ or $\{x,y\}\in\mathcal P(B)$.
If $\{x,y\}\in\mathcal P(A)$ then $y\in B$, and because $y$ was arbitrary we have that $B\subseteq A$. In an analogous way if $\{x,y\}\in\mathcal P(B)$ we have that $A \subseteq B$.
EDIT: well, we need to formalize this proof a bit more to be a valid proof. Suppose an arbitrary $X\subseteq\mathcal P(A\cup B)$. If $A=B$ we are done. If $A\neq B$ then we divide four cases for $X$:


*

*$X$ is empty

*$X$ is composed of points exclusively of $A$

*$X$ is composed of points exclusively of $B$

*$X$ is composed of points of $A$ and points of $B$


Now we can check that when the four cases hold then the statement of the question hold.
