Preimages of intersections/unions Let $f(x) = x^2$ and suppose that $A$ is the closed interval $[0, 4]$ and $B$ is the closed interval $[−1, 1]$. In this case find $f^{−1}(A)$ and $f^{−1}(B)$. 
Does $f^{−1}(A\cap B) = f^{−1}(A) \cap f^{−1}(B)$ in this case? 
Does $f^{−1}(A \cup B) = f^{−1}(A) \cup f^{−1}(B)$?
 A: I will assume that $f: \mathbb{R} \rightarrow \mathbb{R}$.
If $F: U \rightarrow V$ and $Y \subset U$ then $f^{-1}(Y)= \{x \in U : F(x) \in Y \}$
So $f^{-1}(A)=f^{-1}([0,4])=\{x \in \mathbb{R} : x^2 \in [0,4] \}= [-2,2]$
and $f^{-1}(B)=f^{-1}([-1,1])=\{x \in \mathbb{R} : x^2 \in [-1,1] \}= [0,1]$ as complex solutions are not allowed by assumption on $f$.
Also $f^{-1}(A \cap B)=f^{-1}([0,1])=\{x \in \mathbb{R} : x^2 \in [0,1] \}= [0,1]$
and $f^{-1}(A \cup B)=f^{-1}([-1,4])=\{x \in \mathbb{R} : x^2 \in [-1,4] \}= [-2,2]$ again as complex solutions not allowed.
Therefore $f^{-1}(A) \cap f^{-1}(B)= [-2,2] \cap [0,1]=[0,1] = f^{-1}(A \cap B)$
$f^{-1}(A \cup B)=f^{-1}([-1,4])=\{x \in \mathbb{R} : x^2 \in [-1,4] \}= [-2,2] = f^{-1}(A) \cup f^{-1}(B)$
A: Hint: 
The statements $f^{-1}(A\cup B)=f^{-1}(A)\cup f^{-1}(B)$ and $f^{-1}(A\cap B)=f^{-1}(A)\cap f^{-1}(B)$ are always true. Also if you are dealing with infinite unions/intersections (and much more). 
It can be proved on base of: $$x\in f^{-1}(U)\iff f(x)\in U$$
Prove this in general and you do not have to bother about 'cases' anymore.
