Proof for inverse functions If you could help me out with this I will be so glad. I have been tring to come up with some proof, but I haven't found one yet. 
Let $X$ and $Y$ be sets, and let $f : X \to Y$ be a function defined in $X$ with values in $Y$. Prove that if $A,B \subset Y$; then
$$f^{-1}(A \setminus B) = f^{-1}(A) \setminus f^{-1}(B)$$
 A: Hint: You need to prove $f^{-1}(A \setminus B) \subset f^{-1}(A) \setminus f^{-1}(B)$ and $ f^{-1}(A) \setminus f^{-1}(B)\subset f^{-1}(A \setminus B)  $. 
A: This kind of problem is often very easily handled by ‘element-chasing’. You need to show that $$f^{-1}[A\setminus B]\subseteq f^{-1}[A]\setminus f^{-1}[B]\tag{1}$$ and $$f^{-1}[A]\setminus f^{-1}[B]\subseteq f^{-1}[A\setminus B]\;.\tag{2}$$
For $(1)$, start with an arbitrary element $x\in f^{-1}[A\setminus B]$, and try to show that it must be an element of $f^{-1}[A]\setminus f^{-1}[B]$. This is largely just a matter of unravelling definitions. If $x\in f^{-1}[A\setminus B]$, then $f(x)\in A\setminus B$. This means that $f(x)\in A$ and $f(x)\notin B$. Since $f(x)\in A$, $x\in f^{-1}[A]$, and since $f(x)\notin B$, $x\notin f^{-1}[B]$, in both cases just by the definition of $f^{-1}$. But then $x\in f^{-1}[A]\setminus f^{-1}[B]$, which was exactly what we wanted to show.
Now apply the same kind of reasoning to show that $(2)$ is also true, and conclude that $$f^{-1}[A\setminus B]=f^{-1}[A]\setminus f^{-1}[B]\;.$$
