# How to approach proving $f^{-1}(B\setminus C)=A\setminus f^{-1}(C)$?

Let $A,B,C$ be sets such that $C\subseteq B$. Let $f: A \to B$ be a function. Prove that $f^{-1} (B\setminus C)=A\setminus f^{-1} (C).$

I really need help with this proof problem. I'm not sure where to begin or what strategy to consider using.

• If $\;A=\emptyset\;$ the claim's blatantly false (unless $\;B=C\;$_. Perhaps you forgot to add some condition on $\;A\;$ ? – DonAntonio Nov 5 '13 at 19:52
• This may depend on $f$ ? What is $f$ ? – Dietrich Burde Nov 5 '13 at 19:52
• Why vote down? People can't ask questions? – Kaa1el Nov 5 '13 at 20:10

You seem to be leaving out the assumption that $\;f : A \to B\;$.
The defining property of $\;\cdot^{-1}[\cdot]\;$ (yes, I'm using a slightly different notation for clarity) is $$(0) \;\;\; x \in f^{-1}[Y] \;\equiv\; f(x) \in Y$$ for any $\;f : A \to B\;$, $\;x \in A\;$, and $\;Y \subseteq B\;$.
With this, the left hand side is rewritten like this, for any $\;x \in A\;$: \begin{align} & x \in f^{-1}[B \setminus C] \\ \equiv & \;\;\;\;\;\text{"property $(0)$"} \\ & f(x) \in B \setminus C \\ \equiv & \;\;\;\;\;\text{"definition of $\;\setminus\;$"} \\ & f(x) \in B \land f(x) \not\in C \\ \equiv & \;\;\;\;\;\text{"the range of $\;f\;$ is $\;B\;$, so $\;f(x) \in B\;$ is true"} \\ & f(x) \not\in C \\ \end{align} Now do something similar with the right hand side, and draw your conclusion.