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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.

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

Here is one strategy: go from the set level to the element level by expanding the definitions (or defining properties), and then use the laws of logic to simplify.

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.

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