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

functions elementary-set-theory

Solution 1:

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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