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I have a question about inverse images and containment.

Let $\;f: A \rightarrow B, D \subseteq A$, and $E \subseteq B$. Show that $\; f^{-1}(B-E) \subseteq A-f^{-1}(E)$.

I have started by letting $x \in f^{-1}(B-E)$ so $f(x) \in (B-E)$, but I am unsure where to go from here.

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To finish the proof, you need to show that $x \in A-f^{-1}(E)$. That is, $x \in A$ and $x \notin f^{-1}(E)$. The first one is automatic since $A$ is the domain of $f$. To show $x \notin f^{-1}(E)$, use the fact you found: $f(x) \in B-E$.

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The two sets are equal. For every $x$ we have $$x\in f^{-1}(B-E)\iff(x\in A\land f(x)\in B - E)\iff$$ $$\iff (x\in A\land f(x)\in B\land f(x)\not \in E)\iff$$ $$\iff (x\in A\land f(x)\not \in E)\iff$$ $$\iff(x\in A\land x \not \in f^{-1}(E))\iff x\in A-f^{-1}(E).$$

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