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

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

Proof: Let $x \in f^{-1}(B-E)$, then $x \in f^{-1}(B)$ and $x \notin f^{-1}(E)$... ...

I know I want to show that $x \in A - f^{-1}(E)$, but how can I show that $x \in A$?

• what about $D$? why did u mention it at all? – Fibo Kowalsky Nov 22 '13 at 1:58
• – Marnix Klooster Nov 22 '13 at 7:04

Hint: $A$ is the domain of $f$. If it's anywhere, it's in there.
Also, a stronger result is true: $f^{-1}(B\setminus E)=f^{-1}(B)\setminus f^{-1}(E)$.