$f^{-1}(D-C)=f^{-1}(D)-f^{-1}(C)$

Question

Please can you give me feedback on this proof?

Result: Let $f:A \rightarrow B$ be a function. Let $C$, $D \subseteq B$. Then $f^{-1}(D-C)=f^{-1}(D)-f^{-1}(C)$.

Proof: To show that $f^{-1}(D-C)=f^{-1}(D)-f^{-1}(C)$, it is sufficient to show that the set in each side is a subset of the other.

Let $x \in f^{-1}(D-C)$. By definition, we see that $f(x) \in D-C$. Hence, $f(x) \in D$ and $f(x) \notin C$. We deduce that $x \in f^{-1}(D)$ and $x \notin f^{-1}(C)$. Then $x \in f^{-1}(D) - f^{-1}(C)$. Therefore $f^{-1}(D-C) \subseteq f^{-1}(D) - f^{-1}(C)$.

Now, let $y \in f^{-1}(D) - f^{-1}(C)$. Then $y \in f^{-1}(D)$ and $y \notin f^{-1}(C)$. By definition, we see that $f(y) \in D$ and $f(y) \notin C$. From here we see that $f(y) \in D-C$. Then, by definition, $y \in f^{-1}(D-C)$. Therefore $f^{-1}(D)-f^{-1}(C) \subseteq f^{-1}(D-C)$.

This ends the proof.

Thank you for your attention!

Answer 1

Your proof is correct. You can alternatively combine the two parts for a shorter proof as follows:

$$\begin{align*}x \in f^{-1}(D-C)&\iff f(x) \in D-C\\ &\iff f(x) \in D \text{ and } f(x) \notin C\\ &\iff x \in f^{-1}(D) \text{ and } x \notin f^{-1}(C)\\ &\iff x \in f^{-1}(D)-f^{-1}(C). \end{align*}$$