In my study of functions, I found this result in ”Proofs and Fundamentals” by Ethan D. Bloch that I’m attempting to prove. First, I already now that $X \subseteq f^{-1}(f(X))$ and $f(f^{-1}(Y)) \subseteq Y $ and I’m using this two results in my proof.
Result: Let $f:A \rightarrow B$ a map and let $X \subseteq A$ and $Y \subseteq B$. Then $X = f^{-1}(f(X))$ if and only if $X = f^{-1}(Z)$ for some $Z \subseteq B$.
My proof came as following.
Proof: $\impliedby$. Suppose that there exists a set $Z \subseteq B$ such that $X = f^{-1}(Z)$. Let $Z_0$ be that set. By the result mentioned above, we have that $X \subseteq f^{-1}(f(X))$. Let $x_0 \in f^{-1}(f(X))$. By definition, $f(x_0) \in f(X)$. Since $X = f^{-1}(Z_0)$, we see that $f(x_0) \in f(f^{-1}(Z_0)).$ By the second result mentioned above, we conclude that $f(x_0) \in Z_0$. By definition, we have that $x_0 \in f^{-1}(Z_0)$. Hence $x_0 \in X$. By definition of equality of sets we conclude that, in these conditions, $X = f^{-1}(f(X))$.
$\implies$. Suppose that $X = f^{-1}(f(X))$ and let $Z_1$ be the set defined by $Z_1 = f(X)$. By definition, $f(X) =$ {$b \in B$ | $b = f(x)$ for some $x \in X$}. Hence $f(X) \subseteq B$. From here we deduce that $Z_1 \subseteq B$. By hypothesis, we have that $X = f^{-1}(f(X))$, therefore $X = f^{-1}(Z_1)$. We have shown that there exists a subset of $B$ such that the inverse image of this set is $X$.
MY PROBLEM:
To me, the first part of the proof seems right but I would like to get some feedback.
The second part is making me uncomfortable. It just don’t seem right to me. Is it right? Is there any other approach to prove the second part?
- In the book, Bloch gives some hints to some exercises. And for this one, he suggests the use of the following theorem: “Let $f:A \rightarrow B$ be a map. Let $S, T \subseteq B$. If $S \subseteq T$, then $f^{-1}(S) \subseteq f^{-1}(T)$”. Although i don’t see the point is using this theorem here. Do you have any idea?
Thank you for your attention.
Your proof of the second part is correct. Indeed, this is the "easy" part of the problem. If $X=f^{-1}(f(X))$ then certainly $X$ is of the form $f^{-1}(Z)$ for some $Z\subseteq Y$. Just let $Z=f(X)$ as you have done.
As for the hint, you could argue the first part as follows.
Suppose $X=f^{-1}(Z)$. Then $f(X)\subseteq Z$. So $f^{-1}(f(X))\subseteq f^{-1}(Z)$ (by the hint). This says $f^{-1}(f(X))\subseteq X$ by assumption that $X=f^{-1}(Z)$. So $f^{-1}(f(X))= X$ since $X\subseteq f^{-1}(f(X))$ is always true (as you note).
(This is more or less the same as your proof, just using fewer words. You implicitly use the hint in your argument too.)