Zorn's lemma to prove existence of maximal element Let $f : A \rightarrow A$ be bijection. I have to prove that exists maximal subset $B \subset A$ such $B \subset f(A \setminus B)$ using Zorn's lemma.I have observed that $f(A \setminus B) = A \setminus f(B)$. I guess $(\{B : B \subset A, B \subset f(A \setminus B)\}, \subset)$ is a chain and it has upper bound (mybe $\bigcup{B}$). Can somebody help me?

Let $\Sigma$ be the set of all subsets of $A$ with that property. $\Sigma$ is not empty because $\emptyset \in \Sigma$. Take now a chain $C \subseteq \Sigma$. Our goal is to prove that $\cup C \in \Sigma$. Suppose not. So $\cup C \nsubseteq f(A \backslash (\cup C)) = A \backslash f(\cup C)$. So there exist $x \in \cup C$ such that $x \notin A \backslash f(\cup C) \Rightarrow x \in f(\cup C)$. Therefore $x = f(y)$, where $y \in \cup C$. $x \in \cup C \Rightarrow x \in L \in C$. Same with $y$ : $y \in M \in C$. Since $C$ is a chain then $L \subseteq M$ or $M \subseteq L$. WLOG we can say that $L \subseteq M$, so $x, y \in M$. Then $x \in M$, but $x = f(y) \in f(M)$, so $x \notin A \backslash f(M) = f(A \backslash M)$, which is not possible since $M \in \Sigma$. This contradiction proves that $\cup C \in \Sigma$. Now you can use the lemma.
