X is union of 2 sets with special properties Let $f: X \rightarrow X$ be an injection. Show that $X = A\cup B$ for some subsets $A, B \subset X$ such that $\bigcap_{i=1}^{\infty} f^{n} (A) = \emptyset$ and $f(B)=B$. Here, $f^n$ denotes the $n$-fold composite of $f$ with itself.
I tried to using properties of the injection such as for an injection $f$, we have $f(A \cap B) = f(A) \cap f(B)$ for $A, B \subset X$. I'm not sure whether $A$ and $B$ should be disjoint or not but I tried to work in this direction, but to no avail. I currently really have no idea which direction I should tackle this question in. Any help is greatly appreciated!
 A: If you choose an element $x$ of $X$ you can look at $\ldots f^{-3}(x),f^{-2}(x),f^{-1}(x),x, f(x), f^2(x), f^3(x) \ldots$.  There are three possibilities. 
 There may be finite $n$ such that $f^n(x)=x$ or not.  If there is, let $x$ be in $B$.  The chain may stop in the negative direction because there is no $k$ such that $f^{-k}(x)$ exists.  If so, let $x$ be in $A$.  If the negative chain continues forever, put $x$ in $B$.  $B$ will consist of a bunch of finite cycles, and a bunch of orders like the integers, so every element in $B$ has a predecessor and $f(B)=B$.  $A$ will consist of a bunch of linear chains of the form $a, f(a), f^2(a), f^3(a), \ldots$ that do not repeat any element and do not include an element from another chain.  We will never have a case where $f$ applied to some element of a chain is one of the elements of $B$ because $f$ is an injection and all the elements of $B$ have predecessors in $B$.  Each element of a chain is at some finite distance $k$ from the beginning.  It is not $f^{k+1}$ applied to any element of $X$, so $\bigcap_{i=1}^{\infty} f^{n} (A) = \emptyset$ 
Added:  Another way to characterize the same $A,B$ is to find all the elements in $X$ that are not the image of any element of $x$.  Call the set of them $C$.  $A=\bigcup_{i=0}^\infty f^i(C)$ and $B=X\setminus A$
