Suppose that $f : X \to X$ is an injective function. Define by recursion a sequence $Z_0$, $Z_1$, $Z_2$, $\ldots$ of subsets of $X$ as follows. $$Z_0 =X- ran(f),$$ $$Z_{n+1} = f(Z_n)$$ for any $n\geq 0$.
Prove by induction on $n$ that for any $k\geq n + 1$ the sets $Z_n$ and $Z_k$ are disjoint.
I am lost. Could anyone give me a hint, please? Thanks
Let us see that $Z_0\cap Z_k = \varnothing$, whenever $k>0$:
if $x \in Z_0\cap Z_k$ then $x \notin f(X)$ and $x = f(y)$, for some $y \in Z_{k-1}$, a contradiction.
Suppose now (induction hypothesis) that $Z_n\cap Z_k=\varnothing$ for $k>n$ is true whenever $n < n_0$.
Let $k >n_0$ and suppose that $x \in Z_{n_0}\cap Z_k$. Then $$x=f(y), \;\text{with}\; y \in Z_{n_0-1}\quad\text{and} x=f(z),\;\text{with}\;z \in Z_{k-1}.$$ Since $f$ is injective, we get $y=z$, and so $Z_{n_0-1}\cap Z_{k-1}\neq\varnothing$, a contradiction with the induction hypothesis because $k-1>n_0-1$.