$f : X \to X$ is an injective function.
The recursive sequence of sets $Z_0, Z_1, Z_2,\dots$ is defined by the following rules:
1) $Z_0 = X \setminus f(X)$
2) $Z_{n+1} = f(Z_n)$ 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 familiar with the principles behind the proof by induction and I still cannot prove even the base case (k=0). I am unsure how are we certain that the function of the previous set is surely producing a new set without any common entries ($f(Z_m) ∩ f(Z_{m+1}) = \emptyset$).
Note that $Z_n=f^n(Z_0)$ hence $Z_0 =X\setminus f(X)$ is clearly disjoint from $Z_n$ for all $n$. That gives you the base case for the induction.
Assume that $Z_n$ is disjoint from all $Z_{n+k}$ for $k>0$. For contradiction assume that $Z_{n+1} $ and $Z_{n+l}$ share the point $z$ for $l>1$. But since $f$ is injective it will have an inverse on its image, hence $f^{-1}(z)$ will be in both $Z_n$ and $Z_{n+l-1}$ which contradicts the induction hypothesis.