A statement w.r.t. the injection of $\mathbb{Z^+}$ Let $v$ be an injection form $\mathbb{Z^+}$ to $\mathbb{Z^+}$ without any fixed points. Denote the image of $v$ as $v\left(\mathbb{Z^+}\right)$. Take a subset $S \in v\left(\mathbb{Z^+}\right)$ s.t. $S$ and $v\left(\mathbb{Z^+}\right) \setminus S$ are both countably infinite. Is $\left(\mathbb{Z^+}\setminus S\right)\cap \left(\mathbb{Z^+}\setminus v^{-1}\left(S\right)\right)$ also countably infinite? 
Also I am not sure if "$v$ does not have fixed points" helps to prove or disprove this statement.

Here is a counter example (I mentioned it in the comment)：
Define $v: \mathbb{Z^+} \leftrightarrow \mathbb{Z^+}$: $v\left(n\right) = n+1, \text{ for } n \text{ odd}; v\left(n\right) = n-1\text{ for } n \text{ even} $. Take $S$ to be the set of all odd numbers, then $\mathbb{Z^+}\setminus S$ is exactly the even number set. However, $v^{-1}\left(S\right)$ is the even number set by definition so it has complement odd number set. Now the contradiction is clear.
