Infinite Set is Equivalent to Proper Subset/Proof 2

Theorem
A set is infinite if and only if it is equivalent to one of its proper subsets.

Proof
Let $S$ be a set.

Suppose $S$ is finite.

From Proper Subset of Finite Set No Bijection we have that $S$ can not be equivalent to one of its proper subsets.

Suppose $S$ is infinite.

From Infinite Set has Countably Infinite Subset, we can construct $v: \N \to S$ such that $v$ is an injection.

We now construct the mapping $h: S \to S$ as follows.


 * $h \left({x}\right) = \begin{cases}

v \left({n + 1}\right) & : \exists n \in \N: x = v \left({n}\right)\\ x & : x \notin \operatorname{Im} \left({v}\right) \end{cases}$

It is clear that $h$ is an injection.

But we have that $v \left({0}\right) \notin \operatorname{Im} \left({h}\right)$ and so $\operatorname{Im} \left({h}\right) \subsetneq S$.

The result follows from Injection to Image is Bijection.