Is injective function $f:A \to A$ always surjective? Ok so while browsing a book(namely Herbert Endertons book "Elements of set theory") I have stumbled upon a curiosity which provoked me to try to prove this.Here is how I went about it,but I do not think my solution is correct.
All answers as well as corrections are more then welcome.
Proof:
Since $f: A\to A$ and f is injective we have $$(x,y)\in f \implies x \in A  \;\;\land \;\; y\in A \implies (y,z)\in f \;\; \land \;\; z\neq x $$
by using this step repeatedly,we will eventually exhaust set A of members,and thus range is equal to domain.
However I am not sure that I have covered a case where there is element which is in relation with x.Also I am not sure what would happen if the domain and range are set of real numbers(as I have only managed to study up to natural numbers so far).
All input is highly appreciated

You statement works for finite $A$. If 
$$
f: A\to A
$$
is injective then it is surjective. 
This is not true when $A$ is infinite. Consider for example the funcstion $f : \mathbb{Z} \to \mathbb{Z}$ given by $f(x) = 2x$. This function is injective, but not surjective,
