Axiom of choice needed? finite sets

Question

I have to show that the following 2 quotes are equivalent:
a) The set A is finite
b) Every injective function from A to A is also surjective

The direction a) ==> b):
I've worked with induction on the cardinality of A and just definied the function

The direction b) ==> a):
I've got no idea! I would begin with: Presume that A is infinite.. But I'dont see what to do.

Thanks in advance
Silke

Answer 1

Yes, the axiom of choice is in fact needed for this equivalence.

We say that a set $A$ is Dedekind-finite if every injection from $A$ into $A$ is a bijection. And it is consistent that the axiom of choice fails and there are infinite Dedekind-finite sets.

For example, it is consistent that such set is dense in the real numbers.

One useful equivalent is that $A$ is Dedekind-infinite (i.e. not Dedekind-finite) if and only if it has a countably infinite subset. In particular, when we assume the axiom of choice every infinite set has a copy of $\Bbb N$ inside. From this copy we can generate an injection which is not a bijection.