# Can it be shown that a set X is infinite if and only if there exists some $F:X\to X$ that is an injection but not a surjection?

If the function is not surjective, then at least one element of the codomain has no pre-image. However, because F is a function, every element in the domain is mapped to something in the codomain. But since we know that F is injective, then we can't use an element in the codomain more than once in our mapping. That must mean that there is an infinite set of elements in the codomain if we can "skip" one and still be injective. I'm having trouble, however, translating my thoughts into a proof format.

The Axiom of Choice is not available to me for the purposes of this proof. Infinite for this question would be that it as has a countably infinite subset.

Thank you to all who addressed my question, both for your help and patience.

If you have defined "infinite" as "having a countably infinite subset", then there is no need to use the axiom of choice. Use the fact that $n\mapsto n+1$ is an injection which is not a surjection, and conclude the existence of such map; in the other direction let $f$ be a non-surjective injection and let $a$ be an element not in its range, show that $f^n(a)$ is a countably infinite subset of $A$, and conclude that $A$ is infinite. This sort of definition is called Dedekind-infinite.
Not without some part of the axiom of choice. It’s consistent with $\mathsf{ZF}$ that there be infinite, Dedekind-finite sets, i.e., sets that are not finite but that do not admit an injection to a proper subset.