Maxwell Rosenlicht claims in "Introduction to analysis" that a set is infinite if and only if it may be placed into one-to-one correspondence with a proper subset of itself.
He says this is self-evident because a finite set cannot be placed into a one-to-one correspondence with a proper subset of itself (because it has fewer elements), and whilst this is reasonable – I cannot follow Rosenlicht in that "the above therefore follows obviously". Why must a set be infinite just because of some property of finite sets?
Best Answer
In standard terminology, a "finite" set means one whose cardinality is a natural number, or in other words a set what is in bijective correspondence with $\{i\in\mathbb N\mid i<n\}$ for some $n\in \mathbb N$.
A set that is not in bijective correspondence with any proper subset of itself is called Dedekind-finite.
As you note, it is obvious that a finite set is also Dedekind-finite. But you're right that it is not obvious that every Dedekind-finite set is finite. In fact this is not necessarily true if we're working in a set theory without the Axiom of Choice.
If we do have the Axiom of Choice, however, we can prove easily that every set is either finite or contains a subset (not necessarily proper) that is in bijective correspondence with $\mathbb N$. In the latter case we can prove using the Hilbert's-Hotel construction that the set is not Dedekind-finite.
(Proof sketch. Let $A$ be a set and assume $A$ is not finite. Fix a choice function on the set of nonempty subsets of $A$. Construct by induction a function $f:\mathbb N\to A$ such that $f(n)$ is the chosen element of $A_n = A\setminus f(\{0,1,2,\ldots,n-1\})$. Because $A$ is not finite, $A_n$ is never empty. Then $f$ is a bijection between $\mathbb N$ and a subset of $A$.)