A set is infinite iff there is a one-to-one correspondent with one of its proper subsets? 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?
 A: 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 $A$.)
A: This is a relatively subjective issue -- we're talking about what it means, exactly, for a set to be infinite.
That said, if you agree that a finite set is, by definition, a set that cannot be put into 1-1 correspondence with itself and a set that's not finite is infinite, you get
$$ \left(A \mbox{ finite} \iff A \mbox{ cannot be put into 1-1 correspondence with a proper subset}\right) \implies \left(\lnot(A \mbox{ finite}) \iff \lnot(A \mbox{ cannot be put into 1-1 correspondence with a proper subset})\right) \implies \left(A \mbox{ infinite} \iff A \mbox{ can be put into 1-1 correspondence with a proper subset}\right) $$
A: I agree with you. The author gave a reason why a finite set cannot have a one-to-one correspondence with a proper subset of itself, but he did not give a reason why a set that does not have a one-to-one correspondence with a proper subset of itself must necessarily be finite.  Therefore the only thing that he can logically assert is that if a set has a one-to-one correspondence with a proper subset of itself, it must be infinite.  He gives no justification for the if-and-only-if relationship.
In other words, the author asserts that
$$(A \implies \lnot B) \implies (\lnot A \iff B)$$
Which is not sound.  Having said that, the author's statement is true, despite the fact that his logic does not back it up.
A: The given definition is usually taken as the definition of "Dedekind finite", whereas "finite" is usually taken to mean "equivalent to a natural number/finite ordinal". A finite set is always Dedekind-finite. However, in set theory without the axiom of choice, it's consistent for an infinite set to be Dedekind-finite!
Here's a proof, using the axiom of countable choice, that every infinite set has a countably infinite subset, allowing Hilbert's hotel to finish the argument.
A: This is known as
the Dedekind definition
of a set being infinite.
Here is more:
http://en.wikipedia.org/wiki/Dedekind-infinite_set
As an exercise,
you might try to show that
this is equivalent 
to the definition
stating that
the set,
or some subset of it,
can be placed into a
1-1 correspondence with
the positive integers.
A: If set $S$ is finite then there is no such proper subset. This means that if there is such proper subset for $S$ then $S$ is infinite. It's just $(A \implies B) \iff (\neg B \implies \neg A)$ for $A = [\text{$S$ is finite}]$ and $B = [\text{no such proper subset exists}]$.
A: This answer might be a bit late but here we go. 
Statement 1: A set is either finite or infinite. (Law of excluded middle)
Statement 2: A finite set cannot be put into one-to-one correspondence with a proper subset of itself. Proof.
Statement 3: So, by statement 1, if there exists a set which can be put into one-to-one correspondence with a proper subset of itself, then it should be an infinite set. 
See this link for more descriptive proof!
A: 
Theorem For every set $A$, the following conditions are equivalent (assuming AC$_{\omega}$ only for $\text{(a)}\Longrightarrow\text{(b)}$):
$\text{(a)}\;A$ is an infinite set
$\text{(b)}\;A$ has a countably infinite subset
$\text{(c)}\;A$ has a proper subset to which it is equinumerous

We will prove, first, the chain of implications $\text{(a)}\Longrightarrow\text{(b)}\Longrightarrow\text{(c)}$, to then prove the chain $\text{(c)}\Longrightarrow\text{(b)}\Longrightarrow\text{(a)}$.
$\text{(a)}\Longrightarrow\text{(b)}$: Let $A$ be an infinite set. For each $n\in\omega$, let $A_n$ be the set of injective functions from $n$ to $A$. For each $n\in\omega$, the set $A_n$ is not empty, since $A$ is an infinite set. By AC$_{\omega}$, for the countable set $\{A_n|\;n\in\omega\}$, the cartesian product $\prod_{n\in\omega}A_n$ is not empty, and there exists a function $h$ of domain $\omega$ such that for each $n\in\omega$, $h(n)=h_n$ is an injective function from $n$ to $A$.
Then, we can form a $\omega$-sequence of distinct elements of $A$, first putting the value of $h_1$, then the values of $h_2$, etc.:
$$h_1(0);h_2(0),h_2(1);h_3(0),h_3(1),h_3(2);h_4(0),h_4(1),h_4(2),h_4(3);\dots$$
And then deleting each occurence of any element $a$ of $A$ after appearing for the first time. This sequence is actually infinite, because for any $n\in\omega,\;h_n$ has $n$ distinct terms, and we have deleted at most $n-1$ of them, if $n\geq 1$.
The set of all the terms of this sequence is a countably infinite subset of $A$.
$\text{(b)}\Longrightarrow\text{(c)}$: Let $C$ be a countably infinite subset of $A$. Let $f$ be a bijective function from $\omega$ onto $C$. Consider the function $g:A\longrightarrow A$ defined by: for each $a\in A$,
$$g(a)=\begin{cases}
  f(n+1)\quad\text{if }a\in C\text{ and }a=f(n)\text{ for some }n\in\omega\\
  a\qquad\qquad\,\text{if }a\in A\setminus C
\end{cases}$$
The function $g$ s injective from $A$ to $A$, so $A\approx\text{Im}(g)$, and since $f(0)\not\in\text{Im}(g),\;\text{Im}(g)$ is a proper subset of $A$ to which it is equinumerous
$\text{(c)}\Longrightarrow\text{(b)}$: Let $B$ a proper subset of $A$ such that $A\approx B$, and let $f$ be a bijective function from $A$ ontro $B$. Since $B$ is proper, let $a_0$ be an element of $A\setminus B$. Let $g:\omega\longrightarrow A$ be the sequence defined by
$$\begin{cases}
  g(0)=a_0\\
  g(n+1)=f(g(n))
\end{cases}$$
We will prove that $g$ is injective, and therefore $\text{Im}(g)$ will be a countably infinite subset of $A$.
Suppose that $g$ is not injective. Then there exists two natural numbers $m,n$ with $m<n$ wuch that $g(m)=g(n)$. Therefore, the set
$\mathcal{N}=\{n\in\omega|\;\text{ there exists }m>n\text{ such that }g(m)=g(n)\}$
is a nonempty subset of $\omega$. Let $n_0$ be its first element.

*

*In first place, $n_0\not=0$, because $g(0)=a_0,\;a_0\not\in B$, and if $n>0,\;g(n)\in\text{Im}(f)=B$

*Since $n_0>0$, we have that if $m\in\mathcal{N}$ is such that $m>n_0$ and $g(n_0)=g(m)$, then

$$g(n_0)=f(g(n_0-1))=f(g(m-1))=g(m)$$
And, since $f$ is injective
$$g(n_0-1)=g(m-1)$$
Which contradicts the choice of $n_0$ as the first element of $\mathcal{N}$
$\text{(b)}\Longrightarrow\text{(a)}:$ Suppose that $A$ has a countably infinite subset $B$, and let $f:\omega\longrightarrow B$ be a bijection. If $A$ was finite, we would have that $A\approx n$ for some $n\in\omega$, for some bijective function $g:A\longrightarrow n$. The composition $g\circ f:\omega\rightarrowtail n$ would be injective, and its restriction to $n+1$ an injective function from $n+1$ to $n$, in contrary to the pidgeonhole principle.
