Q: How does the Schroder-Bernstein Theorem imply this? I'm reading a proof of the "Fundamental Theorem of Cardinal Arithmetic," that is, that for any  infinite set A, A is equinumerous to A x A (assuming the AC, as necessary). 
Now, the background information doesn't seem relevant, but anyway, here it is. The proof proceeds as follows.  The author lets E be the set of bijections from subsets S of A onto S x S .  Then E is closed under chain unions, so by a maximal principle, E contains a maximal element f.  Now let M be the domain of f.  Then the author claims that A - M must be strictly smaller than M. (meaning, of course, that there is an injection from A - M onto a proper subset of M, but that there is no bijection from A - M onto M.)
Now's where I get lost.  To prove that A - M is smaller than M, the author argues by contradiction.  So suppose A - M were not smaller than M. The author states that "by the Bernstein-Schroder Theorem 2.1, A - M would include a subset M' of the same size as M." (I.e., M <= A - M, or M can be mapped injectively into A - M)
Maybe I'm stupid, but I don't see how this follows from the Schroder-Bernstein Theorem at all.  I know I could prove M <= A - M by using an AC argument--and of course, AC is assumed anyway--but this still bothers me.
The author seems to be asserting this proposition:
If B is a subset of A, then if A - B is not smaller than B, then B <= A - B.

A: Instead of $A \setminus M$ we may as well consider an arbitrary subset $B$ of $A$.
The statement $B \prec A$, that $B$ is strictly smaller than $A$, is usually defined in the absence of AC to mean that $B \preceq A$ and $A \not \preceq B$.  That is, there is an injection $B \to A$ but no injection $A \to B$.
If it is not the case that $B \prec A$, then because $B \preceq A$ holds (as $B$ is a subset of $A$) we must have $A \preceq B$.  Therefore $B \approx A$ by the Cantor–Bernstein Theorem.
