Let $A$ and $B$ be sets, not necessarily finite. Let $f: A\rightarrow B$ be injective, and let $g:B\rightarrow A$ be injective.
The Cantor–Schröder–Bernstein theorem then says there must be a bijection between $A$ and $B$.
I once got the proof of this assigned on a problem set, and when I presented the following proof, the TA said that it was insufficient. Take $\vert X\vert$ to be the cardinality of a set $X$.
$f$ injective $\implies \vert A\vert\le\vert B\vert$
$g$ injective $\implies \vert A\vert\ge\vert B\vert$
Therefore, $\vert A\vert=\vert B\vert$, and there is a bijection between $A$ and $B$. $\square$
I have never understood why this proof is insufficient but assume it has something to do with $\le$, $\ge$, and $=$ when it comes to comparing infinite cardinalities.
What is inadequate about that proof?
Best Answer
You must prove that $|A| \le |B|$ and $|A| \ge |B|$ implies $|A| = |B|$. That is the whole point of the theorem. You cannot simply assume that because we use the symbol $\le$ that it has the properties we expect.