Let E and F be two sets. if $f\colon E\longrightarrow F$, $g\colon F\longrightarrow E$ are two injections, prove that there exists a bijection $h\colon E\longrightarrow F$.
I found an interesting question about an altenate proof of Cantor-Bernstein theorem (see Alternate proof of Cantor-Bernstein Theorem).
Consider: $\mathcal L:=\left\{C\subseteq E\mid g\left(F-f(C)\right)\subseteq E-C\right\}$
1. Prove that $\mathcal L$ is not empty and that $K \in \mathcal L$, where $K:=\bigcup\limits_{C\in \mathcal L}C$.
2. Let $H:=E-g\left(F-f(K)\right)$. Prove that $K\subseteq H$.
It is not difficult to prove that the first statement is true. And the second is particulary self-evident if the first is true.
My question is: How can we construct an alternate proof of Cantor-Bernstein theorem, using these statements?
I've been trying to prove $E = K$, but it's not going well.
Best Answer
Let $h(x) = f(x)$ if $x\in K$, and let $h(x)=g^{-1}(x)$ if $x\notin K$. There is some work to do to show that this is well defined and is a bijection, but I think you will be able to figure it out from here.