Cantor-Bernstein-Schröder Theorem/Proof 6

Theorem
Let $A$ and $B$ be sets.

Let $f: A \to B$ and $g: B \to A$ be injections.

Then there is a bijection $h: A \to B$ with the following property:


 * For all $\left({a, b}\right) \in h$, either $b = f \left({a}\right)$ or $a = g \left({b}\right)$.

Proof
Let $\mathcal P \left({A}\right)$ be the power set of $A$.

Define a mapping $E: \mathcal P \left({A}\right) \to \mathcal P \left({A}\right)$ thus:


 * $E \left({S}\right) = A \setminus g \left({B \setminus f \left({S}\right)}\right)$

$E$ is increasing
Let $S, T \in \mathcal P \left({A}\right)$ such that $S \subseteq T$.

Then:

That is, $E \left({S}\right) \subseteq E \left({T}\right)$.

By the Knaster-Tarski Lemma, $E$ has a fixed point $X$.

By the definition of fixed point:
 * $E \left({X}\right) = X$

Thus:
 * $A \setminus \left({A \setminus g \left({B \setminus f \left({X}\right)}\right)}\right) = A \setminus X$

Since $g$ is a mapping into $A$:


 * $g \left({B \setminus f \left({X}\right)}\right) = A \setminus X$

Let $f' = f \restriction_{X \times f \left({X}\right)}$ be the restriction of $f$ to $X \times f \left({X}\right)$.

Similarly, let $g' = g \restriction_{\left({B \setminus f \left({X}\right)}\right) \times \left({A \setminus X}\right)}$.

$f'$ and $g'$ are both bijections.

Define a relation $h: A \to B$ by $h = f' \cup {g'}^{-1}$.

We will show that $h$ is a bijection from $A$ onto $B$.

The domain of $f'$ is $X$, which is disjoint from the codomain, $A \setminus X$, of $g'$.

The domain of $g'$ is $B \setminus f \left({X}\right)$, which is disjoint from the codomain, $f \left({X}\right)$, of $f'$.

Thus by There and Back Again:


 * $f' \cup {g'}^{-1}$ is a bijection from $X \cup \left({A \setminus X}\right)$ onto $f \left({X}\right) \cup \left({B \setminus f \left({X}\right)}\right)$.

That is, $f' \cup {g'}^{-1}$ is a bijection from $A$ onto $B$.