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 $(a, b) \in h$, either $b = f(a)$ or $a = g(b)$.

Proof
For the purposes of this proof, if $f$ is a mapping and $S$ is a set, then the image of $S$ under $f$ will be denoted $f[S]$.

Let $\mathcal P(A)$ be the power set of $A$.

Define a mapping $E: \mathcal P(A) \to \mathcal P(A)$ thus:


 * $E(S) = A \setminus g[B \setminus f[S]]$

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

Then:

That is, $E(S) \subseteq E(T)$.

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

By the definition of fixed point, $E(X) = X$.

Thus $A \setminus (A \setminus g[B \setminus f[X]]) = A \setminus X$.

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


 * $g[B \setminus f[X]] = A \setminus X$

Let $f' = f \restriction_{X \times f[X]}$ be the Definition:restriction of $f$ to $X \times f(X)$.

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

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

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