Cantor-Bernstein-Schröder Theorem/Proof 3

Theorem
Let $S$ and $T$ be sets, such that:
 * $\exists f: S \to T$ such that $f$ is an injection
 * $\exists g: T \to S$ such that $g$ is an injection.

Then there exists a bijection from $S$ to $T$.

Proof
Let $S, T$ be sets.

Let $\powerset S, \powerset T$ denote their power sets.

Let $f: S \to T$ and $g: T \to S$ be injections that we know to exist between $S$ and $T$.

Consider the relative complements of elements of $\powerset S$ and $\powerset T$ as mappings:


 * $\complement_S: \powerset S \to \powerset S: \forall X \in \powerset S: \relcomp S X = S \setminus X$
 * $\complement_T: \powerset T \to \powerset T: \forall Y \in \powerset T: \relcomp T Y = T \setminus Y$

which follow directly from the definition of relative complement.

Let $\alpha$ and $\beta$ denote the direct image mappings of $f$ and $g$, respectively.

Consider the mapping $z: \powerset S \to \powerset S$ defined by the composition:
 * $z = \complement_S \circ \beta \circ \complement_T \circ \alpha$

$z$ is an increasing mapping
Let $A, B \in \powerset S$ with $A \subseteq B$.

We have:

By the Knaster-Tarski Lemma: Power Set, there is a $\mathbb G \in \powerset S$ such that:


 * $\map z {\mathbb G} = \mathbb G$

From Relative Complement of Relative Complement we have that $\complement_S \circ \complement_S$ is the identity mapping on $\powerset S$.

Thus we obtain:

At this stage, a diagram can be helpful:
 * Cantor-Bernstein-Schroeder.png

Let $h: S \to T$ be the mapping defined as:
 * $\forall x \in S: \map h x = \begin {cases}

\map f x & : x \in \mathbb G \\ \map {g^{-1} } x & : x \in \relcomp S {\mathbb G} \end {cases}$

From the above, we have that:
 * $\relcomp S {\mathbb G} \subseteq g \sqbrk T$

Therefore, as $g$ is an injection, it follows that the preimage $\map {g^{-1} } x$ is a singleton.

So $h$ is a bijection by dint of the injective nature of both $f$ and $g^{-1}$.