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, and let $\mathcal P \left({S}\right), \mathcal P \left({T}\right)$ be 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 $\mathcal P \left({S}\right)$ and $\mathcal P \left({T}\right)$ as mappings:

which follow directly from the definition of relative complement.
 * $\complement_S: \mathcal P \left({S}\right) \to \mathcal P \left({S}\right): \forall X \in \mathcal P \left({S}\right): \complement_S \left({X}\right) = S \setminus X$
 * $\complement_T: \mathcal P \left({T}\right) \to \mathcal P \left({T}\right): \forall Y \in \mathcal P \left({T}\right): \complement_T \left({Y}\right) = T \setminus Y$

Let $\alpha$ and $\beta$ denote the mappings induced on $\mathcal P \left({S}\right)$ and $\mathcal P \left({T}\right)$ by $f$ and $g$, respectively.

Consider the mapping $z: \mathcal P \left({S}\right) \to \mathcal P \left({S}\right)$ defined by the composition:
 * $z = \complement_S \circ \beta \circ \complement_T \circ \alpha$

Consider $A \subseteq B \subseteq S$, so $A, B \in \mathcal P \left({S}\right)$.

We have:

Now consider the set:
 * $\mathbb F = \left\{{X \in \mathcal P \left({S}\right): X \subseteq z \left({X}\right)}\right\}$

Now consider the union $\mathbb G$ of $\mathbb F$:
 * $\displaystyle \mathbb G = \bigcup \mathbb F$

Don't lose sight of the fact that:
 * $\mathbb F \subseteq \mathcal P \left({S}\right)$;
 * $\mathbb G \subseteq S$.

From Subset of Union: General Result, we have that $\forall X \in \mathbb F: X \subseteq \mathbb G$.

Thus:
 * $X \subseteq z \left({X}\right) \subseteq z \left({\mathbb G}\right)$

From Union Smallest: General Result, it follows that:
 * $\mathbb G \subseteq z \left({\mathbb G}\right)$

and so from above:
 * $z \left({\mathbb G}\right) \subseteq z \left({z \left({\mathbb G}\right)}\right)$

So $z \left({\mathbb G}\right) \in \mathbb F$ and so $z \left({\mathbb G}\right) \subseteq \mathbb G$.

So from:
 * $z \left({\mathbb G}\right) \subseteq \mathbb G$

and:
 * $\mathbb G \subseteq z \left({\mathbb G}\right)$

we have from Equality of Sets:
 * $z \left({\mathbb G}\right) = \mathbb G$

From Relative Complement of Relative Complement we have that $\complement_S \circ \complement_S$ is the identity mapping on $\mathcal P \left({S}\right)$.

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: h \left({x}\right) = \begin{cases}

f \left({x}\right) & : x \in \mathbb G \\ g^{-1} \left({x}\right) & : x \in \complement_S \left({\mathbb G}\right) \end{cases}$

From the above, we have that $\complement_S \left({\mathbb G}\right) \subseteq g \left({T}\right)$.

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

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