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.
 * $$\complement_S: \mathcal P \left({S}\right) \to \mathcal P \left({S}\right): \forall X \in \mathcal P \left({S}\right): \complement_S \left({X}\right) = \mathcal P \left({S}\right) \setminus X$$
 * $$\complement_T: \mathcal P \left({T}\right) \to \mathcal P \left({T}\right): \forall Y \in \mathcal P \left({T}\right): \complement_T \left({Y}\right) = \mathcal P \left({T}\right) \setminus Y$$

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

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

We have:

$$ $$ $$ $$ $$

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

As Empty Set Subset of All, we note that $$\varnothing \in \mathbb F$$, so $$\mathbb F \ne \varnothing$$.

Now consider the union of all the sets in $$\mathbb F$$:
 * $$\mathbb G = \bigcup \left\{{X: X \in \mathbb F}\right\}$$

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

From Subset of Union, we have that $$X \in \mathbb F \implies X \subseteq \mathbb G$$.

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

From Union Smallest: Generalized 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:

$$ $$ $$

Now consider the sets $$S$$ and $$T$$ in light of the fact that the relative complement forms a partition.

We have that:
 * $$\left\{{\mathbb G, \complement_S \left({\mathbb G}\right)}\right\}$$ forms a partition of $$S$$;
 * $$\left\{{f \left({\mathbb G}\right), \complement_T \left({f \left({\mathbb G}\right)}\right)}\right\}$$ forms a partition of $$T$$.

Thus we see that we can set up the mapping $$h: S \to T$$ 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}$$

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

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