Cantor-Bernstein-Schröder Theorem/Lemma/Proof 2

Proof
Define a mapping $E: \mathcal P \left({S}\right) \to \mathcal P \left({S}\right)$ as:


 * $E \left({K}\right) = S \setminus \left({T \setminus f \left[{K}\right]}\right)$

where $f \left[{K}\right]$ is the image of $K$ under $f$.

Then:
 * $E \left({K}\right) = \left({S \setminus T}\right) \cup f \left[{K}\right]$

By Image of Subset under Relation is Subset of Image and the corollary to Set Union Preserves Subsets, $E$ is increasing.

Thus by Knaster-Tarski Lemma: Power Set, $E$ has a fixed point $X$.

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

That is:
 * $\left({S \setminus \left({T \setminus f \left[{X}\right]}\right)}\right) = X$

Taking the set difference from $S$:


 * $T \setminus f \left[{X}\right] = S \setminus X$

Let $f'$ be the restriction of $f$ to $X \times f \left[{X}\right]$.

By Injection to Image is Bijection, $f'$ is a bijection.

By Identity Mapping is Bijection, the identity mapping $I_{S \mathop \setminus X}$ from $S \setminus X$ to $T \setminus f \left[{X}\right]$ is a bijection.

Thus by Union of Bijections with Disjoint Domains and Codomains is Bijection, $g = f' \cup i$ is the bijection we seek.