Theorem: Cantor-Bernstein-Schröder Theorem

If a subset of one set is equivalent to the other, and a subset of the other is equivalent to the first, then the two sets are themselves equivalent:
:$\forall S, T: T \sim S_1 \subseteq S \land S \sim T_1 \subseteq T \implies S \sim T$

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:
{{begin-eqn}}
{{eqn | l = \map \alpha A
      | o = \subseteq
      | r = \map \alpha B
      | c = Image of Subset under Mapping is Subset of Image
}}
{{eqn | ll= \leadsto
      | l = \map {\paren {\complement_T \circ \alpha} } A
      | o = \supseteq
      | r = \map {\paren {\complement_T \circ \alpha} } B
      | c = Set Complement inverts Subsets
}}
{{eqn | ll= \leadsto
      | l = \map {\paren {\beta \circ \complement_T \circ \alpha} } A
      | o = \supseteq
      | r = \map {\paren {\beta \circ \complement_T \circ \alpha} } B
      | c = Image of Subset under Mapping is Subset of Image
}}
{{eqn | ll= \leadsto
      | l = \map {\paren {\complement_S \circ \beta \circ \complement_T \circ \alpha} } A
      | o = \subseteq
      | r = \map {\paren {\complement_S \circ \beta \circ \complement_T \circ \alpha} } B
      | c = Set Complement inverts Subsets
}}
{{eqn | ll= \leadsto
      | l = \map z A
      | o = \subseteq
      | r = \map z B
      | c = Definition of $z$
}}
{{end-eqn}}
{{qed|lemma}}
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:
{{begin-eqn}}
{{eqn | l = \relcomp S {\mathbb G}
      | r = \map {\paren {\complement_S \circ z} } {\mathbb G}
      | c = 
}}
{{eqn | r = \map {\paren {\complement_S \circ \complement_S \circ \beta \circ \complement_T \circ \alpha} } {\mathbb G}
      | c = Definition of $z$
}}
{{eqn | r = g \sqbrk {\map {\paren {\complement_T \circ \alpha} } {\mathbb G} }
      | c = as $\complement_S \circ \complement_S = I_{\powerset S}$
}}
{{end-eqn}}
At this stage, a diagram can be helpful:
:500px
Let $h: S \to T$ be the mapping defined as:
:<nowiki>$\forall x \in S: \map h x = \begin {cases}
\map f x & : x \in \mathbb G \\
\map {g^{-1} } x & : x \in \relcomp S {\mathbb G}
\end {cases}$</nowiki>
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 $g^{-1}$.
{{qed}}