Cantor-Bernstein-Schröder Theorem/Proof 5

Theorem

Let $S$ and $T$ be sets.

Let $f: S \to T$ and $g: T \to S$ be injections.

Then there exists a bijection $\phi: S \to T$.

Proof

By Injection to Image is Bijection:

$g: T \to g \sqbrk T$ is a bijection.

Thus $T$ is equivalent to $g \sqbrk T$.

By Composite of Injections is Injection $g \circ f: S \to g \sqbrk T \subset S$ is also an injection (to a subset of the domain of $g \circ f$).

Then by Cantor-Bernstein-Schröder Theorem: Lemma:

There exists a bijection $h: S \to g \sqbrk T$.

Thus $S$ is equivalent to $g \sqbrk T$.

We already know that $T$ is equivalent to $g \sqbrk T$.

Thus by Set Equivalence behaves like Equivalence Relation, $S$ is equivalent to $T$.

By the definition of set equivalence:

There is a bijection $\phi: S \to T$.

$\blacksquare$