# Zorn's Lemma Implies Axiom of Choice

## Theorem

If Zorn's Lemma is true, then so must the Axiom of Choice be.

## Proof

Let $X$ be a set.

Let $\mathcal F$ be the set of mappings defined as:

$f \in \mathcal F \iff \begin{cases} \operatorname{Dom} \left({f}\right) \subseteq \mathcal P \left({X}\right) & \ \\ \operatorname{Im} \left({f}\right) \subseteq X & \ \\ \forall A \in \operatorname{Dom} \left({f}\right): f \left({A}\right) \in A & \ \end{cases}$

Let $\preceq$ be the relation defined on $\mathcal F$ as:

$\forall f_1, f_2 \in \mathcal F: f_1 \preceq f_2 \iff f_2$ is an extension of $f_1$.

Straightforwardly, $\preceq$ is a partial ordering on $\mathcal F$.

Suppose Zorn's Lemma holds.

Then there exists a maximal element of $\mathcal F$.

We then show that if $g$ is such a maximal element, then:

$\operatorname{Dom} \left({g}\right) = \mathcal P \left({X}\right) \setminus \varnothing$