Zorn's Lemma Implies Zermelo's Well-Ordering Theorem

Theorem
Zorn's Lemma implies the Well-Ordering Theorem.

Proof
Let $X$ be a set.

If $X = \O$ the theorem holds vacuously.

Assume $X$ is not empty.

Let $\WW \subseteq \powerset X$ be the set of all well-ordered subsets of $X$.

Then by Subset Relation on Power Set is Partial Ordering, $\WW$ is partially ordered by $\subseteq$.

To apply Zorn's Lemma, we need to show that every chain in $\WW$ has an upper bound.

Let $\CC \subseteq \WW$ be such a chain.

Then $\bigcup \CC \in \WW$ and $\bigcup \CC$ is an upper bound for $\CC$.

Thus the hypotheses of Zorn's Lemma hold and we can conclude that $\WW$ has a maximal element.

Let $E$ be the maximal well-ordered subset of $X$.

Suppose that $E \ne X$.

Then there exists $x_0 \in X \setminus E$.

Let $\preceq$ be a well-order on $E$.

Define an ordering $\preceq'$ on $E \cup \set {x_0}$ as follows:


 * $x \preceq' y$ $x = x_0$ or $x, y \in E$ and $x \preceq y$.

This is a well-order on $E \cup \set {x_0}$, which properly includes $E$.

This contradicts that $E$ is the maximal well-orderable subset of $X$.

So $E = X$.

So $X$ is well-orderable.