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 $\subseteq$.

Furthermore, there is an upper bound for every chain in $\WW$: the union of all chains in $\WW$.

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

that $\WW \subsetneqq X$.

Let $\struct {E, \subseteq}$ be the maximal well-ordering on $X$, as defined above, and assume $x_0 \in X \setminus E$.

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


 * $\preceq' := \begin{cases} x \preceq y & : x, y \in E \\ x_0 \preceq x & : x \in E, x_0 \in X \setminus E \end{cases}$

Then $\preceq'$ properly includes $\preceq$, contradicting the maximality of $\preceq$.

Thus the well-ordering on $\WW$ is a well-ordering on all of $X$.