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 $X = \varnothing$ the theorem holds vacuously, so assume $X$ is not empty.

Let $\mathcal W \in \mathcal P(X)$ be the set of all well-ordered subsets of $X$.

Then $\mathcal W$ is partially ordered by $\subseteq$.

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

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

Seeking a contradiction suppose that $\mathcal W \subsetneqq X$.

Let $(E,\subseteq)$ be the maximal well ordering on $X$, as defined above, and assume $x_0 \in X \setminus E$.

Define an ordering $\preceq'$ on $X \cup \{ x_0 \}$ 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 $\mathcal W$ is a well ordering on all of $X$.