Order-Extension Principle/Proof 1

Theorem
Let $S$ be a set.

Let $\preceq$ be an ordering on $S$.

Then there exists a total ordering $\le$ on $S$ such that:
 * $\forall a, b \in S: a \preceq b \implies a \le b$

Proof
Let $\preceq$ be an ordering on the set $S$. If $\preceq$ is a total ordering, the result is trivial.

Suppose, then, that $\preceq$ is not a total ordering.

Let $T$ be the set of orderings on $S$ that extend $\preceq$, ordered by inclusion.

Let $C$ be a chain in $T$. By Union of Nest of Orderings is Ordering, $\bigcup C$ is an ordering.

Thus every chain in $T$ has an upper bound in $T$.

By Zorn's Lemma, $T$ has a maximal element, $\le$.

$\le$ is a total ordering:

Suppose that $a, b \in S$, $a \not\le b$, and $b \not\le a$.

Let $\le'$ be the relation defined as:
 * $\le' \mathop{:=} \le \cup \left\{ {(a,b)} \right\}$

Let $\le'^-$ be the transitive closure of $\le'$.

Then $\le'^-$ is an ordering by One-Point Extension of Ordering is Ordering.

But $\le'^- \mathop {\supsetneq} \le$, contradicting the maximality of $\le$. Thus, $\le$ is a total ordering.