Order-Extension Principle

Theorem
Every ordering on a set can be extended to a total ordering on that set.

More precisely, given a set $S$ and an ordering $\preceq$ on $S$,

there exists a total ordering $\le$ on $S$ such that for all $a,\,b \in S$, $a \preceq b \implies a \le b$.

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 transitive closure of $\le'=\le \cup \left\{ {(a,b)} \right\}$.

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

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