Order-Extension Principle/Proof 1

Proof
Let $\preceq$ be an ordering on the set $S$.

If $\preceq$ is a total ordering, the result is complete.

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

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

Let $C$ be a chain in $T$.

By Union of Chain of Orderings is Ordering, $\bigcup C$ is an ordering.

Thus every chain in $\TT$ has an upper bound in $\TT$.

By Zorn's Lemma, $\TT$ has a maximal element, $\RR$.

$\RR$ is seen to be the total ordering whose existence is to be demonstrated, as follows:

Suppose that:
 * $\exists a, b \in S: \tuple {a, b} \notin \RR \land \tuple {b, a} \notin \RR$

Let $\RR'$ be the relation defined as:
 * $\RR' := \RR \cup \set {\tuple {a, b} }$

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

Then by Ordering can be Expanded to compare Additional Pair, $\RR'^-$ is an ordering.

But $\RR'^- \supsetneq \RR$, contradicting the maximality of $\RR$.

Thus, $\RR$ is a total ordering.