Order-Extension Principle/Strict

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 $\prec$ be the reflexive reduction of $\preceq$.

We will show that $\prec$ can be extended to a strict total ordering, $<$, on $S$.

Let $T$ be a finite subset of $S$.

By Order-Extension Principle/Finite Set, there exists a strict total ordering $<_T$ on $T$ such that:
 * $\forall a, b \in T: a \prec b \implies a \mathop{<_T} b$

Let $M$ be the set of partial functions $f$ from $S \times S$ to $\{0, 1\}$ such that for all $x, y, z \in S$:


 * $(a): \quad \left({x, y}\right), \left({y, z}\right), \left({x, z}\right) \in \operatorname{Dom} \left({f}\right) \implies \left({f \left({x, y}\right) = f \left({y, z}\right) = 1 \implies f \left({x, z}\right) = 1}\right)$
 * $(b): \quad \left({x, y}\right), \left({y, x}\right) \in \operatorname{Dom} \implies \left({f \left({x, y}\right) = 1 \iff f \left({y, x}\right) = 0}\right)$

We will apply the Cowen-Engeler Lemma to show that there is an element of $M$ whose domain is $S \times S$.

It will then follow that that element of $M$ is the characteristic function of a strict total ordering on $S$.

For each $g \in M$ and each $x \in \operatorname{Dom} \left({g}\right)$, $g \left({x}\right) \in \left\{{0, 1}\right\}$, so requirement $(1)$ of the Cowen-Engeler Lemma is trivially satisfied.

Let $F$ be a finite subset of $S$.

Let $S_F = \operatorname{Dom} \left({F}\right) \cup \operatorname{Img} \left({F}\right)$.

Since $F$ is finite, so is $S_F$.

Thus there is a strict total ordering $<_F$ on $S_F$ which is compatible with $\prec$.

Let $f_F$ be the restriction of the characteristic function of $<_F$ to $F$.

Then $f_F \in M$ and $\operatorname{Dom} \left({f_S}\right) = F$.

As such an $f_F$ exists for each such $F$, $M$ satisfies requirement $(2)$ of the Cowen-Engeler Lemma.

It remains only to show that $M$ has finite character.

Let $f$ be a partial function from $S \times S$ to $\left\{{0, 1}\right\}$.