Order-Extension Principle/Strict

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

First we show that for each finite subset $T$ of $S$ there exists a strict total ordering $<_T$ on $T$ such that for $a,b \in T$, $a \prec b \implies a <_T b$.

We proceed by induction on the number of elements in $T$.

If $|T| = 0$, then $\varnothing$ is a strict total ordering meeting the requirements.

Suppose that a suitable strict total ordering exists for all $T$ of cardinality $n$.

Let $|T| = n+1$.

Let $t \in T$.

By Cardinality Less One, $|T \setminus \{t\}| = n$.

Thus there is a strict total ordering $<_{T\setminus\{t\}}$ compatible with $\prec$.

Define $<_T$ extending $<_{T\setminus\{t\}}$ by letting $a <_T b$ iff one of the following holds:


 * $a, b \in T \setminus \{t\}$ and $a <_{T\setminus\{t\}} b$
 * $a \prec b$
 * $a \nprec b$, $a \in T \setminus \{t\}$ and $b = t$

$<_T$ is clearly compatible with $\prec$.

We must show that it is antireflexive and transitive.

Antireflexive: Suppose that $a <_T b$.

Then one of the three conditions above holds.

In the first case, $a <_{T\setminus\{t\}} b$.

Since $<_{T\setminus\{t\}}$ is a strict total ordering of $T \setminus \{t\}$, $a \ne b$.

In the second case, $a \prec b$.

Since $\prec$ is antireflexive, $a \ne b$.

In the third case, $a \in T \setminus \{t\}$ and $b = t$, so $a \ne b$ by the definition of set difference.

Thus $<_T$ is a strict total ordering of $T$.

For each partial function $f$ from $S \times S$ to $\{0, 1\}$, let $S_f$ be the intersection of all sets $T$ such that $\operatorname{Dom}(f) \subseteq T \times T$.

Let $M$ be the set of partial functions $f$ from $S \times S$ to $\{0, 1\}$ such that there exists a mapping $f': S_f \times S_f$ such that $f \subseteq f'$ and $f'$ is the characteristic function of a strict total ordering of $S_f$ compatible with $\prec$.

We will apply the Cowen-Engeler Lemma to show that there is an element of $M$ whose domain is $S \times S$, and which is therefore the characteristic function of a strict total ordering on $S$.

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

As shown above, each finite subset of $S$ can be strictly totally ordered.

For each finite subset $F$ of $S \times S$, $