Order-Extension Principle/Strict/Proof 2

Theorem
Let $S$ be a set.

Let $\prec$ be a strict ordering on $S$.

Then there exists a strict total ordering $<$ on $S$ such that:
 * $\forall a, b \in S: a \prec b \implies a < b$

Proof
For the purposes of this proof, a relation $<_U$ on a subset $U$ of $S$ will be considered compatible with $\prec$ iff:


 * $\forall a, b \in U: a \prec b \implies a < b$

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


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


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


 * $(c): \quad \left({x,y}\right) \in \operatorname{Dom} \left({f}\right) \implies \left({x \prec y \implies f(x, y) = 1}\right)$

We will first show that for some $\phi \in M$, the domain of $f$ is $S \times S$.

We will then show that $\phi$ is the characteristic function of a strict total ordering on $S$ that extends $\prec$.

We will apply the Cowen-Engeler Lemma to show that $\phi$ exists, so we must show that $M$ satisfies the premises of that theorem. Specifically, we must show that:


 * $(1): \quad \left\{{f \left({x, y}\right): f \in M}\right\}$ is finite for each $\left({x, y}\right) \in S \times S$


 * $(2): \quad$ For each finite subset $F$ of $S \times S$, there exists an element $f \in M$ such that the domain of $f$ is $F$


 * $(3): \quad M$ has finite character. That is, a mapping $f$ from some subset of $S \times S$ to $\left\{{0, 1}\right\}$ is an element of $M$ iff its restriction to each finite subset of $\operatorname{Dom} \left({f}\right)$ is in $M$.

$M$ satisfies premise $(1)$
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 $(1)$ is trivially satisfied.

$M$ satisfies premise $(2)$
Let $F$ be a finite subset of $S \times S$.

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

That is, $S_F$ is the set of all $x$ such that $x$ is either the first or second component of some element of $F$.

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

Thus by Order-Extension Principle/Finite Set 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 since $F \subseteq S_F \times S_F$, $\operatorname{Dom} \left({f_F}\right) = F$ by the definition of restriction.

Then $f_F \in M$:

If $\left({x, y}\right), \left({y, z}\right), \left({x, z}\right) \in \operatorname{Dom} \left({f_F}\right) = F$ and $f_F(x,y) = 1 \land f_F(y,z) = 1$, then $x <_F y \land y <_F z$.

Thus since $<_F$ is transitive, $x <_F z$.

Thus $f_F(x,z) = 1$.

Similarly, if $(x,y), (y,x) \in \operatorname{Dom}(f_F) = F$ then since $<_F$ is asymmetric and connected, $x <_F y \iff \neg (y <_F x)$.

Thus $f_F(x,y) = 1 \iff f_F(y,x) = 0$.

If $(x,y) \in \operatorname{Dom}(f_F)$, then since $<_F$ is compatible with $\prec$, $x \prec y \implies f(x,y) = 1$.

So $f_F \in M$.

As such an $f_F$ exists for each such $F$, $M$ satisfies premise $(2)$.

$M$ satisfies premise $(3)$
We must show that $M$ has finite character.

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

Suppose that $f \in M$.

Let $F$ be a finite subset of $\operatorname{Dom}(f)$.

Let $x, y, z \in S$.

If $(x,y), (y,z), (x,z) \in \operatorname{Dom}(f \restriction F)$ then $(x,y), (y,z), (x,z) \in \operatorname{Dom}(f)$.

Thus since $f \in M$:
 * $f \left({x, y}\right) = 1 \land f \left({y, z}\right) = 1 \implies f \left({x, z}\right) = 1$.

But $f \restriction F \left({x, y}\right) = f \left({x, y}\right)$, $f \restriction F \left({y, z}\right) = f \left({y, z}\right)$ and $f \restriction F \left({x, z}\right) = f \left({x, z}\right)$ by the definition of restriction.

Thus $f \restriction F \left({x, y}\right) = 1 \land f \restriction F \left({y, z}\right) = 1 \implies f \restriction F \left({x, z}\right) = 1$.

So $f \restriction F$ satisfies $(a)$.

A precisely similar argument shows that $f \restriction F$ satisfies $(b)$ and $(c)$, so $f \restriction F \in M$.

Suppose instead that for every finite subset $F$ of $\operatorname{Dom}(f)$, $f \restriction F \in M$.

Let $x, y \in S$.

Suppose that $(x, y), (y, x) \in \operatorname{Dom})(f)$.

Then $K = \left\{{ (x, y), (y, x)}\right\}$ is a finite subset of $\operatorname{Dom}(f)$.

Thus $f \restriction K \in M$.

So $f \restriction K (x, y) = 1 \iff f \restriction K (y, x) = 0$.

Since $f \restriction K (x, y) = f(x, y)$ and $f \restriction K (y, x) = f(y, x)$:
 * $f(x, y) = 1 \iff f(y, x) = 0$.

Thus $f$ satisfies $(b)$ for all pairs.

A precisely similar argument shows that $f$ satisfies $(a)$ for all pairs.

Thus $f \in M$.

Since $M$ satisfies $(1)$, $(2)$, and $(3)$, by the Cowen-Engeler Lemma there is a $\phi \in M$ whose domain is $S \times S$.

$\phi$ is the characteristic function of a strict total ordering on $S$ compatible with $\prec$
Since $\operatorname{Dom}(\phi) = S \times S$ and $\phi \in M$, for all $x, y, z \in S$:


 * $(a): \quad \phi \left({x, y}\right) = 1 \land \phi \left({y, z}\right) = 1 \implies \phi \left({x, z}\right) = 1$


 * $(b): \quad \phi \left({x, y}\right) = 1 \iff \phi \left({y, x}\right) = 0$


 * $(c): \quad x \prec y \implies \phi \left({x, y}\right) = 1$

By an argument similar to that given for premise $(2)$, above, $\phi$ is the characteristic function of a strict total ordering on $S$ compatible with $\prec$.