Order-Extension Principle/Strict

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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:

$\forall a, b \in S: a \prec b \implies a < b$


Finite Set

Let $T$ be a finite set.

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


Then there exists a strict total ordering $<$ on $T$ such that:

$\forall a, b \in T: \left({a \prec b \implies a < b}\right)$


Proof 1

Let $\mathcal A$ be the set of relations $A$ on $S$ with the property that the transitive closure $A^+$ of $A$ is a strict ordering of $S$.

For each $\left({x, y}\right) \in S \times S$, let $\left({x, y}\right)' = \left({y, x}\right)$.


Let $A \in \mathcal A$.

Let $\left({x, y}\right) \in S \times S$.

Let $\left({x, y}\right) \in A^+$.

Then:

$\left({A \cup \left\{ {\left({x, y}\right)}\right\} }\right)^+ = A^+$

so:

$A \cup \left\{ {\left({x, y}\right)}\right\} \in \mathcal A$


Let $\left({y, x}\right) \in A^+$.

Then:

$\left({ A \cup \left\{ {\left({y, x}\right)}\right\} }\right)^+ = A^+$

so:

$A \cup \left\{ {\left({x, y}\right)'}\right\} = A \cup \left\{ {\left({y, x}\right)}\right\} \in \mathcal A$


Otherwise, $x$ and $y$ are non-comparable by $A^+$.

So by Strict Ordering can be Expanded to Compare Additional Pair:

$A \cup \left\{ {\left({x, y}\right)}\right\} \in \mathcal A$

Thus it has been shown that for each $A \in \mathcal A$ and each $\left({x, y}\right) \in S$, either:

$A \cup \left\{ {\left({x, y}\right)}\right\} \in \mathcal A$

or:

$A \cup \left\{ {\left({x, y}\right)'}\right\} \in \mathcal A$

$\Box$


Let $A \subseteq S \times S$.

Let $A \in \mathcal A$.

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

Since $A^+$ is a strict ordering, it is asymmetric.

Since Transitive Closure is Closure Operator, $F^+ \subseteq A^+$.

So $F^+$ is also asymmetric.

Since $F^+$ is also transitive, it is a strict ordering.

So $F \in \mathcal A$.


Suppose instead that each finite subset of $A$ is in $\mathcal A$.

We must show that $A^+$ is antireflexive.

Suppose for the sake of contradiction that for some $x \in S$, $\left({x, x}\right) \in A^+$.

Then by the definition of transitive closure, there are elements $y_0, \dots, y_n$ such that $x = y_0 = y_n$ and:

$\left({y_0, y_1}\right), \left({y_1, y_2}\right), \dotsc, \left({y_{n - 1}, y_n}\right) \in A$

Let $F = \left\{ {\left({y_0, y_1}\right), \left({y_1, y_2}\right), \dotsc, \left({y_{n - 1}, y_n}\right)}\right\}$.

Then $F$ is a finite subset of $A$.

But $\left({x, x}\right) \in F^+$, contradicting the fact that $F \in \mathcal A$.

Thus we see that $A^+$ is antireflexive, and thus a strict ordering of $S$.

Therefore, $\mathcal A$ has finite character.

$\Box$


Note that ${\prec} = {\prec^+}$, so ${\prec} \in \mathcal A$.

By the Restricted Tukey-Teichmüller Theorem (Strong Form), there exists an $R \in \mathcal A$ such that:

${\prec} \subseteq R$
For each $\left({m, n}\right) \in S \times S$, either $\left({m, n}\right) \in R$ or $\left({n, m}\right) = \left({m, n}\right)' \in R$.

Then:

$R^+$ is a strict total ordering of $S$.
$\forall a, b \in S: a \prec b \implies a < b$

$\blacksquare$


Proof 2

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

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

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

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



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.


$M$ 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.

$\Box$


$M$ satisfies premise $(2)$

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

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


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


Thus by the Order-Extension Principle for Finite Sets 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 $f_F \in M$:

Let:

$\left({x, y}\right), \left({y, z}\right), \left({x, z}\right) \in \operatorname{Dom} \left({f_F}\right) = F$
$f_F \left({x, y}\right) = 1 \land f_F \left({y, z}\right) = 1$

Then $x <_F y \land y <_F z$.

Thus since $<_F$ is transitive:

$x <_F z$

Thus:

$f_F \left({x, z}\right) = 1$

Similarly, let $\left({x, y}\right), \left({y, x}\right) \in \operatorname{Dom} \left({f_F}\right) = F$.

Then since $<_F$ is asymmetric and connected:

$x <_F y \iff \neg (y <_F x)$

Thus:

$f_F \left({x, y}\right) = 1 \iff f_F \left({y, x}\right) = 0$.

If $\left({x, y}\right) \in \operatorname{Dom}(f_F)$, then since $<_F$ is compatible with $\prec$:

$x \prec y \implies f \left({x, y}\right) = 1$

So $f_F \in M$.

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

$\Box$


$M$ satisfies premise $(3)$

We must show that $M$ has finite character.

That is, a mapping $f$ from some subset of $S \times S$ to $\left\{{0, 1}\right\}$ is an element of $M$ if and only if:

its restriction to each finite subset of $\operatorname{Dom} \left({f}\right)$ is in $M$.

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

Suppose that $f \in M$.

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

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

Let $\left({x, y}\right), \left({y, z}\right), \left({x, z}\right) \in \operatorname{Dom}(f \restriction F)$.

Then:

$\left({x, y}\right), \left({y, z}\right), \left({x, z}\right) \in \operatorname{Dom} \left({f}\right)$

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 by the definition of restriction:

$f \restriction F \left({x, y}\right) = f \left({x, y}\right)$
$f \restriction F \left({y, z}\right) = f \left({y, z}\right)$
$f \restriction F \left({x, z}\right) = f \left({x, z}\right)$

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 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 $\operatorname{Dom} \left({f}\right)$, $f \restriction F \in M$.

Let $x, y \in S$.

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

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

Thus $f \restriction K \in M$.

So:

$f \restriction K \left({x, y}\right) = 1 \iff f \restriction K \left({y, x}\right) = 0$

Since $f \restriction K \left({x, y}\right) = f \left({x, y}\right)$ and $f \restriction K \left({y, x}\right) = f \left({y, x}\right)$:

$f \left({x, y}\right) = 1 \iff f \left({y, x}\right) = 0$

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

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

Thus $f \in M$.

$\Box$


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 Characteristic Function which is Compatible with $\prec$

Since $\operatorname{Dom} \left({\phi}\right) = S \times S$ 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 $\prec$.

$\blacksquare$