Order-Extension Principle/Strict/Proof 1

Proof
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 $(x,y) \in S \times S$, let $(x,y)' = (y,x)$.

For each $A \in \mathcal A$ and each $(x,y) \in S$, either $A \cup \{(x,y)\} \in \mathcal A$ or $A \cup \{(x,y)'\} \in \mathcal A$
Suppose that $A \in \mathcal A$ and $(x,y) \in S \times S$.

Then if $(x,y) \in A^+$, $\left({ A \cup \{(x, y)\} }\right)^+ = A^+$, so $A \cup \{(x,y)\} \in \mathcal A$.

If $(y,x) \in A^+$, $\left({ A \cup \{(y, x)\} }\right)^+ = A^+$, so $A \cup \{(x,y)'\} = A \cup \{(y,x)\} \in \mathcal A$.

Otherwise, $x$ and $y$ are non-comparable by $A^+$, so by Strict Ordering can be Expanded to Compare Additional Pair, $A \cup \{(x,y)\} \in \mathcal A$.

$\mathcal A$ has finite character
Let $A \subseteq S \times S$.

Suppose that $A \in \mathcal A$

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 $x \in S$, $(x,x) \in A^+$.

Then by the definition of transitive closure, there are elements $y_0, \dots, y_n$ such that $x = y_0 = y_n$ and $(y_0, y_1), (y_1, y_2), \dots, (y_{n-1}, y_n) \in A$.

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

Then $F$ is a finite subset of $A$, but $(x,x) \in F^+$, 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.

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 $(m,n) \in S \times S$, either $(m, n) \in R$ or $(n,m) = (m,n)' \in R$.

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