Order-Extension Principle/Strict/Proof 1

Proof
Let $\AA$ 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 $\tuple {x, y} \in S \times S$, let $\tuple {x, y}' = \tuple {y, x}$.

Let $A \in \AA$.

Let $\tuple {x, y} \in S \times S$.

Let $\tuple {x, y} \in A^+$.

Then:
 * $\paren {A \cup \set {\tuple {x, y} } }^+ = A^+$

so:
 * $A \cup \set {\tuple {x, y} } \in \AA$

Let $\tuple {y, x} \in A^+$.

Then:
 * $\paren {A \cup \set {\tuple {y, x} } }^+ = A^+$

so:
 * $A \cup \set {\tuple {x, y}'} = A \cup \set {\tuple {y, x} } \in \AA$

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

So by Strict Ordering can be Expanded to Compare Additional Pair:
 * $A \cup \set {\tuple {x, y} } \in \AA$

Thus it has been shown that for each $A \in \AA$ and each $\tuple {x, y} \in S$, either:
 * $A \cup \set {\tuple {x, y} } \in \AA$

or:
 * $A \cup \set {\tuple {x, y}'} \in \AA$

Let $A \subseteq S \times S$.

Let $A \in \AA$.

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 $F \in \AA$.

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

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

that for some $x \in S$, $\tuple {x, x} \in A^+$.

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

Let $F = \set {\tuple {y_0, y_1}, \tuple {y_1, y_2}, \dotsc, \tuple {y_{n - 1}, y_n} }$.

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

But $\tuple {x, x} \in F^+$, contradicting the fact that $F \in \AA$.

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

Therefore, $\AA$ has finite character.

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

By the Restricted Tukey's Theorem (Strong Form), there exists an $R \in \AA$ such that:
 * ${\prec} \subseteq R$
 * For each $\tuple {m, n} \in S \times S$, either $\tuple {m, n} \in R$ or $\tuple {n, m} = \tuple {m, n}' \in R$.

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