Ordering on Cuts satisfies Trichotomy Law

Theorem
Let $\alpha$ and $\beta$ be cuts.

Then exactly one of the following applies:

where $<$ and so $>$ denote the strict ordering of cuts:
 * $\alpha < \beta \iff \exists p \in \Q: p \in \alpha, p \notin \beta$

Hence the ordering of cuts $\le$ is a total ordering.

Proof
Let $\alpha = \beta$.

By definition of equality of cuts:
 * $p \in \alpha \iff p \in \beta$

By definition of strict ordering of cuts it follows that neither $\alpha < \beta$ or $\alpha > \beta$.

both $\alpha < \beta$ and $\alpha > \beta$.

Because $\alpha < \beta$ there exists $p \in \Q$ such that:
 * $p \in \beta, p \notin \alpha$

Because $\alpha > \beta$ there exists $q \in \Q$ such that:
 * $q \in \alpha, q \notin \beta$

By Rational Number Not in Cut is Greater than Element of Cut:
 * $p \in \beta$ and $q \notin \beta$ implies $p < q$.
 * $q \in \alpha$ and $p \notin \alpha$ implies $p > q$.

But from Rational Numbers form Totally Ordered Field, $p < q$ and $p > q$ is not possible.

Hence by Proof by Contradiction, it is not possible for both $\alpha < \beta$ and $\alpha > \beta$.

Suppose either $\alpha < \beta$ or $\alpha > \beta$.

Then by definition of strict ordering of cuts it follows that $\alpha \ne \beta$.

Thus mutual exclusivity of the three conditions has been demonstrated.

That is, at most one of the three conditions holds.

It remains to be shown that at least one of the three conditions holds.

Suppose $\alpha \ne \beta$.

Then by definition of equality of cuts, either:
 * $\exists p \in \alpha: p \notin \beta$

in which case $\alpha > \beta$, or:
 * $\exists q \in \beta: q \notin \alpha$

in which case $\alpha < \beta$.

Hence the result.