Ordering on Cuts is Total

Theorem
Let $\CC$ denote the set of cuts.

Let $<$ denote the strict ordering on cuts defined as:
 * $\forall \alpha, \beta \in \CC: \alpha < \beta \iff \exists p \in \Q: p \in \alpha, p \notin \beta$

Then $<$ is a (strict) total ordering on $\CC$.

Proof
Let $\struct {\CC, <}$ denote the relational structure defined from the above.

From Ordering on Cuts is Transitive, $<$ is a transitive relation on $\CC$.

From Ordering on Cuts satisfies Trichotomy Law, we have that:
 * $\alpha < \beta \implies \lnot \paren {\beta < \alpha}$

demonstrating that $<$ is asymmetric.

Hence, by definition, $<$ is a strict ordering.

Also by Ordering on Cuts satisfies Trichotomy Law, $\struct {\CC, <}$ has no non-comparable pairs:


 * $\forall \alpha, \beta \in \CC: \alpha \ne \beta \implies \alpha < \beta \lor \beta < \alpha$

The result follows by definition of strict total ordering.