Ordering on Cuts is Transitive

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

Let:

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

Then:
 * $\alpha < \gamma$

Hence the ordering of cuts $\le$ is a transitive relation.

Proof
We have that:
 * $\alpha < \beta$

By definition of strict ordering of cuts:
 * $\exists p \in \Q: p \in \beta, p \notin \alpha$

Similarly, we have that:
 * $\beta < \gamma$

and so:
 * $\exists q \in \Q: q \in \gamma, q \notin \beta$

We have by definition of a cut that:
 * $p \in \beta$ and $q \notin \beta$ implies that $p < q$

Together with $p \notin \alpha$, this implies that $q \notin \alpha$.

Thus we have:
 * $q \in \gamma$ and $q \notin \alpha$

from which by definition of strict ordering of cuts:
 * $\alpha < \gamma$