Sum of Rational Cuts is Rational Cut

Theorem
Let $p \in\ Q$ and $q \in \Q$ be rational numbers.

Let $p^*$ and $q^*$ denote the rational cuts associated with $p$ and $q$.

Then:
 * $p^* + q^* = \paren {p + q}^*$

Thus the operation of addition on the set of rational cuts is closed.

Proof
From Sum of Cuts is Cut, $p^* + q^*$ is a cut.

Let $r \in p^* + q^*$.

Then:
 * $r = s + t$

where $s < p$ and $t < q$

Thus:
 * $r < p + q$

and so:
 * $r \in \paren {p + q}^*$

Hence:
 * $p^* + q^* \subseteq \paren {p + q}^*$

Let $r \in \paren {p + q}^*$.

Then:
 * $r < p + q$

Let:
 * $h = p + q - r$
 * $s = p - \dfrac h 2$
 * $t = q - \dfrac h 2$

Then:
 * $s \in p^*$
 * $t \in q^*$

Hence:
 * $r = s + t$

and so:
 * $r \in p^* + q^*$

So:
 * $\paren {p + q}^* \subseteq p^* + q^*$

Hence the result by definition of set equality.