Sum of Cuts is Cut

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

Let $\gamma$ be the set of all rational numbers $r$ such that:
 * $\exists p \in \alpha, q \in \beta: r = p + q$

Then $\gamma$ is also a cut.

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

Proof
By definition of cut, neither $\alpha$ nor $\beta$ are empty.

Hence there exist $p \in \alpha$ and $q \in \beta$.

Hence there exists $r = p + q$ and so $\gamma$ is likewise not empty.

Let $s, t \in \Q$ such that $s \notin \alpha$ and $t \notin \beta$, where $\Q$ denotes the set of rational numbers.

Such $s$ and $t$ are bound to exist because by definition of cut, neither $\alpha$ nor $\beta$ equal $\Q$.

We have:

Thus it is demonstrated that $\gamma$ does not contain every rational number.

Thus condition $(1)$ of the definition of a cut is fulfilled.

Let $r \in \gamma$.

Let $s \in \Q$ such that $s < r$.

Then $r = p + q$ for some $p \in \alpha, q \in \beta$.

Let $t \in \Q$ such that $s = t + q$.

Then $t < p$.

Hence $t \in \alpha$.

Hence by definition of $\gamma$, $t + q = s \in \gamma$.

Thus we have that $r \in \gamma$ and $s < r$ implies that $s \in \gamma$.

Thus condition $(2)$ of the definition of a cut is fulfilled.

$r \in \gamma$ is the greatest element of $\gamma$.

Then $r = p + q$ for some $p \in \alpha, q \in \beta$.

By definition of a cut, $\alpha$ has no greatest element:

Hence:
 * $\exists s \in \Q: s > p: s \in \alpha$

But then $s + q \in \gamma$ while $s + q > r$.

This contradicts the supposition that $r$ is the greatest element of $\gamma$.

Hence $\gamma$ itself can have no greatest element.

Thus condition $(3)$ of the definition of a cut is fulfilled.

Thus it is seen that all the conditions are fulfilled for $\gamma$ to be a cut.