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.

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:

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

$\Box$

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:

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

$\Box$

Hence the result by definition of set equality.

$\blacksquare$

1964: Walter Rudin: Principles of Mathematical Analysis (2nd ed.) ... (previous) ... (next): Chapter $1$: The Real and Complex Number Systems: Dedekind Cuts: $1.28$. Theorem: $\text {(a)}$