Addition of Cuts is Commutative

Theorem

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

Let the operation of $\alpha + \beta$ be the sum of $\alpha$ and $\beta$.

Then:

$\alpha + \beta = \beta + \alpha$

Proof

$\alpha + \beta$ is the set of all rational numbers of the form $p + q$ such that $p \in \alpha$ and $q \in \beta$.

Similarly, $\beta + \alpha$ is the set of all rational numbers of the form $q + p$ such that $p \in \alpha$ and $q \in \beta$.

From Rational Addition is Commutative we have that:

$p + q = q + p$

The result follows.

$\blacksquare$

Sources

• 1964: Walter Rudin: Principles of Mathematical Analysis (2nd ed.) ... (previous) ... (next): Chapter $1$: The Real and Complex Number Systems: Dedekind Cuts: $1.14$. Theorem $\text {(a)}$