Addition of Cuts is Commutative
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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 $\text {(a)}$