Set of Cuts under Addition forms Abelian Group

Theorem
Let $\CC$ denote the set of cuts.

Let $\struct {\CC, +}$ denote the algebraic structure formed from $\CC$ and the operation $+$ of addition of cuts.

Then $\struct {\CC, +}$ forms an abelian group.

Proof
In the below, $\alpha$, $\beta$ and $\gamma$ denote arbitrary cuts.

Taking the abelian group axioms in turn:

From Sum of Cuts is Cut:


 * $\forall \alpha, \beta \in \CC: \alpha + \beta \in \CC$

Thus $\struct {\CC, +}$ is closed.

From Addition of Cuts is Associative:


 * $\paren {\alpha + \beta} + \gamma = \alpha + \paren {\beta + \gamma}$

Thus the operation $+$ of addition of cuts is associative on $\struct {\CC, +}$.

$\text C$: Commutativity
From Addition of Cuts is Commutative:


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

Thus the operation $+$ of addition of cuts is commutative on $\struct {\CC, +}$.

Consider the rational cut $0^*$ associated with the (rational) number $0$:
 * $0^* = \set {r \in \Q: r < 0}$

From Identity Element for Addition of Cuts:
 * $\alpha + 0^* = \alpha$

for all $\alpha \in \CC$.

From Addition of Cuts is Commutative it follows that:
 * $0^* + \alpha = \alpha$

That is, $0^*$ is the identity element of $\struct {\CC, +}$.

We have that $0^*$ is the identity element of $\struct {\CC, +}$.

From Existence of Unique Inverse Element for Addition of Cuts, there exists a unique cut $-\alpha$ such that:
 * $\alpha + \paren {-\alpha} = 0^*$

From Addition of Cuts is Commutative it follows that:
 * $\paren {-\alpha} + \alpha = 0^*$

Thus every element of $\struct {\CC, +}$ has an inverse $-\alpha$.

All the abelian group axioms are thus seen to be fulfilled, and so $\struct {\CC, +}$ is a group.