Addition, negation, and subtraction of Dedekind cuts

Let $A,B\in \mathbb{R}$ be Dedekind cuts of $\mathbb{Q}$ (Dedekind cut, The real numbers $\mathbb{R}$ as Dedekind cuts).

Sum. The sum is the Minkowski sumset in $\mathbb{Q}$: $A+B:=\{a+b:a\in A,b\in B\}\mathrm{.}$

Additive identity. $0^{\ast }:=\{q\in \mathbb{Q}:q<0\}$, the cut of the rational $0$ under the embedding $q\mapsto q^{\ast }=\{r\in \mathbb{Q}:r<q\}$ (The real numbers $\mathbb{R}$ as Dedekind cuts).

Additive inverse. For a cut $A$, $-A:=\{p\in \mathbb{Q}:\mathrm{\exists }r\in \mathbb{Q},r>0,\text{ with }-p-r\mathrm{\notin }A\}\mathrm{.}$ Equivalently, $p\in -A$ iff there is a rational $s\mathrm{\notin }A$ with $s<-p$ (set $s=-p-r$; conversely $r=-p-s>0$). Intuitively $-p$ is bounded away from $A$ from below: some rational strictly beneath $-p$ already fails to lie in $A$.

Subtraction. $A-B:=A+(-B)$.

The sum $A+B$ is again a cut, and $(\mathbb{R},+)$ is an abelian group with identity $0^{\ast }$: closure, commutativity, associativity, and the identity law are Cut addition: $A+B$ is a cut, commutative and associative, with identity $0^{\ast }$, and existence of inverses is For a cut $A$, $-A$ is a cut and $A+(-A)=0^{\ast }$.

The $r>0$ slack in the definition of $-A$ is essential and is not cosmetic. Neither $\{-a:a\in A\}$ nor $\{-a:a\mathrm{\notin }A\}$ is a cut in general: the first need not be downward closed, and the second can acquire a greatest element. Excising the boundary rational (the "$-p-r$ with $r>0$" clause) makes $-A$ a genuine cut with no greatest element and forces the exact identity $A+(-A)=0^{\ast }$, not merely $A+(-A)⊊0^{\ast }$ (For a cut $A$, $-A$ is a cut and $A+(-A)=0^{\ast }$).