The real numbers $\mathbb{R}$ as Dedekind cuts

The real numbers are defined to be the set of all Dedekind cuts of $\mathbb{Q}$ (Dedekind cut): $\mathbb{R}:=\{A\subseteq \mathbb{Q}:A\text{ is a Dedekind cut}\}\mathrm{.}$ Elements of $\mathbb{R}$ are written $A,B,C,\dots$; each is a subset of $\mathbb{Q}$ satisfying (C1)–(C3).

The rationals embed into $\mathbb{R}$ by the rational embedding: for $q\in \mathbb{Q}$ set $q^{\ast }:=\{r\in \mathbb{Q}:r<q\},$ the cut of all rationals strictly below $q$. Each $q^{\ast }$ is a Dedekind cut (Each rational cut $q^{\ast }$ is a Dedekind cut ↗), and $q\mapsto q^{\ast }$ sends $\mathbb{Q}$ into $\mathbb{R}$. The images of $0$ and $1$ are written $0^{\ast }$ and $1^{\ast }$; being cuts they lie in $\mathbb{R}$ and serve as its additive and multiplicative identities.

A cut $A$ is exactly the set of rationals lying below a putative real point; $\mathbb{R}$ is thus built by naming each point through the downward gap of rationals it determines. Where $\mathbb{Q}$ has a genuine rational point $q$, the cut $q^{\ast }$ recovers it, but the construction also admits cuts $A$ with no largest excluded rational and no rational boundary at all, such as $\{q:q<0\text{ or }{q}^2<2\}$. These are precisely the missing limits of $\mathbb{Q}$: the cut convention manufactures a real number wherever $\mathbb{Q}$ leaves a hole, which is why $\mathbb{R}$ is complete while $\mathbb{Q}$ is not.

The order on $\mathbb{R}$ is set inclusion, $A\le B:\iff A\subseteq B$ (Order on the Dedekind reals). That $q\mapsto q^{\ast }$ is an order-preserving ring embedding, and that its image is dense, is recorded in The rational cuts embed densely in $\mathbb{R}$, preserving sums, products, $0$, $1$ and the order; the arithmetic and order structure making $\mathbb{R}$ a complete ordered field is developed in The Dedekind reals form a totally ordered field and Dedekind completeness: the least-upper-bound property.