Proposition 1

The relation $y - x \in \overline{\mathbf{Q}}_+$ is an ordering on $\mathbf{R}$ which makes $\mathbf{R}$ into a linearly ordered set; is compatible with the additive group structure on $\mathbf{R}$, and induces the ordering $x \leq y$ on $\mathbf{Q}$.

We begin by showing that the relations $y - x \in \overline{\mathbf{Q}}_+$ and $z - y \in \overline{\mathbf{Q}}_+$ imply $z - x \in \overline{\mathbf{Q}}_+$. Indeed, the function $x + y$ is continuous on $\mathbf{R} \times \mathbf{R}$, and therefore by (8) we have $\overline{\mathbf{Q}}_+ + \overline{\mathbf{Q}}_+ \subset \overline{\mathbf{Q}}_+$ (Chapter I, § 2, no. 1, Theorem 1). Next, we shall show that the relations $y - x \in \overline{\mathbf{Q}}_+$ and $x - y \in \overline{\mathbf{Q}}_+$ imply $x = y$; this will establish that $y - x \in \overline{\mathbf{Q}}_+$ is an ordering on $\mathbf{R}$. It is enough to show that $\overline{\mathbf{Q}}_+ \cap (-\overline{\mathbf{Q}}_+) = \{0\}$. Now the functions $x \to x^+$ and $x \to x^-$ are uniformly continuous on $\mathbf{Q}$ and can therefore be extended by continuity to $\mathbf{R}$ (Chapter II, , no. 6, Theorem 2); let $f$ and $g$ be their respective extensions. By extension we have $x = f(x) - g(x)$ for all $x \in \mathbf{R}$; if $x \in \overline{\mathbf{Q}}_+$ then $g(x) = 0$, and since $-\overline{\mathbf{Q}}_+$ is the closure of $-\mathbf{Q}_+$ by the continuity of $-x$, it follows that if $x \in -\overline{\mathbf{Q}}_+$ then $f(x) = 0$. Hence if $x \in \overline{\mathbf{Q}}_+ \cap (-\overline{\mathbf{Q}}_+)$ we have $f(x) = g(x) = 0$ and therefore $x = 0$

By (10), we have $\overline{\mathbf{Q}}_+ \cup (-\overline{\mathbf{Q}}_+) = \mathbf{R}$, and hence $\mathbf{R}$ is linearly ordered by the ordering $y - x \in \overline{\mathbf{Q}}_+$.

Furthermore, since the relations $y - x \in \overline{\mathbf{Q}}_+$ and $(y + z) - (x + z) \in \overline{\mathbf{Q}}_+$ are equivalent, the ordering $y - x \in \overline{\mathbf{Q}}_+$ is compatible with the additive group structure of $\mathbf{R}$.

Finally, if $x$ and $y$ belong to $\mathbf{Q}$ the relations $y - x \in \overline{\mathbf{Q}}_+$ and $y - x \in \mathbf{Q}_+$ are equivalent, and therefore the relation $y - x \in \overline{\mathbf{Q}}_+$ induces the relation $x \leq y$ on $\mathbf{Q}$. This completes the proof.

The relation $y - x \in \overline{\mathbf{Q}}_+$ is again denoted by $x \leq y$. The set $\overline{\mathbf{Q}}_+$ is the set of all $x \geq 0$ in $\mathbf{R}$ and is denoted by $\mathbf{R}_+$; it is a closed set. The set of all $x > 0$ is denoted by $\mathbf{R}_+^*$; it is the complement of $-\mathbf{R}_+$ and is therefore open in $\mathbf{R}$.