Exercise 1, § 1

A topology $\mathfrak{T}$ on an ordered commutative group $G$ is said to be compatible with the ordered group structure of $G$ if it is compatible with the group structure of $G$ and if in addition the set $G_+$ of elements $x \geq 0$ in $G$ is closed with respect to $\mathfrak{T}$.

a) Let $\mathfrak{T}$ be a Hausdorff topology compatible with the ordered group structure of $G$, and let $\hat{G}$ be the completion of $G$ (with respect to $\mathfrak{T}$). In the group $\hat{G}$, the closure $P$ of $G_+$ is such that $y - x \in P$ is a pre-ordering compatible with the group structure of $\hat{G}$.

b) Let $\theta$ be an irrational number, and consider the ordering on $\mathbf{Q}^2$ for which the set of positive elements consists of $(0, 0)$ and all pairs $(x, y)$ such that $y - \theta x \geq 0$. With respect to this ordering, $\mathbf{Q}^2$ is a linearly ordered group, and the topology on $\mathbf{Q}^2$ which is the product of the topology of the rational line by itself is compatible with this ordered group structure. But on $\hat{G} = \mathbf{R}^2$, the relation $y - x \in P = \overline{G}_+$ is not an ordering.

c) On a linearly ordered group $G$, the topology $\mathfrak{T}_0(G)$ (Chapter I, § 2, Exercise 5) is the coarsest of all topologies compatible with the ordered group structure (distinguish two cases, according as the set of all $x > 0$ has or has not a smallest element). Every topology on $G$ which is compatible with the group structure of $G$ and is finer than $\mathfrak{T}_0(G)$ is compatible with the ordered group structure of $G$, and the mapping $x \to x^+$ is continuous with respect to such a topology. This mapping is uniformly continuous with respect to $\mathfrak{T}_0(G)$, but not necessarily with respect to a group topology $\mathfrak{T}$ strictly finer than $\mathfrak{T}_0(G)$ [see b)].