Exercise 3, § 1

Let $G$ be a lattice-ordered commutative group, and let $M$ be the semigroup of major subsets of $G$. Every subset of $M$ which is bounded above (resp. below) has a least upper bound (resp. greatest lower bound), and $G$ can be canonically identified with a subset of $M$. Let $G'$ denote the largest subgroup of $M$ (the set of all symmetrizable elements of $M$).

a) Consider a topology $\mathfrak{T}$ on $G$ defined by the process of Exercise 2 a); let $V_\alpha$ be the set of all pairs $(z, z')$ in $M \times M$ such that

show that the $V_\alpha$ form a fundamental system of entourages of a uniformity $\mathfrak{U}$ on $M$, and that the topology $\mathfrak{T}'$ induced by $\mathfrak{U}$ induces $\mathfrak{T}$ on $G$. If $\mathfrak{T}$ is Hausdorff, then $\mathfrak{U}$ is a Hausdorff uniformity (remark that the elements of $M$, considered as subsets of $G$, are then closed), and the uniform space $M$ thus defined is complete. Show that the mapping $(z, z') \to z + z'$ of $M \times M$ into $M$ is uniformly continuous. Supposing that $\mathfrak{T}$ is Hausdorff, show that in $M$ the set of upper (resp. lower) bounds of any subset of $M$ is closed with respect to $\mathfrak{T}'$ (prove this first for the set of upper or lower bounds of an element of $M$, by considering the elements of $M$ as subsets of $D$), and that the mapping

of $M \times M$ into $M$ is uniformly continuous (same method).

b) Suppose from now on in this exercise that $\mathfrak{T}$ is Hausdorff. Then the closure $\overline{G}$ of $G$ in $M$ is a lattice-ordered group which is complete in the topology induced by $\mathfrak{T}'$. The group $G'$ is closed in $M$ (remark that its closure is a subgroup of $M$) and the topology induced by $\mathfrak{T}'$ on $G'$ is compatible with the ordered group structure of $G'$. For $G'$ to be equal to $\overline{G}$ it is sufficient that every non-empty open interval $]a, b[$ in $G'$ should contain an element of $G$, and this is always the case if $G$ is linearly ordered. In this case the topology induced on $G'$ by $\mathfrak{T}'$ is $\mathfrak{T}_0(G')$.

c) Take $G$ to be the ordered group $Q^2$, the product of the linearly ordered group $Q$ by itself; then $G' = M = R^2$ with the product ordering. Show that there are only three distinct non-discrete Hausdorff topologies on $G$ defined by the procedure of Exercise 2 a), and that they are obtained by taking the set of elements $c_\alpha$ to be either the set of pairs $(x, y)$ such that $x > 0$ and $y > 0$, or the set of pairs $(x, 0)$ such that $x > 0$, or the set of pairs $(0, y)$ such that $y > 0$. For the first of these three topologies we have $\overline{G} = G'$, although there are non-empty open intervals of $G'$ which contain no element of $G$; for the other two topologies, we have $\overline{G} \neq G'$. For each of these three topologies, the set of elements $z > 0$ in $G$ is not open.

d) Take $G$ to be the group $Q^2$ endowed with the lexicographic order (Set Theory, Chapter III, § 2, no. 6); $G$ is a non-Archimedean linearly ordered group. Then $\overline{G} = G' \neq M$; $M$ is linearly ordered but the topology $\mathfrak{T}'$ on $M$ is distinct from the topology $\mathfrak{T}_0(M)$; $G'$ is isomorphic to the group $R \times Q$ with the lexicographic order; in $G'$ the subgroup $H = R \times \{0\}$ is open and isolated, but on the quotient group $G'/H$ the quotient topology is distinct from $\mathfrak{T}_0(G'/H)$.