Exercise 30, § 1

Let G be an ordered group. For every subset A of G let $m(A)$ (resp. $M(A)$) denote the set of lower (resp. upper) bounds for A in G ; then $m(A) = -M(-A)$. a) The relation $A \subset B$ implies $m(B) \subset m(A)$ and $M(m(A)) \subset M(m(B))$. b) We have $A \subset M(m(A))$ and $M(m(M(A))) = M(A)$. c) If $(A_i)$ is an arbitrary family of subsets of G then $m\left( \bigcup_i A_i \right) = \bigcap_i m(A_i)$. d) Every set $M(B)$, where B is a nonempty subset bounded above in G, is called a major set in G. For any nonempty subset A bounded below in G, the set $M(m(A))$ is the smallest major set containing A ; denote it $(A)$ ; if $A \subset B$ then $(A) \subset (B)$ ; if A has an infimum a in G, then $(A) = M(a)$, which we also write (a). e) If A and B are two nonempty subsets bounded below in G, then $(A + B) = ((A) + B)$. Deduce that, in the set $\mathfrak{M}(G)$ of major sets in G, the map $(A, B) \mapsto (A + B)$ is an associative and commutative composition law, for which $(0) = P$ is an identity, that the order relation $A \supset B$ is compatible with this law, and that $\mathfrak{M}(G)$ is a completely semi-lattice ordered monoid (Ex. 29) under this structure if G is filtered. Moreover, the map $x \mapsto (x)$ from G to $\mathfrak{M}(G)$ is an isomorphism from the ordered group G onto a subgroup of the monoid $\mathfrak{M}(G)$. f) If an element A of $\mathfrak{M}(G)$ is invertible under the composition law of this monoid, then its inverse is $M(-A) = -m(A)$ (notice that if B is the inverse of A, then the relations $B \subset M(-A)$ and $A + M(-A) \subset (0)$ hold, whence $\langle A + B \rangle = \langle A + M(-A) \rangle$). g) For an element A of $\mathfrak{M}(G)$ to be invertible, it is necessary and sufficient that $x + A \subset A$ imply $x \geq 0$ (write down an expression for $0 \in (A + M(-A))$).