Proposition 9

— a) For each element $x$ of a lattice ordered group $G$ we have $x = x^+ - x^-$ and $\inf(x^+, x^-) = 0$.

b) For every expression of $x$ as the difference of two positive elements, say $x = u - v$, we have $u = x^+ + w$ and $v = x^- + w$ with $w = \inf(u, v)$. In particular if $\inf(u, v) = 0$ then $u = x^+$ and $v = x^-$.

c) The relation « $x \leq y$ » is equivalent to « $x^+ \leq y^+$ and $x^- \geq y^-$ ».

d) We have $|x| = x^+ + x^- \geq 0$.

e) For any $x$ and $y$ in $G$, we have the inequality $|x + y| \leq |x| + |y|$, and more generally $\left| \sum_{i=1}^n x_i \right| \leq \sum_{i=1}^n |x_i|$ for any finite family $(x_i)$ of elements of $G$.

f) For any $x$ and $y$ in $G$ we have $||x| - |y|| \leq |x - y|$.

We will prove a) and b) simultaneously. If $x = u - v$ with $u$ and $v$ positive, then $u$ a $x$, so $u \geq \sup(x, 0) = x^+$, and $w = u - x^+$ is positive. On the other hand we have

from which it follows that $x = x^+ - x^-$, and $v - x^- = w$. If $z \leq x^-$ then $z \leq x^+ - x$, and so $x \leq x^+ - z$; if also $z \leq x^+$, then $x^z - z$ is positive, and so $x^i \leq x^+ - z$ by definition of $x^i$. Hence we have $z \leq 0$, which implies $\inf(x^+, x^-) = 0$, whence by translation $\inf(u, v) = w$.

c) The relation $x \leq y$ implies $\sup(y, 0) \geq x$ and $\sup(y, 0) \geq 0$, hence $x^i \leq y^i$; similarly if $-y \leq -x$ we deduce $x^- \geq y^-$. The converse implication follows immediately from $x = x^+ - x$ and $y = y^+ - y$.

d) Since $x \leq x^+$ and $-x \leq x^-$, it is clear that

Conversely, if $a \geq x$ and $a \geq -x$ then we deduce from c) that $a^+ \geq x^+$, $a^+ \geq x^-$, $a^- \leq x^-$ and $a^- \leq x^+$; since $a^-$ is positive and $\inf(x^i, x^-) = 0$, the last two inequalities imply $a^- = 0$ and $a = a^+$; the first two inequalities then give $a \geq \sup(x^+, x^-)$, and $\sup(x^+, x^-)$ is equal to $x^+ + x^-$ by a) and by Prop. 7 of VI, p. 10.

e) Since $x$ and $y$ s $|x|$ and $y$, we have $x + y \leq |x| + |y|$; since $-x \leq |x|$ and $-y \leq |y|$, we have $-x - y \leq |x| + |y|$; whence the first inequality. The second follows by induction on n.

f) Replacing x and y by y and x - y in e), we obtain

similarly we have $|y| - |x| \leq |y - x| = |x - y|$; whence the stated result.