Proposition 13

— In a lattice ordered group $G$, let $(x_\alpha)$ be a family having an infimum (resp. supremum), and let $z$ be an arbitrary element of $G$; then the family $(\sup(z, x_\alpha))$ (resp. $(\inf(z, x_\alpha))$) has an infimum (resp. supremum) and

respectively.

Suppose that the family $(x_\alpha)$ has an infimum $y$. We will show that $\sup(z, y)$ is an infimum of the family $(\sup(z, x_\alpha))$.

Indeed $\sup(z, x_\alpha) = z + (x_\alpha - z)^+$ and by translating we can reduce to the case $z = 0$, that is we must show that $(x_\alpha^+)$ has an infimum equal to $y^+$. Since $y \leq x_\alpha$ we have $y^+ \leq x_\alpha^+$ for all $\alpha$ (VI, p. 12, Prop. 9, c)). Conversely, if $a \leq x_\alpha^+$ for all $a$, then it follows that $a \leq x_\alpha + x_\alpha^-$ (Prop. 9, a)); now $y^- \geq x_\alpha^-$ since $y \leq x_\alpha$; hence $a \leq x_\alpha + y^-$ for all $a$, that is $a \leq y + y^- = y^+$.

The other formula follows by changing to the opposite order relation.