Example

— 1) If A is any set, the space $\mathbf{R}^A$ of real-valued functions defined on A is fully lattice-ordered, the supremum in $\mathbf{R}^A$ of a family that is bounded above being its upper envelope (GT, IV, §5, No. 5).

2) Let F be any set; the space $\mathscr{B}(F)$ of bounded real-valued functions on F, equipped with the order structure induced by that of $\mathbf{R}^F$, is fully lattice-ordered. However, if F is a topological space, the space $\mathscr{C}(F)$ of continuous real-valued functions on F (equipped with the order structure induced by that of $\mathbf{R}^F$) is a Riesz space that is not in general fully lattice-ordered (cf. Exer. 13). Consider for example the case that $F = \mathbf{R}$; let I be the interval ]0, 1[, $\varphi_I$ the characteristic function of I, and let H be the set of continuous functions $x(t)$ such that $x \leq \varphi_I$; it is clear that H is bounded above in $\mathscr{C}(F)$. The function $\varphi_I$ is the upper envelope of the $x \in H$, but it is not their supremum in $\mathscr{C}(F)$, since $\varphi_I$ is lower semi-continuous but not continuous. Let us show that, in fact, H has no supremum in $\mathscr{C}(F)$; it suffices to prove that if $u$ is a continuous function such that $u \geq \varphi_I$, then there exists a continuous function $v \neq u$ such that $u \geq v \geq \varphi_I$. Now, $u(0) \geq 1$, therefore there exists a number $\alpha > 0$ such that $u(t) > 0$ for $-\alpha \leq t \leq 0$; if w is a continuous function that is zero outside of the interval ]$-\alpha, 0$ and is such that $0 < w(t) < u(t)$ on this interval, then the function $v = u - w$ meets the requirements.