Proposition 3

Let $(E_\iota)_{\iota \in I}$ be a family of ordered vector spaces. For the product space $E = \prod_{\iota \in I} E_\iota$ to be a Riesz space (resp. a fully lattice-ordered space), it is necessary and sufficient that each of the spaces $E_\iota$ be a Riesz space (resp. a fully lattice-ordered space).

Let us restrict ourselves to examining the case of fully lattice-ordered spaces. Suppose that all of the $E_\iota$ are fully lattice-ordered; let $A$ be a nonempty subset of $E$ that is bounded above and let $a = (a_\iota)$ be an upper bound for $A$. For every $\iota \in I$, $\operatorname{pr}_\iota A$ is bounded above by $a_\iota$, hence has a supremum $b_\iota$ in $E_\iota$; it is clear that $b = (b_\iota)$ is the supremum of $A$ in $E$.

Conversely, suppose $E$ is fully lattice-ordered. Let $A_\kappa$ be a subset of $E_\kappa$ that is bounded above, $A'_\kappa$ the subset of $E$ formed by the $x = (x_\iota)$ such that $x_\kappa \in A_\kappa$ and $x_\iota = 0$ for $\iota \neq \kappa$. It is immediate that $A'_\kappa$ is bounded above in $E$, hence has a supremum $b = (b_\iota)$; by the definition of the product order relation, necessarily $b_\iota = 0$ for $\iota \neq \kappa$, and $b_\kappa$ is the supremum of $A_\kappa$, which completes the proof.