Let $(P,\leq)$ be a poset. We define the *order convergence topology*, denoted by $\tau_o(P)$. By a *set filter* $\mathcal{F}$ on $P$ we mean a collection of subsets of $P$ such that:

- $\emptyset \notin \mathcal{F}$;
- $A, B\in \mathcal{F}$ implies $A\cap B\in \mathcal{F}$;
- $U\in \mathcal{F}$, $U'\subseteq P$ and $U'\supseteq U$ implies $U'\in \mathcal{F}$.

If $S\subseteq P$ we define $S^u= \{x\in P: x\geq s\text{ for all } s\in S\}$, and $S^l= \{x\in P: x\leq s\text{ for all } s\in S\}$. If $\cal{F}$ is a set filter, then we set ${\cal F}^u = \bigcup\{F^u: F\in \cal{F}\}$ and define ${\cal F}^l$ similarly. For $x\in P$ and ${\cal F}$ a set filter on $P$ we write $${\cal F}\to x \textrm{ iff } \bigwedge\cal{F}^u = x = \bigvee \cal{F}^l.$$

Then we set $\tau_o(P)=\{U\subseteq P: \textrm{ for any } x\in U \text{ and any filter }\mathcal{F} \text{ with } \mathcal{F}\to x \text{ we have } U\in \mathcal{F}\}$. It is not hard to verify that this defines a topology.

If $(P_i)_{i\in I}$ is a family of posets, is $\tau_o(\prod_{i\in I} P_i)$ equal to the product topology of the product of $(P_i, \tau_o(P_i))_{i\in I}$?