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 $(P\_i, \tau\_o(P\_i))\_{i\in I}$?



Product topology and order convergence topology
Please have a look at "Topologies on products of partially ordered sets I,II,III" by Marcel Erne. ([MR602017](http://www.ams.org/mathscinet-getitem?mr=602017), [MR602018](http://www.ams.org/mathscinet-getitem?mr=602018), [MR0631406](http://www.ams.org/mathscinet-getitem?mr=631406) (82m:54028c))


