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 $(P_i, \tau_o(P_i))_{i\in I}$?


Please have a look at "Topologies on products of partially ordered sets I,II,III" by Marcel Erne. (MR602017, MR602018, MR0631406 (82m:54028c))


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.