Throughout this post, let $(P,\leq)$ be a poset. The interval topology $\tau_i(P)$ on $P$ is generated by $$\{P\setminus\downarrow x : x\in P\} \cup \{P\setminus\uparrow x : x\in P\},$$ where $\downarrow x = \{y\in P: y\leq x\}$ and $\uparrow x = \{y\in P: y\geq x\}$.

Now 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 $U'\supseteq U$ imply $A\cap B\in \mathcal{F}$.

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.

Question: Given any poset $(P,\leq)$, does $\tau_i(P)\subseteq \tau_o(P)$ hold?