On any poset $(P, \leq)$ we can consider two different topologies that arise directly from the ordering relation.
1) Order convergence topolog $\tau_o(P)$ : 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.
2) Interval topology $\tau_i(P)$ : This is the topology generated by the subbasis $$ \Big\{P\setminus\{x\}^u: x\in P\Big\}\cup \Big\{P\setminus\{x\}^u: x\in P\Big\}.$$
Finally, we define an equivalence relation on ${\cal P}(\omega)$ by letting $A, B\subseteq \omega$ be equivalent if and only if their symmetric difference $(A\setminus B) \cup (B\setminus A)$ is finite. By $\newcommand{\Pfin}{{\cal P}(\omega)/(\text{fin})}\Pfin$ we define the collection of equivalence classes modulo the above equivalence relation, and we order it by the relation $\subseteq^*$ where for $A, B\in {\cal P}(\omega)$ we say that $[A]\subseteq^*[B]$ if $A\setminus B$ is finite. It is not hard to prove that this is well-defined (independent of the representative of the equivalence classes) and it is a partial order on $\Pfin$.
Question. Do we have $\tau_o(\Pfin) = \tau_i(\Pfin)$?
