On most partially ordered sets, the order-convergence topology (defined below) is often highly disconnected, often even discrete or [extremally disconnected].1

However, the order-convergence topology is connected for $\mathbb{R}$ and coincides with the Euclidean topology.

Suppose that $(P,\leq)$ is a poset such that $|P|\geq 2$ and connected order-convergence topology. Does this imply that there are $a<b\in P$ such that $\{x\in P: a<x<b\}$ is order-isomorphic to $\mathbb{R}$? If yes, that would make $\mathbb{R}$ kind of a "primary" poset amongst the posets with connected order-convergence topology.

**Definition of the order-convergence topology.** Let $(P,\leq)$ be a poset. We define the *order convergence topology*, denoted by $\tau_o(P)$ on $P$. 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.

totallyordered.. $\endgroup$ – Pietro Majer Jan 27 '17 at 10:193more comments