On most partially ordered sets, the order-convergence topology (defined below) is often highly disconnected, often even discrete or [extremally disconnected].[1](https://en.wikipedia.org/wiki/Extremally_disconnected_space) 


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.




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**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.



$\mathbb{R}$ and the order-convergence topology
I will show below that for any linearly ordered set $(L, \leq)$, the order-convergence topology coincides with the usual order topology. Thus the answer to the OP´s question is no, since there are linear continua (e.g. a nowhere-separable Aronszajn continuum) that do not contain copies of $\mathbb{R}$.


Fix $a \in L$. To show that $(-\infty,a)$ belongs to the order-convergence topology, let $x \in (-\infty,a)$ and let $\mathcal{F}$ be a filter with $\mathcal{F} \to x$. Since $x<a$ and $x=\bigwedge\cal{F}^u$, there must be an $F \in \mathcal{F}$ and $b \in F^u$ with $b<a$. But then $F \subseteq (-\infty,b] \subseteq (-\infty,a)$ and hence $(-\infty,a) \in \mathcal{F}$. In a similar fashion we can show that $(a, +\infty)$ is also open in the order-convergence topology and therefore we get that the order-convergence topology contains the usual order topology.


Now fix $U \subseteq L$ open in the order-convergence topology. To show that $U$ is open in the usual topology, let $x \in U$ and consider the filter $\mathcal{F}$ of all subsets of $L$ which have $x$ as an interior point in the sense of the order topology. We will be done if we show that $U \in \mathcal{F}$, and for this it is enough to show that $\mathcal{F} \to x$. If $x$ has an immediate successor or if $x$ is the maximum of $L$ then $(-\infty,x] \in \mathcal{F}$ and hence $\mathcal{F}^u=[x,\infty)$ so $\bigwedge\cal{F}^u=x$. Otherwise since $\mathcal{F}$ contains every interval of the form $(-\infty,a)$ with $a>x$, we have that $\mathcal{F}^u=(x,\infty)$ and again we get $\bigwedge\cal{F}^u = x$. In a similar way we can verify that $\bigvee \cal{F}^l=x$ and hence $\mathcal{F} \to x$.


