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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.

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  • $\begingroup$ I don't see why an interval $\{x\in P : a < x < b\}$ should be totally ordered.. $\endgroup$ – Pietro Majer Jan 27 '17 at 10:19
  • $\begingroup$ Intuitively speaking, the order-convergence topology tends to become disconnected quickly if you have a lot of incomparable elements. $\endgroup$ – Dominic van der Zypen Jan 27 '17 at 12:22
  • $\begingroup$ What about the lexicographic square? It does not contain open interval order isomorphic to $\mathbb R$. Is the order topology of the lexicographic square connected? $\endgroup$ – Taras Banakh Jan 27 '17 at 22:04
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    $\begingroup$ The definition of order-convergence topology is a bit hard for me to understand. If $P$ is totally ordered, does it reduce to the usual order topology? Is the answer to your question known in that case: i.e. is it true that any totally ordered set which is connected in the order topology contains an open interval order-isomorphic to $\mathbb{R}$? $\endgroup$ – Nate Eldredge Jan 28 '17 at 16:29
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    $\begingroup$ @NateEldredge I don't understand the order-convergence topology either, but the answer to your second question is no. There are nowhere-separable Aronszajn continua, for instance. $\endgroup$ – Ramiro de la Vega Jan 28 '17 at 17:23
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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 $\mathcal{F} \to x$.

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  • $\begingroup$ Very nice, thanks Ramiro! Despite having thought quite a bit about order-convergence, I still have trouble getting my head around it. Your answer is very helpful. $\endgroup$ – Dominic van der Zypen Jan 30 '17 at 20:47
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Since I haven't checked the details, this should really be a comment but I don't have sufficient brownie points. I think that the long line, also known as the Alexandroff line, might be a counterexample to your claim. In general, to get the real line via a collections of conditions involving the order, one requires some kind of cardinality restriction, e.g., separability in the topological or order-theoretical sense.

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    $\begingroup$ Aren't there points $a,b$ on the Alexandrov line such that $[a,b]$ looks like $[0,1]$, viewed just as ordered sets? $\endgroup$ – Dominic van der Zypen Jan 28 '17 at 14:53
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    $\begingroup$ @Dominic Yes, actually all of them. (But I guess your question was rhetorical.) $\endgroup$ – Mathieu Baillif Jan 30 '17 at 14:31
  • $\begingroup$ So - the Alexandroff line is definitely not a counterexample $\endgroup$ – Dominic van der Zypen Jan 30 '17 at 15:30

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