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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 2
    $\begingroup$ @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

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 $\mathcal{F} \to x$.

  • $\begingroup$ 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

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.

  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.