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This SE question invites proofs of the fact that any open subset of $\mathbb R$ is a countable union of open intervals. The highest-voted answer there is of Brian M. Scott where they give the following argument:

Let $U$ be a non-empty open subset of $\Bbb R$. For $x,y\in U$ define $x\sim y$ iff $\big[\min\{x,y\},\max\{x,y\}\big]\subseteq U$. It’s easily checked that $\sim$ is an equivalence relation on $U$ whose equivalence classes are pairwise disjoint open intervals in $\Bbb R$. (The term interval here includes unbounded intervals, i.e., rays.) Let $\mathscr{I}$ be the set of $\sim$-classes. Clearly $U=\bigcup_{I \in \mathscr{I}} I$. For each $I\in\mathscr{I}$ choose a rational $q_I\in I$; the map $\mathscr{I}\to\Bbb Q:I\mapsto q_I$ is injective, so $\mathscr{I}$ is countable.

They then say that the

arguments generalize to any LOTS (= Linearly Ordered Topological Space)...

However, I don't agree that the argument indeed generalizes to arbitrary LOTS. For instance, consider $\mathbb Q$ under the topology due to the usual order. Now, take $U = (-\infty, \pi)_{\mathbb R}\cap\mathbb Q$. It is clear that there is only one equivalence class, namely $U$ itself. But $U$ is not an interval of the form $(-\infty, q)_{\mathbb Q}$ for some $q\in\mathbb Q$! (This problem does not arise in the case of $\mathbb R$ since $\mathbb R$ is order-complete so that all the convex sets are indeed intervals of the form $(a, b)$, $(-\infty, a]$, etc.)

Question: Is my concern valid, making the proof erroneous, and non-extensible to all the LOTS?

Atom
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    Yeah, you have to use the term "interval" more loosely, not implying endpoints that are in the LOTS. We often define interval in a linear ordering as just a subset $I$ of the set such if $x,y\in I$ with $x<y$ then any $z$ between $x,y$ are also in $I.$ This definition includes closed intervals and open intervals, of course (and also includes empty sets, and singletons, but you can add further conditions if you want to eliminate those cases.) it also doesn't require the "endpoints" to be i; the set. – Thomas Andrews Aug 17 '23 at 17:38
  • Why is it relevant that $U$ is not of the form $(-\infty, q)$ for some $q\in\mathbb Q$? –  Aug 17 '23 at 17:39
  • This definition doesn't require the "endpoints" to be in the set. But that is akin to the condition that allows infinite rays in $\mathbb R,$ since $\pm\infty$ are not in $\mathbb R.$ – Thomas Andrews Aug 17 '23 at 17:41
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    @StinkingBishop Cuz Brian says in their answer that the equivalence classes are intervals. – Atom Aug 17 '23 at 17:52
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    @ThomasAndrews So the topology being considered here is generated by the convex sets you talk of? If so, then $[a, b]$ will also be open, and then Brian's question definitely doesn't answer what was being asked in the post, namely, if open sets of $\mathbb R$ (under the standard topology, under which $[a, b]$ is not open) are countable unions of open sets. – Atom Aug 17 '23 at 17:54
  • Yes, your example shows that the statement in the last paragraph of that answer is (was) not correct, as it is not the comment by Peter L. Clark. The statement should be "disjoint union of intervals with endpoints in the completion" The proof works in $\mathbb{R}$, since this is complete. – NDB Aug 17 '23 at 18:30
  • No, the basis is still intervals of the form $(x,y)$ but that then generates all open intervals. For example, if $a_n$ is an increasing sequence of rationals that converge to $\pi,$ then $\bigcup (-n,a_n)$ is your interval. So $(-\infty,\pi)\cap \mathbb Q$ is open. – Thomas Andrews Aug 17 '23 at 19:41
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    The distinction is that $(-\infty,\pi)$ is not an interval of the LOTS $\mathbb{Q}$, but an interval of the completion. This distinction is important, for example, when defining the topology. One uses intervals of the ordered set, not intervals of its completion. And also in the linked statement. It proves a disjoint union of intervals of the completion, not intervals of the LOTS. – NDB Aug 17 '23 at 19:48
  • I deleted my answer after once more having read Brian M. Scott's answer. He allows intervals to have endpoints in the completion of the linear order on $X$. Thus your example $U = (-\infty, \pi)_{\mathbb R}\cap\mathbb Q$ is not a rational interval in the proper sense, but an interval in the "completion sense". – Paul Frost Aug 18 '23 at 14:26
  • @PaulFrost Yes, now he does. – Atom Aug 19 '23 at 05:56
  • @PaulFrost: That was an edit made by NDB. It works, but I prefer what I actually had in mind and should have made clear: in the general case one wants to talk about order-convex sets rather than intervals. – Brian M. Scott Aug 31 '23 at 06:56

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Here's a direct port of the argument in your link that applies to any separable LOTS. To do this, we need to define (as Thomas Andrews suggests in their comment) that an interval is a set such that $x,y\in I$ implies $[\min\{x,y\},\max\{x,y\}]\subseteq I$. This allows e.g. $(0,\pi)\cap\mathbb Q$ to be a subinterval of $\mathbb Q$ even though $\pi\not\in\mathbb Q$ (and avoids any discussion of completions).

Let $U$ be a non-empty open subset of a separable LOTS $X$ ordered by $<$ with a countable dense subset $Q$. For $x,y\in U$ define $x\sim y$ iff $\big[\min\{x,y\},\max\{x,y\}\big]\subseteq U$. It’s immediate that that $\sim$ is an equivalence relation on $U$ whose equivalence classes are pairwise disjoint [open] intervals in $X$. Let $\mathscr{I}$ be the set of $\sim$-classes. Then $U=\bigcup_{I \in \mathscr{I}} I$. For each $I\in\mathscr{I}$ choose $q_I\in I\cap Q$; the map $\mathscr{I}\to Q$ defined by $I\mapsto q_I$ is injective, so $\mathscr{I}$ is countable.

What I think isn't actually particularly clear, even in the original, but especially without assuming $X=\mathbb R$, is that these equivalence classes are open. For example, if $X=[0,1]\cup[2,3]\setminus\mathbb Q$ as a suborder of $\mathbb R$, and we consider the whole space to be our open set, then we have the equivalence classes $[0,1]$ and $[2,3]\setminus\mathbb Q$. Then to show $1$'s equivalence class is open, you must observe something like that the $X$-interval $(0.5,2)=(0.5,1]$ is a basic open neighborhood of $1$ which is a subset of $1$'s equivalence class $[0,1]$.

Openness certainly can be proven in general, but there is some work to be shown that was hand-waved in your link.

Steven Clontz
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  • $(0.5,2)$ is not an open nbhd of $1$ in $X$ at all in the order topology on the set $X$. It’s open in the subspace topology on $X$, but that’s irrelevant here. – Brian M. Scott Aug 31 '23 at 06:47