Is a regular open set a scattered open set?

Question

Recall that an open set $U$ is regular if $\operatorname{int}(\operatorname{cl}(U))=U$.

Let us talk about the real numbers now. Given an open set, it can be written uniquely as the countable union of pairwise disjoint intervals. We say that an open set $U\subseteq\Bbb R$ is scattered if the midpoints of these intervals form a scattered subset of $\Bbb R$.

For example $(0,1)\cup(1,2)$ is an irregular open set (as $1$ witnesses) but it is well-ordered since $\{0.5,1.5\}$ is a subset of $\Bbb R$ which is well-ordered, and therefore scattered, in the natural order of the reals.

Is every regular open set a scattered open set? And are there natural conditions which guarantee that an open set is not scattered?

Answer 1

Let us construct open sets $A, B$ iteratively. We set $A_0$ to be an open interval of diameter $\tfrac14$ around each integer, and $B_0$ an open interval of diamater $\tfrac12$ around each integer. We define $A_i,B_i$ for $i \in \mathbb N$, so that they are, restricted an interval $[n,n+1]$, a union of finitely many open intervals. We define $A_{n+1}$ to be $A_n$ together with for each interval $[s,t]$ of length $l = [t - s]$ in the complement of $B_n$ the open set $(s+\frac l3, t-\frac l3)$, and we define $B_{n+1}$ to be $B_n$ together with, for such $[s,t]$, the interval $(s+\frac l4, t - \frac l4)$.

Now let $A = \cup_{i \in \mathbb N} A_i$, and we see that $A$ is regular, because the open sets all have "padding"; meanwhile, the order type of $A$ is clearly that of a dense unbounded linear order.

Answer 2

Construct pairwise disjoint open intervals $I_q\subset\mathbb R\ (q\in\mathbb Q)$ so that $q,r\in\mathbb Q,\ q\lt r\implies\sup I_q\lt\inf I_r.$

Partition $\mathbb Q$ into two disjoint dense subsets $\mathbb Q_0$ and $\mathbb Q_1.$

Define $U=\bigcup_{q\in\mathbb Q_0}I_q.$

It's clear that $U$ is open and not scattered.

A little consideration will show that $\operatorname{int}(\operatorname{cl}(U))=U.$