Example of a dense open subset that is not of the form $U \cup {\sim} U$ for $U$ regular If $U\subseteq X$ is an open set in a topological space $X$, let $\newcommand{\negation}{\mathop{\sim}}\negation U$ denote the largest open set $V$ such that $V\cap U = \varnothing$ (i.e., the set of points having a neighborhood disjoint from $U$, i.e., the interior of the complement of $U$, i.e., the complement of the closure of $U$).
Note that $U$ is dense iff $\negation U = \varnothing$, or equivalently, $(\negation\negation U) = X$.  At the other extreme, open sets such that $(\negation\negation U) = U$ (equivalently, those of the form $\negation V$ for some $V$ open) are known as “regular” open sets.
Clearly, every open set of the form $U \cup \negation U$ is dense.  Also, every dense open set is of this form (just take $U$ to be the set itself).  But can we do this with $U$ regular?  In other words:
Question: Is there an example of a dense open set in $X = \mathbb{R}^n$ that is not of the form $U \cup (\negation U)$ for $U$ regular open?  (Equivalently, not of the form $(\negation V) \cup (\negation\negation V)$ for $V$ open.)  Or is every dense open set of this form?
 A: Counterexample in every dimension $n\geq 2$. Let $D$ be the complement of a point $p$ in any dimension $\newcommand\R{\mathbb{R}}\R^n$ with $n\geq 2$. This is a dense open set, but I claim it cannot be represented as $U\cup\sim U$ for any regular open set $U$. Suppose toward contradiction that $D=U\cup\sim U$ for such a $U$. But $D$ is connected, and so we cannot write it as the union of two disjoint open sets, unless one of those is empty. If $U$ is empty, however, then $\sim U$ will be all of $\R^n$. And if $\sim U$ is empty, then we would need $D=U$, but this is not regular open.
But true in dimension $n=1$. In the line, every dense open set does have that form.
Theorem. Every dense open set $D\subseteq\R$ has the form $D=U\cup\sim U$ for some regular open set $U$.
Proof. We may write $D$ as the union of a disjoint collection of intervals. The order of $\R$ induces an order on those intervals. Let $U$ be the union of a collection of intervals, with no two of them adjacent in that ordering, and furthermore, not containing all the intervals in any nontrivial interval of intervals, such that the intervals not in $U$ also have no two adjacent and don't contain all the intervals in any nontrivial interval of intervals. We can form such a $U$ by taking every-other interval in any discrete chain of intervals in the induced order, and then in the densely ordered parts, apportioning some to $U$ and others not to $U$ in a way to fulfill the requirement. The set $U$ will thus be regular open, because of the generalized no-adjacency requirement, and $\sim U$ will be all the other intervals, since $D$ is open dense. So $D=U\cup\sim U$, as desired. $\Box$
