Let $(X,\tau)$ be a topological space. Recall that a subset $A\subseteq X$ is called regular open iff $A=\mathrm{Int}(\overline{A})$. Obviously it follows from the definition that a regular open set is open (being the interior of something). It is also clear that a clopen set is regular open. Now here's my question:
- Suppose that $(X,\tau)$ is a topological space with the property that every open set is regular open. That is, $\forall A\in\tau,A=\mathrm{Int}(\overline{A})$. Does this imply that every open set is also closed?
Clopen sets satisfy the definition of being regular open in a sort of trivial way in that both the interior and closure operators act as the identity. So a topology in which every set is clopen is, a fortiori, a topology with only regular open sets. I was wondering if there is any more interesting example. At the extreme end, one might wonder if there exists a connected topological space in which every open set is regular open (besides the indiscrete topology of course).
The problem is making sure that $\tau$ is closed under union and finite intersections. For example, the interval $(0,1)\subseteq \mathbb{R}$ is regular open, but the open set $(0,1)\cup (1,2)\subseteq \mathbb{R}$ is not because $\mathrm{Int}\Big(\overline{(0,1)\cup (1,2)}\Big)=(0,2)$. Any ideas for a proof/counter-example would be greatly appreciated!
