The Galois connection between topological closure and topological interior [Update: I changed the question so that $-$ is only applied to closed sets and $\circ$ is only applied to open sets.]
Let $X$ be a topological space with open sets $\mathcal{O}\subseteq 2^X$ and closed sets $\mathcal{C}\subseteq 2^X$. Consider the pair of maps $-:\mathcal{O}\leftrightarrows\mathcal{C}:\circ$ where $-$ is the topological closure and $\circ$ is the topological interior. Under what conditions will it be true that for all $A\in\mathcal{O}$ and $B\in\mathcal{C}$ we have $$A\subseteq B^\circ \Longleftrightarrow A^-\subseteq B \,\,?$$
[Update: Darij showed that this condition holds for all topological spaces.] If this condition does hold then by general nonsense (the theory of abstract Galois connections), we obtain an isomorphism of lattices $$-:P\approx Q:\circ$$
where $P$ is the the lattice of sets $A\in\mathcal{O}$ such that $(A^-)^\circ=A$ and $Q$ is the lattice of sets $B\in\mathcal{C}$ such that $(B^\circ)^-=B$, where both $P$ and $Q$ are partially ordered by inclusion.
The existence of this lattice isomorphism makes me wonder: is there a nice characterization of the elements of $P$ and $Q$? Certainly not every open set is in $P$. For example, if $X=\mathbb{R}$ with the usual topology then the set $(0,1)\cup (1,2)$ is open, but $$(((0,1)\cup(1,2))^-)^\circ = ([0,2])^\circ = (0,2) \supsetneq (0,1)\cup(1,2).$$
[Update: I found the answer. See below.]

Apparently the elements of $P$ are called regular open sets and the elements of $Q$ are called regular closed sets. Interestingly, the lattices $P$ and $Q$ are Boolean with the function $A^\bot:=X-A^-$ acting as a complement on $P$ and the function $B^\top:=X-B^\circ$ acting as a complement on $Q$.
http://planetmath.org/regularopenset
I didn't find a full characterization of these sets, but apparently every convex open set in $\mathbb{R}^n$ is regular:
Is convex open set in $\mathbb{R}^n$ is regular?
