Playing with closure and interior to get the maximum number of sets Can you find $A \subset \mathbb R^2$ such that $A, \overline{A}, \overset{\circ}{A}, \overset{\circ}{\overline{A}}, \overline{\overset{\circ}{A}}$ are all different?
Can we get even more sets be alternating again closure and interior?

You can start by looking at $B = [0,1]^2\cap \Bbb Q^2$. In that case 
$$
\overline B = [0,1]^2\\
\overset{\circ} B = \emptyset\\
\overset{\circ}{\overline{B}} = (0,1)^2\\
\overline{\overset{\circ}{B}} = \emptyset
$$
If you let $A$ be the union of $B$ and, say, the square $[3, 4]^2$, this would let you tell the difference between $\overset\circ A$ and $\overline{\overset{\circ}{A}}$, since the latter contains $(3,3)$ and $(4,4)$, while the former does not.
