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?

Well, given $A$, we know that $\overline A$ is not equal to $A$ whenever $A$ is not closed, and we that ${\rm int}\, A\neq A$ whenever $A$ is not open (where ${\rm int}\, A$ is the interior of $A$ in $\mathbb R$). So the set we are looking for is neither open nor closed.
Now, one can quite easily construct a non-open, non-closed set in the plane. But can you arrange for the other two sets to be different? 
Note: arranging for ${\rm int}\, A=\emptyset$ is not good enough if you require $\overline{{\rm int}\, A}\neq{\rm int}\, A$. So the interior of $A$ must have some interior.
I leave you to think about the final two properties.
