What subsets $A$ of $\mathbb{R}^2$ are such that $\partial(A)=\partial(\partial(A))$? What subsets $A$ of $\mathbb{R}^2$ are such that $\partial(A)=\partial(\partial(A))$?
Is it necessary for a set to have an empty boundary for this property to hold?
 A: A guess at what's going through your head:
It is important to distinguish from the topological notion of a border of a subset of some ambient topology from a somewhat more intuitive (and in some settings more useful) sense of a border that you might have in mind.
For example, consider a line segment $L \subset \Bbb R^2$ under the usual topology. The intuitive result might be that $\partial L$ should consist of the endpoints of the line segment.  However, with a careful application of the definition, you should find that $\partial L = L$.
Now, a hint towards classification: remember that $\partial B = \bar B \setminus \mathrm{int}(B)$.  What can we say if $B = \bar B$ (i.e. if $B$ is closed)? If $B = \partial A$, must it be closed?
A: It is necessary that the boundary have empty interior. It's also sufficient actually.
I assume the definition $\partial A =\overline{A}-A^\circ$.
Then we have 
$$\partial\partial A=\overline{\left(\overline{A}-A^\circ\right)}-\left(\overline{A}-A^\circ\right)^\circ=\left(\overline{A}-A^\circ\right)-\left(\overline{A}-A^\circ\right)^\circ$$
Since $\left(\overline{A}-A^\circ\right)^\circ$ is a subset of $\left(\overline{A}-A^\circ\right)$, if we have $\partial A=\partial\partial A$ then $\left(\overline{A}-A^\circ\right)^\circ=\left(\partial A\right)^\circ=\varnothing$.
Conversely, if $\left(\partial A\right)^\circ=\varnothing$ then $\partial\partial A=\partial A$.
A: The circle is a counterexample, as in fact if $S^1$ is the circle $\partial(S^1) = S^1$.
