Boundary in the topological space For a topological space $X$ and $A\subset X$ can topological boundary of $A$ contain an open set?
 A: Denote by $bd(S)$ the boundary of $S$.
Take $X=\Bbb{R}$ and $A=[0,1] \cap \Bbb{Q}$. Then $bd(A)=[0,1]$ which contains many open sets.
The statement turns true if we are speaking of open sets. Then if $A$ is an open set of $X$, we can write the disjoint decomposition $X=A \cup bd(A) \cup ext(A)$, where $ext(A)$ is the interior of $X\setminus A$. Suppose there exists $U \subset bd(A)$ open. Then pick $x \in U \subset bd(A)$. This makes $U$ an open neighborhood for $x$, which must intersect both $A$ and $ext(A)$, since $x$ is a boundary point. Contradiction with $U\subset bd(A)$.
A: Yes. Of course in general it contains the boring open set $\emptyset$. For a more exciting example, 
$\partial \mathbb{Q}=\mathbb{R}$.
Indeed, the boundary of a dense set with empty interior is the whole space.
A: Yes, the boundary $\partial A$ of a set $A$ can have a non-empty interior (the interior being the largest contained open set). This also means that for such sets the boundary of the boundary of $A$ is different than the boundary of $A$ (because the boundary is disjoint to the interior), that is, $\partial\partial A\ne \partial A$.
The interior of the boundary consists of all those points which have a neighbourhood in which both $A$ and its complement are dense.
Note however that the boundary of the boundary always has empty interior. This is because the boundary is always closed (it is the complement of the union of two open sets, the interior and the exterior of $A$), and the boundary of a closed set always has empty interior.
A consequence of this is that while for some sets, $\partial\partial A\ne\partial A$, we always have $\partial\partial\partial A=\partial\partial A$.
