Can the boundary of a subset be open? Let $X$ be a topological space and $A$ a subset of $X$. Can $\partial A$ be an open subset of $X$?
Of course the answer is no if $A$ were closed; that is $\partial A\subset A$, because in this case we would have $int(\partial A)\subset int(A)$ hence by definition of $\partial A$ no point of $\partial A$ can be in $int(\partial A)$ because in that case we would have a point in $\partial A$ which is also in $int(A)$, something that is not allowed. 
Now what if $A$ is not closed, that is there  are boundary points in $X$ to $A$ that are not contained in $A$?  

Recall that the boundary $\partial A$ is defined to be $\overline A - \operatorname{int}(A) = \overline A \cap \operatorname{int}(A)^c$. Since $\operatorname{int}(A)$ is open, its complement is closed. Hence the boundary of any set is an intersection of closed sets, which is closed.
Edit: As pointed out in the comments, I haven't addressed your question. As I'm sure you're aware, sets can be both open and closed in a topological space. In fact, there are always at least two sets that are both open and closed: the entire space and the empty set. With this in mind, consider $\mathbb Q \subset \mathbb R$. 
