Show that the boundary of a set equals the boundary of its complement $\newcommand{\bdy}{\operatorname{bdy}}$

I'm trying to show that $\bdy(A) = \bdy(A^c)$.

I know that $\bdy(A) = \operatorname{closure} A \setminus \operatorname{int}(A) = (\operatorname{int}(A^c))^c \setminus \operatorname{int}(A)$, but I don't know where to go from there.

Any help or hints would be very much appreciated.
If you know that $\mathrm{closure}(A)=(\mathrm{int}(A^c))^c$, then you also have  $\mathrm{closure}(A^c)=(\mathrm{int}(A))^c$ because $(A^c)^c=A$.  Therefore 

$$\mathrm{boundary}(A)=\mathrm{closure}(A)\cap(\mathrm{int}(A))^c = \mathrm{closure}(A)\cap \mathrm{closure}(A^c).$$

The last expression is symmetric in $A$ and $A^c$.
To show that the boundary of a set A is equal to the boundary of it complement , it suffices to show that there exist an element x in between the boundaries of $A$ and $A^c$ . 
So if $x$ belongs to boundary of $A$, 
This implies there exist $y$ and $z$ such that $y$ belongs to $A$ and $z$ belongs to $A^c$ .
This implies bd $(A)=x=bd (A^c)$. Hence $bd (A)=bd(A^c)$ .