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.
 A: A point in X is said to be a boundary point of a set A if each neighborhood of the point intersects both A and X\A. If you replace A with the complement of A in the statement, you get the same statement. So very simply both the sets have the same boundary. 
A: 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 $A^c$.
A: Note that for a metric space $(X,d)$, and $A \subset X$, $bdy(A)$ is $cl(A) \cap$ $cl(X \backslash A)$. To show that $bd(A)$ is a subset of $bd(X \backslash A)$ (a similar argument can show the reverse inclusion), it really just suffices to show that $X \backslash (X \backslash A)$  is $A$.
Note that an equivalent definition of the boundary of A is
$cl(A) \backslash int(A)$. If you’re using this definition, I suggest you prove that the two definitions are indeed equivalent.
A: 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 $bd (A)=bd(A^c)$ .
