Boundary is equal to its closure Let $(X,T)$ be a topological space and let $S$ be a subset of $X$.  Prove $Bd(S) = Cl(Bd(S))$.
My initial thought is that the boundary is a closed set of points and the closure of a closed set is going to equal itself.
Am I approaching this problem correctly?
 A: Yes you are right, the boundary of $S$ is defined as
$$Bd(S)=\overline{S}\setminus Int(S)=\overline{S}\cap (X\setminus Int(S))$$ which is the intersection of two closed sets and thus closed.
Therefore, $Cl(Bd(S))=Bd(S)$ like what you claimed.
A: This can also be proved directly from the definitions of closure and boundary.  A point $Q$ is in $bd(S)$ if every open set containing $Q$ contains some point $x\in S$ and some point $y$ not in $S$. A point $P$ is in $cl(bd(S))$ provided every open set containing $P$ contains a point of $bd(S)$. Every set in $bd(S)$ has this property, so $ bd(S)\subset cl( bd(S))$. So one must then show that $cl(bd(S))\subset bd(S)$. But if $P\in cl(bd(S))$ then it must contain some point $Q$ in $bd(S)$ and therefore also some point $x\in S$ and some point $y$ not in $S$. Therefore, $P$ must be in $bd(S)$. So $cl(bd(s))=bd(S)$.
A: Yes you are right. You could also use the fact that $Bd(Bd(A))  \subseteq Bd(A)$ holds for all $A\in (X,T)$ as I do below:
To prove the equality we show that both sides are subsets of each other. 
$Bd(S)\subseteq Cl(Bd(S))$ clearly holds.
The other direction follows from $Cl(Bd(S)) = Bd(Bd(S)) \cup Bd(S)  \subseteq Bd(S) \cup Bd(S) = Bd(S)$
Thus the equality holds. 
