Topology of open and closed sets Since any open set A is basically the interior of the set A itself, and the boundary of a set is basically the end points of the set being that it could be in the set A or the complement of A, then is A-BdA = IntA and is A-IntA= BdA
 A: I'm answering in the context suggested in the post (that $A$ is an arbitrary subset.). The title makes it a little confusing as to whether or not closed sets are involved.
No, the closure $\overline{A}$ is the disjoint union of the boundary $\partial A$ and the interior $A^\circ$.  It is perfectly possible for a boundary point to be outside of $A$, so you only have $A\setminus A^\circ\subseteq \partial A$.  Just consider the half-open interval $(0,1]$ in $\mathbb R$, for example. Of course, if $A$ is closed then you do have equality.
Somewhat paradoxically, the other equation does happen to be true: $A^\circ = A\setminus \partial A=\overline{A}\setminus\partial A$. This is something you can do as an easy exercise.
A: $\partial (A)=Fr(A)=Bd(A)=\bar A\cap \overline {A^c},$ where $A^c$ is the complement of $A.$
[ $Fr$ is for Frontier. The LaTex for $\partial$ is \partial$\;$.]
If $A$ is open then $A^c$ is closed, so $A^c=\overline {A^c}$, and so $$\bar A\setminus A=\bar A\cap A^c=\bar A\cap \overline {A^c}=\partial (A).$$ 
If $A$ is any subset of any space $X,$ the trio $Int(A),Int(A^c), \partial (A)$ are pair-wise disjoint and their union is $X,$ where $Int(A)$ denotes the interior of $A$, also written $int(A)$ or $A^o$.
