Let $(X, \mathfrak T)$ be a topological space and let $A$ but a subset of $X$ then $Int(Bd(A)) = \emptyset$ Let $(X, \mathfrak T)$ be a topological space and let $A$ but a subset of $X$ then $Int(Bd(A)) = \emptyset$ 
I need to decide if this is true or not.  I have done a little research and some contemplating and I believe this statement is true for open sets but I am not sure if it is true for closed sets or if I am overthinking and do not need to consider both sets. I am wondering if I am overthinking and this is true and a proof by contradiction would work? 
My definition of interior is: Let $(X, \mathfrak T)$ be a topological space and let $A \subset X$ is the set of all points $x \in X$ for which there exists an open set $U$ such that $x \in U \subseteq A$.  
My definition of boundary is: Let $(X,\mathfrak T)$ be a topological space and let $A \subseteq X$. A point $x \in X$ is in the boundary of A if every open set containing $x$ intersects both $A$ and $X−A$.

Take $X=\Bbb R$ and $A= \Bbb Q$, then $\operatorname{Bd}(A)=\operatorname{Bd}(\Bbb Q)=\Bbb R$ so $\operatorname{Int}(\operatorname{Bd}(A))=\operatorname{Int}(\Bbb R)\neq \emptyset$, so the proposition doesn't hold for every set $A$.
