Let $(X, \mathcal{T})$ be a topological space. Suppose $A \in \mathcal{T}$. We define the interior of $A$, the closure of $A$ and the boundary of $A$ as follows:
$\operatorname{int}A=\lbrace x \in A\mid A \in \gamma(x) \rbrace$, where $\gamma(x)$ denotes the neighborhood filter of $x$.
$\operatorname{cl}A=\lbrace x \in X\mid \forall V \in \gamma(x): V \cap A \neq \emptyset \rbrace$
$\partial A= \operatorname{cl}A \cap \operatorname{cl}(X\setminus A)$
How do you prove that $\operatorname{int}(\partial A) = \emptyset$? I already have that $\operatorname{int}(\partial A)=\operatorname{int}(\operatorname{cl}A \setminus A)$, but how do you continue from this point?
When A is open
int $\partial$A = int (cl A $\cap$ cl $A^c$)
= int cl A $\cap$ int cl A$^c$ = int cl A - cl int A
= int cl A - cl A = empty set.
The crutial step uses the important
self dual theorem of topological duality.
cl K$^c$ = (int K)$^c$; int K$^c$ = (cl K)$^c$