Let \(X\) be a topological space and \(A\subseteq X\). Then
\begin{equation*} \bar{A} = \Int A \cup \partial A. \end{equation*}
Proof
We have \(\Int A \subseteq \bar{A}\). Furthermore (0x677aec01) and (0x66a8bd2a) implies \(\partial A \subseteq \bar{A}\). Therefore \(\Int A \cup \partial A \subseteq \bar{A}\).
On the contrary let \(x\in \bar{A}\). Then every neighbourhood of it contains a point of \(A\). If there is a neighbourhood which is contained in \(A\), the point \(x\) is in \(\Int A\). Otherwise it is in \(\partial A\), due to the characterization of the boundary.