Closure of the interior of another closure Let $X$ be a topological space and let $A \subset X$.
Is it true that $\overline{\rm{Int}(\overline{A})}=\overline {A}$?
This question arose when I try to show$\overline{X-\overline{\rm{Int}(\overline{A})}}=\overline{X-\overline{A}}$ 
 A: The statement is false in general. Take $A$ to be a closed, nowhere dense set; then $\operatorname{cl}A=A$, but $\operatorname{cl}\operatorname{int}\operatorname{cl}A=\operatorname{cl}\operatorname{int}A=\operatorname{cl}\varnothing=\varnothing.$ In the space $\Bbb R^n$, for instance, any closed, discrete set provides a counterexample, as does any Cantor set.
Added: To show that 
$$\overline{X-\overline{\rm{Int}(\overline{A})}}=\overline{X-\overline{A}}\;,$$
note first that you already know that 
$$\overline{X-\overline{\rm{Int}(\overline{A})}}\supseteq\overline{X-\overline{A}}\;.$$
Suppose that $x\notin\operatorname{cl}\left(X\setminus\operatorname{cl}A\right)$, and let $V$ be an open nbhd of $x$ disjoint from $X\setminus\operatorname{cl}A$; $V\subseteq\operatorname{cl}A$, so $V\subseteq\operatorname{int}\operatorname{cl}A\subseteq\operatorname{cl}\operatorname{int}\operatorname{cl}A$, and therefore $V\cap (X\setminus\operatorname{cl}\operatorname{int}\operatorname{cl}A)=\varnothing$, i.e., $x\notin\operatorname{cl}(X\setminus\operatorname{cl}\operatorname{int}\operatorname{cl}A)$. It follows that $\operatorname{cl}(X\setminus\operatorname{cl}\operatorname{int}\operatorname{cl}A)\subseteq\operatorname{cl}\left(X\setminus\operatorname{cl}A\right)$, and we’re done.
