To prove or disprove $\operatorname{int}(\operatorname{cl}(A))\subseteq \operatorname{cl}(\operatorname{int}A))$. Let $(X,\tau)$ be a topological space, $A$ be a subset of $X$, $\operatorname{cl}(A)$ and $\operatorname{int}(A)$ denote the closure and interior of $A$, respectively. I want to find a relation between $\operatorname{int}(\operatorname{cl}(A))$ and $\operatorname{cl}(\operatorname{int}A))$. By an example I found that $\operatorname{cl}(\operatorname{int}A))\nsubseteq \operatorname{int}(\operatorname{cl}(A))$, but I could not prove or disprove $\operatorname{int}(\operatorname{cl}(A))\subseteq \operatorname{cl}(\operatorname{int}A))$. Any hint to prove or disprove it?
 A: Hint: Take $X$ the set of real numbers and $A$ the subset of rational numbers. 
A: We are going to prove the following property .
Property :
If $\,S\,$ is a subset of a topological space $\,X\,$ then
$$\operatorname{int}\!\big(\!\operatorname{cl}(S)\big)\subseteq\operatorname{cl}\!\big(\!\operatorname{int}(S)\big)\iff\delta(S)\subseteq\delta\big(\!\operatorname{int}(S)\big)\cup\delta\big(\!\operatorname{cl}(S)\big)$$
where $\,\operatorname{int}(\cdot)\,,\,\operatorname{cl}(\cdot)\,$ and $\,\delta(\cdot)\,$ are respectively the interior , the clousure and the boundary operators.
Proof :
First we will prove the logical implication $\implies.$
Since $\;\operatorname{cl}\!\big(\!\operatorname{cl}(S)\big)=\operatorname{cl}(S)\;$ and $\;\operatorname{int}\!\big(\!\operatorname{cl}(S)\big)\subseteq\operatorname{cl}\!\big(\!\operatorname{int}(S)\big)\;\,,$
it follows that
$\begin{align}
\delta(S)&=\operatorname{cl}(S)\setminus\operatorname{int}(S)=\operatorname{cl}\!\big(\!\operatorname{cl}(S)\big)\setminus\operatorname{int}(S)=\\
&=\big[\!\operatorname{int}\!\big(\!\operatorname{cl}(S)\big)\cup\delta\big(\!\operatorname{cl}(S)\big)\big]\setminus\operatorname{int}(S)\subseteq\\
&\subseteq\big[\!\operatorname{cl}\!\big(\!\operatorname{int}(S)\big)\cup\delta\big(\!\operatorname{cl}(S)\big)\big]\setminus\operatorname{int}(S)=\\
&=\big[\!\operatorname{cl}\!\big(\!\operatorname{int}(S)\big)\cup\delta\big(\!\operatorname{cl}(S)\big)\big]\setminus\operatorname{int}\!\big(\!\operatorname{int}(S)\big)\subseteq\\
&\subseteq\big[\!\operatorname{cl}\!\big(\!\operatorname{int}(S)\big)\setminus\operatorname{int}\!\big(\!\operatorname{int}(S)\big)\big]\cup\delta\big(\!\operatorname{cl}(S)\big)=\\
&=\delta\big(\!\operatorname{int}(S)\big)\cup\delta\big(\!\operatorname{cl}(S)\big)\;\;,
\end{align}$
hence ,
$\delta(S)\subseteq\delta\big(\!\operatorname{int}(S)\big)\cup\delta\big(\!\operatorname{cl}(S)\big)\;.$
Now we will prove the logical implication $\;\Longleftarrow\;.$
Since $\;\operatorname{cl}\!\big(\!\operatorname{cl}(S)\big)=\operatorname{cl}(S)\;$ and $\;\delta(S)\subseteq\delta\big(\!\operatorname{int}(S)\big)\cup\delta\big(\!\operatorname{cl}(S)\big)\;,$
it follows that
$\begin{align}
\operatorname{int}\!\big(\!\operatorname{cl}(S)\big)&=\operatorname{cl}\!\big(\!\operatorname{cl}(S)\big)\setminus\delta\big(\!\operatorname{cl}(S)\big)=\operatorname{cl}(S)\setminus\delta\big(\!\operatorname{cl}(S)\big)=\\
&=\big(\!\operatorname{int}(S)\cup\delta(S)\big)\setminus\delta\big(\!\operatorname{cl}(S)\big)\subseteq\\
&\subseteq\big[\!\operatorname{int}(S)\cup\delta\big(\!\operatorname{int}(S)\big)\cup\delta\big(\!\operatorname{cl}(S)\big)\big]\setminus\delta\big(\!\operatorname{cl}(S)\big)=\\
&=\big[\!\operatorname{int}\!\big(\!\operatorname{int}(S)\big)\cup\delta\big(\!\operatorname{int}(S)\big)\cup\delta\big(\!\operatorname{cl}(S)\big)\big]\setminus\delta\big(\!\operatorname{cl}(S)\big)=\\
&=\big[\!\operatorname{cl}\!\big(\!\operatorname{int}(S)\big)\cup\delta\big(\!\operatorname{cl}(S)\big)\big]\setminus\delta\big(\!\operatorname{cl}(S)\big)\subseteq\\
&\subseteq\operatorname{cl}\!\big(\!\operatorname{int}(S)\big)\;\;,
\end{align}$
hence ,
$\operatorname{int}\!\big(\!\operatorname{cl}(S)\big)\subseteq\operatorname{cl}\!\big(\!\operatorname{int}(S)\big)\;.$
