Does interior of closure of open set equal the set?

Would you help me to solve this question. Is it true that if A is open set then $A=\operatorname{int}(Cl(A))$ where Cl(A) denote the closure of A. I already prove that $A\subseteq\operatorname{int}(Cl(A)) $ only using definition of closure and interior, but have no idea about proving $\operatorname{int}(Cl(A))\subseteq A$ or give a counter example.

Solution 1:

HINT: See what happens with $A=(0,1)\cup(1,2)$.

Solution 2:

Let $\{r_n\}$ an enumeration of rational numbers and $O_{\varepsilon}:=\bigcup_{n=1}^{\infty}(r_n-\varepsilon 2^{-n},r_n+\varepsilon 2^{-n})$. It is an open dense set: hence the interior of its closure is $\Bbb R$ (for the usual topology). But $O_{\varepsilon}$ is "small", as its Lebesgue measure is $\leq\varepsilon$.