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.
Does interior of closure of open set equal the set?
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real-analysis
general-topology
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0@anon: fair enough. +1 for the example itself. – 2012-10-09
2 Answers
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HINT: See what happens with $A=(0,1)\cup(1,2)$.
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0@Prism: You’re welcome — and thanks! – 2015-02-12
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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 $\leq\varepsilon$.