# 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.

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The complement of a subset of $\Bbb R$ with vanishing derivative (i.e. no limit points) should always be a counterexample. –  anon Oct 9 '12 at 12:07
@anon: I've never heard the word derivative used this way. This is more usually called a discrete set, is it not? –  Grumpy Parsnip Oct 9 '12 at 12:18
Set which is equal to interior of its closure is called regular open. –  Martin Sleziak Oct 9 '12 at 12:18
@JimConant It's used in the topology class I'm taking right now (alongside the alternate "derived set"). I'm not familiar enough with topology to know what's standard terminology here, which is why I put the definition of the term I was using in parentheses... –  anon Oct 9 '12 at 18:59
@anon: fair enough. +1 for the example itself. –  Grumpy Parsnip Oct 9 '12 at 22:38

## 2 Answers

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

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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 $\leq\varepsilon$.

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