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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1The complement of a subset of $\Bbb R$ with vanishing derivative (i.e. no limit points) should always be a counterexample. – 2012-10-09
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1@anon: I've never heard the word derivative used this way. This is more usually called a discrete set, is it not? – 2012-10-09
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4Set which is equal to interior of its closure is called [regular open](http://www.proofwiki.org/wiki/Definition:Regular_Open). – 2012-10-09
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0@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... – 2012-10-09
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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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0Does there exist such counterexample with $A$ connected? It seems no, to me. – 2015-02-12
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10@Prism: Not on the line, but $$\{\langle x,y\rangle\in\Bbb R^2:0
is a connected example in the plane: it’s open, but the interior of its closure includes the origin. – 2015-02-12 -
0Of course, the punctured disk. Thanks a lot, Brian. You are too awesome :) – 2015-02-12
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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 $\leq\varepsilon$.