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In the quiz of a class in MIT OCW, there is a T/F problem :

For any open subset $A \subseteq \mathbb{R}$, $\operatorname{int}(\overline{A})=A$?

The hompage of the class also provided a answer, and I saw the answer of the above.

The answer is False, because (the writer said) $\operatorname{int}(\overline{A})$ does not contain all isolated points of $A$.

But I think the reason is incorrect, because A is a open subset of $\mathbb{R}$. (Every open subset of $\mathbb{R}$ consists of interior points.)

Although the original answer of the above question may be False, but is the reason that writer said incorrect?

Michael Hardy
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    Yes, the reason is incorrect. If $A$ is open, then $\operatorname{int}(\overline{A})$ contains all isolated points of $A$ (and for $A \subseteq \mathbb{R}$, there are none, but that's not important). $\operatorname{int}(\overline{A})$ in general contains points of $\mathbb{R}\setminus A$, that's why the equality does not generally hold. – Daniel Fischer Oct 17 '13 at 17:49
  • You and Daniel are correct. In a question like this, a specific counterexample is probably better than an answer like the answerer of this question attempted. – Stefan Smith Oct 18 '13 at 00:20

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False. $A=(0,1)\cup(1,2)$, $A$ is open, but $\overline{A}=[0,2]$ and $\operatorname{int}(\overline{A})=(0,2)$. It may be the case that the real reason is, in fact, that $\operatorname{int}(\overline{A})$ may contain isolated points of the complement of $A$.

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