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Let $X$ be a topological space and let $A \subset X$. Is it true that $\overline{\rm{Int}(\overline{A})}=\overline {A}$?

This question arose when I try to show$\overline{X-\overline{\rm{Int}(\overline{A})}}=\overline{X-\overline{A}}$

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Are you sure that this is true? If I take $A = \{0\}$ inside of $X = \mathbf R$ then $\bar A = A$ has empty interior, which seems to contradict the statement. –  Dylan Moreland Jun 12 '12 at 1:41
    
It would be true if $A$ were open. –  Jonas Meyer Jul 11 '13 at 6:26
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1 Answer

up vote 3 down vote accepted

The statement is false in general. Take $A$ to be a closed, nowhere dense set; then $\operatorname{cl}A=A$, but $\operatorname{cl}\operatorname{int}\operatorname{cl}A=\operatorname{cl}\operatorname{int}A=\operatorname{cl}\varnothing=\varnothing.$ In the space $\Bbb R^n$, for instance, any closed, discrete set provides a counterexample, as does any Cantor set.

Added: To show that

$$\overline{X-\overline{\rm{Int}(\overline{A})}}=\overline{X-\overline{A}}\;,$$

note first that you already know that

$$\overline{X-\overline{\rm{Int}(\overline{A})}}\supseteq\overline{X-\overline{A}}\;.$$

Suppose that $x\notin\operatorname{cl}\left(X\setminus\operatorname{cl}A\right)$, and let $V$ be an open nbhd of $x$ disjoint from $X\setminus\operatorname{cl}A$; $V\subseteq\operatorname{cl}A$, so $V\subseteq\operatorname{int}\operatorname{cl}A\subseteq\operatorname{cl}\operatorname{int}\operatorname{cl}A$, and therefore $V\cap (X\setminus\operatorname{cl}\operatorname{int}\operatorname{cl}A)=\varnothing$, i.e., $x\notin\operatorname{cl}(X\setminus\operatorname{cl}\operatorname{int}\operatorname{cl}A)$. It follows that $\operatorname{cl}(X\setminus\operatorname{cl}\operatorname{int}\operatorname{cl}A)\subseteq\operatorname{cl}\left(X\setminus\operatorname{cl}A\right)$, and we’re done.

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Thanks! But then how to show the original equality? –  user31899 Jun 12 '12 at 1:53
    
As wrote above by @Dylan Moreland, the result is not true, in general. So the answer to your questions is 'no'. –  Sigur Jun 12 '12 at 1:56
    
@user31899: I’ve expanded the answer to include a proof of the missing half of the original equality. –  Brian M. Scott Jun 12 '12 at 2:06
    
I see,that's prove by contrapostion with nbh definition of closure, thanks! –  user31899 Jun 12 '12 at 2:17
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