# $A\subset (X,d)$ nowhere dense $\iff$ $(\overline{A})^o=\emptyset$

Show $A\subset (X,d)$ nowhere dense $\iff$ $(\overline{A})^o=\emptyset$.

My attempt: $A$ nowhere dense $\implies$ $(\overline{A})^c$ is dense in $X$. Then, $(A^c)^o$ is dense in $X$ (previously proven equivalence). Then, $\overline{((A)^c)^o}=X\implies\overline{((A)^c)^o}^c=\emptyset$. Then I have to show $\overline{((A)^c)^o}^c=(\overline{A})^o$. Am I going about this the right way?

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Maybe you should include your definition of nowhere dense. –  Sam Mar 4 '12 at 23:19
Here is convenient to specify what definition of nowheredense are you using. The claim in the title is often used as the definition of nowhere dense set. –  leo Mar 4 '12 at 23:21
$A$ nowhere dense $\implies \bar{A}^c$ dense in $X$. –  Emir Mar 5 '12 at 1:45
Suppose that $\overline{A}$ contains an open set $U$. But because $A$ is nowhere dense there is open $V\subseteq U$ so that $V\cap A =\emptyset$. Hence no point of $V$ is a limit point of $A$, and hence no point of $V$ is in the closure of $A$. This is a contradiction.