TITLE: If $A$ is dense in $X$ then $X\setminus A$ is nowhere dense
QUESTION [1 upvotes]: Suppose $A$ is dense in $X$. Let $B^o$and $\bar{B}$ denote the interior ad closure of a set $B$. Then
$$\left(\overline{X\setminus A}\right)^o=(X\setminus A^o)^o=X\setminus \bar{A}=\emptyset$$
Hence $X\setminus A$ is nowhere dense. 
Is my proof correct? And is the converse true?

REPLY [2 votes]: Extending William Elliot's comment: Your second equation is wrong, as you can see in his example:
$$\left({\mathbb{R}\setminus\mathbb{Q}^o}\right)^o=\left(\mathbb{R}\setminus \emptyset\right)^o=\mathbb{R},$$
but $${\mathbb{R}\setminus\overline{\mathbb{Q}}}=\mathbb{R}\setminus\mathbb{R}=\emptyset.$$
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