TITLE: How to show the that a set $A$ nowhere dense is equivalent to the complement of $A$ containing a dense open set?
QUESTION [28 upvotes]: I was reading a textbook and saw an alternative formulation of nowhere dense. I am not sure how to prove this alternate formulation below:
The Normal Nowhere Dense Statement:
Let $X$ be a metric space. A subset $A ⊆ X$ is called nowhere dense in $X$ if the interior of
the closure of $A$ is empty, i.e. $(\overline{A})^{\circ} = ∅$. Otherwise put, $A$ is nowhere dense iﬀ it is contained in a closed set with empty interior. 
Alternate Formulation:
"Passing to complements, we can say equivalently that $A$ is nowhere dense iﬀ its complement contains a dense open set."
Does anyone know how I can prove this? It seems rather painfully straightforward but I am not sure how to show it exactly. Thank you!

REPLY [36 votes]: First, you should know that, for any $B\subseteq X$, $X\setminus\overline{B}=(X\setminus B)^\circ$ and that $X\setminus B^\circ=\overline{X\setminus B}$. Now
\begin{align*}
A\text{ nowhere dense }&\iff\left(\overline{A}\right)^\circ=\varnothing\\
&\iff X\setminus(\overline{A})^\circ=X\\
&\iff\overline{X\setminus \overline{A}}=X\\
&\iff\overline{(X\setminus A)^\circ}=X\\
&\iff (X\setminus A)^\circ\text{ is dense in }X\\
&\iff(X\setminus A)\text{ contains a dense open subset}.
\end{align*}
The last equivalence may not be so obvious if you're not very used to metric spaces. See below, if necessary:

If $(X\setminus A)^\circ$ is dense in $X$, then $(X\setminus A)^\circ$ is a dense open subset of $X\setminus A$.Conversely, if $(X\setminus A)$ contains a dense open subset $D$, then $D\subseteq (X\setminus A)^\circ$, so $(X\setminus A)^\circ$ is dense as well.