# How to show the that a set $A$ nowhere dense is equivalent to the complement of $A$ containing a dense open set?

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!

• Which direction do you find difficult: From "normal" to "alternate" or vice versa? – Moishe Kohan Jun 10 '14 at 20:46
• From normal to alternate. I just cant get the intuitive feel. Thanks! – user123276 Jun 10 '14 at 21:03

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
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.