# 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? Jun 10, 2014 at 20:46
• From normal to alternate. I just cant get the intuitive feel. Thanks! Jun 10, 2014 at 21:03
• Using another definition of nowhere dense, the proof will be much easier, almost obvious. The definition: a set X is called nowhere dense iff each nonempty open set contains a nonempty open subset disjoint from X. Aug 24, 2022 at 12:06
• @Michael Sorry to bother up so late but May I know the proof using the definition that you mentioned ? Oct 18, 2023 at 12:49
• @user-492177 X is nowhere dense <=> each nonempty open set contains a nonempty open subset disjoint from X <=> each nonempty open set contains a nonempty open subset in the complement of X <=> Let B be the union of all such open sets in the last clause, then B is a dense open set contained in the complement of X. Oct 25, 2023 at 12:29

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*}
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.