Real Analysis – How to Show a Set A Nowhere Dense is Equivalent to the Complement of A Containing a Dense Open Set

Question

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!

Answer 1

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.