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