A set $A\subseteq X$ is said to be nowhere dense if $\operatorname{Int}\left(\overline{A}\right)=\emptyset$, where the bar denotes closure.
I want to prove that $A$ is nowhere dense if and only if $A\subseteq \overline{X\backslash \overline{A}}$.
To prove sufficiency note that if $A$ is nowhere dense,
$$\emptyset=\operatorname{Int}\left(\overline{A}\right)=X\backslash \left(\overline{X\backslash \overline{A}}\right)$$
which implies
$$\overline{X\backslash \overline{A}}=X$$ and evidently $$A\subseteq X=\overline{X\backslash \overline{A}}$$
If, on the other hand, $A\subseteq \overline{X\backslash \overline{A}}$, then
$$X\backslash \overline{\left(X\backslash\overline{A}\right)}\subseteq X\backslash \overline{A} \subseteq X\backslash A$$
thus
$$\operatorname{Int}\overline{A}\subseteq X\backslash \overline{A} \subseteq X\backslash A$$
In order to complete the proof I would like one of the expressions on the right to be empty, but I can't see how that could happen.
And that's when I get stuck. How do I proceed? Am I barking up the wrong tree here? Is there a different and more simple approach to this?
Best Answer
For simplicity, I'll denote the interior of a subset $E$ by $E^\circ$. Recall that $\overline{X\setminus E}=X\setminus E^\circ$ for every $E\subset X$. Then, it follows that
$$\begin{align} A \textrm{ is nowhere dense} \quad &\Leftrightarrow \quad (\overline{A})^\circ = \varnothing \\ &\Leftrightarrow \quad X\setminus (\overline{A})^\circ = X \\ &\Leftrightarrow \quad \overline{X\setminus \overline{A}} = X \end{align}$$
So, the implication $(\Rightarrow)$ is already done (since $A \subset X$).
Now, let's see the other direction. By the previous observation, it is sufficient to show that $$A \subset \overline{X\setminus\overline A} \quad \Rightarrow \quad X=\overline{X\setminus\overline A}$$
Clearly, $\overline{X\setminus\overline A} \subset X$. Let $y\in X$. If $y\in \overline A$ then $y\in \overline{X\setminus\overline A}$ (since $A \subset \overline{X\setminus\overline A}$). If not, $y\in X\setminus\overline A$ and then $y\in \overline{X\setminus\overline A}$ (since $X\setminus\overline A \subset \overline{X\setminus\overline A}$).
In any case $y\in \overline{X\setminus\overline A}$, which means $X\subset \overline{X\setminus\overline A}$ as we want to show.