# $E$ is a nowhere dense set iff $\overline{E}$ is a nowhere dense set.

I know of several characterization for nowhere dense sets, but how do I prove this if I start with the definition that $E$ is nowhere dense in a topological space $X$ iff $X \setminus E$ is dense in $X$? The backward implication is easy to prove since $X \setminus \overline{E} \subseteq X \setminus E$. How do I prove the forward implication? Is it possible to prove using only the definition given?

• You don't, since it doesn't work for that definition. ​ ​ – user57159 May 13 '17 at 8:58

This is false as $\mathbb{Q}$ has a dense complement but its closure is $\mathbb{R}$ which is not nowhere dense, nor does it have a dense complement.
If you amend the definition to $E$ is nowhere dense if $X \setminus \overline{E}$ is dense, then this is correct, as $X\setminus \overline{E}$ is dense iff $\operatorname{int}(\overline{E})= \emptyset$, which is the usual definition. Then $\overline{E}$ is nowhere dense iff $E$ is,as $\overline{\overline{E}} = \overline{E}$.
The latter correct definition is also equivalent to $E$ is nowhere dense iff for every non-empty open set $O$ there is a non-empty open set $O'$ such that $O' \subseteq O$ and $O' \cap E = \emptyset$ (i.e. $E\cap O$ is not dense in $O$ in the subspace topology, which is the origin of its name: it's not dense in any open subset).
• Okay, I think I just realized that I have an error in my lecture notes. Does it work if I define it this way: $E$ is nowhere dense iff $X \setminus \overline{E}$ is dense in $X$? – Kurome May 13 '17 at 9:15
• If you define it that way, what do you think of $\overline{\overline{E}}$ ? – Max May 13 '17 at 10:08
• @Max $\bar{\bar{E}}=\bar{E}$. – egreg May 13 '17 at 10:35