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?

  • $\begingroup$ You don't, since it doesn't work for that definition. ​ ​ $\endgroup$ – 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).

  • $\begingroup$ 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 $X$? $\endgroup$ – Kurome May 13 '17 at 9:15
  • $\begingroup$ If you define it that way, what do you think of $\overline{\overline{E}}$ ? $\endgroup$ – Max May 13 '17 at 10:08
  • $\begingroup$ @Max $\bar{\bar{E}}=\bar{E}$. $\endgroup$ – egreg May 13 '17 at 10:35
  • $\begingroup$ @egreg I know that, I was giving Kurome a hint $\endgroup$ – Max May 13 '17 at 10:36

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