Suppose $(S,d)$ is a metric space and $E\subset S$. Prove that $E$ is nowhere dense if and only if $S\setminus \bar{E}$ is dense.
I have yet been able to prove the $\Leftarrow$ part. Suppose $E$ is not nowhere dense, this implies that $\bar{E}^\circ\neq\emptyset$. So, $\exists x\in\bar{E}$ and $\exists\epsilon>0$ such that $B(x,\epsilon)\in\bar{E}$. But this contradicts that $S\setminus \bar{E}$ is dense.
Now the $\Rightarrow$ part. Suppose $S\setminus\bar{E}$ is not dense, which implies that there is an non-empty open set $V$ such that $V\cap S\setminus\bar{E}=\emptyset$. This implies there exist $x\in V$ such that $x\in \bar{E}$, so there exists a sequence $(x_n)_{n \in\mathbb{N}}$ converging to $x$.
How can I proceed from here?
Best Answer
Indeed, if $S \setminus \overline E$ is not dense, then we find an open $V$ such that $V \cap S \setminus \overline E = \emptyset$. Now $$ V \cap S \setminus \overline E = S \setminus (\overline E \cup (S \setminus V)) $$ and we get $\overline E \cup (S \setminus V) = S$. Thus, $V \subseteq \overline E$.