Let $(X, \tau)$ be a topological space and let $A \subset X$. The following are equivalent:
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(i) $A$ is nowhere dense: $\operatorname{int}(\overline{A}) = \varnothing$.
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(ii) The complement of the closure is dense: $\overline{X \setminus \overline{A}} = X$, i.e., the open set $X \setminus \overline{A}$ is [dense](/page/Dense%20Subset) in $X$.
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(iii) Every nonempty open set contains a nonempty open subset disjoint from $A$: for every nonempty $G \in \tau$, there exists a nonempty $H \in \tau$ with $H \subset G$ and $H \cap A = \varnothing$.
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(iv) Every nonempty open set contains a nonempty open subset disjoint from $\overline{A}$: for every nonempty $G \in \tau$, there exists a nonempty $H \in \tau$ with $H \subset G$ and $H \cap \overline{A} = \varnothing$.