# A set $A$ is nowhere dense if and only if every non-empty open set has a non-emtpy open subset disjoint from $\overline{A}$

$$A$$ is nowhere dense if $$\overline{A}$$ has empty interior. Show that $$A$$ is nowhere dense if and only if every non-empty open set in $$(X,T)$$ has a non-emtpy subset disjoint from $$\overline{A}$$.

Attempt:

$$(\leftarrow)$$

Let $$A$$ be nowhere dense. Then $$\overline{A}$$ contains no non-empty open set. Let $$U$$ be open in $$(X,T)$$. Then we must have $$U\cap(X-\overline{A})\neq \emptyset$$ or else $$\overline{A}$$ would contain an open set. So $$U\cap(X-\overline{A})=V$$ is a non empty subset of $$U$$ disjoint from $$\overline{A}$$.

$$\rightarrow$$ Let every $$U$$ open have a non empty subset disjoint from $$A$$. Say $$int(\overline{A})$$ is not empty and contains a point $$x$$. Then there must be $$V$$ such that $$x \in V \subset \overline{A}$$. But $$V$$ is open so it must contain a non empty subset disjoint from $$A$$ so we cannot have such a $$V$$ or $$x$$ and so $$A$$ is nowhere dense.

• For clarity, $T$ is the topology of open sets in $X$? Commented May 23, 2019 at 21:03

The proof is essentially correct. In the second step you might want to explain why $$V$$ does not have a non-empty open subset disjoint from $$A$$, you just claim it (though it is true).
because if $$y \in W \subseteq V \subseteq \overline{A}$$, then $$W$$ is an open neighbourhood of a point of $$\overline{A}$$ and so intersects $$A$$.