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$A$ 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:

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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.

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  • $\begingroup$ For clarity, $T$ is the topology of open sets in $X$? $\endgroup$ – Robert Shore May 23 at 21:03
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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 $A$.

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Your condition for A to be nowhere dense is just that every open set has a point not in the closure of A;in other words that no open non empty set is a subset of the closure of A ,in other words that the closure of A has no (nonempty) interior

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