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For the first direction I assume that $A$ is nowhere dense, let $U$ be a nonempty set hence $\overline A$ doesn't intersect $U$ so that $U$ intersects the complement of the closure of $A$ so $U$ is a subset of the complement of the closure of $A$ .. Now I am trying to find the final result but I couldn't .. I don't know if there is something wrong in my proof ??

For the other direction I assume that $U\setminus\overline A$ is nonempty . Then $x$ is in $U$ and $x$ is not in $\overline A$ so $U$ intersects the complement of the closure so $U$ doesn't intersect $\overline A$ and since $U$ is arbitrary it follows that $A$ is nowhere dense.. does the proof for this direction right??

José Carlos Santos
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S.N.A
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3 Answers3

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If $A$ is nowhere dense, let $U$ be a non-empty oepn set. If $U\setminus\overline A=\emptyset$, then $U\subset\overline A$. But then, sense $U\neq\emptyset$, since $U$ is open, and since $\mathring{\overline A}$ is largest open subset of $\overline A$, $\mathring{\overline A}\neq\emptyset$, which is impossible, because $A$ is nowhere dense.

On the other hand, is $A$ is not nowhere dense, then $\mathring{\overline A}\neq\emptyset$ is a non-empty open set and $\mathring{\overline A}\setminus\overline A=\emptyset$.

José Carlos Santos
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I prove it by contradiction. Suppose that $int(cl(A))\neq \emptyset,$ It means that there is $x\in int(cl(A)).$ Consequently, there is open set $U$ which $x\in U$, such that $x\in U \subseteq cl(A)$. Contradiction.

Maurten Erik
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Denoting $X$ the whole space we have to show that $A\subseteq X$ nowhere dense $\iff \forall U\subseteq X, U^\circ =U:U\setminus \overline A\neq \varnothing$.

$(\Rightarrow)$ Let $U$ an open subset of $X$. If $U\setminus \overline A=\varnothing$, then there is no point of $U$ that does not exist in $\overline A$, which means that every point of $U$ is in $\overline A$, thus $U\subseteq \overline A$, which means that $(\overline A)^\circ\neq\varnothing$ . This is a cotradiction, because $A$ is nowhere dense, so $(\overline A)^\circ=\varnothing$.

$(\Leftarrow)$ Let $A$ is not nowhere dense; then $(\overline A)^\circ\neq\varnothing$, which means that there is an open subset of $X$, let $U$, which is contained in $\overline A$. Then $U\setminus \overline A=\varnothing$, which contradicts to our main hypothesis. q.e.d.

SK_
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