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I learned that the following two definitions of nowhere dense set are equivalent.

(1) $E$ is nowhere dense in $X$ if the closure of $E$ in $X$ has an empty interior.

(2) For any open set $V\subset X$, the set $V \cap E$ is NOT dense in $V$ (in the subspace topology).

I am able to show that (1) implies (2), but I am not able to show (2) implies (1).

My attempt: Assuming (2), I assumed to the contrary that there is an non-empty open set $J$ such that $J \subset closure(A)$. Then, I tried to show $J\cap E$ is dense in $J$, but I am stuck.

How should I proceed?

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Let $J$ be a nonempty open subset of $\bar{E}$. Then $J \subseteq \mathring{E}$. Take $V = J \cap \mathring{E} = J$. $V \cap E (= J)$ is dense in $V (=J)$, contradicting (2).

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  • $\begingroup$ Why is $V\cap E = J$? $\endgroup$ – nan Mar 8 '18 at 20:23
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    $\begingroup$ @nan $V \cap E = (J \cap \mathring{E}) \cap E = J \cap (\mathring{E} \cap E) = J \cap \mathring{E} = J$ $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Mar 8 '18 at 20:25
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    $\begingroup$ crystal clear! got it! $\endgroup$ – nan Mar 8 '18 at 20:29

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