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Let $X$ be a topological space, and let $A$ be a subset. Show that $\overline{A} = X\ ($the subset $A$ is dense in $X) \Leftrightarrow$ for every nonempty open set $U$, $A∩U\neq\varnothing$.

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What did you try? – Asaf Karagila Feb 6 '13 at 0:51
You can try a Proof by Contradiction, suppose there exists an open set $U \neq \varnothing$, such that $A \cap U = \varnothing$, then is $A$ still dense in $X$? – user49685 Feb 6 '13 at 1:04

HINT: For any $x\in X$, $x\in\operatorname{cl}A$ if and only if every open nbhd of $x$ contains a point of $A$.

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You have to prove that $e\in X\setminus\bar{A}$, if an only if there exists a neighbouhood $U$, for that $A\cap U= \varnothing $.

$\Rightarrow $ If $e\in X\setminus\bar{A}$, then it's clear that the nonempty set $X\setminus\bar{A}$ is open and contains $e$. So there exists a neighbourhood $U$ of $e$, s.t. $U\subset X\setminus \bar{A}$ and $A\cap U= \varnothing $.

$\Leftarrow $ Suppose that there is a neighbourhood $U$ of $e$ in $X$, s.t $A\cap U= \varnothing $, then we get $C= X\setminus U$ must be closed, $e\in X\setminus C$ and $A\subset C$. You know that $\bar{A}$ is the smallest closed set countaining $A$, so we have $\bar{A}\subset C$ and thus $e\in X\setminus \bar{A}$.

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Hint : $\overline{A}$ is the intersection of all the closed subsets containing $A$.

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