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In a Hausdorff topological space, $(X, T)$, with non-empty subset $A \subset X$, how can we prove that an open set $U \in T$ has non-empty intersection with $A$ iff $U$ has non-empty intersection with $\overline{A}$?

The $\Leftarrow$ direction is obvious, but I'd be interested in seeing how we can prove the opposite relation holds.

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    Assume there is $x \in \bar{A} \cap U$. Since $U$ is open, what can you say? And what can you say considering $x$ is in the closure of $A$?2012-02-13
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    If $U \cap A = \phi$, then $A \subseteq U^c$, but $U^c$ is closed. What can you say about $\overline{A}$? Can you complete the argument?2012-02-13
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    @zulon: I have actually completed the argument in that direction, apologies. I mistyped the implication arrow in the wrong direction, I am looking for assistance with the $\Rightarrow$ case.2012-02-13
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    $A\subseteq \overline A$.2012-02-13

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This pretty much follows straight from the definition of closure.

Let's say $U$ and $\overline{A}$ meet at some point $x$. If $U$ does not intersect $A$, then $\overline{A} - U$ is a closed set containing $A$ but not containing $x$. This is a contraction from the definition of closure, so $U$ intersects $A$ as desired.

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    Thanks, I actually used the wrong implication direction, I'm looking for help with the $\Rightarrow$ case. I'll edit the post accordingly.2012-02-13
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    @MiamiMath: That direction follows because $A \subseteq \overline{A}$.2012-02-13
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    note also that Hausdorffness is not needed2012-02-13