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Let A be a set in a metric space $(X,\rho)$. Prove that for each open set $O$ in $(X,\rho)$

$O$ $\cap$ $\bar{A}$ $\neq$ $\emptyset$ $\iff$ $O$ $\cap$ $A$ $\neq$ $\emptyset$

I'm trying to prove this directly as you can tell from below I'm probably nowhere close to doing so.

$\implies$ Let $x\in O \cap \bar{A}$ consider the open ball of radius $\epsilon$ centered at $x$, $B(x,\epsilon)$ Since $x\in O \cap \bar{A}$ $\implies x\in \bar{A}$ So x is a limit point of A

then $B(x,\epsilon) \cap A$ $\neq$ $\emptyset$

any help is appreciated thanks.

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    $\begingroup$ Right to left direction is immediate. For the other direction it suffices to consider a limit point of $A$ and use the definition. $\endgroup$ – Kal S. Jan 31 '17 at 21:13
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The "$\Leftarrow$" direction is immediate. Indeed, since every set is a subset of its closure, $A \subset \bar{A} \Rightarrow A \cap O \subset \bar{A} \cap O$.

To show the $"\Rightarrow"$ direction, let $x \in \bar{A} \cap O$. If $x \in A$, then we are done. If $x \notin A$, then $x$ is a limit point of $A$. Since $x \in O$ as well, there exists a neighborhood around $x$ contained in $O$. Since $x$ is a limit point of $A$, there exists a point $y$ in this neighborhood that is also in $A$. Thus $y \in A \cap O$ so that $A \cap O$ is nonempty.

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