I'm trying to recreate the proof that: If $Y $ is a subset of a metric space $ X $, then the closure of $ Y $ is closed. But I cannot provide a proof of the following: Let $ x \in \overline{\overline Y} $so that $ x $ is adherent to $ \overline Y $ and $ r> 0$ then there exist a $ z \in B(x,r/2) \cap \overline Y $. How can I prove that $ z $ is adherent to $ Y $ ?

My definition of an adherent point is that: $ x $ is an adherent point if for all $ r$ the set the open ball $ B(x,r) \cap Y $ is nonempty. (Here Y is a subset of a metric space).


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    $\begingroup$ I think you mean $ B(x,r) \cap Y $ in the last sentence. $\endgroup$ – Kaj Hansen Oct 19 '14 at 7:02

Okay I got it, so The proof is as follows: Since $ z$ also belongs to $ \bar Y $ and this set is the collection of all the points that are adherent to $ Y $ then $z$ is adherent to $Y$


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