TITLE: The closure of A is closed in X
QUESTION [1 upvotes]: I am studying for my point-set topology test and want to see if I did this proof right. We are able to assume basic properties of closure...
A$\subset$X and (X,d) a metric space 
Prove that $\bar{A}$ is closed in X. 
This is what I have so far: 
Let x$\in$ $\bar{A}$. Then by definition $\exists$ $\epsilon$>0 such that the neighborhood around x when intersected with A is non-empty. By definition, A is closed as well.  
Where do I go from here?

REPLY [1 votes]: Recall that $\bar A$ is the union of $A$ and its limit points.  
Prove that the complement of $\bar A$ is open in X by contradiction, i.e. assume that $X -\bar A$ is not open, meaning that there is a point $y$ in $X -\bar A$ with the property that there is NO $\epsilon > 0$ such that if $x \in X$ and $d(x,y) < \epsilon$, then $x \in X - \bar A$.  What does this mean for $y$?
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