The closure of A is closed in X 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? 
 A: 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 $y$?
A: Hint: assume that $x\not\in \overline{A}$, and prove that for some $\epsilon>0$, $B(x,\epsilon)\cap \overline{A}\neq\emptyset$. (I guess this hint is based on one definition of "closed", though of course there are others.) I don't know what definition of $\overline{A}$ you are using, but depending on that, this hint may or may not lead to the most natural proof of your claim.
I'm not sure how to fix your proof attempt, because you are simply restating a consequence of the definition of $\overline{A}$; it is true that for all $x\in\overline{A}$, there exists $\epsilon>0$ such that $B(x,\epsilon)\cap A\neq\emptyset$. However, the stronger claim that for all $\epsilon>0$, $B(x,\epsilon)\cap A\neq\emptyset$ is true, for $x\in\overline{A}$, so using simply the consequence is essentially weakining the definition of $\overline{A}$.
Hope this helps!
