Show that $\overline{\overline{X}} = \overline{X}$ for all subsets $X$ of a metric space $S$. I have to prove this and I am not sure how to go about it. Help please
 A: Hint: $\overline X$ is the intersection of all closed sets containing $X$; in particular $\overline X$ is closed. Having that, what is the closure of a closed set?
A: $\newcommand{\cl}{\operatorname{cl}}$How you prove this will depend on what you already know about closed sets. If you know already know that for any set $A$,
$$\cl A=\bigcap\{F\subseteq X:F\supseteq A\text{ and }F\text{ is closed}\}\;,$$
then you should use Matemáticos Chibchas’s hint. If not, one approach is to prove the hint and then use it. Another, though, is to work directly with whatever definition you have.
For instance, you may have defined $\cl A$ to be the set of all points $x\in X$ such that every open nbhd of $x$ contains a point of $A$. Then in order to show that $\cl\cl A=\cl A$, you need to show that if $x\in X$, and $U\cap\cl A\ne\varnothing$ for every open nbhd $U$ of $x$, then $U\cap A\ne\varnothing$ for every open nbhd $U$ of $X$. To do this, suppose that $U\cap\cl A\ne\varnothing$ for every open nbhd $U$ of $x$, and let $U$ be any open nbhd of $x$. By hypothesis there is some $y\in U\cap\cl A$. Clearly $U$ is an open nbhd of $y$, and $y\in\cl A$ so $U$ must contain ... what?
A: The following supposes that your definition of $x \in \overline A$ is something to the effect of $B ( x ; \epsilon ) \cap A \neq \emptyset$ for all $\epsilon > 0$ (or, equivalently, for each $\epsilon > 0$ there is a $y \in A$ with $d ( x,y) < \epsilon$).  If so, the following fact may be helpful:

Given $x \in S$ and $\epsilon > 0$, for each $y \in B ( x ; \epsilon )$ there us a $\delta > 0$ such that $B ( y , \delta ) \subseteq B ( x , \epsilon )$.

