Showing that the closure of the closure of a set is its closure I have the feeling I'm missing something obvious, but here it goes...
I'm trying to prove that for a subset $A$ of a topological space $X$, $\overline{\overline{A}}=\overline{A}$.  The inclusion $\overline{\overline{A}} \subseteq \overline{A}$ I can do, but I'm not seeing the other direction.
Say we let $x \in \overline{A}$.  Then every open set $O$ containing $x$ contains a point of $A$.  Now if $x \in \overline{\overline{A}}$, then every open set containing $x$ contains a point of $\overline{A}$ distinct from $x$.  My problem is: couldn't $\{x,a\}$ potentially be an open set containing $x$ and containing a point of $A$, but containing no other point in $\overline{A}$?
(Also, does anyone know a trick to make \bar{\bar{A}} look right?  The second bar is skewed to the left and doesn't look very good.)

This depends on what is admissible. The closure of a closed set is itself. So just show the closure is closed.
Btw, remark on Kris's answer: The fact that closure of closure is closure is the same thing as saying either of these

*

*the intersection of all closed sets containing $A$ is equal to the intersection of all closed sets containing $\overline{A}$


*the intersection of all closed sets containing $A$ is equal to the intersection of all closed sets containing (the intersection of all closed sets containing $A$)
