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.)