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.)

The condition you want to check is
\[
x \in \bar A \quad \Leftrightarrow \quad \text{for each open set $U$ containing $x$, $U \cap A \neq \emptyset$}
\]
This definition implies, among other things, that $A \subset \bar A$. Indeed, with the notation above we always have $x \in U \cap A$. Is it clear why this implies the remaining inclusion in your problem?
If you instead require that $U \cap (A - \{x\}) \neq \emptyset$ then you have defined the set $A'$ of limit points of $A$. We have $\bar A = A \cup A'$. Simple examples such as $A = \{0\}$ inside of $\mathbb R$, for which $\bar A = A$ but $A' = \emptyset$, can be helpful in keeping this straight.
