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.)
How do you define the closure of a set $A$? If it's not already your definition, it might be a useful thing to prove that the closure of a set is precisely the intersection of all closed sets containing it.
A set is closed iff every convergent sequence in the set has its limit in that set. The closure operation adds a set's limit points to the set. A closed set already contains its limit points, so repeating the operation doesn't change the answer (idempotent).



For Example:

The sequence 1/n; 0 &lt; n &lt; inf goes like this..

1, 1/2, 1/3, 1/4, ... , 1/465674564, ...

It's limit is 0, which is not part of the sequence, so closure would add the 0.

NOTE - The above example is just meant to illustrate how closure works.  As far as I know, this sequence is not a subset of any metric space.