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.)
 A: 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: 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.
A: Suppose $x\in\overline{\overline{A}}$. Let $U$ be an open set containing $x$; we want to show that $U\cap A\neq\varnothing$. We know that $U\cap\overline{A}\neq\varnothing$, so there exists $y\in \overline{A}$ such that $y\in U$. But since $U$ is an open set that contains $y$ and $y\in\overline{A}$, then...
A: 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 $A$)
A: It's clear that $\overline{A}\subset{\overline{\overline{A}}}$.
So we only have to prove that $\overline{\overline{A}}\subset{\overline{A}}$.
Let $x\in\overline{\overline{A}}$, left to prove that $\forall{\epsilon\gt0}:U_{\epsilon}(x)\cap{A}\neq\emptyset$.
Assume we have a $\epsilon\gt{0}$, find a $y\in\overline{A}\cap{U_{\frac{\epsilon}{2}}(x)}$.  Since we have $y\in\overline{A},\text{ we can find } z\in{A}\cap{U_{\frac{\epsilon}{2}}(y)}$
$$\implies|z-x|\le|z-y|+|y-x|\lt\frac{\epsilon}{2}+\frac{\epsilon}{2}$$
$$\implies{z}\in{U_{\epsilon}(x)\cap{A}\neq\emptyset}$$
A: 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 < n < 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.
