Is the closure of a set equal to the closure of the closure of that set? I know it seems silly, but is it possible for the closure of a set to have limit points that are not limit points of the original set?
 A: Yes. $\overline{\overline A}$ is the smallest closed set containing $\overline A$.   The smallest closed set containing $\overline A$ is $\overline A$ since $\overline A$ is closed.
A: The closure $\mbox{Cl}(A)$ of a a subset $A$ of a topological space is the intersection of all closed sets that contain $A$. If $A$ is closed, then $\mbox{Cl}(A) = A$, because $A$ is itself a closed set that contains $A$ and there can be no smaller closed set containing $A$. $\mbox{Cl}(A)$ is closed for any $A$, because it is the intersection of a family of closed sets. Hence $\mbox{Cl}(\mbox{Cl}(A)) = \mbox{Cl}(A)$.
A: Yes. The closure of a set $S$, also known as $\lim S$, is equal to its closure. That is, $\lim\lim S = \lim S$, so $\lim$ is an idempotent operation.
A: In terms of limit points, suppose $x\in X$ is a limit point of the closure $\overline{S}$ of the set $S$.  Then every open neighborhood $U$ of $x$ intersects $\overline{S}$, so we may choose $y \in U\cap \overline{S}$.  So $y$ is a limit point of $S$, and since $U$ is an open neighborhood of $y$, $U\cap S$ is nonempty.  That is, every open neighborhood of $x$ intersects $S$, so $x$ is a limit point of $S$.  So every limit point of $\overline{S}$ is also a limit point of $S$, meaning that $\overline{\overline{S}}\subset \overline{S}$.  Can you prove the reverse inclusion?
A: Yes. $\bar{A} = \bar{\bar{A}}$. We can show it, too, by showing that $\bar{A}$ contains all of its limit points (and hence equals its closure). Let $A'$ denote the set of all limit points of $A$, such that $\bar{A} = A \cup A'$.
Let $x$ be a limit point of $\bar{A}$. Then every open ball $B_r(x), r \gt 0$ contains either a point $p \in A$ or a point $p \in A'$, where $p \not= x$. 
Suppose $p \in A$. then $x$ is a limit point of $A$, hence $x \in \bar{A}$.
Now suppose $p \in A'$. Then $p$ is a limit point of $A$, so every ball $B_h (x), h \gt 0$ contains a point $y \in A, y \not=p$.
We need to show that $y \in B_r(x)$. We have that $d(x,p) \lt r$, so that $r - d(x,p) \gt 0.$ Set $h = r - d(x,p).$ Then $\exists y \in B_h(p)$, where $y \in A$,  by the nature of $p$ being a limit point of $A$. We have that $d(x,y) \le d(x,p) + d(p,y) \lt d(x,p) + h = d(x,p) + r - d(x,p) = r$. Thus $y \in B_r(x)$, so that every open ball about $x$ contains a point of $y$ of $A$. This makes $x$ a limit point of $A$, hence $x \in A' \subset \bar{A}$.
Thus $\bar{A}$ contains all of its limit points, so $\bar{\bar{A}} = \bar{A} \cup \bar{A}' = \bar{A}$, since $\bar{A}' \subset \bar{A}$.
A: By the definition of closure, A is a subset of Cl(A), thus it follows that that Cl(A) is a subset of the Cl(Cl(A)).
On the other hand Since Cl(A) is a subset of Cl(A) (as every set is a subset of itself) and Cl(A) is closed.
now since Cl(Cl(A)) is the smallest closed subset containing Cl(A) thus it is concluded that Cl(Cl(A)) is a subset of Cl(A).
Therefore Cl(Cl(A))= Cl(A).
