Prove that $\text{int(intA)=int(A)}$? I want to prove that $\text{int(intA)=int(A)}$ (and we are in metric space). I have two questions regarding this. 
(1). I came up with this proof but don't know if it's correct or not. First I use one of the results in the book, which claims that "set $\text{A}$ is open if and only if $\text{A=intA}$. A set is an open set if every points in that set is an interior points, so $\text{int(A)}$ is an open set. So $\text{int(intA)=int(A)}$.
(2). Now, I'm thinking about using open balls to do this. It seems fairly easy to prove that $\text{int(intA)}\subset \text{int(A)}$. But I don't know how to go the other way around. 
 A: Your proof is fine. You might want to clean it up a little. 
You might say this: 


*

*$U = int(A)$ is open. 

*From a theorem in class, U is open if and only if $int(U) = U$

*Thus $int(U) = int(int(A)) = U = int(A)$. 
A: So you agree that int(int($A$)) $\subseteq \text{int}(A)$.
Suppose $x \in \text{int}(A)$.  Since $\text{int}(A)$ is open, there is some $\epsilon > 0$ such that $B(x, \epsilon) \subseteq \text{int}(A)$.  But this precisely means $x \in \text{int}(\text{int}(A))$ by the definition of the interior.  Thus, $\text{int}(A) \subseteq \text{int}(\text{int}(A))$, as desired.
Note: I used the characterization of the interior of a set $B$ as the set of elements of $B$ such that you can find a ball around the element contained in $B$.  Specifically, $\text{int}(B) = \{ x \in B \mid \exists \epsilon_{x} > 0 \text{ with } B(x, \epsilon) \subseteq B \}$. 
A: By definition (since you tag as general topology and not as metric spaces) $\operatorname{int}(A)$ is the union of all open subsets of $A$,
$$ \operatorname{int}(A)=\bigcup_{U\subseteq A, U\text{ open}}U.$$
Then $\operatorname{int}(\operatorname{int}(A))$ is the union of: all open sets $U$ that are contained in the union of all open sets contained in $A$. But as each such open set $U$ is contained in $A$, and any open subset contained in $A$ is also contained in the union $\operatorname{int}(A)$, we see that $\operatorname{int}(\operatorname{int}(A))$ is just the union of all open sets contained in $A$, i.e. it equals $\operatorname{int}(A))$.
A: If $\tau$ denotes the topology then $\operatorname{int}\left(B\right):=\bigcup\mathcal{U}_{B}$
where $\mathcal{U}_{B}:=\left\{ U\in\tau\mid U\subseteq B\right\} $. 
As a union of open sets $\operatorname{int}\left(B\right)$ is an open set itself. 
If $B$ is open then $B\in\mathcal{U}_{B}$ and consequently $$\operatorname{int}\left(B\right)=B$$
Applying that on $B=\operatorname{int}(A)$, which - as said - is open, gives: $$\operatorname{int}\left(\operatorname{int}\left(A\right)\right)=\operatorname{int}\left(A\right)$$
