If $A\subset\mathbb{R}^n$, show that int $A$ is open If $A\subset\mathbb{R}^n$ then int $A$ is open.
I am first trying to show that int $A$ is open. A set $X$ is open iff $X=$ int $X$. So if we want to show that int $A$ is open, by using that definition, we need to show that int $A =$ int(int$(A)$). This is where it confuses me, what is actually int(int$(A)$) ?
Is there any other way to approach this problem?
I also tried this:
Let $A\subset\mathbb{R}^n$ and consider any $x\in$ int $A$. Then $B_r(x)\subset A$ for some $r>0$ since $x$ is an interior point. My other doubt is that if I show that $B_r(x)\subset$ int $A$, is it helpful in showing the original statement?
Thanks in advance!
 A: You started correctly. Indeed try to show that $B_r(x) \subset \operatorname{int}(A)$, which then shows that $x \in \operatorname{int}(\operatorname{int}(A))$ as required.
To do so, for any $y \in B_r(x)$, find $r' > 0$ such that $B_{r'}(y) \subset B_r(x)$. (Draw a picture to understand the idea!)
A: int(int(A)) would be the collection of all interior points of int(A).  If you can show that if x is an interior point of int(A) it is also an interior of A you'd get int(A) $\subset$ int(int(A)) and as int(int(A)) $\subset$ int(A), int(A) = int(int(A)).
Hey, you asked!
It's easier to use the definition of open "A is open if every point of A is an interior point of A".  (It's the same thing but the wording is clearer for this example.)
If x $\in$ int(A) then x is an interior point of A.  So there exists a $N(x, \epsilon)$ that is a subset of A.  Let y $\in N(x, \epsilon)$.  Let $\delta < \epsilon - d(x,y)$ then for any point z $\in N(y, \delta)$, $d(x,z) \le d(x,y) + d(y,z) < d(x,y) + \delta = d(x,y) + \epsilon - d(x,y) = \epsilon$.  So all y $\in N(x, \epsilon)$ are limit points of A.  So $N(x, \epsilon) \subset int(A)$ so x is a limit point of int(A).  So every point of int(A) is a limit point of int(A).  So int(A) is open.  
A: Sometimes, it is good to see the definition from different angles. 
Interior of A: Interior of a set $A$ is the biggest open set contained in $A$. 
Or, you can think of $Int(A)$ is the union of all open sets contained in $A$. 
Look, form any of these two definitions $Int(A)$ is an open set is very clear. 
