For any set $A\subseteq\mathbb{R}^n$, we have $ \overline{A^{\circ}} = \overline{\overline{A^{\circ}}^{\,\circ}}$ I have to prove that for any set $A\subseteq\mathbb{R}^n$,
$$ \large\overline{A^{\circ}} = \overline{\overline{A^{\circ}}^{\,\circ}} $$
This is what I got so far: for any set $A$ I'm using these definitions:
Interior:
$$\exists r > 0\text{ such that }\{x \mid B_r(x) \subseteq A\}$$
Closure:
$$\{x \in \mathbb{R}\mid \exists (X_n) \subseteq A \land X_n \rightarrow x\}$$
Now what I don't get is, I think the right part of the equal, because, I have the interior of $A$, that is all the points that have a ball that is included in $A$, using this I know using the definition of closure that I can pick a sequence that converges to them (using $r$ and decreasing it with $\frac{r}{n}$,  $n\to\infty$ for example). But then I don't know how to take the interior of that, I mean what I'm getting at, is that the closure of the interior is the interior, and then the right part of the equation is trivial, as it is the same (m the interior of the interior is the interior, and its closure its the interior)
I think I'm missing something..

Observation 1: for $U$ open, $\bar{U}^o$ is the largest open set contained in $\bar{U}$, so since $\bar{U}$ contains the open set $U$ we have $U \subset \bar{U}^o$. Observation 2: for $C$ closed, $\bar{C^o}$ is the smallest closed set containing $C^o$, so since the closed set $C$ contains $C^o$ we have $C\supset \bar{C^o}$. 
Apply observation 1 to $U=A^o$ to get $A^o\subset \bar{A^o}^o$, and take closure on both sides to get $\bar{A^o}\subset \bar{\bar{A^o}^o}$. Now apply observation 2 to $C=\bar{A^o}$ to get the other inclusion.
