Any open set in $\Bbb{R}^n$ is a countable union of closed sets I don't know how to directly approach it. If we were looking at an open ball, WLOG $B(0,r)$ in $\Bbb{R}^n$, I suppose I would be looking sets of the form $\overline{B}_n(0,r-{1\over n})$ (supposing $r\ge 1$). But is there a way to generalize it to all open sets using balls? I am focusing on the Euclidean space, in Analysis and not in Topology, so I would appreciate if you could refer in terms of basic Topology and not advanced one...
 A: For any point $x$ in opens set of $R^n$, you can find an open ball $B_\epsilon(x)$ centered at $x$, so that $B_\epsilon(x)$ is in open set. Clearly closed ball $ \bar{B_{\epsilon/2(x)}} $ is in open ball $B_\epsilon(x)$. The union of all those closed balls is in the openset, and covers all points in the open set.
Here is for the countable union of closed balls:
First, we notice that, for each open ball, $B_\epsilon(x)$, we can represent the open ball as the union of closed balls $\bar{B}_{(1-\frac{1}{n+1})\epsilon}(x)$ for $n = 1, 2, ...$. Each open set in $R^n$ is the union of countable some open balls. We already know it is each open ball is the union of countable closed balls. So the open set in $R^n$ is the union of countable of closed balls as we choose. This proves it is the union of countable closed balls.
A: Hint By taking complements you need to show that every closed set is a countable intersection of open sets.
Let $X$ be any closed set.
Show that 
$$X_n= \bigcup_{x \in X} B_{\frac{1}{n}}(x) $$
is open and 
$$X =\bigcap_n X_n$$
A: The natural way to generalize your approach could be: 


*

*consider the border of your open set $\partial A=\overline A \backslash A$ 

*inflate it by $\frac 1 n$ defining $B_n=\cup_{x \in \partial A}B(x,\frac 1 n)$, that is open

*define $A_n=A\backslash B_n$, that are closed


Then if $x \in A$ you have a ball $B(x,r)$ inside $A$ and so for $\frac 1 n <r$ you have $x \in A_n\subset \cup_k A_k$.
