# 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...

• An infinite union? Is there anything stopping you from, say, writing $S=\bigcup_{s\in S}\{s\}$? – Milo Brandt Mar 9 '16 at 22:52
• I forgot to require countability, sorry. – Meitar Abarbanel Mar 9 '16 at 22:54
• Do you know how to show that every open set is a union of countably many open balls? – Eric Wofsey Mar 9 '16 at 22:57
• – Henricus V. Mar 9 '16 at 22:58

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 $x \in A_n\subset \cup_k A_k$.

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 $$X =\bigcap_n X_n$$

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.

• Only countably many closed balls are permitted. – Jonathan Y. Mar 9 '16 at 23:16
• I believe the main idea will also serve the concept of countable union, but I am a little confused as for how to arrive at it clearly enough. – Meitar Abarbanel Mar 9 '16 at 23:42
• I added countable part into the proof. It needs the open set is the union of countable open ball in $R^n$. – runaround Mar 9 '16 at 23:45