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...
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1An infinite union? Is there anything stopping you from, say, writing $S=\bigcup_{s\in S}{s}$? – Milo Brandt Mar 09 '16 at 22:52
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I forgot to require countability, sorry. – Meitar Abarbanel Mar 09 '16 at 22:54
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3Do you know how to show that every open set is a union of countably many open balls? – Eric Wofsey Mar 09 '16 at 22:57
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1See http://math.stackexchange.com/questions/318299/any-open-subset-of-bbb-r-is-a-at-most-countable-union-of-disjoint-open-interv – Henricus V. Mar 09 '16 at 22:58
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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.
runaround
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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 09 '16 at 23:42
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I added countable part into the proof. It needs the open set is the union of countable open ball in $R^n$. – runaround Mar 09 '16 at 23:45
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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 $$X =\bigcap_n X_n$$
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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 $x \in A_n\subset \cup_k A_k$.
Marco Disce
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