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Is it true in general that every open set in $\mathbb{R}^2$ is a union of countably many disjoint balls (allowing infinite radius)?

My thoughts so far:

This doesn't seem true in general. For example, if we take $\mathbb{R}^2 - \{0\}$, which is open, we exclude the possibility of having $B(\{0, 0\}, +\infty)$ cover our set. Thus, any ball in our covering of $\mathbb{R}^2 - \{0\}$ must have finite radius. However, it remains to show that any covering by disjoint balls, each with finite radius, is uncountable.

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So, you mean closed balls? I think $B(x,r)$ usually denotes an open ball. –  Stefan Hamcke Sep 8 '13 at 17:12
    
Sorry, I mean open balls. –  user93779 Sep 8 '13 at 17:13
    
I am used to $B(x,r)$ or $B_r(x)$ denoting an open ball. On the other hand, one can guess that you mean closed balls since open balls would not work due to connectedness. –  Stefan Hamcke Sep 8 '13 at 17:16
    
So, you are sure you mean open balls? –  Stefan Hamcke Sep 8 '13 at 17:17
    
No, but I agree with your reasoning if we assume that we're working with open balls. –  user93779 Sep 8 '13 at 17:19

1 Answer 1

No, even with uncountably many disjoint open balls it won't be possible. The reason is that if $\mathcal B$ is such a cover for the open set $U$, then take a ball $B\in\cal B$. This $B$ is open, but so is its complement as the union $\bigcup(\mathcal B\setminus\{B\})$ of all balls distinct from $B$. This implies that $U$ is disconnected, unless it can be covered by a single open ball.

Edit: Sorry, I realized that there can't even exist uncountably many disjoint balls. For if $\mathcal A$ is a family of disjoint sets, each with a non-empty interior, then we could choose an element $q_A$ in $\Bbb Q\times \Bbb Q\ \cap\ \text{int} A$ for each $A\in\cal A$. This gives an injection from $\mathcal A$ to $\Bbb Q\times\Bbb Q$, hence $\cal A$ can at most be countable.

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