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I'm trying to prove the following result.

Prove or disprove: every open set in $\mathbb{R}^2$ is a union of countably many disjoint open balls $B_r (x)$, where we allow $r = \infty$.

The use of "disjoint" open balls is throwing me off. If $U \subset \mathbb{R}^2$ is an open set, then for every $u \in U$, I can find an open ball $B_{r_u} (u) \subset U$. Then $U = \bigcup\limits_{u \in U} B_{r_u} (u)$, and I've written $U$ as a union of disjoint open balls. I'm not sure how to "shrink" the radii of the open balls. I could perhaps add a condition that the radii become progressively decreasing, e.g., I start with some radius $r$, then take $\frac{r}{2}$, then $\frac{r}{3}$, and so forth.

I'd appreciate some help on how to show this.

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    $\begingroup$ An open rectangle cannot be written as a disjoint union of open balls by connectedness. $\endgroup$
    – Kavi Rama Murthy
    May 29, 2022 at 23:12
  • $\begingroup$ As an immediate consequence of Yuval's answer and the comment by @KaviRamaMurthy , the claim is false for all connected open sets. $\endgroup$
    – Matheus Andrade
    May 29, 2022 at 23:22
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    $\begingroup$ @MatheusAndrade Well, it is true for open balls. $\endgroup$
    – Theo Bendit
    May 29, 2022 at 23:31
  • $\begingroup$ @TheoBendit thanks for correcting me. I was wrong indeed. $\endgroup$
    – Matheus Andrade
    May 30, 2022 at 19:00

1 Answer 1

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Consider a countable collection $\{B_j\}_{j \ge 1}$ of disjoint open balls in the the open unit square $Q=(0,1)^2$. At most 4 of the points on the boundary of $B_1$ are outside $Q$, and none of the remaining balls $B_2, B_3, \ldots$ can cover any point of $\partial B_1$. Thus $\{B_j\}_{j \ge 1}$ cannot cover $Q$.

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  • $\begingroup$ It seems that the remaining balls can't even cover a single point of $\del B_1$. If you agree, I think modifying the answer in this way would make it clearer. $\endgroup$
    – Greg Martin
    May 29, 2022 at 23:21
  • $\begingroup$ Just as a complement: your answer actually proves the claim is false for any connected open set, since open sets always contain rectangles $(u, u + \varepsilon)^2$ for $\varepsilon > 0$ small enough. $\endgroup$
    – Matheus Andrade
    May 29, 2022 at 23:21
  • $\begingroup$ Obviously, a duplicate question. $\endgroup$
    – Kavi Rama Murthy
    May 29, 2022 at 23:27
  • $\begingroup$ @GregMartin Thanks, corrected $\endgroup$
    – Yuval Peres
    May 30, 2022 at 8:22
  • $\begingroup$ @MatheusAndrade: that generalization doesn't follow by this argument, because the balls would not then be constrained to lie inside the rectangle. And indeed, a ball itself can be covered by a countable collection of (one) open ball. $\endgroup$
    – Greg Martin
    May 30, 2022 at 16:18

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