# Prove or disprove: an open set in $\mathbb{R}^2$ is a union of disjoint open balls [duplicate]

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.

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

## 1 Answer

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 $$Q$$.

• 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. May 29, 2022 at 23:21
• 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. May 29, 2022 at 23:21
• Obviously, a duplicate question. May 29, 2022 at 23:27
• @GregMartin Thanks, corrected May 30, 2022 at 8:22
• @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. May 30, 2022 at 16:18