Is every open set a disjoint union of open balls?

Question

I know that in $\mathbb R$, every open set is a disjoint union of open intervals, i.e., the basic open balls. Is there a similar result that holds for arbitrary metric spaces?

Suppose $X$ is a metric space and $U$ be any open set of $X$. Can I write $U$ as the following : $$U=\bigcup_n B_n\ \ s.t.\ \ \\I )B_n\cap B_m=\Phi,\text{ if } m\neq n \\\text{ and } \\ii)\ r(B_n)\le \epsilon \text{ where $r$ denotes the radius of a ball }$$ ? I can see it can be written as this union easily if I omit the condition $i) B_n\cap B_m=\Phi$ but what happens if I want to retain both conditions?

A little relaxation on the condition $ii)$ is that each ball does not need to have radius exactly $\epsilon.$ It's just the given $\epsilon$ is the highest possible radius of a ball, could be less anything.

So, is such a decomposition possible?

Do you require only a countable number of disjoint open balls? — MaudPieTheRocktorate, May 22 '17 at 10:11
@MaudPieTheRocktorate : ok. So ,if I say uncountable number of such open balls are acceptable then? Can it be possible if $U=\cup_{\alpha}B_{\alpha}:{\alpha\in \lambda }?$ — user118494, May 22 '17 at 10:22
Look at https://math.stackexchange.com/questions/195437/connected-metric-spaces-with-disjoint-open-balls — MaudPieTheRocktorate, May 22 '17 at 10:30

score 11 · Answer 1 · 2017-06-13T01:10:32.187

By definition, a connected open set cannot be written as the union of (more than one) disjoint open sets. In particular, a connected open set cannot be represented as a disjoint union of open balls unless it is an open ball.

The case of $\mathbb{R}$ is exceptional because every bounded connected open set in $\mathbb{R}$ is an open ball. This is not so in $\mathbb{R}^n$ for $n>1$, or pretty much any metric space other than $\mathbb{R}$.

Actually, even in $\mathbb{R}$ some open sets are not disjoint unions of open balls. For example, $(0,\infty)$ is not.

To look at this another way, the problem with trying to represent $U$ as the disjoint union of open balls $B_n$ is that you can't cover $U\cap \partial B_n$ with anything: an open ball that intersects this set will also intersect $B_n$. So, it's necessary that $U\cap \partial B_n = \emptyset$ for all $n$. This means $U$ has to be a very special kind of a set, nearly divided into open balls.

score 4 · Answer 2 · answered Jun 13 '17 at 01:23

This result is already false for $\mathbb{R}^2$. A disjoint union of open balls in $\mathbb{R}^2$ has the property that if it is connected it must consist of a single open ball. But e.g. $\mathbb{R}^2 \setminus \{ (0, 0) \}$ is not an open ball; in fact it is not even homotopy-equivalent to an open ball.

Is every open set a disjoint union of open balls?

2 Answers2