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?
Best Answer
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.