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?

  • $\begingroup$ Do you require only a countable number of disjoint open balls? $\endgroup$ – MaudPieTheRocktorate May 22 '17 at 10:11
  • $\begingroup$ @MaudPieTheRocktorate : yeah, that will do. $\endgroup$ – user118494 May 22 '17 at 10:12
  • $\begingroup$ @MaudPieTheRocktorate : ok. So ,if I say uncountable number of such open balls are acceptable then? Can it be possible if $U=\cup_{\alpha}B_{\alpha}:\{\alpha\in \lambda \}?$ $\endgroup$ – user118494 May 22 '17 at 10:22
  • 1
    $\begingroup$ Look at math.stackexchange.com/questions/195437/… $\endgroup$ – MaudPieTheRocktorate May 22 '17 at 10:30

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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.