Is it possible for any open set $U \in \mathbb{R^n}$ to be written as countable union of open balls? For example this is true when $n=1$. I would like to know if there is similar result in higher dimensions. Thanks

  • $\begingroup$ In $\mathbb R$ it is even possible for any open set to be written as countable union of disjoint open intervals. I don't know what happens in higher dimensions, though ... $\endgroup$ Mar 19, 2013 at 20:14
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    $\begingroup$ @Stefan: Even in the plane you can’t necessarily do it with disjoint open Euclidean balls: you can’t get $(0,1)^2$, for instance, because of the corners. $\endgroup$ Mar 19, 2013 at 20:16
  • $\begingroup$ @BrianM.Scott Isn't it primarily because such a union would not be connected. I didn't think about connectedness earlier ;-) $\endgroup$ Mar 19, 2013 at 20:25
  • $\begingroup$ @Stefan: That’s a large part of it, but not the whole story: after all, there are non-connected open sets in the plane! $\endgroup$ Mar 19, 2013 at 20:27
  • $\begingroup$ @BrianM.Scott Oh, I see, that is exactly what you mean because you only use one single ball for each connected component ;-) $\endgroup$ Mar 19, 2013 at 20:27

1 Answer 1


Yes, because for each $n$ the space $\Bbb R^n$ has a countable base for the topology consisting entirely of open balls. For example, the open balls of rational radius whose centres are in $\Bbb Q^n$ are such a base.

  • $\begingroup$ They might cover the set. But can we find open balls whose union is exactly the given open set? $\endgroup$
    – chandu1729
    Mar 19, 2013 at 23:01
  • $\begingroup$ @chandu1729: Yes, of course: that’s an immediate consequence of the fact that they’re a base for the topology. $\endgroup$ Mar 19, 2013 at 23:04

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