Suppose we have a metric space in which every open set is expressible as a countable union of balls. Is this space necessarily separable?

Thank you.

  • $\begingroup$ A starting point might be that $X$ is separable iff any open cover has a countable subcover, but I'm not quite sure how to proceed. $\endgroup$ Commented Dec 26, 2011 at 21:40
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    $\begingroup$ @MihaHabič a large discrete space is not a counterexample; an uncountable number of points cant necessarily be expressed as countable union of open balls. $\endgroup$
    – yoyo
    Commented Dec 26, 2011 at 22:08

1 Answer 1


If the space has at most countably many isolated points, then the answer is yes. To see this, argue as follows. For each rational $r\gt 0$, let $E_r$ be a maximal subset of the space of pairwise distance between points at least $r$. Such a set exists by Zorn's lemma. The complement of $E_r$ is open, and thus is a countable union of balls. By the maximality of $E_r$, we may assume that all of these balls have radius at most $r$. Let $D_r$ be the centers of such a choice of balls, together with the isolated points of $E_r$. Notice that every point of the space has distance $r$ to a point in $E_r$, which is either isolated, or which is very close to points in the complement of $E_r$, which are within $r$ of a point in $D_r$. So every point is within $3r$ of a point in $D_r$. It follows that $D=\bigcup_r D_r$ is a countable dense subset of the space, which is therefore separable.

If the continuum hypothesis holds, then the answer to the question is yes. The reason is that under CH, the comment of Srivatsan below shows that the space can have at most countably many isolated points. To see this, suppose that there were uncountably many isolated points. Let $S$ consist of $\omega_1$ many of them. Every subset of $S$ is open, and hence by hypothesis is a countable union of balls. Thus, every subset of $S$ is determined by a countable subset of $S$, the centers of the balls, and the radii to be used for the ball at each such center. Thus, we have a one-to-one function from $2^{\omega_1}$ into $(\omega_1\times\mathbb{R})^\omega$. The latter set has size continuum, and the former set has size $2^{\omega_1}$, which under CH is larger than the continuum, a contradiction. So there can be at most countably many isoated points, and so the argument at top establishes an affirmative answer to the question under CH.

The argument actually uses somewhat less than CH, and it suffices merely that $2^{\omega_1}$ is strictly larger than the continuum, a hypothesis that follows from but is strictly weaker than CH.

Update. This question recently came up on MathOverflow, and the need for any CH-type assumption has now been omitted. See my answer there, and also the argument of Ashutosh, which specifically handles the case of uncountably many isolated points.

So the full answer is yes, without any qualifications. If a metric space has every open set realized as the countable union of balls, then it is separable.

  • $\begingroup$ I had thought at first that it was obvious that there must be at most countably many isolated points, but now I don't quite see this anymore, and so I have edited. Does this follow? $\endgroup$
    – JDH
    Commented Dec 26, 2011 at 23:40
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    $\begingroup$ Does the following observation help at all? Let $S$ be the set of isolated points. Then every $U \subseteq S$ is open, so every $U$ can be written as countable union of balls; moreover each of these balls is centered at some point in $S$. That is, for each $U \subseteq S$, we can write down a sequence of pairs of the form (center, radius) with center in $S$, from which we can uniquely recover $U$. This gives an injection from $2^S$ to $(S \times \mathbb R)^{\mathbb N}$. I do not know how to go from here. $\endgroup$
    – Srivatsan
    Commented Dec 27, 2011 at 0:43
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    $\begingroup$ Srivatsan, that seems to do it, under CH! We may assume that $S$ has size $\omega_1$, since just use $\omega_1$ many. Now, under CH, the right hand set has size $\omega_1$, but the left side is larger. Great! $\endgroup$
    – JDH
    Commented Dec 27, 2011 at 0:49
  • $\begingroup$ I have edited to give this argument. One actually can do it with something slightly less than CH. $\endgroup$
    – JDH
    Commented Dec 27, 2011 at 1:05
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    $\begingroup$ Thanks. Indeed it will be quite interesting if the original question is independent of ZFC and needs some form of CH. :) $\endgroup$
    – Srivatsan
    Commented Dec 27, 2011 at 1:12

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