If a metric space is separable, then any open set is a countable union of balls. Is the converse statement true?


UPDATE1. It is a duplicate of the question here
<https://math.stackexchange.com/questions/94280/if-every-open-set-is-a-countable-union-of-balls-is-the-space-separable/94301#94301>


UPDATE2. Let me summarize here the positive answer following Joel David Hamkins and Ashutosh. It is a matter of taste, but I omit using ordinals and use Zorn lemma instead, which may be more usual for most mathematicians (at least, it is for me). 


Lemma 1. If $(X,d)$ is non-separable metric space, then for some $r>0$ there exists an uncountable subset $X\_1\subset X$ such that $d(x,y)>r$ for any two points $x\ne y$ in $X\_1$.


Proof. For each $r=1/n$ consider the maximal (by inclusion) subset with such property. If it is countable, then $X$ has a countable $1/n$-net for each $n$, hence it is separable.


Now consider two cases. Define $X\_2\subset X\_1$ as a set of points $x\in X\_1$ for which there exist a point $y\_x\in X$ such that $0<d(x,y\_x)<r/10$. Consider two cases.


1) $X\_2$ is uncountable. Consider the union of open balls $U=\cup\_{x\in X\_2} B(x,d(x,y\_x))$. Consider any open ball $B(z,a)$ containing in $U$. We have $z\in U$, so $d(z,x)<d(x,y\_x)$ for some $x$, but $r/5>2d(x,y\_x)\geq d(x,z)+d(x,y\_x)\geq d(z,y\_x)>a$ since $y\_x\notin B(z,a)$. It implies that $B(z,a)$ is contained in a unique ball $B(x,d(x,y\_x))$, hence we need uncountably many such balls to cover whole $U$.


2) $X\_3=X\setminus X\_2$ is uncountable. For any $x\in X\_3$ define $R(x)>0$ as a radius of maximal at most countable open ball centered in $x$. Clearly $R(x)\geq r/10$ for any $x\in X\_3$. For any $x\in X\_3$ define a star centered in $x$ as a union $D=x\cup C$, where $C=\{z\_1,z\_2,\dots\}\subset X\_3$ is a countable sequence of points with $d(x,z\_i)\rightarrow R(x)+0$. Choose a maximal disjoint subfamily of stars. Clearly it is uncountable, else we may easily increase it. Denote by $U$ the set of centers of chosen stars. It is open (as any subset of $X\_3$), assume that it is a countable union of balls $U=\cup\_{i=1}^{\infty} B(x\_i,r\_i)$, $x\_i\in U$. We have $r\_i> R(x\_i)$ for some $i$, else $U$ is at most countable. But then $B(x\_i,r\_i)$ contains infinitely many points of the star $D$ centered in $x\_i$, while by our construction $U\cap D=\{x\_i\}$. A contradiction.



If any open set is a countable union of balls, does it imply separability?
Towards a contradiction, let us assume that we have a metric space $X = \{x\_i : i < \omega\_1\}$ in which any two points are at least unit distance apart and every subset of $X$ is the union of a countable family of open balls. Let $r\_i$ be the supremum of all $r > 0$ such that $B(x\_i, r)$ is countable. Construct $\{C\_i : i < \omega\_1\}$ such that


(1) Each $C\_i$ is countable and the infimum of $\{d(x\_i, y): y \in C\_i\}$ is $r\_i$


(2) If $i < j$, then $x\_i \notin C\_j$


Now let $I \in [\omega\_1]^{\omega\_1}$ be such that whenever $i < j$, $x\_j \notin C\_i$. Suppose $Y = \{x\_i : i \in I\}$ can be covered by a family $F$ of countably many balls. Let $i \in I$ be least such that $B(x\_i, r) \in F$ for some $r > r\_i$. Pick $y \in C\_i \cap B(x\_i, r)$. So $y \in \bigcup F = Y$ so that $y = x\_j$ for some $j \in I$ which is impossible.


