Pre-compact balls of a separable metric space Let $(X,d)$ be a complete metric space (I am not assuming that the metric is finite, there could be points in $X$ with infinite distance). Assume that each Ball in $X$ is pre-compact i.e. $\forall x \in X, \forall R > 0$ the closed Ball $B_{R}(x)$ is pre-compact (or if its closed then compact). Does it follow that $X$ is separeble? If not, does one need to make additional assumptions ?
Greetings
Nina
 A: If $d$ is an extended metric, taking values in $\Bbb R^+\cup\{\infty\}$, then the result is false, as Valerio Capraro noted in the comments. Let $X$ be any set, and for $x,y\in X$ define
$$d(x,y)=\begin{cases}0,&\text{if }x=y\\\infty,&\text{if }x\ne y\;.\end{cases}$$
Then $\langle X,d\rangle$ is complete, because every Cauchy sequence is eventually constant. However, $d$ induces the discrete topology on $X$, so the space is separable if and only if $X$ is countable. In particular, if $X$ is uncountable, then the space is not separable.
If $d$ is an ordinary metric, and you simply meant that the diameter of $X$ is not necessarily finite, then the result is true. Fix a point $p\in X$. For each $n\in\Bbb Z^+$ let $K_n=\operatorname{cl}B_n(p)$, the closure of the open ball of radius $n$ centred at $p$. By hypothesis each $K_n$ is compact, and a compact metric space is separable, so each $K_n$ has a countable dense subset $D_n$. Let $D=\bigcup_{n\in\Bbb Z^+}D_n$; clearly $D$ is countable. Finally, $X=\bigcup_{n\in\Bbb Z^+}K_n$, so $D$ is dense in $X$, and $X$ is therefore separable.
