Let $(X,d)$ be a separable metric space. Can $X$ always be covered by a sequence of balls $B(x\_i,r\_i) (i=1,2,\dots)$ s.t. radii $r\_i$ tend to 0? 



covering a separable metric space by small balls
The answer is no for the Banach space $c\_0$. Suppose $B(x\_i,r\_i)$ is a sequence of balls with $r\_i\to 0$ and WLOG $x\_i$ is supported in $[1,N\_i]$ with 
$N\_1<N\_2<...$. Consider a point $x$ in $c\_0$ whose $N\_i+1$ coordinate is $2 r\_i$. 


I think the answer is no for any separable Banach space: IIRC, for any separable Banach space $X$ and any increasing sequence $E\_n$ of finite dimensional subspaces and any sequence of positive $r\_n\to 0$, there is a vector $x$ in $X$ s.t. the distance from $x$ to $E\_n$ is larger than $r\_n$ (in fact, even equal to $2r\_n$ if $r\_n$ is decreasing).


ADDED 10/12/10: It is not hard to check what I said in the second paragraph of my answer, from which it follows that the answer is no for any infinite dimensional Banach space. Is the answer no for any infinite dimensional linear metric space? 


Can you characterize the metric spaces for which the answer is yes? I suspect that the reason Fedor is interested in the property is that a modification of the proof of the Vitali covering theorem yields that if $X$ is covered by such a sequence of balls, then there are DISJOINT balls $B(y\_n,t\_n)$ with $t\_n\to 0$ s.t. $B(y\_n,5t\_n)$ covers $X$.


