# covering a separable metric space by small balls

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?

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Yes if $X$ is $\sigma$-compact, which means that at least we should not be looking at very ordinary spaces for counterexamples. – Pete L. Clark Oct 11 '10 at 23:22
of course, but, say, Banach spaces are ordinary too) – Fedor Petrov Oct 11 '10 at 23:30
Let X be the space $c_0$ of sequences tending to 0 with the uniform norm. Now just use the Cantor diagonal argument to create a sequence $x$ such that $x$ and $x_j$ differ by at least $r_j$ in the $j$-th position... – fedja Oct 12 '10 at 0:42
@Pete: while I take your point, I think it would be a stretch to describe $c_0$ as a "not very ordinary space", just as I would hesitate to call the $p$-adic integers a pathological topological space... – Yemon Choi Oct 12 '10 at 6:23
Let X be the disjoint union of R^d as d runs through the naturals. Can a decreasing covering sequence of r's be found, ideally be constructed, with the r's tending toward 0, for this space? Gerhard "Ask Me About System Design" Paseman, 2010.10.12 – Gerhard Paseman Oct 12 '10 at 7:49

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).
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 $X$.
I think, the answer is yes for $c_{00}$ (finitely supported sequnces with $\max$-norm), see the covering procedure in my comment to Gerhard above. – Fedor Petrov Oct 12 '10 at 22:01