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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
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of course, but, say, Banach spaces are ordinary too) –  Fedor Petrov Oct 11 '10 at 23:30
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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

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up vote 12 down vote accepted

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 $X$.

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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
    
Right. I meant COMPLETE infinite dimensional linear metric space in my added question. As you note, any countable dimensional linear metric space can be covered by a sequence of balls whose radii goes to zero. –  Bill Johnson Oct 13 '10 at 1:00

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