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?
$\begingroup$
$\endgroup$
7
asked Oct 11, 2010 at 23:00
-
3$\begingroup$ Yes if $X$ is $\sigma$-compact, which means that at least we should not be looking at very ordinary spaces for counterexamples. $\endgroup$– Pete L. ClarkCommented Oct 11, 2010 at 23:22
-
1$\begingroup$ of course, but, say, Banach spaces are ordinary too) $\endgroup$– Fedor PetrovCommented Oct 11, 2010 at 23:30
-
3$\begingroup$ 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... $\endgroup$– fedjaCommented Oct 12, 2010 at 0:42
-
$\begingroup$ @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... $\endgroup$– Yemon ChoiCommented Oct 12, 2010 at 6:23
-
$\begingroup$ 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 $\endgroup$– Gerhard PasemanCommented Oct 12, 2010 at 7:49
|
Show 2 more comments
1 Answer
$\begingroup$
$\endgroup$
2
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 $X$.
answered Oct 12, 2010 at 0:38
-
$\begingroup$ 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. $\endgroup$ Commented Oct 12, 2010 at 22:01
-
$\begingroup$ 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. $\endgroup$ Commented Oct 13, 2010 at 1:00