My question is probably naive but an positive answer would be useful for me. Let $X$ be a separable metric space, $x\in X$, $\delta>0$ and $r_1, r_2,...>\delta$ are given. Is it possible to choose points $x_1,x_2,... \in X$ such that $\bigcap_{k=1}^{\infty} B(x_k,r_k)\subset B(x,\delta)\;?$ (where $B$ stand for a closed balls) It seems to hold in $\mathbb{R}^d$ but I am not sure if it is true in an arbitrary separable metric space. I would be greatfull for help.
A representation of a ball as the intersetion of balls with larger radius
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general-topology
metric-spaces
1 Answers
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Here’s a counterexample. Let $d$ be the metric on $\Bbb N$ given by $d(n,n)=0$ and $d(m,n)=1$ if $m\ne n$; this metric generates the discrete topology. Since $\Bbb N$ is countable, the space is certainly separable. Note that if $B(n,r)=\{m\in\Bbb N:d(m,n)\le r\}$ is the closed ball of radius $r$, then $B(n,r)=\Bbb N$ if $r\ge 1$, and $B(n,r)=\{n\}$ if $r<1$. Thus, if $\delta=1/2$ and $r_k=1$ for each $k$, there is no way to choose the points $x_k$ so that $\bigcap_kB(x_k,1)=B(0,1/2)$: the lefthand side can only be $\Bbb N$, while the righthand side is $\{0\}$.
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0Thank you very much for the response – 2012-09-15