Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $(X,d)$ be a connected compact metric space. Does there exist a finite family $B_1,...,B_n$ of open balls in $X$ with a given radius $R>0$ such that $B_i \cap B_{i+1} \neq \emptyset$ for $i=1,\ldots ,n-1$ and $X=\bigcup_{k=1}^n B_i$.

Thanks.

share|improve this question
    
Since $X$ is compact, it is totally bounded. This means you can find a finite cover made up of balls for any $R$. Finiteness and connectedness should then allow you to order the balls in such a way to satisfy your intersection condition. –  Miha Habič Mar 31 '12 at 12:55
    
In fact, since $X$ is totally bounded, it is bounded, so there exists an $R>0$ so that $X$ is contained in a ball $B$ of radius $R$. The family $\{B\}$ satisfies the conditions you want. (Edit: I may be mixing up quantifiers, though -- I'm assuming the question is $\exists (\{B_i\} \mbox{ and } R>0)$ such that ... .) –  Neal Mar 31 '12 at 12:58
3  
Maybe first show that given any two points in the space, there is a finite collection of overlapping balls of radius $R$ "starting" at one point and ending at the other. Then from a finite cover of $X$, "connect the centers" of the balls with such chains in some order. –  David Mitra Mar 31 '12 at 13:04
    
@Neal I assumed that $R>0$ is given. –  Richard Mar 31 '12 at 13:08
2  
See here and here. –  Martin Sleziak Mar 31 '12 at 13:35

1 Answer 1

up vote 3 down vote accepted

Let ${\cal B}=\{B_1,B_2,\ldots, B_n\}$ be an open cover of $X$ of balls of radius $R$.

Claim: Let $x$ and $y$ be in $X$. Then there is a "chain" from $x$ to $y$ consisting of elements of $\cal B$. That is, there is a sequence of sets $B_{n_1},B_{n_2},\ldots, B_{n_k}$ from $\cal B$ with $x\in B_{n_1}$, $y\in B_{n_k}$ and $B_{n_i}\cap B_{n_{i+1}}\ne\emptyset$ for each admissable $i$.

To see why the claim is true, you can use the argument presented in the second link mentioned by Martin Sleziak in the comments.

So, between any two points $x$ and $y$ in $X$, there is a chain from $x$ to $y$ consisting of open balls of radius $R$. Now let $x_1$, $x_2$, $\ldots\,$, $x_n$, be the centers of the open balls in $\cal B$. One obtains the desired finite family of open balls by constructing chains from $x_1$ to $x_2$, then from $x_2$ to $x_3$, and so on (note that this usually won't give the smallest collection).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.