# About some open covering of compact connected space

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.

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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
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
See here and here. – Martin Sleziak Mar 31 '12 at 13:35

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 $i$.
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).