Expansion Constant of a Metric Space From Wikipedia 

The expansion constant of a metric space is the infimum of all
  constants $\mu$ such that whenever the family
  $\left\{\overline{B}(x_\alpha,r_\alpha)\right\}$ intersects pairwise,
  the intersection $$\bigcap_\alpha\overline{B}(x_\alpha,\mu r_\alpha)$$
  is non empty. A metric space is complete if and only if its expansion
  constant is $\leq2$.

Any intuitions? Can you show why the subspace $[0,1]$ has expansion constant smaller than 2 for example?
I really don't get it. 
 A: *

*As for $[0,1]$ with the Euclidean metric, the expansion constant is $1$, which means that every family of pairwise-intersecting closed intervals has nonempty intersection. This is a consequence of the following: given two families $\{a_i\}_{i\in I}$ and $\{b_j\}_{j\in J}$ of real numbers such that $a_i\leq b_j$  $ \forall i,j$ then $\sup_{i\in I} a_i\leq \inf_{j\in J} b_j$. Apply this to any family of intervals $[a_i,b_i]$ to find that any point in $[\sup a_i,\inf b_j]\neq \emptyset$ is in the intersection.

*Assuming that the space $X$ is compact (or just that bounded subsets are relatively compact), there is a simple proof of $\mu\leq 2$ using the finite intersection property: we just need to prove that any finite subfamily $\{\overline B(x_1,r_1),\ldots,\overline B(x_n,r_n)\}$ is such that $\bigcap_{i=1}^n \overline B(x_n,2r_n)\neq\emptyset$. To this aim use the triangle inequality to prove that the center of a ball of least radius among these is in the intersection. 

*For the general case, if a metric space $X$ is not complete then it does not admit any finite expansion constant.
Indeed, suppose $X$ is not complete. Then there is a non-convergent Cauchy sequence $(x_n)_n$. Consider the family of balls $\{\overline B(x_n,r_n)\}_n$ with $r_n=\sup_{m\geq n}d(x_n,x_m)$. The pairwise intersection is nonempty because of any pair the biggest ball contains the center of the other one. Assume by contradiction that there is a point $x\in\bigcap_n \overline B(x_n,\mu r_n)$. Then $d(x,x_n)\leq \mu r_n\to 0$ which means that $x_n\to x$, contradiction. This proves that a non-complete space can not have a finite expansion constant.
I don't know about the optimality of $\mu =2$. For Euclidean spaces the worst case seems to be $\mu=\sqrt 2$.
