Connectivity of metric measure space implies that doubling dimension is greater or equal $1$ Currently I'm reading this article: Hajlasz, Koskela, Sobolev met Poincaré. They consider metric measure spaces $(X, d, \mu)$, where $X$ is a metric space with metric $d$ and measure $\mu$ satisfies doubling condition
$$
\mu (2B) \le C \mu(B).
$$
Iterating this inequality we can obtain that
$$
\frac{\mu B}{\mu B_0} \ge c \Big( \frac{r}{r_0}\Big)^s, \; r < r_0, \; B = B(x,r), \; x\in B_0
$$
for some constant $c$ (the exact value does not really matter) and some exponent $s$ (see inequality (21) on page 22 for details). $s$ is called a doubling dimension.
Remark after theorem 6.1 on page 29 states that last condition cannot hold with $s < 1$, when space is connected. Currently I have no idea how it can be deduced from connectivity. Why doubling dimension of connected spaces is $\ge 1$? 
 A: They implicitly assume that the space contains more than one point. Another implicit assumption that is frequently omitted in the discussion of doubling measures is that $0<\mu(B)<\infty$ for every ball; e.g., the counting measure on $\mathbb{R}$ is not considered a doubling measure, even though one could say that $\mu(2B)\le C\mu(B)$ by virtue of both sides being infinite. 
So, fix distinct points $x,y\in X$ and let $R=d(x,y)$.  Because $X$ is connected, for every $r\in [0,R]$ there exists $z\in X$ such that $d(x,z)=r$. (Otherwise, $d(x,z)>r$ and $d(x,z)<r$ determine a partition of $X$ into two disjoint open sets.) Incidentally, we can use this to make sure $R<r_0$ where $r_0$ is as in (21).
By letting $r=(k-1)R/n$ where $k = 1,\dots,n$, we obtain points $z_1,\dots, z_n$ such that $d(z_k, x) = (k-1)R/n$. The balls $B_k = B(z_k, R/(2n))$ are disjoint and contained in $B(x,R)$.  Thus
$$
\mu(B(x,R))\ge \sum_{k=1}^n \mu(B(z_k, 1/(2n))) \ge c\mu(B(x,R))\sum_{k=1}^n \frac{1}{(2n)^s}  
$$
When $s<1$, the right hand side can be made arbitrarily large by choosing $n$ large, which yields a contradiction.
