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Let $(X,d)$ be a compact metric space. Given $r>0$ I can obviously cover any ball $B(x,r)$ by finitely many balls $B(x_j,r/2)$, $j=1,...,k(r)$, by compactness. My question is whether there is always a finite number $k$, which doesn't depend on the radius $r$ and is such that any $r$-ball in $(X,d)$ can be covered by $k$-many $r/2$-balls?

Actually, what I would like to know is the following and I thought the previous question might be something simpler which also happens to answer my question; Let $X$ be a compact abelian group with Haar measure $m$. Is there a constant $c>0$, such that for any $r>0$ we have $m(B(x,r)) \leq c \cdot m(B(x,r/2))$? (again, the point is that $c$ wouldn't depend on the radius $r$)

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  • $\begingroup$ In $R^n$ a ball has measure of order $r^n$ so comparing volumes at least $2^n$ balls are needed to cover the larger one. Maybe one can take a product of such spaces (with growing number of half-balls needed) to see that, in general there is no such $k$. $\endgroup$
    – Salcio
    Commented Feb 1 at 16:40

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In a compact metric space the general answer is no.

In fact, your condition has a name. A metric space $X$ such that every ball of radius $r$ can be covered by $k$-many balls of radius $\frac{r}{2}$ is called doubling. It turns out the doubling condition is equivalent to having finite Assouad dimension, which is stronger than having finite Hausdorff dimension, which in turn is stronger than having finite topological dimension.

Also any space with a doubling measure $(\mu(2B)\leq C\mu(B)$ for some constant $C$, for every ball $B$) must have a doubling metric, which is an easy exercise.

Therefore any metrizable compact abelian group with infinite topological dimension will do as a counter-example. For example, the countably infinite product of the circle with itself $\left(\mathbb R/\mathbb Z\right)^{\mathbb N}$ cannot have a doubling metric, thus its Haar measure cannot be doubling with respect to an invariant metric (or any compatible metric, for that matter).

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  • $\begingroup$ That was very much unexpected; thanks. $\endgroup$
    – User
    Commented Feb 2 at 18:22

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