Given a metric space, $(X,d)$ with finite doubling dimension $ddim(X)$ (Namely, for every $r>0$, every ball of radius $r$ can be covered by $2^{ddim(X)}$ balls of radius $\frac{r}{2}$ ), and a set $A\subseteq X$.
We want to bound the size of $A$, and we know that for each $x,y\in A$ s.t. $x\neq y$ we have $d(x,y)\geq \alpha$, and $diam(X)=\beta$. We have the following claim:
$|A| \leq {\lceil}\frac{\beta}{\alpha}{\rceil}^{ddim(X)+1}$
I tired to find a proof for this claim, but I can only show that
$|A| \leq (2\cdot{\lceil}\frac{\beta}{\alpha}{\rceil})^{ddim(X)}$
Here is my proof:
We need at most $(2^{ddim(X)})^i$ balls of radius $\frac{\beta}{2^{i}}$ to cover $A$ . Hence, we need at most $(2^{ddim(X)})^{{\lceil}log\frac{\beta}{\alpha}{\rceil}}$ balls of radius $\alpha$ to cover $A$ .
Now, $(2^{ddim(X)})^{{\lceil}log\frac{\beta}{\alpha}{\rceil}} \leq$
$(2^{ddim(X)})^{(log{\lceil}\frac{\beta}{\alpha}{\rceil}+1)} = $
$(2^{(log{\lceil}\frac{\beta}{\alpha}{\rceil}+1))^{ddim(X)}}$ =
$(2\cdot{\lceil}\frac{\beta}{\alpha}{\rceil})^{ddim(X)}$
Now, since $(2\cdot{\lceil}\frac{\beta}{\alpha}{\rceil})^{ddim(X)}$ balls of radius $\alpha$ covers $A$, and the minimum interpoint distance in $A$ is at least $\alpha$, then $A$ contains at most $(2\cdot{\lceil}\frac{\beta}{\alpha}{\rceil})^{ddim(X)}$
How can I get a tighter bound? Thanks!