Upper bound on covering multiplicity Suppose $M^n$ is a manifold with $Ric(M)\geq(n-1)k$ for some $k \in \mathbb{R}$. For a given point $p \in M$, let $B_p(r)$ denote the metric ball of radius $r$. 
Given $B_p(r)$ and $\varepsilon > 0$, show that There exists a $c = c(n,k\varepsilon^2) \in \mathbb{N}$ and covering $\mathcal{F}$ = $\lbrace B_{p_i}(\varepsilon) \rbrace$ of $B_p(r)$ such taht a for arbitrary $x \in B_p(r)$, there are at most $c = c(n,k\varepsilon^2)$ balls containing $x$ (where the constant $c$ depends only on $n$ and $k\varepsilon^2$)
I think I can consider this problem as a topological problem.(Consider as a metric space)
But where could I use the condition of  $Ric(M)\geq(n-1)k$?
Thank you very much!
 A: The result you want is often called Gromov's Packing Lemma. A more precise version states 

Lemma Let $M$ be an $n$-dimensional manifold satisfying $\mathrm{Ric} \geq (n-1)k$. Given $r,\epsilon > 0$, there are two constants $C_1 = C_1(n, kr^2, r/\epsilon)$ and $C_2 = C_2(n, k\epsilon^2)$ such that for any $p\in M$, the metric ball $B(p,r)$ can be covered by balls $B(p_i,\epsilon)$ with $p_i\in B(p,r)$ such that $\#\{p_i\} \leq C_1$ and the covering multiplicity $$\sup_{y\in B(p,r)} \#\{ p_i | y\in B(p_i,\epsilon)\} \leq C_2$$

This lemma is a direct consequence of the Bishop-Gromov volume comparison theorem, one form of which being

Theorem (Gromov's relative volume) If $M$ is an $n$-dimensional manifold with $\mathrm{Ric} \geq (n-1)k$ and $N_k$ is the model space (simply connected $n$-dimensional manifold with constant curvature $k$). Let $p \in M$ and $q\in N$. Then for $r \geq R > 0$ we have
  $$ \frac{B(p,r;M)}{B(p,R;M)} \leq \frac{B(q,r;N)}{B(q,R;N)} $$
  where $B(p,r;M)$ denotes the metric ball in $M$ centered at $p$ with radius $r$. 

Rmk: notice that the right hand side of the inequality is independent of $q$, as $N$ is a symmetric space. 
Proof of Lemma from Theorem:
Choose a maximal set of points $\{p_i\}\subset B(p,r- \epsilon_2)$ such that $d(p_i,p_j) \geq \epsilon / 2$ for $i\neq j$. (We can choose this inductively, fix an arbitrary point $p_0$ (which we can even take to be $p$), choose $p_{i + 1}$ arbitrarily as an element of $B(p,r-\epsilon/2) \setminus \left(\cup_{j = 0}^i B(p_j,\epsilon/2)\right)$.) 
By construction $\{ B(p_i,\epsilon/2)\}$ covers $B(p,r - \epsilon/2)$. We also have that $B(p_i,\epsilon/2) \subset B(p,r)$ by the triangle inequality. We furthermore have that $\{B(p_i,\epsilon) \}$ covers $B(p,r)$ also by the triangle inequality. It remains to show that the estimates hold.  
Now, by construction the balls $\{ B(p_i,\epsilon/4)\}$ are pairwise disjoint, and all entirely contained in $B(p,r)$. Furthermore, we have that $B(p,r)\subset B(p_i,2r)$ for any $p_i$ by the triangle inequality again. So we have that
$$ \#\{p_i\} \leq \frac{|B(p,r)|}{\inf_i |B(p_i,\epsilon/4|} \leq \sup_{i} \frac{|B(p_i,2r)|}{|B(p_i,\epsilon / 4)|} $$
applying the Gromov comparison theorem, we have that (denoting $\phi_k(r)$ the volume of the radius $r$ ball in the model geometry with constant curvature $k$)
$$ \#\{p_i\} \leq \frac{\phi_k(2r)}{\phi_k(\epsilon/4)} = C_1(n,kr^2, r/\epsilon) $$
the last equality can be seen by a scaling argument (scaling the metric by a fixed factor $r_0^2$ changes the volume and the curvature accordingly). 
Similarly, for the multiplicity bound, observe that if $B(p_i,\epsilon)\cap B(p_j\epsilon)\neq \emptyset$, we must have that $d(p_i,p_j) < 2\epsilon$. This implies that $B(p_j,\epsilon / 4) \subseteq B(p_i,3\epsilon)$. Using that the $B(p_j,\epsilon/4)$ balls are pairwise disjoint again, we have that for a fixed $i$
$$ \#\{ p_j | j\neq i, B(p_i,\epsilon)\cap B(p_j,\epsilon) \} \leq \frac{|B(p_i,3\epsilon)|}{\inf_j |B(p_j,\epsilon / 4)|} $$
Taking the sup over all $i$ we have that 
$$ \text{multiplicity} \leq \sup_i \frac{|B(p_i,3\epsilon)|}{|B(p_i,\epsilon/4)|} \leq \frac{\phi_k(3\epsilon)}{\phi_k(\epsilon/4)} = C_1(n,\frac94 k\epsilon^2,12) = C_2(n,k\epsilon^2)$$
as desired. 
