Number of balls inside a ball - metric space version Suppose we have a metric space $(X,d)$ and let $a>1$. Also, let $x\in X$ and $r>0$ small enough.
Is there an upper bound $N(a)> 0$ (independent of $x,r$) so that there can fit at most $N(a)$ disjoint balls of radius $r/a$ inside the ball $B(x,r)$?
I can see that this happens if $X=\mathbb R^n$... but I use volume of balls to check this. What can we say in the general case? What about more geometric $X$? I see that this is also true, for small balls, if $X$ is a manifold.
Thanks!
 A: Consider a Hilbert space with an orthonormal basis $(e_n)$. Let $\epsilon >0$. Take $a=2$ and $ r=\epsilon $. Consider  the balls $B(\frac {\epsilon e_i} {3}, \frac {\epsilon } {3\sqrt 2})$. These balls are disjoint and they are all contained in $B(0,r)$. So $N(a)$ need not be finite.
A: No. E.g., consider the space $X$ comprising the coordinate axes in the space $\Bbb{R}^* = \bigcup_{n=1}^\infty\Bbb{R}^n$. $X$ looks like countably many copies of the real line identified at the origin metrized such that each pair of lines has the same metric as the union of the coordinate axes in $\Bbb{R}^2$. The open unit ball centred at the origin contains infinitely many pairwise disjoint open balls of radius $\frac{1}{2}$ centred at the points at distance $\frac{1}{2}$ from the origin. Likewise the closed unit ball centred at the origin contains infinitely many pairwise disjoint closed balls of radius $\frac{1}{3}$.
A: Here's an easy construction where you can have infinity:
Let $X$ be a metric space, and let $d(x, y)=1 \forall x \neq y \in X$. Of course, $d(x,x) = 0$. It's easy to check that $d$ is a metric.
Now, $B(x, 1) = X$. On the other hand, $B(x, a) = {x}$ if $a < 1$, so for any $a < 1$, $\cup_{y \in X} B(y,a) \subset B(x, 1)$ and the union is disjoint. So if $X$ is not finite, you can pack infinitely many balls.
A: In infinite-dimensional spaces, there’s a limit to the radii at which balls of unit radius can fit infinitely many non-overlapping balls of those radii; precisely, let $(X,\|\cdot\|)$ be Banach space with unit ball centered at the origin $B_X=B(0,1)$ and define the packing constant
$$P(X):=\sup\{r>0:B_X\text{ contains infinitely many non-overlapping } B(x,r)\}$$ then we have a formula given by $$P(X)=\frac{K(X)}{2+K(X)}\,$$ where $K(X)$ is the Kottman separation constant defined as $$K(X)=\sup\left\lbrace r:\|x_n-x_m\|\ge r\,,\forall n\ne m\,, \{x_n\}_n\subset B_X\right\rbrace\,.$$
For some spaces, the Kottman constant values are well-known; for instance, given the sequence space $\ell^p$ one obtains $$K(X)=2^{1/p}\,,~\,~\,~\,~\text{if $1\le p<\infty$}\,.$$ Thus, as example, the space $\ell^1$ of absolutely summable sequences can contain infinitely many non-overlapping balls of radius $r<\frac{1}{2}$ in its unit ball.
(For further reading, see for instance, https://pdfs.semanticscholar.org/f857/7f978843efd0b7b96cd7cd7b94770907032a.pdf).
