# Packing a Riemannian manifold with disjoints balls

Let $$M$$ be a smooth Riemannian manifold with riemannian measure $$\mu$$. I don't suppose that $$M$$ is complete. Can we find a finite or countable disjoint collection of open (or closed) and relatively compact geodesic balls$$(B_n)_{n\in\mathbb{N}}$$ such that :

$$\mu\left(M\setminus\bigsqcup_{n\in\mathbb{N}}B_n\right) = 0$$ ? We might assume that $$M$$ has a bounded curvature. The $$M$$ I'm interested in is the intersection of a (complete) submanifold of $$\mathbb{R}^d$$ with a ball. The compactness restriction ensures that each ball is a "real" ball, and hasn't been cropped by the "edge" of $$M$$.

• It seems to me we should be able to actually cover $M$ by a countable collection of closed (compact) balls. – Kevin Casto Apr 4 at 21:14
• I forgot the disjoint hypothesis in my question.. – Pii_jhi Apr 5 at 8:16
• In the case of $E^n$'s, such packings do exist. For instance, Apollonian packing does the job for the Euclidean plane. Not sure what happens if one requires zero Hausdorff dimension of the residual set. – Moishe Kohan Apr 5 at 8:49
• The residual set will contain the boundary of the ball (if we consider open balls), so it will be of Hausdorff dimension $d-1$ at least and I think it has to be strictly greater than $d-1$. In the case of Apollonian packing, which feels close to an"optimal" packing, has Hausdorff dimension $\simeq 1.3$.. – Pii_jhi Apr 5 at 10:26
• @Pii_jhi: Of course, I meant closed balls, as in your question. In all packing constructions I know, the residual set has positive H.D. – Moishe Kohan Apr 7 at 4:45

If you follow the proof using an open cover of $$M$$ by Riemannian normal coordinate charts, you should end up with a countable basis of precompact geodesic balls of $$M$$ so that the union of all such balls is equal to $$M$$. Bounded curvature and completeness are not necessary for the proof.
I'm not sure what you mean by the last two sentences, but the fact that $$M$$ is topologically embedded into $$\mathbb{R}^d$$ should imply that the intersection of $$M$$ with a ball in $$\mathbb{R}^d$$ is an open set in $$M$$.