Prove: $m$ balls in $\mathbb{R^3}$ cut $\mathbb{R^3}$ into less than $m^3$ connected components.

Question

I need to prove or at least to understand why $m$ balls in $\mathbb{R^3}$ cut $\mathbb{R^3}$ into less than $m^3$ connected components. But I've no idea how to deal with it. I even tried to draw it but that didn't help me much because I only could explain $m^2$ but one $m$ was still missing. I hope you can help me and thanks a lot for doing so!

Answer 1

It seems the following.

To make our arguments more clear, we shall proceed by the dimension $n$.

$n=1$. A line $\Bbb R^1$ is cut by $m$ segments into at most $2m+1$ parts. This is so, because a cut is done by their endpoints, and there are at most $2m$ endpoints.

$n=2$. Assume we have $m$ circles placed at the plane $\Bbb R^2$. Assume there are $k$ circless which don’t intersect with others, and consider the plane partition by others $m-k$ circles. Similarly to the previous we can show that each of the circles is partitioned by the others into at most $$2(m-k)-2$$ parts, which we shall call arcs. Since each part of the plane partition is bounded by at least two arcs and each arc bounds at most two parts, we have that the total number of parts is not greater than $$2m(m-k)^2-2m(m-k).$$ Not intersecting $k$ circles add $k$ to this number. So the total number of parts is not greater than $$2m(m-k)^2-2m(m-k)+k\le 2m^2-2m.$$ In fact, the maximal number of parts is equal to $$m^2-m+2,$$ see this answer or "Plane Division by Circles" by Eric W. Weisstein.

$n=3$. Assume we have $m\ge 2$ spheres placed at the space $\Bbb R^3$. Assume there are $k$ spheres which don’t intersect with others, and consider the plane partition by others $m-k$ spheres. Since each two (non-trivially) intersecting spheres have a circle as their intersection, similarly to the previous we can show that each of the spheres is partitioned by the others into at most $$(m-k-1)^2-(m-k-1)+2=(m-k)^2-3(m-k)+4$$ parts, which we shall call spherical faces. Since each part of the space partition is bounded by at least two spherical faces and each spherical face bounds at most two parts, we have that the total number of parts is not greater than $$(m-k)^3-3(m-k)^2+4(m-k).$$ Not intersecting $k$ spheres add $k$ to this number. So the total number of parts is not greater than $$(m-k)^3-3(m-k)^2+4(m-k)+k\le m^3-3m^2+4m,$$ which is less than $m^3$ provided $m>1$. In fact, the maximal number of parts is equal to $$m(m^2-3m+8)/3,$$ see "Space Division by Spheres" by Eric W. Weisstein.