TITLE: An upper bound on the number of regions in $\mathbb{R}^3$ from the interection of unit spheres
QUESTION [9 upvotes]: If you place $n$ unit spheres in $\mathbb{R}^3$, then the number of regions formed by the intersection of the spheres should be bounded above by $n^3$.
This is something that a book handwaved as "easy to show", but I can't get a handle on how to prove this.
It's possible to place 4 unit spheres in a configuration like a tetrahedron to get $2^4 = 16$ regions, but beyond that, we can't place another unit sphere that intersects every single region. Of course, 16 is well below $4^3 = 64$. The bound of $n^3$ only becomes nontrivial at $n = 10$, because $2^{10} > 10^3$. I don't know how I would begin counting the number of regions that each new unit sphere can intersect to create new regions.

REPLY [3 votes]: Claim 1: $n\ge1$ circles drawn on a sphere divide the sphere into at most $n(n-1)+2$ regions.
Proof: For $n=1$ this is obvious. Suppose we have $k-1$ circles , and investigate what happens when you draw a new circle $c$.
The new circle $c$ may intersect the previous circles at no more than $2(k-1)$ points; so the previous circles divide $c$ into at most $2(k-1)$ arcs. Each of these arc split a region in two. So the increase in the number of regions will be at most
$2(k-1)$; the total number of regions cannot exceed $2+\sum\limits_{k=2}^n 2(k-1)=n(n-1)+2$.
The proof shows that this bound is sharp; if every two circles intersect and the intersection points do not collide then we have exactly $k(k-1)+2$ regions.
Claim 2: $n$ spheres divide the space into at most
$2+\sum\limits_{k=2}^{n} \big((k-1)(k-2)+2\big)= \frac{n^3-3n^2+8n}3$
regions.
(You can repeat the same proof.)