Given convex polytopes P and Q in R^n, say P "pierces" Q if Q\P is connected but not simply connected. Are there convex polyhedra P,Q in R^3 that pierce each other? (I don't think so but no proof leaps to mind.) Are there cubes P,Q,R in R^3 such that P pierces Q, Q pierces R, and R pierces P? (One might conceive of "mutually ruperting" cubes subject to some rules that flout, in various mathematically specific ways, the fact that two physical objects cannot occupy the same space at the same time.) Perhaps Pechenik and Shultz already consider such questions? Jim Propp -- 1. Pechenik & Shultz are the same person. She kept changing her name. The worst instance I saw of that was this Frenchwoman named "Yvonne" who proved some good results in general relativity. Her last name(s) kept changing plus there also was a combinatorial explosion of different hyphenation possibilities which she also eagerly employed at random. Finally it dawned on me: ignore all her names except for Yvonne, there was only one Yvonne in France doing general relativity during 1900-2000. 2. It is not possible for P to pierce Q and Q simultaneously to pierce P, (both convex 3D objects) if P\Q and Q\P each have the topology of a solid torus. Because then Q must "drill a hole thru" P and the two outer surfaces of this hole, which are topological disks, each must cut Q disconnecting it. So any example (if one exists) must have a more complex topology. More generally the intersection of Q with boundary(P) is a collection of connected components and it is not permitted for any of them to be a topological disk, because that disk would cut Q. Anyhow, my competence in such topological arguments is nil, and here I am inherently depending on some kind of Jordan-curve-like theorem for 3D convex objects, which I presume is already known. Some real topologists should do better... 3. I can sketch a proof there are M equal cubes in R^3 such that cube(j) pierces cube(j+1 mod M) for all j. For some modulus M>=3, whose value I do not know. You keep applying appropriate rotation matrices one at a time until the product of all M of them equals identity matrix. We know each step of this process is always possible with a certain amount of "free play" freedom in each rotation matrix, and since the process cannot continue forever without two rotation matrices getting arbitrarily near one another, near enough to be within the free play zone, there must be a "cycle", which proves my claim for some M>=3. This argument works not only for cubes, but actually anything "Rupertable (or piercable) with a positive radius of free play" in any finite dimension.