Trapping a convex body by a finite set of points In $\mathbb{R}^n$, let $K$ be a convex body and $T$ a finite set of points disjoint from the interior of $K$. Say that $T$ traps $K$ if there is no continuous motion of $K$ carrying $K$ arbitrarily far away from its original position during which no point of $T$ penetrates the interior of $K$. Is there a finite number $k=k(n)$ such that each $n$-dimensional convex body $K$ can be trapped by a set of at most $k$ points? If so, what is the smallest such $k$?  Observe that the $3$-cube can be trapped by six points, but not by five (for the $n$-cube it's $2n$ points, but not $2n-1$). The $n$-ball can be trapped by $n+1$ points, and it seems that no $n$-dimensional convex body can be trapped by $n$ points.
The main question is: Can every convex body in $\mathbb{R}^3$ be trapped by six points?
More generally: Can every convex body in $\mathbb{R}^n$ be trapped by $2n$ points?
Footnote 1. Here is a variation of the problem, perhaps easier to handle: restrict the motions of $K$ to parallel translations.
Footnote 2. For $n=2$, a closely related problem, namely of immobilizing the body with a finite set of points, has been studied and solved: four points always suffice. Reference will be provided upon request (I will have to look it up).
 A: I will assume we are not allowing re-orientation of the body (rotation).
If one wishes to determine if some arbitrary rigid body $K$ can pass some obstacle $T$ without touching it, one can reduce the problem to determining if some point $k \in K$ can pass by the Minkowski sum of $K$ and $T$ (which I will denote $K+T$) without touching it. In this case, if we wish to determine if $K$ can escape to infinity, assuming $K$ is bounded and finite, we need only determine if $K+T$ disconnects $\mathbb{R}^n$ into at least two connected components.
Since $T$ is a set of points, $K + T$ is simply taking $|T|$ copies of $K$, where each copy is centered at a point in $T$ (you can pick any point in $K$ to be the "center" of $K$, WLOG). Our question then becomes: what is the shape that maximizes the number of copies we would need to union together (without rotations) to disconnect $\mathbb{R}^n$ into two non-empty sets, and how many copies would that be?
I am fairly certain that proving that the $n$-cube is this shape, and that it requires $2n$ copies is a hop and a skip away from this, but I'm at a loss for a sufficiently formal argument. Something about how the faces are orthogonal and can only restrict one direction in one dimension at a time or something, which is somehow the worst-case. There's probably a similar argument for n+1 being a lower bound and tetrahedra. Unfortunately, this is where my knowledge stops.
de Berg et al's Computational Geometry: Algorithms and Applications covers the Minkowski Sum stuff very well in Chapter 13: Robot Motion Planning.
A: This is not an answer, just a reference on
immobilizing in $\mathbb{R}^3$:

 



Bracho, J., Fetter, H., Mayer, D., & Montejano, L. (1995). Immobilization of solids and mondriga quadratic forms. Journal of the London Mathematical Society, 51(1), 189-200.
  (ResearchGate link)

They cite one W. Kuperberg from a 1990 presentation. :-)
