What is known about Lebesgue's universal covering problem in three dimensions? Lebesgue's universal covering problem is a relatively well-known open problem in geometry, asking for the convex set of minimal area which contains all planar sets of diameter 1. While the problem is not yet fully resolved, we have pretty good bounds on its size (between $0.832$ and $0.84409$) and, as I understand it, a decent sense of what an optimal solution ought to look like (some minor trimming-off from Pál's 1920 solution).
                                          
What progress has been made on the analogous problem in three dimensions? As a very crude starting point, we can take a prism of height $1$ whose base is any solution to the covering problem in two dimensions, but this is almost certainly suboptimal; for instance, given any starting solution with an axis of symmetry (such as the one above), we can find two congruent regions of positive volume whose points are all of distance $1$ from each other (by cutting along planes normal to a sort of space diagonal), and then remove one of those regions.
Are there more sophisticated modifications we can make, or entirely different regions to use? (After all, the best 2D solutions do not look much like the 1D solution crossed with a unit interval.)
 A: After perusing section 11.4 of Research Problems in Discrete Geometry, I came across a result of Makeev, Hausel, Makai, Szucs, and G. Kuperberg that the rhombic dodecahedron with unit inradius, which has volume $\frac{\sqrt{2}}2$, is a universal cover for all sets in $\mathbb{R}^3$ of unit diameter.
One can then observe that opposite degree-4 vertices of this rhombic dodecahedron are at distance $\sqrt{2}$. Consider a slab of width 1 perpendicular to the line connecting these vertices whose center coincides with that of the rhombic dodecahedron. There are two square pyramids of height $\frac{\sqrt{2}-1}2$ lying outside the slab; since any two points in opposite pyramids are at distance at least $1$ from each other, a set of unit diameter has points in at most one of them (WLOG, say the bottom one, since we can flip the set within the vertically-symmetric rhombic dodecahedron.) Then we can remove the top pyramid to obtain a slightly smaller universal cover.
In fact, we can repeatedly apply this procedure to each of the three mutually orthogonal axes between two degree-4 vertices on the rhombic dodecahedron, so we can ultimately remove 3 of these pyramids. This leaves us with a universal cover of volume
$$\frac72-2\sqrt{2}\approx 0.67157$$
(It is also mentioned in this book that the regular octahedron of inradius $1$ serves as a universal cover, but as this has volume $\frac{\sqrt{3}}2\approx 0.866$, it is not a promising candidate.)
