Take any four points in space. Draw all lines connecting pairs of them. Then draw all lines connecting pairs of points on those lines.

Can the resulting set of points cover all of space?

(In reply to going much further by Tristan)

The three pairs of parallel planes, each containing untouched points from opposite skew lines, would be the planes containing the faces of a cube that contains the tetrahedron, with one edge of the tetrahedron on each face of the cube.

The eight points are the vertices of this cube.  Four of them are also the vertices of the tetrahedron.  As such, they are reachable--part of the final set of points.

The other four vertices of the cube are not reachable, and are the only four points in space that are not part of the final set.

When the tetrahedron is not regular, an affine transformation of space (one consisting of linear stretching and shears, to preserve straight lines as straight lines) will transform it into one.  Looked at the other way, a regular tetrahedron can be transformed into any irregular one.  When that happens, what had been a cube becomes a parallelepiped but the tetrahedron still stretches one of its edges along each diagonal of the parallelepiped, meeting at four of its vertices, leaving the other four vertices out of the final set of points.

Posted by Charlie on 2005-02-06 19:24:15