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?

Now, if we had 4 non coplanar points in the shape of a regular tetrahedron, and drew all lines connecting pairs of them, we would end up with three pairs of skew lines.

Each pair of skew lines would go in perpendicular directions, but be in two different, parallel planes.  Now if we did the "connect the dots" function on just one pair of skew lines, it would not cover all of space.  It would leave out the two parallel planes.  To make clearer why this is, I'll draw a cross-section.

       .  This dot represents one point on the line

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Now, if we did the "connect the dots" function with just the cross-section above, we would cover the whole plane except the parallel line going through that dot.  This occurs similarly with all the points on the upper line, so all but two parallel planes are covered.  However, the planes aren't completely left out; the skew lines remain.

So it seems that we will have three pairs of parallel planes.  The intersection of all three pairs--8 points?--will be all that which is left out.  But do these 8 points coincide with the lines that are already there?  Or do they coincide with the 4 faces of the tetrahedron, which should be extended infinitely?

Lastly, if the tetrahedron we began with were not regular, it becomes even harder to visualize.  The skew line pairs would no longer go in perpendicular directions.  What happens then?

Edited on February 6, 2005, 5:47 pm

Edited on February 6, 2005, 5:51 pm

Posted by Tristan on 2005-02-06 17:43:25