Matt's answer is correct, but at an even simpler level: If you take two generic line segments in a compact subset of R^2, they'll intersect with positive probability. If you take two generic line segments in R^3, they'll intersect with probability 0. This isn't a proof by any means, but it's the simplest conceptual reason I know of. If instead of edges we wanted surfaces, we'd have to go up to dimension 5.
::sigh:: Okay, here's a constructive example of an infinite set of points such that no straight line segments between any two of them intersect. Take any two real numbers, say 2 and pi, that are algebraically independent. Then I claim that the set of points (n,2n,pin) is such a set.
Why? Suppose the line segment between the points with x=a and x=b intersected the line segment between x=c and x=d, parameterizing the line segments so the equations:
(a,2a,pia)+lambda(b−a,2b−2a,pib−pia)
(c,2c,pic)+gamma(d−c,2d−2c,pid−pic)
give us the same point for some choice of the variables.
Looking at the first two components tells us that lambda,gamma are both rational. But then the third component gives us a polynomial with rational coefficients that has a root equal to pi, which is impossible since pi is transcendental. So none of these line segments can intersect.
No comments:
Post a Comment