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Radon's theorem

In geometry, Radon's theorem on convex sets states that any set of d + 2 points in Rd can be partitioned into two (disjoint) sets whose convex hulls intersect.

For example, in the case d = 2, the set, call it X, would consist of four points. Depending on the set, it might be possible to partition X, into a triple and a singleton, where the convex hull of the triple (a triangle) contains the singleton, or it would be possible to partition X, into two pairs of points such that the line segments with these points as endpoints intersect. The latter situation would be the case if X, consists of the vertices of a rectangle.

Proof

The proof of Radon's theorem is not too difficult. Suppose X = \{x_1,x_2,\dots,x_{d+2}\}\subset \mathbf{R}^d. The system of equations

\sum_{j=1}^{d+2} a_j x_j=0

\sum_{j=1}^{d+2} a_j=0

can be thought of as a system of d + 1 equations in d + 2 unknowns a_1,a_2,\dots,a_{d+2}, since the first equation is really a vector equation in dimension d. There must therefore be a nonzero solution. Fix some nonzero solution a_1,a_2,\dots,a_{d+2}. Let I be the set of i=1,2,\dots,d+2 such that ai > 0, and let J be the set of i=1,2,\dots,d+2 such that ai < 0. The required partition of X is X_1=\{x_i:i\in I\} and X_2=\{x_i:i\in J\}. One point in the intersection of the convex hull of X1 and that of X2 is

\frac{\sum_{i\in I} a_i x_i}{\sum_{i\in I} a_i}.

It is clear that this point is in the convex hull of X1 and it is an easy consequence of the above equations satisfied by a_1,\dots,a_{d+2} that this point is also in the convex hull of X2. This completes the proof.

See also

References:

  • H. G. Eggleston, Convexity, Cambridge Univ. Press.
  • S. R. Lay, Convex Sets and Their Applications, Wiley.
Last updated: 05-27-2005 06:51:38
10-26-2009 08:16:03
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