proof of ham sandwich theorem

This proof uses the Borsuk-Ulam theorem, which states that any continuous functionMathworldPlanetmath from Sn to n maps some pair of antipodal points to the same point.

Let A be a measurable boundedPlanetmathPlanetmathPlanetmath subset of n. Given any unit vectorMathworldPlanetmath n^Sn-1 and s, there is a unique n-1 dimensional hyperplaneMathworldPlanetmathPlanetmath normal to n^ and containing sn^.

Define f:Sn-1×[0,) by sending (n^,s) to the measure of the subset of A lying on the side of the plane corresponding to (n^,s) in the direction in which n^ points. Note that (n^,s) and (-n^,-s) correspond to the same plane, but to different sides of the plane, so that f(n^,s)+f(-n^,-s)=m(A).

Since A is bounded, there is an r>0 such that A is contained in Br¯, the closed ballPlanetmathPlanetmath of radius r centered at the origin. For sufficiently small changes in (n^,s), the measure of the portion of Br¯ between the different corresponding planes can be made arbitrarily small, and this bounds the change in f(n^,s), so that f is a continuous function.

Finally, it’s easy to see that, for fixed n^, f(n^,s) is monotonically decreasing in s, with f(n^,-s)=m(A) and f(n^,s)=0 for s sufficiently large.

Given these properties of f, we see by the intermediate value theorem that, for fixed n^, there is an intervalMathworldPlanetmathPlanetmath [a,b] such that the set of s with f(n^,s)=m(A)/2 is [a,b]. If we define g(n^) to be the midpointMathworldPlanetmathPlanetmathPlanetmath of this interval, then, since f is continuous, we see g is a continuous function from Sn-1 to . Also, since f(n^,s)+f(-n^,-s)=m(A), if [a,b] is the interval corresponding to n^, then [-b,-a] is the interval corresponding to -n^, and so g(n^)=-g(-n^).

Now let A1,A2,,An be measurable bounded subsets of n, and let fi,gi be the maps constructed above for Ai. Then we can define h:Sn-1Rn-1 by:


This is continuous, since each coordinateMathworldPlanetmathPlanetmath function is the composition of continuous functions. Thus we can apply the Borsuk-Ulam theorem to see there is some n^Sn-1 with h(n^)=h(-n^), ie, with:


where we’ve used the property of g mentioned above. But this just means that for each Ai with 1in-1, the measure of the subset of Ai lying on one side of the plane corresponding to (n^,gn(n^)), which is fi(n^,gn(n^)), is the same as the measure of the subset of Ai lying on the other side of the plane, which is fi(-n^,-gn(n^)). In other words, the plane corresponding to (n^,gn(n^)) bisects each Ai with 1in-1. Finally, by the definition of gn, this plane also bisects An, and so it bisects each of the Ai as claimed.

Title proof of ham sandwich theorem
Canonical name ProofOfHamSandwichTheorem
Date of creation 2013-03-22 16:40:29
Last modified on 2013-03-22 16:40:29
Owner Statusx (15142)
Last modified by Statusx (15142)
Numerical id 5
Author Statusx (15142)
Entry type Proof
Classification msc 54C99