proof of ham sandwich theorem
Define by sending to the measure of the subset of lying on the side of the plane corresponding to in the direction in which points. Note that and correspond to the same plane, but to different sides of the plane, so that .
Since is bounded, there is an such that is contained in , the closed ball of radius centered at the origin. For sufficiently small changes in , the measure of the portion of between the different corresponding planes can be made arbitrarily small, and this bounds the change in , so that is a continuous function.
Finally, it’s easy to see that, for fixed , is monotonically decreasing in , with and for sufficiently large.
Given these properties of , we see by the intermediate value theorem that, for fixed , there is an interval such that the set of with is . If we define to be the midpoint of this interval, then, since is continuous, we see is a continuous function from to . Also, since , if is the interval corresponding to , then is the interval corresponding to , and so .
Now let be measurable bounded subsets of , and let be the maps constructed above for . Then we can define by:
This is continuous, since each coordinate function is the composition of continuous functions. Thus we can apply the Borsuk-Ulam theorem to see there is some with , ie, with:
where we’ve used the property of mentioned above. But this just means that for each with , the measure of the subset of lying on one side of the plane corresponding to , which is , is the same as the measure of the subset of lying on the other side of the plane, which is . In other words, the plane corresponding to bisects each with . Finally, by the definition of , this plane also bisects , and so it bisects each of the as claimed.
|Title||proof of ham sandwich theorem|
|Date of creation||2013-03-22 16:40:29|
|Last modified on||2013-03-22 16:40:29|
|Last modified by||Statusx (15142)|