# proof of ham sandwich theorem

This proof uses the Borsuk-Ulam theorem, which states that any continuous function from $S^{n}$ to $\mathbb{R}^{n}$ maps some pair of antipodal points to the same point.

Let $A$ be a measurable bounded subset of $\mathbb{R}^{n}$. Given any unit vector $\hat{n}\in S^{n-1}$ and $s\in\mathbb{R}$, there is a unique $n-1$ dimensional hyperplane normal to $\hat{n}$ and containing $s\hat{n}$.

Define $f:S^{n-1}\times\mathbb{R}\rightarrow[0,\infty)$ by sending $(\hat{n},s)$ to the measure of the subset of $A$ lying on the side of the plane corresponding to $(\hat{n},s)$ in the direction in which $\hat{n}$ points. Note that $(\hat{n},s)$ and $(-\hat{n},-s)$ correspond to the same plane, but to different sides of the plane, so that $f(\hat{n},s)+f(-\hat{n},-s)=m(A)$.

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

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

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

Now let $A_{1},A_{2},...,A_{n}$ be measurable bounded subsets of $\mathbb{R}^{n}$, and let $f_{i},g_{i}$ be the maps constructed above for $A_{i}$. Then we can define $h:S^{n-1}\rightarrow R^{n-1}$ by:

 $h(\hat{n})=(f_{1}(\hat{n},g_{n}(\hat{n})),f_{2}(\hat{n},g_{n}(\hat{n})),...f_{% n-1}(\hat{n},g_{n}(\hat{n})))$

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 $\hat{n}\in S^{n-1}$ with $h(\hat{n})=h(-\hat{n})$, ie, with:

 $f_{i}(\hat{n},g_{n}(\hat{n}))=f_{i}(-\hat{n},g_{n}(-\hat{n}))=f_{i}(-\hat{n},-% g_{n}(\hat{n}))$

where we’ve used the property of $g$ mentioned above. But this just means that for each $A_{i}$ with $1\leq i\leq n-1$, the measure of the subset of $A_{i}$ lying on one side of the plane corresponding to $(\hat{n},g_{n}(\hat{n}))$, which is $f_{i}(\hat{n},g_{n}(\hat{n}))$, is the same as the measure of the subset of $A_{i}$ lying on the other side of the plane, which is $f_{i}(-\hat{n},-g_{n}(\hat{n}))$. In other words, the plane corresponding to $(\hat{n},g_{n}(\hat{n}))$ bisects each $A_{i}$ with $1\leq i\leq n-1$. Finally, by the definition of $g_{n}$, this plane also bisects $A_{n}$, and so it bisects each of the $A_{i}$ as claimed.

Title proof of ham sandwich theorem ProofOfHamSandwichTheorem 2013-03-22 16:40:29 2013-03-22 16:40:29 Statusx (15142) Statusx (15142) 5 Statusx (15142) Proof msc 54C99