# 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^{} $\widehat{n}\in {S}^{n-1}$ and $s\in \mathbb{R}$, there is a unique $n-1$ dimensional hyperplane^{} normal to $\widehat{n}$ and containing $s\widehat{n}$.

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

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 $(\widehat{n},s)$, 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 $f(\widehat{n},s)$, so that $f$ is a continuous function.

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

Given these properties of $f$, we see by the intermediate value theorem that, for fixed $\widehat{n}$, there is an interval^{} $[a,b]$ such that the set of $s$ with $f(\widehat{n},s)=m(A)/2$ is $[a,b]$. If we define $g(\widehat{n})$ 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 $f(\widehat{n},s)+f(-\widehat{n},-s)=m(A)$, if $[a,b]$ is the interval corresponding to $\widehat{n}$, then $[-b,-a]$ is the interval corresponding to $-\widehat{n}$, and so $g(\widehat{n})=-g(-\widehat{n})$.

Now let ${A}_{1},{A}_{2},\mathrm{\dots},{A}_{n}$ 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 $h:{S}^{n-1}\to {R}^{n-1}$ by:

$$h(\widehat{n})=({f}_{1}(\widehat{n},{g}_{n}(\widehat{n})),{f}_{2}(\widehat{n},{g}_{n}(\widehat{n})),\mathrm{\dots}{f}_{n-1}(\widehat{n},{g}_{n}(\widehat{n})))$$ |

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

$${f}_{i}(\widehat{n},{g}_{n}(\widehat{n}))={f}_{i}(-\widehat{n},{g}_{n}(-\widehat{n}))={f}_{i}(-\widehat{n},-{g}_{n}(\widehat{n}))$$ |

where we’ve used the property of $g$ mentioned above. But this just means that for each ${A}_{i}$ with $1\le i\le n-1$, the measure of the subset of ${A}_{i}$ lying on one side of the plane corresponding to $(\widehat{n},{g}_{n}(\widehat{n}))$, which is ${f}_{i}(\widehat{n},{g}_{n}(\widehat{n}))$, is the same as the measure of the subset of ${A}_{i}$ lying on the other side of the plane, which is ${f}_{i}(-\widehat{n},-{g}_{n}(\widehat{n}))$. In other words, the plane corresponding to $(\widehat{n},{g}_{n}(\widehat{n}))$ bisects each ${A}_{i}$ with $1\le i\le n-1$. 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 |
---|---|

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 |