Home > Geometry, shape optimization > Sets of finite perimeter which do not contain any open ball

## Sets of finite perimeter which do not contain any open ball

Some time ago, I wondered myself if I could find a very small ball included in a set of finite perimeter (see definition for set of finite perimeter here and here). It turns out, that the answer is no, there mustn’t be any open ball included in a set of finite perimeter. My teacher gave me the following nice example.

Example: Take ${ D=B(0,1)}$, the unit ball in ${ \Bbb{R}^2}$ and denote

$\displaystyle S=D\cap \mathbb{Q}^2 = (x_n)_{n \geq 0}$

. Then, we can find a sequence of positive real numbers ${(r_n)}$ such that

$\displaystyle \sum_{n=1}^\infty 2\pi r_n < \infty$

$\displaystyle \sum_{n=1}^\infty \pi r_n^2 < \pi$

Define $\displaystyle B=\bigcup_{n=1}^\infty B(x_n,r_n)$.
Then ${C=D\setminus B}$ has the desired property. Indeed, ${|C|=|D|-|B|>0}$, and ${Per_D(C)=Per_D(B)<\infty}$.

If ${C}$ would contain an open ball then that ball would intersect ${S \subset B}$, which is not possible.