3
$\begingroup$

Denote $B_n\subset\Bbb R^n$ to be unit ball at origin.

Denote $S\subset B_n$ to region of type $\mathsf I$ if it satisfies $$s\in S\iff\forall t\in S, s+t\in S\mbox{ or }s-t\in S$$

I am convinced $\mu(S)=0$ and likely is just a discrete set.

Denote $R\subset B_n$ to region of type $\mathsf{II}$ if it satisfies $$s\in R\iff\forall t\in R, s+t\in mB_n\mbox{ or }s-t\in mB_n$$

where $mB_n$ is unit ball scaled $m$ times where $m\in\Bbb R$.

Minimum $m$ needed so that $R=B_n$ is $\sqrt{2}$.

Among all such regions of type $\mathsf{II}$, what is $R$ with maximum volume with general $m\in[1,\sqrt{2}]$?

How many disjoint $R$ do you need to cover $p\%$ of $B_n$ when $m\in[1,\sqrt{2}]$?

Is there a simple geometric description such sets $R$ at least for those with maximum volume?

$\endgroup$
8
  • $\begingroup$ It seems to me that $B_n$ is of type $I$ so that is the answer to all your questions. $\endgroup$
    – Rbega
    Aug 18, 2015 at 16:51
  • 2
    $\begingroup$ Not for $n>1$. For example, if $N=2$, $s=(1,0)$, and $t=(0,1)$, neither $s+t$ nor $s-t$ is in the ball. Actually, do you have an example of a set with positive volume that is type I? $\endgroup$ Aug 18, 2015 at 18:54
  • $\begingroup$ I must be missing something here. Suppose (just for simplicity) we're in $\mathbb{R}^2$ and $S$ is type $1$. WLOG we can assume the furthest point from the origin in the set is of the form $(x,0)$. Now consider the set of all $a$ such that $S$ has positive (one dimensional Lebesgue) measure along the line $x=a$. If we take a near maximal (or near minimal) such $a$, then there must be two points whose sum isn't in $a$ (the set doesn't have positive measure along $x=2a$). But the difference is on the $y$ axis, contradicting the type I property when combined with $(x,0)$'s maximal norm $\endgroup$ Aug 18, 2015 at 19:48
  • 1
    $\begingroup$ @KevinP.Costello Perhaps there is no point at maximal distance from the origin? $\endgroup$
    – Ben Barber
    Aug 20, 2015 at 16:44
  • 1
    $\begingroup$ It states that for measurable sets $A,B\subset \mathbb{R}^n$, if $A+B$ is meausurable, then $\mu(A+B)^{1/n}\geq \mu(A)^{1/n}+\mu(B)^{1/n}$. In particular, if $A=-A=A-A$ then $\mu(A)=0$. $\endgroup$ Aug 21, 2015 at 22:35

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.