Given $n$ points of general position in $\mathbb{R}^d$ (say, $n>d$ and no $d+1$ lie in a hyperplane.) We want to draw $k$ hyperplanes not passing through those points so that they all are in different open regions of the complement of the drawn planes. What is the minimal value $k(n,d)$ for which it may be always done? Case $d=2$ is solved by D. Gerbner, G. Tóth, `Separating families of convex sets' (arxiv:1211.2982). The answer is $\lceil n/2\rceil$, the upper estimate holds for points in convex position, the lower is proved by rotating the line which halves the number of points. For $d$, I may only show lower estimate like $(n-1)/d$ by considering points on the moment curve $(t,t^2,\dots,t^d)$ (and, of course, $k(n,d)\leq k(n,d-1)$ by projection argument). 



separating points in $\mathbb{R}^d$ by minimal number of planes
$\lceil n/d\rceil+d-2$ hyperplanes always suffice. To achieve this, we first choose $H\_1,\dots,H\_{d-1}$ in a manner that $H\_1$ separates $\lceil n/d\rceil$ points from the rest, $H\_2$ separates $\lceil n/d\rceil$ of those rest ones, and so on. Thus we get $d$ sets $S\_1,\dots,S\_d$ separated from each other, each consisting of at most $k=\lceil n/d\rceil$ points. 


We prove by induction on $k$ that such $d$ sets can be separated completely by $k-1$ hyperplanes. To perform the induction step, we implement the ham-sandwich theorem to bisect att $S\_1,\dots,S\_d$. We may shift this bisecting hyperplane a bit, so that on one side we get a collection of sets $S\_1^1,\dots,S\_d^1$ with at most $\lceil k/2\rceil$ points in each, and on the other side we have a collection of $S\_1^2,\dots,S\_d^2$ with at most $\lfloor k/2\rfloor$ points in each. Now apply the induction hypothesis to each of these collections separately. 


Perhaps, the `$+d$' term can be improved by choosing the initial hyperplanes in a smarter way?


