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Suppose $A_1, \dots, A_k$ are sets of hyperplanes in $\mathbb{R}^d$ such that within any one set $A_i$, the hyperplanes are parallel. Moreover, suppose $|A_i| \leq s$ for all $i \in \{1, \dots, k\}$.

I'd like to know an upper bound on the number of connected components in $\mathbb{R}^d \setminus \left(\cup_{i = 1}^k A_i\right)$. In other words, we can think of $\cup_{i = 1}^k A_i$ as partioning $\mathbb{R}^d$ into a set of cells, and I'd like to know an upper bound on the number of cells there are in this partition.

Ignoring the fact that many of the hyperplanes are parallel, I know from this post that the number of connected components is at most $\sum_{i = 0}^d {sk \choose i}$. I wonder if there's a better bound that takes into account the fact that many of the hyperplanes are parallel.

Alex Ravsky
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1 Answers1

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Given natural $d$, $k$, and $s$, denote the maximal number of connected component in $\mathbb{R}^d \setminus \left(\cup_{i = 1}^k A_i\right)$ by $(d,k,s)$. Clearly, $(1,k,s)=ks+1$ and $(d,1,s)=s+1$.

To see that $(2,k,s)\le \sum_{i=0}^2 {k \choose i}s^i$ by induction, remark that adding to an arrangement of $k$ families of $s$ parallel lines each, we have that each of $s$ new lines crosses at most $ks$ lines, increasing a number of connected components by at most $ks+1$. That is

$$(2,k+1,s)\le (2,k,s)+ {k \choose 2}s^2+{k\choose 1}s^1+{k\choose 0}s^0+ks^2+s^1=$$ $${k+1 \choose 2}s^2+{k+1\choose 1}s^1+{k+1\choose 0}s^0.$$

Similarly for $d\ge 2$ we obtain an inequality $(d,k+1,s)\le (d,k,s)+s(d-1,k,s)$, which implies by induction with respect to $d$ and $k$, $$(d,k,s)\le \sum_{i=0}^d {k \choose i}s^i.$$

Alex Ravsky
  • 90,434