A general partitioning principle

Introduction

In the proof of DHJ(3), a key step was to show that a dense set of complexity 1 could be almost entirely partitioned into subspaces of dimension m, for some m that tends to infinity with n. The argument turned out not to depend too heavily on the precise definition of "set of complexity 1," which makes it easy to generalize. In this article we present a generalization that can be used in the proof of DHJ(k).

Definitions and statement of result

Let [math]\displaystyle{ \mathbf{K} }[/math] be a collection of subsets of [math]\displaystyle{ \bigcup_{n=1}^\infty[k]^n. }[/math] We shall say that [math]\displaystyle{ \mathbf{K} }[/math] is hereditary if [math]\displaystyle{ \mathcal{A}\cap S\in\mathcal{K} }[/math] whenever [math]\displaystyle{ \mathcal{A}\in\mathbf{K} }[/math] and S is a combinatorial subspace of [math]\displaystyle{ [k]^n. }[/math] We shall call it subspace rich if for every m and every [math]\displaystyle{ \delta }[/math] there are constants [math]\displaystyle{ M=M(m,\delta) }[/math] and [math]\displaystyle{ c=c(m,\delta)\gt 0 }[/math] such that the following statement holds for all sufficiently large n.

Richness hypothesis. Let [math]\displaystyle{ \mathcal{A}\in\mathbf{K} }[/math] have density [math]\displaystyle{ \delta }[/math] in [math]\displaystyle{ [k]^n. }[/math] Choose a random M-dimensional subspace [math]\displaystyle{ S_0 }[/math] of [math]\displaystyle{ [k]^n }[/math] by randomly fixing all coordinates outside a randomly chosen set Z of size M. Next, choose a subspace [math]\displaystyle{ S_1\subset S_0 }[/math] of dimension m, uniformly at random from all such subspaces. Then [math]\displaystyle{ S_1\subset\mathcal{A} }[/math] with probability at least [math]\displaystyle{ c. }[/math]

To be continued

 [math]\displaystyle{  }[/math][math]\displaystyle{  }[/math][math]\displaystyle{  }[/math][math]\displaystyle{  }[/math][math]\displaystyle{  }[/math][math]\displaystyle{  }[/math][math]\displaystyle{  }[/math][math]\displaystyle{  }[/math][math]\displaystyle{  }[/math][math]\displaystyle{  }[/math][math]\displaystyle{  }[/math][math]\displaystyle{  }[/math][math]\displaystyle{  }[/math][math]\displaystyle{  }[/math][math]\displaystyle{  }[/math][math]\displaystyle{  }[/math][math]\displaystyle{  }[/math][math]\displaystyle{  }[/math][math]\displaystyle{  }[/math][math]\displaystyle{  }[/math][math]\displaystyle{  }[/math][math]\displaystyle{  }[/math][math]\displaystyle{  }[/math][math]\displaystyle{  }[/math][math]\displaystyle{  }[/math][math]\displaystyle{  }[/math][math]\displaystyle{  }[/math][math]\displaystyle{  }[/math][math]\displaystyle{  }[/math]