DHJ(k) implies multidimensional DHJ(k)

Introduction

This is a result that will be needed if the proof of DHJ(3) is correct and we want to push it through for DHJ(k).

The proof

Let [math]\displaystyle{ \mathcal{A} }[/math] be a density-[math]\displaystyle{ \delta }[/math] subset of [math]\displaystyle{ [k]^n }[/math] and let M be large enough so that every subset of [math]\displaystyle{ [k]^M }[/math] of density at least [math]\displaystyle{ \theta }[/math] contains a combinatorial line. Now split [math]\displaystyle{ [k]^n }[/math] up into [math]\displaystyle{ [k]^M\times[k]^{n-M}. }[/math] For a proportion at least [math]\displaystyle{ \delta/2 }[/math] of the points y in [math]\displaystyle{ [k]^{n-M} }[/math] the set of [math]\displaystyle{ x\in[k]^M }[/math] such that [math]\displaystyle{ (x,y)\in\mathcal{A} }[/math] has density at least [math]\displaystyle{ \delta/2. }[/math] Therefore, by DHJ(k) (with [math]\displaystyle{ \theta=\delta/2 }[/math]) we have a combinatorial line. Since there are fewer than [math]\displaystyle{ (k+1)^M }[/math] combinatorial lines to choose from, by the pigeonhole principle we can find a combinatorial line [math]\displaystyle{ L\subset[k]^M }[/math] and a set [math]\displaystyle{ \mathcal{A}_1 }[/math] of density [math]\displaystyle{ \delta/2(k+1)^M }[/math] in [math]\displaystyle{ [k]^{n-M} }[/math] such that [math]\displaystyle{ (x,y)\in\mathcal{A} }[/math] whenever [math]\displaystyle{ x\in L }[/math] and [math]\displaystyle{ y\in\mathcal{A}_1. }[/math] And now by induction we can find an (m-1)-dimensional subspace in [math]\displaystyle{ \mathcal{A}_1 }[/math] and we're done.

A weak multidimensional DHJ(k) implies DHJ(k)

It is also true that a weak multidimensional DHJ(k) implies DHJ(k). We will show that the following statement is equivalent to DHJ(k):

“There is a constant, c < 1 that for every d there is an n that any c-dense subset of [math]\displaystyle{ [k]^n }[/math] contains a d-dimensional subspace.”

We should show that the statement above implies DHJ(k). As before, write [math]\displaystyle{ [k]^n }[/math] as [math]\displaystyle{ [k]^r\times[k]^s }[/math] , where s is much bigger than r. For each [math]\displaystyle{ y\in [k]^s }[/math] , define [math]\displaystyle{ \mathcal{A}_y }[/math] to be [math]\displaystyle{ \{x\in[k]^r:(x,y)\in\mathcal{A}\} }[/math] . Let Y denote the set of [math]\displaystyle{ y\in [k]^s }[/math] , that \mathcal{A}_y</math> is empty. Suppose that [math]\displaystyle{ \mathcal{A} }[/math] is large, line-free, and its density is [math]\displaystyle{ \delta =\Delta-\epsilon }[/math] where [math]\displaystyle{ \Delta }[/math] is the limit of density of line-free sets and [math]\displaystyle{ \epsilon \lt (1-c)\Delta }[/math] . We can also suppose that no [math]\displaystyle{ \mathcal{A}_y }[/math] has density much larger than [math]\displaystyle{ \Delta }[/math] as that would guarantee a combinatorial line. But then the density of Y is at most 1-c, so there is a c-dense set, [math]\displaystyle{ Z=[k]^s-Y }[/math], such that any element is a tail of some elements of [math]\displaystyle{ \mathcal{A} }[/math] . For every [math]\displaystyle{ y \in Z }[/math] choose an [math]\displaystyle{ x\in [k]^r:(x,y)\in\mathcal{A} }[/math] . This x will be the colour of y. It gives a [math]\displaystyle{ [k]^r }[/math] colouring on Z. By the initial condition Z contains arbitrary large subspaces, so by HJ(k) we get a line in [math]\displaystyle{ \mathcal{A} }[/math] .