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.