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

We should show that the statement above implies DHJ(k). As before, write [math] [k]^n[/math] as [math] [k]^r\times[k]^s [/math] , where s is much bigger than r. For each [math] y\in [k]^s[/math] , define [math] \mathcal{A}_y[/math] to be [math] \{x\in[k]^r:(x,y)\in\mathcal{A}\}[/math] . Let Y denote the set of [math] y\in [k]^s[/math] such that [math]\mathcal{A}_y[/math] is empty. Suppose that [math] \mathcal{A} [/math] is large, line-free, and its density is [math] \delta =\Delta-\epsilon[/math] where [math] \Delta[/math] is the limit of density of line-free sets and [math] \epsilon \lt (1-c)\Delta[/math] . We can also suppose that no [math] \mathcal{A}_y[/math] has density much larger than [math] \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] Z=[k]^s-Y[/math], such that any element is a tail of some elements of [math] \mathcal{A}[/math] . For every [math] y \in Z[/math] choose an [math] x\in [k]^r:(x,y)\in\mathcal{A}[/math] . This x will be the colour of y. It gives a [math] [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] \mathcal{A}[/math] .