View source for DHJ(k) implies multidimensional DHJ(k)

==Introduction==

This is a result that will be needed if the [[A_second_outline_of_a_density-increment_argument|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 < (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> .