DHJ(k) implies multidimensional DHJ(k): Difference between revisions

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

Revision as of 12:18, 23 April 2009



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] such that [math]\displaystyle{ \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] .