DHJ(1,3): Difference between revisions
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*Let <math>\mathcal{U}, \mathcal{V}</math> and <math>\mathcal{W}</math> be three collections of subsets of <math>[n]</math> and let <math>\mathcal{A}</math> be the set of all sequences <math>x\in[3]^n</math> such that the 1-set, 2-set and 3-set of <math>x</math> belongs to <math>\mathcal{U}, \mathcal{V}</math> and <math>\mathcal{W},</math> respectively. If <math>\mathcal{A}</math> is dense, does it follow that it contains a combinatorial line? | *Let <math>\mathcal{U}, \mathcal{V}</math> and <math>\mathcal{W}</math> be three collections of subsets of <math>[n]</math> and let <math>\mathcal{A}</math> be the set of all sequences <math>x\in[3]^n</math> such that the 1-set, 2-set and 3-set of <math>x</math> belongs to <math>\mathcal{U}, \mathcal{V}</math> and <math>\mathcal{W},</math> respectively. If <math>\mathcal{A}</math> is dense, does it follow that it contains a combinatorial line? | ||
The answer is of course yes, since it is just DHJ(3) with an extra restriction on the dense set that one would like to contain a combinatorial line. The reasons for considering the question are (i) that it ought to be amenable to the kinds of techniques used to prove Sperner's theorem, and (ii) that sets of this special kind need to be understood, as they appear to be an important class of [[obstructions to uniformity]] in DHJ(3). Since Sperner's theorem is most naturally proved using [[equal-slices measure]] on <math>2^n,</math> we shall use it here. | The answer is of course yes, since it is just DHJ(3) with an extra restriction on the dense set that one would like to contain a combinatorial line. The reasons for considering the question are (i) that it ought to be amenable to the kinds of techniques used to prove Sperner's theorem, and (ii) that sets of this special kind need to be understood, as they appear to be an important class of [[obstructions to uniformity]] in DHJ(3). Since Sperner's theorem is most naturally proved using [[equal-slices measure]] on <math>[2]^n,</math> we shall use it here. | ||
Since any element <math>x\in[3]^n</math> is determined by just its 1-set and its 2-set, it is natural to consider the yet simpler variant of DHJ(1,3) where <math>\mathcal{W}</math> consists of all subsets of <math>[n].</math> This problem can be solved by an easy adaptation of one of the proofs of Sperner's theorem. | Since any element <math>x\in[3]^n</math> is determined by just its 1-set and its 2-set, it is natural to consider the yet simpler variant of DHJ(1,3) where <math>\mathcal{W}</math> consists of all subsets of <math>[n].</math> This problem can be solved by an easy adaptation of one of the proofs of Sperner's theorem. | ||
To be continued. | To be continued. |
Revision as of 03:26, 18 February 2009
Recall that DHJ(1,3) is the following problem.
- Let [math]\displaystyle{ \mathcal{U}, \mathcal{V} }[/math] and [math]\displaystyle{ \mathcal{W} }[/math] be three collections of subsets of [math]\displaystyle{ [n] }[/math] and let [math]\displaystyle{ \mathcal{A} }[/math] be the set of all sequences [math]\displaystyle{ x\in[3]^n }[/math] such that the 1-set, 2-set and 3-set of [math]\displaystyle{ x }[/math] belongs to [math]\displaystyle{ \mathcal{U}, \mathcal{V} }[/math] and [math]\displaystyle{ \mathcal{W}, }[/math] respectively. If [math]\displaystyle{ \mathcal{A} }[/math] is dense, does it follow that it contains a combinatorial line?
The answer is of course yes, since it is just DHJ(3) with an extra restriction on the dense set that one would like to contain a combinatorial line. The reasons for considering the question are (i) that it ought to be amenable to the kinds of techniques used to prove Sperner's theorem, and (ii) that sets of this special kind need to be understood, as they appear to be an important class of obstructions to uniformity in DHJ(3). Since Sperner's theorem is most naturally proved using equal-slices measure on [math]\displaystyle{ [2]^n, }[/math] we shall use it here.
Since any element [math]\displaystyle{ x\in[3]^n }[/math] is determined by just its 1-set and its 2-set, it is natural to consider the yet simpler variant of DHJ(1,3) where [math]\displaystyle{ \mathcal{W} }[/math] consists of all subsets of [math]\displaystyle{ [n]. }[/math] This problem can be solved by an easy adaptation of one of the proofs of Sperner's theorem.
To be continued.