DHJ(2.7): Difference between revisions

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New page: '''DHJ (2.7)''': For all <math>\delta>0</math> there exists <math>n=n(\delta)</math> such that if <math>A\subset [3]^n</math> with <math>|A|> \delta 3^n</math>, there exists <math>\alpha\i...
 
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'''DHJ (2.7)''': For all <math>\delta>0</math> there exists <math>n=n(\delta)</math> such that if <math>A\subset [3]^n</math> with <math>|A|> \delta 3^n</math>, there exists <math>\alpha\in {\mathbf N}</math>, <math>|\alpha| ??? 0</math>.  
'''DHJ (2.7)''': For all <math>\delta>0</math> there exists <math>n=n(\delta)</math> such that if <math>A\subset [3]^n</math> with <math>|A|> \delta 3^n</math>, there exists <math>\alpha\subset [n]</math>, and variable words <math> w_{01}, w_{12}, w_{20}</math> such that <math>\{n: w_{xy}=3\} =\alpha, \; xy\in \{01,12,02\},</math> and <math>\big\{w_{xy}(x),w_{xy}(y):xy\in \{01,12,02\}\big\}\subset A.</math> (Remark: <math>w(x)</math> is just <math>w</math> with <math>x</math> substituted for 3.)


In other words, DHJ(2.7) asserts that in a dense set of <math>[3]^n</math>, one can find three ''parallel'' [[combinatorial line]]s, which intersect the set in the 0 and 1, 1 and 2, and 2 and 0 positions respectively.  This is slightly stronger than [[DHJ(2.6)]] and is implied by [[DHJ(3)]].
'''Proof'''. Let <math>\delta_0</math> be the infimum of the set of <math>\delta</math> for which the conclusion holds and assume for contradiction that <math>\delta_0>0</math>.
Let <math>m>{3\over \delta_0}</math> and choose by Ramsey’s theorem an <math>r</math> such that for any 8-coloring of the 2-subsets of <math>[r]</math>, there is an <math>m</math>-subset
Let <math>m>{3\over \delta_0}</math> and choose by Ramsey’s theorem an <math>r</math> such that for any 8-coloring of the 2-subsets of <math>[r]</math>, there is an <math>m</math>-subset
<math>B\subset [r]</math> all of whose 2-subsets have the same color. Let <math>\delta=\delta_0-{\delta_0\over 4\cdot 3^r}</math> and put <math>n_1=n(
<math>B\subset [r]</math> all of whose 2-subsets have the same color. Let <math>\delta=\delta_0-{\delta_0\over 4\cdot 3^r}</math> and put <math>n_1=n(
\delta_0+{\delta_0\over 4\cdot 3^r})</math>. Finally put <math>n=r+n_1</math> and suppose <math>A\subset [3]^n</math> with <math>|A|>\delta 3^n</math>. For each <math>v\in [3]^r</math>,
\delta_0+{\delta_0\over 4\cdot 3^r})</math>. Finally put <math>n=r+n_1</math> and suppose <math>A\subset [3]^n</math> with <math>|A|>\delta 3^n</math>. For each <math>v\in [3]^r</math>,
let <math>E_v=\{w\in [3]^{n_1}:vw\in A\}</math>. If <math>|E_v|> (\delta_0+{\delta_0\over 4\cdot 3^r})3^{n_1}</math> for some <math>v</math> we are done; otherwise
let <math>E_v=\{w\in [3]^{n_1}:vw\in A\}</math>. If <math>|E_v|> (\delta_0+{\delta_0\over 4\cdot 3^r})3^{n_1}</math> for some <math>v</math> we are done; otherwise
<math>|E_v| > {\delta_0\over 3}</math> for ''every'' <math>v\in [3]^r</math>.
<math>|E_v| > {\delta_0\over 3}3^{n_1}</math> for ''every'' <math>v\in [3]^r</math>.


Some notation: for <math>i\in [r]</math> and <math>xy\in \{10,21,02\}</math>, let <math>v_i^{xy}\in [3]^r</math> be the word  
Some notation: for <math>i\in [r]</math> and <math>xy\in \{10,21,02\}</math>, let <math>v_i^{xy}\in [3]^r</math> be the word  
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:<math>z_1z_2\cdots z_r</math>,
:<math>z_1z_2\cdots z_r</math>,


where <math>z_a=x</math> if <math>0\leq a\leq i</math> and <math>z_a=y</math> otherwise. Color <math>\{1,j\}</math>, <math>0\leq i<j<r</math>, according to whether or not the sets
where <math>z_a=x</math> if <math>0\leq a\leq i</math> and <math>z_a=y</math> otherwise. Color <math>\{i,j\}</math>, <math>0\leq i<j<r</math>, according to whether or not the sets
<math>E_{v_i^{xy}}\cap E_{v_j^{xy}}</math> are empty or not, <math>xy\in \{10,21,02\}</math>. (This is an 8-coloring.) By choice of <math>r</math> we can find
<math>E_{v_i^{xy}}\cap E_{v_j^{xy}}</math> are empty or not, <math>xy\in \{10,21,02\}</math>. (This is an 8-coloring.) By choice of <math>r</math> we can find
<math>0\leq k_1<k_2<\cdots <k_m<r</math> such that <math>\big\{\{ k_i,k_j\}: 0\leq i<j<m \big\}</math> is monochromatic for this coloring. By pigeonhole,
<math>0\leq k_1<k_2<\cdots <k_m<r</math> such that <math>\big\{\{ k_i,k_j\}: 0\leq i<j<m \big\}</math> is monochromatic for this coloring. By pigeonhole,

Latest revision as of 22:29, 23 February 2009

DHJ (2.7): For all [math]\displaystyle{ \delta\gt 0 }[/math] there exists [math]\displaystyle{ n=n(\delta) }[/math] such that if [math]\displaystyle{ A\subset [3]^n }[/math] with [math]\displaystyle{ |A|\gt \delta 3^n }[/math], there exists [math]\displaystyle{ \alpha\subset [n] }[/math], and variable words [math]\displaystyle{ w_{01}, w_{12}, w_{20} }[/math] such that [math]\displaystyle{ \{n: w_{xy}=3\} =\alpha, \; xy\in \{01,12,02\}, }[/math] and [math]\displaystyle{ \big\{w_{xy}(x),w_{xy}(y):xy\in \{01,12,02\}\big\}\subset A. }[/math] (Remark: [math]\displaystyle{ w(x) }[/math] is just [math]\displaystyle{ w }[/math] with [math]\displaystyle{ x }[/math] substituted for 3.)

In other words, DHJ(2.7) asserts that in a dense set of [math]\displaystyle{ [3]^n }[/math], one can find three parallel combinatorial lines, which intersect the set in the 0 and 1, 1 and 2, and 2 and 0 positions respectively. This is slightly stronger than DHJ(2.6) and is implied by DHJ(3).

Proof. Let [math]\displaystyle{ \delta_0 }[/math] be the infimum of the set of [math]\displaystyle{ \delta }[/math] for which the conclusion holds and assume for contradiction that [math]\displaystyle{ \delta_0\gt 0 }[/math]. Let [math]\displaystyle{ m\gt {3\over \delta_0} }[/math] and choose by Ramsey’s theorem an [math]\displaystyle{ r }[/math] such that for any 8-coloring of the 2-subsets of [math]\displaystyle{ [r] }[/math], there is an [math]\displaystyle{ m }[/math]-subset [math]\displaystyle{ B\subset [r] }[/math] all of whose 2-subsets have the same color. Let [math]\displaystyle{ \delta=\delta_0-{\delta_0\over 4\cdot 3^r} }[/math] and put [math]\displaystyle{ n_1=n( \delta_0+{\delta_0\over 4\cdot 3^r}) }[/math]. Finally put [math]\displaystyle{ n=r+n_1 }[/math] and suppose [math]\displaystyle{ A\subset [3]^n }[/math] with [math]\displaystyle{ |A|\gt \delta 3^n }[/math]. For each [math]\displaystyle{ v\in [3]^r }[/math], let [math]\displaystyle{ E_v=\{w\in [3]^{n_1}:vw\in A\} }[/math]. If [math]\displaystyle{ |E_v|\gt (\delta_0+{\delta_0\over 4\cdot 3^r})3^{n_1} }[/math] for some [math]\displaystyle{ v }[/math] we are done; otherwise [math]\displaystyle{ |E_v| \gt {\delta_0\over 3}3^{n_1} }[/math] for every [math]\displaystyle{ v\in [3]^r }[/math].

Some notation: for [math]\displaystyle{ i\in [r] }[/math] and [math]\displaystyle{ xy\in \{10,21,02\} }[/math], let [math]\displaystyle{ v_i^{xy}\in [3]^r }[/math] be the word

[math]\displaystyle{ z_1z_2\cdots z_r }[/math],

where [math]\displaystyle{ z_a=x }[/math] if [math]\displaystyle{ 0\leq a\leq i }[/math] and [math]\displaystyle{ z_a=y }[/math] otherwise. Color [math]\displaystyle{ \{i,j\} }[/math], [math]\displaystyle{ 0\leq i\lt j\lt r }[/math], according to whether or not the sets [math]\displaystyle{ E_{v_i^{xy}}\cap E_{v_j^{xy}} }[/math] are empty or not, [math]\displaystyle{ xy\in \{10,21,02\} }[/math]. (This is an 8-coloring.) By choice of [math]\displaystyle{ r }[/math] we can find [math]\displaystyle{ 0\leq k_1\lt k_2\lt \cdots \lt k_m\lt r }[/math] such that [math]\displaystyle{ \big\{\{ k_i,k_j\}: 0\leq i\lt j\lt m \big\} }[/math] is monochromatic for this coloring. By pigeonhole, for, say, [math]\displaystyle{ xy=10 }[/math], there are [math]\displaystyle{ i\lt j }[/math] such that [math]\displaystyle{ E_{v_{k_i}^{xy}}\cap E_{v_{k_j}^{xy}}\neq \emptyset }[/math], hence non-empty for all [math]\displaystyle{ i,j }[/math] by monochromicity and similarly for [math]\displaystyle{ xy=21,02 }[/math]. Now for [math]\displaystyle{ xy=10,21,02 }[/math], pick [math]\displaystyle{ u_{xy}\in E_{v_{k_1}^{xy}}\cap E_{v_{k_2}^{xy}} }[/math] and put [math]\displaystyle{ q_{xy}=s_1s_2\cdots s_r }[/math], where [math]\displaystyle{ s_i=x }[/math], [math]\displaystyle{ 0\leq i\lt k_1 }[/math], [math]\displaystyle{ s_i=3 }[/math], [math]\displaystyle{ k_1\leq i\lt k_2 }[/math], and [math]\displaystyle{ s_i=y }[/math], [math]\displaystyle{ k_2\leq i\lt r }[/math]. Finally put [math]\displaystyle{ w_{xy}=q_{xy}u_{xy} }[/math]. Then [math]\displaystyle{ w_{xy}(x)=v^{xy}_{k_2} u_{xy} }[/math] and [math]\displaystyle{ w_{xy}(y)=v^{xy}_{k_1} u_{xy} }[/math] are in [math]\displaystyle{ A }[/math], [math]\displaystyle{ xy\in \{10,21,02\} }[/math]. Hence [math]\displaystyle{ n=n(\delta) }[/math], contradicting [math]\displaystyle{ \delta\lt \delta_0 }[/math].