on set of integers containing no k elements in arithmetic progression e. szemerédi acta...

On set of integers containing no k elements in arithmetic progression

E. SzemerédiActa Arithmetica, 1975

Shuchi Chawla

Shuchi Chawla, Computer Science 2

The Problem and some History

[van der Waerden, 1926] Let N = S1[S2. Then either S1 or S2 contains arbitrarily long APs.

Erdós and Türan [1936] defined rk(n) – the largest l for which an l -element sequence 2 {1,…,n} does not contain a k-term AP.

How big can rk(n) be? How does it grow with n?



We show [Sz., 1975] that:

limn!1 rk(n)/n = 0

This was the motivation behind the Regularity Lemma




limn!1 rk(n)/n = 0

In other words:

For every and k, 9 N(k,), such that for all n¸N(k,), rk(n)· n




limn!1 rk(n)/n = 0

In other words:


Equivalently,

For every set R with +ve upper density, R contains arbitrarily long APs




limn!1 rk(n)/n = 0

In other words:


Equivalently,

For every set R with limn!1|RÅ{1,…,n}|/n > 0, R contains arbitrarily long APs

Not obvious!!


From 1 sets to a bound on rk(n)

Assume 9 n1·n2·… and Riµ[0,ni) with |Ri|>n and Ri contains no k-term AP

Let {n’i} be a subseq with n’i+1¸3n’i and di=j<in’j

R’ = [i (Rn’i+di) Note that each R gets mapped to a disjoint set

R’ has +ve U.D. ) it contains a sequence of 3k-terms. Say A = {a+di | 0·i·3k}

Let Rn’l be the last set in this. Then either this or the second last one must contain k terms.


The Plan

Regularity Lemma

Definitions

Main Proof


A “few” definitions

Configurations of order m B(l1,…,lm) X2B(l1,…,lm), then, X=[iXi Xi2B(l1,…,lm-1) eg. (1,2, 5,6, 9,10)2 B(2,3)

t1,…,tm – numbers arising from reglem Saturated and Perfect configurations

S(l1,…,lm)½B(l1,lm) sm(X) = #i : Xi2S(l1,…,lm-1) gm(l) = max {sm(X) : X2B(t1,…,tm-1,l)} – rate of convergence of g(l)/l ; – distance from

pm, fm, m and m defined analogously


Saturated Sets

S(;) = B(;) = {{n} : n2N}

S(t1,…,tm) = {X : sm(X)¸( m-m )tm and pm(X)¸( m-(..) )tm}

A large fraction of Xi are saturated and a large fraction are perfect


R-equivalence and Perfect Sets

X and Y are R-equivalent if: for corresponding elements x2 X and y2 Y, x2 R , y2 R

P(t1,…,tm) is the “largest” equivalence class

P(;) = {{n} : n2R}


More Definitions

C(t1,…,tm,l) = {X2B(t1,…,tm,l) : sm+1(X)=l} i.e., all Xi are saturated

Fact: For appropriate choice of tis, S, P and C are non empty.

Di(t1,…,tm,K) = {X2C : all j<i have Xj2P}

Main Theorem: Dk-1(k) is non empty.

Proof by induction on i that for fixed k and any m, Di(t1,…,tm,K);


Further More Definitions

E(t,K) = all K-term APs with each term·t Fact: Given X2 B(t1,…,tm,K),

{ji} 2 E(tm,K) , [i<KXi,ji2 B(t1,…,tm,K)

E(t,K,j,i) = all APs in E(t,K) with j as the ith element e(t,K,j,i) = |E(t,K,j,i)|

Fact: e(t,K,j,i)·t and if t/4·j·3t/4, e(t,K,j,i)¸ t/K2


More Definitions (Last)

F(X,j,i,s) = all APs in E(tm,K,j,s) such that Xi’,j’2Di(t1,…,tm,K)

f(X,j,i,s) = |F(X,j,i,s)|

Gi(t1,…,tm,K) = those X in C, such that for every s:

f(X,j,i,s) · 2imitm fails for · 2im

K(1-m)tm indices j, j · tm, and,

f(X,j,i,s) ¸ 1/K2 ½imitm fails for · 2im

K(1- m)tm indices j, tm/4 · j · 3tm/4


Finally… the Graph!

This sequence is in F(X,j,i,s) and

ji=j’or

All these sets are perfect

B A

X0,0 … Xi,0 … Xs,0 …X0,1 … Xi,1 … Xs,1 …

X0,tm-1 … Xi,tm-1 … Xs,tm-1 …

Xi,j’

Xs,j…

……

…

I(X,i,s) :


The Proof: Part 1

X is well-saturated if for all s, | pm(Xi,C,(s)) - m|C,(s)| |· |C,(s)|

(whenever C,(s) is large enough)

Lemma 4: If X2Gi is well saturated, then X2Gi+1

Proof uses regularity


The Proof: Part 2

Lemma 5: Suppose for all <l, X()2 Gi(t1,…,tm,K) and X()

j and X()j are R-equivalent for all

,,j<i, [<l X()i 2 C, then one of them is in

Gi+1

Proof by contradiction: Show that one of them has to be well saturated.


The Proof: Part 3

Lemma 6: X() are as before. Then, there exists a sequence of lm APs in E(tm,K) such that their ith elements form an AP and for each X and for each AP, [i’<K Xi’j’ 2 Di

Proof: Consider any X. Define Z to be the set of indices j such that f(j,i,i) is non empty. Show that this set contains an AP of length lm. Since Xs are R-equivalent, the conditions will hold for every X.


One more definition…

If Y2 B(t1,…,tm,K) and X2 B(t1,…,tm’,K), and, Yi is a subconfig of Xi, we write Y|X

If Y|X and Y’|X’, we say that the position of Y in X is the same as that of Y’ in X’ if for each i, Yi is the jith subconfig of Xi and same for Y’i.


The Proof: Part 4

Fact 12: For every m, and m’¸ h(m,i) and X()2 Di(t1,…,tm’,K), there exist Y()2 Gi(t1,…,tm,K) such that Y()|X() and the position of the Ys are the same in the Xs

Proof: induction on i


The Proof: Part 5

Theorem: 8 m,i,K Di(t1,…,tm,K);

Proof: induction on i


Summary of results

Lem 4: Gi & well-sat ) Gi+1

Lem 5: Gi & R-equiv for j· i ) Gi+1

Lem 6: Gi & R-equiv ) Di

Fact 12: Di ) Gi

Theorem: non empty Di ) non empty Di+1


Concluding Remarks

Paper by Tim GrowersA new proof of Szemerédi’s

Theoremhttp://www.dpmms.cam.ac.uk/~wtg10/papers.html

(129 pages!!!)

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