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On set of integers containing no k elements in arithmetic progression

On set of integers containing no k elements in arithmetic progression. E. Szemerédi Acta Arithmetica, 1975 Shuchi Chawla. The Problem and some History. [van der Waerden, 1926] Let N = S 1 [ S 2 . Then either S 1 or S 2 contains arbitrarily long APs.

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On set of integers containing no k elements in arithmetic progression

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  1. On set of integers containing no k elements in arithmetic progression E. Szemerédi Acta Arithmetica, 1975 Shuchi Chawla

  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? Shuchi Chawla, Computer Science

  3. The Problem and some History • We show [Sz., 1975] that: limn!1 rk(n)/n = 0 This was the motivation behind the Regularity Lemma Shuchi Chawla, Computer Science

  4. The Problem and some History • We show [Sz., 1975] that: limn!1 rk(n)/n = 0 • In other words: For every e and k, 9 N(k,e), such that for all n¸N(k,e), rk(n)·en Shuchi Chawla, Computer Science

  5. The Problem and some History • We show [Sz., 1975] that: limn!1 rk(n)/n = 0 • In other words: For every e and k, 9 N(k,e), such that for all n¸N(k,e), rk(n)·en • Equivalently, For every set R with +ve upper density, R contains arbitrarily long APs Shuchi Chawla, Computer Science

  6. Not obvious!! The Problem and some History • We show [Sz., 1975] that: limn!1 rk(n)/n = 0 • In other words: For every e and k, 9 N(k,e), such that for all n¸N(k,e), rk(n)·en • Equivalently, For every set R with limn!1|RÅ{1,…,n}|/n > 0, R contains arbitrarily long APs Shuchi Chawla, Computer Science

  7. From 1 sets to a bound on rk(n) • Assume 9 n1·n2·… and Riµ[0,ni) with |Ri|>en and Ri contains no k-term AP • Let {n’i} be a subseq with n’i+1¸3n’i and di=åj<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. Shuchi Chawla, Computer Science

  8. Regularity Lemma Definitions Main Proof The Plan Shuchi Chawla, Computer Science

  9. 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)} b – rate of convergence of g(l)/l ; m – distance from b • pm, fm, am and em defined analogously Shuchi Chawla, Computer Science

  10. Saturated Sets • S(;) = B(;) = {{n} : n2N} • S(t1,…,tm) = {X: sm(X)¸( bm-mm )tm and pm(X)¸( am-(..) )tm} A large fraction of Xi are saturated and a large fraction are perfect Shuchi Chawla, Computer Science

  11. 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} Shuchi Chawla, Computer Science

  12. 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)¹; Shuchi Chawla, Computer Science

  13. 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 Shuchi Chawla, Computer Science

  14. 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) · 2iamitm fails for · 2iamK(1-bm)tm indices j, j · tm, and, f(X,j,i,s) ¸ 1/K2 ½iamitm fails for · 2iamK(1- bm)tm indices j, tm/4 · j · 3tm/4 Shuchi Chawla, Computer Science

  15. A B I(X,i,s) : X0,0 … Xi,0 … Xs,0 … X0,1 … Xi,1 … Xs,1 … … … Xs,j … … Xi,j’ This sequence is in F(X,j,i,s) and ji=j’ or All these sets are perfect X0,tm-1 … Xi,tm-1 … Xs,tm-1 … Finally… the Graph! Shuchi Chawla, Computer Science

  16. The Proof: Part 1 • X is well-saturated if for all s, | pm(Xi,Cm,n(s)) - am|Cm,n(s)| |·d|Cm,n(s)| (whenever Cm,n(s) is large enough) • Lemma 4: If X2Gi is well saturated, then X2Gi+1 Proof uses regularity Shuchi Chawla, Computer Science

  17. The Proof: Part 2 • Lemma 5: Suppose for all x<l, X(x)2 Gi(t1,…,tm,K) and X(x)j and X(h)j are R-equivalent for all x,h,j<i, [x<l X(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. Shuchi Chawla, Computer Science

  18. The Proof: Part 3 • Lemma 6: X(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. Shuchi Chawla, Computer Science

  19. 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. Shuchi Chawla, Computer Science

  20. The Proof: Part 4 • Fact 12: For every m, and m’¸ h(m,i) and X(x)2 Di(t1,…,tm’,K), there exist Y(x)2 Gi(t1,…,tm,K) such that Y(x)|X(x) and the position of the Ys are the same in the Xs • Proof: induction on i Shuchi Chawla, Computer Science

  21. The Proof: Part 5 • Theorem: 8 m,i,K Di(t1,…,tm,K)¹; • Proof: induction on i Shuchi Chawla, Computer Science

  22. 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 Shuchi Chawla, Computer Science

  23. Concluding Remarks • Paper by Tim Growers A new proof of Szemerédi’s Theorem http://www.dpmms.cam.ac.uk/~wtg10/papers.html (129 pages!!!) Shuchi Chawla, Computer Science

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