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

2. Shuchi Chawla, Computer Science2 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.  Erdós and Türan [1936] defined r k (n) – the largest l for which an l -element sequence 2 {1,…,n} does not contain a k-term AP.  How big can r k (n) be? How does it grow with n?

3. Shuchi Chawla, Computer Science3 The Problem and some History  We show [Sz., 1975] that: lim n !1 r k (n)/n = 0 This was the motivation behind the Regularity Lemma

4. Shuchi Chawla, Computer Science4 The Problem and some History  We show [Sz., 1975] that: lim n !1 r k (n)/n = 0  In other words: For every  and k, 9 N(k,  ), such that for all n ¸ N(k,  ), r k (n) ·  n

5. Shuchi Chawla, Computer Science5 The Problem and some History  We show [Sz., 1975] that: lim n !1 r k (n)/n = 0  In other words: For every  and k, 9 N(k,  ), such that for all n ¸ N(k,  ), r k (n) ·  n  Equivalently, For every set R with +ve upper density, R contains arbitrarily long APs

6. Shuchi Chawla, Computer Science6 The Problem and some History  We show [Sz., 1975] that: lim n !1 r k (n)/n = 0  In other words: For every  and k, 9 N(k,  ), such that for all n ¸ N(k,  ), r k (n) ·  n  Equivalently, For every set R with lim n !1 |R Å {1,…,n}|/n > 0, R contains arbitrarily long APs Not obvious!!

7. Shuchi Chawla, Computer Science7 From 1 sets to a bound on r k (n)  Assume 9 n 1 · n 2 · … and R i µ [0,n i ) with |R i |>  n and R i contains no k-term AP  Let {n’ i } be a subseq with n’ i+1 ¸ 3n’ i and d i =  j<i n’ j  R’ = [ i (R n’i +d i )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 R n’l be the last set in this. Then either this or the second last one must contain k terms.

8. Shuchi Chawla, Computer Science8 The Plan Regularity Lemma Definitions Main Proof

9. Shuchi Chawla, Computer Science9 A “few” definitions  Configurations of order m B(l 1,…,l m ) X 2 B(l 1,…,l m ), then, X= [ i X i X i 2 B(l 1,…,l m-1 ) eg. (1,2, 5,6, 9,10) 2 B(2,3)  t 1,…,t m – numbers arising from reglem  Saturated and Perfect configurations  S(l 1,…,l m ) ½ B(l 1,l m )s m (X) = #i : X i 2 S(l 1,…,l m-1 ) g m (l) = max {s m (X) : X 2 B(t 1,…,t m-1,l)}  – rate of convergence of g(l)/l ;  – distance from   p m, f m,  m and  m defined analogously

10. Shuchi Chawla, Computer Science10 Saturated Sets  S( ; ) = B( ; ) = {{n} : n 2 N}  S(t 1,…,t m ) = {X : s m (X) ¸ (  m -  m )t m and p m (X) ¸ (  m -  (..) )t m } A large fraction of X i are saturated and a large fraction are perfect

11. Shuchi Chawla, Computer Science11 R-equivalence and Perfect Sets  X and Y are R-equivalent if: for corresponding elements x 2 X and y 2 Y, x 2 R, y 2 R  P(t 1,…,t m ) is the “largest” equivalence class  P( ; ) = {{n} : n 2 R}

12. Shuchi Chawla, Computer Science12 More Definitions  C(t 1,…,t m,l) = {X 2 B(t 1,…,t m,l) : s m+1 (X)=l} i.e., all X i are saturated  Fact: For appropriate choice of t i s, S, P and C are non empty.  D i (t 1,…,t m,K) = {X 2 C : all j<i have X j 2 P}  Main Theorem: D k-1 (k) is non empty. Proof by induction on i that for fixed k and any m, D i (t 1,…,t m,K)  ;

13. Shuchi Chawla, Computer Science13 Further More Definitions  E(t,K) = all K-term APs with each term · t  Fact: Given X 2 B(t 1,…,t m,K), {j i } 2 E(t m,K), [ i<K X i,ji 2 B(t 1,…,t m,K)  E(t,K,j,i) = all APs in E(t,K) with j as the i th 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/K 2

14. Shuchi Chawla, Computer Science14 More Definitions (Last)  F(X,j,i,s) = all APs in E(t m,K,j,s) such that X i’,j’ 2 D i (t 1,…,t m,K) f(X,j,i,s) = |F(X,j,i,s)|  G i (t 1,…,t m,K) = those X in C, such that for every s: f(X,j,i,s) · 2 i  m i t m fails for · 2i  m K (1-  m )t m indices j, j · t m, and, f(X,j,i,s) ¸ 1/K 2 ½ i  m i t m fails for · 2i  m K (1-  m )t m indices j, t m /4 · j · 3t m /4

15. Shuchi Chawla, Computer Science15 Finally… the Graph! This sequence is in F(X,j,i,s) and j i =j’ or All these sets are perfect B A X 0,0 … X i,0 … X s,0 … X 0,1 … X i,1 … X s,1 … X 0,tm-1 … X i,tm-1 … X s,tm-1 … X i,j’ X s,j … … … … I(X,i,s) :

16. Shuchi Chawla, Computer Science16 The Proof: Part 1  X is well-saturated if for all s, | p m (X i,C , (s)) -  m |C , (s)| | ·  |C , (s)| (whenever C , (s) is large enough)  Lemma 4: If X 2 G i is well saturated, then X 2 G i+1 Proof uses regularity

17. Shuchi Chawla, Computer Science17 The Proof: Part 2  Lemma 5: Suppose for all  <l, X (  ) 2 G i (t 1,…,t m,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 G i+1  Proof by contradiction: Show that one of them has to be well saturated.

18. Shuchi Chawla, Computer Science18 The Proof: Part 3  Lemma 6: X (  ) are as before. Then, there exists a sequence of l m APs in E(t m,K) such that their i th elements form an AP and for each X and for each AP, [ i’<K X i’j’ 2 D i  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 l m. Since Xs are R-equivalent, the conditions will hold for every X.

19. Shuchi Chawla, Computer Science19 One more definition…  If Y 2 B(t 1,…,t m,K) and X 2 B(t 1,…,t m’,K), and, Y i is a subconfig of X i, 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, Y i is the j i th subconfig of X i and same for Y’ i.

20. Shuchi Chawla, Computer Science20 The Proof: Part 4  Fact 12: For every m, and m’ ¸ h(m,i) and X (  ) 2 D i (t 1,…,t m’,K), there exist Y (  ) 2 G i (t 1,…,t m,K) such that Y (  ) |X (  ) and the position of the Ys are the same in the Xs  Proof: induction on i

21. Shuchi Chawla, Computer Science21 The Proof: Part 5  Theorem: 8 m,i,K D i (t 1,…,t m,K)  ;  Proof: induction on i

22. Shuchi Chawla, Computer Science22 Summary of results  Lem 4: G i & well-sat ) G i+1  Lem 5: G i & R-equiv for j · i ) G i+1  Lem 6: G i & R-equiv ) D i  Fact 12: D i ) G i  Theorem: non empty D i ) non empty D i+1

23. Shuchi Chawla, Computer Science23 Concluding Remarks  Paper by Tim Growers A new proof of Szemerédi’s Theorem http://www.dpmms.cam.ac.uk/~wtg10/papers.html (129 pages!!!)