STABLE GRAPHS BENJAMIN OYE

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1 STABLE GRAPHS BENJAMIN OYE Abstract. In Reguarity Lemmas for Stabe Graphs [1] Maiaris and Sheah appy toos from mode theory to obtain stronger forms of Ramsey's theorem and Szemeredi's reguarity emma for stabe graphs," graphs which admit a uniform nite bound on the size of an induced sub-haf-graph. This paper provides a background to the rst theorem of that, an improved form of Ramsey's theorem for stabe graphs without mode theory as a prerequisite. Contents 1. Introduction 1 2. Denitions 2 3. Order and Indiscernibes 3 4. Random Graphs 3 5. Ordery graphs 4 6. Stabe Simpe Graphs 5 7. Stabe Hypergraphs 9 Acknowedgments 12 References Introduction Ramsey theory studies the conditions under which order arises in mathematica structures. Whie Ramsey probems are often stated in the anguage of graphs, the connection to ogic has been evident since F.P Ramsey's 1930 On a Probem of Forma Logic which proved the centra theorem now known as Ramsey's theorem. The theorem which wi be deveoped in this paper was proved in a paper utimatey aimed at improving Szemerédi's Reguarity Lemma for a certain cass of graphs. Theorem 1.1 (Szemerédi's Reguarity Lemma). For every ɛ, m there exist N = N(ɛ, m), m = m(ɛ, m) such that given any nite graph X of at east N vertices, there is a k with m k m and a partition X = X 1 X k satisfying: (1) X i X j 1 for a i, j k and (2) a but at most ɛk 2 of the pairs (X i, X j ) are ɛ-reguar. Whie it was not known for some time after the origina proof whether the ɛk 2 non ɛ-reguar pairs were necessary, mutipe researchers independenty noted that the exampe of the haf graph demonstrates that they are necessary [4]. In essence, the haf graph cannot be partitioned nicey using the Szemeredi method. In [1] Maiaris and Sheah show that haf graphs are essentiay the ony dicuty in 1

2 2 BENJAMIN OYE creating reguar we-behaved partitions. First they show that Ramsey's theorem works better than usua for stabe graphs (graphs which admit a bound on the size of an induced haf-graph), then use this improved Ramsey theorem to construct a variety of improved reguarity emmas for stabe graphs. A haf-graph of height 6 The idea for the work comes from a coincidence with mode theory: the haf graph witnesses the mode theoretic order property. Its presence aigns with mode theoretic instabiity and in its absence the toos of stabiity theory can be appied. The mode theoretic notions invoved in the construction of the reguarity emmas are beyond the scope of this paper. Rather, we wi be concerned ony with the improved Ramsey emma, extending this emma to hypergraphs, and the exposition uses a minimum of mode theory. After a brief description of Ramsey's theorem, we wi examine the orderiness of random or average graphs. After a brief reection on casses of ordery graphs and the Erd s Hájna conjecture, we wi discuss the orderiness of stabe graphs and nay of stabe hypergraphs. 2. Definitions We dene graphs in the mode theoretic way. Denition 2.1. An hypergraph is an ordered pair consisting of a set and a coection of truth vaued formuas on that set (G, ) where a formuas in are invariant under permutation of the variabes and the interpretation G consists of a X G connected by an edge. For our purposes, hypergraphs are nite. Note that whie this denition may ook dierent from the traditiona hypergraph denition, it is equivaent. Denition 2.2. An hypergraph is m-uniform if every foruma in the theory is m-ary. We then write (G, R) for the graph. For our purposes, hypergraphs are uniform. Denition 2.3. A 2-uniform hypergraph is caed a simpe graph or just a graph. In this case we sometimes write the reation R(x, y) as xry. Denition 2.4. An induced subgraph of a graph (G, R) is a graph (H, R) where H G. If G has H as an induced subgraph, then G induces H. If G has no subgraph isomorphic to H, then G omits H. Denition 2.5. An m reguar hypergraph (G, R) is homogeneous if there exists a truth vaue t such that for a X G, X = m, R(X) = t. Homogeneous sets, in the anguage of simpe graphs, are either compete or discrete.

3 STABLE GRAPHS 3 Denition 2.6. A sequence a i : i I in some mode is indiscernibe with respect to a coection of formuas Γ if for every n, and formua ϕ(x 1,..., x n ) Γ, every i 1 < < i n I, j 1 < < j n I, ϕ(a i1... a in ) ϕ(a j1... a jn ). Remark 2.7. In the theory of m-uniform hypergraphs, the m-ary edge reation with equaity, an indiscernibe sequence a i : i < n is a either a compete or discrete induced subgraph graph. Denition 2.8 (Privieged symbos). Let a, k, m, n < ω. 3. Order and Indiscernibes The two centra theorems of Ramsey theory are often phrased in the foowing way: Theorem 3.1 (Cooring Innite Ramsey"). Every k-edge-cooring of an innite a-uniform compete hypergraph has an innite monochromatic compete hypergraph as an induced subgraph. Theorem 3.2 (Cooring Finite Ramsey"). There exists R = r(a, k, n) < ω such that every k-edge-cooring of a compete a-uniform hypergraph on at east R vertices has a monochromatic compete graph of size n as an induced subgraph. In the two coor case, we can think of one coor as designating the edges in the graph and the other coor designating the edges in the graph's compement, giving the foowing readings. Theorem 3.3 (Innite Ramsey). Every innite graph has an innite homogeneous set. Theorem 3.4 (Finite Ramsey). For a s < ω there exists R(s) such that every graph with at east R(s) vertices has a homogeneous set of size s. The reation between homogeneity and indiscernibe sequences is aready cear: for graphs and hypergraphs they are one and the same. In this way we can think of Finite Ramsey as showing the existence of a ower bound on the ength of a edge-reation-indiscernibe sequence of vertices. Homogeneous sets are quite ordery in the sense that they are uniform. Using this informa denition, Finite Ramsey's theorem states that arge ordery subgraphs become inevitabe in sucienty arge graphs, or compete disorder is impossibe. Whie it is trivia to construct the maximay ordery graphs (i.e. homogeneous graphs) the construction of minimay ordery graphs has ong proved chaenging, and the best bounds have been obtained not by expicit construction, but using the probabiistic method, initiated by Pá Erd s. 4. Random Graphs A random graph is a graph in which each edge has probabiity 1/2 of being in the graph. The study of these random graphs can give us insight into the orderiness of norma graphs. Ramsey's theorem shows that R(s) is we dened, but the cacuation of expicit vaues is dicut: R(s) is ony known for s 4. Nonetheess, the order of growth is known. Here, we wi give some bounds for the size of homogeneous subset of random graphs, proving Ramsey's theorem aong the way.

4 4 BENJAMIN OYE Denition 4.1. Let the the Ramsey function R(s) be the smaest positive integer such that every graph on R(s) nodes has a homogeneous set of size s. Theorem 4.2. R(s) 2 2s 3 Proof. Let n = 2 2s 3. Consider a random graph G(V, E) on n vertices where each edge is independenty in G or not in G with probabiity 1/2. We construct a sequence of neighborhoods V i recursivey. Step 1: Fix a vertex v 1 V. We have that {v V : vrv 1 } {v V : (vrv 1 )} = V so one set has size at east 2 2s 3 1. Let this set be V 1. Give the truth function t(v 1 ) = 1 if V 1 = {v V : vrv 1 } and 0 otherwise. Step i+1: Suppose that, after i steps, we have V i of size 2 2s 3 i. Fix a vertex v i+1 V i. We have that {v V i : vrv i+1 } {v V i : (vrv i+1 )} = V i so at east one of the sets has size 2 2s 3 (i+1). Let V i+1 be this set, and give the truth function t(v i+1 ) = 1 if V i+1 = {v V i : vrv i+1 } and 0 otherwise. We end this process in 2s 3 steps. Note that V 2s 3 2 2s 3 (2s 3) = 1. Let v 2s 2 V 2s 3. We have a sequence of 2s 3 subsets of V, each of which is assigned a truth vaue of 0 or 1, so by the pigeonhoe principe there must be a subsequence of s 1 neighborhoods a of which have the same truth vaue. V j1, V j1... V js 1 consider the sequence v j0, v j1,..., v js 1, v 2s 2. If the truth vaue of the sequence is 1, then each vertex in the sequence is connected to every subsequent vertex. If it is 0, each vertex is connected to none of the subsequent vertices. So G has a homogeneous set of size s [6]. Remark 4.3. Finite Ramsey theorem foows from the previous resut. Theorem 4.4. For s 3, R(s) > 2 s/2 Proof. Let n = 2 s/2. Consider a graph G on n vertices where each edge is independenty in the graph with probabiity 1/2. The probabiity that S G, S = s is homogeneous is 2 2 (s 2), the factor of two because S coud be compete or discrete. G has ( n s) distinct subgraphs of size s, and ( n s ) = n(n 1)... (n s 1) s! < ns s! so, using that n = 2 s/2 and s 3, the probabiity that there is a homogeneous set of size s is ( ) n 2 s 2 (s 2) < 2ns 21+s/2 = < 1 s!2s(s 1)/2 s! So with nonzero probabiity, a graph on 2 s/2 vertices has no homogeneous set of size s. Thus, R(s) > 2 s/2 [7]. Coroary 4.5. A random graph on n vertices has a homogeneous set of size Θ(og(n)). Proof. Let G be a random graph on more than six vertices. By the previous two theorems we have that G must contain a homogeneous set of size s where (og 2 (n) 3)/2 s < 2 og 2 (n) Remark 4.6. The constants have been improved by ater work.

5 STABLE GRAPHS 5 5. Ordery graphs We have given a tight ogarithmic ower bound for the growth of homogeneous sets of random graphs. Ceary, we cannot expect to do better than inear growth (no graph on n vertices can have a homogeneous set arger than n), so it makes sense to consider that ordery graphs as those which have a homogeneous set which grows poynomiay in the size of the graph. Erd s and Hajna made an important conjecture regarding ordery graphs, motivating the foowing denition. Denition 5.1. A graph H has the Erd s-hajna property if there exists a constant δ(h) > 0 such that every graph on n vertices which omits H has a homogeneous set of size at east n δ(h) Conjecture 5.2 (Erd s-hajna conjecture). Every graph has the Erd s-hajna property. The conjecture can be interpreted as the statement orderiness can be expressed in many ways, i.e. through the omission of any graph. For exampe, two vertex graphs ceary have the EH property, as do homogeneous sets by Ramsey's theorem. [3] 6. Stabe Simpe Graphs In this section, we consider stabe simpe graphs: graphs which admit a nite bound on the size of an induced haf-graph. The main resut of this section is the proof that the haf-graph has the Erd s Hajna property. The method of the proof, using the mode theoretic notion of type trees, can aso be appied to prove the Erd s Hajna property for other graphs. This presentation is based on the one found in [2]. Sheah's [9] is the standard exposition of stabiity theory. For the purposes of this paper, Theorem 6.1 (Sheah). A theory is unstabe i there is some formua ϕ(x, y) and {a i : i < ω} and {b k : j < ω} and ϕ(a i, b i ) i i < j. Denition 6.2. A graph is caed k-edge stabe if there do not exist a 1,... a k b 1,..., b k such that R(a i, b k ) i i < j. Note that the statement that G is k-edge stabe is equivaent to the statement that G does not have an induced haf-graph of height of k or greater. A key eement in the proof is the idea of indexing a graph's nodes with a tree. Denition 6.3. A tree is a partia order (P, ) such that for each p P, the set {q P : p q} is a we order under. Given an integer, n 2, dene 2 <n = {0, 1} i, n 1 i=0 where {0, 1} 0 = is the empty string and for i > 0 {0, 1} i is the usua cartesian product. This set has a natura tree structure given by η η if and ony if η = or η is an initia segment of η. Given η {0, 1} i, et η = i denote the ength of η (the ength of the empty string is 0).

6 6 BENJAMIN OYE An important concept is to take a graph G = (V, E) and arrange G into a tree by indexing its vertex set with the eements of 2 <n. Suppose G(V, E) is a graph and we have an indexing V = {a η : η X} of the vertices of G by some X 2 <n. Denition 6.4. Given a graph G = (V, E) on n vertices and A 2 <n, we say that an indexing V = {a η : η A} of V by eements of A is a type tree if for each η A the fowing hods. If η 0 A, then a η 0 is not adjacent to a η. If η 1 A, then a η 1 is adjacent to a η. If η 0 and η 1 are both in A, then for a η η, a η 1 is adjacent to a η if and ony if a η 0 is adjacent to a η. Note that we coud have repaced the rst condition with the foowing If η 1 η 2 then for a η η 1, R(η, η 1 ) R(η, η 2 ) an equivaence which wi be usefu ater. Lemma 6.5. Every nite graph G = (V, E) can be arranged into a type tree. Proof. Suppose that V = n. We arrange the vertices of G into a type tree indexed by a subset of 2 <n. Step 1: Choose any eement of G to be a, and set A 0 = {a }. Set X 1 = N(a ), the neighbors of a, and X 0 = V \ ({a } N(a )), the nonneighbors of a. Note that X 1, X 0 partition V \ A 0. Step m+1: Suppose we've dened eements of the tree up to height m 0 and for each 0 i m, A i is the set of vertices of height i. Suppose further that we have a coection of sets of vertices {X η i : η A m, i {0, 1}} which partition the unassigned vertices V \ m i=1 A i and such that for each η A m, X η 1 N(a η ) and X η 0 V \ {N(a η ) {a η }. Then for each η A m and i {0, 1}, if X η i, choose a η i to be any eement of X η i. Dene A m+1 to be the set of these a η i. Now for each vertex of height m + 1, a v A m+1 and i {0, 1} et X v 1 = N(a v ) X v and X v 0 = (V \ (N(a v ) {a v })) X v. By assumption, {X v : v A m+1 } is a partition of V \ m i=1 A i, and by construction, for each v A m+1, {X v 1, X v 0 } is a partition of X v \ A m+1. Therefore, {X v i : v A m+1, i {0, 1}} is a partition of V \ m+1 i=1 A i A eements of V wi be chosen after at most n steps. So we obtain an indexing of V by a subset of 2 <n which is a type tree by construction. Denition 6.6. Suppose G = (V, E) is a nite graph. (1) The tree rank of G, denoted t(g), is the argest integer t such that there is a subset V V and an indexing V = {a η : η 2 <t } which is a type tree. (i.e V is a fu binary type tree of height n.) (2) The tree height of G, denoted h(g) is the smaest integer h such that every indexing of V which is a type tree has a branch of ength h. Lemma 6.7. Suppose t, h are integers, and G = (V, E) is a nite graph with tree rank t and a tree height h. Then G contains a compete graph or independent set of size max{t, h/2}.

7 STABLE GRAPHS 7 Proof. By the denition of tree rank, there is a V V and an indexing V = {a η : η 2 <t } which is a type tree. By the denition of the standard type tree, I 1 = {a, a 0,..., a 0t 1 } is an independent set of size t. Aternativey, {a, a 1,..., a 1t 1 } is a compete graph of size t. On the other hand, by the denition of tree height and the emma which states that every graph can be indexed by a type tree, there is an indexing V = {a η : η B 2 <n } which is a standard type tree which has a branch J of ength h. Let a τ be the ast eement of J. If N(a τ ) J J /2 then et I 2 = N(a τ ) J. Otherwise, et I 2 = (V \ N(a τ )) J. In either case, we have that I 2 J /2 = h/2. Consider eements x, y I 2. Because x and y are on the same branch, we can assume without oss of generaity that x y. By the denition of I 2, either a z I 2 are adjacent to a τ or none of them are, so x is adjacent to a τ if and ony if y is adjacent to a τ. Because x precedes y, by construction of the type tree we have that a τ is adjacent to x if and ony x is adjacent to y. If I 2 = N(a τ ) J then a τ is adjacent to a x and hence a members of I 2 are adjacent to each other, so I 2 is a compete graph. If I 2 = (V \ N(a τ )) J then a τ is not adjacent to any x and hence no members of I 2 are adjacent to each other, so I 2 is a discrete graph. So G contains a compete or independent set of size max{ I 1, I 2 } = max{t, h/2}. Denition 6.8. Suppose G = (V, E) is a graph, A 2 <n, and V = {a η : η A} is a type tree. (1) Given an eement a η V, we say there is a fu binary tree of height k beow a η if there exists a set V {a σ : a η a σ } and a bijection f : V 2 <k with the property that a σ precedes a σ if and ony if f(a σ ) f(a σ ) in 2 <k. (2) The tree rank of an eement a η V, denoted t(a η ), is the argest k such that there is a fu binary tree of height k beow a η. Let p(a η ) denote the immediate predecessor to a η Theorem 6.9. Suppose that n 2 is an integer and G = (V, E) is a graph of size n. Then h(g) (n/t(g)) 1 t(g)+1 2 Proof. Note that the minimum tree height is attained when branching is maxima. Thus, a ower bound on tree height which hods for maximay branching trees wi hod for a type trees of a graph. Suppose that A 2 <n and V = {a η : η A} is a type tree with maxima branching. Under these conditions, given a η with t(a η ) = s, at most one of the immediate successors to a η has t(a η i ) = s and the other t(a η i ) = s 1 or both have t(a η i ) = s 1. Let h be the ength of the ongest branch of this tree, and et t = max{t(a η ): η A}. Note that t t(g), the maximum tree height attained by any indexing of G. Given a xed and s, Z s = {a η V : t(a η ) = s, ht(a η ) = } X s = {a η Z s : t(p(a η )) = s}, and Y s = {a η Z 2 : t(p(a η )) = s + 1}. Using words, Z s : the nodes in the indexing at height which admit a compete binary tree of at most height s beneath them.

8 8 BENJAMIN OYE X s: the nodes in Zs whose predecessors have a compete binary tree of at most height s beneath them. Y s: the nodes in Zs whose predecessors have a compete binary tree of at most height s + 1 beneath them. and et N s = Zs, xs = Xs and ys = Y s. Note that if a η admits a tree of height s beneath it, then the predecessor to a η, p(a η ) admits a tree of at east height s beneath it, the tree beneath a η, and at most a tree of height s + 1. Thus, Z s = Xs Y s and N s = xs + ys. Note aso that every node must admit a tree of height between 0 and t, so we can count the nodes by n = h t =0 s=0 N s. We caim that the foowing facts hod (i) For a s t and, x s +1 N s. (By maxima branching) (ii) For a s < t and a, y+1 s s+1 2N. (By maxima branching and every node must have at most two direct successors.) (iii) For a s < t and a, N+1 s N s s+1 + 2N. (Sum of (i) and (ii)) (iv) For a 1 s t, N0 t s = 0. (The ony eement of height 0 is a ) (v) For a, N t 1. (If N 2 2 then we woud have t(a ) t + 1 which is a contradiction.) (vi) For a 0 s t, N1 t s 2. (The tree is binary, so the second eve has at most two eements.) We now show that for each 0 s t and 9 < h, N t 1 +1 (2( + 1))s. If s = 0 this foows immediatey from (v). Case s = 1: We wish to show that 0 < h, N t 1 +1 (2( + 1))s. The case where = 0 is done by (vi). Let > 0 and suppose by induction that N t 1 < 2. By (iii), (v) and the induction hypothesis, we have that N t 1 +1 N t 1 + 2N t = (2( + 1)) 1 Case s > 1: Suppose by induction that for a 0 s < s, that for a 0 < h, N t s +1 (2( +. We wish to show that for a 0 < h, N t s 1))s +1 (2( + 1))s. The case of = 0 is done by (vi). Let > 0 and suppose for the sake of induction that for a 0 <, M t s +1 (2( + 1)) s. Then, by (iii) and the induction hypothesis, we have that ( + 1 N t s +1 N t s + 2N t s+1 (2) s + 2(2) s 1 = (2) s ) (2( + 1)) s Now suppose that for a 0 <, N t 1 +1 (2( + 1)) s. Having competed this nested induction we concude that for a 0 < h, N +1 N+1 s (2( + 1)) s t(2( + 1)) t t(2h) t. Thus, 0 s t n = N <h 0 s t N <h t(2h) t t(2h) t+1.

9 STABLE GRAPHS 9 Rearranging, we obtain that and since t t(g), and h h(g), competing the proof. (n/t) 1 t+1 2 (n/t(g)) 2 1 t(g)+1 h h(g) Observation If G is k-stabe, then t(g) < k. Proof. Let G be k-stabe and suppose that some indexing of G contained a compete binary tree a η : η 2 <k. Consider the branches a 0, a 00,..., a 0 k 1 and a 01, a 001,..., a 0 k 1 1. These form a haf graph of height k, which is a contradiction. Thus, t(g) < k Observation If G contains no homogeneous set of size k, t(g) < k. Proof. Let G be a graph with no homogeneous set of size k. Suppose that some indexing of G contained a compete binary tree a η : η 2 <k. Consider the branches a 0, a 00,..., a 0 k 1 and a 1, a 11,..., a 1 k 1. These are homogeneous sets of size k, which is a contradiction. Thus, t(g) < k. Observation Suppose that G(n) is a famiy of graphs that admit a nite bound on their tree height, t(g(n)). Then there exists c such that for a n, G(n) contains a homogeneous set arger than n c. Proof. Suppose that for a n < ω, t(g(n)) < k. Then we have that G(n) contains an independent set of size at east concuding the proof. (n/k) 1 k+1 4 Concusion 1. The haf-graph, the compete graph, and the discrete graph have the Erd s-hajna property. Proof. By the previous observations. 7. Stabe Hypergraphs In this section we present the proof of the anaogous statement for nite hypergraphs. Denition 7.1. A hypergraph is k-stabe if for a partitions of the variabes, there do not exist sequences of tupes a 1,... a k b 1,..., b k such that R(a i, b k ) i i < j. Note that the tupes may be of any ength. Fact 7.2. The stabiity rank gives a nite bound for the height of a fu binary tree which can be embedded in a type tree This is a direct consequence case of the Unstabe Theory Lemma. See [8] for a presentation. Fact 7.3. Let (G, ) be a graph. There exists r such that for any A G, A 2, we have that S (A) A r.

10 10 BENJAMIN OYE For a proof of the above, see [9] Theorem II.4.10(4) and II.4.11(4) p.74. Again, we appy the idea of arranging a sequence of nodes into a type tree with the intuition that trees with bounded branching and a bound on the size of an induced compete binary tree must have ong branches, which equate to indiscernibe sequences. Theorem 7.4 (for hypergraphs). If G is an k-reguar, stabe hypergraph then each A G A = n has a homogeneous subgraph of size at east f k (n) where f(x) = 1 x tr+t+1 k and r, t are constants depending ony on the theory of G. t Proof. We prove by induction on m k that there are u m n such that I u m+1 f( u m ) where f(x) = 1 x t(r+1) t k where r and t are dened beow II If i 0 < < i k 1, and j 0 < < j k 1 are from u m and for a < k m, i = j, then R(a i0,..., a ik 1 ) R(a j0,..., a jk 1 ) Note that this condition is weaker than indiscernibiity for m < k. Case m=0: Let u 0 = n. This sequence of ength n wi serve as the base case for condition (I). Condition (II) is triviay satised for i 0 < < i k 1, and j 0 < < j k 1 from u 0 with i = j for a < k, R(a i0,..., a ik 1 ) R(a j0,..., a jk 1 ) Case m+1: Let u m be given, and suppose that u m = m. Let m = {R(x 0,..., x k m, a m m,..., a m 1)}, i.e. the edges where the ast m eements are the ast m eements of u m. Step 0: (Arranging the eements of u m into a type tree) We wish to arrange u m into a type tree as we did in the simpe graph case, however this time the types are from m and thus the tree wi not necessariy be binary. These type trees are order isomorphic to downward-cosed subsets of ω <ω, the nite strings of naturas, where v p if v is an initia segment of p. By induction on < m seect eements a η from u m with a tree order such that If i j, then for a A {a η : η i}, R(A, η 1 ) R(A, η 2 ). If (i j) and (j i) then there exists A {a η : (η i) (η j)} such that R(A, i) R(A, j) Note that these conditions are equivaent to the conditions for a type tree in the case of a binary reationship. In the case of a non binary reationship, greater branching is possibe. Consider two distinct successors to the same node. They must satisfy dierent m formuas over their common branch. In the case of the binary reationship, this means that the must satisfy dierent formuas over their immediate common predecessor. However in the nite branching tree, they can disagree about any subset of the eements in their common branch. Thus, in a type tree for a k-reguar hypergraph, a node at height h coud be as much as ( h k 1). Nonetheess, as we wi show, branching is bounded for stabe trees.

11 STABLE GRAPHS 11 Step 1: (Branches of the type tree suce) Consider the ongest branch of the type tree. Let this branch be u m+1, with the order inherited from the type tree. Now consider i 0 < < i k 1, j 0 < < j k 1 from u m+1 such that a but the ast m eements are equa, i.e. ( < k m = i = j ). Without oss of generaity, suppose that j k m < i k m. Reca that in m we repace the ast m variabes in the formua with the ast m eements of u m, thus for ϕ and γ m R(a i0,..., a ik 1 ) R(a i0,..., a ik m, a m m,..., a m 1) by the inductive hypothesis, the rst k m 1 indices agree, so R(a i0,..., a ik m, a m m,..., a m 1) R(a j0,..., a jk m 1, a ik m, a m m,..., a m 1) By the construction of u m, a ik m, a jk m are on the same branch so they reaize the same m type over their common initia segment, and by supposition j k m < i k m, so a jk m, a ik m satisfy the same m formuas over a j0,..., a jk m 1. Thus, R(a j0,..., a jk m 1, a ik m, a m m,..., a m 1) R(a j0,..., a jk m, a m m,..., a m 1) which by the denition of m gives R(a j0,..., a ik m, a m m,..., a m 1) R(a j0,..., a jk 1 ) Thus, we have shown that ϕ(a i0,..., a ik 1 ) ϕ(a j0,..., a jk 1 ) in other words, part II of the induction. Step 2: (Lower bounds on the ength of a branch) We now must show that sucienty ong branches exist. The rationae is the same: trees that have a bound on the size of an induced binary tree must have ong branches. However we must be more carefu in counting, as our type tree is no onger stricty binary. Step 2B: (Bound on branching) By fact 7.7 we have that at height h, branching is at most (h + m) r as immediate successors to the same node must satisfy dierent m types over their h node ong common initia segment and over the m variabes specied in m. We note, as above, that the shortest trees are obtained when branching is maxima. As successors of the same rank must ie aong the same branch, and rank successors cannot have a greater rank than their predecessors, this attains when at most one successor to a node i of rank s has rank s and the rest have rank s 1. Step 2C: (Counting nodes) We count the nodes in our type tree by their tree rank. Reca that t(a i ) is the height of the argest binary tree which can be after a i in the type tree. Because G is stabe, its theory has a stabiity rank t, which is a bound on the tree height embeddabe in a type tree of G. Given a xed and s, dene as before Z s = {a η V : t(a η ) = s, ht(a η ) = } X s = {a η Z s : t(p(a η )) = s}, and Y s = {a η Z 2 : t(p(a η )) = s + 1}.

12 12 BENJAMIN OYE N t (s+1) and et N s = Zs, xs = Xs and ys = Y s, and Zs = Xs Y s so N s = xs + ys. Note aso that every node must admit a tree of height between 0 and t, so we can count the nodes by n = h t =0 s=0 N s. We caim that the foowing facts hod (i) For a s t and, x s +1 N s, by maxima branching (ii) For a s < t and a, y+1 s N s+1 ( + m) r, by the bound on branching. (iii) Thus for a s < t and a, N+1 s N s + N s+1 ( + m) r. (iv) For a 1 s t, N0 t s = 0, as the ony eement of height 0 is the root. (v) For a, N t 1. Step 2D: (Induction) We observe by induction that N t s +1 (+m)s(r+1). Case s = 1: By a nested induction on using (i) and (iii) N t 1 +1 (j + m) r ( + m) r+1 j Case s + 1: By induction on using (iii) and (iv) we have that +1 N t (s+1) +N t s (+m) r (+m) s(r+1) (j+m) r (+m) (s+1)(r+1) j So, the tota number of nodes of tree height + 1, N +1 is N +1 s t N t s +1 ( + m) s(r+1) t( + m) t(r+1) s t Noting that N 0 = 1, the tota number of nodes in the tree is at most N = 1 + <h N h t( + m) t(r+1) < t(h + m) t(r+1) Step 2E: (Part (I) of the induction) Suppose for the sake of contradiction that the branch we chose (i.e. u m+1 the ongest branch in the type tree of u m ) had ength h satisfying t(h + m) t(r+1) < u m. We have shown that this is a contradiction, as such a type tree woud not be arge enough to exhaust a the nodes of u m. Thus, h f( u m ) where 1 1 f(x) = x/t t(r+1) k x/t t(r+1) m where here we subtract k instead of m for uniformity. Thus, after k steps we have extracted a sequence of indices u for an sequence which is edge indiscernibe, and this sequence has ength at east f k (n), after k appications of f Acknowedgments. Thanks to Maryanthe Maiaris for iuminating conversations, Lászó Babai for suggesting the topic, Xiao Grin" Wang for paper guidance, and Peter May for organizing the REU.

13 STABLE GRAPHS 13 References [1] M. Maiaris and S. Sheah. Reguarity emmas for stabe graphs. Trans. Amer. Math. Soc. 366 (2014), mem/978-revised.pdf. [2] M. Maiaris and C. Terry. On Unavoidabe Induced Subgraphs in Large Prime Graphs. arxiv: [math.lo]. [3] Maria Chudnovsky. The Erd s-hajna Conjecture - A Survey. arxiv: [math.co]. [4] János Komós and Mikós Simonovits. Szemerédi's Reguarity Lemma and its appications in graph theory. ehudf/courses/ramsey04/reguaritysurvey.ps [5] F.P. Ramsey. On a probem of forma ogic. Proc. London Math. Soc. Ser (1930), [6] David Conon. Graph Ramsey Theory. dc340/ramseylecture1.pdf [7] Jacob Fox. Ramsey theory: continued. fox/mat307-ecture06.pdf [8] Wiifrid Hodges. Mode Theory. Cambridge University Press. [9] S. Sheah. Cassication Theory Esevier.