Erdős-Hajnal-type results for monotone paths

Article history: Received 16 April 2020 Available online xxxx


Introduction
Erdős and Hajnal [11] proved that graphs avoiding some fixed induced subgraph or subgraphs have very favorable Ramsey-theoretic properties. In particular, they contain surprisingly large homogeneous (that is, complete or empty) subgraphs and bipartite subgraphs. According to the celebrated Erdős-Hajnal conjecture, every graph G on n vertices which does not contain some fixed graph H as an induced subgraph, has a clique or an independent set of size at least n c , where c = c(H) > 0 is a constant that depends only on H. There is a rapidly growing body of literature studying this conjecture (see, e.g., [1,2,5,6,8,12,14,16,24]).
For any graph G and any disjoint subsets A, B ⊂ V (G), we say that A is complete to B if ab ∈ E(G) for every a ∈ A, b ∈ B. If |A| = |B| = k and A is complete to B, then A and B are said to form a bi-clique of size k. Denote the maximum degree of the vertices in G by Δ(G). Following [14], a family of graphs G is said to have the Erdős-Hajnal property if there exists a constant c = c(G) > 0 such that every G ∈ G has either a clique or an independent set of size at least |V (G)| c . The family G has the strong Erdős-Hajnal property if there exists a constant b = b(G) > 0 such that for every G ∈ G, either G or its complement G has a bi-clique of size b|V (G)|. It was proved in [1] that if a hereditary family (that is, a family closed under taking induced subgraphs) has the strong Erdős-Hajnal property, then it also has the Erdős-Hajnal property.
The aim of this paper is to discuss Erdős-Hajnal type problems for ordered graphs. An ordered graph is a graph with a total ordering on its vertex set. With a slight abuse of notation, in every ordered graph, we denote this ordering by ≺. If the vertex set of G is a subset of the integers, then ≺ stands for the natural ordering. An ordered graph H is an ordered subgraph (or simply subgraph) of G if there exists an order preserving embedding from V (H) to V (G) that maps edges to edges. If, in addition, non-edges are mapped into non-edges, then H is called an induced ordered subgraph of G. If G does not have H as induced ordered subgraph, then we say that G avoids H. The ordered path with vertices 1, . . . , k and edges {i, i + 1}, for i = 1, . . . , k − 1, is called a monotone path of size k.
Our main result is the following.
Theorem 1. For any positive integer k, there exists c = c(k) > 0 with the following property. If G is an ordered graph on n vertices such that neither G nor its complement contains an induced monotone path of size k, then G has either a clique or an independent set of size at least n c .
One can deduce from our proof that c(k) = k −5−o (1) suffices, but in order to make the paper more readable, we will not include the computations. Our theorem obviously implies the analogous statement for unordered graphs, which was first established by Bousquet, Lagoutte, and Thomassé [5]. The idea of their proof was the following. We call a family of graphs, H, lopsided if there exists a constant c = c(H) > 0 with the following property: any graph G on n vertices which does not contain any element of H as an induced subgraph, and for which Δ(G) < cn, the complement of G has a bi-clique of size at least cn. If H consists of a single graph H, then H is called lopsided. They proved that the (unordered) path of size k is lopsided. It follows from the arguments of Bousquet et al. that if H is lopsided, then the family of all graphs which avoid every element of H as an induced subgraph, and whose complements also avoid them, has the strong Erdős-Hajnal and, thus, the Erdős-Hajnal property.
Since then, this idea has been exploited to prove the Erdős-Hajnal conjecture for various other families of graphs: the family of graphs avoiding a tree T and its complement [8], the family of graphs avoiding all subdivisions of a graph H and the complements of these subdivisions [9], the family of graphs avoiding a graph H as a vertex minor [7], families of graphs avoiding a fixed cycle as a pivot minor [17], etc.
However, for ordered graphs, this method does not work even in the simplest case: for monotone paths. A construction of Fox [13] shows that, for every n and δ > 0, there exists an ordered graph G with |V (G)| = n and Δ(G) < n δ which avoids the monotone path of size 3, and whose complement does not contain a bi-clique of size larger than cn log n , for a suitable constant c = c(δ) > 0. Hence, using the above terminology, the monotone path of size at least 3 is not lopsided.
Although monotone paths are not lopsided, they satisfy a somewhat weaker property, as is shown by the following theorem of the authors.
Theorem 2. ( [20]) For any positive integer k, there exists a constant c = c(k) > 0 with the following property. If G is an ordered graph on n vertices that does not contain an induced monotone path of size k, and Δ(G) < cn, then the complement of G contains a bi-clique of size at least cn log n .
Unfortunately, Theorem 1 cannot be deduced from this weaker property. Our approach is based on a technique in [25], where it was shown that the family of string graphs has the Erdős-Hajnal property.
Recently, Seymour, Scott, and Spirkl [24] extended our Theorem 2 from monotone paths to all ordered forests T , albeit with a weaker bound n 1−o(1) in place of cn log n . They proved that for any 0 < c < 1, there exists = (T, c) > 0 with the following property. If G is an ordered graph on n vertices that does not contain T as an induced ordered subgraph and Δ(G) < n, then the complement of G contains a bi-clique of size at least n 1−c . Therefore, if we want to guarantee a bi-clique of size n 1−o (1) in G, we need to assume that the maximum degree of G is o(n). This is definitely a stronger condition than the one we had for monotone paths.
Our next construction shows that this stronger condition is indeed necessary. We also provide new examples of ordered trees T (that do not contain a monotone path of size 3), for which one cannot expect to find linear size bi-cliques. For any positive integer n ≥ n 0 , there is an ordered graph G with n vertices and Δ(G) ≤ n such that the size of the largest bi-clique in G is at most n 1−δ , and G does not contain either of the following ordered trees as an induced ordered subgraph: S: The investigation of bipartite variants of the problems considered in this paper were initiated in [18]; see also [3,23].
Our paper is organized as follows. In Section 2, we introduce the key concept needed for the proof of Theorem 1 and reduce Theorem 1 to another statement (Theorem 6). Sections 3 and 4 are devoted to the proof of this latter statement. The construction proving Theorem 3 will be presented in Sections 5.
Throughout this paper, we use the following notation, which is mostly conventional. For any graph G and any subset U ⊂ V (G), we denote by stand for the graph obtained from G by deleting the vertex v. Also, if G is an ordered graph, the forward neighborhood of a vertex v ∈ V (G), denoted by N + G (y) = N + (y) is the set of neighbors y such that x ≺ y.
For easier readability, we omit the use of floors and ceilings, whenever they are not crucial.

The quasi-Erdős-Hajnal property
After introducing some notation and terminology, we outline our proof strategy for Theorem 1.
For any k ≥ 3, let P k denote the family of all ordered graphs G such that neither G nor its complement contains a monotone path of size k as an induced subgraph. Instead of proving that P k has the Erdős-Hajnal property, we prove that it has the quasi-Erdős-Hajnal property. This concept was introduced by the second named author in [25], in order to show that the family of string graphs has the Erdős-Hajnal property.
It was proved in [25] that in hereditary families, the quasi-Erdős-Hajnal property is equivalent to the Erdős-Hajnal property. We somewhat relax the definition of the quasi-Erdős-Hajnal property, and with a slight abuse of notation, we overwrite the previous definition as follows.
Definition 5. A family of graphs, G, has the quasi-Erdős-Hajnal property if there are two constants, α, β > 0, with the following property. For every G ∈ G with at least 2 vertices, there exist t ≥ 2 and t pairwise disjoint subsets X 1 , . . . , It is easy to verify that the two definitions are in fact equivalent. If G satisfies Definition 4, then, obviously, it also satisfies Definition 5. In the reverse direction, setting Therefore, it is enough to show that P k has the quasi-Erdős-Hajnal property. The advantage of the quasi-Erdős-Hajnal property over the Erdős-Hajnal property is that it allows us to establish the following lopsided statement, which will imply Theorem 1.
Theorem 6. For every positive integer k, there exist two constants , α > 0 with the following property.
Let G be an ordered graph on n vertices with maximum degree at most n such that G does not contain a monotone path of size k as an induced subgraph. Then there exist t ≥ 2 and t pairwise disjoint subsets , and there is no edge between X i and X j for 1 ≤ i < j ≤ t.
Our proof shows that = 2 −O(k) and α = 2 −O(k) suffice. In the inequality t ≥ α( n |X i | ) 1/2 , the exponent 1/2 has no significance: the statement remains true with any 0 < β < 1 instead of 1/2 (with the cost of changing and α). However, it is not true with β = 1, as it would contradict the aforementioned construction of Fox [13].
In the rest of this section, we show how Theorem 6 implies Theorem 1. Very similar ideas were used in [5,8,9]. The next two sections are devoted to the proof of Theorem 6.
By a classical result of Rödl [21], any graph G avoiding some fixed graph H contains a linear size subset that is either very dense or very sparse. A quantitatively stronger version of this result was proved by Fox and Sudakov [15].

For any graph G with n vertices that does not contain H as an induced subgraph, there is a subset
Lemma 7 applies to unordered graphs, but it can be easily extended to ordered graphs, using the following statement. Lemma 8. [22] For every ordered graph H, there exists an unordered graph H 0 with the property that introducing any total ordering on V (H 0 ), the resulting ordered graph H 0 always contains H as an induced ordered subgraph.
By the combination of these two lemmas, we obtain the following.
Lemma 9. For every ordered graph H and > 0, there exists δ > 0 with the following property.
For any ordered graph G with n vertices that does not contain H as an induced ordered subgraph, there exists a subset U ⊂ V (G) such that |U | ≥ δn, and either Δ( Proof. By Lemma 8, there exists a graph H 0 such that introducing any total ordering on V (H 0 ), the resulting ordered graph H 0 contains H as an induced ordered subgraph. Let 0 = 2 , and let δ 0 be the constant given by Lemma 7 with respect to H 0 and 0 .
Let G be an ordered graph with n vertices that does not contain H as an induced ordered subgraph. Then the underlying unordered graph of G does not contain H 0 as an induced subgraph. Hence, there exists U ⊂ V (G) such that |U | ≥ δ 0 n, and either , the other case can be handled similarly. Let W be the set of vertices in U whose degree in G[U ] is larger than 2 0 |U |. Then Hence, δ = δ 0 2 will suffice.
After this preparation, it is easy to deduce from Theorem 6 that P k has the quasi-Erdős-Hajnal property and, therefore, the Erdős-Hajnal property.
Proof of Theorem 1. Let , α > 0 be the constants given by Theorem 6, and let δ > 0 be the constant given by Lemma 9, where H is the monotone path of size k.
Let G be an ordered graph on n vertices such that neither G nor its complement contains a monotone path of length k as an induced subgraph. Then there exists U ⊂ V (G) such that |U | ≥ δn, and either Δ( ) < |U |, the other case can be handled similarly. Applying Theorem 6 to G[U ], we obtain that there exist t ≥ 2 and t pairwise disjoint sets and there is no edge between X i and X j for 1 ≤ i < j ≤ t.
In the next two sections, we present the proof of Theorem 6.

The main lemma
The backbone of the proof of Theorem 6 is the following technical lemma, whose proof is already contained in [25], within the proof Lemma 7. For convenience and to make this paper self-contained, it is also included here. Recently, our lemma was also utilized by Chudnovsky et al. [10], who provided a different proof as well.
Lemma 10. There exist two constants 0 < , α < 1 4 with the following property. Let H be a bipartite graph with vertex classes A and B, |A| = |B| = n. Then at least one of the following three conditions is satisfied.
Let us briefly outline the idea of the proof. We want to find an induced subgraph H of H with vertex classes A ⊂ A and B ⊂ B such that • H is almost bi-regular, more precisely, the degree of every vertex in A is at most Δ, and the degree of every vertex in B is within a constant factor of some d, • |A | and |B | are large with respect to d and Δ.
If we can find such an H , we construct our sets W 1 , . . . , W t and X 1 , . . . , X t as follows. By a probabilistic argument, we find S ⊂ A such that Ω(|B |) vertices in B have exactly one neighbor in S. Then we can group the vertices in S into sets W 1 , . . . , W t such that X i := N H (W i ) has size roughly Δ for i ∈ [t]. The 2t sets W 1 , . . . , W t and X 1 , . . . , X t will satisfy (i). We find a suitable H algorithmically: we either found our desired A or B , or there are too few vertices with too large degrees, in which case we remove these vertices and continue. We show that if we cannot find H , then at least one of (ii) or (iii) must hold. In what follows, we describe an algorithm, which will be referred to as the main algorithm. It will output (i)' either an integer t ≥ 2 and 2t pairwise disjoint sets W 1 , . . . , W t ⊂ A and (ii)' or two subsets X ⊂ A and Y ⊂ B such that |X|, |Y | > n 2 and there is no edge between X and Y .
We declare the following constants for the main algorithm. Let J 0 := log 2 n + 1, and for j = 1, . . . , J 0 , let t j := n 1/2 2 j/2 . Then Also, let A * 0 := ∅ and B * 0 := ∅. In the q-th step of the main algorithm, we define A q , A * q , B q , B * q , J q . We will think of A * q and B * q as a set of "leftovers". That is, we get A q and A * q by transferring certain elements from A q−1 to A * q−1 , and we get B q and B * q by transferring certain elements from B q−1 to B * q−1 . Also, J q will keep track of the maximum degree in B q , and it will decrease after each step. We make sure that the following properties are satisfied: Note that by (1) and conditions 1 and 2, we have |A q |, |B q | ≥ n 2 . Also, the conditions 1-3 are certainly satisfied for q = 0. Next, we describe the q-th step of our main algorithm.
Main algorithm. If J q−1 = 0, then stop the main algorithm, and output X := A q−1 , Y := B q−1 . In this case, there is no edge between A q−1 and B q−1 by condition 3, and |A q−1 |, |B q−1 | ≥ n 2 by condition 2. This output satisfies condition (ii)'. Suppose next that J q−1 ≥ 1.
where the first inequality follows from condition 2, and the first equality is the consequence of condition 1. Comparing the left-hand and right-hand sides, and using (1), we get |V 0 | ≥ n/2. In this case, stop the algorithm and output X := A q−1 and Y := V 0 . This output satisfies condition (ii)'. Suppose that there exists k with the desired property. Let B q,0 = B q−1 \ ( i=k+1 t i holds. Also, set J q := k, A q,0 = A q−1 and A * q,0 = A * q−1 . Note that properties 1-3 are satisfied with A q,0 , A * q,0 , B q,0 , B * q,0 , J q instead of A q , A * q , B q , Bq * , J q , respectively. Now we shall run a sub-algorithm. Let Z 0 = V k . With help of the sub-algorithm, we construct a sequence Z 0 ⊃ · · · ⊃ Z r satisfying the following properties. During the -th step of the sub-algorithm, we either find an output satisfying (i)', or we will transfer certain elements of A q, −1 to A * q, −1 , resulting in the sets A q, and A * q, . At the end of the -th step of this algorithm, Z will be the set of vertices in B q,0 that still have at least 2 k−1 neighbors in A. We stop the algorithm if Z is too small. Let us describe the -th step of the algorithm.

Sub-algorithm. Suppose that
Move to the next step of the main algorithm. Note that conditions 1 and 3 are satisfied, and B * q satisfies condition 2. Later, we will see that A * q satisfies 2. as well. On the other hand, if |Z −1 | ≥ 2t k , we define Z as follows. Let x = and let K be the set of heavy vertices. Counting the number of edges f between K and Z −1 in two ways, we can write where the equality holds by the definition of Δ . Set A q, := A q, −1 \ K and A * q, = A * q, −1 ∪ K . Examine how the degrees of the vertices in Z −1 changed, and consider the following two cases: Pick each element of A q, with probability p = 2 −k , and let S be the set of selected vertices. We say that v ∈ T is good if |N (v) ∩ S| = 1, and let D be the set of good vertices. We have 12 . Therefore, there exists a choice for S such that |D| ≥ |Z −1 | 12 . Let us fix such an S. For each v ∈ S, let D v be the set of elements w ∈ D such that N (w) ∩ S = {v}. Also, note that In other words, the sets D v for v ∈ S partition D into sets of size at most Δ . Here, we have By the choice of , the right-hand side is always at least 6. But then we can partition S into t ≥ |D| 3Δ ≥ 2 parts W 1 , . . . , W t such that the sets X i = v∈W i D v have size at least Δ for i = 1, . . . , t. The integer t and the resulting sets X 1 , . . . , X t satisfy that Stop the main algorithm, and output t and the 2t pairwise disjoint sets W 1 , . . . , W t and X 1 , . . . , X t . By the choice of α, this output satisfies (i)'.
In this case, define Z as the set of elements of Z −1 with at least 2 k−1 neighbors in A q, (then Z is the set of all elements in B q,0 with at least 2 k−1 neighbors in A q, as well). Also, move to the next step of the sub-algorithm.
We show that conditions 1-3 are still satisfied for A q, , A * q, , B ,0 , B * ,0 , J q instead of A q , A * q , B q , B * q , J q . Conditions 1 and 3 are clearly true, and 2 holds for B * q,0 . It remains to show that 2 holds for A * q, as well. Note that, as |Z j | ≤ Hence, condition 2 is also satisfied.
As we have J 0 > J 1 > · · · ≥ 0, the main algorithm will stop after at most J 0 steps. When the algorithm stops, its output will satisfy either (i)' or (ii)'.
Let us remark that if (i) holds, then the 2t sets W 1 . . . , W t and X 1 , . . . , X t have the additional property that every vertex in t i=1 X i has exactly one neighbor in

The proof of Theorem 6
Now we are in a position to prove Theorem 6. Let G be an ordered graph. The transitive closure of G is the ordered graph G on the vertex set V (G) in which x and y are connected by an edge if and only if there exists a monotone path in G with endpoints x and y.
Let us briefly outline the proof idea. We assume that for G there is no integer t ≥ 2 and t sets X 1 , . . . , X t with the desired properties. Then we show that G contains a monotone path x 1 , . . . , x k with the following additional property. For s = 1, . . . , k, there are Ω(n) vertices in G that can be reached by a monotone path from x s , which avoids all the neighbors of x 1 , . . . , x s−1 . This additional property lets us do induction on s, allowing us to find x 1 , . . . , x k one-by-one.
Proof of Theorem 6. Let 0 < 1 , α 1 < 1 4 be the constants given by Lemma 10 as , α, respectively. Furthermore, define the following constants: c 1 = 1 2 , c i+1 = 1 c i 4 (for i = 1, 2, . . . ), = c k 2 , and α = Let G be an ordered graph on n vertices such that 1. the maximum degree of G is at most n, 2. there exists no t ≥ 2 such that for some pairwise disjoint sets X 1 , . . . , X t ⊂ V (G) we have t ≥ α( n |X i | ) 1/2 and there is no edge between X i and X j for 1 ≤ i < j ≤ t.
Then we show that G contains a monotone path of size k as an induced subgraph. In particular, we find k vertices x 1 ≺ · · · ≺ x k with the following properties. For s = 1, . . . , k, (a) x 1 , . . . , x s form an induced monotone path.
, and let G s be the transitive closure of G s . Then the forward degree of x s in G s is at least c s n.
First, we find a vertex x 1 with the desired properties, that is, if G is the transitive closure of G, then the forward degree of x 1 must be at least c 1 n. Let A 0 be the set of the first n/2 elements of V (G), and set B 0 = V (G) \ A 0 . Also, let H 0 denote the bipartite subgraph of G with parts A 0 and B 0 . By Lemma 10, at least one of the following three conditions is satisfied.
(i) There exist t ≥ 2 and 2t pairwise disjoint sets W 1 , . . . , W t ⊂ A 0 and X 1 , . . . , X t ⊂ B 0 such that (ii) There exist X ⊂ A 0 and Y ⊂ B 0 such that |X|, |Y | ≥ n 8 , and there is no edge between X and Y .
As the non-edges of G are also non-edges of G, (ii) cannot hold. Otherwise, t = 2 and X 1 = X, X 2 = Y contradicts property 2 of G. Suppose that (i) holds. Note that there is no edge between X i and X j in G, for 1 ≤ i < j ≤ t. Suppose for contradiction that x ∈ X i and y ∈ X j are joined by an edge in G, for some x ≺ y. Then there exists w ∈ W i such that wx ∈ E(G ), but wy / ∈ E(G ). This is a contradiction, as this means that there is a monotone path from w to x in G, so there is a monotone path from w to y as well. Hence, there is no edge between X i and X j for 1 ≤ i < j ≤ t, which contradicts 2. Therefore, (iii) must hold: there exists a vertex x 1 ∈ V (G) whose forward degree in G = G 1 is at least c 1 n.
Suppose that we have already found x 1 , . . . , x s with the desired properties, for some 1 ≤ s ≤ k − 1. Then we define x s+1 as follows. Let F s be the forward neighborhood of x s in G s , let K s be the forward neighborhood of x s in G s , and let L s = K s \ F s . As |F s | ≤ n and |K s | ≥ c s n, we have |L s | ≥ c s 2 n. Let A s be the set of the first |L s | 2 elements of L s with respect to ≺, and let B s = L s \ A s . A monotone path in G s is said to be good if none of its vertices, with the possible exception of the first one, belongs to F s . For every v ∈ A s , there exists at least one element x ∈ F s such that v ∈ N + G s (x); assign the largest (with respect to ≺) such element x to v. Then there is a good monotone path from x to v.
Define a bipartite graph H s between A s and B s as follows.
If v ∈ A s and y ∈ B s , and x ∈ F s is the vertex assigned to v, then join v and y by an edge if there is a good monotone path from x to y. Applying Lemma 10 to H s , we conclude that at least one of the following three statements is true.
(i) There exist t ≥ 2 and 2t pairwise disjoint sets W 1 , . . . , W t ⊂ A s and X 1 , . . . , X t ⊂ B s such that (ii) There exist X ⊂ A s and Y ⊂ B s such that |X|, |Y | ≥ |A s | 4 ≥ c s n 16 , and there is no edge between X and Y in H s .
Suppose first that (i) holds. Then, as before, we show that there is no edge between X i and X j in G for 1 ≤ i < j ≤ t. Suppose that u ∈ X i and w ∈ X j are joined by an edge in G, for some u ≺ w. Then there exists v ∈ W i such that vu ∈ E(H s ), but vw / ∈ E(H s ). Let x ∈ F s be the vertex assigned to v. Then we can find a good monotone path from x to u. Since uw is an edge of G, there is a good monotone path from x to w, contradicting the assumption vw / ∈ E(H s ). Therefore, there cannot be any edge between X i and X j in G, which means that (i) contradicts 2.
Suppose next that (ii) holds. Again, we can show that there is no edge between X and Y in G, which then contradicts 2. by setting t = 2 and X 1 = X, Y 1 = Y . Suppose that v ∈ X and y ∈ Y are joined by an edge in G, and let x ∈ F s be the vertex assigned to v. There is a good monotone path from x to v in G s , so there is a good monotone path from x to y, contradicting the assumption that vy is not an edge of H s . Therefore, we can assume that (iii) holds. Let v ∈ A s be a vertex of degree at least c s+1 n in H s , and let x s+1 ∈ F s be the vertex assigned to v. We show that x s+1 satisfies the desired properties. We have U s+1 = U s \ F s , and the forward degree of x s+1 in G s+1 is exactly the number of vertices y such that there is a good monotone path from x s+1 to y. That is, the forward degree of x s+1 is at least |N H s (v)| ≥ c s+1 n, as required. This completes the proof.

The construction-Proof of Theorem 3
In this section, we present our construction for Theorem 3. The construction involves expander graphs, which are defined as follows.
The closed neighborhood of U in a graph H is defined as U ∪ N H (U ), and is denoted by The graph H is called an (n, d, λ)-expander if H is a d-regular graph on n vertices, and for every U ⊆ V satisfying |U | ≤ |V |/2, we have |N H [U ]| ≥ (1 + λ)|U |. By a well-known result of Bollobás [4], a random 3-regular graph on n vertices is a (n, 3, λ 0 )-expander with high probability for some absolute constant λ 0 > 0. In the rest of this section, we fix such a constant λ 0 . For explicit constructions of expander graphs see, e.g., [19].
For any positive integer r, let H r denote the graph with vertex set V (H) in which two vertices are joined by an edge if there exists a path of length at most r between them in H. Here we allow loops, so that in H r every vertex is joined to itself. We need the following simple property of expander graphs.
Claim 11. Let H be an (n, d, λ)-expander graph and let r ≥ 1. For any subsets X, Y ⊆ V (H) such that there is no edge between X and Y in H r , we have |X| ·|Y | ≤ n 2 (1 +λ) −r .
2. if x, y, z ∈ V such that x ∈ A a , y ∈ A b , z ∈ A c for some a < b < c, and xy, xz ∈ E(G), then yz ∈ E(G) as well; 3. for any a = b and any pair of subsets X ⊂ A a and Y ⊂ A b that are not connected by any edge of G, we have |X| · |Y | ≤ m 2 (1 + λ 0 ) −f .
Proof. Let H be an (m, 3, λ 0 )-expander. Let φ : V → V (H) be an arbitrary function such that φ is a bijection when restricted to the set A i , for i = 1, . . . , k. Define the graph G, as follows. Suppose that x ∈ A a and y ∈ A b for some a < b. Join x and y by an edge if there exists a path of length at most f 2 a−1 between φ(x) and φ(y) in H. By Claim 12, the maximum degree of G is at most , so that G has property 1. To see that G also has property 2, consider x ∈ A a , y ∈ A b , z ∈ A c such that a < b < c and xy, xz ∈ E(G). We have to show that yz ∈ E(G). By definition, there exists a path of length at most f 2 a−1 between φ(x) and φ(y) in H, and there exists a path of length at most f 2 a−1 between φ(x) and φ(z). But then there exists a path of length at most f 2 a ≤ f 2 b−1 between φ(y) and φ(z), so yz is also an edge of G.
It remains to verify that G has property 3. If 1 ≤ a < b ≤ k and X ⊂ A a and Y ⊂ A b are not connected by any edge in G, then there is no edge between φ(X) and φ(Y ) in H f 2 a−1 . By Claim 11, we have |X| · |Y | ≤ m 2 (1 + λ 0 ) −f 2 a−1 ≤ m 2 (1 + λ 0 ) −f . Now we are in a position to prove Theorem 3.
Proof of Theorem 3. Let k = 2 , f = log 2 n 2·2 k , and m = n k . We show that the theorem holds with δ = log 2 (1+λ 0 ) 2 k . Let A 1 , . . . , A k be pairwise disjoint sets of size m. By Lemma 13, there exists a graph G 0 on V = m i=1 A i satisfying conditions 1-3 with the above parameters. Define the ordered graph G on the vertex set V as follows. Let ≺ be any ordering on V satisfying A 1 ≺ · · · ≺ A k . For any x ∈ A a and y ∈ A b , join x and y by an edge of G if xy ∈ E(G 0 ), or a = b. Then the maximum degree of G is at most n k +Δ(G 0 ) ≤ n. Notice that the complement of G does not contain a bi-clique of size n 1−δ . Indeed, if (X, Y ) is a bi-clique in G, then there exists a = b such that |X ∩A a | ≥ |X| k and |Y ∩A b | ≥ |Y | k = |X| k . Thus, which implies that |X| ≤ n 1−δ .
It remains to show that G contains neither S, nor P as an induced ordered subgraph. Let us start with S. Suppose that there are four vertices, , v 2 ∈ A c , and v 3 ∈ A d , then a ≤ b ≤ c ≤ d. If c = a, then b = a, which implies v 1 v 2 ∈ E(G), contradiction. Therefore, a < c ≤ d. As v 2 v 3 / ∈ E(G), we must have c < d as well. But then the three vertices v 0 , v 2 , v 3 contradict property 2, so that G does not contain S an induced ordered subgraph.
To show that G does not contain P , we can proceed in a similar manner. Suppose for contradiction that there are four vertices, v 0 ≺ v 1 We have a < b, otherwise v 0 v 1 ∈ E(G). In the same way, c < d, otherwise v 2 v 3 ∈ E(G). Therefore, a < c < d, and the vertices, v 0 , v 2 , and v 3 , contradict condition 2 of Lemma 13.