Turán-Type Results for Complete h-Partite Graphs in Comparability and Incomparability Graphs

We consider an h-partite version of Dilworth’s theorem with multiple partial orders. Let P be a finite set, and let <1,...,<r be partial orders on P. Let G(P, <1,...,<r) be the graph whose vertices are the elements of P, and x, y ∈ P are joined by an edge if x<iy or y<ix holds for some 1 ≤ i ≤ r. We show that if the edge density of G(P, <1, ... , <r) is strictly larger than 1 − 1/(2h − 2)r, then P contains h disjoint sets A1, ... , Ah such that A1 <j... <jAh holds for some 1 ≤ j ≤ r, and |A1| = ... = |Ah| = Ω(|P|). Also, we show that if the complement of G(P, <) has edge density strictly larger than 1 − 1/(3h − 3), then P contains h disjoint sets A1, ... , Ah such that the elements of Ai are incomparable with the elements of Aj for 1 ≤ i < j ≤ h, and |A1| = ... = |Ah| = |P|1−o(1). Finally, we prove that if the edge density of the complement of G(P, <1, <2) is α, then there are disjoint sets A, B ⊂ P such that any element of A is incomparable with any element of B in both <1 and <2, and |A| = |B| > n1−γ(α), where γ(α) → 0 as α → 1. We provide a few applications of these results in combinatorial geometry, as well.


Introduction
Let k and n be positive integers. A weak version of the widely used Dilworth's theorem [2] states that every partially ordered set with n elements either contains a chain of size k or an antichain of size n/k . Applying Dilworth's theorem multiple times, one can easily deduce the following result. Let P be an n element set, and let < 1 , ..., < r be partial orders on P . There exists H ⊂ P such that |H | ≥ r+1 √ n, and H is either a < i -chain for some 1 ≤ i ≤ r or any two elements of H are incomparable in any of the partial orders < 1 , ..., < r .
Bipartite versions of Dilworth's theorem have been considered in a series of papers by Fox, Pach and Tóth. Before we state their results, we introduce some notation.
Let < 1 , ..., < r be partial orders on a set P . If a, b ∈ P , write a ⊥ i b if a and b are incomparable in < i . Also, write a ⊥ b if a ⊥ i b holds for i = 1, ..., r. If A, B ⊂ P and 1 ≤ i ≤ r, let A < i B if for every a ∈ A and b ∈ B we have a < i b. Define A ⊥ i B and A ⊥ B analogously.
In [4], Fox proved the following theorem for a single partial order.
Theorem 1 ([4]) There exists n 0 such that for all n > n 0 and for all partially ordered sets (P , <) on n elements, there exist A, B ⊂ P such that A and B are disjoint, |A| = |B| > n 4 log 2 n , and either A < B or A ⊥ B.
In [5], Fox and Pach generalized this result for multiple partial orders.
Theorem 2 ( [5]) Let r be a fixed positive integer and let < 1 , ..., < r be partial orders on the n element set P . There exist A, B ⊂ P such that A and B are disjoint, |A| = |B| > n 2 (1+o(1))(log 2 log 2 n) r , and either A < i B holds for some 1 ≤ i ≤ r or A ⊥ B.
In [6], Fox, Pach and Tóth proved a Turán-type version of these results. Before we state it we introduce some further notation. If < 1 , ..., < r are partial orders on the set P , let G(P , < 1 , ..., < r ) be the graph whose vertex set is P and in which two elements a, b ∈ P are joined by an edge if a < i b or b < i a holds for some 1 ≤ i ≤ r. Call this graph the rcomparability graph of (P , < 1 , ..., < r ), and call the complement of G(P , < 1 , ..., < r ) the r-incomparability graph of (P , < 1 , ..., < r ). Similarly, the directed comparability graph of (P , < 1 , ..., < r ) is − → G (P , < 1 , ..., < r ), in which − → xy is an edge if x < i y for some 1 ≤ i ≤ r. We note that it is allowed to have both − → xy and − → yx in the directed edge set. For positive integers h, r, n, m, define f C r,h (n, m) and f I r,h (n, m) as follows. Let f C r,h (n, m) be the maximal s such that if P is an n element set with partial orders < 1 , ..., < r , and G(P , < 1 , ..., < r ) has exactly m edges, then there exist 1 ≤ i ≤ r and A 1 , ..., A h ⊂ P pairwise disjoint subsets such that |A 1 | = ... = |A h | = s, and A 1 < i ..
Similarly, let f I r,h (n, m) be the maximal s such that if P is an n element set with partial orders < 1 , ..., < r , and the incomparability graph of (P , < 1 , ..., < r ) has exactly m edges, then there exist A 1 , ..., A h ⊂ P pairwise disjoint subsets such that |A 1 | = ... = |A h | = s, and A j ⊥ A l for all 1 ≤ j < l ≤ h.
The aim of this paper is to generalize the previous theorem and to understand the behavior of the functions f C r,h and f I r,h . Let us note a few things about Theorem 3. The functions f I 1,2 and f C 1,2 behave quite differently. As we can see, f C 1,2 (n, m) has a large jump at m/n 2 = 1/4, and for m/n 2 > 1/4 the function f C 1,2 (n, m) is linear in n. We show that f C r,h has a similar behavior.
However, as we shall see, f I 1,h also jumps at some value of m/n 2 for h > 2. Our paper is organized as follows. In the next section, we prove bounds on f C r,h for arbitrary r, h positive integers. We show that if α = 1/2 − 1/2(2h − 2) r , the function f C r,h (n, m) jumps at the point m/n 2 = α. If m/n 2 is strictly below the threshold α, then f C r,h (n, m) is O(log n), while above this point f C r,h (n, m) becomes linear in n. An h-partite graph is balanced if its classes have the same size. In Section 3, we investigate the largest balanced h-partite graph of the 1-incomparability graph. We show that f I we have f I 1,h (n, m) = n 1−o(1) . In Section 4, we investigate the largest balanced bipartite graph of the 2-incomparability graph. As we shall see, f I 2,2 behaves quite differently as f I 1,2 . We show that f I 2,2 (n, m) is approximately n α for some α satisfying α → 1 as m/n 2 → 1/2.
In the last section, we provide applications of these results for two problems in combinatorial geometry.
Before we start, we introduce some of the standard notation we use. As usual, is the subgraph of G induced on X, and G[X, Y ] is the induced bipartite subgraph of G with vertex classes X and Y . Also, K s is the complete graph on s vertices and K s,t is complete bipartite graph with vertex classes having sizes s and t.
A linear extension of a partial order < is a total order < * such that x < y implies x < * y.
To avoid clutters, we omit floors and ceilings whenever they are not crucial.

The r-Comparability Graph
In this section, generalizing part (i) and (ii) of Theorem 3, we prove the following result about the behaviour of f C r,h .
(ii) There exists a constant c(r, h, ) > 0 such that f C r,h n, Also, for h = 2, we have f C r,2 n, Proof of (i). Let G = (A, B, E) be a bipartite graph with and |E| > |A||B|(1 − ) such that G does not contain K t,t with t > 2 −1 log n. A random bipartite graph, where the edges are chosen with probability (1 − /2), has this property with high probability, see [1].
Define (P , < 1 , ..., < r ) as follows. Let {P t } t∈[2h−2] r be a partition of the n-element set P into (2h − 2) r equal sized parts, and let f t : P t → A and g t : P t → B be arbitrary bijections. Let t = (t 1 , ..., t r ) and u = (u 1 , ..., u r ) be two different elements of [2h − 2] r and suppose that the first coordinate they differ in is the q-th coordinate. Without loss of generality, t q < u q . If t q + 1 < u q , let x < q y for all x ∈ P t and y ∈ P u . If t q + 1 = u q , let x < q y if f t (x)g u (y) ∈ E.
One can easily check that the relations < 1 , ..., < r we have defined are partial orders. Also, G(P , < 1 , ..., < r ) contains at least edges. Suppose that A 1 , ..., A h are disjoint subsets of P such that |A 1 | = ... = |A h | = t and A 1 < q ... < q A h with some q ∈ [r]. Then there exist t 1 , ..., t h such that for i = 1, ..., h, we have Also, the q-th coordinates of t 1 , ..., t h are strictly monotone increasing. Hence, there exists 1 ≤ j < h such that the difference between the q-th coordinate of t j and t j +1 is 1. But then f t j (A j ∩P t j ) and g t j +1 (A j +1 ∩P t j +1 ) span a complete bipartite graph in G, so t/(2h−2) r < 2 −1 log n. Hence, f C r,h n, In the rest of this section, we shall prove part (ii) of the theorem. We are going to deduce part (ii) from a Turán-type result for multicolored directed graphs. But first, we need some definitions.
Call the vertex a the bottom of D and a the top of D. Call Call a 1 the root and |B 1 | the width of the rooted spiral ( Fig. 1). It is clear that if the directed comparability graph of a partially ordered set (P , <) contains an h-part rooted spiral with classes B 1 , ..., B h , then B 1 < ... < B h . Hence, it is enough to find an h-part spiral with large width in the directed comparability graph. To prove such a result, we need the following lemma first.
The sets {V H } H ⊂[q] partition V into 2 q parts. The number of edges connecting two different parts in this partition is at most Hence, there exists F ⊂ [q] such that G[V F ] contains at least n 2 /2 q edges. Let E be the set of edges in G[V F ] whose color is in F . Note that for every x ∈ V F there are at most qλn edges e containing x such that χ(e) ∈ F . Thus, There are at least λn|A| > 2 n 2 q 2 2 2(q+1) edges of color p connecting an element of A with an element of V , as every element of A has at least λn edges of color p containing it. Hence, there exists a ∈ V with Then the vertex set {a, a } ∪ D A p (a ) spans a p-colored k-diamond with k > 2 n q 2 2 2(q+1) . Now we are ready to prove our key result about spirals.

Theorem 6
Let r, h be positive integers and > 0. There exists c(r, h, ) > 0 with the following property. Let G = (V , E) be a directed graph with |V | = n and and let χ : E → [r] be an r-coloring of the edges of G. Then G contains a monochromatic h-part spiral of width at least c(r, h, )n.
Proof Let λ be the unique solution of the quadratic equation satisfying λ < /h r . We shall prove that G contains an h-part spiral of width at least λn.
Suppose to the contrary that G does not contain an h-part spiral of width at least λn. For W ⊂ V , x ∈ V and i ∈ [r], define U W i (x) and D W i (x) as in the previous proof. For x ∈ V and i ∈ [r], let l i (x) be the largest l such that G contains an l-part rooted spiral with root x and width λn in color i. Note that if there exists x ∈ V and i ∈ [r] with l i (x) ≥ h, we are done as an h-part rooted spiral of width λn trivially contains an h-part spiral of width λn. Hence, we can suppose that 0 The sets {V t } t∈{0,...,h−1} r partition V into h r parts. Let n t = |V t |. Also, let edges. First of all, this forces n t to be at least √ n, as G[V t ] has more than n 2 edges. If and an (h − 1)-part rooted spiral with root x and width λn, we get an h-part spiral of width λn.
Hence, the number of edges in G[V t ] with color in I (t) is at least Applying Lemma 5 with q = |I (t)|, we get that there exists a monochromatic k-diamond Let a, a , b 1 , ..., b k ∈ V t be the vertices of this k-diamond, where the vertex a is the bottom and a is the top of the diamond. Let S be a t p -part rooted spiral with root a and width λn, then taking the union of this k-diamond and S, we get a p colored t p + 1-part rooted spiral with root a and width λn, contradicting l p (a) = t p . So far, we showed that the graph induced on V t can contain at most edges. Hence, the complement of G contains at least edges. Using the Cauchy-Schwarz inequality, we have edges, which is a contradiction.
Solving the quadratic equation in the beginning of the proof yields However, in the case h = 2, we can get a better bound. In this special case, while we repeat the previous proof, we do not need to use Lemma 5 at any point. We can deduce the following result.

Proposition 7 Let r be a positive integer and
Any r coloring of the edges of G contains a monochromatic 2-part spiral of width at least n/r2 r+1 .
Proof We shall proceed similarly as in the previous proof and in the proof of Lemma 5. Let partitions V into 2 r parts. Thus, the number of edges connecting two different parts is at most After these preparations, the proof of Theorem 4 is immediate.
Proof of Theorem 4, part (ii). Let < 1 , ..., < r be partial orders on the n element set P . Define the directed graph G = (P , E) and the coloring χ : E → [r] as follows: if x, y ∈ P are comparable in at least one of the partial orders < 1 , ..., < r , then choose one of them, say < i . Without loss of generality, x < i y. Let − → xy ∈ E and χ( − → xy) = i. By Theorem 6, there exists a color p such that the directed graph G contains a p-colored h-part spiral of width c(r, h, )n, let its vertex set be In case h = 2, we repeat the proof of (*), but we use Proposition 7 instead of Theorem 6. This yields f C r,2 n,

Balanced Complete h-Partite Subgraph in the Incomparability Graph
In this section, we prove a result about large balanced complete h-partite subgraphs in the incomparability graph of (P , <). Note that if P is the disjoint union of h − 1 chains, each of size n/(h − 1), then there is no K h in the incomparability graph of (P , <). Hence, the incomparability graph of (P , <) needs to have density at least 1 − 1/(h − 1) if we hope to find a large balanced complete h-partite graph in it. Our next result shows that if we are slightly above this density, we do find a large balanced complete h-partite graph in the incomparability graph.

Theorem 8 Let h ≥ 2 be a positive integer and let
In the proof, we shall use the following easy corollary of Theorem 3 and Theorem 4.

Proposition 9
Let h, n be positive integers. Let s be the smallest integer such that h ≤ 2 s . There exist c(h) > 0 with the following property. Let < be a partial order on the n element set P . If n is sufficiently large, then either Proof Let c = c(1, 3, 1/16), where c(r, h, ) is the constant defined in Theorem 4. If the comparability graph of a poset (Q, <), with |Q| = m has more than 7m 2 /16 edges, then by Theorem 4 there exists Hence, we can suppose that the comparability graph of P does not contain a subgraph of size at least n/(log n) s with edge density larger than 7/8, otherwise (ii) holds if c(h) < c. But then, applying Theorem 3, every subgraph of size n > n/(log n) s contains two sets, A and A such that |A| = |A | > c 0 n /(log n ) with a suitable constant c 0 > 0, and A ⊥ A . For k = 0, ..., s and i = 1, ..., 2 k , we shall define the sets X k,1 , ..., X k,2 k ⊂ P with the following properties: X 0,1 = P ; |X k,1 | = ... = |X k,2 k | > c k 0 n/(log n) k , and X k,i ⊥ X k,j for 1 ≤ i < j ≤ 2 k . Suppose that X k,1 , ..., X k,2 k are already defined satisfying those properties. We define X k+1,1 , ..., X k+1,2 k+1 as follows. As |X k,i | > c k 0 n/(log n) k > n/(log n) s if n is sufficiently large, there exist X k+1,2i−1 , X k+1,2i ⊂ X k,i such that and X k+1,2i−1 ⊥ X k+1,2i . Then X k+1,1 , ..., X k+1,2 k+1 also satisfy the properties. Set A i = X s,i for i = 1, ..., h. Then (i) holds.
Proof of Theorem 8. We shall prove part (ii) of the theorem. Let (P , <) be a partially ordered set on n elements such that Let k = 2 −1 . Let < be any linear extension of <, and let x 1 < ... < x n be the enumeration of the elements of P by < . Partition P into k equal < intervals P 1 , ..., P k .
Let c 0 = c(h) be the constant defined in Proposition 9, and set c(h, ) = c 0 /k. Also, let z = c(h, )n/(log n) s . Suppose that P does not contain A 1 , ..., A h disjoint sets such that and A i ⊥ A j for 1 ≤ i < j ≤ h. By Proposition 9, every subset of P of size at least n/k contains three sets B 1 , B 2 , B 3 of size z such that B 1 < B 2 < B 3 . Let Suppose H has d edges. If (i, j ) and (i , j ) are joined by an edge, where i < i , then G(P , <) contains every edge between B i,j,1 and B i ,j ,3 . This is true as there exists x ∈ B i,j,2 and y ∈ B i ,j ,2 with x < y, so for any x ∈ B i,j,1 and y ∈ B i ,j ,3 , we have Here, kz 2 m 2 < n 2 /18k < n 2 /2. Hence, we have Thus, using Eq. 1, we get Applying Turán's theorem [13] to H there is a complete graph on h vertices in the complement of H . Let the vertices of this K h be (i 1 , j 1 ), ..., (i h , j h ). For l = 1, ..., h, let A l = B i l ,j l ,2 . Then |A 1 | = ... = |A h | = c(h, )n/(log n) s , and A l ⊥ A l for 1 ≤ l < l ≤ h, which is a contradiction.

Conjecture 10
Let h be a positive integer, > 0. There exists c(h, ) > 0 such that holds.

Balanced Complete Bipartite Graph in the 2-Incomparability Graph
In this section, we investigate the size of the largest balanced complete bipartite graph in the 2-incomparability graph of (P , < 1 , < 2 ). Fix a positive integer h. By our previous results, if the edge density of the incomparability graph of (P , <) exceeds some threshold strictly less than 1, we have a complete balanced h-partite graph of size n 1−o(1) in the incomparability graph. However, as we shall see, this is no longer true for the 2-incomparability graph, or in general, for the r-incomparability graph, where r ≥ 2.
However, we show that if the incomparability graph of (P , < 1 , < 2 ) has edge density (1− +o(1)), there is a complete balanced bipartite graph of size n β( ) , where β( ) → 1 as → 0. This is still much larger than the size of the largest balanced complete bipartite graph of a random graph, whose edges are chosen with probability 1 − . With high probability, such a graph has edge density (1 − + o(1)), and its largest balanced bipartite graph has size O( −1 log n).
We prove the following result.
Theorem 11 (i) For every 0 < < 1 and positive integer k ≥ 2, we have f I 2,2 n, (ii) For every δ > 0, if n is a sufficiently large positive integer, there exists γ (δ) > 0 such that f I 2,2 n, The proof of part (i) is a probabilistic construction. We shall only briefly sketch the idea, the reader can find more about random graphs in [1].
Proof of (i). Our task is to construct partial orders < 1 , < 2 on an n element set P , such that the complement of G(P , < 1 , < 2 ) does not contain a large complete bipartite graph.
For any positive integer N , let G N = (X N , Y N , E N ) be a bipartite graph with the following properties: (4) G N has a complete matching M N .
Let A 1 , ..., A k be disjoint sets of size n/k, and let P = A 1 ∪ ... ∪ A k . Let < 1 be any partial order such that A 1 , ..., A k are < 1 -chains, and A i ⊥ 1 A j for 1 ≤ i < j ≤ k. Now define < 2 as follows: for i = 1, ..., k, let f i : A i → X n/k and g i : A i → Y n/k be arbitrary bijections. Define the relation < * 2 such that for any a ∈ A i and b ∈ A i+1 , where Let < 2 be the partial order induced by the relation < * 2 . First of all, we shall bound the number of edges of G(P , < 1 , < 2 ) from above. Note that e(G(P , < 1 )) = k n/k 2 < n 2 2k .
Also, e(G(P , < 2 )) < n 2 . This is true as for every 1 ≤ i < j ≤ k and x ∈ A i , y ∈ A j , we have x < y iff there exists a sequence x 0 , ..., x j −i such that x 0 = a, x j −i = y, x l ∈ X i+l for l = 1, ..., j − i − 1, and f i+l (x l )g i+l +1 (x l +1 ) ∈ E(G n/k ) for l = 0, ..., j − i − 1.
As every vertex in G n/k has degree less than N 1/(k−1) , the number of such sequences with given x 0 is at most Hence, for every x ∈ P there are at most n elements y ∈ P such that x < 2 y. Thus, e(G(P , < 2 )) < n 2 .
We deduce that e(G(P , < 1 , < 2 )) < (1/2k + )n 2 . Also, let X, Y ⊂ P be disjoint sets such that X ⊥ Y and |X| = |Y |. Then, there exist positive integers t and u such that 1 ≤ t, u ≤ k, |X ∩ A t | ≥ |X|/k and |Y ∩ A u | ≥ |Y |/k. We cannot have t = u, otherwise, there exist x ∈ X ∩ A t and y ∈ Y ∩ A t with x < 1 y or y < 1 x, contradicting X ⊥ Y . Hence, t = u. Without loss of generality, suppose that t < u.
Let H be the bipartite subgraph of G(P , < 2 ) induced on A t ∪ A u . We show that H contains a subgraph isomorphic to G n/k . Let x ∈ A t arbitrary, and let a 0 (x), ..., a u−t−1 (x) be the unique sequence such that a 0 (x) = x, a l (x) ∈ A t+l for l = 1, ..., u − t − 1, and f t+l (a l (x))g t+l +1 (a l +1 (x)) ∈ M n/k . As M n/k is a complete matching, every a l : A t → A t+l is a bijection. Also, the subgraph of G(P , < 2 ) induced on A u−1 ∪ A u is isomorphic to G n/k . If x ∈ A u−1 and x ∈ A u with x < 2 x , then a −1 u−1 (x ) < 2 x . Hence, the subgraph of G(P , < 2 ) induced on A t ∪ A u contains a subgraph isomorphic to G n/k . Thus, the complement of H does not contain K t,t with t > 2 −1 N 1−1/(k−1) log n, so |X| = |Y | < 2kN 1−1/(k−1) log n < 2 −1 kn 1−1/(k−1) log n.
Our next aim is to prepare the proof of part (ii) of Theorem 11. It turns out, our proof would be simpler if < 1 and < 2 had a common linear extension, which is not the case in general. However, the next lemma shows that we can find a constant number of large subsets in our poset such that between these subsets < 1 and < 2 behave as if they had a common linear extension.
, and x and y are comparable in < s , then x < α s s y.

Claim 13
Let p and r be positive integers. There exists c (p, r) > 0 with the following property. Let < 1 , ..., < r be total orders on the n element set P . There exist B 1 , ..., B p ⊂ P pairwise disjoint subsets such that (i) |B 1 | = ... = |B p | > c (p, r)n; (ii) for s = 1, ..., r and 1 ≤ i < j ≤ r, we have either B i < s B j or B j < s B i .
Proof We shall proceed by induction on r. In case r = 1, the statement is trivial with c (p, 1) = 1/p. Let r ≥ 2 and suppose the statement holds for r − 1 instead of r. Let C 1 , ..., C p ⊂ P be disjoint sets such that and for every 1 ≤ i < j ≤ p and s = 1, ..., r − 1, we have C i < s C j or C j < s C i .
C i , and for x ∈ P , let τ (x) be the position of x in the order < r in P . For i = 1, ..., p, let We have D i < r D j for any 1 ≤ i < j ≤ p. Our B 1 , ..., B p are going to be suitable subsets of C 1 , ..., C p and D 1 , ..., D p .
Let S, T be two disjoint copies of [p], and define the bipartite graph G = (S, T , E) as follows: for i ∈ S and j ∈ T , let ij ∈ E if We show that G has a complete matching. By Hall's theorem [7], we only need to check if Hall's condition holds. Let X ⊂ [p] be arbitrary and let (X) denote the set of neighbours of X in G. Let U = i∈X D i , then Also, the elements of (X) cover at most | (X)||P |/p elements in U , while the elements not in (X) cover at most p|X|(|P |/p 2 (p +1)) = |P ||X|/p(p +1) elements in U . Hence, Thus, we have But |X| and | (X)| are integers not larger than p. Hence, |X| ≤ | (X)| also holds. So, Hall's condition is satisfied and there exists a a complete matching in G. Let the edge set of such a matching be {ix i : i ∈ S}. Setting B i = C i ∩ D x i and c (p, r) = c (p, r − 1)/p 2 (p + 1), we have both (i) and (ii) satisfied.
Let p = (h − 1) 2 r−1 + 1 and let B 1 , ..., B p ⊂ P be disjoint sets such that |B 1 | = ... = |B p | > c (p, r)n, and for 1 ≤ i < j ≤ p and 1 ≤ s ≤ r, we have either B i < s B j or B j < s B i . Define the partial orders {≺ v } v∈{0,1} r−1 on [p] as follows: for i, j ∈ [p] and v ∈ {0, 1} r−1 , let i ≺ v j if B i < r B j , and for s = 1, ..., r −1, we have B i < s B j in case v s = 0, and B j < s B i in case v s = 1. Then any two different elements of [p] are comparable in at least one of the partial orders {≺ v } v∈{0,1} r−1 . Hence, by repeated applications of Dilworth's theorem, there exist w ∈ {0, 1} r−1 and C ⊂ [p] such that and C is a ≺ w chain. Let i 1 ≺ w ... ≺ w i h be h elements of this chain, and for j = 1, ..., h, let A j = B i j . Also, for s = 1, ..., r, let α i = w i . Finally, let c(r, h) = c (r, p). Then the conditions of the theorem are satisfied.
Before we start the proof of part (ii) of Theorem 11, we still need the following two lemmas.
Lemma 14 Let A 0 , .., A k be pairwise disjoint sets of size m, and let Let < be a partial order on P such that whenever x < y for some x ∈ A i and y ∈ A j , then i < j. Suppose that G(P , <)[A 0 , A k ] has less than m 2 /4 edges. There exist 0 ≤ l ≤ k − 1 and X ⊂ A l , Y ⊂ A l+1 such that |X|, |Y | > m 1−1/k , and X ⊥ Y .
We show that we can find a decreasing sequence of sets B ⊇ B 1 ⊇ .... ⊇ B k with the following properties: |B i | = 2 k−i m 1−i/k , and |U i (B i )| > m/2. Note that B k is a one element set. Hence, writing x for that one element, we have contradicting x ∈ B, finishing our proof. We shall define our sets B 1 , ..., B k recursively. Let B 1 be any subset of B of size

Claim 15 For any positive integer t ≤ |B
Proof Let x 1 , ..., x p be the elements of B i . Let S 1 , ..., S p be a partition of U(B i ) such that S j ⊂ U i ({x j }) for j = 1, ..., p. Without the loss of generality, Setting t = 2 k−i−1 m 1−(i+1)/k , we get a set C such that Then, we have X ⊥ Y and |X|, |Y | > m 1−1/k , which is a contradiction. Hence, |U i+1 (C)| > m/2, and B i+1 = C satisfies our conditions. We also need the following easy corollary of Theorem 2, which we shall state without proof.
Proposition 16 Let < 1 , < 2 be partial orders on the n element set P . At least one of the following holds: Proof of Theorem 11,(ii). We have to prove that there exists a constant γ (δ) such that if P is a set with n elements, and < 1 , < 2 are partial orders on P satisfying e(G(P , < 1 , < 2 )) < γ (δ)n 2 , then P contains two disjoint subsets A, B of size at least n 1−γ such that A ⊥ B. For simplicity, let G 1 = G(P , < 1 ) and G 2 = G(P , < 2 ).
Suppose that P does not contain two disjoint subsets A, B of size at least n 1−δ such that A ⊥ B. Let k = 2/δ and h = 128k, and let c 1 = c (2, h), where c(r, h) is the constant defined in Lemma 12. Then there exist L 1 , ..., L h ⊂ P pairwise disjoint sets with the following properties: |L 1 | = ... = |L h | = c 1 n; replacing < 2 with its dual if necessary, if x ∈ L i and y ∈ L j for some 1 ≤ i < j ≤ h, and x, y are comparable in < 1 or < 2 , then x < 1 y or x < 2 y, respectively.
Let m = n 1−δ/2 . By Proposition 16, if n is sufficiently large, every subset of P of size at least c 1 n/2 contains three disjoint subsets B 1 , B 2 , B 3 of size m such that B 1 < 1 B 2 < 1 B 3 or B 1 < 2 B 2 < 2 B 3 . Hence, we can cover at least half of L i with disjoint triples of subsets such that each set has size m and each triple spans a balanced complete 3-partite graph in G 1 or in G 2 .
More precisely, let s = c 1 n/2m. Then, for i = 1, ..., h, there is a system of disjoint sets 3 , and call it type 2 otherwise. Without the loss of generality, we can suppose that there are at least sh/2 type 1 pairs in [h] × [s], and let S be the set of such pairs.
Let H = (S, E) be the complete graph on S, and let w be a weight function defined on E as follows. Let (i, j ), (i , j ) ∈ S, and let f be the edge joining (i, j ) and (i , j ). If i = i , or there exist x ∈ B i,j,2 and y ∈ B i ,j ,2 such that x < 1 y or y < 1 x, then let w(f ) = 1; otherwise, let Note that if there exist x ∈ B i,j,2 and y ∈ B i ,j ,2 such that x < 1 y, then B i,j,1 < 1 B i ,j ,3 . Hence, there are at least m 2 edges between B i,j, Let w(E) = f ∈E w(f ). Then the number of edges of G(P , < 1 , < 2 ) is at least Let t be the number of edges f ∈ E such that w(f ) ≤ 1/4. We show that Suppose that t > |S| 2 (1/2 − 1/2k). Consider the graph H with vertex set S, and edge set E = {f ∈ E : w(f ) ≤ 1/4}. By Turán's theorem [13], there exists T ⊂ S of size k + 1 such that H [T ] is a complete graph. Let (i 0 , j 0 ), ..., (i k , j k ) be the elements of T and suppose that i 0 < ... < i k . First, note that A i l ,j l ,2 ⊥ 1 A i l ,j l ,2 for all 0 ≤ l < l ≤ k, as the weight of the edge {(i l , j l ), (i l , j l )} is less than 1.
Set A l = B i l ,j l ,2 for l = 0, ..., k. Then e(G 2 [A 0 , A k ]) < m 2 /4. Hence, by Lemma 14, there exist 0 ≤ l ≤ k − 1 and X ∈ A l , Y ∈ A l+1 such that |X| = |Y | = m 1−1/k , and X ⊥ 2 Y . But then X ⊥ Y , and contradiction. Thus, we must have Then where the last inequality holds if n is sufficiently large. Plugging this result in Eq. 2, we get the following lower bound on the number of edges of G(P , < 1 , < 2 ): Thus, setting γ (δ) = 256c 2 1 δ −1 finishes the proof of the theorem. We remark that if < 1 and < 2 have a common linear extension, which is often the case in applications, then we do not need to use Lemma 12 in the previous proof and we can simply write 1/h instead of c 1 . Then we get the bound γ (δ) = δ/256, which almost matches the constant of part (i) in Theorem 11. However, we conjecture that an even stronger bound holds in general. We also conjecture that f I r,h (n, m) has a similar growth as f 2,2 (n, m) for r ≥ 3 or r = 2 and h ≥ 3, but we cannot even quantify a precise conjecture for these cases.

Applications
Partial orders naturally arise in some geometric problems. The intersection graph of a set system C is the graph G = G(C, E), where A, B ∈ C forms an edge if A ∩ B = ∅. The intersection graph of convex sets in the plane was investigated in a series of paper. Larman et. al. [8], and Pach and Törőcsik [10] showed that the intersection graph of convex sets is a 4-incomparability graph. Hence, by an immediate application of Dilworth's theorem yields that amongst n convex sets there are always at least n 1/5 such that they are pairwise disjoint, or any two of them intersects. Also, as it was noted in [5], Theorem 2 implies a bipartite version of this theorem, namely that if C is a family of n convex sets, then there are A, B ⊂ C of size n 1−o(1) such that for any A ∈ A and B ∈ B, we have A ∩ B = ∅, or for any A ∈ A and B ∈ B, we have A ∩ B = ∅. In [6], this result was improved, showing that we can find two linear sized families A, B ⊂ C with the same property.
Call a set in R 2 vertically convex, if every vertical line intersects the set in an interval. The intersection graph of connected, vertically convex sets is also a 4-incomparability graph.
Hence, Theorem 2 also implies the existence of a complete bipartite graph of size n 1−o (1) either in the intersection graph or its complement. However, we can no longer guarantee a linear sized complete bipartite graph in the intersection graph or in its complement. In [9], it is shown that for any > 0, there is a collection of n continuous functions on [0, 1] such that the largest bipartite graph in the intersection graph has size O(n/ log n), and the largest complete bipartite graph in its complement has size O(n ).
Nevertheless, Theorem 4 immediately implies that if the intersection graph of vertically convex sets is sparse enough, then we have a linear sized complete bipartite graph in its complement.
Theorem 18 Let > 0 and let C be a collection of n connected, vertically convex sets in the plane. If the number of unordered pairs {A, B} ∈ C (2) with A ∩ B = ∅ is less than , Proof As we aim for the self-containment of the paper, we shall define the 4-partial orders on C, whose incomparability graph is the intersection graph. For any C ∈ C, let l(C) = inf{x ∈ R : ∃y : (x, y) ∈ C} and let r(C) = sup{x ∈ R : ∃y : (x, y) ∈ C}.
Define the relations ≺ 1 , ≺ 2 , ≺ 3 on C as follows: Note that ≺ 1 , ≺ 2 , ≺ 3 are not partial orders, as it is possible that C ≺ i D and D ≺ i C both hold.
However, define the relations < 1 , < 2 , < 3 , < 4 on C as follows: One can easily check that < 1 , < 2 , < 3 , < 4 are partial orders on C, and C and D are comparable in some < i if and only if C and D are disjoint. Now, if there are less than (1/32 − )n 2 unordered pairs {A, B} ∈ C (2) such that A and B intersect, then G(P , < 1 , < 2 , < 3 , < 4 ) has more than We note that these results have no analogue in higher dimensions: Tietze [11] proved that any graph can be realized as the intersection graph of 3-dimensional convex sets.
We also show an application of Theorem 4 for a variant of a classical problem of Erdős: let α < π be a positive real. Let g d (α) be the smallest integer m such that in any configuration of m points in the d-dimensional space there is an angle larger than α. Erdős and Szekeres [3] proved that g 2 π − π r + = 2 r + 1, if r ≥ 2 is an integer and > 0 is sufficiently small. They also proved that for any 0 < β < π. The author of this paper [12] considered the following generalization of this problem: given 0 < α < π and positive integers m and d, what is the maximal s such that any configuration of m points in the d-dimensional space contains s points, where every triangle has an angle larger than α. It was proved that any configuration of t r + 1 points in the plane contains t + 1 points, where every triangle has an angle larger than π − π/r. We prove a tripartite version of this result.

Theorem 19
Let 0 < α < π and let d be a positive integer. There exists a constant c(α, d) > 0 with the following property. Suppose that n is a sufficiently large positive integer and S is a configuration of n points in the d-dimensional space. There exist three pairwise disjoint sets A, B, C ⊂ S such that |A| = |B| = |C| > c(α, d)n, and for every X ∈ A, Y ∈ B, Z ∈ C, the angle XY Z∠ is larger than α.
Proof Let s = 1/(π −α) , and let V be a finite set of unit vectors with the property that for any w ∈ R d , there exists v ∈ V such that the angle of v and w is less than (π − α)/2. Such V trivially exists. For each v ∈ V , define the relation < v on R d as follows: if X, Y ∈ R d , then X < v Y if the angle of v and − → XY is less than (π − α)/2. Then < v is a partial order: X < v Y is equivalent to the inequality v, − → XY > | − → XY | sin(α/2).
Also, if X < v Y < v Z holds for some X, Y, Z ∈ R d , then by elementary geometry, the angle XY Z∠ is larger than α.
Let S ⊂ R d , |S| = n. Then G(S, {< v } v∈V ) is the complete graph on n vertices, because we choose V such that for any X, Y ∈ R d , there exists v ∈ V with X < v Y . Let Hence, by Theorem 4, there exist v ∈ V and A, B, C ⊂ S such that A, B, C are pairwise disjoint, |A| = |B| = |C| > c(α, d)n, and A < v B < v C. But then, for any X ∈ A, Y ∈ B and Z ∈ C, we also have XY Z∠ > α.
Consider the case d = 2 and α < π − π/r. In the proof above, we can choose V to be a 2r element set, so using the bound in the remark after Theorem 6, we can show that c(α, 2) > e −cr with some constant c > 0. However, we conjecture that an even stronger bound holds.
Conjecture 20 Let 0 < α < π−π/r and let n be a positive integer. Let S be a set of n points in the plane. There exist A, B, C ⊂ S disjoint subsets such that |A| = |B| = |C| = (n/r), and for every X ∈ A, Y ∈ B, Z ∈ C, we have XY Z∠ > α.
Taking S to be the [ √ n] × [ √ n] square grid, one can easily show that the dependence on r in the conjecture cannot be improved.