Ôåäåðàëüíîå ãîñóäàðñòâåííîå àâòîíîìíîå îáðàçîâàòåëüíîåó÷ðåæäåíèå âûñøåãî îáðàçîâàíèÿ

Ìîñêîâñêèé ôèçèêî-òåõíè÷åñêèé èíñòèòóò(ãîñóäàðñòâåííûé óíèâåðñèòåò) (ÌÔÒÈ)

Íà ïðàâàõ ðóêîïèñè

Àíäðåé Áîðèñîâè÷ Êóïàâñêèé

ÑÅÌÅÉÑÒÂÀ ÌÍÎÆÅÑÒÂ Ñ ÇÀÏÐÅÙÅÍÍÛÌÈ ÊÎÍÔÈÃÓÐÀÖÈßÌÈÈ ÏÐÈËÎÆÅÍÈß Ê ÄÈÑÊÐÅÒÍÎÉ ÃÅÎÌÅÒÐÈÈ

05.13.17 òåîðåòè÷åñêèå îñíîâû èíôîðìàòèêè

Äèññåðòàöèÿ íà ñîèñêàíèå ó÷åíîé ñòåïåíèäîêòîðà ôèçèêî-ìàòåìàòè÷åñêèõ íàóê

Ìîñêâà 2019

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Federal state autonomous educationalorganisation of higher education

Moscow institute of physics and technology(state university) (MIPT)

On the rights of a manuscript

Andrey Borisovich Kupavskii

FAMILIES OF SETS WITH FORBIDDEN CONFIGURATIONSAND APPLICATIONS TO DISCRETE GEOMETRY

05.13.17 theoretical foundations of computer science

Dissertation for the degreeof doctor of physical and mathematical sciences

Moscow 2019

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Ïîñâÿùàåòñÿ ìîèì ðîäèòåëÿì: Áîðèñó è Íàäåæäå.

Dedicated to my parents Boris and Nadezhda.

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Contents

Notations 8

Introduction 10

1. Intersecting families of sets 311.1. Introduction and statement of results . . . . . . . . . . . . . 31

1.1.1. Stability for the ErdosKoRado theorem . . . . . . . 321.1.2. Diversity . . . . . . . . . . . . . . . . . . . . . . . . . 331.1.3. Counting intersecting families . . . . . . . . . . . . . . 341.1.4. Product-type inequalities . . . . . . . . . . . . . . . . 351.1.5. Cross s-intersecting families . . . . . . . . . . . . . . . 371.1.6. t-intersecting, cross s-intersecting families . . . . . . . 37

1.2. Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . 381.2.1. Shifting . . . . . . . . . . . . . . . . . . . . . . . . . . 381.2.2. Lex order and the KruskalKatona theorem . . . . . . 39

1.3. Switchings via regular bipartite graphs and theorems on inter-secting families . . . . . . . . . . . . . . . . . . . . . . . . . 411.3.1. Cross-intersecting families . . . . . . . . . . . . . . . . 431.3.2. Proof of Theorem 1.5 . . . . . . . . . . . . . . . . . . 46

1.4. Proof of Theorem 1.7 . . . . . . . . . . . . . . . . . . . . . . 481.5. Diversity for 2k ≤ n ≤ 3k . . . . . . . . . . . . . . . . . . . . 51

1.5.1. Intersecting families with the largest diversity are notshifted . . . . . . . . . . . . . . . . . . . . . . . . . . 52

1.6. Counting intersecting families . . . . . . . . . . . . . . . . . . 581.6.1. Cross-intersecting families . . . . . . . . . . . . . . . . 581.6.2. Intersecting families . . . . . . . . . . . . . . . . . . . 61

1.7. Product inequalities . . . . . . . . . . . . . . . . . . . . . . . 631.7.1. Preliminaries . . . . . . . . . . . . . . . . . . . . . . . 631.7.2. Short proof of Theorem 1.11 . . . . . . . . . . . . . . 641.7.3. Proof of Theorem 1.12 modulo Theorem 1.13 . . . . . 651.7.4. Proof of Theorem 1.13 . . . . . . . . . . . . . . . . . 65

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1.8. Proof of Theorem 1.15 . . . . . . . . . . . . . . . . . . . . . 701.8.1. Preliminaries . . . . . . . . . . . . . . . . . . . . . . . 701.8.2. Proof of Theorem 1.15 . . . . . . . . . . . . . . . . . 71

1.9. t-intersecting, cross s-intersecting families . . . . . . . . . . . 751.9.1. Proof of Theorem 1.43 . . . . . . . . . . . . . . . . . 76

2. Intersecting families of vectors 802.1. Introduction and statement of results . . . . . . . . . . . . . 80

2.1.1. Vectors with a xed number of +1's and −1's . . . . . 812.1.2. Vectors of xed length . . . . . . . . . . . . . . . . . . 84

2.2. Proofs of Theorems 2.5 and 2.6 . . . . . . . . . . . . . . . . . 852.2.1. Summary . . . . . . . . . . . . . . . . . . . . . . . . . 852.2.2. Comparing the constructions . . . . . . . . . . . . . . 872.2.3. Auxiliaries from extremal set theory . . . . . . . . . . 89

2.2.3.1. Shifting . . . . . . . . . . . . . . . . . . . . 892.2.3.2. Shadows . . . . . . . . . . . . . . . . . . . . 902.2.3.3. General form of Katona's circle method . . . 912.2.3.4. An inequality for cross-intersecting families of

sets . . . . . . . . . . . . . . . . . . . . . . . 922.2.3.5. Analysis of the KruskalKatona's Theorem . 932.2.3.6. A sharpening of Theorem 2.16 . . . . . . . . 97

2.2.4. Proof of Theorem 2.6 . . . . . . . . . . . . . . . . . . 992.2.5. Proof of Theorem 2.5 in the case 3k ≤ n ≤ k2 . . . . . 100

2.3. The proof of Theorem 2.9 . . . . . . . . . . . . . . . . . . . . 1042.4. Proof of Theorem 2.8 . . . . . . . . . . . . . . . . . . . . . . 1052.5. Vectors of xed length . . . . . . . . . . . . . . . . . . . . . . 106

2.5.1. Simple properties of F (n, k, l) . . . . . . . . . . . . . 1062.5.2. The case k = 3 . . . . . . . . . . . . . . . . . . . . . . 1072.5.3. Preliminaries . . . . . . . . . . . . . . . . . . . . . . . 1082.5.4. Proof of Theorems 2.10 and 2.11 . . . . . . . . . . . . 1082.5.5. Proof of Theorem 2.37 . . . . . . . . . . . . . . . . . 112

3. Families with small matching number 1153.1. Introduction and statement of results . . . . . . . . . . . . . 1153.2. Proof of Theorem 3.4 for s ≥ 5 . . . . . . . . . . . . . . . . . 119

3.2.1. Averaging over partitions . . . . . . . . . . . . . . . . 1193.2.2. Calculations . . . . . . . . . . . . . . . . . . . . . . . 1213.2.3. Hilton-Milner-type result for Erdos Matching Conjecture1233.2.4. Proof of Theorem 3.4 (i) for s ≥ 5 . . . . . . . . . . . 126

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3.2.5. Proof of Theorem 3.4 (ii), (iii) . . . . . . . . . . . . . 1303.3. Proof of Theorem 3.4 (i) for s = 3, 4 . . . . . . . . . . . . . . 134

3.3.1. Calculations . . . . . . . . . . . . . . . . . . . . . . . 1343.3.2. Proof of Theorem 3.4 for s = 3 . . . . . . . . . . . . . 134

3.3.2.1. The case m ≤ 2, s = 3 . . . . . . . . . . . . 1433.3.3. Proof of Theorem 3.4 for s = 4 . . . . . . . . . . . . . 1433.3.4. The case m ≤ 2, s = 4 . . . . . . . . . . . . . . . . . . 148

3.4. Inequalities for k-dependent families . . . . . . . . . . . . . . 1493.4.1. The proof of (3.55) . . . . . . . . . . . . . . . . . . . 1503.4.2. The proof of (3.56) . . . . . . . . . . . . . . . . . . . 1513.4.3. Applications . . . . . . . . . . . . . . . . . . . . . . . 154

3.5. Discussion . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156

4. Random Kneser graphs 1614.1. Introduction and statement of results . . . . . . . . . . . . . 161

4.1.1. The bounds . . . . . . . . . . . . . . . . . . . . . . . 1644.2. Proof of Theorem 4.1 . . . . . . . . . . . . . . . . . . . . . . 166

4.2.1. Proof of Theorem 1 . . . . . . . . . . . . . . . . . . . 1684.3. Proof of Theorem 4.2 . . . . . . . . . . . . . . . . . . . . . . 171

4.3.1. Basic approach . . . . . . . . . . . . . . . . . . . . . . 1714.3.1.1. Coloring random subgraphs of blow-ups of

hypergraphs . . . . . . . . . . . . . . . . . . 1724.3.1.2. Numerical Corollaries for Kneser hypergraphs 173

4.3.2. The approach rened . . . . . . . . . . . . . . . . . . 1754.3.3. Simple lower bounds . . . . . . . . . . . . . . . . . . . 179

5. Coverings and ε-nets 1815.1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . 1815.2. Proof of Theorem 5.3 using Lemma 5.4 . . . . . . . . . . . . 1875.3. Proof of Lemma 5.4 . . . . . . . . . . . . . . . . . . . . . . . 1885.4. Proof of Theorem 5.6 . . . . . . . . . . . . . . . . . . . . . . 190

6. Applications to distance problems 1966.1. Introduction and statement of results . . . . . . . . . . . . . 196

6.1.1. Borsuk's problem . . . . . . . . . . . . . . . . . . . . 1966.1.2. Distance graphs with high girth . . . . . . . . . . . . 197

6.2. Borsuk's problem . . . . . . . . . . . . . . . . . . . . . . . . 2016.2.1. Proofs of Theorems 6.1 and 6.5 . . . . . . . . . . . . . 202

6.2.1.1. Construction . . . . . . . . . . . . . . . . . . 202

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6.2.1.2. Transforming Ω′ into an Ω ⊂ Sd−1r . . . . . . 203

6.2.1.3. Lower bound for f(Ω) . . . . . . . . . . . . . 2046.2.1.4. Proof of Lemma 6.8 . . . . . . . . . . . . . . 206

6.2.2. Proof of Theorem 6.6 . . . . . . . . . . . . . . . . . . 2076.2.3. Proof of Theorem 6.7 . . . . . . . . . . . . . . . . . . 2106.2.4. Improving Construction from Section 6.2.1.1 is hard . 210

6.3. Distance graphs of high girth . . . . . . . . . . . . . . . . . . 2136.3.1. Preliminaries . . . . . . . . . . . . . . . . . . . . . . . 2136.3.2. Proof of Theorem 6.2 . . . . . . . . . . . . . . . . . . 215

Bibliography 219

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Notations

N the set of all natural numbers;

Z the set of all integer numbers;

R the set of all real numbers;

[n] the set 1, . . . , n of rst n integers;

2[n] the power set of [n];([n]k

) the set of all k-element subsets of [n];

|A| the cardinality of a set A;

bac the lower integer part of a;

dae the upper integer part of a;

〈x,y〉 the scalar product of vectors x and y;

Kr a complete graph on r vertices;

Ks,t a complete bipartite graph with parts of sizes s and t;

Pr[E] the probability of an event E;

E[X] the expectation of a random variable X;

f(n) = o(g(n)) for any c > 0 there exists n0, such that for any n > n0

we have |f(n)| ≤ c|g(n)|;

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f(n) = O(g(n)) there exists C > 0, such that for any n ∈ N we have|f(n)| ≤ C|g(n)|;

f(n) = Θ(g(n)) there exist c, C > 0, such that for any n ∈ N we havec|g(n)| ≤ |f(n)| ≤ C|g(n)|;

f(n) ∼ g(n) for any ε > 0 and any suciently big n ∈ N we have(1− ε)g(n) ≤ f(n) ≤ (1 + ε)g(n);

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Introduction

Introduction to Extremal Combinatorics.

Extremal combinatorics questions typically ask to maximize a certain func-tion over a class of objects. This is one of possible approaches to analyze agiven class of objects, and is dierent from, say, structural approach, whenthe goal is to nd a compact characterization of the objects belonging tothe class. Let us illustrate it on the following two famous results on planargraphs. Recall that a graph is a pair (V,E) where V is a set and E is a col-lection of pairs of elements from V . A graph is planar if, roughly speaking,one may draw it on the plane without crossing edges.

• One consequence of the famous Euler's formula states that any planargraph with n vertices has at most 3n− 6 edges. This is may be seen asan extremal graph theory result: we upper bound the number of edgesin the graphs belonging to a certain class.

• On the other hand, Kuratowski's theorem states that a graph is planar ifand only if it does not contain a subdivision ofK5 orK3,3 as a subgraph.Without giving denition of a subdivision, this result gives a compactcombinatorial characterisation of all planar graphs. Thus, this resultbelongs to structural graph theory.

These two results (and two ways of thinking about classes of combinatorialobjects) complement each other.

Arguably the most famous topic in extremal graph theory is to maximizethe number of edges in a graph that does not contain some other graph Has a subgraph. The two most known examples are when H = Kr or Ks,t.Mantel's theorem, going back to 1907, states that the n-vertex completebipartite graph with almost equal parts has the largest number of edgesamong all triangle-free n-vertex graph. This result was then generalized toKr-free graphs by Turan. We state it in a slightly simplied form.

Theorem (Turan). For any xed r ≥ 3, any n-vertex Kr-free graph has atmost (1− 1

r−1)n2

2 edges.

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The number of edges in triangle-free (and Kr-free for any r > 3) graphsis quadratic in n. The situation with the Ks,t-free graphs is very dierent: ifs ≤ t are xed, then the famous KovariSosTuran theorem states that anyKs,t-free graph has at most tn2−1/s edges, which is subquadratic.

Extremal questions are studied for many objects that include hypergraphsor families, words, subsets of integers, collections of vectors, arrangements ofpoints and hyperplanes etc. As an example, the celebrated result of Greenand Tao on existence of arbitrarily long arithmetic progressions in primesor, more generally, in dense subsets of integers, is an extremal combinatoricsresult.

The probabilistic method. In combinatorics, seemingly very dif-ferent subjects share common methods, and it is sometimes very useful tothink about methods, rather than subjects, in combinatorics. There aremany methods used in extremal combinatorics: the probabilistic method,the algebraic and eigenvalue methods, the topological methods, the meth-ods of discrete Fourier analysis and the analysis of Boolean functions, thequasirandom method etc. Some of these methods will be discussed later inthe introduction.

One of the most important methods, pioneered by P. Erdos, is the prob-abilistic method. The main principle behind the method is to show that acertain object exists by developing a right probability space and proving thata random object in this space has the desired properties with positive prob-ability, rather than to construct the object explicitly. Let me illustrate it.In coding theory, one is often confronted with the problems of the followingtype.

Question. How many unit vectors in Rd one may select so that the scalarproduct of any of them is at most 0.01?

It is not easy to provide a good explicit construction. However, one caneasily show that there exist cd such vectors with c > 1 using the probabilisticmethod: it is sucient to simply choose the vectors at random, bound theprobability that a pair has scalar product bigger than 0.01, and nally takea union bound over all such pairs. (To bound the probability, we use thefollowing principle: the surface area of a sphere in high dimension is concen-trated around (any) diametral sphere.) One big advantage of probabilisticmethod is that it is very exible.

Let us revisit the bound e(G) ≤ 3v(G) − 6, valid for any planar graph.In an attempt to measure the representation complexity of dierent graphs,

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it is a natural question to ask, how many crossing pairs of edges one mayexpect for the best drawing of an n-vertex graph G with e edges on theplane. There is such a bound, called the crossing lemma, sharp up to aconstant factor. In particular, it tells us that the number of crossing pairsof edges in any representation of Kn or Kn,n on the plane is at least Cn4 forsome absolute C > 0:

Theorem (Crossing lemma). In any drawing of G with e edges and v ver-tices, e ≥ 4v, on the plane, there are at least e3

64v2 pairs of crossing edges.

A very short and beautiful proof of this theorem uses probabilistic method.From Euler's formula, one can easily deduce a bound linear in e on thenumber of crossings (deleting one edge, one deletes at least one crossing,until nally making the graph planar). Using probabilistic method, we candeduce the crossing lemma by using this linear bound for a subgraph of G,formed by including each vertex of G independently with probability p!

The major inherent drawback of probabilistic method is that it supplieswith existential proofs rather than explicit constructions. In some cases, onecan deal with it using derandomization, but in most cases there is a big gapbetween the objects (or algorithms) constructed probabilistically and deter-ministic constructions. I will illustrate it with an example from a closelyrelated eld: Ramsey theory.

Ramsey theory. The underlying principle in Ramsey theory is thatin any suciently large object one can nd a highly structured part. Forexample, the famous van der Vaerden theorem from 1927 states that, nomatter how one colors natural numbers in 2 colors, one can always nd amonochromatic arithmetic progression of length l, for any integer l. However,the theory got its name from a 1930 result1 due to an English logician Ramsey,who proved the following.

Theorem (Ramsey). For any k there exists n, such that in any 2-coloringof the edges of the complete graph Kn one always nds k-vertices, such thatall edges between these k vertices have the same color (in other words, onecan nd a monochromatic copy of Kk).

The same result was later rediscovered by Erdos and Szekeres. One mayrestate it in terms of arbitrary graphs, rather than colorings: for sucientlylarge n = n(k), any n-vertex graph contains either a complete graph Kk oran independent set of size k. This theorem, and Ramsey theory in general,

1Probably, the rst Ramsey-type result was proven by Schur in 1913.

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plays an important role in computer science. In particular, it is used in theproof of hardness of approximating the chromatic number of graphs. (Recallthat the chromatic number of a graph is the minimum number of colorsneeded to color the vertices so that no edge has monochromatic endpoints.Chromatic number plays important role in many areas of computer science,in particular, in scheduling.)

Naturally, one may ask for the behaviour of n = n(k) in the theoremabove. The original proof yields the bound n(k) ≤ 4k. The lower boundof roughly n ≥ 2k was shown by Erdos, again using simple probabilisticmethod: color each edge of Kn with one of the two colors at random. Thenshow that the expected number of monochromatic Kk is strictly smallerthan 1 for appropriately chosen parameters. This implies that there exists acoloring with no monochromatic Kk. Although the proof of Erdos essentiallyimplies that almost all colorings of the complete graph on 2k vertices have nomonochromatic Kk (we may state it as follows: the random graph G(2k, 1/2)does not contain neither a Kk nor an independent set of size k with highprobability), it is extremely hard to come up with an explicit construction ofsuch a graph. The best known such construction gives a graph on roughly2√k vertices with on monochromatic Kk. It is due to Frankl and Wilson,

and it is coming from extremal set theory, the eld that is in the focus of mydissertation.

Introduction to Extremal Set Theory

Extremal set theory is a eld started by Erdos, Kleitman, Sperner and oth-ers, which typically searches for the largest systems of sets (families) undercertain restrictions. A family is simply a collection of sets.2 Many results andnotions from extremal set theory have real-life applications: measuring thecomplexity of an arrangement of objects, the algorithms for ecient encodingand error-free decoding of information.

Below, I state several key results from the area that motivated my research.One of the rst results in the eld belongs to Sperner and is as follows.

Theorem (Sperner [228]). The largest size of a family F ⊂ 2[n] such thatno F1 ∈ F is strictly contained in F2 ∈ F , is

(nbn2 c).

Any family F with this property is called an antichain, referring to an2Recall that a pair (V,E), where V is a set of vertices and E is a collection of subsets of E. A

hypergraph is k-uniform if all sets in E have size k. Thus, a 2-uniform hypergraph is a graph. When theset V is not important then we use the term family (of sets) instead. In that case, we typically assumethat V = [n] := 1, . . . , n.

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antichain in the containment order on the poset on 2[n]. The bound is sharpand is attained on the family of all sets of size bn2c (or d

n2e). This result

has numerous applications, in particular, to the famous HardyLittlewoodproblem on the maximum number of distinct sums of numbers a1, . . . , an > 1that all fall into some interval of length 1.

Another cornerstone of extremal set theory is the following theorem ofErdos, Ko and Rado, proved in 1938, but published only 23 years later, in1961. We say that a family is intersecting if any two of its sets intersect.

Theorem (Erdos, Ko, Rado [73]). Any intersecting family F of k-elementsubsets of [n] has size at most

(n−1k−1

), provided n ≥ 2k.

It is tight for a star: a family of all sets containing a xed element. Ithas several very dierent proofs, and many classical extremal set theorymethods, such as shifting, cycle method, the eigenvalues method etc. wereintroduced in these proofs. Its generalizations, such as the famous FranklWilson theorem and the Complete t-Intersection Theorem due to Ahlswedeand Khachatrian, have several applications elsewhere in discrete mathematicsand computer science, some of which we will mention in the dissertation.

The following result plays a central role in many applications of extremalset theory to dierent threshold phenomena in combinatorics and computerscience. For a family F of k-element sets, let ∂F stand for the collection ofall (k − 1)-element sets contained in at least one set from F . Recall that(xk

):= x(x−1)...(x−k+1)

k! for any real x ≥ k. The KruskalKatona theorem, inthe form found by Lovasz, states that

Theorem (Kruskal [159]; Katona [138]). Given a family F of k-element setswith |F| =

(xk

)for some x ≥ k, we have |∂F| ≥

(xk−1

).

In particular, it was used by Bollobas and Thomason [35] to show thatmonotone properties always have a threshold. It is also directly related tothe famous KahnKalaiLinial theorem [135] from the analysis of Booleanfunctions.

Let us conclude the introduction to extremal set theory with the VapnikChervonenkisSauerShelah lemma, which is important in the areas of sta-tistical learning and computational geometry. For a family F ⊂ [n], itsVC-dimension is the size of the largest subset X ⊂ [n], such that X is shat-tered, that is, F ∩X : F ∈ F = 2X . (In plain words, X is shattered if anysubset of X, including the empty set and X itself, can be obtained by inter-secting X with some member of F .) Many families that arise in geometrical

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or statistical applications, have bounded VC-dimension. If this is the case,then the size of a family is polynomial in n:

Theorem (Vapnik, Chervonenkis [231]; Sauer [222]; Shelah [226]). If F ⊂[n] has VC-dimension at most d, then |F| ≤

∑di=0

(ni

)= O(nd).

This lemma is crucial for proving the existence of small ε-nets, which wediscuss later in the introduction.

In what follows, I discuss the development of several topics within extremalset theory. These are the topics of this dissertation.

Families with forbidden intersection patterns

The ErdosKoRado theorem, stated above, is tight on a full star: the collec-tion of all k-element subsets of [n] containing a given element. Any subfamilyof a full star is a star. Erdos, Ko and Rado asked, what is the next largestmaximal intersecting family. This question was answered by Hilton and Mil-ner (cf. Theorem 1.3), but more general questions lead to a development ofa whole eld of research. One perspective on this eld is as follows. The goalis to determine, how stable is the full star: in a given metric, how largeis the largest intersecting family F(d) ⊂

([n]k

), that is at distance at least d

from a star in that metric. Or, seen from a dierent perspective, we xa certain parameter of a family and ask for the largest intersecting familywith this parameter falling in a certain range. Below, I shall discuss severalresults of this type.

Frankl [81] determined the size of the largest intersecting family F ⊂(

[n]k

)in terms of its maximum degree ∆(F): the maximum over the elements ofthe ground set of the number of sets from F containing it. Together withZakharov [176], we got a stronger, dual version of Frankl's result in termsof diversity γ(F), where γ(F) = |F| −∆(F). More precisely, we found thelargest size of an intersecting family given the lower bound on the Hammingdistance from it to the closest star (see Theorem 1.5). Similar, but weaker,stability results were obtained by Keevash and Mubayi, Keevash, Das andTran, Keevash and Long (see [54, 144, 145, 147]).

Another approach was proposed in the works of Friedgut [115] and Dinurand Friedgut [60] (see Theorem 1.30) and recently greatly developed in theworks of Keller, Lifshitz and coauthors (see, e.g. [66, 148]). There, using themachinery of discrete Fourier transform, they proved the following type ofresults: any intersecting family, up to a small remainder, is contained in an

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intersecting junta. Here, a j-junta J is a family such that whether F ∈ Jor not depends on an intersection of F with a xed set of size at most j. Inthese terms, a full star (as well as its complement) is a 1-junta.

This type of results has proved to be very useful for dierent problems inextremal set theory. I will discuss some of them later. I will also mentionthat this type of results falls in the same framework as, say, the famousSzemeredi regularity lemma: a combinatorial object is decomposed into ahighly structured part, a quasirandom part, and a small remainder (for moredetails, see the paper by Tao [230]). In a recent work with Frankl [106],we have obtained essentially best possible junta approximation results forshifted families (see Section 1.2.1). Without giving precise denitions, thisis an extremely important class of families since in many of the extremal settheory problems is sucient to deal with shifted families.

Another possible distance is the covering number τ(F) of F , that is, thesize of the smallest set that intersects all sets in F . One may summarize itin the following problem.

Problem. Find the largest intersecting family in(

[n]k

)with τ(F) ≥ t.

For t = 2, the answer is given by the same result of Hilton and Milner,for t = 3, 4 and n > n0(k) it was solved by Frankl [78] and Frankl, Otaand Tokushige [109], respectively. In a recent paper [166], I have used acombination of the junta method and the bipartite exchange trick (whichI and Zakharov introduced in [176]) to resolve the question for t = 3 andn > Ck. With some work, the same could be done for the t = 4 case.However, the case t ≥ 5 remains wide open even for large n.

The particular case t = k of this problem was raised and studied in aclassic paper of Erdos and Lovasz [74]. Since each set from an intersectingfamily F is a cover for the family, we have τ(F) ≤ k for any intersectingfamily. Erdos and Lovasz showed that the largest size of such a family is atmost kk (interestingly, the bound does not depend on the size of the groundset). Recently, this bound was improved by Frankl [87], but still has theform k(1+o(1))k. The lower bound, due to Frankl, Ota and Tokushige [110], isroughly (k/2)k.

In the study of intersecting families, it is often very helpful to have someanalogous results for several families. We say that families A,B are cross-intersecting, if for any A ∈ A, B ∈ B, A and B intersect. The study ofcross-intersecting families goes back to the aforementioned work of Hilton andMilner. There are two types of questions that were studied quite extensively:

16

sum-type inequalities [38, 97, 113, 128, 176], in which the goal is to bound|A| + |B|, given some restrictions on one of the families, and product-typeinequalities [91, 191, 206] where the goal is to bound |A||B|. For the rsttype of inequalities, we refer to Theorem 1.14, and for the second type werefer to Theorem 1.11. We note that the latter theorem, in particular, impliesthe ErdosKoRado theorem.

Of course, one may ask stability-type questions for cross-intersecting fam-ilies. However, this turns out to be much more dicult. In a recent work[105], we have obtained an analogue of the result of [176] for cross-intersectingfamilies. This is directly related to the question of stability in the KruskalKatona theorem, which is a much harder question, see [105, 144, 146].

A more general question, addressed by Erdos, Ko, and Rado, is to ndthe largest t-intersecting family. We say that a family is t-intersecting if anytwo sets from the family intersect in at least t elements. In [73], the authorshave shown that the size of a t-intersecting family in

([n]k

)is at most

(n−tk−t),

provided n > n0(k, t). For n ≥ k > t ≥ 1 and n ≥ 2k − t dene

Ai(n, k, t) :=A ⊂

([n]

k

): |A ∩ [t+ 2i]| ≥ t+ i

, 0 ≤ i ≤ k − t.

These families are t-intersecting. It was conjectured by Frankl [76], proved inmany cases by Frankl and Furedi [89], and nearly 20 years later proved in fullgenerality by Ahlswede and Khachatrian [1] that the largest t-intersectingfamily in

([n]k

)for any n ≥ 2k − t (excluding n = 2k, t = 1) must be

isomorphic to one of Ai(n, k, t). This theorem found an application to theproblem of hardness of approximation of vertex cover [61]. Of course, it isnatural to ask for the cross-intersecting analogues of this result (cf. [92, 107,234]), as well as for stability results (cf. [67, 115]).

So far we have been talking only about families of k-element sets. Similarquestions for non necessarily uniform families are in general much easier. Inparticular, it is an easy exercise to show that the largest intersecting familyF ⊂ 2[n] has size 2n−1. As for the t-intersecting families, Katona [137] (seeTheorem 2.1.2) determined the maximum size for all n, t. An interestinganalogue of k-uniform setting is p-biased setting, in which a set of size ` getsweight p`(1 − p)n−` and we aim to maximize the sum of weights of sets ina family (cf. [115]). We do not dwell on this since we are mostly concernedwith k-uniform families.

Erdos and Sos in 1971 asked the following question: how large can F ⊂

17

([n]k

)be, provided no two sets in F intersect in exactly t elements? Recently, a

big progress on this question was made by Ellis, Keller and Lifshitz [67], whoshowed that for any xed t and a very wide range of k = k(n), the answerto this question is the same as in the theorem of Ahlswede and Khachatrian.In this dissertation, however, I am particularly interested in the case whent is comparable to n. In this case, there is a beautiful result of Frankland Wilson [114] (see Theorem 2.1), which shows that for a certain choiceof parameters, the size of such family is exponentially in n smaller than(

[n]k

). Interestingly, this theorem require certain numbers to be prime powers.

Without that restriction, the bounds in the theorem are false. However, thereis a qualitatively similar result of Frankl and Rodl [111] for a pair of families(cf. its supersaturation corollary Theorem 6.10) that, albeit gives worsequantitative bounds, is much more exible and, in particular, does not haveany algebraic restrictions. Recently, Keevash and Long [145] managed todeduce the result of Frankl and Rodl from the result of Frankl and Wilsonusing the modern probabilistic technique of dependent random choice.

We have already mentioned one application of the results of Frankl andWilson: explicit constructions for Ramsey numbers. We are particularlyinterested in its applications in combinatorial geometry to the problems ofnding the chromatic number of Euclidean space and Borsuk's problem. Thechromatic number of Rd is the minimum number of colors needed to color allpoints of Rd so that no two points at unit distance apart receive the samecolor. This quantity was studied quite extensively (see the surveys [209],[229]).

The result of Frankl and Wilson was a breakthrough for the chromaticnumber of the space, for the following reasons. One way of looking at fam-ilies is to associate with each set R ⊂ [n] its characteristic vector v(F ) :=(x1, . . . , xn) with xi = 1 for i ∈ F and xi = 0 for i /∈ F . Then each familycorresponds to a collection of 0, 1-vectors. Moreover, one forbidden inter-section corresponds to one forbidden scalar product, which, given that allvectors have the same length, corresponds to one xed forbidden distance.Scaling this distance to 1, we obtain a nite subset X of Rn (which cor-responds to

([n]k

)), for which we need exponential in n number of colors to

properly color: the FranklWilson theorem states that any subset X ′ ⊂ Xwithout distance 1 (without intersection of size t) must be exponentially inn smaller than X.

The situation is similar with Borsuk problem [41], which asks for the min-imum number of parts needed to partition each body in Rn of unit diameter

18

into parts of strictly smaller diameter. There, using the result of Frankl andWilson, Kahn and Kalai [134] managed to obtain a subexponential in n lowerbound on the number of parts.

Raigorodskii [209] succeeded in improving the bounds in the aforemen-tioned geometric problems by enlarging the scope of vectors from 0, 1-vectors to 0,±1-vectors and proving FranklWilson type theorems for suchvectors (see also [48], [162], [195], [205], [209], [215]). Motivated by this, ina series of works [93, 97, 99] I and Frankl initiated systematic studies ofintersecting theorems for families of 0,±1-vectors. We note here that in-tersecting theorems were studied for other objects, such as subspaces andpermutations, but this is beyond the scope of this dissertation.

Let us return to the rst question we have discussed in this section: sta-bility of the ErdosKoRado theorem. Another way of looking at stabilityis by studying the so-called transference results in Kneser graphs. Let usrst give some denitions. A Kneser graph KGn,k is a graph with vertexset(

[n]k

)and edge set formed by all pairs of disjoint sets from

([n]k

). Note

that the ErdosKoRado theorem determines the size of the largest indepen-dent set in KGn,k. Similarly to the classical random graph model G(n, p)(cf., e.g., [13, 31]), dene the random Kneser graph KGn,k(p) as follows:V (KGn,k(p)) = V (KGn,k) and the set of edges of KGn,k(p) is a subsetof the set of edges of KGn,k, obtained by including each edge from KGn,k

independently and with probability p.Returning to transference, in general, we speak of transference if a certain

combinatorial result (with high probability) holds with no changes in therandom setting. Studying this phenomenon in the context of the indepen-dence number and the chromatic number of generalized Kneser graphs wassuggested by Bogolyubskiy et. al. in [28]. One example of such theorem isdue to B. Bollobas, B. Narayanan and A. Raigorodskii [33]. They studiedthe size of maximal independent sets in KGn,k(p), and showed that for awide range of parameters the independence number of KGn,k(p) is exactlythe same as that ofKGn,k:

(n−1k−1

). Later on, their result was further strength-

ened in [19, 54, 59]. Random subgraphs of generalized Kneser graphs werestudied in [27, 29, 204, 205, 216].

Finally, let us mention the following surprising (in the words of Erdos)conjecture of Chvatal [49]. Recall that a down-set is a family G, such thatG ⊂ F and F ∈ G imply G ∈ G.

19

Conjecture (Chvatal [49]). For any down-set G, one of the largest intersect-ing subfamily F ⊂ G is the family of all sets from G containing the elementof the largest degree.

Thus, the simple result of Erdos, Ko and Rado stating that the largestintersecting family F ⊂ 2[n] has size 2n−1 actually proves the conjecturefor G = 2[n]. Chvatal himself proved it for shifted families. Recently, aninteresting connection with correlation inequalities was established in [116].For some further results see [39, 40, 227].

Families with forbidden congurations

We say that a family F has matching number s if s is the maximum numberof pairwise disjoint sets in F . Intersecting families have matching number 1,and for the edge sets of graphs this denition corresponds to the standardnotion of a matching. Erdos and Kleitman in the 60's suggested the followingtwo problems that generalize the ErdosKoRado theorem. The rst one isas follows.

Problem (Erdos, Kleitman [151]). Determine the size of the largest familyF ⊂ 2[n] with matching number s.

Kleitman resolved the problem above for n = sm− 1, sm, where m is aninteger. Queen [200] resolved the only remaining case for s = 3: n = 3m+1.For a while, there was no progress. In a recent work with Frankl [94, 101] wemanaged to resolve this problem in many new cases.

The more famous question on matchings deals with uniform families.

Problem (Erdos Matching Conjecture [71]). The size ek(n, s) of the largestfamily F ⊂

([n]k

)with matching number s is max

(nk

)−(n−sk

),(sk+k−1

k

).

The Erdos Matching Conjecture, or EMC for short, is trivial for k = 1and was proved by Erdos and Gallai [72] for the graph case k = 2 (this resulthas stimulated a series of papers in graph theory, cf. e.g., the recent paperof Luo [185]). It was settled in the case k = 3 and n ≥ 4s in [112], for k = 3,all n and s ≥ s0 in [184], and, nally, it was completely resolved for k = 3in [86]. The case s = 1 is the Erdos-Ko-Rado theorem.

In his original paper, Erdos proved the conjecture for n ≥ n0(k, s). Hisresult was sharpened by Bollobas, Daykin and Erdos [32], who veried it forn ≥ 2k3s. Subsequently, Hao, Loh and Sudakov [130] proved the EMC forn ≥ 3k2s. Their proof relies in part on the multipartite version of the

20

following universal bound from [80]:

ek(n, s) ≤ s

(n− 1

k − 1

).

If n = k(s + 1) then the right hand side of the equation above is equal to|A0(s, k)|. For this case, the EMC was implicitly proved by Kleitman [151] inhis studies of the non-uniform problem. This was extended very recently byFrankl [85], who showed that ek(n, s) ≤

(k(s+1)−1

k

)for all n ≤ (s+ 1)(k+ ε),

where ε depends on k. Improving the aforementioned bounds, Frankl [83]resolved the conjecture for n ≥ (2s+ 1)k − s. In a recent work with Frankl[104], we managed to get the current best bound for the EMC: we showedthat it is valid for any s ≥ s0 and n > 5

3sk −23s.

The Erdos Matching Conjecture takes a central place in extremal com-binatorics and attracted a lot of attention recently, in particular, due toconnections with Dirac thresholds, theory on deviations of sums of randomvariables and some computer science problems. For details, see [8] and [104].We will just briey overview the questions on Dirac thresholds. This activearea of research in extremal combinatorics stems from the famous Dirac's cri-terion for Hamiltonicity: any n-vertex graph with minimum degree at leastn/2 contains a Hamilton cycle. The generic question is as follows: what isthe condition on the minimum degree3 for a family in order for it to containa certain spanning (that is, covering the ground set) structure. Probably themost basic structure is a perfect matching, and it turns out that the resultsfor the EMC imply the results on Dirac thresholds for perfect matchings.For details on this connection, we refer to [8, 104], and for more on Diracthresholds we refer to a recent survey [236].

One may think of intersecting families or families with matching numbers as of families with forbidden congurations: in the rst case, the forbiddenconguration is 2 disjoint sets, and in the second case it is s + 1 pairwisedisjoint sets. One may similarly interpret the other restrictions on intersec-tions. Having this perspective, it is natural to ask, what could we say aboutother potential forbidden congurations.

The family F ⊂ 2[n] is called partition-free if there are no A,B,C ∈ Fsatisfying A ∩ B = ∅ and C = A ∪ B. Half a century ago Kleitman [152]proved the following beautiful result.

3We note here that it may be the degree of an element, or a collective degree of a d-tuple of elements,that is, the number of sets from the family containing that d-tuple

21

Theorem (Kleitman [152]). Suppose that n = 3m + 1 for some positiveinteger m. Let F ⊂ 2[n] be partition-free. Then

|F| ≤2m+1∑t=m+1

(n

t

).

He asked, what is the answer for n = 3m, 3m+ 2. In a recent work withFrankl [95] we used the methods developed for the rst problem from thissection and completely resolved this problem. Frankl [87] generalized theresult of Kleitman to s-partitions: s pairwise disjoint sets and their union.

Kleitman [153] considered the following related problem. What is themaximum size u(n) of a family F ⊂ 2[n] without three distinct memberssatisfying A ∪ B = C. The dierence with partition-free families is thatone does not require A and B to be disjoint. Kleitman proved u(n) ≤(

nbn/2c

)(1 + c

n) for some absolute constant c.An abstract version of this problem was solved by Katona and Tarjan

[143]. Let v(n) denote the maximum size of a family F without three distinctmembers A,B,C such that A ⊂ C and B ⊂ C. Katona and Tarjan provedthat v(2m+ 1) = 2

(2mm

).

This result was the starting point of a lot of research. The central problemmight be stated as to determine the largest size of subsets of the boolean lat-tice without a certain subposet. We refer the reader to the survey [122]. Oneof the important recent advancements in the topic was the result of [192],where the authors showed that for any nite poset there exists a constantC, such that the largest size of a family without an induced copy of thisposet has size at most C

(nbn/2c

). However, the value of C is unknown in most

cases, including the diamond poset: four sets A,B,C,D, where A ⊂ B ⊂ D,A ⊂ C ⊂ D.

A structure of completely dierent spirit was suggested by Chvatal. Ad-simplex is a family of d + 1 sets that have empty intersection, such thatthe intersection of any d of them is nonempty. Chvatal conjectured that ford ≤ k ≤ d

d+1n the largest family in(

[n]k

)not containing a d-simplex is a full

star. Recently, great progress in this problem was obtained in [148]. We referto that paper and the survey [196] for the discussion of a more general classof forbidden conguration problems, called Turan problems for expansion.The techniques of [148] are based on junta approximations, discussed in theprevious section in the context of intersecting families.

22

Methods

I wanted to survey the methods that were used in the extremal set theoryproblems mentioned in the previous sections. Most of these methods areemployed in this dissertation.

One of the basic methods in extremal set theory is shifting. It is an opera-tion that replaces larger elements by smaller elements (in the sense of naturalorder on [n]) in all sets in the family simultaneously. Its precise denitionis given in Section 1.2.1. It allows to reduce the study of a certain classof extremal set theory problems to the class of shifted families: the familiesthat, given that they contain a certain set F , would also contain any set Gthat is obtained from F by replacing some elements in F by smaller ones.Typically, one can restrict oneself to shifted families, given that the problemis monotone: i.e., shifting increases the sizes of minimum intersections, de-creases the sizes of unions and the matching numbers, decreases the size ofa shadow etc. In a certain sense, shifted families are similar to down-sets(or up-sets). Moreover, there is a similar combinatorial operation to pro-duce down-sets from a family, which is, e.g., used in Frankl's proof of theVapnikChervonenkisSauerShelah lemma.

Shifting is particularly ecient if combined with induction: one wouldoften decompose a shifted family into two families, typically all the sets con-taining x and all the sets not containing x, where x is either the rst orthe last element, and then apply inductive hypothesis plus the properties onthese two families combined with shiftedness. One of the most beautiful ex-amples of proofs of this kind is Frankl's proof of the KruskalKatona theorem[79]. The limitation of this approach is that it is dicult/impossible to applyfor subtler problems, such as forbidden one intersection, or when we imposeadditional restrictions on the family (such as not being a star).

Another classical extremal combinatorics method is Katona's circle method.It gained its name from the proof for the ErdosKoRado theorem, given byKatona [139]. The idea of that proof was as follows: x a permutation andconsider only k-sets that consist of k consecutive elements in that permuta-tion (the circle). There are n such sets, but it is not dicult to show thatat most k can belong to an intersecting family F (which is a k/n-fractionof these n sets). This is true for any permutation. Then average over thechoice of the permutation, concluding that at most a k/n-fraction of all setsis contained in F .

More generally, one can x more elaborate structures over which one av-

23

erages. This approach eventually allowed us to progress in the questions onnon-uniform families with matching number s and on partition-free families[94, 95]. The main diculty (and the limitation of the method) is to choosea structure over which one averages: it should be suciently simple to an-alyze and it should capture the features of the problem. For example, thebest known bound for the EMC that is obtained using this approach is theinequality displayed in the previous section. We simply do not know of anygood conguration that could possibly give the correct bound.

The KruskalKatona theorem itself is a very strong tool in extremal com-binatorics. E.g., this is the main ingredient in proving that any monotoneproperty has a threshold [35]. Here we only focus on its applications tointersecting and cross-intersecting families. Thanks to the KruskalKatonatheorem, some questions may be reduced to the so-called lexicographicalfamilies, i.e., families consisting of the initial segment of sets in the lexico-graphical order (for the precise statement, cf. Section 1.2.2). Thus, one mayrestrict to a very small and quite well-understood class of families. How-ever, in many cases (e.g., in proving stability results) such reduction wouldpotentially destroy certain other properties. For this purpose, one may em-ploy certain exchange operations, which would transform any given familyinto a lexicographical family gradually (which allows to have better controlof some parameters). One such operation, generalizing shifting, was foundby Daykin [55], and another one was developed by myself and Zakharov [176].

The methods we have discussed so far work for any values of parametersn, k. The next method, called the Delta-system, or sunower method, albeitbeing very powerful, works for n > n0(k) only. Erdos and Rado showed thatin any suciently large family of k-sets there is a sunower with t petals:sets F1, . . . , Ft, such that ∪i∈[t]Fi = Fj ∩ Fl for any j, l ∈ [t]. The commonintersection of all the sets in the sunower is called a core. The method worksroughly as follows: given a family F , we nd minimal cores of sunowers inF . Moreover, we have control over the number of cores of given size: if thereare too many, we would have found a smaller common core. Each set in Fcontains an element of the core, and it is easy to see that the vast majorityof sets in F contains small cores (typically, of size 2). Then, one is basicallyleft to understand the possible structure of the family of cores of small size(size 2), which is often easy. This method was developed in the early worksof Frankl (see, e.g., [78]).

24

Another method, which we have already discussed above, is the so-calledjunta method. The idea is to approximate a family with given properties bya j-junta with the same properties, where j-junta is a family that dependson at most j coordinates, called the center of the junta.4 Once such an ap-proximation is obtained, one may again argue that most sets would intersectthe center in few coordinates, and then, as in the case with the Delta-systemmethod, it is sucient to nd the optimal arrangement of small sets. Here,as in the case with the Delta-system method, stability results play a crucialrole: one rst gets that the approximating junta must have a specic form,and then uses the result that states that such junta is locally optimal. Werefer to a recent paper of Keller and Lifshitz [148] for some general junta-type results. This method is applicable for n > Ck, where C is independentof k, but depends on the problem. E.g., in the case of the Erdos MatchingConjecture, it depends on s. Thus, it has a wider range of applicability thanthe Delta-system method, but still requires n to be large. Keevash, Lifshitz,Long, and Minzer are currently working on junta-type results that would beapplicable for n > Csk with C, independent of s.

Finally, let us say a couple of words about the algebraic methods. Oneof the methods is the eigenvalue method. One potential application goes asfollows: interpret a family with a certain property as an independence setin a certain graph (e.g., for intersecting families, the relevant graph is theKneser graph), and then use Homan's bound for the largest independent setin terms of eigenvalues of the adjacency matrix. Thus, one requires to ndthe eigenvalues. This approach works for all values of parameters, however,it is often very dicult to nd the right interpretation of the problem, sothat the bound would be sharp, and/or to nd the eigenvalues/eigenspaces.The examples include the proof of the ErdosKoRado theorem (cf. Lovasz'famous paper [183] and Wilson's result [235] on t-intersecting families. Moreadvanced applications use Delsarte's linear programming bound.

Frankl and Wilson's result [114] is most conveniently proved using poly-nomial method. The idea behind the method is to associate polynomialswith each set in the family, and, using the restricted intersections property,show that these polynomials are independent over a certain eld. Then, onebounds the size of the family by the dimension of the space of the polyno-mials (taking into account the possible monomials that may have appeared).We refer to an unnished book of Babai and Frankl [17] for a systematic

4The presence of any set is determined by its intersection with a xed set of size at most j.

25

treatment of this topic.

ε-nets

Extremal set theory has important applications in computational geometryand statistical learning, thanks to the notions of ε-nets and VC-dimension.

Let X be a nite set and let F be a system of subsets of an X. The pair(X,F) is usually called a range space. A subset X ′ ⊆ X is called an ε-net for(X,F) if X ′∩F 6= ∅ for every F ∈ F with at least ε|X| elements. (In words,ε-net is a hitting set for all large sets from the range space.) The use of small-sized ε-nets in geometrically dened range spaces has become a standardtechnique in discrete and computational geometry, with many combinatorialand algorithmic consequences. In most applications, ε-nets precisely andprovably capture the most important quantitative and qualitative propertiesthat one would expect from a random sample. Typical applications includethe existence of spanning trees and simplicial partitions with low crossingnumber, upper bounds for discrepancy of set systems, LP rounding, rangesearching, streaming algorithms; see [187, 201].

A remarkable result due to Haussler andWelzl, based on a ground-breakingwork of Vapnik and Chervonenkis [231], states that if V C(F) ≤ d then theminimum size of an ε-net is O(dε log 1

ε ). It was shown by Komlos, Pach, andWoeginger that this bound is essentially tight for random range spaces. Overthe past two decades, a number of specialized techniques have been devel-oped to show the existence of small-sized ε-nets for geometrically denedrange spaces [15, 16, 45, 46, 47, 50, 155, 186, 190, 198, 207, 232, 233]. Basedon these successes, it was generally believed that in most geometric scenariosone should be able to substantially strengthen the ε-net theorem, and obtainperhaps even a O

(1ε

)upper bound for the size of the smallest ε-nets. Simi-

lar questions were raised in the statistical learning community. However, in2012 Alon [5] gave a slightly superlinear lower bound for the size of ε-nets forrange spaces induced by points and lines In a recent breakthrough, Baloghand Solymosi [22] applied the recently developed container method (comingfrom extremal combinatorics) to give the lower bound of Ω(1

ε log1/3−o(1) 1ε )

for the smallest size of an ε-net for range spaces induced by lines and points.Shortly after Alon's paper [5], Pach and Tardos [202] constructed a range

space induced by points and halfspaces in R4 with the smallest ε-net of (theworst possible) size Ω(1

ε log 1ε ).

On the other hand, following the work of Clarkson and Varadarajan [50], it

26

has been gradually realized that if one replaces the condition that the rangespace (X,R) has bounded VC-dimension by a more rened combinatorialproperty, one can prove the existence of ε-nets of size o(1

ε log 1ε ). This lead

to the introduction of the new complexity measure: shallow cell complexity(cf. the denition in Chapter 5). A series of elegant results [15, 46, 197, 233]illustrate that if the shallow-cell complexity of a set system is ϕ(n) = o(n),then it permits smaller ε-nets than what is guaranteed by the ε-net theorem.

The questions studied in the dissertation

In this dissertation, we address the following questions.

• How stable is the ErdosKo-Rado theorem? More precisely, how largecould an intersecting family be in terms of its distance from the closeststar?

• How far could an intersecting family be from a star?

• What is the structure of a typical intersecting family?

• How big could a pair of cross-intersecting or cross s-intersecting familiesbe?

• Are there analogues of the ErdosKoRado and AhlswedeKhachatriantheorems for families of 0,±1-vectors?

• What is the largest size of a family with no matching of size s? Whatkind of structure does it have?

• What are the analogues of the HiltonMilner theorem for the ErdosMatching Conjecture?

• How does the chromatic number of random Kneser graphs and hyper-graphs behave?

• What is the size of the smallest ε-net for range spaces, dened by half-spaces in Rd?

• How big could the smallest ε-net be, given a bound on the shallow cellcomplexity of a range space?

• Are there geometric obstructions to nding counterexamples to Borsuk'sconjecture?

• Are there distance graphs with large chromatic number and with nolarge cliques or large girth?

27

The structure of the dissertation

The dissertation consists of the Introduction chapter, 6 chapters in which Ipresent my results, and a bibliography of 237 items.

In Chapter 1, I present the results on families and pairs of families withrestricted intersections. In Chapter 2, I present the restricted intersections-type results for the families of 0,±1-vectors. In Chapter 3, I present myresults on the families with matching number s. In Chapter 4, I discuss myresults on random Kneser graphs and hypergraphs. In Chapter 5, I presentmy results on applications of extremal set theory to ε-nets. In Chapter 6, Ipresent the applications of extremal set theory to the combinatorial geometryproblems: the chromatic number of the space and Borsuk's problem.

Each of the chapters has its own introduction, in which the area is surveyedand the main results are stated. The proofs are presented in the later sections.

The main results of the dissertation

• I have found an essentially sharp upper bound on the size of an inter-secting family in terms of its distance from the closest star (diversity).

• I have shown that the diversity of an intersecting family is at most(n−3k−2

),

provided n > Ck.

• I have showed that a typical intersecting family is a star, provided n >2k + 2

√k log k.

• I have obtained analogues of the ErdosKoRado and the AhlswedeKhachatrian theorems for families of 0,±1-vectors.

• I have resolved Kleitman's problem on the size of the largest family withmatching number s in many new cases. For that, we have developed avery powerful averaging method.

• I have obtained a HiltonMilner type theorem for the Erdos MatchingConjecture.

• I studied the chromatic number of random Kneser graphs and showedthat it stays almost the same as the chromatic number of Kneser graphsin many scenarios.

• I have showed that the smallest ε-net for range spaces, dened by halfs-paces in Rd, has size Ω(dε log 1

ε ) in the worst case, thus giving a conclu-

28

sive answer to the question on whether the geometric range spaces aresimpler then abstract range spaces.

• I have obtained sharp lower bounds on the smallest ε-nets for rangespaces as a function of their shallow cell complexity.

• I have showed that there are counterexamples to Borsuk's conjecturelying on the spheres of radius arbitrarily close to 1/2.

• I have showed that there distance graphs with exponential in the di-mension chromatic number and with large girth.

My stability result for the ErdosKoRado theorem has already found ap-plications in the studies of intersecting families (some of which are presentedhere), and we believe that it would nd more. More generally, the bipartiteswitching trick that I have introduced seems to be a powerful tool.

The situation is similar with the stability result for the Erdos MatchingConjecture. In the paper [102], we have applied it to advance on an anti-Ramsey type problem. The averaging technique I developed for Kleitman'sproblem on families with matching number s is very powerful and should ndapplications elsewhere, in particular, for problems on forbidden subposets.

My results on ε-nets have already found an application in statistical learn-ing: in a work with Csikos and Mustafa, [52], we proved that the constructiongiven in [168] allows to resolve a long-standing question on the VC-dimensionof a k-fold union F∪k of range spaces dened by halfspaces. The notion ofshallow cell complexity seems to be very useful in dierent areas where ran-dom sampling is employed, and it is of importance to understand the strengthand the limitations of this measure.

Finally the extremal set theory techniques that I introduced in study-ing distance problems enhance the connections between these two areas andshould open the possibilities for some new applications.

Acknowledgements

First of all, I would like to heartily thank my parents Nadezhda and Borisand my elder brothers Alexey and Alexandr, who raised me with a lot oflove, nourished my interest in the world and in mathematics in particular,and later always supported me in my projects. I specially thank my fatherBoris for pushing me to write this dissertation.

29

My wife Aria gave and gives me constant inspiration and support beyondany limit. It is thanks to her plan to do studies in Paris, I had to staysomewhere close, which has stimulated me to advance in my research.

I am greatly indebted to my MSc and PhD thesis supervisor Andrei Mi-hailovich Raigorodskii for introducing me to the area of combinatorics andteaching me how to do research. His help was essential in establishing myselfas a mathematician, as well as in making the defense of this thesis happen.

Janos Pach played a crucial role in my personal and career development.Thanks to his invitation to EPFL, I spend great 3.5 years in Lausanne, foundgreat new friends and colleagues, my wife, and learned a lot of new mathe-matics. He became my close friend and gave me a great deal of invaluableadvice, as well as helped to develop my taste in movies.

Peter Frankl is probably the most important person for this thesis from themathematical point of view. I started my research with the FranklWilsontheorem and then was so happy and excited to work on a common projectwith him following his visit to Lausanne in 2015. This was the beginning ofa very fruitful collaboration and a great friendship. Virtually everything Iknow about extremal set theory I learned from him.

I am very grateful to my coauthors and colleagues (many of whom arefriends), from whom I learned so much and who helped me at dierent mo-ments of my life: Noga Alon, Nora Frankl, Stefan Glock, Gleb Gusev, FelixJoos, Grigoriy Kabatianskii, Gil Kalai, Jaehoon Kim, Sergey Kiselev, DanielaKuhn, Noam Lifshitz, Nikolai Moshchevitin, Nabil Mustafa, Marton Nas-zodi, Deryk Osthus, Liudmila Ostroumova, Guillem Perarnau, Will Perkins,Rom Pinchasi, Sasha Polyanskii, Natan Rubin, Gabor Tardos, Prasad Tetali,Dmitriy Shabanov, Shakhar Smorodinsky, Andrew Suk, Tibor Szabo, KonradSwanepoel, Istvan Tomon, Emo Welzl, Dima Zakharov, Maxim Zhukovskii,as well as those whom I have forgotten to mention.

I thank my friends for all the many great moments, interesting discussions,support and inspiration: my Moscow friends Yulia, Andrey, Boris, Zhamal,Max and Max, Sasha, Roma, Igor, Osad, and my Lausanne friends Andres,Igore, Mariann and Samy, Grisha and Vasilisa, Alfonso, Manuel, Chillu, Filip.I also thank the family of Aria for being so welcoming and especially herfather Michel for his very useful and reasonable advice.

30

Chapter 1.

Intersecting families of sets

My results presented in this chapter (some joint with P. Frankl and with D.Zakharov) are published in [91, 92, 93, 96, 98, 164, 176].

1.1. Introduction and statement of results

In this chapter, we discuss one of the classical topics of extremal set theory:properties of intersecting families.

Denition 1.1. For integer t ≥ 1, a family F ⊂ 2[n] is called t-intersecting,if for any F1, F2 ∈ F we have |F1 ∩ F2| ≥ t. If t = 1, then we call suchfamilies intersecting.

In this chapter, we will mostly work with families F ⊂(

[n]k

), that is,

k-uniform families.The rst theorem on intersecting families is due to Erdos, Ko and Rado:

Theorem 1.2 (Erdos, Ko, Rado, [73]). Let n ≥ 2k > 0. Then for anyintersecting family F ⊂

([n]k

)one has |F| ≤

(n−1k−1

).

They proved this theorem back in 1938, but the publication for variousreasons, notably the Second World War, was postponed until 1961. By now,this became one of the cornerstones of extremal combinatorics.

It is easy to give an example of an intersecting family, on which the boundfrom Theorem 1.2 is attained: take the family of all k-element sets containingelement 1.

If in a family all sets contain a xed element, then we call it triviallyintersecting or a star. We call any inclusion-maximal star a full star.

Let us also note that if we do not require uniformity, that is, we considerintersecting families F ⊂ 2[n], then it is easy to see that |F| ≤ 2n−1 and thisbound is attained.

31

1.1.1. Stability for the ErdosKoRado theorem

Erdos, Ko, and Rado asked, how large can an intersecting family be, providedthat it is not trivially (or nontrivially) intersecting? For n = 2k it is easyto construct exponentially many intersecting families of size

(2k−1k−1

), most of

which are not stars. Indeed, it suces to choose exactly one k-set out ofeach pair of complementary sets. For n > 2k the answer is given by theHilton-Milner theorem.

Theorem 1.3 (Hilton, Milner, [128]). Let n > 2k and F ⊂(

[n]k

)be a

nontrivially intersecting family. Then |F| ≤(n−1k−1

)−(n−k−1k−1

)+ 1.

Let us mention that the upper bound is attained on the family Hk, whereHu for integer u ∈ [2, k] is dened below.

Hu :=A ∈

([n]

k

): [2, u+ 1] ⊂ A

∪

A ∈(

[n]

k

): 1 ∈ A, [2, u+ 1] ∩ A 6= ∅

. (1.1)

The HiltonMilner theorem shows that the example for the tightness ofthe ErdosKoRado theorem is unique, moreover, that the ErdosKoRadotheorem is stable in the following sense: any intersecting family that is largeenough must be a star. We also remark here that, while the size of a full staris(n−1k−1

), the size of the HiltonMilner family is less than k

(n−2k−2

)= o(

(n−1k−1

))

provided k2 = o(n).Later, a much stronger stability result was obtained by Frankl [81] in

terms of the maximum degree. For a family F ⊂ 2[n], the degree δi(F) ofan element i ∈ [n] is the number of sets from F containing i. We denote by∆(F) the largest degree of an element: the maximum of δi over i ∈ [n].

Theorem 1.4 ([81]). Let n > 2k > 0 and F ⊂(

[n]k

)be an intersecting

family. Then, if ∆(F) ≤(n−1k−1

)−(n−u−1k−1

)for some integer 3 ≤ u ≤ k, then

|F| ≤(n− 1

k − 1

)+

(n− u− 1

n− k − 1

)−(n− u− 1

k − 1

). (1.2)

The bound from Theorem 1.5 is sharp for every admissible value of u, aswitnessed by the example (1.1).

One can deduce the HiltonMilner theorem from the u = k case of Theo-rem 1.5. On a high level, Theorem 1.4 provides us with an upper bound on|F| in terms of the size of the largest star in F .

32

My rst contribution that I cover in this thesis starts with the followingnotion. For a family F , the diversity γ(F) is the quantity |F| − ∆(F).One may think of diversity as of the distance from F to the closest star.This notion allows us to state a stronger, dual version of Theorem 1.4.The following strengthening of Theorem 1.4 was obtained by myself andZakharov [176].

Theorem 1.5 ([176]). Let n > 2k > 0 and F ⊂(

[n]k

)be an intersecting

family. If γ(F) ≥(n−u−1n−k−1

)for some real 3 ≤ u ≤ k, then

|F| ≤(n− 1

k − 1

)+

(n− u− 1

n− k − 1

)−(n− u− 1

k − 1

). (1.3)

We note that the HiltonMilner theorem, as well as Theorem 1.4, is im-mediately implied by Theorem 1.5. The deduction of Theorem 1.5 fromTheorem 1.4 for integer values of u is possible, but not straightforward. The-orem 1.5 is the strongest known stability result for the ErdosKoRado the-orem for large intersecting families, more precisely, for the families of size atleast

(n−2k−2

)+ 2(n−3k−2

). There are several other stability results for the Erdos

KoRado theorem, see, e.g. [54, 66, 115]. Many of them are implied by thetheorem above.

One important insight in Theorem 1.5 is that diversity, rather than themaximum degree, is the correct measure to look at when working with in-tersecting families. But, more importantly than the result itself, we havedeveloped a simple but very powerful and exible technique to work withintersecting families, which may be called the bipartite switching trick and isbased on matchings in regular bipartite graphs. Using it, we managed to ob-tain a unied proof for several theorems on intersecting and cross-intersectingfamilies.

Denition 1.6. For positive integer s, two families A,B ⊂ 2[n] are said tobe cross s-intersecting if |A ∩ B| ≥ s holds for all A ∈ A and B ∈ B. Ifs = 1 then we call such pairs of families cross-intersecting.

These results are presented in Section 1.3, including the proof of Theo-rem 1.5. Later, we shall discuss some applications of these results.

1.1.2. Diversity

What happens when the diversity in Theorem 1.5 is large? Interestingly,the families H3 and H2 have the same cardinality:

(n−1k−1

)−(n−4k−1

)+(n−4k−3

)=

33

(n−1k−1

)−(n−3k−1

)+(n−3k−2

)=(n−2−2

)+ 2(n−3k−2

). However, the latter has a much

larger diversity:(n−3k−2

). Thus, one cannot hope for a upper better bound on

the size of an intersecting family F than that in Theorem 1.5, as long asγ(F) ≤

(n−3k−2

). But can we say something if γ(F) >

(n−3k−2

)?

The following problem was suggested by Katona and addressed by Lemonsand Palmer [179]: what is the maximum diversity of an intersecting familyF ⊂

([n]k

)? They found out that for n > 6k3 we have γ(F) ≤

(n−3k−2

), with the

equality possible only for H2 and some of its subfamilies. Recently, Frankl[84, Theorem 2.4] proved that γ(F) ≤

(n−3k−2

)for all n ≥ 6k2, and conjectured

that the same holds for n > 3k.My main contribution here is the following theorem.

Theorem 1.7 ([164]). There exists a constant C, such that for any n >

Ck > 0 any intersecting family F ⊂(

[n]k

)satises γ(F) ≤

(n−3k−2

). Moreover,

if γ(F) =(n−3k−2

), then F is a subfamily of an isomorphic copy of H2.

We note that a somewhat similar proof strategy, which rst uses resultson Boolean functions to obtain a rough structural result for the extremalfamilies, and then uses combinatorics to obtain a precise result, was recentlyused by Keller and Lifshitz [148] in a much more general setting.

The proof of Theorem 1.7 is presented in Section 1.4. In Section 1.5,we present an interesting counterexample to a very natural conjecture thatgeneralizes Theorem 1.7. The ideas in Section 1.5 rely on the discrete Fouriertransform methods in the Analysis of Boolean Functions.

We also note that recently, Hao Huang provided a counterexample to theaforementioned Frankl's conjecture in the range 3k ≤ n ≤ (2 +

√3)k [129].

1.1.3. Counting intersecting families

In this section, we are going to present an application of Theorem 1.5 to thefollowing question: how many dierent intersecting subfamilies of

([n]k

)are

there?Let us take compare the extremal families in the ErdosKoRado and the

HiltonMilner theorem. For n = 2k+ 1 the dierence between the sizes of afull star and Hk is only k − 1. However, as n− 2k increases, this dierencegets much larger quickly. The number of subfamilies of F is 2|F|, and thusthe ratio between the number of subfamilies of the ErdosKoRado familyand that of the HiltonMilner family is 2k−1 for n = 2k + 1 and grows veryfast as n− 2k increases. This serves as an indication that most intersectingfamilies are stars.

34

In an important recent paper, Balogh, Das, Delcourt, Liu, and Sharifzadeh[20] proved this in the following quantitative form. Let I(n, k) denote thetotal number of intersecting families F ⊂

([n]k

).

Theorem 1.8 (Balogh, Das, Delcourt, Liu, and Sharifzadeh [20]). If n ≥3k + 8 log k then

I(n, k) = (n+ o(1))2(n−1k−1), (1.4)

where o(1)→ 0 as k →∞.

One of the main tools in the proof of (1.4) is a bound on the number ofmaximal (i.e., non-extendable) intersecting families (see Lemma 1.36). Theyobtain this bound using the following fundamental result of Bollobas.

Theorem 1.9 (Bollobas [30]). Suppose that A ⊂(

[n]a

),B ⊂

([n]b

)with A :=

A1, . . . , Am, B := B1, . . . , Bm satisfy Ai ∩Bj = ∅ i i = j. Then

m ≤(a+ b

a

). (1.5)

Note that the bound (1.5) is independent of n. In [30] it is proved in amore general setting, not requiring uniformity. The uniform version (1.5)was rediscovered several years later by JaegerPayan [131] and Katona [141].

For an integer t, denote by I(n, k, t) (I(n, k,≥ t)) the number of inter-secting families with diversity t (at least t). In particular, I(n, k,≥ 1) is thenumber of non-trivial intersecting families. In [98], I together with Franklobtained the following renement of Theorem 1.8.

Theorem 1.10 ([98]). For n ≥ 2k + 2 + 2√k log k and k →∞ we have

I(n, k) =(n+ o(1))2(n−1k−1), (1.6)

I(n, k,≥ 1) =(1 + o(1))n

(n− 1

k

)2(n−1k−1)−(n−k−1k−1 ). (1.7)

It is easy to see that (1.7) implies (1.6) (see Section 1.6.2 for details). Theproof uses a counting result for cross-intersecting families. This theorem,along with the cross-intersecting version, is proved in Section 1.6.

1.1.4. Product-type inequalities

If F is intersecting then A := F , B := F are cross-intersecting. Thereforethe following result implies the ErdosKoRado theorem.

35

Theorem 1.11 (Pyber [206]). Suppose that A,B ⊂(

[n]k

)are cross-intersecting

and n ≥ 2k. Then

|A||B| ≤(n− 1

k − 1

)2

holds. (1.8)

We have already seen that the notion of cross-intersecting families is notjust a natural extension of the notion of intersecting, but is also useful forobtaining results for one family. As a matter of fact, this was already used inthe paper of Hilton and Milner [128]. It explains the interest in two-familyversions of intersection theorems (cf. e.g. [38], [191], [218]).

My contribution (joint with Frankl) in this direction is two-fold. First weprovide a very short proof of (1.8) [91, 96]. Then we use the ideas of thisproof and some counting based on the KruskalKatona Theorem [159], [138]to obtain the following sharper, best possible bounds.

Example 1. Let i be an integer and dene Bi := B ∈(

[n]k

): 1 ∈

B∪B ∈(

[n]k

): 1 /∈ B, [2, i] ⊂ B, Ai := A ∈

([n]k

): 1 ∈ A, [2, i]∩A 6= ∅.

Note that Ai,Bi are cross intersecting with

|Ai| =(n− 1

k − 1

)−(n− ik − 1

), |Bi| =

(n− 1

k − 1

)+

(n− i

k − i+ 1

).

The inequalities (1.9) and (1.10) given below show that the pair (Ai,Bi) isextremal in the corresponding range.

Theorem 1.12 ([91]). Let A,B ⊂(

[n]k

)be cross-intersecting, n > 2k > 0

and suppose |A| ≤(n−1k−1

)≤ |B| and ∩B∈BB = ∅. Then

|A||B| ≤((n− 1

k − 1

)+ 1)((n− 1

k − 1

)−(n− k − 1

k − 1

))holds. (1.9)

Theorem 1.13 ([91]). Let A,B ⊂(

[n]k

)be cross-intersecting, n ≥ 2k > 0

and suppose that |B| ≥(n−1k−1

)+(n−ik−i+1

)holds for some 3 ≤ i ≤ k + 1. Then

|A||B| ≤((n− 1

k − 1

)+

(n− i

k − i+ 1

))((n− 1

k − 1

)−(n− ik − 1

)). (1.10)

Note that plugging in i = k + 1 into (1.10) gives (1.9) except for the case|B| =

(n−1k−1

)which we treat separately. In the proof of Theorem 1.13, we give

a short computation-free proof of an important special case of an inequalitydue to Frankl and Tokushige [113], based on the Konig-Hall theorem [156],[125]. These results are proved in Section 1.7.

36

1.1.5. Cross s-intersecting families

As we have mentioned, in their seminal paper [128], Hilton and Milner dealtwith pairs of cross-intersecting families. They proved the following inequality:

Theorem 1.14 (Hilton and Milner [128]). Let A,B ⊂(

[n]k

)be non-empty

cross-intersecting families with n ≥ 2k. Then |A|+ |B| ≤(nk

)−(n−kk

)+ 1.

This inequality was generalized by Frankl and Tokushige [113] to the caseA ⊂

([n]a

), B ⊂

([n]b

)with a 6= b, and with more general constraints on the

sizes of A,B. A simple proof of the theorem above may be found in [91].In this section I present the following generalization of the HiltonMilner

theorem, obtained in a joint work with Frankl (cf. Denition 1.6). Denethe family C := C ⊂

([n]k

): |C ∩ [k]| ≥ s.

Theorem 1.15 ([92]). Let k > s ≥ 1 be integers. Suppose that A,B ⊂(

[n]k

)are non-empty cross s-intersecting families. Then for n > 2k − s we have

maxA,B|A|+ |B| = |C|+ 1. (1.11)

The proof of this theorem is given in Section 1.8. Some time after the resulthas been published, we found out that a more general result was obtainedby Wang and Zhang [234] using dierent methods.

1.1.6. t-intersecting, cross s-intersecting families

In the proof of Theorem 2.10 from Chapter 2, we needed to estimate thesum of sizes of two t-intersecting families that are cross s-intersecting. Thismotivated the studies presented in this section. As a matter of fact, thisproblem in the case of non-uniform families was solved by Sali [221] (cf. [82]for an extension with a simpler proof).

To state our results for the k-uniform case let us make some denitions.

Denition 1.16. For n ≥ k > t ≥ 1 and n ≥ 2k−t dene the Frankl-familyAi(n, k, t) :

Ai(n, k, t) :=A ⊂

([n]

k

): |A ∩ [t+ 2i]| ≥ t+ i

, 0 ≤ i ≤ k − t.

Note that for A,A′ ∈ Ai(n, k, t) one has |A∩A′∩[t+2i]| ≥ t, in particular,Ai(n, k, t) is t-intersecting.

The following result was conjectured by Frankl [76], proved in many casesby Frankl and Furedi [89], and nearly 20 years later proved in full generalityby Ahlswede and Khachatrian [1].

37

Theorem 1.17 (Complete Intersection Theorem [1]). Suppose that F ⊂(

[n]k

)is t-intersecting, n ≥ 2k − t. Then

|F| ≤ max0≤i≤k−t

|Ai(n, k, t)| =: AK(n, k, t) holds. (1.12)

Moreover, unless n = 2k, t = 1 or F is isomorphic to Ai(n, k, t), the in-equality is strict.

In Section 1.9, we formulate and prove a cross-version of this result fort-intersecting, cross s-intersecting families, provided n is suciently large.

1.2. Preliminaries

In this section, we are going to introduce several combinatorial results andconcepts which are heavily used in this and following chapters.

We use the following standard notation. Given a family F ⊂ 2[n] andi, j ∈ [n], the family F (ij) is dened in the following way:

F (ij) := F − j : F ∈ A, j ∈ F, i /∈ F.

More generally, for any disjoint I, J ⊂ [n] dene

F(IJ) := F \ J : F ∈ F , J ⊂ F, F ∩ I = ∅ ⊂(

[n] \ (I ∪ J)

k − |J |

).

1.2.1. Shifting

Shifting is an important combinatorial operations on families of sets.For a given pair of indices 1 ≤ i < j ≤ n and a set A ∈ 2[n] dene

its (i, j)-shift Si,j(A) as follows. If i ∈ A or j /∈ A, then Si,j(A) := A. Ifj ∈ A, i /∈ A, then Si,j(A) := (A − j) ∪ i. That is, Si,j(A) is obtainedfrom A by replacing j with i.

Next, dene the (i, j)-shift Si,j(F) of a family F ⊂ 2[n]:

Si,j(F) := Si,j(A) : A ∈ F ∪ A : A, Si,j(A) ∈ F.We call a family F shifted, if Si,j(F) = F for all 1 ≤ i < j ≤ n. (Note

also that we always shift to the left, replacing the larger element with thesmaller one.)

Shifting is a powerful tool in dealing with many extremal set theory ques-tions. On the one hand, it preserves many properties of families. In par-ticular, shifting does not alter the cardinality of the set or the family. It

38

also preserves many combinatorial properties, for example, being intersect-ing. For completeness, we give a proof of this fact. For much more on shiftingand its use in extremal set theory we refer to the survey of Frankl [80].

Proposition 1.18. If F is intersecting then Sij(F) is intersecting.

Proof. Take any two sets F1, F2 ∈ Sij(F). If both F1, F2 belong to F thenthey intersect. Moreover, if or both belong to Si,j(F) \ F , then they bothcontain i and thus intersect. Assume next that F1 ∈ F and F2 ∈ Si,j(F)\F .The former means that F1 ∩ i, j = j, moreover, the set Si,j(F1) belongsto F . The latter means that the set Sj,i(F2) belongs to F . But Si,j(F1) ∩Sj,i(F2) = F1 ∩ F2, and, since the former pair belong to F , both pairs areintersecting.

1.2.2. Lex order and the KruskalKatona theorem

Let us dene the lex (lexicographic) and the colex (colexicographic) orders.

Denition 1.19. For given A,B ∈(

[n]k

)we say that A lexicographically

precedes B, or A ≺ B, if the smallest element of A \B is smaller than thatof B \ A.

For example, 1, 100 ≺ 2, 3.

Denition 1.20. For given A,B ∈(

[n]k

)we say that A colexicographically

precedes B, or A≺cB, if the largest element of A \ B is smaller than thatof B \ A.

Thus, 2, 3≺c1, 100. Note that 1, 3 precedes 2, 4 in both orders.For 0 ≤ m ≤

(nk

)let L(m, k) denote the initial segment of k-sets of length

m, i.e., the rstm k-sets in the lexicographic order. Note that L((

n−1k−1

), k)

=

F ∈(

[n]k

): 1 ∈ F.

Similarly, C(k)(m) denotes the family of rst m k-sets in the colex order.Note that for k ≤ s ≤ n one has C(k)

((sk

))=(

[s]k

).

Let us state the KruskalKatona Theorem [159], [138], which is one of themost important results in extremal set theory. It was proved independentlyby Kruskal [159] and Katona [138]. For 0 ≤ t ≤ k dene the t-shadow ∂(t)Fof a family F ⊂

([n]k

)by

∂(t)F :=S ∈

([n]

t

): ∃F ∈ F , S ⊂ F

.

39

Theorem 1.21 (KruskalKatona [159], [138]). The inequality

|∂(t)F| ≥ |∂(t)C(|F|, k)| (1.13)

holds for all F ⊂(

[n]k

), n ≥ k ≥ t ≥ 0.

Computationwise, the bounds arising from the KruskalKatona Theoremare not easy to handle. Lovasz [182] found the following slightly weaker butvery handy form.

Theorem (Lovasz, [182]). If n ≥ k ≥ t ≥ 0, F ⊂(

[n]k

)and |F| =

(xk

)for a

real number x ≥ k then

|∂(t)F| ≥(x

t

)holds. (1.14)

Note that for x ≥ k − 1 the polynomial(xk

)is a monotone increasing

function of x. Thus x is uniquely determined by |F| and k.The KruskalKatona theorem is ubiquitous in studying cross-intersecting

families. One reason is the following observation of Hilton, who found a wayto use the KruskalKatona theorem to handle cross-intersecting families. Heobtained the following corollary of the KruskalKatona theorem.

Theorem 1.22 (Hilton's Lemma [127]). If A ⊂(

[n]a

)and B ⊂

([n]b

)are

cross-intersecting then L(|A|, a) and L(|B|, b) are cross-intersecting as well.

We remark that it implies immediately that for any such cross-intersectingfamilies either |A| ≤

(n−1a−1

), or |B| ≤

(n−1b−1

). Since the manuscript with

Hilton's lemma was never published, we present the proof here for complete-ness.

Proof. We may clearly assume that n ≥ a + b. Consider the family Bc ofcomplements of sets from B. We have |Bc| = |B| and Bc ⊂

([n]n−b). Since A

and B are cross-intersecting, then there are no Bc ∈ Bc, A ∈ A, such thatA ⊂ Bc. In other words, A and B are cross-intersecting if and only if A and∂(a)Bc are disjoint.

Next, let us dene the reversed colex order. It is the colex order for [n],in which the order of elements is inverted: n is the rst element, then n− 1,etc. Formally, F ≺rc G in the reversed colex order if minF \G > minG\F.It is not dicult to see the following three things. First, if an a-set A is thei-th in the lexicographical order, then it is (

(na

)− i)-th in the reversed colex

order. Second, the complement [n]−A is ((na

)− i)-th in the lexicographical

order and, thus, i-th in the reversed colex order. Finally, we note that if a

40

family F ⊂(

[n]f

)form an initial segment in the reversed colex order, then

for any 0 < f ′ < f its shadow ∂(f ′)F also forms an initial segment in thereversed colex order.

Returning to the proof of the theorem, we note that |∂(a)(Bc| ≥ |∂(a)Lcb|,where Lb := L(|B|, b), due to (1.13) and the considerations in the previousparagraph, which give that Lcb in an initial segment in the reversed colexorder. We only point out that we can apply (1.13) with both colex andreversed colex orders.

Assume that La := L(|A|, a) and Lb are not cross-intersecting. ThenLa and ∂(a)Lcb intersect. The rst family consists of the rst sets in thelexicographical order, while the second is the initial segment in the reversedcolex order, and so consists of the last elements in the lexicographical order(see two paragraphs above). Therefore, we get that |La| + |∂(a)Lcb| >

(na

).

But then |A| + |∂(a)Bc| >(na

), which implies that A and ∂(a)Bc intersect.

This means that A,B are also not cross-intersecting, a contradiction.

Finally, let us state a version of Hilton's lemma in Lovasz' form.

Theorem 1.23 (Lovasz [182]). Let n ≥ a+ b, and consider a pair of cross-intersecting families A ⊂

([n]a

), B ⊂

([n]b

). If |A| =

(x

n−a)for a real number

x ≥ n− a, then

|B| ≤(n

b

)−(x

b

)holds. (1.15)

The following proposition is its easy corollary.

Proposition 1.24. Let a, b, n be integers, n ≥ a+b. Suppose that A ⊂(

[n]a

)and B ⊂

([n]b

)are cross-intersecting. Then either |A| ≤

(n−1a−1

)or |B| ≤

(n−1b−1

)holds.

Indeed, it suces to substitute x = n− 1 in Theorem 1.23.

1.3. Switchings via regular bipartite graphs and theo-

rems on intersecting families

We start this section by illustrating our new technique and giving a uniedproof of Theorems 1.2 and 1.3. The results on cross-intersecting families arediscussed in Section 1.3.1, and Theorem 1.5 is discussed in Section 1.3.2.

Consider a family F of maximal size, satisfying the conditions of Theorem1.3 or 1.2. First we prove that we may suppose that F is shifted. We showed

41

in Proposition 1.18 that an intersecting family stays intersecting after shifts.This is already sucient for us for the ErdosKoRado theorem. In the caseof the HiltonMilner theorem we must make sure that the shifted family weobtain is nontrivially intersecting. For that it is sucient to do several shiftsof F and renumberings of the ground set, that result in

([k+1]k

)⊂ F . The

subfamily(

[k+1]k

)is invariant under shifts, thus its presence guarantees that

F will stay nontrivially intersecting after any subsequent shifts.Do the shifts of F until either the family is shifted or any new shift will

result in F becoming trivially intersecting. If the former happens, we end uphaving a nontrivially intersecting shifted family. And if the latter happens,then after the last shift all the sets of F intersect a certain two-element subsetx, y. Renumber the elements of [n], so that this pair becomes a pair 1, 2.Since F has maximal size, it contains all the k-element sets, containing both1 and 2. Due to this, the (i, j)-shifts, where i, j > 2, do not aect theproperty of F to be nontrivially intersecting. Therefore, we may assumethat F is invariant under these shifts. Since not all sets in F contained 1(and not all sets contained 2), then F must contain the sets 2, . . . , k + 1and 1, 3, . . . , k + 1. The condition

([k+1]k

)⊂ F is now fullled, and thus

no shift can aect the nontriviality of the family.Next we pass to the part of the proof where the bipartite graphs are

employed. The order of a k-element set is its number in the lexicographicorder on

([n]k

)(see Denition 1.19). The order of a family F ⊂

([n]k

)is

the maximal order among its elements. Among the families of maximal sizesatisfying the conditions of Theorem 1.2 or 1.3 choose a shifted family F ofminimal order. Consider the set F ∈ F of maximal order. By the denitionof order, if F is non-trivial, then F does not contain 1.

Note that for some l ≥ 1 one has |F ∩ [2l − 1]| = l, otherwise one of theimages of F under several shifts is disjoint from F (this property was rstnoted by Frankl in [76]). Take maximal such l, and put L = [2l] \ F . Notethat |L| = l. We remark that, unless F is trivial, we have l ≥ 2. Considerthe families A := A ∈

([n]k

): A ∩ [2l] = F ∩ [2l] and B := B ∈

([n]k

):

B ∩ [2l] = L. The following lemma is the key to the proof.

Lemma 1.25. Let n > 2k > 0 and F be a nontrivially intersecting shiftedfamily. Suppose that F ∈ F has maximal order. Then, in the notationsabove, the family F ′ := (F \ A) ∪ B is intersecting and has smaller orderthan F . Moreover, |F ′| ≥ |F|.

Proof. Let us rst show that for any A,B, where B ∈ B, A ∈ F \A, we haveA∩B 6= ∅. The contrary may happen only for A ∈ F satisfying A∩L = ∅.

42

Since F has the maximal order in F , we have F∩[2l] ⊂ A, and, consequently,A ∈ A. Therefore, F ′ is an intersecting family.

Since all sets from B contain 1, it is clear that the order of F ′ is smallerthan that of F . Finally, we have |A| =

(n−2lk−l)

= |B|. Consider a bipartitegraph with parts A,B, and edges connecting disjoint sets. Independent setsin this graph correspond to intersecting subfamilies of A ∪ B. This graphis regular, and, therefore, it contains a perfect matching. Consequently, thelargest independent set has size |B|. Therefore |F ′| ≥ |F|.

In the case of the ErdosKoRado theorem, since the family F had thesmallest order among the families of maximal size, we conclude that F mustbe trivial.

In the case of the HiltonMilner theorem we obtain a contradiction be-tween the properties of F and Lemma 1.25, unless F = 2, . . . , k + 1.Indeed, if F 6= 2, . . . , k + 1, then 2, . . . , k + 1 ∈ F ′, that is, F ′ is non-trivially intersecting. Therefore, F = 2, . . . , k + 1, we have l = k, and allsets in F , dierent from F , contain 1. Both theorems are proved.

We remark that bipartite graphs have been used many times in ExtremalSet Theory, in particular, in the proof the famous Sperner theorem. TheSperner theorem states that the size of the largest family on [n] with no twosets containing each other is

(ndn/2e

). Sperner used matchings in biregular

bipartite graphs between(

[n]k

)and

([n]k+1

), where the edges connected sets,

one of which contained the other. With the proof above, both cornerstonesof extremal set theory have proofs based on matchings in regular bipartitegraphs.

1.3.1. Cross-intersecting families

One of the classical types of problems concerning cross-intersecting familiesis to bound the sum of the cardinalities of the two families. We will use theresults on cross-intersecting families in the proof of Theorem 1.5 in Section1.3.2. Here, we prove the following general theorem of this type.

Theorem 1.26. Let n ≥ a+b, and suppose that families F ⊂(

[n]a

),G ⊂

([n]b

)are cross-intersecting. Fix an integer j ≥ 1. Then the following inequalityholds in three dierent assumptions, listed below:

|F|+ |G| ≤(n

b

)+

(n− ja− j

)−(n− jb

). (1.16)

43

1. If a < b and we have |F| ≥(n−ja−j);

2. If a ≥ b and we have(n−ja−j)≤ |F| ≤

(n+b−1−a

b−1

)+(n+b−2−a

b−1

);

3. If a ≥ b and we have(n−a+b−3

b−3

)+(n−a+b−4

b−3

)≤ |F| ≤

(n−ja−j).

It is not dicult to come up with an example of a cross-intersecting pairon which the bound (1.16) is attained: take F := F ∈

([n]a

): [j] ⊂ F,G :=

G ∈(

[n]b

): G ∩ [j] 6= ∅. We also give the intuition for the bounds on |F|

in Points 2 and 3. If F = L(a, |F|), then the family F of size attaining theupper bound in Point 2 consists of all sets containing [a− b+ 1] and all thesets that contain [a− b] and a− b+ 2. Similarly, such family F attainingthe lower bound in Point 3 consists of all sets containing [a− b + 3] and allsets containing [a− b+ 2] and a− b+ 4.

We give the proof of Theorem 1.26 in the end of this section. The followingtheorem including Point 1 and some cases of Point 2 of Theorem 1.26 wasproven in [113].

Theorem 1.27 (Frankl, Tokushige, [113]). Let n > a + b, a ≤ b, andsuppose that families F ⊂

([n]a

),G ⊂

([n]b

)are cross-intersecting. Suppose

that for some real number α ≥ 1 we have(n−αn−a)≤ |F| ≤

(n−1n−a). Then

|F|+ |G| ≤(n

b

)+

(n− αn− a

)−(n− αb

). (1.17)

On the one side, in Theorem 1.27 the parameter α may take real values,while in Theorem 1.26 the parameter j is bound to be integer. On the otherside, Theorem 1.26 deals with the case a > b and, more importantly, evenfor a = b the restriction on the size of F is weaker. This plays a crucial rolein the proof of Frankl's theorem in the next section.

Let us obtain a useful corollary of Theorem 1.27, overlooked by Frankland Tokushige, which allows to extend it to the case a > b.

Corollary 1.28. Let n > a+ b, a > b, and suppose that F ⊂(

[n]a

),G ⊂

([n]b

)are cross-intersecting. Suppose that for some real α ≥ a − b + 1 we have(n−αn−a)≤ |F| ≤

(n−a+b−1n−a

). Then (1.17) holds.

Proof. Applying Theorem 1.22, we may suppose that F = L(|F|, a),G =L(|G|, b). Because of the restriction on the size of F we have [a−b+1] ⊂ F forany F ∈ F . Consider two families F ′ := F ′ ∈

([a−b+1,n]

b

): F ′∪ [a−b] ∈ F,

G ′ := G′ ∈(

[a−b+1,n]b

): G ∈ G. We have |F ′| = |F|, and we may apply

Theorem 1.27 for F ′,G ′ with n′ := n − a + b, α′ := α − a + b. We get that|F|+ |G ′| ≤

(n−a+b

b

)+(n−αn−a)−(n−αb

). We have |G \ G ′| ≤

(nb

)−(n−a+b

b

), and

44

summing this inequality with the previous one we get the statement of thecorollary.

Proof of Theorem 1.26. The proof of Point 1 of the theorem is a simpliedversion of the proof of Point 2, and thus we rst give the proof of Point 2,and then specify the parts which are dierent in the proof of Point 1.

The proof of the theorem uses bipartite graph considerations similar tothe ones embodied in Lemma 1.25. We give two slightly dierent versions ofit in the proof of Point 2 and Point 3.

Point 2. Let F ,G be a cross-intersecting pair with minimal cardinality ofF among the pairs satisfying the conditions of Point 2 and having maximalpossible sum of cardinalities.

Due to Theorem 1.22 we may suppose that F = L(|F|, a),G = L(|G|, b).Let F ∈ F be the set with the largest order in F . The bound on thecardinality of |F| implies F ⊃ 1, . . . a−b and F ∩a−b+1, a−b+2 6= ∅.Take the largest l ≥ 1, for which |F ∩ [a− b+2l]| = a− b+ l is satised. PutL := [a−b+2l]\F . Consider the families A := A ∈

([n]a

): A∩ [a−b+2l] =

F ∩ [a− b+ 2l] and B := B ∈(

[n]b

): B ∩ [a− b+ 2l] = L.

The considerations in this and the next paragraph follow closely the ar-gument from Lemma 1.25. For any B ∈ B, F ′ ∈ F \A we have F ′ ∩B 6= ∅.Indeed, take F ′′ ∈ F , F ′′ ∩ L = ∅. Since F has the largest order in F , wehave F ′′ ∩ [a− b+ 2l] ⊂ F , and, consequently, F ∈ A. This implies that thepair F \ A,G ∪ B is cross-intersecting.

Next, |A| =(n−a+b−2l

b−l)

= |B|. Consider a bipartite graph with parts A,Band edges connecting disjoint sets. Independent sets in this graph correspondto the cross-intersecting pairs of subfamilies of A and B. This is a regulargraph, and, consequently, the maximal independent set has the size |B|.Therefore |F \ A|+ |G ∪ B| ≥ |F|+ |G|.

Finally, if |F| >(n−ja−j), then F + [j], and thus all the sets containing

[j] belong to F \ A. Therefore,(n−ja−j)≤ |F \ A| < |F|, and we obtain a

contradiction with the minimality of F .

Point 1. Let F ,G be a cross-intersecting pair with minimal cardinality ofF among the pairs satisfying the conditions of Point 2 and having maximalpossible sum of cardinalities.

The bound on the cardinality of |F| implies F ⊃ [1, j]. Take the largestl ≥ 1, for which |F∩[2l]| = l is satised. Put L := [2l]\F . Consider the fam-iliesA := A ∈

([n]a

): A∩[2l] = F∩[2l] and B := B ∈

([n]b

): B∩[2l] = L.

45

The rest of the proof is the same.

Point 3. Let F ,G be a cross-intersecting pair withmaximal cardinality ofF among the pairs of maximal sum of cardinalities satisfying the conditionsof Point 3. We again w.l.o.g. suppose that F = L(a, |F|),G = L(b, |G|).

Let F ∈(

[n]a

)\ F have the smallest order. Due to the lower bound on

|F|, one of the following conditions hold: either F ∩ [a− b+ 4] = [a− b+ 2]or for some integer 1 ≤ i ≤ a − b + 1 one has F ∩ [i + 1] = [i]. Indeed, ifthe latter condition does not hold, then F ⊃ [a − b + 2]. But all the setscontaining [a−b+2] and at least one of a−b+3, a−b+4 are in F , and soF ∩ [a−b+4] = [a−b+2]. Denote by t the expression a−b+4 in the formercase, and i+ 1 in the latter case. Put L := [t] \F and consider two families:A := A ∈

([n]a

): A∩ [t] = F ∩ [t] and B := B ∈

([n]b

): B∩ [t] = L. As in

the previous point, the pair F ∪A,G \B is cross-intersecting. Moreover, it iseasy to see that a−|F∩[t]| ≥ b−|L| and n−t ≥ a−|F∩[t]|+b−|L| = a+b−t.Consequently, |A| =

(n−t

a−|F∩[t]|)≥(n−tb−|L|

)= |B|. Therefore, arguing as in the

previous point, we conclude that |F| =(n−ja−j).

1.3.2. Proof of Theorem 1.5

Below we give proof of Theorem 1.5, which is much simpler than the proof ofthe proof of Theorem 1.4 in [81]. The proof makes use of the KruskalKatonatheorem and the methods developed earlier in this section.

Proof of Theorem 1.5. Let F have maximal cardinality among the familiessatisfying the condition of the theorem.

Case 1: γ(F) ≤(n−4k−3

)γ(F) ≤

(n−4k−3

)γ(F) ≤

(n−4k−3

). Let 1 be the most popular element in F . Con-

sider the families F(1) and F(1). It is clear that |F| = |F(1)| + |F(1)|.These two families are cross-intersecting and, moreover, we have

(n−u−1n−k−1

)≤

|F(1)| ≤(

n−4n−k−1

). Using Corollary 1.28 for n′ := n − 1, a := k, b := k − 1,

we get the statement of the theorem.

Case 2:(n−4k−3

)< γ(F) ≤

(n−3k−2

)+(n−4k−2

)(n−4k−3

)< γ(F) ≤

(n−3k−2

)+(n−4k−2

)(n−4k−3

)< γ(F) ≤

(n−3k−2

)+(n−4k−2

). This case is treated analogously,

with the only dierence that instead of Corollary 1.28 we use Point 2 of The-orem 1.26 with n′ := n− 1, a := k, b := k − 1, j := 3.

Case 3: γ(F) >(n−3k−2

)+(n−4k−2

)γ(F) >

(n−3k−2

)+(n−4k−2

)γ(F) >

(n−3k−2

)+(n−4k−2

). When the diversity is large, we cannot

proceed as in Part 2 of Theorem 1.26, since we cannot guarantee that, after

46

passing to the lexicographically minimal elements, the set L of maximal orderwill satisfy the necessary condition |L∩ [2l]| = l for some l ≥ 1. (We cannotneither apply nor imitate the proof of Part 2 of Theorem 1.26 in this case.)We will proceed dierently. First we show that we may assume that F isshifted.

Lemma 1.29. For any intersecting family F ⊂(

[n]k

)and any i, j, satisfying

|F(i)| ≥ |F(j)|, we have γ(F)− γ(Si,j(F)) ≤(n−3k−2

).

Proof. Indeed, it is easy to see that the families F(ij),F(ji) are cross-intersecting. The assumption of the lemma implies |F(ij)| ≥ |F(ji)|, andthus by Theorem 1.22 we have |F(ji)| ≤

(n−3k−2

)(see the remark after the the-

orem). On the other hand, after the (i, j)-shift the degree of i cannot increaseby more than |F(ji)|, since it is only in the sets from F(ji) that the elementj may be replaced by i. Therefore, γ(F)−γ(Si,j(F)) ≤ |F(ji)| ≤

(n−3k−2

).

Rearrange the elements in the order of decreasing degree, and do consec-utively the (1, j)-shifts, j = 2, 3, . . ., then the (2, j)-shifts, j = 3, 4, . . . etc.,until either the family becomes shifted, or the diversity of the family becomesat most

(n−3k−2

)+(n−4k−2

). We denote the obtained family by F again. In the

latter case by Lemma 1.29 we have γ(F) ≥(n−4k−2

)= n−k−1

k−2

(n−4k−3

)>(n−4k−3

).

Therefore, this case is reduced to Case 2.

Finally, what if F is shifted? We may suppose that1. γ(F) >

(n−3k−2

)+(n−4k−2

),

2. F is shifted,3. F has the smallest diversity among the families F ′ of maximal size withγ(F ′) ≥

(n−3k−2

).

The last inequality may look strange, since it does not coincide with theinequality dening Case 3. However, since we know that the theorem holdsfor families with diversity between

(n−3k−2

)and

(n−3k−2

)+(n−4k−2

)with u = 3, we

may include all potential families of maximal size with such diversity to theclass of families in question.

The element 1 is the most popular among the sets in F . Find the setF ∈ F of maximal order and, in terms of the proof from Section 1.3, applyLemma 1.25 to get a family F ′ := (F \ A) ∪ B of smaller diversity and atleast as large as F . Note that, again in terms of Lemma 1.25, l ≥ 2 andthus |A| =

(n−2lk−l)≤(n−4k−2

). Therefore, we have γ(F ′) ≥ γ(F)− |A| ≥

(n−3k−2

),

which is a contradiction with the choice of F .

47

1.4. Proof of Theorem 1.7

The following theorem, proven by Dinur and Friedgut [60], is the main ingre-dient in the proof. We say that a family J ⊂ 2[n] is a j-junta, if there existsa subset J ⊂ [n] of size j (the center of the junta), such that the membershipof a set in F is determined only by its intersection with J , that is, for somefamily J ∗ ⊂ 2J (the dening family) we have F = F : F ∩ J ∈ J ∗.

Theorem 1.30 ([60]). For any integer r ≥ 2,there exist functions j(r), c(r),such that for any integers 1 < j(r) < k < n/2, if F ⊂

([n]k

)is an intersecting

family with |F| ≥ c(r)(n−rk−r), then there exists an intersecting j-junta J with

j ≤ j(r) and

|F \ J | ≤ c(r)

(n− rk − r

). (1.18)

We start the proof of the theorem. Choose C suciently large (its choicewill become clear later), n > Ck > 0 and an intersecting family F ⊂

([n]k

).

Then, applying Theorem 1.30 with r = 5, we get that there exists a j-juntaJ , j ≤ j(5), such that |F \ J | ≤ c(5)

(n−5k−5

)<(n−5k−4

), where the second

inequality holds provided C is large enough.The rst step is to show that, unless J = A2, we have γ(F) <

(n−3k−2

).

Proposition 1.31. Consider an intersecting j-junta J ⊂ 2[n], with centerJ ⊂ [n], |J | = j, and dened by an intersecting family J ∗ ⊂ 2J . Then Jsatises one of the two following properties:

• J is contained in a family isomorphic to A2.

• There exists i ∈ J , such that all sets in J ∗ of size at most 2 contain i.

Proof. Note that the intersecting families of ≤ 2-element sets which cannotbe pierced by a single element are isomorphic to

([3]2

). Therefore, the junta

that does not fall into the second category must have the center of size 3 andbe dened by a family containing all 2-element subsets of the center. Thenwe are only left to observe the fact that A2 is a junta with center J = [3]and dened by the family J ∗ =

([3]2

)∪ [3]:

A2 =A ∈

([n]

k

): |A ∩ [3]| ≥ 2

.

48

Assume that J is not isomorphic to A2. Then, as it follows from theproposition above, γ(J ) ≤ 2j

(n−jk−3

). If C = n/k is suciently large, then

2j(n− jk − 3

)≤ 2j

C

(n

k − 2

)<

2j+1

C

(n− 3

k − 2

)≤ 1

2

(n− 3

k − 2

).

Moreover,(n−5k−4

)< 1

2

(n−3k−2

)for any n ≥ 2k. Therefore, in this case we can

conclude that

γ(F) ≤ γ(J ) + |F \ J | <(n− 3

k − 2

).

From now on we suppose that J = A2. For i = 1, 2, 3 consider the familiesFi := F ∈ F : F ∩ [3] = i. W.l.o.g., assume that F1 has the largestsize among Fi. We will use the following obvious bound: γ(F) ≤ |F| − δ1.Consider the following three families on [4, n]:

G :=F ∩ [4, n] : F ∈ F , F ∩ [3] = 2, 3

,

H1 :=F ∩ [4, n] : F ∈ F1

,

H2 :=F : F ∈ F , F ⊂ [4, n]

.

Clearly, G ⊂(

[4,n]k−2

), H1 ⊂

([4,n]k−1

), H2 ⊂

([4,n]k

). Most importantly,

γ(F) ≤ |G|+ |F2|+ |F3|+ |H2| ≤ |G|+ 2|F1|+ |H2| = |G|+ 2|H1|+ |H2|.

Therefore, to conclude the proof of the theorem, it is sucient to show thefollowing two inequalities:

|G|+ 4|H1| ≤(n− 3

k − 2

), (1.19)

|G|+ 2|H2| ≤(n− 3

k − 2

). (1.20)

Summing these two inequalities with coecients equal to 1/2, we get thatγ(F) ≤

(n−3k−2

).

There are two important properties that we are going to use. The rstone is that F \J = F1∪F2∪F3∪H2, and thus, using |H1| = |F1|, we have|H1|, |H2| ≤ |F \ J | ≤

(n−5k−4

). The second one is that the pair of families

G,H1 as well as G,H2 are cross-intersecting (see Denition 1.6).In what follows we show that (1.19), (1.20) hold in a more general form.

Similar inequalities appeared in [176], [97] and [166].

49

Lemma 1.32. Consider a set [m] and two cross-intersecting families A ⊂([m]a

),B ⊂

([m]b

). Assume that m > (C ′ + 1) · maxa, b for some constant

C ′. Assume also that |B| ≤(m−(b−a+1)

a−1

). Then

|A|+ C ′|B| ≤(m

a

). (1.21)

Before proving the lemma, let us deduce the inequalities (1.19), (1.20) outof (1.21) and thus conclude the proof of Theorem 1.7. For (1.19) we need tosubstitute A := G, B := H1, a := k − 2, b := k, C ′ := C, [m] := [4, n].Then we conclude that (1.19) holds even with 4 replaced by C. The deduc-tion of (1.20) is similar. Moreover, we get that a pair of families may achieveequality in (1.19) and (1.20) only if H1 = H2 = ∅ (and therefore Fi = ∅ fori ∈ [3]). Therefore, if γ(F) =

(n−3k−2

), then F ⊂ A2.

Proof of Lemma 1.32. Using Theorem 1.22, we may w.l.o.g. assume thatA = L(|A|, a),B = L(|B|, b). Due to the restriction on the size of B, any setin it contains [b− a+ 1]. Consider the families

B0 :=B \ [b− a+ 1] : B ∈ B,A0 :=A : A ∈ A, A ∩ [b− a+ 1] = ∅.

Put Y := [b − a + 2,m] (if b < a, put Y := [1,m]). Note that |Y | =minm − (b − a + 1),m. Clearly, B0 ⊂

(Ya−1

)and A0 ⊂

(Ya

). Consider a

bipartite graph G with parts(Ya

),(Ya−1

), and edges connecting disjoint sets.

Then the intersection of A ∪ B with the parts of the graph is A0 ∪ B0, andit forms an independent set in G0. Thus, we have

|A0|(|Y |a

) +|B0|( |Y |a−1

) ≤ 1.

We have(|Y |a

)/( |Y |a−1

)= (|Y | − a)/a = minm− a,m− b− 1/a ≥ C ′. This

implies

|A0|+ C ′|B0| ≤ |A0|+(|Y |a

)|B0|( |Y |

a−1

) ≤(|Y |a

).

The lemma follows from the fact that

|A|+ C ′|B| ≤(m

a

)−(|Y |a

)+ |A0|+ C ′|B0| ≤

(m

a

).

50

1.5. Diversity for 2k ≤ n ≤ 3k

Under the same assumption that we make in Theorem 1.7, it is possible toprove certain Hilton-Milner type stability results for diversity (using moreelaborate versions of Lemma 1.32). However, we think that it is more inter-esting to resolve the problem for any n > 3k and show that the family withthe maximum possible diversity must be isomorphic to a subfamily of A2, orthe two out of three family. When 2k < n < 3k, then other families havelarger diversity. They can be described as r + 1 out of 2r + 1 families:

Dr :=D ∈

([n]

k

): |D ∩ [2r+ 1]| ≥ r+ 1

, r = 1, . . . , k− 1. (1.22)

Positive answer to the following question seems to be a reasonable con-jecture at a rst glance. It was actually a conjecture in the earlier version ofthe manuscript.

Question. Fix n ≥ 2k > 0 and consider an intersecting family F ⊂([n]k

). Is it true that if for some r ∈ Z≥0 we have (k− 1)

(2 + 1

r+1

)+ 1 ≤ n ≤

(k − 1)(2 + 1

r

)+ 1, then γ(F) ≤ γ(Dr)?

Substituting r = 0 in the question, it asks whether γ(F) ≤(n−3k−2

)for any

n ≥ 3k − 2. Let us explain what stands behind this question. Assume thatthe element with the highest degree in F is 1. Then the conjecture is juststating that, if one restricts the attention to the family F ′ := F ∈ F :1 /∈ F, F ⊂

([2,n]k

), then the size of F ′ is at most the size of the largest

2-intersecting family on [2, n]. We say that a family is t-intersecting, if anytwo sets from the family intersect in at least t elements. The exact formulasgiven in the question come from the famous Complete Intersection Theoremby Ahlswede and Khachatrian [1].

The families D′r ⊂(

[2,n]k

),D′r := D ∈ Dr : 1 /∈ D are 2-intersecting.

And it comes as no surprise. Indeed, the same must be true for any shiftedintersecting family F (see Section 1.2.1).

For any shifted family δ1(F) = ∆(F) and, if F is intersecting, then F ′must be 2-intersecting. Indeed, if there are two sets F1, F2 ∈ F , such thatF1, F2 ⊂ [2, n] and F1∩F2 = x, then, by shiftedness, F ′1 := F1 \ x∪1also belongs to F , and we have F ′1 ∩ F2 = ∅, a contradiction. Consequently,the question has positive answer for such F).

Therefore, the question above asks if any intersecting family should behaveas shifted intersecting families with respect to diversity. Shifting preservesthe property of a family to be intersecting, but, unfortunately, it does notallow to control the diversity of a family. This is why the general case cannot

51

be directly reduced to shifted case. In fact, it cannot be reduced to theshifted case at all: the answer to the question is negative for families thatare not shifted!

1.5.1. Intersecting families with the largest diversity are not shifted

Here we present a counterexample to the conjecture from the earlier ver-sion of the paper found and communicated to us by Noam Lifshitz. As thecounterexample shows, at least in some cases the extremal value of γ(F)is attained on the families that are not shifted, which is unexpected for aproblem concerning intersecting families.

We use the notions and results coming from the analysis of Boolean func-tions. We give all the necessary denitions, and all the standard results usedhere may be found in [65]. For a real number 0 < p < 1 and a set F ⊂ [n]and F ⊂ 2[n], dene the p-biased measure µp(F ) := p|F |(1 − p)n−|F | andµp(F) :=

∑F∈F µp(F ). The inuence Ipi (F) of coordinate i in F is

Ipi (F) := µp(F : |F, F∆i| ∩ F = 1

),

and the total inuence is Ip(F) :=∑

i Ipi (F). In case if F is closed upwards,

we have

Ipi (F) = p−1µp(F ∈ F : i ∈ F

)− (1− p)−1µp

(F ∈ F : i /∈ F

). (1.23)

Fix a suciently large r and even bigger k ≥ k0(r), n ≥ n0(r), sat-isfying the conditions on n from the conjecture. Put p := k

n . That is,p = 1

2 − (1 + o(1))1r .

Intersecting family with low inuences. In what follows, we de-scribe the family Tr, whose restriction Tr ∩ J (see below) provides an exam-ple showing that the Kahn-Kalai-Linial inequality [135] is sharp for indicatorfunctions of intersecting families. The family Tr has larger diversity than Dr.The example is taken from Gil Kalai's post on MathOverow [136], however,since the explanation of the necessary properties in the post was very brief,we expand the exposition here, hopefully providing all the necessary details.

Consider an intersecting family Tr ⊂(

[n]k

), which is a (2r + 1)-junta with

center J := [2r+ 1], and Tr ∩ J is the following intersecting family. Arrangethe elements of J on the circle, and for each set S ⊂ 2J form a sequenceu := (u1, u2, . . .), where ui is the length of the i-th longest run of consecutive1's; similarly, z := (z1, z2, . . .) is the sequence, in which zi is the i-th longest

52

run of consecutive 0's. Form Tr∩J by including all sets, for which its sequenceu is lexicographically bigger than z (we denote it u z). Note that, since|J | is odd, we cannot have equality between the sequences. Therefore, wehave

|Tr ∩ J | = 2|J |−1

since if T ⊂ J is in Tr ∩ J , then its complement J \ T is not, and vice versa.Let us show that Tr ∩ J is an intersecting family. Assume the contrary,

and let T1, T2 be two disjoint sets in Tr ∩ J . Let ui, zi, i = 1, 2, be thecorresponding one and zero runs sequences. Then, clearly, z1 u2 andz2 u1, otherwise, it would be impossible to t the runs of 1's of T1 insidethe runs of 0's of T2 (and the same with the roles of T1, T2 interchanged).However, if, say, u1 u2, then by transitivity z2 u2, a contradiction.

In what follows, all logarithms have base 2 and all asymptotic notationsare with respect to r →∞.

Lemma 1.33. For each i ∈ J , we have I1/2i (Tr ∩ J) = O

(log rr

).

Proof. The proof requires a somewhat tedious analysis of the typical sets inthe family. The family is clearly transitive on J , and thus it is sucientto show that I1/2(Tr ∩ J) = O(log r). The total inuence I1/2(Tr ∩ J) isthe average number of pivotal coordinates in a randomly chosen set from J

according to µ1/2, that is, the number of coordinates which change results inthe set passing from Tr ∩ J to its complement or vice versa.

Choose a random set T ∈ J according to µ1/2 and denote its zero andone runs sequences z := (z1, z2, . . .) and u := (u1, u2, . . .), respectively. Thecoordinate is pivotal, if after its change the lexicographical order of z and uis reversed.

Using rst moment, it is easy to see that with probability 1 − o(1/r)the largest run of consecutive 1's in T has size at most (2 + o(1)) log r,and the same for the runs of zeros. Thus, the sequences not satisfying thisproperty contribute o(1) to the total inuence. In what follows we ignoresuch sequences.

In what follows, we assume that u z. The other case is treated analo-gously. Choose ρ such that zρ = uρ for j ≤ ρ and uρ+1 > zρ+1. The lex orderis reversed if at least one of the two happens: either (u1, . . . , uρ+1) is replacedby a lexicographically smaller sequence, or (z1, . . . , zρ+1) is replaced by a lex-icographically larger sequence. Denote the number of the former and lattertypes by s1 and s2, respectively. The number of pivotal coordinates is atmost s1 +s2, but we will bound just s1 instead. A pivotal coordinate k of thesecond type is a pivotal coordinate of the rst type for the set T \ k, and,

53

since µ1/2(T ) = µ1/2(T \k), the average value of s1 is not more than twicesmaller than s1 + s2, Thus if E[s1] = O(log r), then E[s1 + s2] = O(log r).

From the above, we clearly have s1 ≤∑ρ+1

j=1 uρ. Moreover, ui = O(log r)for each i = 1 . . . , ρ+ 1. Consequently, we can bound

E[s1] ≤ E[ ρ+1∑j=1

O(log r)]

= O(log r)∞∑k=1

Pr[ρ ≥ k].

We use a slight variation of this bound. Since the number of pivotal coordi-nates is at most 2r + 1, we can bound

E[s1] ≤ O(log r)r0.1∑k=1

Pr[ρ ≥ k] + (2r + 1) Pr[ρ ≥ r0.1]. (1.24)

To complete the proof that E[s1] = O(log r) (and thus the proof of thelemma), it is sucient to show the validity of the following lemma

Lemma 1.34. For any ρ ≤ r0.1 we have Pr[ρ ≥ k] ≤ e−α log2 k + o(1/r) forsome α < 1. In particular, Pr[ρ ≥ rβ] = o(1/r) for any xed β > 0.

We defer the proof of this lemma to the end of the section, and nish theproof of Lemma 1.33. From it, we have Pr[ρ ≥ r0.1] = o(1/r), and the righthand side of the inequality (1.24) is at most O(log r)

∑∞k=1 e

−α log2 k + o(1) =O(log r).

Consider the family T ↑r := T ⊂ [n] : T ∩ J ∈ Tr ∩ J. Then wehave µ1/2(T ↑r ) = 1/2. Similarly, dene the family D↑r based on Dr. Again,µ1/2(D↑r) = 1/2. Dene the p-biased diversity ofF as γp(F) := mini∈[n] µp(F ∈F : i /∈ F).

Using the result of Dinur and Safra [61], for suciently large n = n(r) wehave ∣∣∣γp(T ↑r )− γ(Tr)(

nk

) ∣∣∣ ≤ 1

r2, (1.25)

and the same for Dr and D↑r . (Note that we use the fact that both familiesTr ∩ T ⊂ J : 1 /∈ T and Dr ∩ T ⊂ J : 1 /∈ T are 2r + 1-juntas.)Using the Margulis-Russo lemma, for any upwards closed family F ⊂ 2J andp0 ∈ (0, 1), we have

dµp(F)

dp|p0 = Ip0(F).

Using large deviation estimates, it is not dicult to see, that for any p0 ∈[p, 1/2] the contribution of sets from F of size not in [r − r2/3, r + r2/3] to

54

the inuence is negligible (since the measure of such sets is negligible). Onthe other hand, for any F of size in [r − r2/3, r + r2/3], we have µp0(F ) =(1 + o(1))µ1/2(F ) for any p0 ∈ [p, 1/2]. Thus, Ip0(F) = (1 + o(1))I1/2(F)for any such p0. We conclude that we have

µ1/2(F)− µp(F) = (1 + o(1))(1/2− p)Ip(F).

At the same time, for a symmetric, closed upward family F ⊂ 2J and forany i ∈ J we have pIpi (F) + 1

1−pγp(F) = µp(F). Therefore,

γp(F) = (1− p)(µp(F)− p

|J |Ip(F)

)=

(1− p)µ1/2(F)− (1− p+ o(1))p+

(12 − p

)|J |

|J |Ip(F).

We know that Ip(Dr∩J) = Ω(√r), since Dr∩J is the majority function.

On the other hand, using Lemma 1.33, we have Ip(Tr∩J) = O(log r). Usingthe displayed formula above, we get

γp(D↑r) = γp(Dr ∩ J) =1− p

2− Ω

( 1√r

)and

γp(T ↑r ) = γp(Tr ∩ J) =1− p

2−O

( log r

r

).

Combining the formulas above with (1.25), we conclude that γ(Dr) < γ(Tr).

Proof of Lemma 1.34. How do we express the condition that the vectors ofzero and one runs of a random sequence share the rst k coordinates? LetN(t) be a random variable counting the number of runs of length at least t inthe sequence. Choose an integer t0 such that N(t0) ≤ k, but N(t0− 1) > k.In order to have ρ ≥ k, exactly a half of each N(t), t ≥ t0, must be zeroruns, and a half must be one runs. In what follows, we analyze the behaviourof N(t).

Take an integer t and x the value of N(t). For each run of length at leastt, reveal the values of the rst t coordinates that belong to the run (in theclockwise order), as well as the value that precedes the run clockwise. E.g.,for t = 3 the sequence may look like xx0111xxx1000xx1000x . . ., where xstands for the coordinates that are not revealed. Let us denote Lj, j ∈ [N(t)],the intervals of unrevealed coordinates between the revealed coordinates. Ifsome Lj has length smaller than r1/10, then reveal its coordinates, otherwisekeep it intact. Let us denote S the class of all possible subsequences that

55

can be xed (revealed) in this way. Each subsequence S ∈ S gives rise to afamily C(S) of sequences containing S as a subsequence. Fix any subclassS ′ ⊂ S. Then for a randomly chosen cyclic sequence R we have

Pr[ρ ≥ k] ≤∑

S∈S\S′Pr[ρ ≥ k|R ∈ C(S)] Pr[R ∈ C(S)] +

∑S∈S′

Pr[R ∈ C(S)]. (1.26)

That is, one may think of S ′ as a small set of exceptional classes. Wewill show that the latter term on the right hand side is o(1/r), while Pr[ρ ≥k|R ∈ C(S)] ≤ e−α log2 k for each S ∈ S\S ′. Since

∑s∈S\S ′ Pr[R ∈ C(S)] ≤ 1,

this will conclude the proof of the lemma.We note that the expected value of N(t) is r2−t (we choose the starting

point, choose arbitrarily the coordinate x before the starting point, and thenx the t coordinates to be equal to 1 − x). Moreover, using the Talagrandinequality (e.g., in the form of [13, Theorem 7.7.1]), we can show that whenthe expectation is, say, bigger than r1/10, then the value of N(t) is well-concentrated around the expectation (it is equal to (1 + o(1))E[N(t)] withprobability at least 1− r−c for any c > 0). We restrict our attention only onthe values of t such that the expected value of N(t) does not exceed r1/10,and we include in S ′ all sequences for which the value of N(t) exceeds 2r1/10.By the above, there are o(1/r) of those.

Fix some t satisfying the condition E[N(t)] ≤ r1/10 and note that t =Ω(log r). The probability that there is more than one interval Lj of unre-vealed coordinates of length ≤ r1/10, given that N(t) ≤ 2r1/10, is o(1/r).Indeed, compare the number of all possible choices for the starting positionsof the runs of length ≥ t (roughly

(r

N(t)

)) with the number of choices, in

which we rst x N(t) − 2 starting positions of such runs, and then choosethe remaining two at distance at most 2r1/10 from one of the already chosenones (at most

(r

N(t)−2

)· (4r1/5)2). Include all such subsequences in S ′. We

are not going to include any more subsequences in S ′ and we note that thereare o(1/r) sequences that contain subsequences from S ′.

Fix one subsequence S ∈ S \ S ′. We aim to bound Pr[ρ ≥ k|R ∈ C(S)].Consider a uniform distribution over all sequences in C(S). Recall that atleast [N(t)] − 1 of Lj are unrevealed. We note that the only restriction onthe choices of coordinates in Lj is that they cannot contain runs of ones orzeros of length ≥ t, moreover, there is no dependency between the choicesof coordinates in dierent Lj. Next, for each j we reveal the rst coordinatexj of Lj in the clockwise order. We claim that Pr[xj = 1] = (1

2 + o(1)).Indeed, for each admissible sequence starting from xj we change xj to 1−xjand obtain an admissible sequence and vice versa, unless the t− 1 elements

56

immediately following xj have the same value, while xj had the oppositevalue. But this constitutes at most a 1/2t-fraction of all possible admissiblesequences on Lj, and thus only aects the value of Pr[xj = 1] by o(1).

Since the choices for dierent xj are independent, we have the following.First, the expected number E[N(t+ 1)] of surviving runs of length t+ 1 is(1

2 + o(1))N(t). Moreover, it is tightly concentrated around the expectation:using a Cherno-type bound, we have Pr[|N(t + 1) − 1

2N(t)| ≥ 16N(t)] ≤

e−cN(t) for some xed positive constant c.Now we are in position to bound Pr[ρ ≥ k|R ∈ C(S)]. First, we nd t0 as

in the rst paragraph of the proof of the lemma. By the previous paragraph,with probability at least 1 − e−ck for some xed c > 0 we have 3N(t0) ≥ kand 27N(t0+2) ≥ k. In order for ρ ≥ k to hold, a half of runs contributing toN(t0) should be one runs, and the other half should be zero runs. Moreover,the same should be true for N(t) for t ≥ t0 + 1. Thus, we can obtain thefollowing rough bound:

Pr[ρ ≥ k|R ∈ C(S)] ≤ Pr[ρ ≥ N(t0 + 2)|R ∈ C(S)] · P1 ≤ e−ck + Pr[ρ ≥ k

27|R ∈ C(S)

]· P1,

where P1 is the probability that the values xj will be chosen in such a waythat the number of zero and one runs of length at least t0 + 1 is the same.It is easy to see that

P1 ≤ (1 + o(1))2−N(t0)

N(t0)/2∑j=0

(N(t0)/2

j

)2

= Θ(N(t0)

−1/2).

(note here that we have to take into account the interval Lj that was xedand that potentially gave one zero or one run of length t0, but it does notaect the validity of the bound above). Looking at the recursion Pr[ρ ≥ k] ≤e−ck + Θ(k−1/2) Pr[ρ ≥ k/27], it is not dicult to conclude that Pr[ρ ≥ k] ≤e−α log2 k for some positive constant α. Substituting into (1.26), we get theresult.

We remark that the problem treated in Lemma 1.34 seems to be interestingon its own and that it would be desirable to have a fuller understanding of thebehaviour of the probability Pr[ρ ≥ k]. However, we believe that the boundon the probability (modulo o(1/r) and the constant α in the exponent) isessentially sharp.

57

1.6. Counting intersecting families

In the proof of Theorem 1.10 we shall use the following result, interestingin itself, which gives a sharp upper bound on the number of pairs of cross-intersecting families. Let us denote by CI(n, a, b, t) (CI(n, a, b, [t1, t2])) thenumber of pairs of cross-intersecting families A ⊂

([n]a

),B ⊂

([n]b

)with |A| =

t (t1 ≤ |A| ≤ t2). We also denote CI(n, a, b) :=∑

tCI(n, a, b, t).We prove the following bound for the number of pairs of cross-intersecting

families.

Theorem 1.35. Choose a, b, n ∈ N and put c := maxa, b, T :=(n−a+b−1n−a

).

For n ≥ a + b + 2√c log c + 2 max0, a − b, a, b → ∞, and b log a we

have

CI(n, a, b) =(1 + δab + o(1))2(nc), (1.27)

CI(n, a, b, [1, T ]) =(1 + o(1))

(n

a

)2(nb)−(n−ab ), (1.28)

where δab = 1 if a = b, and 0 otherwise.

In fact, (1.28) easily implies (1.27) (see Section 1.6.1 for details).

1.6.1. Cross-intersecting families

We would need the following result, which is a combination of a result ofFrankl and Tokushige [113] (Theorem 1.27 above) and two results of myselfand Zakharov [176] (Part 1 of Theorem 1.26 and Corollary 1.28 above).

Theorem (Frankl, Tokushige, [113], Kupavskii, Zakharov [176]). Let n >a+b and suppose that the families A ⊂

([n]a

),B ⊂

([n]b

)are cross-intersecting.

If for some real number α ≥ 1 we have(n−αn−a)≤ |A| ≤

(n−a+b−1n−a

), then

|A|+ |B| ≤(n

b

)+

(n− αa− α

)−(n− αb

). (1.29)

Note that the upper bound on |A| in this theorem is exactly the same asin (1.28).

We go on to the proof of Theorem 1.35. First we show that (1.28) implies(1.27). We may w.l.o.g. assume for this paragraph that c = b ≥ a. For b > awe have T ≥

(na

)and thus CI(n, a, b, t) = 0 for t > T . Therefore, we have

CI(n, a, b) = CI(n, a, b, 0) + CI(n, a, b, [1, T ]). If a = b, then T =(n−1a−1

),

58

and it follows from Theorem 1.22 that if A,B ⊂(

[n]a

)are cross-intersecting,

then min|A|, |B| ≤(n−1a−1

). Therefore, in the case a = b we have

2CI(n, a, b, 0)− 1 ≤ CI(n, a, b) ≤ 2(CI(n, a, b, 0) + CI(n, a, b, [1, T ])).

The -1 in the rst inequality stands for a pair of empty families, whichis counted twice. At the same time, we have CI(n, a, b, 0) = 2(nb). Thus,in both cases b > a and b = a it is sucient to show that the right-handside of (1.28) is o(2(nb)). We rst note that n − a − b ≥

√n for b ≥ a,

n ≥ a+ b+ 2√b log b, since 4b log b ≥ a+ b+ 2

√b log b. The rest is done by

a simple calculation:

CI(n, a, b, [1, T ])

2(nb)=

(n

a

)2−(n−ab ) ≤ 2n−(b+

√n

b ) = o(1).

Next, discuss the proof of the lower bound in (1.28). To obtain that manypairs of intersecting families, take A := A, A ∈

([n]a

)and B(A) := B ∈(

[n]b

): B ∩A 6= ∅. Next, choose an arbitrary subfamily B ⊂ B(A). We only

need to assure that few of these pairs of subfamilies are counted twice. Actu-ally, we count a pair of families twice only in the case when a = b and both

A,B consist of one set. The number of such pairs is(

[n]a

)2and is negligible

compared to the right hand side of (1.28).

We pass to the proof of the upper bound.

2 ≤ |A| ≤ n− a2 ≤ |A| ≤ n− a2 ≤ |A| ≤ n− a. Applying Theorem 1.22, the size of the (unique) max-imal family B′ that forms a cross-intersecting pair with A is maximized ifA consists of two sets A1, A2 that intersect in a − 1 elements. Therefore,|B′| ≤

(nb

)−(n−a+1

b

)+(n−a−1b−2

). Any other family B that forms a cross-

intersecting pair with A must be a subfamily of B′.So we can bound the number of pairs of cross-intersecting families A,B

with 2 ≤ |A| ≤ n− a as follows:∑n−at=2 CI(n, a, b, t)

2(nb)−(n−ab )≤

n−a∑t=2

((na

)t

)2(nb)−(n−a+1

b )+(n−a−1b−2 )

2(nb)−(n−ab )≤ 2n

2

2−(n−a−1b−1 ) = o(1).

n− a+ 1 ≤ |A| ≤(n−un−a)

n− a+ 1 ≤ |A| ≤(n−un−a)

n− a+ 1 ≤ |A| ≤(n−un−a), where u =

√c log c+ max0, a− b. Note that

n−a+1 =(n−a+1n−a

). In this case the bound is similar, but we use Theorem 1.23

to bound the size of |B|. For A with |A| = t :=(n−u′n−a), where u ≤ u′ ≤ a−1,

we get |B| ≤ 2(nb)−(n−u′

b ), and since((na)t

)≤ 2nt, we have the following bound:

59

CI(n, a, b, t)

2(nb)−(n−ab )≤((n

a

)t

)2(nb)−(n−u

′b )

2(nb)−(n−ab )≤ 2n(

n−u′n−a)2−(n−u

′−1b−1 ).

At the same time we have n ≥ a+ b+ 2u and(n−u′n−a)(

n−u′−1b−1

) =n− u′

b

n−a−b−1∏i=0

n− b− u′ − in− a− i

≤

n

b

n−a−b−1∏i=0

n− a−√c log c− i

n− a− i≤ n

be−√c log c(

∑n−ai=b+1

1i ) ≤ 1

2n(1.30)

for suciently large c. Indeed,√c log c

∑n−ai=b+1

1i ≥√c log c

∑b+2√c log c

i=b+11i ≥

(1 + o(1))2(√c log c)2

c = (2 + o(1)) log c, which justies (1.30) for n ≤ b3/2.For n > b3/2 we have

√c log c

∑n−ai=b+1

1i ≥ (1 + o(1))

√c log c log n

b ≥ (1 +o(1))

√c log c log n2/3 log n, which justies (1.30) for n > b3/2.

We conclude that

(n−un−a)∑t=n−a+1

CI(n, a, b, t)

2(nb)−(n−bb )≤ 2−

12(

n−u′−1b−1 ) = o(1).

(n−un−a)< |A| ≤ T

(n−un−a)< |A| ≤ T

(n−un−a)< |A| ≤ T , where u =

√b log b+max0, a−b. Using the Bollobas

set-pair inequality (Theorem 1.9), it is not dicult to obtain the followingbound on the number of maximal pairs of cross-intersecting families.

Lemma 1.36. The number of maximal cross-intersecting pairs A′ ⊂(

[n]a

),B′ ⊂(

[n]b

)is at most [

(na

)(nb

)](a+ba ).

We note that the proof is very similar to the proof of an analogous state-ment for intersecting families from [20].

Proof. Find a minimal B′-generating familyM ⊂ A′ such that B′ = B ∈([n]b

): B∩M 6= ∅ for all M ∈M.We claim that |M| ≤

(a+ba

). Indeed, due

to minimality, for each set M ′ ∈ M the family B′′ := B ∈(

[n]b

): B ∩M 6=

∅ for all M ∈ M− M ′ strictly contains B′. Therefore, there is a set Bin B′′ \ B′ such that B ∩ M ′ = ∅, B ∩ M 6= ∅ for all M ∈ M − M ′.Applying the inequality (1.5) toM and the collection of such sets B, we getthat |M| ≤

(a+bb

).

Interchanging the roles of A′ and B′, we get that a minimal A′-generatingfamily has size at most

(a+bb

)as well. Now the bound stated in the lemma is

60

just a crude upper bound on the number of ways one can choose these twogenerating families out of

([n]a

)and

([n]b

), respectively.

Combined with the bound (1.29) on the size of any maximal pair of familieswith such cardinalities, we get that

CI(n, a, b, [(n−un−a), T ])

2(nb)−(n−ab )≤[(n

a

)(n

b

)](a+ba ) 2(nb)+(n−un−a)−(n−ub )

2(nb)−(n−ab )≤ 22n(a+bb )2(n−un−a)−(n−u−1

b−1 ). (1.31)

We also have(a+bb

)(n−u−1b−1

) =n− ub

b−1∏i=0

a+ b− in− u− i

≤ n( a+ b

n− u

)b≤ 1

4n.

Indeed, the last inequality is clearly valid for n ≥ (a + b)2, b → ∞. Ifn < (a+ b)2, then the before-last expression is at most

elog n− b(n−a−b−u)n−u ≤ e2 log(a+b)− b(u+max0,a−b)O(a+b) ≤ e2 log(a+b)−Ω(minb,u).

Since by the assumption we have b log(a + b) and also, obviously, u log(a+ b), the last expression is at most e−4 log(a+b) < 1

4n .

Taking into account (1.30), which is valid for u′ = u, we conclude thatthe right-hand side of (1.31) is o(1).

1.6.2. Intersecting families

We go on to the proof Theorem 1.10. We shall use Theorem 1.5 in the proof.Let us rst show that (1.7) implies (1.6). Indeed, using that n−2k−1 ≥

√n

and k →∞ in the assumptions of Theorem 1.10, we get

I(n, k,≥ 1)

2(n−1k−1)∼ n

(n− 1

k

)2−(n−k−1k−1 ) ≤ 2n+log n−(k+

√n

k−1 ) = o(1).

Therefore, I(n, k,≥ 1) = o(2(n−1k−1)). On the other hand, it is easy to see that

I(n, k, 0) = (n+ o(1))2(n−1k−1) (for the proof see [20]).

Let us prove the lower bound in (1.7). For S ∈(

[n]k

), i ∈ [n] \ S dene

the family H(i, S) := S ∪ H ∈(

[n]k

): i ∈ H,H ∩ S 6= ∅. Due to

Theorem 1.3, these families are the largest non-trivial intersecting families.

61

We have |H(i, s)| =(n−1k−1

)−(n−k−1k−1

)+ 1, and each such family contains no

less than2(n−1k−1)−(n−k−1k−1 ) − k2(n−2k−2) = (1 + o(1))2(n−1k−1)−(n−k−1k−1 ) (1.32)

non-trivial intersecting subfamilies, as k → ∞. Indeed, a subfamily ofH(i, S) containing S is non-trivial unless all sets containing i contain also axed j ∈ S. In other words, they must be a subset of a family I(i, j, S) :=S ∪ I ∈

([n]k

): i, j ∈ S. The number of subfamilies of I(i, j, S) contain-

ing S is 2(n−2k−2). Next, we have(n−1k−1

)−(n−k−1k−1

)−(n−2k−2

)≥(n−3k−2

), and thus

the last inequality in the displayed formula above holds since 2(n−3k−2) k.Denote the set of all non-trivial subfamilies of H(i, S) by H(i, S).

Therefore,∑

S∈(nk),i/∈S|H(i, S)| = (1 + o(1))n

(n−1k

)2(n−1k−1)−(n−k−1k−1 ). On the

other hand, the pairwise intersections of these families are small: the fam-ilies from H(i, S) ∩ H(i, S ′) form the set I(n, k, 2), and we do (somewhatimplicitly) show in the proof that I(n, k, 2) = o(I(n, k, 1)). It could also beveried by a simple direct, but somewhat tedious calculation. Therefore, thelower bound is justied.

Next we prove the upper bound. Recall that, for i ∈ [n] and F ⊂(

[n]k

)we

use the standard notation

F(i) :=F − i : i ∈ F ∈ F ⊂(

[n]− ik − 1

),

F (i) :=F ∈ F : i /∈ F ⊂(

[n]− ik

).

Note that if F is intersecting then F(i) and F (i) are cross-intersecting.We count the number of families with dierent diversity separately. The

number of families F with i being the most popular element and γ(F) ≤(n−3k−2

)is at most the number of cross-intersecting pairs F (i), F(i).

Therefore, we may apply (1.28) with n′ := n−1, a := k, b := k−1, and get

that the number of such families F is at most (1 + o(1))(n−1k

)2(n−1k−1)−(n−k−1k−1 ).

Note that n′ ≥ a + b + 2√a log a + 2 and, in terms of Theorem 1.35, we

have T =(n−3k−2

)for our case. Multiplying the number of such families by the

number of choices of i, we get the claimed asymptotic.We are only left to prove that there are few families with diversity larger

than(n−3k−2

). Using the upper bound

(nk

)(2k−1k−1 ) for the number of maximal

intersecting families in(

[n]k

)obtained in [20] (see Lemma 1.36 for the proof

of a similar statement), combined with the bound (1.3) on the size of any

62

maximal family with such diversity, we get that

I(n, k,≥(n−3k−2

))

2(n−1k−1)−(n−k−1k−1 )≤(n

k

)(2k−1k−1 )2(n−1k−1)+(n−4k−3)−(n−4k−1)

2(n−1k−1)−(n−k−1k−1 )≤ 2n(

2k−1k−1 )2(n−4k−3)−(n−5k−2).

(1.33)Putting n = 2k + x, we have

(n−4k−3

)/(n−5k−2

)= (n−4)(k−2)

(n−k−1)(n−k−2) ≤(2k+x)k

(k+x−2)2 ≤1− x2

(k+x)2 ≤ 1− 1k . On the other hand,(

2k−1k−1

)(n−5k−2

) =n− 4

k − 1

k∏i=1

2k − in− 3− i

≤ n( 2k

n− 3

)k≤ 1

2kn,

where the last inequality is clearly valid for n ≥ 2k + 2 + 2√k log k and

suciently large k. We conclude that the right-hand side of (1.33) is at most

212k(

n−5k−2) = o(1).

1.7. Product inequalities

1.7.1. Preliminaries

Recall the KruskalKatona theorem (Theorem 1.21). Depending on the valueof |F|, equality might hold in (1.13) even if F is not isomorphic to C(k)(|F|).However, Furedi and Griggs [118] succeeded in determining which are thevalues of |F| such that C(k)(|F|) is the unique optimal family. Mors [194]proved a stronger inequality under the assumption that ∪F∈FF = [n]. Weneed the following special case of it.

Theorem (Mors, [194]). Suppose that G ⊂(

[n]l

), n ≥ l > t ≥ 1 and |G| =(

n−1l

). If ∪G∈G = [n] holds then

|S(t)(G)| ≥(n− 1

t

)+

(l − 1

t− 1

). (1.34)

We say that D, E ⊂ 2[n] are cross-union if for any two sets D ∈ D, E ∈ Ewe haveD∪E 6= [n]. Looking at the complements the next statement followsdirectly from Hilton's lemma (Theorem 1.22).

Corollary. If D ⊂(

[n]d

)and E ⊂

([n]e

)are cross-union then C(d)(|D|) and

C(e)(|E|) are cross-union as well.

Let us conclude this section with a simple inequality involving binomialcoecients

63

(n− ik − i

)(n

k

)<

(n− i+ 1

k − i+ 1

)(n− 1

k − 1

)holds for n ≥ 2k, i ≥ 2. (1.35)

Proof. For k < i the LHS is 0. Suppose k ≥ i and divide both sides by(nk

)(n−i+1k−i+1

). We obtain k−i+1

n−i+1 <kn , which is obviously true.

1.7.2. Short proof of Theorem 1.11

By symmetry, we suppose |A| ≤ |B|. First note that if |A| ≤(n−2k−2

), then

|A||B| ≤(n− 2

k − 2

)(n

k

)<

(n− 1

k − 1

)2

by (1.35), case i = 2.

From now on we assume that(n−2k−2

)≤ |A| ≤ |B|. By Theorem 1.22 we

suppose that A = L(|A|, k), B = L(|B|, k), i.e., both families are initialsegments in the lexicographic order.

Note that the rst(n−2k−2

)sets in the lexicographic order are all the k-sets

that contain 1 and 2. Since A,B are cross-intersecting, we infer that all theirmembers must contain either 1 or 2. We shall use this fact to prove:

Proposition 1.37. We have

|A|+ |B| ≤ 2

(n− 1

k − 1

). (1.36)

Note that (1.36) implies (1.8) by the inequality between arithmetic andgeometric means. One can even deduce that (1.8) is strict unless |A| = |B| =(n−1k−1

)holds.

Proof of Proposition 1.37. If |B| ≤(n−1k−1

)then (1.36) is obvious. Therefore,

we assume |B| >(n−1k−1

). Note that the rst

(n−1k−1

)members of

([n]k

)are all

the k-sets containing 1. Since A,B are cross-intersecting, 1 ∈ A holds for allA ∈ A. Let B′ be the family of the remaining sets in B, i.e.,

B′ := B ∈ B : 1 /∈ B.

Let C := C ∈(

[n]k

): 1 ∈ C,C /∈ A. To prove (1.36) we need to show that

|C| ≥ |B′| holds.

Recall that all k-sets containing both 1 and 2 are in A and therefore allmembers of B contain 1 or 2. We infer that B ∩ 1, 2 = 2 for all B ∈ B′and C ∩ 1, 2 = 1 for all C ∈ C.

64

Let us now consider a bipartite graph G = (X1, X2, E), where Xi :=Di ∈

([n]k

): Di ∩ 1, 2 = i

and two vertices D1 and D2 are connected

by an edge if and only if D1 ∩D2 = ∅ holds.Note that G is regular of degree

(n−k−1k−1

), C ⊆ X1,B′ ⊆ X2 hold. Moreover,

the cross-intersecting property implies that if D1 and D2 are connected forsome D2 ∈ B′ then D1 ∈ C. In other words, the full neighborhood of B′ inthe regular bipartite graph G is contained in C. This implies |C| ≥ |B′| andconcludes the proof.

1.7.3. Proof of Theorem 1.12 modulo Theorem 1.13

Set l = n − k and consider the families F := Ac,G := Bc ⊂(

[n]l

). Then

F ,G are cross-union. We assume |F| ≤ |G|. Note that for n = 2k = 2lthe cross-intersecting and cross-union conditions are equivalent and simplymean that if, say, F ∈ F then [n] − F /∈ G. Therefore, for an arbitraryF ⊂

([n]l

)the families F and

([n]k

)−F c are cross-union. Moreover, these are

altogether(

2kk

)= 2

(2k−1k−1

)sets. Consequently, in this case there are many

ways to achieve equality in (1.8).The case |B| =

(n−1k−1

)of Theorem 1.12 is somewhat special because re-

placing B by L(k)((

n−1k−1

))would produce the family of all k-sets containing

1, i.e., the family with the intersection of all its members being non-empty.Fortunately, in this case we can apply the theorem of Mors.

Setting G = Bc, by (1.34) we have

|S(k)(G)| ≥(n− 1

k

)+

(n− k − 1

k − 1

), yielding

|A| ≤(n

k

)− |S(k)(G)| ≤

(n− 1

k − 1

)−(n− k − 1

k − 1

),

which proves strict inequality in (1.9). From now on we may assume |B| >(n−1k−1

), and the remaining part of Theorem 1.12 follows from the case i = k+1

of Theorem 1.13, which we prove in the next section.

1.7.4. Proof of Theorem 1.13

We assume w.l.o.g. that |A| ≤ |B|. For the whole proof we assume thatA,B are the rst sets in the lexicographical order. We consider several casesdepending on the size of |A|.

Case 1. |A| ≥(n−2k−2

)+(n−ik−i+1

).

65

Lemma 1.38. If m ≥ 2a are natural numbers, A′,B′ ⊂(

[m]a

)are cross-

intersecting, and for some integer j ≥ 1 we have(m−ja−j)≤ |A′| ≤ |B′|, then

|A′|+ |B′| ≤(m

a

)+

(m− ja− j

)−(m− ja

).

Proof. We assume that A′,B′ are the rst sets in the lexicographical order.For j = 1 the familyA′ contains all sets containing 1. It implies thatboth families must have cardinality

(n−1a−1

), since otherwise B′ contains the set

2, . . . , a + 1, which is disjoint with 1, a + 2, . . . , 2a ∈ A′. At the sametime, the right hand side of the displayed equation above is exactly 2

(n−1a−1

)for j = 1. Therefore, in what follows we may assume that j ≥ 2. Since|A′| ≤ |B′|, we may w.l.o.g. assume that all sets from A′ contain 1 and B′contains the family L of all the sets containing 1. Since |A′| ≥

(m−ja−j), then

A′ contains the family A1 of all a-sets that contain [1, j]. Therefore, each setfrom B′ must intersect [1, j]. We denote by B0 the family of all a-sets thatdo not contain 1 and intersect [2, j]. By A0 we denote the family L \ A1.Note that |A0| =

(m−1a−1

)−(m−ja−j)and |B0| =

(m−1a

)−(m−ja

).

Consider a bipartite graph G with parts A0,B0 and with two sets joinedby an edge if they are disjoint. We know that (B′ \ L) ∪ (A′ \ A1) is anindependent set in G, since any pair of sets from dierent families intersect.

We aim to show that there is a matching ofA0 into B0 inG. We look at thefollowing decomposition: A0 = P1 t . . .tPj−1, where for any 1 ≤ s ≤ j − 1we have

Ps := A ∈ A0 : [2, s] ⊂ A, s+ 1 /∈ A.Analogously, we consider the decomposition for B0 : B0 = Q1 t . . . t Qj−1,where for any 1 ≤ s ≤ j − 1 we have

Qs := B ∈ B0 : B ∩ [2, s] = ∅, s+ 1 ∈ B

.We claim that for each s there is a matching of Ps into Qs. Indeed,

G|Ps,Qs is a biregular graph with |Ps| =(m−s−1a−s

)≤(m−s−1a−1

)= |Qs|, therefore,

it has a matching exhausting the smaller part by the Konig-Hall theorem.The inequality between two binomial coecients holds since m − s − 1 ≥(a− s) + (a− 1) = 2a− s− 1.

Since Qs for dierent s are disjoint, combining matchings of Ps into Qs

we get a matching of A0 in B0. Thus the biggest independent set in G is B0

and, therefore, |B′ \ L|+ |A′ \ A1| ≤ |B0| =(m−1a

)−(m−ja

), yielding

66

|A′|+ |B′| ≤(m− 1

a

)−(m− ja

)+

(m− 1

a− 1

)+

(m− ja− j

)=(

m

a

)+

(m− ja− j

)−(m− ja

).

The proof is complete.

Remark. Actually, Lemma 1.38 was proved in a more general form in[113]. However, the proof that we presented here is shorter and more ele-mentary.

To conclude the proof of the theorem in this case one has to notice thefollowing. Recall that |B| ≥

(n−1k−1

)+(n−ik−i+1

). Consider the cross-intersecting

families B(12),A(12). We remark that |A| =(n−2k−2

)+ |A(12)| and |B| =(

n−1k−1

)+ |B(12)|. Both families are subsets of

([3,n]k−1

), moreover, we know that

both |B(12)|, |A(12)| ≥(n−ik−i+1

). Applying Lemma 1.38 with m = n− 2, a =

k − 1 and j = i− 2, we get that

|B(12)|+ |A(12)| ≤(n− 2

k − 1

)+

(n− i

k − i+ 1

)−(n− ik − 1

).

Therefore, |B| + |A| ≤(n−1k−1

)+(n−2k−2

)+(n−2k−1

)+(n−ik−i+1

)−(n−ik−1

)= 2(n−1k−1

)+(

n−ik−i+1

)−(n−ik−1

). We know that |A| ≤

(n−1k−1

)−(n−ik−1

). Knowing the bound

on |A|+ |B| above, it follows that the product |A||B| is the biggest if |A| ismaximum possible and |B| = max(|A|+ |B|)−max |A|, which gives exactly|B| =

(n−1k−1

)+(n−ik−i+1

)and |A| =

(n−1k−1

)−(n−ik−1

). The proof in this case is

complete.

Remark. The argument above and Lemma 1.38 together show that thebound in (1.10) actually decreases as i decreases.

Case 2. |A| ≤(n−3k−3

). In this case

|A||B| ≤(n

k

)(n− 3

k − 3

)<

(n− 1

k − 1

)(n− 2

k − 2

)<

(n− 1

k − 1

)((n− 1

k − 1

)−(n− ik − 1

)).

The right hand side is obviously less than the right hand side of (1.10). Thesecond inequality follows from (1.35) with i = 3. The proof of (1.10) in thiscase is complete.

67

Case 3.(n−3k−3

)≤ |A| ≤

(n−2k−2

). For this case we are going to pass to

the complements of sets from A,B and to change from cross-intersecting tocross-union families. Set l = n− k. Note that 2l ≥ n. We may assume thatboth A,B consist of the initial segments of l-sets in the colex order.

First we verify that when |A| =(n−2k−2

)=(n−2l

), the inequality (1.10)

holds. Indeed, then |B| ≤(n−1l

)+(n−2l−1

). At the same time, the right hand

side of (1.10) is the smallest when i = 3 and then it is strictly bigger than

f :=(n−1k−1

)((n−2k−2

)+(n−3k−2

))=(n−1l

)((n−2l

)+(n−3l−1

)). Therefore, in this case

f − |A||B| ≥(n− 1

l

)(n− 3

l − 1

)−(n− 2

l

)(n− 2

l − 1

)=((n− 1)l

l(n− 2)− 1)(n− 2

l

)(n− 2

l − 1

)> 0. (1.37)

Next we pass to the case when |B| =(n−1l

)+(n−2l−1

)+(xl−2

)for some

x ≥ l − 2. We remark that x ≤ n − 3 since |A| ≥(n−3l

). In this case, by

(1.14), we have

|A| ≤(n− 2

l

)−(

x

n− l − 2

).

Indeed, A contains sets from(

[n−2]l

)only, while B contains all sets from(

[n−1]l

), all sets from

n ∪ F : F ∈

([n−2]l−1

), as well as

(xl−2

)sets from

n−1, n∪F : F ∈(

[n−2]l−2

). Let us denote the family of sets B\n−1, n :

n − 1, n ⊂ B ∈ B by B(n − 1, n). We have |B(n − 1, n)| =(xl−2

).

We think of this family and the family A as families of subsets of [n − 2].These families are cross-union in [n − 2]. Therefore, A ∩

[n − 2] − S :

S ∈ S(n−l−2)(B(n − 1, n))

= ∅, which, together with (1.14), implies thedisplayed inequality above.

To verify (1.10) in this case it is sucient for us to prove that(x

l − 2

)(n− 2

l

)≤(

x

n− l − 2

)(n− 1

l

). (1.38)

If we do so, then the value of the product for any n − 3 ≥ x ≥ l − 2 is notbigger than the value of the product, calculated for the case |A| =

(n−2l

),

which, in turn, is smaller than the right hand side of (1.10).It is easy to see that, since l − 2 ≥ n − l − 2, the function

(xl−2

)/(

xn−l−2

)is a monotone increasing function. Therefore, it is sucient to verify (1.38)for x = n− 3. In that case (1.38) transforms into

(n−3l−2

)(n−2l

)≤(n−3l−1

)(n−1l

),

which is true since l−1n−2 ·

ln−1 <

ln−2 .

68

Case 4.(n−2k−2

)≤ |A| ≤

(n−2k−2

)+(n−ik−i+1

). Before starting the proof in this

case we remark that for i = k+1 this case is not necessary, since it is coveredby cases 1 and 3. Since for i = 3 the right hand side of (1.10) is the smallestand the interval of values for |A| is the largest, we may w.l.o.g. assume thati = 3.

We remark that for |A| =(n−2l

)+(n−3l−1

)we have |B| =

(n−1l

)+(n−3l−1

)and we

obtain exactly the bound (1.10). Now assume that |B| =(n−1l

)+(n−3l−1

)+(xl−2

).

Then, again using (1.14), we get that |A| ≤(n−2l

)+(n−3l−1

)−(

xn−l−2

). If

x ≤ n− 4, then(

xn−l−2

)≥(xl−2

)and, therefore, the product |A||B| is smaller

than the right hand side of (1.10).The last remaining case is that |A| ≤

(n−2l

)+(n−3l−1

)−(n−4n−l−2

)=(n−2l

)+(

n−4l−1

). In this case we have |A||B| ≤

((n−2l

)+(n−4l−1

))((n−1l

)+(n−2l−1

)). We

claim that this value is less than the value of the right hand side of (1.10)for i = 3:((n− 2

l

)+

(n− 4

l − 1

))((n− 1

l

)+

(n− 2

l − 1

))≤((n− 1

l

)+

(n− 3

l − 1

))((n− 2

l

)+

(n− 3

l − 1

)).

Indeed, we just check for any j1, j2 ∈ 1, 2 that the product of the j1-thsummand from the rst bracket and the j2-th summand from the secondbracket in the left hand side is not bigger than the corresponding product inthe right hand side. For j1 = j2 = 1 it is obvious and for j1 = 1, j2 = 2 it isshown in (1.37). As for the rest, we have(

n− 4

l − 1

)(n− 1

l

)≤ (n− 3)(n− l − 1)

(n− l − 2)(n− 1)

(n− 4

l − 1

)(n− 1

l

)=

(n− 3

l − 1

)(n− 2

l

)since ((n−1)−2)(n− l−1)−(n−1)((n− l−1)−1) = n−1−2(n− l−1) =2l − n+ 1 > 0, and(

n− 4

l − 1

)(n− 2

l − 1

)≤ (n− 3)(n− l − 1)

(n− l − 2)(n− 2)

(n− 4

l − 1

)(n− 2

l − 1

)=

(n− 3

l − 1

)(n− 3

l − 1

)since ((n−2)−1)((n−l−2)+1)−(n−2)(n−l−2) = n−2−(n−l−2)−1 =l− 1 > 0. The last inequality is due to the well-known fact that

(mb

)is a log-

concave function of m, but to make the argument self-contained, we includedthe proof. The proof of the theorem is complete.

69

1.8. Proof of Theorem 1.15

1.8.1. Preliminaries

We start with the following auxiliary statement, which is of independentinterest. Consider a bipartite graph H with parts V1, V2 and a group ofautomorphisms Γ, where Γ is acting on V1 and V2 but respects the parts(that is, ∀γ ∈ Γ and v ∈ Vi γ(v) ∈ Vi holds). Let T1, . . . , Tp and S1, . . . , Sqbe the orbits of the action of Γ, with V1 = T1t . . .tTp and V2 = S1t . . .tSq.Lemma 1.39. There exists an independent set I in H of maximal cardinal-ity, such that for any i either I ⊃ Ti or I ∩ Ti = ∅ and for any j eitherI ⊃ Sj or I ∩ Sj = ∅.

Make an auxiliary bipartite graph W on the set of vertices T1, . . . , Tpand S1, . . . , Sq and with edges between Ti and Sj i there is at least oneedge connecting a vertex of Ti to a vertex of Sj. Put weights |Ti|, |Sj| on thevertices Ti, Sj, correspondingly.

Let B ⊂ V1∪V2 be an independent set in H and dene βi := |B∩Ti|/|Ti|,βp+j := |B ∩ Sj|/|Sj| for i = 1, . . . , p, j = 1, . . . , q.

Claim 1.40. If there is an edge between Ti and Sj in W then

βi + βp+j ≤ 1. (1.39)

Proof. Just note that Γ is transitive on both Ti and Sj. Therefore, Ti, Sjinduce a biregular bipartite graph (i.e., degree is constant on each side) withnonzero degrees. The inequality (1.39) follows easily from the aforementionedregularity, since among the neighbors of a βi-fraction of the vertices from Tithere is at least a βi-fraction of the vertices from Sj.

We call any nonnegative vector (β1, . . . , βp+q) satisfying (1.39) a fractionalindependent set. Note that each vector corresponding to an independent setin W is a fractional independent set with coordinates from 0, 1.

Proof of Lemma 1.39. Take a fractional independent set v := (β1, . . . , βp+q)in W with maximal weight

∑pi=1 βi|Ti| +

∑qj=1 βp+j|Sj|. The vector u :=

(1 − β1, . . . , 1 − βp+q) is a minimal weight fractional vertex cover (that is,for each edge (Ti, Sj) ∈ W we have ui + up+j ≥ 1). It is a standard resultin combinatorial optimization that there exists a minimal weight fractionalvertex cover u in W with integral coordinates (see, e.g., [178]). Thus, thereexists a maximal weight fractional independent set v in W with all coordi-nates integral. This fractional independent set corresponds to the desiredindependent set in the graph H.

70

1.8.2. Proof of Theorem 1.15

We assume that s ≥ 2, as the case s = 1 is covered by Theorem 1.14. Recallthat C = C ⊂

([n]k

): |C ∩ [k]| ≥ s. It is not hard to see that if A,B are

s-cross-intersecting, then so are Si,j(A), Si,j(B). Therefore, we may w.l.o.g.assume that A,B are shifted and thus both contain [k]. It is obvious thatin any case |A| + |B| ≤ 2|C|, since A,B ⊂ C. Consider a bipartite graphG = (C−[k], C−[k], E) with two copies C1, C2 of C−[k] as parts andwith two sets C1, C2 ∈ C from dierent parts being connected by an edge ifand only if |C1∩C2| < s. Our goal is to show that the maximum independentset inG has size |C|−1. Indeed, it is clear that |A−[k]|+|B−[k]| ≤ α(G)and in that case we get the desired bound |A|+ |B| ≤ |C|+ 1.

One way to show that G does not have independent sets larger than itsparts is to exhibit a perfect matching in G. We were unable to do it in gen-eral. We are going to circumvent this problem by doing something dierent.We cover the vertices of a certain weighted graph directly related to G bydisjoint edges and paths of even vertex length in such a way that the totalweight of any independent set in any of these graphs is at most half of thetotal weight of the set of vertices of that subgraph.

We consider the following group of isomorphisms Γ, acting on [n] (andon the vertex set of G). Γ is a product of two groups, one consists of allpermutations of [k] and the other one consists of all permutations of [k+1, n].Each of the parts of G splits into orbits Cji , where j = 1, 2 and correspondsto the index of the part, and i = s, . . . , k − 1 and indicates the size of theintersection of sets from Cji with [k]: Cji = C ∈

([n]k

): |C ∩ [k]| = i. We

refer to these orbits as Ci if we think of them as set families.Take an independent set B in G of largest size. Using Lemma 1.39, we

may w.l.o.g. assume that for each j, i either B ⊃ Cji or B ∩Cji = ∅. Consider

an auxiliary weighted graph W as in the proof of Lemma 1.39. That is, thevertices of W are Cji and the vertices from dierent parts are connected ithere are two sets from the corresponding orbits that intersect in less than selements. We also put weights on each vertex of W equal to the cardinalityof the corresponding orbit. Thus, in view of Lemma 1.39, it is enough for usto show that one part of W has the same weight as the heaviest independentset in W .

Put n = 2k − s + 1 + l, where l ≥ 0. It is easy to check that C1i is

connected to C2t with t ∈ k− i− l, . . . , k− i+ s− 1∩s, . . . , k− 1. Next

71

we study the weights of the interconnected vertices. We call the value of imeaningful if the resulting indices of all families depending on i lie in the sets, . . . , k − 1. We need the following technical lemma.

Lemma 1.41. 1. |Ci| ≥ |Ck−i+s−1| i s ≤ i ≤ (k + s− 1)/2.2. |Cbk−l2 c−i

| ≤ |Cbk+s−12 c+i| for all meaningful positive integer values of i.

Proof. The proof of the lemma is just a careful algebraic manipulation. Weremark that for the rst reading one may omit all the integer parts in thecomputations below to make the verication easier.

1. Assume i ≤ (k + s − 1)/2. We have |Ci| =(ki

)(n−kk−i), |Ck−i+s−1| =(

kk−i+s−1

)(n−ki−s+1

). Therefore,

|Ci||Ck−i+s−1|

=(i− s+ 1)!2(k − i+ s− 1)!(n− k − i+ s− 1)!

(k − i)!2i!(n− 2k + i)!=

(k − i+ s− 1) · · · (k − i+ 1)

i · · · (i− s+ 2)· (n− k − i+ s− 1) · · · (n− 2k + i+ 1)

(k − i) · · · (i− s+ 2)≥ 1,

because k−i+s−1 ≥ i since i ≤ (k+s−1)/2, and n−k−i+s−1 ≥ k−isince n ≥ 2k − s+ 1.

2. Assume rst that bk−l2 c ≥ dk−s+1

2 e. We have

|Cbk+s−12 c+i||Cbk−l2 c−i

|=

(dk+l2 e+ i)!2(bk−l2 c − i)!(n− k − d

k+l2 e − i)!

(dk−s+12 e − i)!2(bk+s−1

2 c+ i)!(n− k − dk−s+12 e+ i)!

=

(bk−l2 c − i) · · · (dk−s+1

2 e − i+ 1)

(bk+s−12 c+ i) · · · (dk+l

2 e+ i+ 1)·

(dk+l2 e+ i) · · · (dk−s+1

2 e − i+ 1)

(n− k − dk−s+12 e+ i) · · · (n− k − dk+l

2 e − i+ 1)=: (∗).

We note that n = 2k − s + 1 + l, and therefore n − k − dk−s+12 e + i =

bk−s+12 c+ l + i ≤ (dk+l

2 e+ i)− ds−1−l2 e.

(∗) ≥(bk−l2 c − i) · · · (d

k−s+12 e − i+ 1)

(bk+s−12 c+ i) · · · (dk+l

2 e+ i+ 1)·

(dk+l2 e+ i) · · · (dk+l

2 e+ i− d s−1−l2 e+ 1)

(n− k − dk+l2 e+ d s−1−l

2 e − i) · · · (n− k − dk+l2 e − i+ 1)

In the rst fraction the number of factors in both denominator and numeratoris the same and is bk−l2 c − d

k−s+12 e, which is at most ds−1−l

2 e. The ratio of

the j + 1-st factors in the rst fraction, that is, the expression bk−l2 c−i−jbk+s−12 c+i−j

for j ∈ 0, . . . , ds−1−l2 e − 1 is at most 1 and thus does not increase as j

increases.

72

In the second fraction the number of factors in both the numerator anddenominator is ds−1−l

2 e. Similarly to the rst fraction, the expression

dk+l2 e+ i− j

n− k − dk+l2 e+ ds−1−l

2 e − i− jfor j ∈ 0, . . . , ds−1−l

2 e − 1 is at least 1 and thus does not decrease as jincreases. Therefore, the product of these two fractions is at least

(dk−s+1

2 e − idk+l

2 e+ i

)d s−1−l2 e(dk+l

2 e+ i

n− k − dk+l2 e+ ds−1−l

2 e − i

)d s−1−l2 e

=

(dk−s+1

2 e − ik − s+ l + 1− dk+l

2 e+ ds−1−l2 e − i

)d s−1−l2 e

≥ 1.

The last inequality holds since without integer parts the denominator isequal to the enumerator, and the possible gain because of the integer partsin the denominator is at most 1/2. However, both numbers are integer, thusthe denominator is not bigger than the enumerator.

Assume now that bk−l2 c < dk−s+1

2 e.

|Cbk+s−12 c+i||Cbk−l2 c−i

|=

(dk+l2 e+ i)!2(bk−l2 c − i)!(n− k − d

k+l2 e − i)!

(dk−s+12 e − i)!2(bk+s−1

2 c+ i)!(n− k − dk−s+12 e+ i)!

=

(dk+l2 e+ i) · · · (bk+s−1

2 c+ i+ 1)

(dk−s+12 e − i) · · · (bk−l2 c − i+ 1)

·(dk+l

2 e+ i) · · · (dk−s+12 e − i+ 1)

(n− k − dk−s+12 e+ i) · · · (n− k − dk+l

2 e − i+ 1)=: (∗).

We note that n = 2k − s + 1 + l, and therefore n − k − dk−s+12 e + i =

bk−s+12 c+ l + i ≤ (dk+l

2 e+ i) + b l−s+12 c.

(∗) ≥(dk+l

2 e+ i) · · · (bk+s−12 c+ i+ 1)

(dk−s+12 e − i) · · · (bk−l2 c − i+ 1)

·

(dk−s+12 e+ b l−s+1

2 c − i) · · · (dk−s+1

2 e − i+ 1)

(n− k − dk−s+12 e+ i) · · · (n− k − dk−s+1

2 e − b l−s+12 c+ i+ 1)

.

Bounding the two fractions similarly to how it was done in the previous case,one can see that this expression is at least(

dk+l2 e+ i+ 1

dk−s+12 e − i+ 1

)b l−s+12 c(

dk−s+12 e − i+ 1

n− k − dk−s+12 e − b l−s+1

2 c+ i+ 1

)b l−s+12 c

=

73

(dk+l

2 e+ i+ 1

k − s+ l + 1− dk−s+12 e − b l−s+1

2 c+ i+ 1

)b l−s+12 c

≥ 1.

As in the previous case, in the last fraction the denominator is equal to theenumerator if one removes all integer parts, and as in the previous case weconclude that the last inequality is valid.

The next crucial step is to decompose the graphW into symmetric chainsof even length. Let (j, j) = (1, 2) or (2, 1). We know that Cji is connectedto C jk+s−1−i for all i. We call these edges edges of the rst type. Next, Cji isconnected to C ji if dk−l2 e ≤ i < k+s−1

2 .We call these edges edges of the second

type. Finally, Cjbk−l2 c−iis connected to C jbk+s−12 c+i for each meaningful i ≥ 1

and s ≥ 2. We call these edges edges of the third type.We claim that the graph W is decomposed into paths of even length

using the three types of edges above. See Fig. 1, where the two parts arerepresented by two horizontal lines of points, and the values near the pointsindicate the size of the intersection of the sets of the corresponding familywith [k]. The arrows on the edges indicate the direction from the vertex ofsmaller weight to the vertex of larger weight. The black two-directional edgesare the edges of the second type and connect the vertices of the same weight.The vertical edges are the edges of the rst type, and the other edges are theedges of the third type.

The middle vertical line corresponds to the size of the intersection with[k] equal to k+s−1

2 in both parts. The two dashed lines correspond to theintersection size dk−l2 e for one of the parts. Between these two lines eachvertex has an edge of the second type. Note that the starting vertex of theedges of the third type is always outside the region bounded between thetwo dashed lines. The edges of the third type may or may not intersect thedashed lines, but they never intersect the black line.

74

Figure 1. Decomposition of the graph W into symmetric chains.

In the middle of each path P there is an edge of the second type, connect-ing two vertices of the same weight. Due to Lemma 1.41, for each path theweight of the vertex does not decrease as we move along the path towardsthe middle edge of the second type. Therefore, it is easy to see that anyindependent set in P has weight less than or equal to the half of the weightof P . This means that any independent set in W cannot have weight biggerthan the half of the weight of W . This concludes the proof of Theorem 1.15.

1.9. t-intersecting, cross s-intersecting families

Let us start with the following denition.

Denition 1.42. For k ≥ 2s− t, 0 ≤ i < s− t dene

Mi(k, s, t) := A ⊂ [k] : |A| ≥ s, |A∩[t+2i]| ≥ t+i∪A ⊂ [k] : |A| ≥ k−s+t.

Note that A ∈(

[k]s

): A ∈ Mi(k, s, t) = Ai(k, s, t) for k > 2s − t and

thatMi(k, s, t) is t-intersecting.For xed k, s, t and 0 ≤ i < s− t let us dene the pair Ai := A ∈

([n]k

):

A ∩ [k] ∈ Mi(k, s, t), Bi := [k]. Then these non-empty t-intersectingfamilies are cross s-intersecting.

Note that As−t(k, s, t) =(

[2s−t]s

). For i = s− t we dene

As−t :=A ∈

([n]

k

): |A ∩ [2s− t]| ≥ s

and

Bs−t :=B ∈

([n]

k

): [2s− t] ⊂ B

.

75

Again the non-empty t-intersecting familiesAs−t and Bs−t are cross s-intersecting.

With this terminology we prove

Theorem 1.43. Let k > s > t ≥ 1 be integers, k ≥ 2s − t. Supposethat A,B ⊂

([n]k

)are non-empty t-intersecting families which are cross s-

intersecting. Then for n ≥ n0(s, t, k) we have

maxA,B

|A|+ |B|

= max

0≤i≤s−t

|Ai|+ |Bi|

. (1.40)

Moreover, unless k = 2s, t = 1 or A,B are isomorphic to Ai,Bi, the equalityabove transforms into a strict inequality.

We note that the conclusive result for t = 0 was obtained by the authorsin the paper [92], and therefore we do not consider this case here.

Remark. The assumption k ≥ 2s − t is in fact not restrictive, since ifA ∈ A and B ∈ B then any two sets from A intersect in at least 2s − k

elements inside B (and analogously for B and A). If 2s − k ≥ t, then thet-intersecting condition is implied by the cross s-intersecting condition, andit reduces to the case studied in [92].

We also note that the problem makes sense only in case s > t. Otherwise,we may just take both A and B to be the same t-intersecting family ofmaximal cardinality. Thus, in case t ≥ s the problem reduces to a trivialapplication of the Complete Intersection Theorem (Theorem 1.17).

1.9.1. Proof of Theorem 1.43

We may assume that the families A,B are shifted and, thus, both contain[k]. Recall that A(T ) := A\T : A ∈ A, A ⊃ T. We call a set T ∈

([n]s

)a

kernel for a set family S, if S(T ) contains k + 1 pairwise disjoint edges. Aninequality proved in [80] states that

|A(T )| ≤ k

(n− s− 1

k − s− 1

), (1.41)

if T is not a kernel for A, and similarly for B(T ).The following claim is an easy application of the pigeon-hole principle.

Claim 1.44. Suppose that C0, . . . , Ck form a sunower with center T , i.e.Ci ∩ Cj = T for all 0 ≤ i < j ≤ k. Suppose D is a k-element set. Thenthere exists an i such that D ∩ Ci = D ∩ T holds.

76

We immediately get the following corollary:

Corollary 1.45. If |T | = s and |D∩Ci| ≥ s for all 0 ≤ i ≤ k, then T ⊂ D.

Lemma 1.46. If T1 and T2 are kernels for A and B respectively, then T1 =T2 holds.

Proof. From Corollary 1.45 it follows that for any B ∈ B we have T1 ⊂ B.Applying it for B1, B2 ∈ B, B1 ∩ B2 = T2, we get that T1 ⊂ B1 ∩ B2 = T2,but since |T1| = |T2|, we have T1 = T2.

From Corollary 1.45 and Lemma 1.46 it follows that if both A and Bhave kernels, then |A| + |B| ≤ 2

(n−sk−s), which is smaller then the bound in

(1.40). From now on we may w.l.o.g. assume that B does not have a kerneland that B contains the set [k] as an element. Then any kernel of A mustbe a subset of [k]. Let J ⊂

([k]s

)be the family of kernels of A. Dene

A := A ∈ A : @T ∈ J , A ⊃ T.

Claim 1.47. We have |A| = O(nk−s−1). Analogously, we have |B| =O(nk−s−1).

Proof. Any set from A must intersect [k] in at least s elements. Thus,|A| ≤

∑T∈([k]

s ) |A(T )|, which by inequality (1.41) is at most(

[k]s

)k(n−s−1k−s−1

)=

O(nk−s−1). The same proof works for B.

Due to Claim 1.44 and the fact that A is t-intersecting, for any set A ∈ Aand any T ∈ J we have |A ∩ T | ≥ t. Moreover, repeating the proof ofLemma 1.46, it is easy to see that for any T1, T2 ∈ J we have |T1 ∩ T2| ≥ t.Therefore, |J | ≤ AK(k, s, t) (see Theorem 1.17). Due to Claim 1.47, wehave

|A|+ |B| ≤ |J |(n− kk − s

)+O(nk−s−1). (1.42)

Indeed, the only additional thing one has to note is that the number of setsfrom A that intersect [k] in at least s+ 1 elements is O(nk−s−1).

On the other hand, it is easy to exhibit an example of a family A that is t-intersecting and is cross s-intersecting with [k], and which has cardinalityAK(k, s, t)

(n−kk−s). For that one just has to take a maximum t-intersecting

family J ′ of s-element sets in [k] and put A = A ∈(

[n]k

): ∃T ∈ J ′ :

A ∩ [k] = T.Therefore, if |J | < AK(k, s, t), then by (1.42) and the previous con-

struction |A| + |B| cannot be maximal for large n. Therefore, from now

77

on we may assume that J = Ai(k, s, t) for some 0 ≤ i ≤ s − t, where|Ai(k, s, t)| = AK(k, s, t).

We remark that out ofAj(k, s, t) the only family that satises |Aj(k, s, t)| =|Aj(k + 1, s, t)| is the family As−t(k, s, t). Indeed, it consists of all the setsthat have all their s elements among the rst 2s − t elements and does notuse elements j with j > 2s− t. For all the other families, as we have alreadymentioned, the degree of each j ∈ [k] is positive.

Suppose rst that 0 ≤ i < s − t and that |As−t(k, s, t)| < AK(k, s, t).Most importantly for us, it means that AK(k− 1, s, t) < AK(k, s, t) due tothe discussion in the previous paragraph. Then it is easy to see that |B| = 1.Indeed, if B1, B2 ∈ B, then due to Corollary 1.45 we know that T ⊂ B1∩B2

for each T ∈ J . Since |B1 ∩ B2| ≤ k − 1, we have |J | ≤ AK(k − 1, s, t) <AK(k, s, t).

We showed that the extremal pair of families satises B = [k]. Tocomplete the proof in this case and to show that the pair of extremal familiesis one of Ai,Bi for 0 ≤ i < s− t, we have to verify that A∩ [k] : A ∈ A ⊂Mi(k, s, t) provided that J = Ai(k, s, t).

Take any set A such that A /∈ Mi(k, s, t), that is, |A ∩ [k]| < k − s + t

and |A ∩ [t + 2i]| < t + i. The following claim completes the proof in thecase AK(k, s, t) is attained onMi(k, s, t) with i < s− t.

Claim 1.48. In the notations above, there is T ∈ J = Ai(k, s, t) such that|T ∩ A| ≤ t− 1.

Proof. W.l.o.g., let A = [1, l] ∪ [m + 1, k], where l ≤ t + i− 1 and l + m <k − s + t. Set T = [l′, l′ + s − 1] with l′ ≤ i + 1 and with l′ chosen in sucha way that |T ∩ A| is minimized. It is easy to see that, rst, T ∈ J and,second, |T ∩ A| = maxl − i, l +m+ s− k < t.

If the only maximal family is As−t(k, s, t), then J = As−t(k, s, t) andA cannot contain a set A that satisfy |A ∩ [2s − t]| < s. Indeed, then Aintersects one of the T ∈ J in less than t elements, which by Claim 1.44means that there exists A′ ∈ A such that |A′ ∩ A| < t, a contradiction.Therefore, the family A is contained in As−t, and we are only left to provethat B ⊂ Bs−t. Assume that B contains B, such that B + [2s− t]. But thenthere is T ∈ J such that |B ∩ T | ≤ s− 1, a contradiction.

The last case that remains to be veried is when both As−t(k, s, t) andAs−t−1(k, s, t) have size AK(k, s, t). But either J = As−t−1(k, s, t) or J =

78

As−t(k, s, t) and we end up in one of the two scenarios described above.It is clear from the proof that the families that we constructed are the

only possible extremal families (excluding the case n = 2k, s = 1), which ismostly due to the uniqueness in Theorem 1.17.

79

Chapter 2.

Intersecting families of vectors

My results presented in this chapter (joint with P. Frankl) are published in[93, 97, 99].

2.1. Introduction and statement of results

Recall that any subset of the power set 2[n] is called a family. One way oflooking at families is to associate with each set R ⊂ [n] its characteristicvector v(F ) := (x1, . . . , xn) with xi = 1 for i ∈ F and xi = 0 for i /∈ F .Then each family corresponds to a collection of 0, 1-vectors, or simply asubset of the Boolean cube.

This association of a family F with a family of vectors v(F ) : F ∈ Fprovides a fruitful connection between some geometric problems in Rn andfamilies of subsets with restrictions on sizes of pairwise intersections. The bynow classical result of Frankl and Wilson [114] is a good example.

Theorem 2.1 ([114]). If n is a prime power and F ⊂(

[4n−1]2n−1

)satises

|F1 ∩ F2| 6= n− 1 for any F1, F2 ∈ F , then the size of F is at most(

4n−1n−1

).

Note that(

4n−1n−1

)is exponentially smaller than

(4n−12n−1

). Due to that, this

theorem has several important implications for the chromatic number of thespace, as well as some results in Euclidean Ramsey theory. Kahn and Kalai[134] gave it a twist to deduce counterexamples to the famous Borsuk con-jecture. These questions are discussed in more details in Chapter 6.

In this chapter, we work with the families of 0,±1-vectors, that is,vectors v := (v1, . . . , vn), where each vi is 0, 1, or −1. Probably the rstnon-trivial extremal result concerning these objects was a result of Deza andFrankl [60] showing that in a certain situation one can prove the same bestpossible upper bound for 0,±1-vectors as for the restricted case of 0, 1-vectors. Raigorodskii [209] succeeded in improving the bounds in the afore-

80

mentioned geometric problems by enlarging the scope of vectors from 0, 1-vectors to 0,±1-vectors. Some further applications of 0,±1-vectors insuch questions one may nd in [48], [162], [195], [205], [209], [215]. Motivatedby this, I together with Frankl initiated a more systematic investigation ofextremal questions for families of 0,±1-vectors. This chapter deals withthe extensions of the classical ErdosKoRado-type theorems (cf. Chapter 1)to this setting.

More precisely, I cover two types of results. The results of the rst typegeneralize the ErdosKoRado theorem to the families of 0,±1-vectorswith a xed number of +1-coordinates and −1-coordinates. The results ofthe second type deal with the families of 0,±1-vectors with a xed numberof non-zero coordinates, that is, of xed length.

2.1.1. Vectors with a xed number of +1's and −1's

We start by dening of the class of vectors that we work with.

Denition 2.2. For 0 ≤ l, k < n dene V(n, k, l) ⊂ Rn to be the set of all0,±1-vectors having exactly k +1-coordinates and l −1-coordinates.

We tacitly assume that n ≥ k + l and also k ≥ l. Indeed, for l > k onecan replace a family V of vectors by −V := −v : v ∈ V and have thenumber of +1's prevail. Note that

|V(n, k, l)| =(n

k

)(n− kl

). (2.1)

With this notation families of k-sets are just subsets of V(n, k, 0).For vectors v,w their scalar product, is denoted by 〈v,w〉. If both v

and w are 0, 1-vectors, then their scalar product is non-negative with〈v(F ),v(G)〉 = 0 i F ∩G = ∅.

For n ≥ 2k the minimum possible scalar product in V(n, k, l) is −2l. Sucha scalar product is achieved on a pair of vectors i the −1's in each of themstand on the positions of +1's in the other, and no two +1's stand on thesame coordinate position.

Denition 2.3. A family V ⊂ V(n, k, l) of vectors is called intersecting ifthe scalar product of any two vectors in V is strictly larger than the minimumscalar product in V(n, k, l).

By analogy with the ErdosKoRado Theorem, dene

m(n, k, l) := max |V| : V ⊂ V(n, k, l),V is intersecting.

81

With this terminology the ErdosKoRado Theorem can be stated as

m(n, k, 0) =

(n− 1

k − 1

)for n ≥ 2k.

My rst contribution in this chapter is described in two theorems, whichtogether give the exact value of m(n, k, 1) for all n, k. One surprising fact isthat the situation is very dierent from the case l = 0, that is, the ErdosKoRado Theorem. Namely, while for n ≤ k2 the ErdosKoRado-typeconstruction is optimal, it is no longer optimal for n > k2.

Denition 2.4. For n ≥ 2k dene

E(n, k, l) := (x1, . . . , xn) ∈ V(n, k, l) : x1 = 1.

Clearly, E(n, k, l) is intersecting with

|E(n, k, l) =

(n− 1

k − 1

)(n− kl

)=k

n|V(n, k, l)|. (2.2)

Theorem 2.5 ([97]). For 2k ≤ n ≤ k2

m(n, k, 1) = k

(n− 1

k

)holds. (2.3)

The proof of (2.3) is much simpler in the range 2k ≤ n ≤ 3k − 1. It canbe done using Katona's Circle Method (cf. [139]) and follows from a moregeneral Theorem 2.8. For n ≥ 3k the proof of (2.3) is much harder and muchmore technical. It is presented in Section 2.2.5.

Next, we discuss the case n > k2. Suppose that V ⊂ V(n, k, 1) is inter-secting. Consider the family P(V) ⊂ V(n+ 1, k, 1), dened as follows:

P(V) := (v, 0) : v ∈ V ∪ (v,−1) : v ∈ V(n, k, 1).

It is easy to see that P(V) is intersecting. This implies the following inequal-ity:

m(n+ 1, k, 1) ≥ m(n, k, 1) +

(n

k

). (2.4)

The next theorem states that the bound (2.4) is tight.

Theorem 2.6 ([97]). For n ≥ k2 one has

m(n+ 1, k, 1) = m(n, k, 1) +

(n

k

). (2.5)

82

As an immediate consequence we obtain the following.

Corollary 2.7. For n > k2 one has

m(n, k, 1) = m(k2, k, 1) +

(k2

k

)+

(k2 + 1

k

)+ . . .+

(n− 1

k

).

The proof of Theorem 2.6 is presented in Section 2.2.4, and Section 2.2contains the necessary preparations.

In the case l > 1 the problem becomes much harder in general, andnding the exact value of m(n, k, l) is still wide open. However, togetherwith Frankl [99], we have obtained progress in some cases. First, we havecompletely resolved the case 2k ≤ n ≤ 3k − l.

Theorem 2.8 ([99]). If k > l ≥ 1 and 2k ≤ n ≤ 3k − l, then

m(n, k, l) = |E(n, k, l)|. (2.6)

This theorem is proved in Section 2.4. We have also solved the problemasymptotically for growing n and xed k.

Example. Let G ⊂ V(n, k, l) consist of those vectors whose last non-zero coordinate is a −1. Then |G| =

(nk+l

)(k+l−1l−1

), and it is easy to see that G

is intersecting. One can also see that G is obtained from V(k+ l−1, k, l−1)by applying operation P n− k − l + 1 times.

Theorem 2.9 ([99]). If k > l ≥ 1 then we have

m(n, k, l) ≤(

n

k + l

)(k + l − 1

l − 1

)+

(n

2l

)(2l

l

)(n− 2l − 1

k − l − 1

). (2.7)

Note that the last term of (2.7) is O(nk+l−1), while the rst term isΩ(nk+l). We also put

(n−2l−1k−l−1

):= 0 for k = l. Thus (2.7) shows that

Example above is asymptotically best possible. Theorem 2.9 is proved inSection 2.3.

We conclude the section by analyzing the two simple cases that are ex-cluded above.

First, assume that k = l. Then a family V ⊂ V(n, k, l) is intersecting ifand only if it contains no two vectors v,w such that vi = 0 i wi = 0 andvi = ±1 i wi = ∓1. Thus, V(n, k, l) is partitioned into such pairs, and V isintersecting i it contains at most one vector out of each pair.

Second, assume that k+ l ≤ n ≤ 2k. Then V ⊂ V(n, k, l) is intersecting ifand only if it contains no two vectors v,w such if vi ∈ 0,−1 then wi = 1

83

and if wi ∈ 0,−1 then wi = 1. Consider the following family of setsF ⊂

([n]n−k):

F :=R ⊂ [n] : R = [n] \ i : vi = 1 for some v ∈ F

.

Then it is not dicult to see that V is intersecting i F is intersecting, andthus the case is reduced to the ErdosKoRado theorem (Theorem 1.2).

2.1.2. Vectors of xed length

Denote by Lk the family of all vectors v from 0,±1n such that 〈v,v〉 = k.Note that |Lk| = 2k

(nk

). This section is mostly devoted to the study of the

quantity below.

F (n, k, l) := max|V| : V ⊂ Lk,∀v,w ∈ V 〈v,w〉 ≥ l. (2.8)

Recall the following theorem of Katona:

Theorem (Katona, [137]). Let n > s > 0 be xed integers. If U ⊂ 2[n] is afamily of sets such that for any U, V ∈ U we have |U ∪ V | ≤ s then

|U| ≤ f(n, s) :=

s/2∑i=0

(n

i

)if s is even,

2

(s−1)/2∑i=0

(n− 1

i

)if s is odd.

(2.9)

Moreover, for n ≥ s+2 the equality is attained only for the following families.If s is even, than it is the family U s of all sets of size at most s/2. If s isodd, then it is one of the families U s

j of all sets that intersect [n]−j in atmost (s− 1)/2 elements, where 1 ≤ j ≤ n.

Given two sets U, V , we denote the symmetric dierence of these two setsby U4V , that is, U4V := U\V ∪ V \U . A theorem due to Kleitman statesthat the bound (2.9) holds for a more general class of families.

Theorem (Kleitman, [150]). If for any two sets U, V from a family U ⊂ 2[n]

we have |U4V | ≤ s, then the bound (2.9) holds for U .

Note that there is no uniqueness counterpart in Kleitman's theorem.The main result presented in this section is the following theorem, which

determines F (n, k, l) for all k, l and suciently large n and which showsthe connection of F (n, k, l) with the above stated theorems of Katona andKleitman.

84

Theorem 2.10 ([93]). For any k and n ≥ n0(k) we have

1. F (n, k, l) =

(n− lk − l

)for 0 ≤ l ≤ k.

2. F (n, k,−l) = f(k, l)

(n

k

)for 0 ≤ l ≤ k.

Note that f(k, 0) = 1 and so the values for l = 0 in part 1 and l = 0 inpart 2 coincide. Using the same technique, we may extend the result of part2 of Theorem 2.10 in the following way:

Theorem 2.11 ([93]). Let V ⊂ Lk be the set of vectors such that for anyv,w ∈ V we have 〈v,w〉 6= −l − 1 for some 0 ≤ l < k. Then we have

maxV|V| = f(k, l)

(n

k

)+O(nk−1).

These results, along with some simple observations concerning F (n, k, l)and the complete solution for the case k = 3, are presented in Section 2.5.

2.2. Proofs of Theorems 2.5 and 2.6

Throughout the subsection we assume that n ≥ 2k and that k > 1.

2.2.1. Summary

This section is meant to summarize the structure of the proof and often refersto the statements in the proof. Therefore, some parts of it may be dicult tofollow before actually reading through some parts of the proof. Thus, ratherthen giving a rough idea for the reader beforehand, it is aimed to help readersto see through the (rather complicated) details when reading the proof.

First, for the whole proof we assume that the families are shifted, i.e.,1's normally appear before 0's and −1's, and 0's appear before −1's amongthe coordinate positions of vectors from our family. The precise denitionof shifting and the proof of the fact that it preserves the property that thefamily is intersecting is discussed in Section 2.2.3.1.

We start with the sketch of the proof of Theorem 2.6. The goal is to showthat the number of the vectors containing +1 or −1 on the last coordinateis at most

(nk

). These vectors fall into two groups: A, with −1 at the end,

85

and B, with 1 at the end. We x another coordinate i < n+ 1 and take thevectors from A, denoted A(i), that have 1 on i-th position, and vectors fromB, denoted B(−i), that have −1 on i-th position.

For each such i we may treat the resulting families of vectors as two setfamilies of (k − 1)-sets, which are cross-intersecting. We remark that, dueto shifting, B(−i) ⊂ A(i) and, in particular, B(−i) is intersecting. Then webound the expression |A(i)|+k|B(−i)| from above using Theorem 2.16 fromSection 2.2.3.4. Finally we average the result over all possible i, obtaining adesired bound on |A|+ |B|. Note that the coecient k is needed so that weget the expression |A|+ |B| after averaging.

Next, we sketch the proof of Theorem 2.5. For 2k ≤ n ≤ 3k − 1 thetheorem follows from Theorem 2.8, and is obtained via a direct applicationof Katona's circle method. The general form of this method is discussed inSection 2.2.3.3. The only trick is to choose a good subset of vertices to whichwe can apply the method.

The case of 3k ≤ n ≤ k2 is the most dicult part of the proof. Again,the idea is to apply the argument with the averaging used for the proof ofTheorem 2.6. However, in this case there is a major complication. If one takesa look at Theorem 2.16 with given parameters (the size of the ground set isn−1, since i and n+1 are not present, and the size of each set in A(i),B(−i)is k−1), the maximum value of the expression |A(i)|+k|B(−i)| is (k+1)

(n−2k−2

)and is attained in the case when both are the trivial intersecting families.However, if we look at the ErdosKoRado-type family, which is expected tobe maximal in this range, and the corresponding setsA(i),B(i), then they areindeed the trivial intersecting families for all i except for i = 1. In that casewe have the other extreme: B(1) is empty and A(1) =

([n−1]k−1

). Therefore,

if we apply the bound from Theorem 2.16 for all i blindly, then after theaveraging we get a worse bound, with the dierence of (k + 1)

(n−2k−2

)−(n−1k−1

)from the size of the ErdosKoRado-type family.

The idea to circumvent it is as follows. If B(−1) is empty, then we arehome. If not, then, due to the fact that the whole family is shifted, we mayconclude that there is a relatively big set I, I ⊂ [n], such that for all i ∈ Ithere are many sets in B(−i) that do not contain element 1. The precisestatement is Proposition 2.29. The next step is to use Theorem 1.4, whichquanties how can we bound the size of the intersecting family from abovegiven that we know that it is far from trivial intersecting family, that is, ifthere are many sets that do not contain the element with the biggest degree.

86

The result of the manipulations presented in the previous paragraph isa non-trivial bound on the size of each B(−i), i ∈ I, provided that B(−1)is non-empty. Finally, we bound the size of each |A(i)| + k|B(−i)| usingTheorem 2.25 from Section 2.2.3.6, which is a rened version of Theorem2.16. Using this bound, we apply the same averaging as in Theorem 2.6 andshow that in all cases the size of |A| + |B| is at most the size of |A| + |B|when B(−1) is empty. Speaking very roughly, the sets in B(−1) force thesets B(−i) to be small, and thus force the whole sum |A|+ |B| to be small.The large part of Section 2.2.5 is devoted to the calculations that ensure thatit is indeed the case.

Theorem 2.25, which gives very ne-grained bounds on |A(i)|+ k|B(−i)|depending on the size of B(−i), is itself one of the complicated parts ofthe proof. First, we do a detailed analysis of the maximal cross-intersectingfamilies using and rening KruskalKatona Theorem in Section 2.2.3.5. Theresults of this section are interesting in their own right, as they provide abetter understanding of the structure of cross-intersecting families (we referto the papers [176], [166], where these and some complementary ideas weredeveloped). The language of truncated characteristic vectors, introduced inSection 2.2.3.5 seems to be very convenient. Lemmas 2.23, 2.24 allow us toreduce the wide array of dierent cross-intersecting families to a few, one ofwhich is guaranteed to be maximum w.r.t. the expression we maximize. Theproof of Theorem 2.25 itself is a more technical counterpart.

2.2.2. Comparing the constructions

To get some intuition for the problem, we start with the comparison of theconstructions of intersecting families briey discussed in the introduction.

The rst intersecting family E(n, k, 1) is the ErdosKoRado-type family,mentioned in the introduction, in which all the vectors have 1 on the rstposition. We have

e(n, k, 1) := |E(n, k, 1)| = k

(n− 1

k

).

Note that we have v(n, k, 1) = |V(n, k, 1)| = (k + 1)(nk+1

). Therefore, we

have e(n, k, 1)/v(n, k, 1) = k/n.The second family P(n, k, 1) consists of all the vectors for which the last

non-zero coordinate is −1. It is easy to see that this is indeed an intersecting

87

family. We have

p(n, k, 1) := |P(n, k, 1)| =(

n

k + 1

).

Therefore, we have p(n, k, 1)/v(n, k, 1) = 1/(k + 1).

Proposition 2.12. The inequality e(n+1, k, 1)−e(n, k, 1) ≥ p(n+1, k, 1)−p(n, k, 1) holds i n ≤ k2. We have equality i n = k2.

Proof. The proof is a matter of simple calculations:

e(n+ 1, k, 1)− e(n, k, 1) = k

(n

k

)− k(n− 1

k

)= k

(n− 1

k − 1

),

p(n+ 1, k, 1)− p(n, k, 1) =

(n+ 1

k + 1

)−(

n

k + 1

)=

(n

k

)=n

k

(n− 1

k − 1

).

The second construction of intersecting families allows for the followinggeneralization, described in the introduction. Assume we are given an inter-secting family F ⊂ V(n, k, 1). Recall that we can construct an intersectingfamily P(F) ⊂ V(n+ 1, k, 1) in the following way:

P(F) = (v, 0) : v ∈ F ∪ w = (w1, . . . , wn+1) : w ∈ V (n+ 1, k, 1), wn+1 = −1.

Since F is intersecting, P(F) is intersecting as well. We have

|P(F)| − |F| =(n

k

)= p(n+ 1, k, 1)− p(n, k, 1). (2.10)

We denote by Ps(F) the result of s consecutive applications of operationP(∗) to the family F . This gives us the following composite construction ofan intersecting family C(n, k, 1) ⊂ V(n, k, 1).

C(n, k, 1) =

E(n, k, 1), if n ≤ k2;

Pn−k2(E(k2, k, 1)

), if n > k2.

(2.11)

We denote the cardinality of C(n, k, 1) by c(n, k, 1). We have c(n, k, 1) =e(n, k, 1) for n ≤ k2 and, due to (2.10), c(n, k, 1) = e(k2, k) − p(k2, k) +p(n, k) for n > k2. By Proposition 2.12, C(n, k, 1) is the biggest intersect-ing family among the ones discussed in this subsection. In what follows weprove that C(n, k, 1) has maximum cardinality among intersecting familiesin V(n, k, 1).

88

Remark. Due to the fact that equality is possible in Proposition 2.12,there is a slightly dierent intersecting family that has exactly the samecardinality as C(n, k, 1). Its denition is almost the same, we only have toreplace k2 by k2 + 1 in (2.11).

2.2.3. Auxiliaries from extremal set theory

In this section we present several auxiliary results and techniques that we willuse in the latter sections. Some of the results presented here are well-known,while the others appear to be new and may be of independent interest.

2.2.3.1. Shifting

We start with shifting (left compression). We have discussed it in the contextof sets in Section 1.2.1. For a given pair of indices i < j ∈ [n] and avector v := (v1, . . . , vn) ∈ Rn we dene an (i, j)-shift Si,j(v) of v in thefollowing way. If vi ≥ vj, then Si,j(v) := v. If vi < vj, then Si,j(v) :=(v1, . . . , vi−1, vj, vi+1, . . . , vj−1, vi, vj+1, . . . , vn), that is, it is obtained from vby interchanging its i-th and j-th coordinate.

Next, we dene an (i, j)-shift Si,j(Q) of Q for a nite system of vectorsQ ⊂ Rn. We take a vector v ∈ Q and replace it with Si,j(v), if Si,j(v) isnot already in Q. If it is, then we leave v in the system. Formally,

Si,j(Q) := Si,j(v) : v ∈ Q ∪ v : v, Si,j(v) ∈ Q.We call a system Q shifted, if Q = Si,j(Q) for all i < j ∈ [n]. Any system

of vectors may be made shifted by means of a nite number of (i, j)-shifts.Here is the crucial lemma concerning shifting:

Lemma 2.13. For any Q ⊂ Rn and any i < j ∈ [n] we have

min〈v,w〉 : v,w ∈ Q ≤ min〈v′,w′〉 : v′,w′ ∈ Si,j(Q).Proof. Take any two vectors v := (v1, . . . , vn),w := (w1, . . . , wn) ∈ Q. Wedenote by v′,w′ the result of the (i, j)-shift in Q applied to v,w (that is, forv we have v′ = v or Si,j(v), depending on whether Si,j(v) is in Q or not). Ifv′ = v and w′ = w, then, obviously, 〈v′,w′〉 = 〈v,w〉. Moreover, we havethe same if both v′ 6= v and w′ 6= w. Therefore, the only nontrivial case weneed to consider is when v′ 6= v and w′ = w.

The reasons for v′ being dierent from v are unambiguous: vi < vj andSi,j(v) /∈ Q. For w′, however, there are two possible reasons not to beshifted. The rst one is that wi ≥ wj and, thus, w = Si,j(w). Then

〈v′,w′〉 − 〈v,w〉 = viwj + vjwi − viwi − vjwj = (vj − vi)(wi − wj) ≥ 0.

89

The second possible reason is that wi < wj, but Si,j(w) ∈ Q. Then

〈v′,w′〉 = 〈Si,j(v),w〉 = 〈v, Si,j(w)〉.

The last scalar product is, in fact, between two vectors from Q. Therefore,in all cases we have exhibited a pair of vectors from Q, that have a scalarproduct smaller than or equal to 〈v′,w′〉.

Applied to our case, the lemma states that, given an intersecting familyof vectors, we may replace it with a shifted family of vectors, and the shiftedfamily is intersecting as well.

2.2.3.2. Shadows

Given a family F ⊂(

[n]k

), we dene its shadow ∂(F) ⊂

([n]k−1

)as a family

of all (k − 1)-element sets that are contained in one of the sets from F (cf.Section 1.2.2). For this section, we will use the following notion, which isslightly dierent from the one given in Section 1.2.2: if l < k, then the l-thshadow ∂l(F) is the set of all (k − l)-element sets that are contained in oneof F ∈ F . Recall that the famous KruskalKatona theorem (Theorem 1.21)gives a sharp lower bound on |∂F| in terms of |F| for a family F of k-sets.We are going to discuss it in the forthcoming paragraphs.

However, we need an analogous relation for the set system and its shadow,but for sets of specic type. Fix a cyclic permutation π of [n]. Considerthe set U(π, k) ⊂

([n]k

)of n k-sets, each of which forms an interval in the

permutation π. That is, they are composed of cyclically consecutive elementsin the permutation.

Lemma 2.14. For any set system F ⊂ U(π, k) we have |∂l(F) ∩ U(π, k −l)| ≥ min|F|+ k − l, n.

Proof. It is clearly sucient to prove that |∂(F)∩U(π, k− 1)| ≥ min|F|+1, n. If |F| = |U(π, k)| = n, then ∂(F) = U(π, k − 1) and the statement isobvious. Therefore, we may assume that U(π, k) \ F 6= ∅.

Split the family F into subfamilies F1, . . . ,Fs, each of which form aninterval (or a tight path). That is, each Fi is a maximal sequence ofdierent sets F 1

i , . . . , Fdi ∈ F , in which each pair of consecutive sets intersect

in a (k − 1)-element set. Clearly, this is a partition of F into equivalenceclasses. Moreover, sets from dierent subfamilies intersect in less than k− 1elements. Therefore, ∂(F) =

⊔i ∂(Fi).

For each subfamily we have |∂(Fi)∩U(π, k−1)| = |Fi|+1. This is due tothe fact that each F ∈ Fi contains two (k−1)-element sets from U(π, k−1),

90

while each F ′ ∈ ∂(Fi) ∩ U(π, k − 1) is contained in either one or two setsfrom Fi, and there are exactly two sets that are contained in one set fromFi. Informally speaking, these are the left shadow of F 1

i and the rightshadow of F d

i . These two shadow sets are dierent, since U(π, k) \ Fi 6= ∅.Knowing the degrees of the sets, we get the desired equality by simpledouble counting.

Finally, putting the statements for dierent i together, we get that |∂(F)∩U(π, k− 1)| ≥ |F|+ s ≥ |F|+ 1. Repeating the argument l times yields theresult.

2.2.3.3. General form of Katona's circle method

For this subsection only we adopt the language of graph theory. Consider agraph G = (V,E), which is vertex-transitive. That is, the group Aut(G) ofautomorphisms acts transitively on V . For a given vertex v ∈ V we denoteby Sv the stabilizer of v in Aut(G), which is a subgroup of all automorphismsof G that map v to itself. A basic observation in group theory states thatthe size of the stabilizer is the same for all the vertices of G. Indeed, if v, ware two vertices of G and σ ∈ Aut(G) maps v to w, then Sv = σ−1Swσand, therefore, |Sw| = |Sv|. Moreover, |Sv| = |Svw|, where Svw is the set ofelements of Aut(G) that maps v into w. We have as well |Aut(G)| = |G||Sv|,where |G| is the number of vertices in G.

We remind the reader that α(G) is the independence number of G, thatis, the maximum number of vertices that are pairwise non-adjacent. Thefollowing lemma is a special case of Lemma 1 from [140].

Lemma 2.15 (Katona, [140]). Let G be a vertex-transitive graph. Let H ⊂G be a subgraph of G. Then α(G) ≤ α(H)

|H| |G|.

Remark. We formulated the lemma for the independence number,since it meets our demands. However, an analogue of it may be formulatedfor some other graph characteristics.

Proof. For any σ ∈ Aut(G) we denote an induced subgraph of G on the set ofvertices σ(V (H)) := σ(v) : v ∈ V (H) by σ(H). The proof of the lemmagoes by simple double counting. Before doing the crucial double countingstep, we remark that the union over all σ ∈ Aut(G) of σ(V (H)) covers eachvertex exactly |H||Sv| times. Take any independent set I in G.

|I| = 1

|H||Sv|∑

σ∈Aut(G)

|I∩σ(H)| ≤ 1

|H||Sv|∑

σ∈Aut(G)

α(H) =|Aut(G)||Sv|

α(H)

|H|=α(H)

|H||G|. (2.12)

91

There is a natural connection between this lemma and intersecting fam-ilies, which goes via Kneser graphs. A Kneser graph KGn,k is graph whichset of vertices is

([n]k

), and two vertices are adjacent i the corresponding sets

are disjoint. By denition the value of α(KGn,k) is the size of a maximumintersecting family in

([n]k

). Using Lemma 2.15 for the so-called Katona's

circle, it is not dicult to show that the independence number of KGn,k isequal to

(n−1k−1

), which is the statement of the ErdosKoRado theorem.

2.2.3.4. An inequality for cross-intersecting families of sets

In this subsection we prove a theorem about cross-intersecting families thatwe need for the proof of Theorem 2.6. We say that two families A, B arecross-intersecting, if for any B ∈ B, A ∈ A we have B ∩ A 6= ∅.

Theorem 2.16. Let n ≥ 2k, c ≥ 1. Consider two cross-intersecting familiesA, B, where B ⊂ A ⊂

([n]k

). Then

|A|+ c|B| ≤ max(n

k

), (c+ 1)

(n− 1

k − 1

). (2.13)

Remark. Informally speaking, the theorem states that the sum is max-imized in one of the two cases: either B is empty and we may take A to be(

[n]k

), or when A = B, and each of them is a trivial intersecting family, that

is, in which all sets contain a xed element.

Proof. The proof is an application of Katona's cyclic permutation method.The following proposition is the key step.

Proposition 2.17. Fix a cyclic permutation π of [n]. Consider the setU(π, k) ⊂

([n]k

)from Section 2.2.3.2. Consider two subfamilies A(π) :=

A ∩ U(π, k) and B(π) := B ∩ U(π, k). Denote a := |A(π)|, b := |B(π)|.Then

a+ cb ≤ maxn, (c+ 1)k. (2.14)

Proof. If the set B(π) is empty, then the statement is trivial, since a ≤|U(π, k)| = n. Henceforth we assume that |B(π)| = s > 0. We pass to thecomplements of the sets from B(π), considering the set B(π) := B : B ∈B(π). On the one hand, we know that for each A ∈ A(π) and B ∈ B(π) wehave A * B. In other words, A /∈ ∂k

(B(π)

). On the other hand, by Lemma

92

2.14 we have∣∣∂k(B(π)

)∣∣ ≥ min|B(π)| + n − 2k, n = n − 2k + s, for if∣∣∂k(B(π))∣∣ = n, then A(π) and, consequently, B(π) is forced to be empty.

Combining these two facts, we get |A(π)| ≤ n − (n − 2k + s) = 2k − s.From the following chain s = |B(π)| ≤ |A(π)| ≤ 2k − s we conclude thats ≤ k. Finally,

|A(π)|+ c|B(π)| ≤ 2k − s+ cs = 2k + (c− 1)s ≤ (c+ 1)k.

Knowing Proposition 2.17, the rest of the proof of the lemma is a standarddouble-counting argument, which was, in fact, carried out in the proof ofLemma 2.15. We take the circle U(π, k) as a subgraph H from Lemma 2.15.In parallel to (2.12), we get

|A|+ c|B| ≤ maxn, (c+ 1)k1

n

(n

k

)= max

(nk

), (c+ 1)

(n− 1

k − 1

).

2.2.3.5. Analysis of the KruskalKatona's Theorem

We note that the ideas of this section have been applied in the context ofintersecting families, see [166].

We start with the following corollary of Theorem 2.16, which demonstratesthe power of the KruskalKatona theorem and Theorem 1.22 in particular.

Corollary 2.18. The statement of Theorem 2.16 holds even if we replacethe condition B ⊂ A by B ⊂

([n]k

), |B| ≤ |A|.

Proof. The proof is straightforward. One has to pass to the families L(|A|, k),L(|B|, k). Then the condition |B| ≤ |A| is equivalent to L(|B|, k) ⊂ L(|A|, k).After we just have to apply Theorem 2.16 to the families L(|A|, k),L(|B|, k).

To avoid trivialities, for the whole section we assume that a+ b ≤ n.We say, that two sets S and T strongly intersect, if they intersect and for

the rst j ∈ S ∩ T we have [j] ⊂ S ∪ T . Let S be a nite s-element set andt ≥ s an integer. Dene the set family L(S, t) by

L(S, t) := T ∈(

[n]

t

): T < S

Proposition 2.19. Let A and B be an a-element and a b-element set, re-spectively. Assume that n ≥ a + b. Then L(A, a) and L(B, b) are cross-intersecting i A and B strongly intersect.

93

Proof. First suppose that L(A, a) and L(B, b) are cross-intersecting. ThenA and B intersect. Let j be the smallest integer contained in both. If A∪Bcontains [j], then A and B strongly intersect. Otherwise, there exists i < jsuch that i /∈ A ∪ B. Since n ≥ a + b, there exists an a-element set Csatisfying C ∩ [i] = A ∩ [i] ∪ i, which is disjoint with B. At the sametime, C < A, contradicting the assumption that L(A, a) and L(B, b) arecross-intersecting.

Now suppose that A and B strongly intersect. Let C < A. We claimthat C and B strongly intersect. If C ∩ [j] = A∩ [j], then it follows directlyfrom the denition. Otherwise, let i be the rst element where they dier.Since C < A, we have i ∈ C\A, i < j. But then i ∈ C ∩ B, [i] ⊂ C ∪ B.Repeating the same argument for any D with D < B gives that C andD strongly intersect. In particular, they have a non-empty intersection.Therefore, L(A, a) and L(B, b) are cross-intersecting.

Denition 2.20. We say that A ⊂(

[n]a

)and B ⊂

([n]b

)form a maximal cross-

intersecting pair, if whenever A′ ⊂(

[n]a

)and B′ ⊂

([n]b

)are cross-intersecting

with A′ ⊃ A and B′ ⊃ B, then necessarily A = A′ and B = B′ holds.

Now we are in a position to prove the following strengthening of Theorem1.22. We believe that this proposition is of independent interest. It wasdenitely of great use in proving Theorem 2.5.

Proposition 2.21. Let a and b be positive integers, a + b ≤ n. Let P andQ be non-empty subsets of [n] with |P | ≤ a, |Q| ≤ b. Suppose that P andQ strongly intersect in their last element. That is, there exists j, such thatP ∩ Q = j and P ∪ Q = [j]. Then L(P, a) and L(Q, b) form a maximalpair of cross-intersecting families.

Inversely, if L(m, a) and L(r, b) form a maximal pair of cross-intersectingfamilies, then it is possible to nd sets P and Q such that L(m, a) = L(P, a),L(r, b) = L(Q, b) and P,Q satisfy the above condition.

Proof. If P and Q satisfy the condition then L(P, a) and L(Q, b) form a pairof cross-intersecting families by Proposition 2.19. We have to show that it isa maximal pair. Let A be the a-set, following L(P, a) in the lexicographicalorder. Then A is the set obtained from P by omitting the element j andadjoining the interval [j+1, p], where p := a−|P |+1. Note that A∩Q = ∅ ,therefore there are members of L(Q, b) containing Q, which are disjoint withA. The same argument applies to the set B, following the initial segmentL(Q, b). Thus, the maximality of the pair L(P, a) and L(Q, b) is proved.

94

Next, let A, B be the last member of L(m, a) and L(r, b), respectively. Letj be the smallest element of A∩B. Note that the cross-intersecting propertyof the two families implies that j is well-dened. Dene P = A ∩ [j], Q =B ∩ [j].

First we prove that P ∪Q = [j]. Suppose the contrary and let i < j be anelement that is not contained in P ∪Q. Then P ′ := (P − j) ∪ i precedesP in the lexicographical order and hence also P ′ < A. Consequently, alla-element supersets of P ′ precede A as well. Thus they are all members ofL(m, a). However, P ′ ∩ B = ∅ and, since a + b ≤ n we have that somesuperset P ′ is disjoint to B, a contradiction.

The proof is almost complete. We have just proved that P and Q arestrongly intersecting. By Proposition 2.19 the families L(P, a) and L(Q, b)form a pair of cross-intersecting families. These families contain L(m, a)and L(r, b), respectively. By the maximality of the pair L(m, a) and L(r, b),L(m, a) = L(P, a), L(r, b) = L(Q, b) must hold.

In what follows we will use the following simple statement:

Proposition 2.22. Let A ⊂(

[n]a

)and B ⊂

([n]b

)be cross-intersecting. Then

we have |A|+ c|B| ≤ max(

na

), c(nb

).

Proof. Consider the bipartite graph with parts(

[n]a

)and

([n]b

), in which two

sets connected by an edge i they are disjoint. Assign weight 1 to each vertexin(na

)and weight c to any vertex in

(nb

). Then the proposition essentially

states that the independent set of the biggest weight in this graph coincideswith one of its two parts. It is an easy consequence of the fact that thegraph is regular in each of its parts. Denote the fraction of vertices from theindependent set in the part

([n]a

)and

([n]b

)by x and y, respectively. To see

that, take an edge in the graph at random. On the one hand, at most onevertex of the independent set may be in the edge. On the other hand, theexpected intersection of the edge with the independent set is x + y (due toregularity on both sides). Therefore, x+ y ≤ 1 and, clearly,

(na

)x+ c

(nb

)y ≤

max(na

), c(nb

).

We are interested in bounding |L(m, a)|+ c|L(r, b)| from above, and thuswe may w.l.o.g. restrict our attention only to maximal intersecting pairs offamilies, which, by Proposition 2.21, are equal to L(P, a),L(Q, b) for someP,Q ⊂ [n] with P ∪ Q = [i], P ∩ Q = i for some i ∈ [n]. We will callsuch P,Q dening sets for L(P, a), L(Q, b), respectively (or that L(P, a) isdened by P ). All the families will be lexicographical.

95

These easy-to-prove lemmas actually give a very useful consequence con-cerning the sizes of maximal cross-intersecting pairs of families L(m, a),L(r, b).

Lemma 2.23. Consider two maximal cross-intersecting families A = L(P, a),B = L(Q, b), such that P ∩Q = i, P ∪Q = [i]. Take any j < i such thatj ∈ Q, j + 1 /∈ Q (and thus j /∈ P, j + 1 ∈ P ).

Consider the following two pairs of cross-intersecting lexicographical fam-ilies. The rst pair is A′,B′, dened by P ′ := (P ∩ [j])∪ j, Q′ := Q∩ [j].The second pair is A′′,B′′, dened by P ′′ := P ∩ [j + 1], Q′′ := (Q ∩ [j +1]) ∪ j + 1.

Then for any c > 0 we have

|A|+ c|B| ≤ max|A′|+ c|B′|, |A′′|+ c|B′′|. (2.15)

For integers i < j we denote [i, j] = i, i + 1, . . . , j. Before going intothe proof of the lemma, we introduce the following notation. For i ∈ [n], aset S ⊂ [n] and a family F ⊂

([n]k

)we dene

Fi(S) := F ′ ∈(

[i+ 1, n]

k − |S ∩ [i]|

): F ′ ∪ (S ∩ [i]) ∈ F.

Let us also denote Z =(

[n]k

).

Proof. We remark that, by denition, we have B′′ ⊂ B ⊂ B′ and A′′ ⊃ A ⊃A′. Moreover, we have B′ − B′′ = Zj+1(Q). Similarly, we have A′′ − A′ =Zj+1(P ). We also have A−A′ = Aj+1(P ) and B − B′′ = Bj+1(Q).

Since P ∩ Q ∩ [j + 1] = ∅, we know that Aj+1(P ) and Bj+1(Q) arecross-intersecting. Therefore, by Proposition 2.22 we have

|Aj+1(P )|+ c|Bj+1(Q)| ≤ max|Zj+1(P )|, c|Zj+1(Q)|= max|A′′ −A′|, c|B′ − B′′|. (2.16)

Using almost identical proof, one may get the following twin of Lemma2.23:

Lemma 2.24. Consider two maximal cross-intersecting families A = L(P, a),B = L(Q, b), such that P ∩ Q = i, P ∪ Q = [i]. Moreover, assume thatfor some j < i [j] ⊂ Q.

Consider the pair of cross-intersecting lexicographical families A′,B′, de-ned by P ′ := j, Q′ := [j].

96

Then for any c > o we have

|A|+ c|B| ≤ max

|A′|+ c|B′|,

(n

a

). (2.17)

Proof. Note that B ⊂ Zj(Q) = B′. Similarly, Aj(P ) = A − A′ ⊂ Zj(P ).The families Aj(P ) and Bj(Q) = B are cross-intersecting. Therefore,

|Aj(P )|+ c|B| ≤ max|Zj(P )|, c|Zj(Q)|.

Therefore,

|A|+ c|B| ≤ max|A′|+ |Zj(P )|, |A′|+ c|Zj(Q)|

≤ max

(n

a

), |A′|+ c|B′|

.

2.2.3.6. A sharpening of Theorem 2.16

In the proof of Theorem 2.28 we need a sharpened version of Theorem 2.16,which proof relies on the material from the previous subsection. For integerj ≥ 2 and c > 0 let us denote

fn,k(j, c) := (c+ 1)

(n− 1

k − 1

)− c(n− jk − 1

)+

(n− j

k − j + 1

). (2.18)

Note that fn,k(j, c) = |L(P, k)|+c|L(Q, k)|, where P = [2, j] andQ = 1, j.

Theorem 2.25. Let k, n ∈ N,where k ≥ 2 and n ≥ 2k. Consider two cross-intersecting families A, B, where B ⊂ A ⊂

([n]k

). Let 2 ≤ i ≤ k + 1 and

|B| ≤(n−1k−1

)−(n−ik−1

). Then for any c ≥ 1

|A|+ c|B| ≤ maxfn,k(2, c), fn,k(i, c),

(n

k

). (2.19)

Proof. If B is empty, then the statement is obvious, therefore, we assumethe opposite for the rest of the proof. By Theorem 1.22, L0 := L(|A|, k)and L1 := L(|B|, k) are cross-intersecting. W.l.o.g. we may assume thatthis is a maximal cross-intersecting pair, and, thus, by Proposition 2.21,L0 = L(P, k) and L1 = L(Q, k), where P ∩ Q = l and P ∪ Q = [l] forsome l ≥ 2. Moreover, due to the restrictions on |B|, we know that 1 ∈ P ,and |P ∩ [i]| ≥ 2. Let i0 be the smallest integer, for which |P ∩ [i0]| ≥ 2.Then, obviously, |P ∩ [i0]| = 2 and i0 ≤ i.

(i) Assume that i0 = 2. Then, applying Lemma 2.24 toA := L0, B := L1

with j := 2 and a := b := k, we get that |L0|+c|L1| ≤ maxfn,k(2, c),

(nk

).

97

(ii) Assume that i0 > 2. If Q = 1, i0, then |L0| + c|L1| = fn,k(i0, c).Assume further that 1, i0 ( Q. Then, in particular, i0−1 ∈ P and i0 /∈ P .Therefore, we may apply Lemma 2.23 with A := L1, B := L0, j := i0 − 1,a := b := k and get that

c(|L1|+1

c|L0|) ≤ cmax

|L(Q′, k)|+ 1

c|L(P ′, k)|, |L(Q′′, k)|+ 1

c|L(P ′′, k)

,

where P ′ := [2, i0 − 1], Q′ := 1, i0 − 1 and P ′′ := [2, i0], Q′′ := 1, i0.

Thus, the right hand side of the displayed inequality above is precisely

maxfn,k(i0 − 1, c), fn,k(i0, c).

To complete the proof, we need to show that for j ≥ 3 we have fk(j, c) ≤fk(j + 1, c). Indeed,

fn,k(j + 1, c)− fn,k(j, c) = c((n− j

k − 1

)−(n− j − 1

k − 1

))−(( n− j

k − j + 1

)−(n− j − 1

k − j

))= c

(n− j − 1

k − 2

)−(n− j − 1

k − j + 1

)≥ 0, (2.20)

where the last inequality holds due to the fact that c ≥ 1 and(n−j−1k−2

)(n−j−1k−j+1

) =

j−2∏l=2

n− k − lk − l

≥ 1,

since n ≥ 2k. This completes the proof of the theorem.

For our purposes it would be more convenient to apply the followingslightly modied version of the theorem:

Corollary 2.26. In the conditions of Theorem 2.25 let 2 ≤ i ≤ k and|B| ≤

(n−1k−1

)−(n−ik−1

)+ x for some natural x. Then

|A|+ c|B| ≤ maxfn,k(i, c) + cx, fn,k(2, c),

(n

k

). (2.21)

Proof. The proof is basically the same, the only thing one has to noticethat, if |B| gets bigger, then |A| can only get smaller. Therefore, if |B| =(n−1k−1

)−(n−i−1k−1

)+ y, 1 ≤ y ≤ x, then one just remove y elements from |B|,

applies the same proof, and puts the elements back, adding cy to the righthand side. Note that we only add cx to the rst expression, since the othertwo appear in the proof only when |B| is relatively small.

98

2.2.4. Proof of Theorem 2.6

For any intersecting family G ⊂ V(n + 1, k, 1) we introduce the followingnotations. By G(i),G(−i),G (i) we denote the subfamilies of G, that have1,−1, 0 as an i-th coordinate, respectively. It is important to mention thatwe consider them as families of vectors on the set of coordinates with the i-thcoordinate excluded. The denition extends in an obvious way on the familyG(I1, I2,−I3), where I1, I2, I3 ⊂ [n] are non-intersecting sets of indices.

Consider a maximum intersecting family F ⊂ V(n + 1, k, 1). Based onthe conclusion of Section 2.2.3.1, we may and will assume that F is shifted.Denote A := F(−(n+ 1)) and B := F(n+ 1).

Proposition 2.27. We have m(n+ 1, k, 1)−m(n, k, 1) ≤ |A|+ |B|.

Proof. First, by denition we have |F| = m(n + 1, k, 1). Second, considera family F(n+ 1). It is a subfamily in V(n, k, 1), moreover, it is intersect-ing. Therefore, |F(n+ 1)| ≤ m(n, k, 1). Finally, m(n + 1, k, 1) = |F| =|F(n+ 1)|+ |A|+ |B|.

For a given i ∈ [n] consider two families of sets B(−i),A(i) (both ofthese families could be seen as sets, since the only −1-coordinate is xed andexcluded in both). We have B(−i),A(i) ⊂

([n]−ik−1

).Moreover, B(−i) ⊂ A(i)

due to shifting, and we may apply Theorem 2.16 to these two families withc = k and obtain

|A(i)|+ k|B(−i)| ≤ max(n− 1

k − 1

), (k + 1)

(n− 2

k − 2

).

Summing this inequality over all i ∈ [n], we get

n∑i=1

|A(i)|+ k|B(−i)| = k(|A|+ |B|) ≤ nmax(n− 1

k − 1

), (k + 1)

(n− 2

k − 2

)

|A|+ |B| ≤ max(n

k

),n(k + 1)

k

(n− 2

k − 2

). (2.22)

Maximum in the right hand side of (2.22) is attained on the rst expressionif (

n

k

)=n(n− 1)

k(k − 1)

(n− 2

k − 2

)≥ n(k + 1)

k

(n− 2

k − 2

),

which is equivalent to n ≥ k2.

99

Proof of Theorem 2.6. The bound m(n + 1, k, 1) ≤ m(n, k, 1) +(nk

)was, in

fact, already proven in this section. In the notations above, consider F ,A,B.On the one hand, by (2.22) and the discussion after this inequality, we have|A|+ |B| ≤

(nk

)for n ≥ k2. Applying Proposition 2.27, we get the bound.

The bound m(n + 1, k, 1) ≥ m(n, k, 1) +(nk

)was already obtained in

Section 2.2.2 (and mentioned in the introduction).

2.2.5. Proof of Theorem 2.5 in the case 3k ≤ n ≤ k2

Looking at equation (2.22) in the case n < k2, we see that m(n + 1, k, 1)−m(n, k, 1) ≤ n(k+1)

k

(n−2k−2

). On the other hand, from Section 2.2.2 we know,

that e(n+ 1, k)− e(n, k) = k(n−1k−1

)= (n−1)k

k−1

(n−2k−2

). We have n(k+1)

k − (n−1)kk−1 =

k2−nk(k−1) . Using Theorem 2.25, we are going to improve the inequality (2.22), sothat it matches the bound given by the construction E(n, k, 1). To completethe proof of Theorem 2.5, it is enough to prove the following theorem.

Theorem 2.28. Let k ≥ 3, 3k − 1 ≤ n < k2. We have m(n + 1, k, 1) −m(n, k, 1) = (n−1)k

k−1

(n−2k−2

)= e(n+ 1, k)− e(n, k).

Note that for k = 2 we have 3k − 1 > k2, thus, there is nothing to provein this case. Before proving the theorem, we state and prove the followingproposition:

Proposition 2.29. Let n ≥ 3k − 1. In terms of Section 2.2.4, consider amaximum intersecting family F ⊂ V(n + 1, k, 1) and families of vectors A,B in 0,±1n. There is a subset I ⊂ [n], |I| ≥ d3k

2 e, such that for everyl ∈ I we have |B(1,−l)| ≥ 1

3 |B(−1)|.

Proof. Take any B ∈ B(−1). Then, due to the fact that F is shifted, if weswap -1, which is on the rst coordinate position, and some of the 0's in B, weobtain a setB′ ∈ B(1). Since there are n−k zeros in eachB ∈ B(−1), we mayobtain n− k sets B′ ∈ B(1) out of each B. Moreover, any two sets obtainedare dierent. Indeed, for B any two vectors obtained out of it have dierentpositions of -1, while any two vectors obtained from dierent B1, B2 ∈ B(−1)have dierent sets of 1's. Thus, |B(1,−2)|+. . .+|B(1,−n)| ≥ (n−k)|B(−1)|.By pigeon-hole principle we get that one of the summands must be at leastn−kn−1 |B(−1)|. This is the rst element from I.

Once we have found the i-th element, which satises the inequality fromthe proposition, we add it to I, and delete this element from all the vectorsfrom B(−1) and from the ground set. The set of already found elements wedenote Ii. After i steps each of the sets from the modied B(−1) has at least

100

n−k− i zeros, while the total number of coordinates is n− i−1. Therefore,the inequality from the previous paragraph, modied for this case, looks like∑

j∈[2,n]\Ii |B(1,−j)| ≥ (n− k − i)|B(−1)|, and by pigeon-hole principle we

can nd an element li such that |B(1,−li)| ≤ n−k−in−i−1 |B(−1)|, which is bigger

than 13|B(−1)| for i+ 1 ≤ d3k/2e and n ≥ 3k − 1.

Proof of Theorem 2.28. We argue in terms used in Section 2.2.4. We con-sider several cases depending on the size of B(−1).

(i) First, assume that B(−1) is empty. Then, applying Theorem 2.16 withc := k, we get |A(i)| + k|B(−i)| ≤ (k + 1)

(n−2k−2

)for i = 2, . . . , n (see (2.22)

and the calculations after it) and |A(1)|+ k|B(−1)| ≤(n−1k−1

). Therefore,

n∑i=1

|A(i)|+ k|B(−i)| = k(|A|+ |B|) ≤(n− 1

k − 1

)+ (n− 1)(k + 1)

(n− 2

k − 2

)

|A|+ |B| ≤( n− 1

(k − 1)k+

(n− 1)(k + 1)

k

)(n− 2

k − 2

)=

(n− 1)k

k − 1

(n− 2

k − 2

)= e(n+ 1, k)− e(n, k).

It is apparent from the calculations above that the theorem follows if wesucceed to show that for some I ⊂ [2, n] we have∑

l∈I

(|A(l)|+ k|B(−l)|

)+k|B(−1)| ≤ |I|(k + 1)

(n− 2

k − 2

).

This is exactly what are we going to do in a range of cases.

(ii) Assume that |B(−1)| ≥ 3(n−4k−3

). Then, by Proposition 2.29, we can

nd I ⊂ [2, n], |I| ≥ d3k/2e ≥ 5, such that we have |B(1,−l)| ≥(n−4k−3

).

For any l ∈ I consider the collection B(−l). Due to shifting, we knowthat the maximum degree d(B(−l)) is equal to the number of sets fromB(−l) that have 1 on the rst coordinate position. Therefore, we know thatγ(B(−l)) = |B(1,−l)| ≥

(n−4k−3

)and, thus, we may apply Theorem 1.4 to

B(−l) with u = 3 and n, k replaced by n− 1, k − 1. We obtain that

|B(−l)| ≤(n− 2

k − 2

)−(n− 4

k − 2

)+

(n− 4

k − 3

).

101

For each l ∈ I we apply Corollary 2.26 to B(−l),A(−l) with c = k,x =

(n−4k−3

)and i = 3 and obtain that

|A(l)|+ k|B(−l)| ≤ maxfn−1,k−1(3, k) + k

(n− 4

k − 3

), fn−1,k−1(2, k),

(n− 1

k − 1

).

If the maximum of the right hand side is attained on the third summand,then for a single l ∈ I

|A(l)|+ k|B(−l)|+ |A(1)|+ k|B(−1)| ≤(n− 1

k − 1

)+ (k + 1)

(n− 2

k − 2

)and we are done. Recall that, by (2.18), fn−1,k−1(2, k) = (k+ 1)

(n−2k−2

)− (k−

1)(n−3k−2

). Note also that |B(−1)| ≤

(n−2k−2

)since B(−1) is intersecting. If the

maximum is attained on the second expression, then, since |I| ≥ 5, for vedierent l1, . . . , l5 we have

5∑j=1

(|A(lj)|+ k|B(−lj)|) + k|B(−1)| − (5k + 5)

(n− 2

k − 2

)

≤ k

(n− 2

k − 2

)− 5(k − 1)

(n− 3

k − 2

)≤ (2k − 5(k − 1))

(n− 3

k − 2

)< 0,

since k ≥ 3, and we are done. Recall that, by (2.18), fn−1,k−1(3, k) =(k + 1)

(n−2k−2

)− k(n−4k−2

)+(n−4k−3

). Finally, if the maximum is attained on the

rst summand, then

1

|I|

(∑l∈I

(|A(l)|+ k|B(−l)|) + k|B(−1)|)− (k + 1)

(n− 2

k − 2

)≤

≤− k(n− 4

k − 2

)+ (k + 1)

(n− 4

k − 3

)+

k

|I|

(n− 2

k − 2

). (2.23)

Before continuing, we need the following two bounds, valid for n ≥ 3k − 1:

(k + 1)(n−4k−3

)k(n−4k−2

) =(k + 1)(k − 2)

k(n− k − 1)≤ 1

2.(

n− 2

k − 2

)=

(n− 2)(n− 3)

(n− k)(n− k − 1)

(n− 4

k − 2

)≤ 9

4

(n− 4

k − 2

).

Using these three bounds and |I| ≥, we get

(2.23) ≤− 1

2k

(n− 4

k − 2

)+

9

4· k

5

(n− 4

k − 2

)< 0.

102

This case is settled.

(iii) Assume that for some 4 ≤ i ≤ k we have

3

(n− i− 1

k − i

)≤ |B(−1)| ≤ 3

(n− i

k − i+ 1

).

Similarly to the previous case, for I ⊂ [n], |I| ≥ 5, we obtain

|B(−l)| ≤(n− 2

k − 2

)−(n− i− 1

k − 2

)+

(n− i− 1

k − i

).

Note that if 0 < |B(−1)| < 3(n−k−1k−k

)= 3 we have B(1,−l) ≥ 1

3|B(−1)| > 0and we get the same bound on |B(−l)| as the one above for i = k, so thiscase is also covered and cases (i), (ii), (iii) altogether cover all possible valuesof |B(−1)|.

We again apply Corollary 2.26 for each B(−l),A(−l) with x =(n−i−1k−i

). If

the maximum of the expression from the inequality (2.21) is attained on oneof the last two expressions, we are done, since we can do exactly the same asin case (ii). Recall that fn−1,k−1(i, k) = (k + 1)

(n−2k−2

)− k(n−i−1k−2

)+(n−i−1k−i

).

If it is attained on the rst one, then

1

|I|

(∑l∈I

(|A(l)|+ k|B(−l)|) + k|B(−1)|)− (k + 1)

(n− 2

k − 2

)≤

≤− k(n− i− 1

k − 2

)+ (k + 1)

(n− i− 1

k − i

)+

k

|I||B(−1)| ≤

≤ − k(n− i− 1

k − 2

)+ (k + 1)

(n− i− 1

k − i

)+

3k

5

(n− i

k − i+ 1

). (2.24)

To proceed further, we need the following bounds, valid for n ≥ 3k − 1:

(k + 1)(n−i−1k−i

)k(n−i−1k−2

) =k + 1

k

i−1∏j=2

k − jn− k − j + 1

≤(1

2

)i−2

≤ 1

4.

(n−ik−i+1

)(n−i−1k−2

) ≤ (n− i)∏i−2

j=2(k − j)∏i−2j=1(n− k − j)

≤ n− 4

n− k − 1

(k − 2

n− k − 2

)i−3≤ 3k − 5

2k − 2· 1

2≤ 3

4.

Now we may conclude.

(2.24) ≤(−3

4k +

9k

20

)(n− i− 1

k − 2

)< 0.

103

2.3. The proof of Theorem 2.9

Let F ⊂ V(n, k, l) be an intersecting family. For a 0,±1-vector v :=(v1, . . . , vn), let S(v) denote its support, i.e.,

S(v) := i ∈ [n] : vi 6= 0.

Dene also

S+(v) := i : vi = +1,S−(v) := i : vi = −1.

Obviously, S(v) = S+(v) t S−(v).Also, for k > l two vectors v,w satisfy 〈v,w〉 = −2l i S−(v) ⊂

S+(w), S−(w) ⊂ S+(v) and S+(v) ∩ S+(w) = ∅ hold simultaneously.Our assumption is that no such pair v,w exist in F . For a pair A,B of

disjoint l-element sets we dene F(A,B) to be the family of those (k − l)-element sets C that the vector u dened by S+(u) = A t C, S−(u) = B isin F .

Lemma 2.30. For disjoint l-subsets A,B ⊂ [n] the two families F(A,B)and F(B,A) are cross-intersecting.

Proof. Suppose the contrary and let C ∈ F(A,B), D ∈ F(B,A) be disjoint(k−l)-sets. Then the vectors v,w determined by S+(v) = AtC, S−(v) = B,S+(w) = B t D, S−(w) = A are both in F . However, 〈v,w〉 = −2l, acontradiction.

This lemma and Proposition 1.24 motivate the following procedure. Forall(n2l

)(2ll

)choices of a pair of disjoint l-sets A and B, if |F(A,B)| ≤

(n−2l−1k−l−1

),

then delete from F all vectors v with S−(v) = B, A ⊂ S+(v).Let F ′ be the collection of remaining vectors and note:

|F ′| ≥ |F| −(n

2l

)(2l

l

)(n− 2l − 1

k − l − 1

). (2.25)

Let us x now a (k+ l)-element set T ⊂ [n] and consider the family B ⊂(Tl

)dened as follows:

B :=B ∈

(T

l

): ∃v ∈ F ′, S(v) = T, S−(v) = B

.

Lemma 2.31. The family B is intersecting.

104

Proof. Suppose for contradiction that A,B ∈ B are disjoint. By the deni-tion of B there are u,v ∈ F ′ satisfying S−(u) = A, S−(v) = B, S(u) =S(v) = T . This implies A ⊂ S+(v), B ⊂ S+(u).

Since both u and v survived the deletion process, we have

|F(A,B)| >(n− 2l − 1

k − l − 1

),

|F(B,A)| >(n− 2l − 1

k − l − 1

).

However, Proposition 1.24 shows that F(A,B) and F(B,A) are not cross-intersecting. This contradicts Lemma 2.30.

Since k ≥ l, the ErdosKoRado Theorem (Theorem 1.2) implies

|B| ≤(k + l − 1

l − 1

).

Consequently,

|F ′| ≤(

n

k + l

)(k + l − 1

l − 1

).

Combining with (2.25), the inequality (2.7) follows.

2.4. Proof of Theorem 2.8

The proof is based on the application of the general Katona's circle method[139] to V(n, k, l). Consider the following subfamily H of V(n, k, l):

H := v = (v1, . . . , vn) : ∃i ∈ [n] : vi = . . . = vi+k−1 = 1, vi−k = . . . = vi−k+l−1 = −1.

We remark that all indices are written modulo n. Note that |H| = n. Forany permutation σ of [n] we dene H(σ) := σ(H) : H ∈ H.

Take an intersecting family F ⊂ V(n, k, l).

Lemma 2.32. For any permutation σ we have |H(σ) ∩ F| ≤ k.

Proof. Denote by F ′ ⊂(

[n]k

)the family S+(F ) : F ∈ F, and, similarly,

H′ := S+(H) : H ∈ H(σ).We claim thatH′∩F ′ is an intersecting family. Assume that there are two

sets F ′1, F′2 ∈ H′∩F ′, that are disjoint. W.l.o.g., F ′2 = [k+1, 2k]. Then F ′1 is

obliged to contain [1, l], since any cyclic interval of length k in [n]\ [k+1, 2k]contains [1, l], provided that n ≤ 3k − l.

105

We conclude that the corresponding vector v1 ∈ F ∩H satises S+(v) ⊃[1, l]. At the same time, by the denition of H, the vector v2 correspondingto F ′2 satises S−(v2) = [1, l]. That is, S−(v2) ⊂ S+(v1). Interchanging theroles of F1, F2, we get that S−(v1) ⊂ S+(v2). Moreover, S+(v1)∩S+(v2) = ∅.This means that v1 and v2 are antipodal, a contradiction.

Therefore, the family H′ ∩ F ′ is intersecting. It is proven in [139] that inthis case |H′ ∩ F ′| ≤ k, but we sketch the proof of this simple fact here forcompleteness. Take a set H ∈ H′ ∩ F ′. Then the 2k − 2 sets from H′ thatintersect H can be split into pairs of disjoint sets. We can take only one setfrom each pair.

The rest of the argument is a standard averaging argument (cf. Sec-tion 2.2.3.3). Let us count in two ways the number of pairs (permutation σ,a vector from H(σ) ∩ F). On the one hand, each vector from F is countednk!l!(n− k − l)! times. On the other hand, for each permutation, there areat most k pairs by Lemma 2.32. Therefore,

|F|n k! l! (n−k− l)! ≤ kn! ⇔ |F| ≤ k

n

n!

k! l! (n− k − l)!=k

n|V(n, k, l)|.

2.5. Vectors of xed length

2.5.1. Simple properties of F (n, k, l)

First we state and prove some simple observations concerning F (n, k, l).

Proposition 2.33. Fix any n ≥ k ≥ 1. Then F (n, k,−k + 1) = 2k−1(nk

)=

|Lk|/2.

Proof. We split vectors from Lk in pairs v,w so that 〈v,w〉 = −k. We cantake exactly one vector out of each pair in the family.

For any v = (v1, . . . , vn) and any vector family V dene S(v) = i : vi 6=0 and V(S) = v ∈ V : S(v) = S. We also dene N(v) = i : vi = −1.

Proposition 2.34. We have F (2t, 2t, 2s) = F (2t, 2t, 2s− 1) = f(2t, t− s)and F (2t+ 1, 2t+ 1, 2s) = F (2t+ 1, 2t+ 1, 2s+ 1) = f(2t+ 1, t− s).

Proof. For any v,w ∈ L2t we have 〈v,w〉 = 2t−2|N(v)4N(w)|. Therefore,the statement of the proposition follows from Kleitman's Theorem.

We say that a family V ⊂ Lk of vectors is homogeneous, if for every i ∈ [n]the i'th coordinates of vectors from V are all non-negative or all non-positive.

106

Proposition 2.35. For n > k we have F (n, k, k−1) = maxk+1, n−k+1.

Proof. First of all note that uivi = −1 forces 〈u,v〉 ≤ k − 2. Thus V ishomogeneous and the condition translates into |S(u) ∩ S(v)| ≥ k − 1.

Since the only (k − 1)-intersecting families of k-sets are stars around a(k − 1)-element set and subsets of

([k+1]k

), we are done.

Given a vector family V , for any i ∈ [n] we denote by di the degree ofi, that is, the number of vectors from V that have a nonzero coordinate onposition i.

Proposition 2.36. For n > k ≥ 2 the following holds:(i) F (3, 2, 0) = 4.(ii) F (n, k, k − 2) =

(nk

)if n = k + 2 or k ≥ 3 and n = k + 1.

(iii) F (n, k, k − 2) = max(

n−k+22

), k(n− k) + 1

for n ≥ k + 3.

Proof. The rst part is very easy to verify. As for the other parts, assumethat ui = −vi for some u,v ∈ V . Then S(u) = S(v) and u and v mustagree in the remaining coordinate positions. Thus, di = 2 in this case.Since F (k, k, k − 2) = 2, for n = k + 1 a vertex of degree 2 would force|V| ≤ 2 + 2 = 4. On the other hand, if we assume that V is homogeneous,then |V| ≤

(k+1k

)for k ≥ 3, as desired. Since

(k+2k

)−(k+1k

)≥ 2 for k ≥ 2,

the same argument works for n = k + 2, k ≥ 2 as well.If n > k+ 2, the same argument as above implies that the family S(u) :

u ∈ V is (k − 2)-intersecting. If |V| is maximal, then it is homogeneous.The bound then follows from the Complete Intersection Theorem (see Section1.1.6).

2.5.2. The case k = 3

For k = 3 the values not covered by Propositions 2.33, 2.35, 2.36 and The-orem 2.10 are F (n, 3, 0) and F (n, 3,−1) for n not too large. We determineF (n, 3, 0) in the following theorem, but rst we need some preparation.

Let us use the notation (a, b, c) for a set a, b, c if we know that a < b < c.Also for (a, b, c) ⊂ [n] let u(a, b, c) = (u1, . . . , un) with ui = 1 for i ∈ a, b, cand ui = 0 otherwise. Further let v(a, b, c) = (v1, . . . , vn) be the vector withva = vb = 1, vc = −1 and vi = 0 otherwise.

Let us show two lower bounds for F (n, 3, 0). Taking all non-negativevectors gives

F (n, 3, 0) ≥(n

3

).

107

The other one is based on the following family:

V(n) = u(a, b, c),v(a, b, c) : |(a, b, c) ∩ [3]| ≥ 2.

Note that 〈v,v′〉 ≥ 0 for all v,v′ ∈ V(n) and the rst few values of |V(n)|are |V(3)| = 2, |V(4)| = 8, |V(5)| = 14, |V(6)| = 20.

Theorem 2.37. We have(1) F (n, 3, 0) = |V(n)| for n = 3, 4, 5.(2) F (6, 3, 0) = 21.(3) F (n, 3, 0) =

(n3

)for n ≥ 7.

As we will see, the proof of (2) is the most dicult.

2.5.3. Preliminaries

In the proof of Theorem 2.10 we use two types of shifting. The rst onepushes bigger coordinates to the left: see Section 2.2.3.

The second is the up-shift : Si(v1, . . . , vn) := (v1, . . . , vn) if vi = 0 or 1 andSi(v1, . . . , vn) := (v1, . . . , vi−1, 1, vi+1, . . . , vn) if vi = −1. The shift Si(W) isdened similarly to Si,j(W):

Si(W) := Si(v) : v ∈ W ∪ v : v, Si(v) ∈ W.

We call a system W shifted, if W = Si,j(W) for all i < j ∈ [n] andW = Si(W) for all i ∈ [n]. Any system of vectors may be made shiftedby means of a nite number of (i, j)-shifts and i-up-shifts. Moreover, it wasshown in [97] and [150] (and is easy to verify directly) that (i, j)-shifts andup-shifts do not decrease the minimal scalar product in W .

2.5.4. Proof of Theorems 2.10 and 2.11

Part 1 of Theorem 2.10

First we show that the left hand side is at least the right hand side in The-orem 2.10 part 1. Take the family of vectors V :=

v = (v1, . . . , vn) : v1 =

. . . = vl = 1, vi ∈ 0, 1. It clearly satises the condition 〈v,w〉 ≥ l for

any v,w ∈ V and has cardinality(n−lk−l).

We proceed to the upper bound. Take any family V of 0,±1-vectorshaving minimal scalar product at least l and dene its subfamilies Vi(ε) :=w ∈ V : wi = ε for ε = ±1.

108

Claim 2.38. If for some i both Vi(−1) and Vi(1) are nonempty, then

|Vi(1)|+ |Vi(−1)| ≤ 2k(k − 1

l + 1

)(n− l − 2

k − l − 2

)= O(nk−l−2).

We call any such i ∈ [n] bad.

Proof. Consider the set families

A+ :=S ∈

([n]− ik − 1

): ∃w ∈ Vi(1) such that S(w) = S ∪ i

,

A− :=S ∈

([n]− ik − 1

): ∃w ∈ Vi(−1) such that S(w) = S ∪ i

.

Both A+,A− are nonempty. Each of them is (l − 1)-intersecting, moreover,they must be (l + 1)-cross-intersecting, since otherwise we would have twovectors in V that have scalar product at most l − 1.

Let A ∈ A+, B ∈ A−. Then |A ∩ B| ≥ l + 1 for all B ∈ A− implies|A−| ≤

( |A|l+1

)(n−l−2k−l−2

)and similarly |A+| ≤

( |B|l+1

)(n−l−2k−l−2

).

The statement of the claim follows from the trivial inequalities |Vi(1)| ≤2k−1|A+|, |Vi(−1)| ≤ 2k−1|A−|.

Consider the set I ⊂ [n] of all bad coordinate positions and dene asubfamily Vb of all vectors that have non-zero values on the bad positions:Vb := v ∈ V : ∃i ∈ I : vi 6= 0. Put Vg := V −Vb. Since Vg is homogeneouson [n] − I and is l-intersecting, we have |Vg| ≤

(n−|I|−lk−l

)for n ≥ k2 due to

the result of [76]. Therefore,

|V| ≤(n− |I| − lk − l

)+ |I|O(nk−l−2) ≤

≤(n− lk − l

)− |I|

((n− |I| − lk − l − 1

)−O(nk−l−2)

)≤(n− lk − l

),

for n large enough, with the last inequality strict in case |I| > 0.

Part 2 of Theorem 2.10 and Theorem 2.11

First we show that there is such a family of vectors for which the inequal-ity is achieved. For each S ⊂

(nk

)we take in V all the vectors from Lk(S)

that have at most l/2 −1's in case l is even and the sets that have at most(l−1)/2 −1's among the rst (k−1) nonzero coordinates in case l is odd (inother words, take the families of vectors w whose corresponding sets N(w)duplicate the examples U l, U lk from the formulation of Katona's Theorem).

109

It is not dicult to see that V satises the requirements of both Theorems2.10 and 2.11 on the size and on the scalar product.

Now we prove the upper bound. The proof is somewhat similar to theprevious case. Take any vector family satisfying the conditions of Theorem2.10 or Theorem 2.11. For any set X ⊂ [n] and any vector w of length ndenote by w|X the restriction of w to the coordinates i, i ∈ X. Obviouslyw|X is a vector of length |X|. We look at all vectors v ∈ ±1l+1 andsplit them into groups of antipodal vectors v, v, that is, vectors that satisfy〈v, v〉 = −l − 1. Fix a set of coordinates I := i1, . . . , il+1 and deneV(I,v) := w ∈ V : w|I = v.

Claim 2.39. If both V(I,v) and V(I, v) are nonempty, then

|V(I,v)|+ |V(I, v)| ≤ 2k−l−1(k − l)(n− l − 2

k − l − 2

)= O(nk−l−2).

We call any such I bad. In particular, the claim gives that the totalcardinality of all such bad subfamilies is O(nk−1).

Proof. Consider the set families

A :=S ∈

([n]− Ik − l − 1

): ∃w ∈ V(I,v) such that S(w) = S ∪ I

,

A :=S ∈

([n]− Ik − l − 1

): ∃w ∈ V(I, v) such that S(w) = S ∪ I

.

Both A, A are nonempty. Moreover, they must be cross-intersecting, sinceotherwise we would have two vectors in V that have scalar product exactly−l − 1. Therefore, by Hilton-Milner theorem,

|A|+ |A| ≤(n− l − 1

k − l − 1

)−(

n− kk − l − 1

)+ 1 ≤ (k − l)

(n− l − 2

k − l − 2

).

The statement of the claim follows from the trivial inequalities |V(I,v)| ≤2k−l−1|A|, and |V(I, v)| ≤ 2k−l−1|A|.

Dene Vb := w ∈ V : ∃I ⊂(

[n]l+1

): V(I,v|I) 6= ∅,V(I, v|I) 6= ∅.

Consider the remaining family of good vectors Vg := V − Vb.First we nish the proof of Theorem 2.11. The family Vg is good in the

following sense. If we x S ∈(

[n]k

)and consider the family of sets A :=

110

N(w) : w ∈ Vg(S), then A satises A1 4 A2 ≤ l for any A1, A2 ∈ A.Therefore, |Vg(S)| ≤ f(k, l) by Kleitman's theorem and

|V| ≤∑S∈([n]

k )

f(k, l) + |Vb| = f(k, l)

(n

k

)+O(nk−1).

The proof of Theorem 2.11 is complete.From now on we suppose that the family V is shifted. Note that, since

shifting may increase scalar products, it is impossible to use it in the caseof Theorem 2.11. It is not dicult to see that the family Vb is up-shifted,since if we deleted some w ∈ V , then we would have deleted all u ∈ V withS(u) = S(w), N(u) ⊃ N(w). One can see in a similar way that Vb is shifted.

Claim 2.40. For every S ∈(

[n]k

)the family Vg(S) satises N(w)∪N(v) ≤ l

for all v,w ∈ Vg(S).

Proof. Suppose for contradiction N(w) ∪ N(u) ≥ l + 1. Consider any setI ⊂ N(w) ∪ N(u), |I| = l + 1. Since Vg is up-shifted, there is a vector u′

in Vg(S) such that 〈w|I ,u′|I〉 = −l − 1, which means that for v = w|I bothVg(I,v) and Vg(I, v) are nonempty, a contradiction.

Claim 2.41. If I ⊂(

[n]l+1

)is bad, then for any S ⊃ I, S ∈

([n]k

), we have

|Vg(S)| ≤ f(k, l)− 1.

Proof. Consider the family A ⊂ 2S of subsets of S, that is dened in thefollowing way: A := N(w) ∩ S : w ∈ Vg(S). For simplicity, we identifyS with [k], preserving the order of elements. In view of the uniqueness partof Katona's theorem, it is sucient to show that A does not contain one ofthe sets from the extremal families: from U l in case l is even and from anyof U li , i = 1, . . . , k in case l is odd.

First consider the case when l is even. If I is bad, it means that thereexists a vector v of length l+1, such that both V(I,v),V(I, v) are nonempty.Therefore, all vectors from these families are not present in Vg. Assumew.l.o.g. that |N(v)| ≤ l/2. Then the set N(v) is not present in A, but atthe same time it belongs to U l.

From now on, assume that l is odd. We remark that, due to the fact thatVg is shifted, the family A is shifted to the right, and, therefore, out of theextremal families A may coincide with U lk only. We argue similarly to theprevious case. If |N(v)| ≤ (l− 1)/2 or |N(v)| ≤ (l− 1)/2, then we are doneas in the previous case. The only remaining case is when |N(v)| = (l+ 1)/2.If k ∈ I, then we are done again: one of N(v), N(v) then contains k andtherefore belongs to U lk.

111

Suppose from now on that k /∈ I. Fix some i ∈ N(v) and take anyw ∈ V such that w|I = v. Since V is shifted, there is a vector u ∈ Vsuch that u|I−i = w|I−i, uk = 1. Assume that A contains the set N ′ :=N(v)−i∪k, and take v′ ∈ Vg(S) such that N(v′) = N ′. Then, puttingI ′ := I − i ∪ k, we get that both V(I ′,v′) and V(I ′, v′) are nonempty:the rst family due to the assumption that N ′ ∈ A, and the second one dueto the fact that u|I ′ = v′. Therefore, N ′ /∈ A. But N ′ ∈ U lk, and thus A doesnot coincide with U lk.

Assume now that there are t bad sets I ⊂(

[n]l+1

). Then the number of

sets S ⊂(

[n]k

)that contain one of the bad sets I is at least t

(n−l−1k−l−1

)/(kl+1

).

Therefore, we have

|V| − f(k, l)

(n

k

)≤ −t

(n−l−1k−l−1

)(kl+1

) +1

2

∑v∈±1l+1

∑bad I

(|V(I,v)|+ |V(I, v)|

)≤

≤ −t((n−l−1

k−l−1

)(kl+1

) − 2k(k − l)(n− l − 2

k − l − 2

))< 0,

provided n > 2kk2(kl+1

). We note that taking n > 4kk2 makes the choice

of n from which the proof works independent of l.

2.5.5. Proof of Theorem 2.37

Let W ⊂ 0,±1n satisfy 〈w,w〉 = 3, 〈w,v〉 ≥ 0 for all w,v ∈ W andsuppose that W is shifted.

Claim 2.42. If S(w) = (a, b, c) for w ∈ W then either w = u(a, b, c) orw = v(a, b, c) holds.

Proof. Assume the contrary and let (x, y, z) be the non-zero coordinate valuesof the vector w (at positions a, b, c). Using the up-shift w.l.o.g. (x, y, z) =(−1, 1, 1) or (1,−1, 1). In both cases by shifting (1, 1,−1) is also in W .However it has scalar product −1 with both, a contradiction.

From the claim F (3, 3, 0) ≤ 2, and F (4, 3, 0) ≤(

43

)× 2 = 8 follow.

From now on n ≥ 5 and we use induction on n to prove the statement ofthe theorem. Set

G := (a, b) ∈(

[n− 1]

2

): u(a, b, n) ∈ W,

112

H := (a, b) ∈(

[n− 1]

2

): v(a, b, n) ∈ W.

By shiftedness, H ⊂ G and by 〈v,w〉 ≥ 0, G and H are cross-intersecting.We may assume that H is non-empty, since otherwise |W| ≤

(n3

)holds due

to shiftedness.

Claim 2.43. If v(2, 3, n) ∈ W then |H| = |G| = 3.

Proof. By shiftedness (1, 2), (1, 3), (2, 3) ⊂ H ⊂ G. The statement followsfrom the fact that no other 2-element set can intersect those three sets.

Since W(n) := w ∈ W : wn = 0 satises |W(n)| ≤ F (n − 1, 3, 0),v(2, 3, n) ∈ W and Claim 2.43 imply

|W| ≤ F (n− 1, 3, 0) + 6, which gives

|W| ≤ 8 + 6 = 14 for n = 5,

|W| ≤ 14 + 6 < 21 for n = 6, and

|W| ≤(n− 1

3

)+ 6 <

(n

3

)for n ≥ 7.

Consequently, we may suppose that the degree of n in W is at least 7, inparticular, H is the star, H = (1, 2), (1, 3), . . . , (1, p) for some p ≤ n− 1.

The following lemma is obvious.

Lemma 2.44. One of the following holds.(i) p = 2, |H| = 1, |G| ≤ (n− 2) + (n− 3) = 2n− 5(ii) p = 3, |H| = 2, |G| ≤ (n− 2) + 1 = n− 1(iii) p ≥ 4, |H| ≤ |G| ≤ n− 2.

Since for n = 5 in all cases |H| + |G| ≤ 6 holds, the proof of part (1) ofTheorem 2.37 is complete. Also, for n ≥ 7 it follows that |H|+ |G| ≤ 2n−4.Since

(n3

)−(n−1

3

)=(n−1

2

)> 2n− 4 for n ≥ 7, the induction step works ne

in this case too.

The only case that remains is n = 6. Using shiftedness and a = 1 for all(a, b, n) with v(a, b, n) ∈ W it follows that a = 1 holds for all (a, b, c) withv(a, b, c) ∈ W .

Dene B = (b, c) ∈(

[6]−12

): v(1, b, c) ∈ W and D = (d, e, f) ∈(

[2,6]3

): u(a, b, c) ∈ W. Note that (b, c) ∈ B and D ∈ D cannot satisfy

D ∩ b, c = c, since otherwise 〈u(d, e, f),v(1, b, c)〉 = −1. Let us notethat one can put intoW all the vectors u(1, b, c) with 2 ≤ b < c ≤ 6 becauseall the vectors with a −1 position have 1 in the rst coordinate.

113

Therefore |W| = 10 + |B|+ |D| holds. We have to prove

|B|+ |D| ≤ 11. (2.26)

Let us rst exhibit the system with 21 vectors showing F (6, 3, 0) ≥ 21.

U6 := u(1, b, c),v(1, b, c) : (b, c) ⊂ [2, 6] ∪ u(2, 3, 4).

We have |U6| = 2×(

52

)+ 1 = 21 and the only non-trivial scalar product to

check is that 〈v(1, b, c),u(2, 3, 4)〉 ≥ 0. It is true automatically if c ≥ 5. Forc = 3, 4 it follows that b ∈ 2, 3 making the scalar product equal to 0 asdesired.

To avoid a tedious case by case analysis it is simplest to give a matchingfrom the 9-element set D ∈

([2,6]

3

): D 6= (2, 3, 4) into the 10-element set

(b, c) : 2 ≤ b < c ≤ 6 such that whenever D and (b, c) are matched,D ∩ (b, c) = c holds. This will prove (2.26). An example of the desiredmatching is exhibited below.

(2, 3, 5) − (4, 5) (2, 4, 5) − (3, 5) (2, 3, 6) − (4, 6)

(2, 4, 6) − (5, 6) (2, 5, 6) − (3, 6) (3, 4, 5) − (2, 3)

(3, 4, 6) − (2, 4) (3, 5, 6) − (2, 5) (4, 5, 6) − (2, 6)

A careful analysis shows that U6 is the only extremal conguration. Theproof is complete.

114

Chapter 3.

Families with small matching number

My results presented in this chapter (joint with P. Frankl) are published in[94, 95, 100, 101].

3.1. Introduction and statement of results

Denition 3.1. Any collection of pairwise disjoint sets is called a matching.The maximum number of pairwise disjoint members of a family F is denotedby ν(F) and called the matching number of F .

Note that ν(F) ≤ n unless ∅ ∈ F . Several important classical results inextremal set theory deal with families of xed matching number.

Denition 3.2. For n ≥ s ≥ 2 dene

e(n, s) := max|F| : F ⊂ 2[n], ν(F) < s

.

Similarly, for positive integers n, k, s ≥ 2, n ≥ ks dene

ek(n, s) := max|F| : F ⊂

([n]

k

), ν(F) < s

.

Note that families with matching number 1 are intersecting and vice versa.Thus, ek(n, 2) is determined in the ErdosKoRado theorem (Theorem 1.2).It is also a simple result of the same authors that e(n, 2) = 2n−1.

For m =⌈n+1s

⌉the family(

[n]

≥ m

):= H ⊂ [n] : |H| ≥ m

does not contain s pairwise disjoint sets. Erdos conjectured that for n =sm− 1 one cannot do any better. Half a century ago Kleitman proved thisconjecture and determined e(sm, s) as well.

115

Theorem (Kleitman [151]). Let s ≥ 2,m ≥ 1 be integers. Then the follow-ing holds.

For n = sm− 1, we have e(n, s) =∑

m≤t≤n

(n

t

), (3.1)

for n = sm, we have e(n, s) =s− 1

s

(n

m

)+

∑m+1≤t≤n

(n

t

). (3.2)

The following matching example for (3.2) was proposed by Kleitman:K ⊂ [sm] : |K| ≥ m+ 1

∪(

[sm− 1]

m

).

(We have(sm−1m

)= s−1

s

(smm

).) Note that e(sm, s) = 2e(sm−1, s). In general,

e(n+ 1, s) ≥ 2e(n, s) is obvious, and, since the constructions of families thatmatch the bounds are easy to provide, (3.1) follows from (3.2). For s = 2both formulae give 2n−1, the easy-to-prove bound mentioned above. In thecase s = 3 there is just one case not covered by the Kleitman Theorem,namely n ≡ 1 ( mod 3). This was the subject of the PhD dissertation ofQuinn [200]. In it a very long, tedious proof for the following equality isprovided:

e(3m+ 1, 3) =

(3m

m− 1

)+

∑m+1≤t≤3m+1

(3m+ 1

t

). (3.3)

Unfortunately, this result was never published and no further progress wasmade on the determination of e(n, s).

Let us rst make a general conjecture.

Denition 3.3. Let n = sm+ s− l, 0 < l ≤ s. Set

P(s,m, l) :=P ⊂ 2[n] : |P |+ |P ∩ [l − 1]| ≥ m+ 1

,

Claim. ν(P(s,m, l)) < s.

Proof. Assume that P1, . . . , Ps ∈ P(s,m, l) are pairwise disjoint. Then

|P1|+ . . .+ |Ps| = |P1 ∪ . . . ∪ Ps| ≤ n = sm+ s− l

and|P1 ∩ [l − 1]|+ . . .+ |Ps ∩ [l − 1]| ≤

∣∣[l − 1]∣∣ = l − 1

hold. However, adding these two inequalities, the left hand side is at leasts(m+ 1) while the right hand side is s(m+ 1)− 1, a contradiction.

116

Conjecture 1. Suppose that s ≥ 2,m ≥ 1, and n = sm + s − l for some0 < l ≤ ds2e. Then

e(sm+ s− l, s) = |P(s,m, l)| holds. (3.4)

Let us mention that for l = 1 the formula (3.4) reduces to (3.1) and fors = 3, l = 2 it is equivalent to (3.3). Unfortunately, Conjecture 1 does notcover the whole range of parameters m, s, l. We discuss reasons for that andgive a meta-conjecture for all values of the parameters in Section 3.5.

My main result in this section (joint with Frankl) is the proof of Conjec-ture 1 in a relatively wide range.

Theorem 3.4 ([94, 101]). e(sm+ s− l, s) = |P(s,m, l)| holds for

(i) l = 2 and s ≥ 3, (ii) m = 1,

(iii) s ≥ lm+ 3l + 3.

The proof of (i) for s ≥ 5 is given in Section 3.2.4. The proof of (iii) isgiven in Section 3.2.5. The proof of (ii) is very easy, but it illustrates ourapproach for the proof of (iii), so we give it in the beginning of Section 3.2.5.Contrary to the intuition, the problem gets easier as s becomes larger, andthus the proof for s = 3, 4 is more intricate. It is given in Sections 3.3.2and 3.3.3, respectively. We discuss possible generalizations and open prob-lems in Section 3.5.

The proof for s = 3, 4 is based on a non-trivial averaging technique some-what in the spirit of Katona's circle method [139] (cf. also Section 2.2.3.3):we choose a certain conguration of sets, show that the intersection of afamily satisfying the conditions of Theorem 3.4 with each such congurationcannot be too large and then average over all such congurations. However,the conguration is quite complicated, the sets in the conguration actuallyhave weights, and, in order to bound the weighted intersection of the familywith each conguration, we use some kind of discharging method.

This method has proved to be very useful and was already used in severalpapers. In a recent paper [103], we applied it to completely resolve thefollowing problem studied by Kleitman: what is the maximum cardinality ofa family F ⊂ 2[n] that does not contain two disjoint sets F1, F2, along withtheir union F1 ∪ F2?

In Section 3.4, we analyze this method and obtain interesting new inequal-ities concerning the structure of the families with small matching number.

We also note that (3.1) and (3.2), along with more general statements,are proved using a simpler version of our technique in [102].

117

Recall that F is called an up-set if for any F ∈ F all sets that contain Fare also in F . When dealing with e(n, s), we may restrict our attention tothe families that are closed upward and shifted (cf. Section 1.2.1), which wetacitly assume in what follows.

The problems of determining e(n, s) and ek(n, s) are, in fact, closely inter-connected. In the proof of Theorem 3.4 we use some a stability result for theso-called Erdos Matching Conjecture, which is interesting in its own right.

There are several natural examples of a familyA ⊂(

[n]k

)satisfying ν(A) =

s for n ≥ (s+ 1)k. Following [80], let us dene the families A(k)i (n, s) :

A(k)i (n, s) :=

A ∈

([n]

k

): |A ∩ [(s+ 1)i− 1]| ≥ i

, 1 ≤ i ≤ k. (3.5)

Conjecture 2 (Erdos Matching Conjecture [69]). For n ≥ (s+ 1)k

ek(n, s+ 1) = max|A(k)

1 (n, s)|, |A(k)k (n, s)|

. (3.6)

The conjecture (3.6) is known to be true for k ≤ 3 (cf. [72, 184, 86]).Improving earlier results of [69], [32], [130] and [108], in [83]

ek(n, s+ 1) =

(n

k

)−(n− sk

)is proven for n ≥ (2s+ 1)k− s. (3.7)

Recently, in a joint work with P. Frankl [104], we have proved that

The equality (3.7) is true for any s ≥ s0 and n ≥ 53sk −

23s.

In the case s = 1 (that is, the case of the ErdosKoRado Theorem) onehas a very useful stability theorem due to Hilton and Milner [128]. Here, wepresent its analogue for the Erdos Matching Conjecture. Let us dene thefollowing families.

H(k)(n, s) :=H ∈

([n]

k

): H ∩ [s] 6= ∅

∪

[s+ 1, s+ k]−

−H ∈

([n]

k

): H ∩ [s] = s, H ∩ [s+ 1, s+ k] = ∅

.

Note that ν(H(k)(n, s)) = s for n ≥ sk and

|H(k)(n, s)| =(n

k

)−(n− sk

)+ 1−

(n− s− kk − 1

). (3.8)

The covering number τ(H) of a hypergraph is the minimum of |T | over all Tsatisfying T ∩H 6= ∅ for all H ∈ H. Recall the denition (3.5). If n ≥ k+ s,

118

then the equality τ(A(k)1 (n, s)) = s is obvious. At the same time, if n ≥ k+s,

then τ(H(k)(n, s)) = s+ 1 and τ(A(k)i (n, s)) > s for i ≥ 2.

Let us make the following conjecture.

Conjecture 3. Suppose that n ≥ (s + 1)k and F ⊂(

[n]k

)satises ν(F) =

s, τ(F) > s. Then

|F| ≤ max|A(k)

i (n, s)| : i = 2, . . . , k, |H(k)(n, s)|

holds. (3.9)

The HiltonMilner theorem (Theorem 1.3) shows that (3.9) is true fors = 1.

Let us mention that for n > 2sk the maximum on the RHS of (3.9) isattained on |H(k)(n, s)|. For n > 2k3s (3.9) was veried by Bollobas, Daykinand Erdos [32]. In the (yet unpublished) paper [102] we verify the conjecturefor n ≥ (2 + o(1))sk. Here I present a weaker, but easier-to-prove result,which we use in the proof of Theorem 3.4.

Theorem 3.5 ([101]). Let n = (u+ s− 1)(k− 1) + s+ k, u ≥ s+ 1. Thenfor any family G ⊂

([n]k

)with ν(G) = s and τ(G) ≥ s+ 1 we have

|G| ≤(n

k

)−(n− sk

)− u− s− 1

u

(n− s− kk − 1

). (3.10)

It is proved in Section 3.2.3. I also note that stability results for n >

C(s)k with unspecied dependence of C(s) on s we proved by Ellis, Kellerand Lifshitz in [66] using the methods of discrete Fourier transform. Later,much more general stability results that lead to progress in many extremalquestions were obtained by Keller and Lifshitz in [148].

3.2. Proof of Theorem 3.4 for s ≥ 5

We start by providing some results necessary for the proof of Theorem 3.4(ii), (iii) and (i) for s ≥ 5.

3.2.1. Averaging over partitions

The exposition in this subsection follows very closely the original proof ofKleitman [151], borrowing a large part of notation and statements from there.

Let n := sm + s − l, 1 ≤ l ≤ s, for this subsection. Consider a familyF ⊂ 2[n], ν(F) < s. Put F := 2[n] −F and y(q) := |F ∩

([n]q

)|.

119

Let π be an ordered partition of a positive integer x, x ≤ n, into s positiveintegers p1, . . . , ps:

∑sr=1 pr = x. In what follows we simply call any such

π a partition. For an s-tuple T = (A1, . . . , As) of disjoint subsets of [n] wesay that T is of type π if |Ar| = pr. We note that, in general, T does notpartition the whole set [n]. Dene the following class of s-tuples:

Ci(π) :=T = (A1, . . . , As) : T is of type π, |r : Ar ∈ F| = i

.

Informally, an s-tuple T of type π belongs to Ci(π) if exactly i sets from thetuple are not contained in F .

The number of s-tuples of type π we denote n(π). Clearly,

n(π) =n!

(n−∑s

r=1 pr)!∏s

r=1 pr!. (3.11)

We set Xi(π) := |Ci(π)|/n(π).

The following simple lemma is essentially stated in [151]:

Lemma 3.6. For any partition π we have

s∑i=0

Xi(π) =s∑i=1

Xi(π) = 1. (3.12)

s∑i=1

iXi(π) =s∑r=1

y(pr)/(n

pr

). (3.13)

The rst statement is evident, while the second one is veried by a sim-ple application of double counting: count in two ways the number of pairs(s-tuple of type π; a subset from F that belongs to the s-tuple).

Denote by πe the partition of x = ms into s equal parts. From (3.13) weimmediately get that

y(m) =1

s

(n

m

) s∑i=1

iXi(πe). (3.14)

The proof of the following lemma relies on the ideas of the proof of ([151],Lemma 2):

Lemma 3.7. For any 1 ≤ u ≤ s− l we have

y(m+ u) ≥ 1

s

(n

m+ u

) s−lu∑i=1

Xi(πe). (3.15)

120

Proof. Consider an s-tuple T that belongs to Ci(πe) for some 1 ≤ i ≤ s−lu .

Distribute some iu elements of [n] \T evenly between the i sets of T belong-ing to F . We have ui ≤

∣∣[n] \ T∣∣ = s− l, and so the number of elements in

[n]\T is sucient. Since ν(F) < s, at least one of the obtained (m+u)-setsmust be in F . We say that any (m + u)-set from F obtained in this way isassociated with T .

Consider a bipartite graph G = (A,B,E), where the part A consistsof the m-sets from T ∩ F , the part B :=

([n]\Tu

)and E contains the edge

connecting an m-set and a u-subset of [n] \ T i their union belongs to F .Take at random a subfamily B′ ⊂

([n]\Tu

)of i pairwise disjoint u-element sets,

and consider a subgraph G′ of G induced on A and B′.Since ν(F) < s, there is no perfect matching in G′: if we distribute the

elements from B′ between the m-sets from T as it is suggested by the perfectmatching, we would get an s-tuple of pairwise disjoint sets with all sets inF . Therefore, the number of edges in G′ is at most i(i− 1). Averaging overall choices of B′, we get that

|E| ≤(∣∣[n] \ T

∣∣u

)(i− 1) =

(s− lu

)(i− 1).

Thus, the number of non-edges in the bipartite graph is at least(s−lu

). Each

non-edge corresponds to an (m+ u)-set associated with T , and so there areat least

(s−lu

)dierent (m+ u)-sets associated with each T .

On the other hand, each set from F ∩(

[n]m+u

)is associated with at most

N s-tuples of the type πe, where

N := s

(m+ u

u

)(n−m− u)!

(m!)s−1(s− l − u)!.

Therefore, by double counting we get that

y(m+ u) ≥(s−lu

)N

s−lu∑i=1

|Ci(πe)|(3.11)=

1

s

(n

m+ u

) s−lu∑i=1

Xi(πe).

3.2.2. Calculations

In this subsection we prove some technical claims necessary for the proof ofTheorem 3.4.

121

Claim. 1. For n = sm+ s− 2 the following inequality holds:(s− 2− 1

s− 2

)(nm

)+

m∑j=1

(s− 1)

(n

m− j

)≤(

n

m+ 1

). (3.16)

2. For n = sm+ s− l with s ≥ 3, s ≥ l ≥ 2 we have

s− l2

(n

m

)≤(

n

m+ 2

). (3.17)

3. For n = sm+ s− l we have

(s− l)(n

m

)≤(

n

m+ 1

). (3.18)

Proof. 1. Indeed, we have( nm−j)

( nm−j−1)

= n−m+j+1m−j > s − 1 for any j ≥ 0.

Therefore,∑m

j=1(s − 1)(

nm−j)≤ (s − 1) 1

1− 1s−1

(n

m−1

)= (s−1)2

s−2

(n

m−1

). On the

other hand, we have

(n

m+ 1

)−(s− 2− 1

s− 2

)(nm

)=[m(s− 1) + s− 2

m+ 1−(s− 2− 1

s− 2

)](nm

)>

s− 1

s− 2

m

m+ 1

(n

m

)=s− 1

s− 2

m

m+ 1

n−m+ 1

m

(n

m− 1

)=s− 1

s− 2

(s− 1)(m+ 1)

m+ 1

(n

m− 1

)=

=(s− 1)2

s− 2

(n

m− 1

).

2. For s = l the statement is obvious, thus we assume that s > l. We have(n

m+ 2

)=

((s− 1)m+ s− l)((s− 1)m+ s− l − 1)

(m+ 1)(m+ 2)

(n

m

)>

> (s− l)(s− 1)m+ s− l − 1

m+ 2

(n

m

)≥ (s− l)(s− 1)m

m+ 2

(n

m

).

The last expression is greater than s−l2

(nm

)for any s ≥ 3,m ≥ 1.

3. We have(

nm+1

)= (s−1)m+s−l

m+1

(nm

)≥ (s− l)

(nm

).

As in the previous subsection, consider a family F ⊂ 2[n] with ν(F) < s.

Claim 3.8. For n = s(m+ 1)− l, s ≥ 3,m ≥ 1, we have

y(m) +1

2y(m+ 1) + y(m+ 2) ≥ 1

s

(n

m

)(s− l+ 1 +

s∑i=s−l+2

Xi(πe)). (3.19)

122

For n = s(m+ 1)− 2, s ≥ 4,m ≥ 1 we have

y(m) +(s− 5/2)

(nm

)(n

m+1

) y(m+ 1) + y(m+ 2) ≥

≥ 1

s

(n

m

)(s− 1 +

s−2∑i=1

(i− 3

2

)Xi(πe) +X1(πe) +Xs(πe)

)(3.20)

Proof. We give only the proof of (3.19), the proof of (3.20) is analogous.Indeed, by (3.18) and (3.15) with u = 1 we have

1

2y(m+ 1) ≥

s−l2

(nm

)(n

m+1

) y(m+ 1) ≥ s− l2s

(n

m

) s−l∑i=1

Xi(πe).

By the inequalities (3.17) and (3.15) with u = 2 we have

y(m+ 2) ≥s−l2

(nm

)(n

m+2

) y(m+ 2) ≥ s− l2s

(n

m

) s−l2∑i=1

Xi(πe).

Adding them up with (3.14) we get that the left hand side of the inequality(3.19) is at least 1

s

(nm

)∑si=1 αiXi(πe), where each coecient αi is at least

s − l + 1, moreover, αi ≥ s − l + 2 for i ≥ s − l + 2. Using (3.12), we get(3.19).

3.2.3. Hilton-Milner-type result for Erdos Matching Conjecture

We conclude the preparations with the proof of 3.5. For s = 1 the inequalityfollows from Theorem 1.3, therefore we may assume that s ≥ 2. Considerany family G satisfying the requirements of the theorem. The proof uses thetechniques developed in [83].

The case of shifted G

First we prove Theorem 3.5 in the assumption that G is shifted. Following[83], we say that the families F1, . . . ,Fs are nested, if F1 ⊃ F2 ⊃ . . . ⊃ Fs.We also say that the families F1, . . . ,Fs are cross-dependent if for any Fi ∈Fi, i = 1, . . . , s, there are two distinct indices i1, i2, such that Fi1, Fi2 inter-sect. The following lemma may be proven by a straightforward modicationof the proof of Theorem 3.1 from [83]:

123

Lemma 3.9 ([83]). Let N ≥ (u + s− 1)(k − 1) and F1, . . . ,Fs ⊂(

[N ]k−1

)be

cross-dependent and nested, then

|F1|+ |F2|+ . . .+ |Fs−1|+ u|Fs| ≤ (s− 1)

(N

k − 1

). (3.21)

We use the following notation. For any p ∈ [n] and a subset Q ⊂ [1, p]dene

G(Q, p) := G \Q : G ∈ G, G ∩ [1, p] = Q.The rst step of the proof of Theorem 3.5 is the following lemma.

Lemma 3.10. Assume that |G|−|G(∅, s)| ≤(nk

)−(n−sk

)−C for some C > 0.

Then

|G| ≤(n

k

)−(n− sk

)− u− s− 1

uC. (3.22)

Proof. Recall the denition of the immediate shadow

∂H :=H : ∃H ′ ∈ H, H ⊂ H ′, |H ′ −H| = 1

.

For every H ∈ ∂G(∅, s+1) we have H ∈ G(s+1, s+1), since G is shifted.Combining this with the inequality s|∂H| ≥ |H| from ([83], Theorem 1.2),valid for any H with ν(H) ≤ s, we get

|G(∅, s+ 1)| ≤ s|G(s+ 1, s+ 1)|. (3.23)

For anyQ ⊂ [1, s+1], |Q| ≥ 2, we haveA(k)1 (n, s)(Q, s+1) =

([s+2,n]k−|Q|

), and

so we have |G(Q, s+1)| ≤ |A(k)1 (n, s)(Q, s+1)|. We also haveA(k)

1 (n, s)(∅, s+1) = ∅ and

∑s+1i=1 |A

(k)1 (n, s)(i, s+1)| = s

(n−s−1k−1

). Using (3.23) and (3.21),

we have

|G(∅, s+ 1)|+s+1∑i=1

|G(i, s+ 1)| ≤s∑i=1

|G(i, s+ 1)|+ (s+ 1)|G(s+ 1, s+ 1)| ≤

≤ s

(n− s− 1

k − 1

)− (u− s− 1)|G(s+ 1, s+ 1)|.

Thus, |A(k)1 (n, s)|−|G| ≥ (u−s−1)|G(s+1, s+1)|

(3.23)

≥ u−s−1s+1 |G(∅, s)|.

On the other hand, the inequality from the formulation of the lemma tellsus that |A(k)

1 (n, s)| − |G| ≥ C − |G(∅, s)|. Adding these two inequalities (thesecond one taken with coecient u−s−1

s+1 ), we get that |A(k)1 (n, s)| − |G| ≥

u−s−1u C.

124

Therefore, to prove Theorem 3.5, we need to show that C ≥(n−s−kk−1

). We

use the following simple observation:

Observation 3.11. If for some C > 0 and B ⊂(

[s+1,n]k−1

)we have

s∑i=1

∣∣G(i, s) ∩ B∣∣ ≤ s|B| − C,

then |G| − |G(∅, s)| ≤(nk

)−(n−sk

)− C.

Since G(∅, s) is non-empty and shifted, we have [s + 1, s + k] ∈ G(∅, s).Put B :=

([s+k+1,n]

k−1

). Denote GB(i, s) := G(i, s) ∩ B, i = 1, . . . , s. Then

the families GB(i, s), i = 1, . . . , s, are cross-dependent and nested. From(3.21) we get the inequality

|GB(1, s)|+ . . .+ |GB(s− 1, s)|+ u|GB(s, s)| ≤ (s− 1)

(n− s− kk − 1

)= s|B| −

(n− s− kk − 1

).

Applying Observation 3.11, the above inequality implies the desired boundon C. The last thing we note is that the condition n ≥ (u+s−1)(k−1)+s+kis exactly the one needed for the proof to work. The proof of Theorem 3.5for shifted families is complete.

The case of not shifted G

Consider an arbitrary family G satisfying the requirements of the theorem.Since the property τ(G) > s is not necessarily maintained by shifting, wecannot make the family G shifted right away. However, each (i, j)-shift, 1 ≤i < j ≤ n, decreases τ(G) by at most 1, and so we perform the (i, j)-shifts(1 ≤ i < j ≤ n) one by one until either G becomes shifted or τ(G) = s + 1.In the former case we fall into the situation of the previous subsection.

Assume w.l.o.g. that τ(G) = s + 1 and that each set from G intersects[s + 1]. Then all families G(i, s + 1), i = 1, . . . , s + 1, are nonempty.Make the family G shifted in coordinates s + 2, . . . , n by performing all the(i, j)-shifts for s + 2 ≤ i < j ≤ n. Denote the new family by G again.Since the shifts do not increase the matching number, we have ν(G) ≤ s andτ(G) ≤ s+ 1. Each family G(i, s+ 1) contains the set [s+ 2, s+ k].

Next, perform all possible shifts on coordinates 1, . . . , s+1, and denote theresulting family by G ′. We have |G ′| = |G|, ν(G ′) ≤ s, and, most importantly,G ′(i, s + 1) are nested and non-empty for all i = 1, . . . , s + 1. The lastclaim is true due to the fact that all of the families contained the same setbefore the shifting.

125

We can actually apply the proof of the previous subsection to G ′. Indeed,the main consequence of the shiftedness we were using is that G ′(i, s +1), i = 1, . . . , s+ 1, are all non-empty and nested. We do have it for G ′. Theother consequence was the bound (3.23), which we do not need in this caseas G ′(∅, s + 1) is empty since each set from G ′ intersects [s + 1]. The proofof Theorem 3.5 is complete.

3.2.4. Proof of Theorem 3.4 (i) for s ≥ 5

Put n := sm + s − 2 for this section. Consider a family F ⊂ 2[n] withν(F) < s. In terms of Section 3.2.1, the statement (i) is equivalent to thefollowing inequality:

n∑r=0

y(r) ≥(n− 1

m

)+

m−1∑r=0

(n

r

). (3.24)

Applying the inequality (3.13) with the partition (m−j,m+1, . . . ,m+1),we get

y(m− j) + (s− 1)

(n

m−j)(

nm+1

)y(m+ 1) ≥(

n

m− j

). (3.25)

Thus, for s ≥ 4, using (3.16), (3.25), and (3.20), we get

n∑r=0

y(r)(3.16)

≥m∑r=0

y(r) +(s− 5

2)(nm

)+ (s− 1)

∑mj=1

(n

m−j)(

nm+1

) y(m+ 1) + y(m+ 2)

(3.25),(3.20)

≥m∑j=1

(n

m− j

)+

1

s

(n

m

)(s− 1 +

s−2∑i=1

(i− 3

2

)Xi(πe) +X1(πe) +Xs(πe)

). (3.26)

Our goal is to prove the following lemma, which is the main ingredient weadd to the technique of [151].

Lemma 3.12. For s ≥ 5 and a family F ⊂ 2[n] with ν(F) < s we have

X1(πe) +s−2∑i=1

(i− 3

2

)Xi(πe) +Xs(πe) ≥

s− 2

n. (3.27)

We rst deduce part (i) from Lemma 3.12. Note that s − 1 + s−2n =

(s−1)(ms+s−2)+s−2n = s(n−m)

n . Taking that and (3.27) into account and contin-uing the chain of inequalities (3.26), we get that

n∑r=0

y(r) ≥m∑j=1

(n

m− j

)+

1

s

(n

m

)(s− 1 +

s− 2

n

)=

m∑j=1

(n

m− j

)+n−mn

(n

m

).

126

Finally,(n−1m

)= n−m

n

(nm

), which concludes the proof of the rst part of The-

orem 3.4.

Remark. We explain the motivation behind Lemma 3.12. The densitiesXi(πe) for i 6= s− 1 have strictly positive coecients in (3.26). It is only thefact that Xs−1(πe) does not appear in (3.26) that prevents us from gettinga better bound on

∑nk=0 y(k) right away. Thus, we want to prove that the

densities Xi(πe), i 6= s − 1, contribute suciently to the expression on theright hand side of (3.26). Moreover, we implicitly say that the contributionof the densities Xi(πe), i 6= s− 1, for any family F ∩

([n]m

)is at least as big

as the contribution of these densities in the case when F ∩(

[n]m

)is a full star.

(We remind the reader that a full star consists of all m-sets that contain agiven element.) In the extremal family P(s,m, 2) the subfamily P ∩

([n]m

)indeed forms a trivial intersecting family, and this partly explains why weobtain tight bounds on e(n, s) in this case.

We are going to derive (3.27) using Katona's circle method. Let σ be anarbitrary permutation of [n]. Think of the vertices σ(1), . . . , σ(n) as beingarranged on a circle: the vertex next to σ(i) in the clockwise order is σ(i+1),with i+ 1 computed modulo n. For an arbitrary i, 1 ≤ i ≤ n, let Di denotethe circular arc σ(i), . . . , σ(i+m−1), with the computations made modulon.

We deal with s-tuples of pairwise disjoint arcs, and so it is natural to lookat the Di in the following order: Dj, Dm+j, D2m+j, . . .. Let d denote thegreatest common divisor of m and s − 2 and put n := n

d . The above chainof Di's will close after n steps, that is, Dj+nm = Dj holds.

Having several chains may look like an additional trouble but actually itis working in our favor. We end up partitioning the n circular arcs of lengthm into d groups of n arcs. Let Dj, Dm+j, . . . , Dj+(n−1)m form any of thesegroups and let us arrange the numbers 0, 1, . . . , n− 1 on a circle and denethe set R := r : Dj+rm ∈ F.

The objects that interest us most are arcs of length s on this circle. LetCr be the arc starting at r. That is, Cr := r, r + 1, . . . , r + s − 1. Itcorresponds to s pairwise disjoint sets Dj+rm, . . . , Dj+(r+s−1)m. The familyof s-tuples of m-sets, represented by Cr, we denote by C(σ). Note that theorder of sets in the tuple corresponding to each Cr is xed: it is also circular.We use this notation in the averaging part of the proof.

Let us dene fb(R) :=∣∣r : 0 ≤ r < n : |Cr ∩ R| = b

∣∣. The following

127

simple claim is the main tool for proving the analogue of (3.27) on the circle.

Claim 3.13. Dene t, 1 ≤ t < s− 1, by the equation n ≡ t(mod s). Thenat least one of the following possibilities holds:

(i) f0(R) ≥ t,

(ii) f1(R) = 0,

(iii) f2(R) ≥ 2.

Proof. We may assume that (ii) does not hold. Let us note that∣∣|Cr ∩R| −

|Cr+1 ∩R|∣∣ ≤ 1, i.e., |Cj ∩R| is continuous. Consequently, if |Cu ∩R| ≥ 3

for some u, then (iii) holds.Indeed, choosing some v satisfying |Cv ∩R| = 1, u and v divide the circle

into two parts and by the continuity of |Cr ∩R| on each part there exists atleast one r satisfying |Cr ∩R| = 2.

From now on we assume fb = 0 for b ≥ 3 and f2(R) ≤ 1. This implies thatany two vertices of R are at least s−1 apart on the circle. If they are exactlys − 1 apart then there is a Cr containing both of them, i.e., |Cr ∩ R| = 2.Therefore, this can occur at most once.

On the one hand, we have

f0(R) + f1(R) + f2(R) = n. (3.28)

On the other hand, every vertex belongs to Cu for exactly s values of u.So, if f2(R) = 0, then, counting the total degree of vertices in R, we getf1(R) = |R|s. Since f1(R) ≤ n ≡ t(mod s), f0(R) ≥ t follows from (3.28).

If f2(R) = 1, then f1(R) + f2(R) = |R|s − 1. Since t < s − 1, we inferf0(R) ≥ t+ 1 from (3.28), concluding the proof of the claim.

Now we are ready to state and prove (3.27) for the arcs of length min the cyclical permutation σ. Let xi denote the number of those s-tuplesDj, Dj+m, . . . , Dj+(s−1)m from which exactly s − i are members of F (thatis, i are members of F).

Lemma 3.14. Let n = sm+ s− 2. In the notations above, for any permu-tation σ we have

x1 +s−2∑i=1

(i− 3

2

)xi + xs ≥ s− 2. (3.29)

Proof. First consider the case n = n, i.e., the greatest common divisor d ofm and s− 2 is equal to 1. In this case xs−i = fi(R) for all i = 0, . . . , s. Letus apply Claim 3.13. Note that, in the denitions of the claim, t = s− 2. In

128

the case (i) we get xs ≥ s− 2 and in case (iii) the left hand side of (3.29) isbounded from below by 2(s−2− 3

2), which is greater than s−2 for s ≥ 5. Inthe remaining case (ii) we have xs−1 = 0. Since in (3.29) all xi for i 6= s− 1have coecient at least 1

2 , the statement follows from∑s

i=1 xi = n ≥ 2s− 2.Now suppose that d ≥ 2. We apply the claim separately to each of the d

disjoint circles of length n. Let Dj := Dj, Dm+j, . . . , Dj+(n−1)m be one ofthese circles. Similarly to xi, let x

ji be the number of s-tuples Dj+um : u ≡

u1, . . . , u1 + s− 1 mod n from which exactly s− i are members of F . Wehave

∑d−1j=0 x

ji = xi for each i = 0, . . . , s. Below we verify that

xj1 +s−2∑i=1

(i− 3

2

)xji + xjs ≥

s− 2

d. (3.30)

It is clear that, summing over j, the inequality (3.30) implies (3.29).Note that n = n

d ≡s−2d (mod s). We apply Claim 3.13 with t = s−2

d . Theinequality (3.30) is satised if xjs ≥ t or if xjs−2 ≥ 1. The only remainingpossibility is xjs−1 = 0. Then xs +

∑s−2i=0 x

ji = n, and we can lower bound

the left hand side of (3.30) by n2 . But then we have n

2 = n2d ≥

2s−22d > s−2

d ,concluding the proof.

Remark. It is not dicult to verify that the argument above works fors = 4 and even m due to the fact that in that case we have d = 2, and eachof the disjoint circles contributes at least 1 to the sum in the left hand sideof (3.29).

We are left to do a standard averaging, always used in the applicationsof Katona's circle method. We sum over all σ the value of the expression inthe left hand side of (3.29) and compute the sum in two ways: grouping thesummands with the same σ, and grouping the ones that belong to the sameclass Ci(πe) of s-tuples. For any σ the left hand side of (3.29) is at least s−2by Lemma 3.14. On the other hand, each s-tuple belongs to the collection

129

C(σ) for n(m!)s(s− 2)! permutations. We have

n!(s− 2)(3.29)

≤∑σ

[|C(σ) ∩ C1(πe)|+

s−2∑i=1

(i− 3

2

)|C(σ) ∩ Ci(πe)|+ |C(σ) ∩ Cs(πe)|

]=

= n(m!)s(s− 2)!

[|C1(πe)|+

s−2∑i=1

(i− 3

2

)|Ci(πe)|+ |Cs(πe)|

](3.11)=

= n(m!)s(s− 2)!n!

(m!)s(s− 2)!

[X1(πe) +

s−2∑i=1

(i− 3

2

)Xi(πe) +Xs(πe)

]=

= nn!

[X1(πe) +

s−2∑i=1

(i− 3

2

)Xi(πe) +Xs(πe)

].

Dividing the rst and the last expression by nn!, we get that X1(πe) +∑s−2i=1 (i− 3

2)Xi(πe) +Xs(πe) ≥ s−2n .

3.2.5. Proof of Theorem 3.4 (ii), (iii)

We restrict our attention to the families that are shifted and up-sets. Thestatement (ii) is equivalent to the following proposition.

Proposition 3.15. Put n := 2s − l for some 1 ≤ l < s. Let F ⊂ 2[n],ν(F) < s. Then

|2[n] −F| ≥ 2(s− l) + 2. (3.31)

Proof. Since F is closed upward, ∅ /∈ F . If there are at there are at mostl − 1 singletons in F , then (3.31) holds. Otherwise, i ∈ F , 1 ≤ i ≤ l.Consider

G := F ∈ F : F ⊂ [l + 1, 2s− l].The family G contains no s− l pairwise disjoint sets, so

|G| ≤ e(2(s− l), s− l) = 2e(2(s− l)− 1, s− l) (3.1)= 2

∑2≤t≤2(s−l)−1

(2(s− l)− 1

t

)=

= 2(22(s−l)−1 − 2(s− l)

)= 22(s−l) − 4(s− l).

Thus |2[n] − F| ≥ |2[l+1,n] − G| ≥ 4(s − l), proving (3.31) in this case aswell.

We go on to the proof of (iii). Put n := sm+s−l for the rest of the section.Consider the maximum family F with ν(F) < s. As before, we denote thecomplementary family 2[n] − F by F . We have |F| ≥ |P(s,m, l)|. Ourstrategy for proving the theorem is to study the subfamilies Fj = F ∩

([n]j

)for j ≤ m+1 and show successively that F is closer and closer to P(s,m, l).

130

Conjecture 1 holds for l ≤ 2, so we assume for the rest of the section thatl ≥ 3.

We start with the following lemma.

Lemma 3.16. ν(F0 ∪ . . . ∪ Fm) ≤ l − 1.

Proof. Assume for contradiction that F1, . . . , Fl ∈ F0∪ . . .∪Fm are pairwisedisjoint. Set T := F1∪ . . .∪Fl and note that |T | ≤ lm. We have |[n]−T | ≥sm+s− l− lm = (s− l)(m+1). Choose a subset U of [n]−T of cardinality(s− l)(m+ 1).

We have ν(F ∩ 2U) < s − l. Recall that y(i) =∣∣F ∩ ([n]

i

)∣∣. Applyingequality (3.14) for the (m+ 1)-element sets of F ∩ 2U we get

y(m+ 1) ≥∣∣∣F ∩ ( U

m+ 1

)∣∣∣ ≥ 1

s− l

((s− l)(m+ 1)

m+ 1

)=

((s− l)(m+ 1)− 1

m

). (3.32)

On the other hand, from (3.19) we get

y(m) +1

2y(m+ 1) + y(m+ 2) ≥ s− l + 1

s

(n

m

). (3.33)

Combining (3.32) and (3.33), we get

n∑k=0

y(k) ≥m+2∑k=m

y(k) ≥ 1

2

((s− l)(m+ 1)− 1

m

)+s− l + 1

s

(n

m

). (3.34)

Assume that for s ≥ lm + 3l + 3 the last expression exceeds∑m

k=0

(nk

).

Then we obtain a contradiction with the assumption that F has maximalpossible cardinality among families with no s pairwise disjoint sets, since∑m

k=0

(nk

)is a crude upper bound on the number of subsets of 2[n] missing

from P(s,m, l).

We have( nk−1)(nk)≤ 1

s−1 for any 1 ≤ k ≤ m, therefore for any q ≤ m

q∑k=0

(n

k

)≤ s− 1

s− 2

(n

q

)≤(1 +

2

s

)(nq

). (3.35)

From (3.35) we get that the right hand side of (3.34) is greater than∑m

k=0

(nk

)if

1

2

((s− l)(m+ 1)− 1

m

)≥ l + 1

s

(n

m

). (3.36)

We have (s− l)(m+ 1)− 1 = n− lm− 1 and(n−lm−1

m

)(nm

) =lm∏i=0

n−m− in− i

≥(

1− m

n− lm

)lm+1

> 1− m(lm+ 1)

n− lm.

131

Therefore, the inequality (3.36) will follow from the inequality

1

2

(1− m(lm+ 1)

n− lm

)≥ l + 1

s⇔ s− 2l − 2

s≥ m(lm+ 1)

n− lm.

It is easy to check that for any s ≥ ml + 3l + 3 we get that s − 2l − 2 >ss−l(ml + 1). We also have n ≥ sm. Therefore, it is sucient to show that

1

s− l≥ m

sm− lm,

which is obviously true.

The inequality of Frankl [80] that bounds the size of i-uniform familieswith no matchings of size l gives |Fi| ≤ (l − 1)

(n−1i−1

)for each i ≤ m, and∑

i≤k

|Fi| ≤ (l − 1)∑i≤k

(n− 1

i− 1

)for any k, k ≤ m. (3.37)

Lemma 3.17. We have Fm ⊂F ∈

([n]m

): F ∩ [1, l − 1] 6= ∅

=: H.

Moreover,

|H − Fm| ≤ (l − 1)s− 1

s− 2

(n− 1

m− 2

). (3.38)

Proof. Note that

|P(s,m, l)| ≥∑j>m

(n

j

)+ |H|.

Since |F| ≥ |P(s,m, l)|, we have

|Fm| ≥ |H| −∑i<m

|Fi|. (3.39)

Using (3.37) with k = m− 1 and the bound (3.35), we get∑i<m

|Fi| ≤ (l − 1)m−1∑i=1

(n− 1

i− 1

)≤ (l − 1)

s− 1

s− 2

(n− 1

m− 2

). (3.40)

On the other hand, we know from [83] that |Fm| ≤ |H|. Moreover, if Fm *H, then we can apply Theorem 3.5 with k = m, s = l − 1 (note that inthis and only this equation s refers to the s from Theorem 3.5), u = s− l.Indeed, we have n = sm+ s− l ≥ sm = ((s− l) + (l− 1)− 1)(m− 1) + s+2m−2 ≥ ((s− l)+(l−1)−1)(m−1)+m+(l−1). Applying Theorem 3.5,we get

|Fm| ≤ |H| −s− 2l

s− l

(n− l −m+ 1

m− 1

). (3.41)

132

Comparing the right hand sides of (3.40) and (3.41), we get:

s−2ls−l(n−l−m+1m−1

)(l − 1)s−1

s−2

(n−1m−2

) =

s−2ls−l

(∏l+m−2i=0

n−m+1−in−i

)(n

m−1

)(l − 1) (s−1)(m−1)

(s−2)n

(n

m−1

) >

>(s− 2l)s

(s− l)l

l+m−2∏i=0

n−m+ 1− in− i

,

where the inequality follows from the fact that n > s(m−1) and (l−1)s−1s−2 <

l. Thus, the right hand side is at least

(s− 2l)s

(s− l)l

(1− m− 1

n−m− l + 2

)l+m−1

≥ (s− 2l)s

(s− l)l

(1−(m− 1)(l +m− 1)

sm

)≥

≥ (s− 2l)(s− l −m)

(s− l)l≥ 1,

provided s ≥ m+ 3l. Therefore, Fm ⊂ H and

|H − Fm| ≤∑i<m

|Fi| ≤ (l − 1)s− 1

s− 2

(n− 1

m− 2

).

The following claim concludes the proof of the statement (iii) of the the-orem.

Claim 3.18. For each i ≤ m−1 and each F ∈ Fm−i we have |F∩[1, l−1]| ≥i+ 1.

Proof. Assume the contrary and choose F ∈ Fm−i such that |F∩[1, l−1]| ≤ i.W.l.o.g., we may suppose that F ∩ [1, l − 1] = [1, i]. Consider the family

F ′m :=(2[i+1,n] ∩ Fm

)∪F \ [1, i]

.

Remark that ν(2[i+1,n] ∩ Fm) ≤ l − 1− i because of Lemma 3.17.If ν(F ′m) ≤ l−1− i as well, then, via an argument repeating the one after

Observation 3.11, we get that |F ′m| ≤(n−im

)−(n−l+1m

)−(n−l−m+2i+1

m−1

)+ 1.

Therefore, |Fm| ≤(nm

)−(n−l+1m

)−(n−l−m+2i+1

m−1

)+ 1. Making calculations

analogous to the ones made in Lemma 3.17, we get that the last inequalitycontradicts the inequality (3.38), provided s ≥ m+ 3l.

If ν(F ′m) ≥ l−i, then necessarily there exist sets Fj ∈ Fm, 1 ≤ j ≤ l−1−i,such that F, F1, . . . , Fl−1−i are pairwise disjoint.

133

Denote T = F ∪⋃l−1−ij=1 Fj and consider U = [n] \ T . We have |U | =

sm + s − l − (m − i) − (l − i − 1)m = (s − l + i)(m + 1). We also havethat ν

(Fm+1 ∩ 2U

)< s − l + i. Therefore, as in the proof of Lemma 3.16,

we apply equality (3.14) for sets of Fm+1 ∩ 2U and get

y(m+1) ≥∣∣∣F ∩( U

m+ 1

)∣∣∣ ≥ 1

s− l + i

((s− l + i)(m+ 1)

m+ 1

)=

((s− l + i)(m+ 1)− 1

m

).

This inequality is stronger than (3.32) and would lead us to the same contra-diction as in the proof of Lemma 3.16. The proof of the claim is complete.

We have thus shown that for each i, 0 ≤ i ≤ n, we have Fi ⊂ P(s,m, l)∩([n]i

), which concludes the proof of Theorem 3.4.

3.3. Proof of Theorem 3.4 (i) for s = 3, 4

3.3.1. Calculations

We are going to use the following inequality in the proofs:

k−1∑j=1

(sm+ s− 2

j

)≤ 1

s− 2

(sm+ s− 2

k

)for any k ≤ m, s ≥ 3. (3.42)

Indeed, we have(sm+s−2

k−j )(sm+s−2k−j−1 )

= sm+s−2−k+j+1k−j ≥ s− 1 for any j ≥ 0 and k ≤ m,

so by the formula for the summation of a geometric progression,

k−1∑j=1

(sm+ s− 2

j

)≤

1s−1

1− 1s−1

(sm+ s− 2

k

)=

1

s− 2

(sm+ s− 2

k

).

3.3.2. Proof of Theorem 3.4 for s = 3

We rst prove the theorem for m ≥ 3. Suppose that m ≥ 3 and put n :=3m+ 1 for this section. Consider a family F ⊂ 2[n] with ν(F) < 3. Take anarbitrary cyclic permutation σ (assumed in what follows to be the identitypermutation for simplicity) and x three disjoint m-element sets that formarcs in that permutation. This is what we call a triple. For x ∈ [3m + 1],the x-triple is the triple of m-sets that do not contain the element x. Itis clear that there is a one-to-one correspondence between the x's and thetriples. For each triple, we dene three groups of sets of sizes 1, . . . ,m + 3

134

and assign them weights. We call this ensemble of sets an x-family. Notethat the arithmetic operations in the denitions of the sets are performedmodulo n.

We dene three groups of sets, indexed by i = 0, 1, 2. In what follows,we dene group i. The i-th m-set H(m)

i (x) in the x-family has the formim + x + 1, . . . , im + x + m. The set H(m−j)

i (x) of size m− j for j < mhas the form im + j + x + 1, . . . , im + x + m. That is, it consists of thelast m − j elements of the m-set, if seen in the clockwise order. The sets∅ ⊂ H

(1)i (x) ⊂ H

(2)i (x) ⊂ · · · ⊂ H

(m)i (x) form a full chain. The denition of

the sets of size ≥ m + 1 is less straightforward. Each of the (≥ m + 1)-setsin the i-th group contains the corresponding m-set. The (m+ 1)-set

H(m+1)i (x;x) := H

(m)i (x) ∪ x

in group i is called central. Note that the extra element it has is the elementthat was left out by the m-sets and so H(m+1)

i (x;x) is disjoint of H(m)j (x) for

j 6= i. The two others

H(m+1)i (j;x) := H

(m)i (x) ∪ mj + x+ 1 for j ∈ 0, 1, 2 − i

are called lateral and are disjoint of the corresponding H(m−1)j (x) and the

remainingm-setH(m)j1

(x), where j1, j, i = 0, 1, 2. For each j ∈ 0, 1, 2−i, we dene two (m+ 2)-element sets: the central set

H(m+2)i (x, j;x) := H

(m)i (x) ∪ mj + x+ 1, x

and the lateral set

H(m+2)i (j, j;x) := H

(m)i (x) ∪ mj + x+ 2,mj + x+ 3.

The former ones are disjoint of H(m−1)j (x) and the m-set from the remaining

j1-th group, where j1, j, i = 0, 1, 2, while the latter ones are disjointof H(m−3)

j (x). Note that H(m+2)j (x, i;x) and H(m+2)

j1(i, i;x) are disjoint for

j1, j, i = 0, 1, 2. Finally, we have one (m+ 3)-set in each group:

H(m+3)i (x) := H

(m)i (x) ∪ x, x+ 1,m+ x+ 1, 2m+ x+ 1.

It is disjoint of the (m− 1)-sets from the other groups.Each set in each group gets a weight. We denote by wk the weight of the

k-element sets, with possible superscripts l, c depending on whether the setis lateral or central, respectively. Put

α :=(3m+ 2)m

4(2m+ 3)(2m+ 1), α′ := 1− 2α. (3.43)

135

Figure 3.1. All types of sets in G(σ) modulo rotation. The elements of the ground set arerepresented by dots, and the red dots represent the elements that are contained in the set.Inside the circles we specify the length of any interval of points that simultaneously eitherbelong or do not belong to the set (excluding the intervals of length 1 and 2). The weightsof k-sets are specied and divided by

(nk

)for shorthand. For (m + 1)- and (m + 2)-sets

we also mention, how many times does a given set appear as a central and lateral set.

Note that α′ > α. The weights are as follows (j ≥ 0):

wm−j :=

(n

m− j

); wlm+1 := α

(n

m+ 1

); wcm+1 := α′

(n

m+ 1

);

wm+3 :=

(n

m+ 3

); wlm+2 :=

1

8

(n

m+ 2

); wcm+2 :=

3

8

(n

m+ 2

). (3.44)

Thus, the total weight of all sets in an x-family is 3∑m+3

k=1

(nk

). Each set

F that appears in several x-families accumulates all the weight w(F ) that itwas assigned. On Fig. 3.1, we listed all the types of sets that are assignednon-zero weights, together with the corresponding weights. We recommendthe reader to verify Fig. 3.1, since we shall use the information provided onthe gure later in the proof! The elements of the ground set are placed onthe circle and the sets are represented modulo rotation. We denote by G(σ)

136

the family of all sets that got non-zero weight for a given permutation σ.Note that, for each j = 1, . . . ,m+ 3, we have∑

G∈G(σ)∩([n]j )

w(G) = 3n

(n

j

).

Claim 3.19. To prove the theorem for s = 3, it is sucient to show thatfor any σ we have

∑F∈F∩G(σ)

w(F ) ≤ 3n((

n− 1

m− 1

)+

m+3∑k=m+1

(n

k

))= 3m

(n

m

)+ 3n

m+3∑k=m+1

(n

k

). (3.45)

Proof. For an event A, denote by I[A] its indicator random variable. Denoteid the identity permutation. Indeed, if we take a permutation σ uniformlyat random, then, for each j = 1, . . . ,m+ 3, we have

Eσ

[ ∑F∈F∩G(σ)∩([n]

j )

w(F )

]=

∑F∈F∩([n]

j )

Eσ

[ ∑G∈G(id)∩([n]

j )

I[σ(F ) = G]w(G)

]=

=∑

F∈F∩([n]j )

∑G∈G(id)∩([n]

j )

Pr[σ(F ) = G]w(G) =∑

F∈F∩([n]j )

∑G∈G(id)∩([n]

j )

w(G)(nj

) =

=∑

F∈F∩([n]j )

3n = 3n∣∣∣F ∩ ([n]

j

)∣∣∣.Therefore, (3.45) implies that

m+3∑j=1

∣∣∣F ∩ ([n]

j

)∣∣∣ =1

3nEσ

[ ∑F∈F∩G(σ)

w(F )

]≤(n− 1

m− 1

)+

m+3∑k=m+1

(n

k

),

which implies the statement of the theorem for s = 3.

Our strategy to prove (3.45) is as follows. For a set F ∈ G(σ), we denethe charge c(F ) to be equal to w(F ) if F ∈ F , and to be 0 otherwise.Clearly,

∑F∈G(σ) c(F ) =

∑F∈F∩G(σ)w(F ). If among the (≤ m − 1)-sets in

G(σ) there are no sets from F , as well as there are at most m m-sets, thenwe are done since each m-set appears in exactly three x-families. Otherwise,certain (≥ m + 1)-sets do not appear in F . Then we transfer (a part of)the charge of the (≤ m)-sets to the (≥ m + 1)-sets that have zero charge.We show that the total charge transferred to each (≥ m + 1)-set is at mostits weight. As a result of this procedure, the (≤ m − 1)-sets will have zero

137

H(m−3)j1

(x) H(m+2)j2

(j1, j1;x) H(m+2)j3

(x, j1;x)

H(m−1)j1

(x) H(m−1)j2

(x) H(m+3)j3

(x)

H(m−1)j1

(x) H(m)j2

(x) H(m+2)j3

(x, j1;x)

H(m−1)j1

(x) H(m+1)j2

(x;x) H(m+1)j3

(j1;x)

H(m)j1

(x) H(m)j2

(x) H(m+1)j3

(x;x)

Table 3.1. The list of all types of triples of pairwise disjoint sets that we employ in theproof. We assume that j1, j2, j3 = 0, 1, 2. The triples are listed in the order theyappear in the proof.

total charge, the m-sets will have total charge at most 3m(nm

), and each

(≥ m + 1)-set will have a charge not greater than its own weight. This willobviously conclude the proof of the theorem.

Next, we design a charging scheme that satises the above requirements.For the sets of size at most m − 1, we transfer their charge within each x-family, assuring that the charge that we transferred to a bigger set in onex-family is smaller than the weight that this bigger set got from this x-family.See Table 1 for all the triples of pairwise disjoint sets we use in the proof.The reader is welcome to verify that all the triples are actually disjoint.

Stage 1. Transferring charge from the (≤m− 3)-sets to (m + 2)-sets.Assume that, for some x ∈ [n] and i ∈ 0, 1, 2, the set H(m−3)

i (x) is in thefamily. Choose j1, j2 such that j1, j2, i = 0, 1, 2. Then at least one setfrom each of the two pairs

(H

(m+2)j1

(i, i;x), H(m+2)j2

(x, i;x)),(H

(m+2)j2

(i, i;x),

H(m+2)j1

(x, i;x))is missing from F . We transfer 1

2 of the charge of the subsets

H(k)i (x), k ≤ m−3, which is at most 1

2

∑m−3k=1 wk, to some two of these missing

sets. We again remark that, for each x, we transfer only the part of the chargeof the sets H(k)

i (x) that they got as the member of the x-family. We have

1

2

m−3∑k=1

wk =1

2

m−3∑k=1

(n

k

)(3.42)

≤ 1

2

(n

m− 2

)=

1

2

3∏j=0

m+ 2− j2m+ j

(n

m+ 2

)<

1

16

(n

m+ 2

).

(We could have put 132 instead of 1

16 , but it does not matter for the calcula-tions.) Since each (m+ 2)-set in the x-family may get this charge from eachof the two groups to which it does not belong, the total charge transferred in

that way is at most 18

(n

m+2

) (3.44)= wl

m+2. The lateral (m+2)-sets are not goingto get any more charge. As for the central (m+2)-sets, we have to make sure

that they will get not more than wcm+2−wl

m+2

(3.44)= 1

4

(n

m+2

)additional charge.

138

We note that the (m− 2)-sets will be discharged together with the corre-sponding (m− 1)-sets.

Stage 2. Transferring charge from pairs of (m− 1)-sets to (m + 3)-sets.Due to the fact that F is an up-set, from now on we may assume that thecharge of any (≤ m−3)-set is 0. Assume that, for some x and i, j1, j2, wherej1, j2, i = 0, 1, 2, both H(m−1)

j1(x) and H(m−1)

j2(x) belong to F . Then the

set H(m+3)i (x) is not in F and, consequently, has zero charge. Transfer the

charge of the sets H(k)j1

(x) and H(k)j2

(x), k = m−2,m−1, to H(m+3)i (x). The

charge transferred is at most 2wm−2 + 2wm−1, which is

2

(n

m− 2

)+ 2

(n

m− 1

)≤ 3

2

(n

m

)=

3(m+ 1)

2(2m+ 1)

(n

m+ 1

)≤(

n

m+ 1

)≤(

n

m+ 3

)(3.44)

= wm+3,

since m ≥ 3. The (m+ 3)-sets are not going to get any more charge.

Stage 3. Transferring charge from (m− 1)-sets paired with m-sets to central (m + 2)-sets.Assume that, for some x ∈ [n] and i, j1, j2, where j1, j2, i = 0, 1, 2,both H

(m−1)j1

(x) and H(m)j2

(x) belong to F . Then the central (m + 2)-set

H(m+2)i (x, j1;x) is not in F and, consequently, received at most wl

m+2 chargewithin the x-family (because of the charge possibly transferred on Stage 1).Transfer the charge of the sets H(k)

j1(x), k = m−2,m−1, to H(m+2)

i (x, j1;x).The charge transferred is at most wm−2 + wm−1, which is(

n

m− 2

)+

(n

m− 1

)=

3m+ 2

2m+ 3

(n

m− 1

)=

3m+ 2

2m+ 3

2∏j=0

m+ j

2m− j + 2

(n

m+ 2

)≤ 1

4

(n

m+ 2

).

The last inequality is valid for any m ≥ 1 and is easy to verify by a directcalculation. The right hand side is exactly wc

m+2 − wlm+2, and so the total

charge of the central (m + 2)-sets is at most wcm+2 in each x-family. The

(m+ 2)-sets are not going to get any more charge.

Stage 4. Transferring charge from single (m− 1)-sets to (m + 1)-sets.After the above redistribution of charges, for each x ∈ [n], there are no(≤ m− 3)-sets and at most one (m− 2)-set and (m− 1)-set with non-zerocharges in the x-family, moreover, we cannot have an (m − 1)-set and twom-sets with non-zero charges in the x-family.

139

Assume that for some x ∈ [n] and i, j1, j2, where j1, j2, i = 0, 1, 2,the set H(m−1)

i (x) belongs to F . Then one set from each of the two pairs(H

(m+1)j1

(x;x), H(m+1)j2

(i;x))and

(H

(m+1)j1

(i;x), H(m+1)j2

(x;x))is missing from

F . We transfer 12 of the charge of the subsets H(k)

i (x), k = m − 2,m − 1,which is at most 1

2(wm−2 + wm−1), to each of these missing sets. We have

1

2(wm−2+wm−1) =

1

2

(n

m− 2

)+

1

2

(n

m− 1

)=

(3m+ 2)m

4(2m+ 3)(2m+ 1)

(n

m+ 1

)(3.44)

= wlm+1. (3.46)

Recall that α′ > α, and, therefore, wcm+1 > wl

m+1, which means that no(m+ 1)-set gets more charge than its weight up to this stage.

Stage 5. Transferring charge from pairs of m-sets to central(m + 1)-sets.Denote the number of m-sets that have non-zero charge (that is, that arecontained in F ∩ G(σ)) by q. If q ≤ m, then we are clearly done since eachm-set appears in exactly three x-families.

Assume that q > m. On the one hand, it makes an extra contribution3(q −m)

(nm

)to the left hand side of (3.45). On the other hand, the number

of triples with two m-sets belonging to F ∩ G(σ) is non-zero. Indeed, if, forj ≤ 2, we denote by zj the number of triples with j m-sets in the family,then, since each m-set participates in three triples, we have z1 + 2z2 = 3q.Since z0+z1+z2 = n, we have z2 ≥ 3q−n. Assume that for some x ∈ [n] andi, j1, j2, where j1, j2, i = 0, 1, 2, both H(m)

j1(x) and H(m)

j2(x) belong to F .

Then the central (m+ 1)-set H(m+1)i (x;x) is not in the family. Moreover, no

charge was transferred to it from the x-family, since we could not have hadtwo m-sets and an (m− 1)-set with non-zero charges at the same time afterStage 3. We transfer 3(q−m)

z2

(nm

)charge to H(m+1)

i (x;x) from the m-sets.

First note that we have transferred z23(q−m)z2

(nm

)charge from the m-sets

to the central (m + 1)-sets, which results in m-sets having total charge of3m(nm

). This is precisely what we needed to have, and we are only left to

verify that we did not overcharge the central (m + 1)-sets. Unfortunately,we can run into problems in this situation, so we have to consider two cases.First, assume that q ≥ m + 2. Then the charge on each central set in eachx-family is at most

3(q −m)

z2

(n

m

)≤ 3(q −m)

3q − n

(n

m

)≤ 6

5

(n

m

)=

6(m+ 1)

5(2m+ 1)

(n

m+ 1

). (3.47)

The second inequality holds due to the fact that for q ≥ m+ 1 the function3(q−m)3q−n decreases as q grows. Therefore, if α′ = (1 − 2α) ≥ 6(m+1)

5(2m+1) , then we

140

are done. Let us verify that this inequality is implied by (3.43). Adding 2αto the right hand side of the inequality, we get

6(m+ 1)

5(2m+ 1)+

(3m+ 2)m

2(2m+ 3)(2m+ 1)=

12(m+ 1)(2m+ 3) + 5(3m+ 2)m

10(2m+ 3)(2m+ 1)=

=39m2 + 70m+ 36

40m2 + 80m+ 30< 1,

where the last inequality holds for any m ≥ 1. Thus, we fullled all therequirements on the charging scheme and we are done in the case q ≥ m+ 2.

In the case q = m + 1, however, we run into trouble: the inequalityα′ ≥ 3(q−m)(m+1)

z2(2m+1) = 3(m+1)z2(2m+1) may not hold. Recall that z2 ≥ 3q − n = 2. We

are still ne if z2 ≥ 3 since the calculations in (3.47) still go through in thatcase. Thus we only need to examine the case z2 = 2, which we assume untilthe end of this section. The equation z2 = 2 means that there are exactly twotriples with two m-sets from F . This is the only part of the proof when weare not going to compare the amount of charge passed to the (m+ 1)-sets tothe portion of its weight inside the x-family. Instead, we compare the chargeto the full weight of the (m+ 1)-set.

We have two possible congurations with z2 = 2. One possibility is thatwe have two m-sets from F forming an interval of length 2m on the circle,and then the two triples contributing to z2 share the same twom-sets. In thiscase, the central (m + 1)-set that we forbid is the same in both triples, andit is of type 1a (see Fig. 3.1). Recall that this set has weight 2(α+α′)

(n

m+1

).

The other possibility is that we have two pairs of m-sets, with each pairseparated on the two sides by a third m-set forming a triple with the pair,and by the element missing from the triple, respectively. In this case, in eachof the corresponding two x-families we forbid a central (m+1)-set of type 1b.Each of these two sets (that are clearly dierent) has weight (α + α′)

(n

m+1

).

In either case, we need to transfer 3(m+1)2m+1

(n

m+1

)amount of charge from the

m-sets to some of the (m + 1)-sets. We transfer this weight to the central(m+ 1)-set(s), possibly overcharging it.

Assume rst that we do not have any (m− 1)-element sets in F ∩ G(σ).Then, in either of the possibilities described above, the central (m + 1)-setshave zero charge before Stage 5. Therefore, we are good if the weight of these(one or two) (m + 1)-sets is greater than the amount of charge we transferto them from the pairs of m-sets. Namely, we are good if

2(α + α′) = 2− 2α ≥ 3(m+ 1)

2m+ 1⇔ 2 ≥ 3(m+ 1)

2m+ 1+

(3m+ 2)m

2(2m+ 3)(2m+ 1). (3.48)

141

Figure 3.2. In each of the two cases the (m − 1)-set is represented by black points, andtwo (m + 1)-sets, which form a matching with the (m − 1)-set, are represented by red(grey) and yellow (light grey) points.

We have

3(m+ 1)

2m+ 1+

(3m+ 2)m

2(2m+ 3)(2m+ 1)=

15m2 + 32m+ 18

8m2 + 16m+ 6< 2,

where the last inequality holds for m ≥ 3. Thus, this case is covered.Finally, assume that there is at least one (m−1)-element set F ∈ F∩G(σ).

Then, as one can see from Fig. 3.2, it forbids at least one (m+ 1)-set of eachof the types 1a and 1b to appear. Denote them by M1,M2. We have seentwo paragraphs above that in either case of the arrangement of the m-setswe forbid sets of the same type (either one of type 1a, or two of type 1b).Therefore, at least one ofM1,M2 that did not get any charge at Stage 5. Weassume that it is a set M1 of type 1b (the other case is easier and is treatedsimilarly). The set M1 appears in two x-families and got some charge onlyat Stage 4. Moreover, M1 got at most α

(n

m+1

)charge from each of the two

x-families. Thus, the charge of M1 after all ve stages is at most 2α(

nm+1

).

This, in turn, means that it has extra capacity of at least (α′ − α)(

nm+1

).

We redistribute some part of the charge from the two (m + 1)-sets (thatappeared in the x-families with the pairs of m-sets from F) to M1. In orderto be able to full the requirements on the charging scheme, we need thetotal capacity of these (m+1)-sets to be greater than the charge we transfer.More precisely, it is sucient if the following inequality holds:

(3α′ − α) ≥ 3(m+ 1)

2m+ 1⇔ 3− 7α ≥ 3(m+ 1)

2m+ 1.

Note that we replaced the capacity of the (one or two) missing central (m+1)-set(s) by 2α′ since the α-part of the charge may have been already used up

142

by the (m − 1)-sets at Stage 4. We have veried above (see (3.48)) thatthe same inequality holds if one replaces 3 − 7α with 2 − 2α. One caneasily see (cf. (3.43)) that α ≤ 3

16 , so we have 3 − 7α > 2 − 2α. The caseq = m + 1 is examined in its entirety, and the proof of the theorem in thecase s = 3,m ≥ 3 is complete.

3.3.2.1. The case m ≤ 2, s = 3

In the argument above, we assumed that m ≥ 3. However, we want to provethe theorem for m ≥ 1, which leaves us with two cases: m = 1 and m = 2.If m = 1, then we have n = 4, and we have to show that at least four sets,including the empty set, are missing from a family F ⊂ 2[4] with ν(F) ≤ 2.If there is at most one singleton in F , then we are done. If there are atleast two singletons, say, 1 and 2, then F ∩ 23,4 is empty, which gives4 missing sets. The case m = 1 is covered.

If m = 2 then n = 7, and we have to show that at least 1+(

71

)+(

62

)= 23

sets are missing from F . If there is at least one singleton in F , say 1, thenF ∩ 2[2,7] is intersecting, and so, by the ErdosKoRado theorem, a half ofthe sets are missing from it. This gives 32 missing sets.

Thus, we may assume that there are no sets of size smaller than 2 in F .Now we may slightly modify the proof for the case m ≥ 3 so that it worksfor m ≥ 2. Namely, among the (m + 1)-sets, we give weights only to thecentral (m+ 1)-sets (the weights on other layers stay the same). Claim 3.19stays true in this case. Since we do not have sets of size smaller than m, wecan go to Stage 5 of the analysis, where we want to show that with the newweights (3.47) holds for any q ≥ m+ 1 for m = 2, n = 7:

3(q −m)

3q − n

(n

m

)≤ 3

2

(n

m

)≤(

n

m+ 1

)⇔ 3

2

(7

2

)=

63

2≤(

7

3

)= 35.

The last inequality obviously holds. Thus, for m = 2 we may terminate theproof right after (3.47). The proof is complete.

3.3.3. Proof of Theorem 3.4 for s = 4

We rst prove the theorem for m ≥ 3. We put n := 4m+ 2 for some m ≥ 3throughout this section. The logic of this proof is very similar to that of theproof in the case n = 3m + 1, and the proof is in a sense even simpler. Wepresent it somewhat more concisely.

143

We x an arbitrary permutation σ of the ground set. For simplicity, weassume that σ is the identity permutation. Quite predictably, dene fourgroups of sets, indexed by i = 0, 1, 2, 3 and forming an x-family. The fourm-sets H(m)

i (x) in an x-family are disjoint and form an interval of length 4m,leaving two contiguous elements x−1, x out (thus, the x-family is indexed bythe last of the two missing elements in the clockwise order). In what follows,we dene the i-th group. The sets in the i-th group of size m − j, j =1, . . . ,m, form a full chain together with H(m)

i (x):

H(m−j)i (x) := x+ 1 + j + im, . . . , x+ (i+ 1)m.

We again have both central and lateral (m + 1)- and (m + 2)-sets. The(m+ 1)-sets

H(m+1)i (x′;x) := H

(m)i (x) ∪ x′ for x′ = x, x− 1

in group i are called central. Note that the extra element in both sets is leftout by the m-sets, and so H(m+1)

i (x′;x) for both x′ = x − 1, x is disjoint ofthe m-set from the j-th group, j 6= i. The three others

H(m+1)i (j;x) := H

(m)i (x) ∪ jm+ x+ 1 for j ∈ 0, . . . , 3 − i

are called lateral and are disjoint of the corresponding H(m−1)j (x) and of the

m-set in the j′-group, j′ 6= i, j.For each j ∈ 0, . . . , 3− i, we dene two lateral (m+ 2)-element sets:

H(m+2)i (x′, j;x) := H

(m+1)i (j;x) ∪ x′ for x′ = x, x− 1,

and for each i we dene one central set:

H(m+2)i (x− 1, x;x) := H

(m)i (x) ∪ x− 1, x.

The former ones are disjoint of the (m − 1)-set from group j and the twom-sets from the remaining groups, while the latter one is disjoint of the threem-sets from the groups j, j 6= i. Finally, we have one (m + 5)-element setfor each i :

H(m+5)i (x) := H

(m)i (x)∪x− 1, x, x+ 1,m+x+ 1, 2m+x+ 1, 3m+x+ 1.

Each set in each group gets a weight. We denote by wk the weight of thek-element sets, with possible superscripts l, c depending on whether the setis lateral or central, respectively. The weights are as follows (j ≥ 0):

wm−j :=

(n

m− j

); wlm+1 :=

m

5(3m+ 2)

(n

m+ 1

); wcm+1 :=

1

2

(n

m+ 1

)− 3

2wlm+1;

wm+5 :=

(n

m+ 5

); wlm+2 :=

1

22

(n

m+ 2

); wcm+2 :=

8

11

(n

m+ 2

). (3.49)

144

It is easy to check that, for each i and k ∈ 1, . . . ,m+ 2 ∪ m+ 5, ineach group the weight of k-element sets sums up to

(nk

)for xed x, and that

wcm+1 is positive.As before, each set F that appears in some x-families accumulates all

the weight w(F ) that it was assigned. We denote by G(σ) the family of allsets that got non-zero weight for a given permutation σ. Analogously toClaim 3.19, to prove the theorem in this case, it is sucient to show that forany σ we have∑F∈F∩G(σ)

w(F ) ≤ 4n((n− 1

m− 1

)+

∑k∈1,2,5

(n

m+ k

))= 4m

(n

m

)+ 4n

∑k∈1,2,5

(n

m+ k

). (3.50)

For a set F ∈ G(σ), we dene the charge c(F ) to be equal to w(F ) ifF ∈ F , and c(F ) := 0 otherwise. Clearly,

∑F∈G(σ) c(F ) =

∑F∈F∩G(σ)w(F ).

We again design a scheme for the transfer of (a part of) the charge of the(≤ m)-sets to the (≥ m + 1)-sets that have zero charge. We show that thecharge transferred to each (≥ m + 1)-set is at most its weight. As a resultof this procedure, the (≤ m− 1)-sets will have zero total charge, the m-setswill have total charge 4m

(nm

), and each (≥ m + 1)-set will have charge not

greater than its own weight. This will obviously conclude the proof of thetheorem.

Next we design a charging scheme that satises the above requirements.For n = 4m + 2 it is sucient in all cases to redistribute the charge withineach x-family, assuring that the charge that we transferred to the larger setin one x-family is smaller than the weight that this bigger set got from thisx-family.

Stage 1. Transferring charge from triples of (m− 1)-sets to(m + 5)-sets.Assume that, for some x ∈ [n] and i1, i2, i3, j, where i1, i2, i3, j = 0, 1, 2, 3,the sets H(m−1)

iu(x) for u ∈ [3], belong to F . Then H(m+5)

j (x) is missing fromF , and, consequently, has zero charge. Transfer all the charge of the setsH

(k)iu

(x), k ≤ m− 1 to the missing (m+ 5)-set.The charge transferred is at most 3

∑m−1k=1 wk, which is

3

(n

m− 1

)+ 3

m−2∑k=1

(n

k

)(3.42)

≤ 9

2

(n

m− 1

)=

9∏5p=0(m+ p)

2∏3p=−2(3m+ p)

(n

m+ 5

)<

(n

m+ 5

)(3.49)

= wm+5,

145

where the last inequality holds for any m ≥ 2. Note that we apply (3.42)for k = m−1 in the rst inequality above (and in several places below). The(m+ 5)-sets are not going to get any more charge.

Stage 2. Transferring charge from pairs of (m− 1)-sets to lateral(m + 2)-sets.Suppose that, for some x ∈ [n] and i1, i2, j1, j2, where i1, i2, j1, j2 =

0, 1, 2, 3, both H(m−1)j1

(x) and H(m−1)j2

(x) belong to F . Then in each of

the four pairs(H

(m+2)i1

(x′, j′;x), H(m+2)i2

(x′′, j′′;x)), where j′, j′′ = j1, j2

and x′, x′′ = x − 1, x, one of the (m + 2)-sets is missing from F , and,consequently, has zero charge. Note that all these (m + 2)-sets are lateral.Transfer one quarter of the charge of the setsH(k)

j1(x) andH(k)

j2(x), k ≤ m−1,

to each of these missing sets.The charge transferred to each lateral (m+ 2)-set is at most 1

2

∑m−1k=1 wk,

which is

1

2

m−1∑k=1

(n

k

)(3.42)

≤ 3

4

(n

m− 1

)=

3m(m+ 1)(m+ 2)

4(3m+ 3)(3m+ 2)(3m+ 1)

(n

m+ 2

)<

1

22

(n

m+ 2

)(3.49)

= wlm+2,

where the last inequality holds for any m ≥ 1. The lateral (m+ 2)-sets arenot going to get any more charge.

Stage 3. Transferring charge from single (m− 1)-sets to (m + 1)-sets.After the above redistribution of charges, we have at most one (m − 1)-setwith non-zero charge in each x-family.

Assume that for some x ∈ [n] and i, j1, j2, j3, where j1, j2, j3, i =

0, 1, 2, 3, the set H(m−1)i (x) belongs to F and still has non-zero charge.

Then one set from each of the six triples(H

(m+1)j1

(π(i);x), H(m+1)j2

(π(x −1);x), H

(m+1)j3

(π(x);x)), where π is a permutation of the set i, x− 1, x, is

missing from F . It is not dicult to see that it means that at least three outof the listed sets are missing from F . Note that among the possible missingsets there are both central and lateral (m+ 1)-sets.

We transfer 13 of the charge of H(k)

i (x), k ≤ m − 1, to each of the threemissing sets. This is at most 1

3

∑m−1k=1 wk, which is

1

3

m−1∑k=1

(n

k

)(3.42)

≤ 1

2

(n

m− 1

)=

m(m+ 1)

2(3m+ 3)(3m+ 2)

(n

m+ 1

)=

m

6(3m+ 2)

(n

m+ 1

)(3.49)< wlm+1,

We are not going to transfer any more weight to the lateral (m+ 1)-sets.

146

Stage 4. Transferring charge from pairs and triples of m-sets.At this stage only the sets of size greater than or equal to m have non-negative charge. Denote the number of m-sets that have non-zero charge(that is, that are contained in F ∩G(σ)) by q. If q ≤ m, then we are clearlydone.

Assume that q > m. On the one hand, it makes an extra contribution4(q −m)

(nm

)to the left hand side of (3.50). On the other hand, the number

of quadruples with two or three m-sets belonging to F ∩ G(σ) is non-zero.Indeed, if we denote by zj the number of quadruples with j m-sets in thefamily, for j ≤ 3, then we have z1+2z2+3z3 = 4q. Since z0+z1+z2+z3 = n,we have

z2 + 2z3 ≥ 4q − n. (3.51)

We proceed as follows.

(i) Triples of m-sets. Assume that, for some x ∈ [n] and j1, j2, j3, i,where j1, j2, j3, i = 0, 1, 2, 3, the sets H(m)

ju(x) belong to F for all u =

1, 2, 3. Then the central (m+ 2)-set H(m+2)i (x− 1, x;x) is not in the family

F . Moreover, it has zero charge. We transfer 8(q−m)4q−n

(nm

)charge to this set.

We have 8(q−m)4q−n ≤ 4 for q ≥ m+1, since this function for q ≥ m+1 decreases

as q grows. Therefore, we have

8(q −m)

4q − n

(n

m

)≤ 4

(n

m

)= 4

(m+ 1)(m+ 2)

(3m+ 2)(3m+ 1)

(n

m+ 2

)≤ 8

11

(n

m+ 2

)(3.49)= wc

m+2. (3.52)

The last inequality holds for m ≥ 3.

(ii) Pairs of m-sets. Assume that for some x ∈ [n] and i1, i2, j1, j2,where j1, j2, i1, i2 = 0, 1, 2, 3, exactly two m-sets H(m)

j1(x) and H(m)

j2(x)

from the x-family belong to F . Then in each of the two pairs of central(m + 1)-sets

(H

(m+1)i1

(x′;x), H(m+1)i2

(x′′;x))for x′, x′′ = x − 1, x one of

the sets is not in the family. Moreover, each of them has received at mostwlm+1 charge within this x-family (they could have received charge only in

Stage 3). We transfer 2(q−m)4q−n

(nm

)charge to each of the two missing central

sets. We have to verify that the charge transferred is at most wcm+1−wl

m+1.Note that 2(q−m)

4q−n ≤ 1. Therefore, it is enough to verify(n

m

)≤ wcm+1−wlm+1

(3.49)⇔ 2

(n

m

)=

2m+ 2

3m+ 2

(n

m+ 1

)≤(

n

m+ 1

)−5wlm+1. (3.53)

147

The last inequality holds (with equality) since by (3.49) we have 5wlm+1 =

m3m+2

(n

m+1

).

Now we only have to make sure that we have transferred enough charge.Indeed, we have transferred a total amount of charge equal to

4(q −m)

4q − n

(n

m

)z2 +

8(q −m)

4q − n

(n

m

)z3 =

4(q −m)

4q − n

(n

m

)(z2 + 2z3)

(3.51)

≥ 4(q −m)

(n

m

).

Therefore, the total amount of charge that is left on the m-sets is at most4m(nm

), moreover, all sets of size not greater thanm−1 have zero charge, and

none of the sets has the charge that is greater than its weight. The inequality(3.50) is veried, and the proof of Theorem 3.4 in the case s = 4,m ≥ 3 iscomplete.

3.3.4. The case m ≤ 2, s = 4

In the argument above, we assumed that m ≥ 3. However, we wang to provethe theorem for m ≥ 1, which leaves us with two cases: m = 1 and m = 2.If m = 1, then we have n = 6, and we have to show that at least 6 sets,including the empty set, are missing from a family F ⊂ 2[6] with ν(F) ≤ 3.If there is at most one singleton in F , then we are done. If there are atleast two singletons, say, 1 and 2, then F ∩ [3, 6] is intersecting, and,consequently, at least 8 sets are missing from F among the sets from 2[3,6].The case m = 1 is covered.

Ifm = 2 then n = 10 and we have to show that at least 1+(

101

)+(

92

)= 47

sets are missing from F . If there is at least one singleton in F , say 1, then,applying (3.2) to F ∩ 2[2,10], we get that at least 1 +

(91

)+(

92

)+ 1

3

(93

)sets are

missing from F , which is more than 47.Thus, we may assume that there are no sets of size smaller than 2 in F .

Now we may slightly modify the proof for the case m ≥ 3 so that it works form ≥ 2. Namely, among the (m+ 1)- and (m+ 2)-sets we give weights onlyto the central (m + 1)- and (m + 2)-sets (each central (m + 1)-set receivesa weight of 1

2

(n

m+1

), each central (m+ 2)-set receives a weight of

(n

m+2

), and

the weights on other layers stay the same). Since F contains no sets of sizesmaller than m, we may go to part 4 of the analysis, where we have to verifythe following analogues of (3.53) and (3.52) for m = 2, n = 10:

2m+ 2

3m+ 2

(n

m+ 1

)≤(

n

m+ 1

), 4

(m+ 1)(m+ 2)

(3m+ 2)(3m+ 1)

(n

m+ 2

)≤(

n

m+ 2

).

148

Both hold for m = 2. The rest of the proof stays the same. The proof iscomplete.

3.4. Inequalities for k-dependent families

We say that F ⊂ 2[n] is k-dependent if ν(F) < k. If F1, . . . ,Fk ⊂ 2[n] are notnecessarily distinct families, recall that we say that they are cross-dependentif there is no choice of Fi ∈ Fi, i = 1, . . . , k, such that F1, . . . , Fk are pairwisedisjoint.

In a paper [102], I and Frankl determined the maximum of |F1|+. . .+|Fk|for cross-dependent families Fi for all values of n ≥ k ≥ 3.

Denition 3.20. For k ≥ 3 and a family F ⊂ 2[n] we say that F is k-partition-free if F contains no k pairwise disjoint members whose union is[n]. Being k-partition-free is slightly less restrictive than being k-dependent.

For 0 ≤ j ≤ n let us use the notations F (j) = F ∩(

[n]j

), f (j) = |F (j)|.

The following inequality is an important discovery of Kleitman [151].

Lemma 3.21. Let F ⊂ 2[n] be k-partition-free and let j1, j2, . . . , jk be non-negative integers satisfying j1 + . . .+ jk = n. Then∑

1≤i≤k

f (ji)(nji

) ≤ k − 1. (3.54)

The proof of (3.54) is an easy averaging argument over all choices ofpairwise disjoint sets G1, . . . , Gk satisfying |Gi| = ji and noting that at leastone of the relations Gi ∈ F fails.

Since the relation j1 + . . . + jk = n is essential for proving (3.54) it israther surprising that in certain cases one can prove the analogous inequalityeven if j1 + . . .+ jk > n.

Let us rst state our inequality for the case k = 3.

Theorem 3.22. Let m > ` > 0 be integers, n = 3m − `. Suppose thatF ⊂ 2[n] is 3-partition-free. Then

|F (m−`)|(n

m−`) +

|F (m)|(nm

) +|F (m+`)|(

nm+`

) ≤ 2. (3.55)

Looking at the family(

[n]m

)∪(

[n]m+`

)shows that (3.55) is best possible.

149

To state our most general result let us say that the families F1, . . . ,Fk ⊂2[n] are cross-partition-free if there is no choice of Fi ∈ Fi, i = 1, . . . , k suchthat F1, . . . , Fk form a partition of [n].

Theorem 3.23. Let m > ` > 0 be integers, n = km − `, k ≥ 3. For1 ≤ i ≤ k let Fi ⊂

([n]m−`)∪(

[n]m

)∪(

[n]m+`

)and suppose that F1, . . . ,Fk are

cross-partition-free. Then∑1≤i≤k

∣∣F (m−`)i

∣∣(n

m−`) +

∣∣F (m)i

∣∣(nm

) + (k − 2)

∣∣F (m+`)i

∣∣(n

m+`

) ≤ (k − 1)k. (3.56)

Note that for k = 3 and F1 = . . . = Fk the inequality (3.56) implies(3.55). The reason that we treat it separately is that both the statementand the proof are simple and hopefully give the reader the motivation to gothrough the more technical result (3.56).

The proofs of (3.55) and (3.56) are based on Katona's cyclic permutationmethod (cf. [139], [142]).

3.4.1. The proof of (3.55)

Let x1, x2, . . . , x3m−`, x1 be a random cyclic permutation of 1, 2, . . . , n (asindicated above, the element after xn is x1). All (n− 1)! cyclic permutationshave the same probability 1/(n− 1)!. Set d = (3m− `,m).

We dene three families, B, A and C. B :=B1, . . . , B(3m−`)/d

where

Br := xj : (r − 1)m < j ≤ rm, r = 1, . . . , (3m− `)/d. Note that the Br

are arcs ofm consecutive elements xj. Moreover, (m, 3m−`) = d guaranteesthat each of the (3m− `)/d arcs Br are distinct and the last element of Bn/d

is xn.Let us partition each Br as Br = Ar∪Dr with Ar being the arc consisting

of the rst m− ` elements. Formally, Ar := xj : (r − 1)m < j ≤ rm− `,Dr := Br \ Ar. Set Cr := Br ∪Dr+1. Dene

A := Ar : 1 ≤ r ≤ n/d,C := Cr : 1 ≤ r ≤ n/d.

Note that Cr is not an arc but the union of two arcs and that it has theimportant property Cr ∪Ar+1 = Br ∪Br+1 that we are going to use withoutfurther reference.

Lemma 3.24. If F ⊂ 2[n] is 3-partition-free then

|F ∩ A|+ |F ∩ B|+ |F ∩ C| ≤ 2n/d. (3.57)

150

Proof of (3.57). Let R := r : Ar ∈ F, S := s : Bs /∈ F, T := t : Ct /∈F. (We consider S and T as sets on distinct ground sets.) To prove (3.57)it is sucient to show

|R| ≤ |S|+ |T |. (3.58)

We prove (3.58) by constructing an injection ϕ from R into S ∪ T .First note that Br, Br+1, Ar+2 form a partition of [n]. This implies that

if r + 2 ∈ R then at least one of r, r + 1 is in S. If r + 1 ∈ S, we setϕ(r + 2) = r + 1. If not then we let provisionally ϕ(r + 2) = r.

The only problem that might occur is that r+1 is also in R and thereforeϕ(r + 2) = ϕ(r + 1) = r.

Noting that Cr, Ar+1, Ar+2 form a partition of [n], r ∈ T follows. Wechange the value of ϕ(r + 2) to the element r in T . This element is notallocated to any other r′ ∈ R and the proof of (3.58) is complete.

To deduce (3.55) from (3.57) is easy averaging. For every B ∈ B theprobability of B ∈ F is |F (m)|/

(nm

)by the uniform random choice of the

permutation. The expected size E(|F ∩ B|) is

|B| |F (m)|/ (n

m

)=|F (m)|(

nm

) · nd.

The same holds for A and C as well. By linearity of expectation and usingthe trivial fact that the expectation never exceeds the maximum, we infer

n

d

(|F (m−`)|(

nm−`) +

|F (m)|(nm

) +|F (m+`)|(

nm+`

) ) = E(|F ∩A|+ |F ∩B|+ |F ∩C|) ≤ 2n

d.

Dividing by nd yields (3.55).

3.4.2. The proof of (3.56)

The proof is similar to that of (3.55) but both notationally and conceptuallymore complicated. Set d := (km− `,m) and n := n/d. Fix a random cyclicpermutation x1, . . . , xn of 1, . . . , n and dene again the n arcs of lengthm, B := B1, . . . , Bn where Br :=

xq : (r − 1)m < q ≤ rm

. The choice

of n guarantees that Bn ends with the element xn. This time we want todistribute these arcs among the k families Fi, 1 ≤ i ≤ k. For this reason letb be the rst positive integer such that k divides bn. Of course, b = k/(n, k).

We let B(p)r be a copy of Br and make a circle of bn sets in the following

order: B(1)1 , B

(2)2 , . . . , B

(k)k , B

(1)k+1, . . . , B

(k)bn . For each pair (r, p) we dene the

151

arc A(p)r ⊂ B

(p)r as the set of the rst m− ` elements of B(p)

r and let D(p)r be

the rest: D(p)r := B

(p)r \A(p)

r . For 1 ≤ j ≤ k−2 we dene the (m+`)-elementsets

C(p)r (j) := B(p)

r ∪D(p+j)r+j (r + j is mod n, p+ j is mod k).

Note that C(p)r (j)∪A(p+j)

r+j = B(p)r ∪B(p+j)

r+j . For 1 ≤ p ≤ k let us dene B(p) :=B

(p)1 , . . . , B

(p)n

, A(p) :=

A

(p)1 , . . . , A

(p)n

and C(p)

j :=C

(p)r (j) : 1 ≤ r ≤

n, 1 ≤ j ≤ k− 2. Note that altogether we dened

(1 + 1 + (k− 2)

)k = k2

families, each of size bn/k. Therefore (3.56) will follow once we prove thatout of these altogether bnk sets at most bn(k − 1) are in the correspondingfamilies Fi. In other words we have to show that at least bn in total aremissing.

Our plan is very simple. Fixing an arbitrary pair (i, r), 1 ≤ i ≤ k,1 ≤ r ≤ n, we want to show that there is an integer, 0 < t ≤ k such thatout of the following tk sets at least t are missing from the corresponding Fi′.

The list isA(i)r , A

(i−1)r−1 , ..., A

(i−t+1)r−t+1 ; B(i−1)

r−1 , ..., B(i−t)r−t ; C(i−1)

r−1 (j), ..., C(i−t)r−t (j),

1 ≤ j ≤ k − 2.To achieve this goal we prove a slightly stronger assertion. Since we do

not need them for this statement, we remove the upper indices and let Cr(j)denote the set C(i)

r (j) and the same with B(i)r , A(i)

r .

Lemma 3.25. Let r be xed and consider the following k groups of sets.G1 := Ar, Br−1, G2 := Ar−1, Br−2, Cr−2(1), . . . , Gk−1 := Ar−k+2,

Br−k+1, Cr−k+1(1), . . . , Cr−k+1(k − 2). Suppose that we have families Hi,Hi ⊂ Gi, 1 ≤ i < k such that we cannot nd k members of H1 ∪ . . . ∪Hk−1

which partition Ar ∪ Br−1 ∪ . . . ∪ Br−k+1. Then there exists t, 1 ≤ t ≤ ksatisfying ∑

1≤s≤t

∣∣Gs \ Hs

∣∣ ≥ t. (3.59)

Proof. First consider H1. If H1 $ G1 then (3.59) holds with t = 1. IfH1 = G1 then the two members Ar and Br−1 partition Ar ∪ Br−1. Arguingindirectly, suppose that (3.59) does not hold and let 1 ≤ t < k be the smallestinteger such that Ar ∪Br−1 ∪ . . .∪Br−t cannot be partitioned using the setsin H1 ∪ . . . ∪Ht.

By our assumptions t exists and the above considerations show t > 1 andAr ∈ H1. The minimality of t implies the existence of members Hi ∈ Hi,1 ≤ i < t such that

Ar ∪H1 ∪ . . . ∪Ht−1 = Ar ∪Br−1 ∪ . . . ∪Br−(t−1)

152

is a partition with Hi ∈ Hi. To conclude the proof we will prove that (3.59)holds for t.

First note that addingBr−t would make a partition ofAr∪Br−1∪. . .∪Br−t,implying Br−t /∈ Ht. To exhibit t − 1 further missing sets let us note thefollowing important feature about the partition H1∪. . .∪Ht−1 = Br−1∪. . .∪Br−(t−1): whenever a set Cr−s(j) occurs it must come together with Ar−s+jand the union of these two sets is Br−s ∪Br−s+j. Consequently, altering theorder of the Hu, we can break up the partition as

Br−1∪ . . .∪Br−(t−1) = Bu1∪ . . .∪Bu`∪(Bu`+1

∪Bu`+2

)∪ . . .∪

(But−2∪But−1

).

To prove (3.59) we show the existence of distinct sets in⋃

1≤s≤tGs \ Hs, one

for Bui and two for Bui ∪Bui+1.

For Bui note Cr−t(ui−r+t)∪Aui = Br−t∪Bui. Since Ar∪Br−1∪. . .∪Br−tcannot be partitioned by members of the Hs, either Cr−t(ui − r + t) or Aui

is missing from the Hs. For the case of Bv ∪ Bw (to simplify notation,r − t < v < w ≤ r − 1) rst note that one of the corresponding sets inH1 ∪ . . . ∪ Ht−1 that partition Bv ∪ Bw is Aw (the other is Cv(w − v)).Consider two partitions of Br−t ∪Bv ∪Bw.

Cr−t(w − r + t) ∪Bv ∪ Aw and

Cr−t(v − r + t) ∪ Av ∪Bw.

Since at least one set must be missing from both, we are done. Noting thatthe exhibited candidates for missing sets are all distinct, the proof of (3.59)is complete.

Equipped with (3.59) it is not hard to prove Lemma 3.25. Starting at anarbitrary r we nd, say, t1 consecutive groups with at least a total of t1missing sets, 1 ≤ t1 ≤ k. Then starting at r − t1 we nd t2 such groups,etc. Going around the circle (of length bn) the last position of the last groupmight not be r + 1. However, since there are only bn members after makingno more than k full rounds we denitely have two sets of groups starting atthe same element, say r′. That is for the tw in between, say ta, ta+1, . . . , ta+q

one has ta + ta+1 + . . . + ta+q = c · bn with c a positive integer. For thesepositions we exhibited altogether at least cbn missing sets and each of themis counted at most c times. Therefore there are at least bn missing sets,proving Lemma 3.25.

Since Lemma 3.25 implies (3.56) by the same averaging argument asLemma 3.24 implied (3.55), the proof of Theorem 3.23 is complete.

153

3.4.3. Applications

Denition 3.26. For positive integers n ≥ k ≥ 3 let p(n, k) denote themaximum of |F| over all F ⊂ 2[n] that are k-partition-free.

Theorem 3.27.

p(km− 1, k) =∑j≥m

(km− 1

j

), (3.60)

moreover the only k-partition-free family achieving equality in (3.60) isG ⊂

[km− 1] : |G| ≥ m.

Let us note that Kleitman [151] proved the same bound for the somewhatstronger restriction that the family is without k pairwise disjoint sets. Also,Kleitman did not prove the uniqueness of the optimal family.

Proof. Since the case m = 1 is trivial, we suppose m ≥ 2. Our main tool isTheorem 3.23 applied with ` = 1, F1 = . . . = Fk =: F . For the k-partition-free family F ⊂ 2[n], n = km− 1 we get from (3.56):

|F (m−1)|(n

m−1

) +|F (m)|(

nm

) + (k − 2)|F (m+1)|(

nm+1

) ≤ k − 1.

Setting y(j) =(nj

)− |F (j)| we obtain

y(m− 1)(n

m−1

) +y(m)(nm

) + (k − 2)y(m+ 1)(

nm+1

) ≥ 1. (3.61)

Note(km−1m

)= (k − 1)

(km−1m−1

)and for further use(

km− 1

m− j + 1

) / (km− 1

m− j

)=

(k − 1)m+ j − 1

m− j + 1> k − 1 for j ≥ 2. (3.62)

Using(km−1m+i

)≥(km−1m

)(valid for i < m) (3.61) yields the following inequal-

ity.

y(m− 1) +1

k − 1y(m) +

k − 2

k − 1y(m+ 1) ≥

(km− 1

m− 1

). (3.63)

Let us apply (3.54) with

(j1, . . . , jk) := (m− `,m,m, . . . ,m,m+ `− 1) for ` = 2, 3, . . . ,m.

Multiplying both sides by(

nm−`)we obtain

y(m− `) +(k − 2)

(k − 1)`y(m) +

1

(k − 1)`y(m+ `− 1) ≥

(km− 1

m− `

)(3.64)

154

(we used (3.62) and

(n

m+`−1

)>(nm

)). We want to add (3.63) and the sum of

(3.64) over 2 ≤ ` ≤ m. For ` > 2 the term y(m + ` − 1) occurs only onceand its coecient is smaller than 1

k−1 < 1. The term y(m+ 1) has coecient

k − 2

k − 1+

1

(k − 1)2<k − 2

k − 1+

1

k − 1= 1 also.

Finally (k − 1)−2 + (k − 1)−3 + . . . = k−1k−2 ·

1(k−1)2 = 1

(k−2) ·1

k−1 . Thus the

total coecient of y(m) will be less than 2k−1 ≤ 1. That is, we obtain an

inequality of the form

y(0) + y(1) + . . .+ y(m− 1) + cmy(m) + . . .+ c2my(2m) ≥∑

0≤j<m

(n

j

)with c(m + i) < 1 for 0 ≤ i ≤ m. Consequently, |F| ≤ 2n −

∑0≤j<m

(nj

)=∑

j≥m

(nm

), as desired. Moreover, in case of equality, y(m + i) = 0 must hold

because of cm+i < 1 for all 0 ≤ i ≤ m. Plugging these values into (3.63) and(3.64), y(m − `) =

(n

m−`)follows for all 1 ≤ ` ≤ m. That is, F := F ⊂

[n] : |F | ≥ m concluding the proof of the uniqueness.

Remark. If we used (3.54) instead of Theorem 3.23 then instead of (3.63)we would have

y(m− 1) + y(m) ≥(km− 1

m

).

Thus adding more equalities to it would make the coecient of y(m) greaterthan 1.

In [102], we proved the following theorem.

Theorem 3.28 ([102]). Suppose that F1, . . . ,Fk ⊂ 2[n], n = mk − 1, arecross-dependent. Then one has:

|F1|+ . . .+ |Fk| ≤ k ·∑j≥m

(mk − 1

j

). (3.65)

One can use Theorem 3.23 to prove (3.65) under the weaker assumptionof being cross-partition-free and show that equality holds only if

F1 = . . . = Fk = F ⊂ [n] : |F | ≥ m.

Let us mention that in [102] the maximum of |F1|+. . .+|Fk| is determinedfor all values of n and k. The methods presented in this paper seem to beinsucient to tackle the cases n 6≡ −1 (mod k).

155

3.5. Discussion

In this section we discuss one possible generalization of the value e(n, s), aswell as Conjecture 1 and some further open problems.

Families with no s pairwise disjoint sets of small total cardinality

Let us say that a family F ⊂ 2[n] has the property D(s, q) or shortly isD(s, q) if

|F1 ∪ . . . ∪ Fs| > q

for all pairwise disjoint F1, . . . , Fs ∈ F . Note that for q ≥ n being D(s, q)for F is equivalent to ν(F) < s. We introduce the function d(n, q, s):

d(n, q, s) := max|F| : F ∈ 2[n],F is D(s, q)

.

In what follows we show that the task of determining d(n, q, s) is in manycases easily reduced to the problem of determining e(q, s) = d(q, q, s).

Claim 3.29. The property D(s, q) is maintained under shifting.

Proof. Let 1 ≤ i < j ≤ n. Consider a family F ⊂ 2[n] that is D(s, q) andthe sets A1, . . . , As ∈ Si,j(F) that are pairwise disjoint. If A1, . . . , As ∈ F ,then we have nothing to prove. Thus we may assume that A1 ∈ Si,j(F)−F .That is, i ∈ A1, j /∈ A1, and A1 := (A1−i)∪j is in F . Note that i /∈ At

for 2 ≤ t ≤ s, and so At ∈ F .If j /∈ A2 ∪ . . .∪As, then A1, A2, . . . , As are pairwise disjoint members of

F , implyings∑i=1

|Ai| = |A1|+s∑i=2

|Ai| > q.

Suppose now that j ∈ A2. By the denition of Si,j, the set A2 := (A2 −j)∪i is also in F . The sets A1, A2, A3, . . . , As ∈ F are pairwise disjoint.Since |A1| = |A1| and |A2| = |A2|, we conclude that

∑si=1 |Ai| > q.

Given a family F ⊂ 2[n], consider the following two families on [n− 1]:

F(n) := A− n : n ∈ A,A ∈ F,F(n) := A : n /∈ A,A ∈ F.

For n ≥ q := s(m + 1)− l, 0 < l ≤ s, dene the analogue of the familiesP(s,m, l):

B(n, q, s) := F ⊂ [n] : |F |+ |F ∩ [l − 1] ≥ m+ 1.

156

Note that if F = B(n, q, s), n > q, then F(n) = B(n − 1, q, s) and F(n) =B(n − 1, q − s, s) hold. The following easy proposition allows us to extendthe results concerning e(n, s) to d(n, q, s).

Proposition 3.30. Fix n ≥ q ≥ s. If f(n − 1, q, s) = |B(n − 1, q, s)| andd(n− 1, q − s, s) = |B(n− 1, q − s, s)| then

d(n, q, s) = |B(n, q, s)| holds.

Proof. W.l.o.g. we assume that F is shifted. It is clear that F(n) is D(q, s).Therefore, it is sucient to show that F(n) is D(q − s, s).

Assume for contradiction that A1, . . . , As ∈ F(n) are pairwise disjointand H := A1 ∪ . . .∪As has size at most q− s. Since n ≥ q, n− (q− s) ≥ s.That is, we can nd distinct elements x1, . . . , xs ∈ [n]−H. Since F is shifted,A1 ∪ x1, . . . , As ∪ xs are pairwise disjoint members of F . Their unionH ∪ x1, . . . , xs has size |H|+ s ≤ q, a contradiction.

Therefore, |F| = |F(n)|+ |F(n)| ≤ |B(n−1, q−s, s)|+ |B(n−1, q, s)| =|B(n, q, s)|.

We get the following corollary:

Corollary 3.31. Let s ≥ 2,m ≥ 0. For n ≥ q ≥ 0 the following holds:

(i) d(n, sm− 1, s) =∑i≥m

(n

i

),

(ii) d(n, sm+ s− 2, s) =

(n− 1

m− 1

)+∑i≥m+1

(n

i

).

Proof. We derive the corollary from Proposition 3.30 by double induction.We apply induction on m, and for xed m the induction on n. We re-mark that in all three cases on the right hand sides we have the cardinality|B(n, q, s)| for the corresponding n, q and s. The equalities f(n, 0, 2) =|B(n, 0, 2)|, f(n, s − 1, s) = |B(n, s − 1, s)|, f(n, s − 2, s) = |B(n, s − 2, s)|are obvious. The equalities in the case when n = q follow from the resultson e(sm− 1, s), e(sm+ s− 2, s), discussed in the introduction.

What about d(n, sm, s) for s > 2, and, more generally, what about allother values of parameters? Interestingly enough, for large n we can deter-mine d(n, s(m+ 1)− l, s) exactly for any l,m, s.

Theorem 3.32. For 1 ≤ l ≤ s and n ≥ maxl(m2 +m+ 2), s(m+ 1) + l+

m+ 3one has

d(n, s(m+ 1)− l, s) = |B(n, s(m+ 1)− l, s)|.

157

The proof of Theorem 3.32 is very similar to the proof of Theorem 3.4,thus we omit most of it, sketching only the key points. Assuming that theclaim of Lemma 3.16 does not hold and arguing as in the proof of Lemma3.16 one can obtain that

y(m+ 1) ≥n− lm− (s− l)(m+ 1)

n− lm

(n− lmm+ 1

)and

y(m) ≥n− smn

(n

m

),

which by analogy with (3.35), (3.36) leads to contradiction if

n− lm− (s− l)(m+ 1)

n

(n− lmm+ 1

)>

(s+ 2)m

n

(n

m

).

The last inequality holds under the conditions imposed on n in Theorem3.32. Next, the statement and the proof of Lemma 3.17 remain the same.Finally, the proof of Claim 3.18 undergoes the same modications as that ofLemma 3.16.

Remark. The conditions on n in the statement of Theorem 3.32 arerather crude and are likely not dicult to improve, especially in the case ofl = s. However, the order of n = Ω(m2l) for general l,m, seems to be moreor less the limit for the present method to work. Thus, it would be desirableto prove Theorem 3.32 for n > csm with some absolute constant c.

Conjecture 1

We believe that Conjecture 1 should actually be true for an even wider rangeof l. However, the equality (3.4) is not true in general, even if we excludethe case n = sm. The value of l needs to be separated from s for P(s,m, l)to be the largest family with no s-matching. We illustrate it for the casen = sm + 1 (the same can be done for n = sm + c for any positive integerc and large enough s,m depending on c). Let s,m ≥ 20 and consider thefamily

W(m, s) := W ∈ 2[n] : |W ∩ [sm− 1]| ≥ m.We remark that this family is obtained as ∪sm−1

t=m

([n]t

), which we close upward,

and that (3.2) shows that it is the largest family for n = sm.We have ν(W(m, s)) = s− 1, and for n = sm+ 1

|W(m, s)| =n∑

r=m+1

(n

r

)+

(sm− 1

m

)−(sm− 1

m− 1

)=

n∑r=m+1

(n

r

)+s− 2

s− 1

(sm− 1

m

).

158

Next, we remark that s−2s−1

(sm−1m

)= s−2

s−1(sm+1−m)(sm−m)

(sm+1)sm

(nm

)> s−3

s

(nm

)≥

0.85(nm

). On the other hand, we have |P(s,m, s−1)| =

∑nr=m+1

(nr

)+(nm

)−(

n−s+1m

)+∣∣P(s,m, s− 1) ∩

([n]≤m−1

)∣∣. We have(nm

)(n−s+1m

) ≤ ( n−mn−m− s+ 1

)m

< e(s−1)m

n−m−s+1 < emm−1 < 3.

Therefore, |W(m, s)| − |P(s,m, s− 1)| > 16

(nm

)−∣∣P(s,m, s− 1)∩

([n]≤m−1

)∣∣.Finally,

∣∣P(s,m, s − 1) ∩(

[n]≤m−1

)∣∣ <∑m−1r=0

(nr

)< s−1

s−2

(n

m−1

)< 1

10

(nm

), which

proves that P(s,m, s− 1) is not the maximal family.

We managed to prove that for m = 2, n = 2s + 1 W(2, s) is indeedthe largest family with no s pairwise disjoint sets. However, already forn = 2s + t with certain values of t we can construct a yet another familywith no s-matching, which is larger than both W (m, s) and P(s,m, s − t).Therefore, it looks dicult to formulate a general conjecture. Still, thereis something common about all the extremal constructions we know. Theyare all dened as the intersection of the boolean cube 0, 1n and a certainhalfspace in Rn! To make it more precise, let us give some denitions.

Let α1 ≥ α2 ≥ . . . ≥ αn ≥ 0 be reals, and suppose that∑

i αi < s. Putααα := (α1, . . . , αn) and dene

F(ααα) := F ∈ 2[n] :∑i∈F

αi ≥ 1.

Then it is easy to see that ν(F(ααα) < s holds. It is also clear that F(ααα) =0, 1n ∩ x ∈ Rn : 〈x,ααα〉 ≥ 1. All the extremal families that were consid-ered in this paper have a form F(ααα) for suitable vectors α. Indeed,

• P(s,m, l) = F(αααp) with αααp := 1m+1

(2, . . . , 2︸ ︷︷ ︸

l−1

, 1, . . . , 1),

• W(m, s) = F(αααw) with αααw := 1m

(1, . . . , 1︸ ︷︷ ︸sm−1

, 0, . . . , 0),

• H(k)(n, s− 1) = F(αααh) ∩(

[n]k

)with αααh :=

(1, . . . , 1︸ ︷︷ ︸

s−2

, 1− 1k,

1

k, . . . ,

1

k︸ ︷︷ ︸k

, 0 . . . , 0).

It motivates the following conjecture.

Conjecture 4. For any n, s the maximum of e(n, s) is attained on the familyF(ααα) for a suitable ααα ∈ Rn.

159

The vector ααα de-facto is the fractional covering for the family F . Giventhat the size of the smallest fractional covering is equal to the size of thelargest fractional matching, the conjecture above basically states that theextremal families for problems on matchings coincide with those for fractionalmatchings.

The same question posed for ek(n, s) is also very interesting. It is knownto be true asymptotically for xed k (cf. [8]).

Truncated boolean lattice

Let n := s(m + 1) − l and F ⊂(

[n]≤r)satisfy ν(F) < s. What is the

minimal value of∑r

i=0

(ni

)− |F|? If we interpret the results concerning

e(n, s) in terms of how many sets from 2[n] are necessarily missing from afamily F with ν(F) < s, then many of them are possible to generalize tothis setting. Namely, the number of missing sets would be exactly the sameas∣∣2[n] \ P(s,m, l)

∣∣, provided that r ≥ m+ 2. Indeed, in the proofs we onlyused the layers of the boolean lattice up to m+ 2.

On the other hand, it is clear that for r = m the family P(s,m, l)∩(

[n]≤m)is

not the optimal one. Indeed, for l = 2, say, the family A ⊂ [n] : A∩[s−1] 6=∅ clearly has a larger cardinality.

So it is natural to ask what happens for r = m+1. We conjecture that thenumber of missing sets remains the same as in the case of the whole booleanlattice.

160

Chapter 4.

Random Kneser graphs

My results presented in this chapter are published in [163, 165].

4.1. Introduction and statement of results

Kneser graphs and hypergraphs are very popular and well-studied objectsin combinatorics. Fix some positive integers n, k, r, where r ≥ 2. The setof vertices of the Kneser r-graph KGr

n,k is the set of all k-element subsets

of [n], denoted by(

[n]k

). The set of edges of KGr

n,k consists of all r-tuplesof pairwise disjoint subsets. Thus, KGr

n,k is non-empty only if n ≥ kr.Substituting k = 1 in the denition gives the complete r-graph on n vertices.When dealing with the graph case r = 2, we omit the superscript in thenotation of Kneser graphs. For a hypergraph H we denote by χ(H) itschromatic number, that is, the minimum number χ such that there exists acoloring of vertices H into χ colors that leaves no edge of H monochromatic.

Naturally, Kneser graphs were studied rst. They earned their namefrom M. Kneser, who investigated them in the paper [154]. He showed thatχ(KGn,k) ≤ n−2k+2 and conjectured that this bound is tight. This conjec-ture (or rather its resolution) played a very important role in combinatorics.It was conrmed by L. Lovasz [181], who, in order to resolve it, introducedtools from algebraic topology to combinatorics.

Once [181] appeared, there was a burst of activity around Kneser graphs.I. Barany [23] gave an elegant alternative proof of Lovasz' result, and severalauthors studied the chromatic number of Kneser (disjointness) graphs ofarbitrary set systems. In particular, there were results due to V. Dol'nikov[62] and A. Schrijver [224].

In [224], Schrijver studied induced subgraphs SGn,k of KGn,k constructedon the family of all k-element stable sets of the cycle Cn. In other words, theunderlying family contains all k-element sets that do not have two cyclically

161

consecutive elements of [n]. Schrijver noticed that a slight modication ofBarany's proof yields a stronger statement: χ(SGn,k) = n − 2k + 2. Inthe harder part of the paper, he also showed that SGn,k is a vertex-criticalsubgraph of KGn,k, i.e. that any proper induced subgraph of SGn,k hasstrictly smaller chromatic number.

A coloring of KGn,k in n − 2k + 2 colors is easy to obtain: for each1 ≤ i ≤ n − 2k + 1 color the sets with minimum element i into color i,and color the remaining sets, forming the family

([n−2k+2,n]

k

), into color 0. A

similar coloring for KGrn,k gives the upper bound χ(KGr

n,k) ≤⌈n−r(k−1)

r−1

⌉:

for 1 ≤ i ≤ n− kr + 1 color the sets with minimum element in [(r − 1)(j −1) + 1, (r− 1)j] into color j, and color the sets from

([n−rk+2,n]

k

)into color 0.

P. Erdos [70] conjectured that this bound is sharp for all r ≥ 2. After somepartial progress it was conrmed in full generality by N. Alon, P. Frankl, andL. Lovasz [9]. The proof again used topological tools.

Generalizing both the result of Dol'nikov [62] and Alon, Frankl, andLovasz [9], I. Krz [157], [158] obtained the bound on the chromatic numberof Kneser hypergraphs of general set families. Later, an elegant alternativeproof was obtained by J. Matousek [188], and some more general results withcombinatorial proofs were obtained by G. Ziegler [237]. We also refer to theamazing book written by J. Matousek on the subject [189].

In [7], N. Alon, L. Drewnowski and T. Luczak applied results on coloringsof Kneser-type hypergraphs for constructing certain ideals in N. The hy-pergraphs they considered are called s-stable Kneser hypergraphs, and theymay be seen as a generalization of Schrijver graphs. The underlying set fam-ily that denes the s-stable Kneser r-hypergraph KGr, s−stable

n,k consists of allk-element subsets i1, . . . , ik ⊂ [n], whose consecutive elements are su-ciently far apart: if 1 ≤ i1 < . . . < ik ≤ n, then for any j = 0, . . . , k − 1 theelements satisfy ij+1− ij ≥ s, as well as i1 + n− ik ≥ s. The case r = s = 2corresponds to Schrijver graphs.

In their paper, Alon, Drewnowski and Luczak proved and applied the fol-lowing result: χ(KGr, r−stable

n,k ) = χ(KGrn,k) for r = 2t, t ∈ N. They have

also stated explicitly the conjecture tracing back to Ziegler's paper [237],which says that the same equality holds for any r. We are going to use thisresult of [7], and we note that it would have improved some of the bounds inthis paper, shall the conjecture be veried. A more general conjecture wasmade by F. Meunier [193]: χ(KGr, s−stable

n,k ) =⌈n−s(k−1)

r−1

⌉for any s ≥ r. It

was veried in some cases, but is still wide open in general.

162

In fact, Kneser was not the rst to ask a question concerning Knesergraphs. The ErdosKoRado theorem (Theorem 1.2) determines the inde-pendence number of KGn,k, that is, the maximum size of a subset of verticesnot containing an edge of the graph. Later, Erdos [69] asked a more generalquestion: what is the size of the largest family of k-element subsets of [n] withno r pairwise disjoint sets? This is the famous Erdos Matching Conjecture,discussed in Section 3.1. This is obviously a question about the independencenumber of KGr

n,k, and, unlike the question on the chromatic number, it doesnot have a complete solution yet. However, the question was resolved for awide range of parameters by P. Frankl [83] and by Frankl and the author[104]. For some recent progress on the subject see [94], [102].

An r-uniform Kneser hypergraph of any k-uniform set system is an in-duced subgraph of KGr

n,k, and thus the results on the chromatic number ofinduced subgraphs of KGr

n,k belong to the class of results discussed above.Instead of restricting to a subset of vertices, in this paper we study, whathappens if we restrict to a subset of edges. The most natural model to studyis the binomial model of a random hypergraph. For a hypergraph H and areal number p, 0 < p < 1, dene the random hypergraph H(p) as follows:the set of vertices of H(p) coincides with that of H, and the set of edgesof H(p) is a subset of the set of edges of H, with each edge from H takenindependently and with probability p. The results on random graphs and hy-pergraphs, roughly speaking, tell us how does a typical subgraph of a given(hyper)graph that contains a p-fraction of edges behave with respect to agiven property. Theory of random graphs and hypergraphs is very rich inboth results and open problems, and by no means we are going to give anoverview of the eld in this paper. We refer the reader to the books [13], [31]for some classical results on the subject.

One class of questions that is particularly relevant for this paper dealswith transference results. In general, we speak of transference if a certaincombinatorial result holds with no changes in the random setting. Studyingthis phenomenon in the context of the independence number and the chro-matic number of generalized Kneser graphs was suggested by Bogolyubskiy,Gusev, Pyaderkin and Raigorodskii in [28]. One example of such theorem isdue to B. Bollobas, B. Narayanan and A. Raigorodskii [33]. They studiedthe size of maximal independent sets in KGn,k(p), and showed that for awide range of parameters the independence number of KGn,k(p) is exactlythe same as that of KGn,k, given by the ErdosKoRado theorem. Lateron, their result was further strengthened by J. Balogh, B. Bollobas, and B.

163

Narayanan [19], S. Das and T. Tran [54] and P. Devlin and J. Kahn [59].In the paper [163], I initiated the studies of the behaviour of the chromatic

number of KGn,k(p) and SGn,k(p), showing the following surprising fact:compared to χ(KGn,k), it does change at most by a small additive term ina very wide range of parameters. This is the rst result of this section (seeTheorem 4.1). Its proof has topological avour.

Random subgraphs of more general graphs K(n, k, l) were earlier investi-gated by L. Bogolyubskiy, A. Gusev, M. Pyaderkin and A. Raigorodskii in[29, 28]. The vertices of K(n, k, l) are the k-element subsets of [n], with twovertices adjacent if the corresponding sets intersect in exactly l elements. In[27, 29, 28, 204, 205, 216] the authors obtained several results concerning theindependence number and the chromatic number of K(n, k, l)(p) and relatedresults.

In the recent paper [4], which motivated in part the paper [165], M. Al-ishahi and H. Hajiabolhassan generalized the results of [163] to the case ofKneser hypergraphs of arbitrary set systems. They have also strengthenedthe results of [163] in the graph case. The proofs of Alishahi and Haji-abolhassan are quite technically involved and not easy to follow. They usetopological methods to obtain their results.

The second main contribution of this section is a purely combinatorial ap-proach to the problem, which allows us to signicantly improve all previouslyknown bounds on the chromatic numbers in the most interesting cases: forrandom subgraphs of (complete) Kneser and Schrijver graphs and Kneser hy-pergraphs. Our method may be extended to more general classes of Kneserhypergraphs. This does not, however, cover all generalized Kneser hyper-graphs, so the result of Alishahi and Hajabolhassan remains best known insome cases.

4.1.1. The bounds

In this section we discuss both the older and the newer quantitative boundson the chromatic number of Kneser and Schrijver graphs and hypergraphs.We do not state the bounds in full generality as they depend on too manyparameters and thus are very dicult to interpret. Instead, we focus onseveral most interesting cases. These cases were discussed in all papers onthe subject, so we can compare the results. The bounds in their full generalityappear in the latter sections.

For the rest of the section, we assume that r ≥ 2, p > 0 are xed. Note

164

that in the case r = 2 we formulate our results for Schrijver graphs SGn,k(p).The same bounds hold for Kneser graphs, since Schrijver graphs are sub-graphs of Kneser graphs.

We henceforth use the notation f(n) g(n) in a slightly unconventionalway. This inequality should be read as: there exists a suciently largeconstant C, where C = C(r, p, k) if k is xed, or C = C(r, p, l) if l isxed, such that f(n) ≥ Cg(n) for all suciently large n. All logarithmswith unspecied base have base e. All statements below hold asymptoticallyalmost surely (a.a.s.) for n → ∞, and so we omit writing a.a.s. in moststatements for brevity.

Returning to the results on colorings, I [163] proved the following.

Theorem 4.1 ([163]). Let p ∈ (0, 1) be xed. Then a.a.s.

(l = 1l = 1l = 1) χ(SGn,k)(p) ≥ χ(KGn,k+1) = χ(KGn,k)− 2 if n− 2k √n; (4.1)

(xed lll) χ(SGn,k)(p) ≥ χ(KGn,k+l) = χ(KGn,k)− 2l if l is xed and k n32l ; (4.2)

(xed kkk) χ(SGn,k)(p) ≥ χ(KGn,k+l) = χ(KGn,k)− 2l if k is xed and l n32k . (4.3)

The proof of this theorem is presented in Section 4.2. Although theseresults are strengthened in Theorem 4.2, I include a separate proof of Theo-rem 4.1 since it has a dierent avour. Note that in the last two k and l aresimply interchanged in the conditions needed for the inequality to hold. Thefollowing bounds were proven by Alishahi and Hajiabolhassan [4]:

(l = 1l = 1l = 1) χ(KGrn,k)(p) ≥ χ(KGr

n,k+1) if n− rk nr−1r ; (4.4)

(xed lll) χ(KGrn,k)(p) ≥ χ(KGr

n,k+l) if l is xed and k nr

lr−1 log1

lr−1 n. (4.5)

We do not express χ(KGrn,k+l) in terms of χ(KGr

n,k), since the formulas aremuch uglier in the hypergraph case. Note that for r = 2 the bound (4.4)coincides with (4.1), while (4.5) improves on (4.2).

In Section 4.3 I prove the following bounds.

Theorem 4.2. Let p ∈ (0, 1) and r ∈ N, r ≥ 2, be xed.If r = 2q for some q ∈ N then a.a.s.

(l = 1, r = 2ql = 1, r = 2ql = 1, r = 2q) χ(KGrn,k)(p) ≥ χ(KGr

n,k+1) if n− rk nr/(r+1) log−1/(r+1) n. (4.6)

If r = 2 then a.a.s.

(xed l, r = 2l, r = 2l, r = 2) χ(KGn,k)(p) ≥ χ(KGn,k+l) if l is xed and k (n log n)1/l; (4.7)

(xed k, r = 2k, r = 2k, r = 2) χ(KGn,k)(p) ≥ χ(KGn,k+l) if k is xed and l (n log n)1/k. (4.8)

165

If r = 3 then a.a.s.

(xed l, r = 3l, r = 3l, r = 3) χ(KG3n,k)(p) ≥ χ(KG3

n,k+l) if l is xed and k log1/(3l−4) n; (4.9)

(xed k, r = 3k, r = 3k, r = 3) χ(KG3n,k)(p) ≥ χ(KG3

n,k+l) if k is xed and l log2/(6k−11) n. (4.10)

If r > 3 then a.a.s.

(xed l, r > 3l, r > 3l, r > 3) χ(KGrn,k)(p) ≥ χ(KGrn,k+l) if l is xed and k log1

r(l−2)−1 n; (4.11)

(xed k, r > 3k, r > 3k, r > 3) χ(KGrn,k)(p) ≥ χ(KGrn,k+l) if k is xed and l log1

r(k−1)− 2r−1r−1 n. (4.12)

We remark that the bounds with r = 2 stated in the theorem hold forSchrijver graphs and, more generally, for r-stable r-uniform Kneser hyper-graphs, when r = 2t for some t ∈ N.

In the graph case, we see that (4.6), (4.7), and (4.8) improve on (4.1),(4.2) and (4.3), respectively.

Probably, the most interesting results in Theorem 4.2 are (4.9)(4.12).They are much stronger than (4.5) and guarantee that the chromatic numberof KGr

n,k(p) drops by no more than a small additive term for already forpolylogarithmic k (this was known before for polynomial k).

One question that is natural to ask in this context is what makes the caser = 2 so dierent from the case r > 2? Can one obtain a bound similarto (4.9)(4.12) for the case r = 2? This question was partially answered inmy recent paper with Kiselev [149], however, the behaviour of the chromaticnumber even for r = 2 and growing k is not fully understood. In [149], wehave pointed out interesting connections of the behaviour of the chromaticnumber of KGn,k(p) and some questions on embeddability of complexes.

4.2. Proof of Theorem 4.1

The main result of this section is the following quite technical theorem.

Theorem 4.3. Let k = k(n), ` = `(n) be integer functions. Assume thatk ≥ 2, ` ≥ 1 and choose p = p(n) such that 0 < p ≤ 1. Put d := n −2k − 2` + 1 and t :=

⌈(`+kk

)/d⌉. If d ≥ 2 and for some ε > 0 we have

(1 − ε)p > t−2n ln 3 + 2t−1(1 + ln d), then the graph SGn,k(p) w.h.p. haschromatic number at least d+ 1.

Remark 1. We formulate and prove all the results for SGn,k(p), since theyyield the same bounds for KGn,k(p). Thus, in Theorem 4.3, as well as inthe forthcoming corollaries, SGn,k(p) may be replaced by KGn,k(p).

166

Remark 2. A formula f(n) ∼ g(n) signies that limn→∞ f(n)/g(n) = 1.We remind the reader that the abbreviation w.h.p. means with high proba-bility, that is, with probability tending to 1 as n tends to innity.

The condition of Theorem 4.3 is quite dicult to interpret, since thereare too many parameters involved. Before proving the theorem we presentseveral corollaries of Theorem 4.3, which follow directly by easy calculations.

Corollary 4.4. Let p be xed, 0 < p ≤ 1. If k n3/4, then w.h.p.χ(SGn,k(p)) ≥ χ(KGn,k) − 4. Moreover, if n − 2k

√n, then w.h.p.

χ(SGn,k(p)) ≥ χ(KGn,k)− 2.

Proof. We start with the rst claim. It is enough to verify the condition onp from Theorem 4.3 for suciently large n and ` = 2. In these assumptions

we have t =⌈(

2+k2

)/d⌉ (n3/4)2/n = n1/2. Thus, we have t−2n ln 3 +

2t−1(1 + ln d) ≤ t−2n ln 3 + 2t−1(1 + lnn) 1, so the condition on p holdsfor suciently large n.

As for the second claim, it is enough to verify the condition on p forsuciently large n and ` = 1. We have d

√n. We again obtain the

same bound on t: t =⌈(

k+11

)/d⌉ n/

√n = n1/2. The rest remains

unchanged.

Corollary 4.5. Fix some function g(n), g(n)→∞ as n→∞. For any k,where 2 ≤ k ≤ n

2 − g(n), and for any xed p, 0 < p ≤ 1 w.h.p. we haveχ(SGn,k(p)) ∼ χ(KGn,k).

This is an intriguing corollary, since Kneser graph KGn,k for k = 1 is justa complete graph Kn, and it is known that w.h.p. χ(Kn(1/2)) ∼ n

2 log2 n

n = χ(Kn). It is clear that our proof cannot give any reasonable boundin the case when k = 1. Indeed, to fulll the condition on p one has tomake t = d(l + 1)/de bigger than

√n, for which, in turn, one has to choose

d ≤ 12

√n. Therefore, the best we can obtain is the square root bound:

χ(Kn(1/2)) ≥ c√n for some constant c > 0.

Proof. Corollary 4.4 justies the assertion of Corollary 4.5 in the case k n3/4. Taking a sequence of k(n), we may split it into at most two innitesequences N1, N2, N1 ∪ N2 = N, such that k(n) n3/4 if we take a limitover n ∈ N1, and k(n) n4/5 if we take a limit over n ∈ N2. The rstsequence we treat as the case of k n3/4. Therefore, w.l.o.g., we may assumethat N2 = N and, thus, k n4/5. In this case we have χ(KGn,k) ∼ n.It is thus sucient to verify the condition on p with some ` that satises

167

n3/4 ` n. As in the proof of Corollary 4.4, we obtain a similar bound

on t: t =⌈(

k+`k

)/d⌉≥⌈(

2+`2

)/n⌉ (n3/4)2/n = n1/2. The rest remains the

same.

The next two corollaries are a bit more general. In particular, we allowfor p to be a function of n. We omit the proof of Corollary 4.7, as it repeatsthe proof of Corollary 4.6.

Corollary 4.6. Fix a natural number `.1. We have χ(SGn,k(p)) ≥ χ(KGn,k)− 2` w.h.p. if

1 ≥ p n3

k2`+n(1 + lnn)

k`.

2. If p is xed, 0 < p ≤ 1, and k n32` , then w.h.p. χ(SGn,k(p)) ≥

χ(KGn,k)− 2`.

Proof. In the assumptions of the corollary we have t =⌈(

l+kl

)/d⌉≥ ckl/n,

where c > 0 is a constant. Thus, we have t−2n ln 3 + 2t−1(1 + ln d) ≤O(n3

k2l+ n(1+lnn)

k`

) p, so the condition on p from Theorem 4.3 holds for

suciently large n.As for the second part, it is enough to note that if k n

32` , then both

n3

k2l 1 and n(1+lnn)

k` 1 and the condition from the rst part is fullled.

Due to the symmetric role that k and ` play in the condition on p, Corol-lary 4.7 is the same as Corollary 4.6, but with k and ` interchanged.

Corollary 4.7. Fix a natural number k.1. We have χ(SGn,k(p)) ≥ χ(KGn,k)− 2` w.h.p. if

1 ≥ p n3

`2k+n(1 + lnn)

`k.

2. If p is xed, 0 < p ≤ 1, and ` n32k , then w.h.p. χ(SGn,k(p)) ≥

χ(KGn,k)− 2`.

4.2.1. Proof of Theorem 1

We rely on the proof of Kneser's conjecture due to I. Barany (it is possibleto obtain the result for Kneser graphs, but not for Schrijver graphs, out ofGreene's proof), as it is presented in the book [189]. First, for the rest of the

168

paper we x a certain mapping f : [n]→ Sd−1, using the following theoremdue to Gale (see [189]):

Lemma 4.8. There exists a mapping f : [n] → Sd−1, such that every openhemisphere in Sd−1 contains an image of a stable (k + `)-element subset of[n].

This is the point where we use the denition of d: d = n − 2k − 2` + 1.We note that such a mapping can be relatively easily constructed using themoment curve. In particular, this mapping is such that no d points lie ona great (hyper)sphere, that is, a sphere that is an intersection of Sd and ahyperplane that passes through the center of Sd. In what follows we omitthe prex hyper. Consider all combinatorially distinct partitions of the setf([n]) by great spheres.

The rst part of the proof is dedicated to an (a bit technical) estimation ofthe probability for a random graph SGn,k(p), which we for shorthand denoteby G, and a given function f , to have a certain partition.

Combinatorially, each partition by a great sphere is determined by thesubsets S+, S− of points from [n], such that f(S+), f(S−) lie into dier-ent open hemispheres (with respect to a given great sphere). Fix a greatsphere. Consider the sets S+

k ,S−k of all stable k-element subsets of S+, S−

respectively. Note that |S+k |, |S

−k | ≥

(k+`k

), since S+, S− contain a (k + `)-

element stable subset, any of which k-subsets is stable as well. Put t(S+) =d|S+

k |/de, t(S−) = d|S+k |/de. We choose a t(S+)-element subset M+ and a

t(S−)-element subsetM− out of S+k ,S

−k respectively. There are

( |S+k |t(S+)

)( |S−k |t(S−)

)choices for a pair of M+,M−, which is at most

(t(S+)dt(S+)

)(t(S−)dt(S−)

).

Below is the description of an event A, that is crucial for the proof. Theevent A holds if there exists a partition of the set f([n]) by a great sphereand subsets M+,M− of the type described above, so that there is no edgebetween them in G. We can bound the probability of A as follows:

Pr[A] ≤∑

possible partitions into S+, S−

(t(S+)d

t(S+)

)(t(S−)d

t(S−)

)(1− p)t(S

+)t(S−) (4.13)

Let us show that for t1 = t2 = d(k+lk

)/de the value of the function

g(t1, t2) =

(t1d

t1

)(t2d

t2

)(1− p)t1t2

169

is maximal among all t1, t2 ≥ d(k+`k

)/de. Since t1, t2 are integer and the

expression is symmetric with respect to t1, t2, it is enough to compare thevalues of the expression for t1, t2 and t1 + 1, t2:

g(t1 + 1, t2)

g(t1, t2)=

((t1 + 1)d)!

(t1 + 1)!((t1 + 1)(d− 1))!

t1!(t1(d− 1))!

(t1d)!(1− p)t2 =

=(t1 + 1)d

∏t1i=1((t1 + 1)(d− 1) + i)

(t1 + 1)∏t1

i=1((t1(d− 1) + i)(1− p)t2 <

< d

(t1 + 1

t1

)t1(1− p)t2 < e1+ln d−pt2.

The last expression is less than 1 if p > (1+ln d)

d(k+`k )/de, which is a corollary of

the condition on p given in the theorem. Therefore, we obtain the followingbound by changing in (4.13) all the summands g(t1, t2) to g(t, t) (in thefollowing calculation we use the notation t := d

(k+`k

)/de, which we used in

the formulation of Theorem 4.3):

Pr[A] < 3n(td

t

)2

(1− p)t2 < 3n exp

2t(1 + ln d)− pt2

=

= expt2(t−2n ln 3 + 2t−1(1 + ln d)− p

)<

< exp−t2εp

< exp−εn ln 3 1.

The rst inequality follows from the fact that the number of possible par-titions is bounded from above by 3n. In the second inequality we use thebound

(tdt

)<(etdt

)t. In the last two inequalities we used the condition on p

from the statement of Theorem 4.3.

The second part of the proof interconnects the event A and the propertyof the graph G to have chromatic number at most d.

To complete the proof of the theorem, we show that in any (not necessaryproper) coloring of SGn,k into d colors we always can nd two such sets ofvertices M+,M−, which additionally are monochromatic. Since P (A) 1,w.h.p. any two such sets in G have at least one edge between them, and itis thus impossible to nd a proper d-coloring of G.

Consider an arbitrary coloring of SGn,k into d colors. Each coloring to-gether with the mapping f described in the beginning of the proof induce apartition of the sphere Sd−1 into sets B1, . . . , Bd. We describe the partitionbelow. A point x of Sd−1 is included into Bi, where 1 ≤ i ≤ d, if in the open

170

hemisphere with the center in x the color i is the most popular color amongall the stable k-sets fully contained in that hemisphere. If there are equallypopular colors, we put the point in all such Bi's. It means that there are atleast t stable k-sets colored in i, lying in the open hemisphere with the centerin x.

We apply the celebrated theorem due to Lusternik, Schnirelman and Bor-suk (see the book [189]), one of which formulations is as follows:

Theorem 4.9. Whenever the sphere Sd−1 is covered by open sets S1, . . . , Sd,there exists i such that Si ∩ (−Si) 6= ∅.

We claim that the sets B1, . . . , Bd are open. Indeed, let us check that B1

is open. For any given point y ∈ f([n]) the set Uy of points x ∈ Sd, suchthat the open hemisphere with the center in x contains y, is open. Next, forany subset S ⊂ f([n]) the set US = ∩y∈SUy is also open. Finally, let S1 bethe set of all subsets S with the property that in the coloring of k-sets of Sthe color 1 is the most popular. Then B1 = ∪S∈S1US, and, consequently, B1

is open.Thus, we can apply Theorem 4.9 and obtain that one of Bi contains two

antipodal points x,−x. Then, by the denition of the set Bi, there are atleast t(S+), t(S−) vertices of color i in the sets S+

k , S−k respectively. Here the

sets S+k , S

−k correspond to the partition of the sphere by the great sphere,

orthogonal to x. We put M+,M− equal to these two monochromatic sets,and, by the rst part of the proof, w.h.p. they have an edge between them,so the proof is complete.

4.3. Proof of Theorem 4.2

In the next section we present the general approach to the problem, and ob-tain inequality (4.6). The approach, which is more adapted to our particularproblem, is presented in Section 4.3.2. The rest of the inequalities from The-orem 4.2 are obtained there. In Section 4.3.3 we prove some simple upperbounds.

4.3.1. Basic approach

In this section we discuss the general method, proposed to us by N. Alon,along with some of its corollaries to the case of Kneser and Schrijver graphsand hypergraphs. We prove (4.6) and reprove (4.4) in this section.

171

4.3.1.1. Coloring random subgraphs of blow-ups of hypergraphs

We start with the following abstract theorem on hypergraph colorings, pre-ceded by the denition of the class of hypergraphs in question. For an r-uniform hypergraph H = (V,E) and a positive integer number m considerthe m-blow-up H[m] of H: H[m] = (V ′, E ′), where V ′ := V × [m], andE ′ :=

v1× i1, . . . , vr× ir : v1, . . . , vr ∈ E, i1, . . . , ir ∈ [m]

. Informally

speaking, we replace each vertex of the original hypergraph with an m-tuple,and each edge with a complete r-partite hypergraph with m vertices in eachpart.

We denote by A(H,m) the class of hypergraphs that can be obtainedfrom H[m] by identifying some vertices that do not belong to the same edgeand do not arise from the same vertex of H. Formally, consider the class Fof functions f : V ′ → [n] for some n, such that:

1. For any v ∈ V and i 6= j we have f(v × i) 6= f(v × j).2. For any e ∈ E ′ and v1, v2 ∈ e we have f(v1) 6= f(v2).3. The function f is surjective.

Then the class of hypergraphs A(H,m) is dened as follows:

A(H,m) :=(f(V ′), Ef

): f ∈ F , Ef :=

f(v1), . . . , f(vr) : v1, . . . , vr ∈ E′

.

We denote byKr[m] the complete r-partite r-uniform hypergraph with partsof size m. For any 0 < p < 1 and a hypergraph H we dene the randomhypergraph H(p), which has the same set of vertices and in which each edgefrom G is taken independently and with probability p.

Theorem 4.10. Let H = (V,E) be an r-uniform hypergraph with χ(H) =d+ 1. Fix a number m ∈ N and consider a hypergraph G ∈ A(H,m). Thenfor any coloring of G into d colors there is a subhypergraph Kr[dmd e] ⊂ Gwith all vertices colored in the same color.

Moreover, for any 0 < p < 1 we have

Pr[χ(G(p)) ≤ d] ≤ |E|(

m

dm/de

)r(1− p)dm/der . (4.14)

Proof. Consider a coloring of G into d colors. We construct a certain coloringof H based on the coloring of G. For each vertex v ∈ H take its blow-upv1, . . . , vm in G and color v in the most popular color among v1, . . . , vm.It is clear that at least dmd e of vi's are colored in this color.

Since χ(H) > d, there is a monochromatic edge of color κ in this coloring.In G this edge corresponds to an r-uniform r-partite subhypergraph Kr[m].

172

Choosing out of each part the vertices colored in color κ, we get the desiredsubhypergraph.

In view of the argument above, the event χ(G(p)) ≤ d may occur onlyif one of the subhypergraphs Kr[dmd e] of the type described above is emptyin the random hypergraph G(p). The number of such subhypergraphs isbounded from above by |E|

(mdm/de

)r, while the probability for each to be

empty in G(p) is (1 − p)dm/der . Thus, inequality (4.14) follows by applyingthe union bound.

4.3.1.2. Numerical Corollaries for Kneser hypergraphs

For n ≥ (k+l)r the hypergraphKGrn,k belongs to the familyA(KGr

k+l,(k+lk

)).

Indeed, to each vertex S ∈(

[n]k+l

)of KGr

n,k+l we correspond the family of sub-

sets(Sk

). Put d := χ(KGr

n,k+l)− 1 and

t :=⌈(k + l

k

)/d⌉. (4.15)

The following lemma gives the rst (but not the strongest) general bound onthe chromatic number of random Kneser hypergraphs.

Lemma 4.11. For n ≥ (k + l)r we a.a.s. have χ(KGrn,k(p)) ≥ d+ 1 if

3r(

(k + l) logn

k+ t log d

)− ptr → −∞ (4.16)

Proof. Remark that the number of edges in KGrn,k+l is at most(

n

k + l

)r≤(nek

)(k+l)r

.

Therefore, applying the bound (4.14), we get that

Pr[χ(G(p)) ≤ d] ≤ |E(KGrn,k+l)|

((k+lk

)t

)r(1− p)tr ≤

(nek

)(k+l)r

(ed)rte−ptr ≤

exp[(k + l)r

(1 + log

n

k

)+ rt(1 + log d)− ptr

]≤ exp

[3r(

(k + l) logn

k+ t log d)

)− ptr

].

The last expression tends to 0 by (4.16), which concludes the proof of thelemma.

The condition in (4.16) is satised for xed p, r and l, provided tr−1 log d and tr k log n

k . Substituting the value of t and doing some tedious

173

calculations (see the proofs of the corollaries in [163] for more details), weget that the following hold a.a.s.:

(l = 1l = 1l = 1) χ(KGrn,k)(p) ≥ χ(KGr

n,k+1) if n− rk nr−1r ; (4.17)

(xed lll) χ(KGrn,k)(p) ≥ χ(KGr

n,k+l) if k nr

lr−1 log1

lr−1 and l is xed. (4.18)

The rst bound is the same as (4.1) and (4.4), while the second one is thesame as the bound (4.5). However, (4.18) is still a long way from the latterbounds in Theorem 4.10.

The way to improve the bound for l = 1, r = 2q is to work with r-stableKneser hypergraphs. As in the case of (complete) Kneser hypergraphs, wehave KGr, r−stable

n,k ∈ A(KGr, r−stablen,k+l ,

(k+lk

)) and, if n − rk = o(n), it has

fewer vertices and edges than KGrn,k.

Proposition 4.12. KGr, r−stablen,k has at most

(n−(r−1)(k−1)

k

)vertices.

Proof. The vertices of KGr, r−stablen,k are the k-subsets i1, . . . , ik of [n] that

satisfy ij+1 − ij ≥ r for each j = 0, . . . , k − 1, as well as i1 + n − ik ≥ r,provided that 1 ≤ i1 < . . . < ik ≤ n. Let us count the number f(n, k) ofk-sets satisfying all these restrictions except i1 + n − ik ≥ r. This numberwill clearly be an upper bound for |V (KGr, r−stable

n,k )|.It is easy to see that this quantity satises the following recursive formula:

f(n, k) = f(n− 1, k) + f(n− r, k− 1), as well as the condition f(r(k− 1) +1, k) = 1. The function

(n−(r−1)(k−1)

k

)satises both the recursive formula

and the initial condition.

Put d := χ(KGr, r−stablen,k+1 ) − 1. From Proposition 4.12 we get that the

number of edges in KGr, r−stablen,k+1 is at most(

n− (r − 1)(k − 1)

k

)r=

(k +O(d)

k

)r=

(O(n)

O(d)

)r= eO(d log n

d ).

Thus, instead of (4.16) it is sucient to show that (recall that p and r arexed)

d logn

d+ t log d tr. (4.19)

We have t = Θ(nd) (see (4.15)) and d = O(n − rk). Doing some routinecalculations again, we get that a.a.s.

(l = 1l = 1l = 1) χ(KGr, r−stablen,k )(p) ≥ χ(KGr, r−stable

n,k+1 ) if n− rk nr/(r+1) log−1/(r+1) n.

Since for q ∈ N and r = 2q we have χ(KGr, r−stablen,k+1 ) = χ(KGr

n,k+1), we get(4.6).

174

4.3.2. The approach rened

The crucial step in the proof of Theorem 4.10 is to get a monochromaticedge of KGr

n,k+l, induced by the coloring of KGrn,k. The main limitation of

the method from the previous section is related to this step. We have toassume that (in the worst case) among the vertices of the m-blow-up of themonochromatic edge all colors are represented in approximately the sameproportion. This is why we can only guarantee the majority color class tohave size at least m

d . On the other hand, in order to get a good bound on theprobability, we need to work with color classes of growing size. Therefore,the approach from the previous section is bound to work only for m d, or,in terms of Kneser hypergraphs, for

(k+lk

) χ(KGr

n,k+l).In this section we are going to partially overcome the aforementioned

diculty. We assume that n k + l and that r ≥ 2 is xed for the rest ofthe section. Put

d := χ(KGrn,k+l)− 1.

Note that d = n− 2k− 2l+ 1 for r = 2. Fix a coloring κ of(

[n]k

)in d colors.

For each subset S ⊂ [n] of size at least k, dene the color of S to be themost popular color among its subsets. We have thus dened the coloring κ′

of KGrn,k′ for all k

′ ≥ k. Put u := blog2n

k+lβc and consider the following

sequence of numbers

q0 := k+l, qi := d2i(k+lβ)e, where i = 1, . . . , u and β =

1 if r = 2

rr−1

if r ≥ 3.(4.20)

Note that, by denition, no qi is bigger than n and that qi ≤ 2lβ−1qi−1 foreach i ∈ [u]. The numbers qi will play the role of the sizes of subsets onwhich we construct our Kneser hypergraphs. Next, for each i = 0, . . . , u,dene the following two numbers:

ti :=⌈(qi

k

)d

⌉, zi :=

⌈ (qik

)2sqi

⌉, where s :=

(2r + 1)lβ−1 if r = 2, 3;

(2r + 1)lβ−1k if r > 3.(4.21)

Note that t0 is equal to t from the previous section. Both ti and zi willplay the roles of the sizes of popular colors among the k-subsets of a certainqi-element set.

The following lemma is central for this section.

Lemma 4.13. For any coloring κ of KGrn,k in d colors there is a color α

for which one of the following holds.

175

(i) There exists i ∈ 0, . . . , u and r pairwise disjoint subsets A1, . . . , Ar ∈([n]qi

), such that at least zi subsets from each

(Ajk

), j = 1, . . . , r, are

colored in α.

(ii) There exists i ∈ 0, . . . , u−1 and r pairwise disjoint subsets A1 ∈(nqi

),

A2, . . . , Ar ∈(

[n]qi+1

), such that at least ti subsets from

(A1

k

)and zi+1

subsets from each(Ajk

), j = 1, . . . , r, are colored in α.

Proof. We start by analyzing the colorings of KGrn,k into d colors. For each

i = 0, . . . , u associate with each set A ∈(

[n]qi

)the set XA of the colors that

are used at least ti2 times in the coloring of

(Ak

)(recall the denitions (4.20),

(4.21)). Note that at least a half of the vertices of(Ak

)is colored by colors

from XA. We have two possibilities for a given i: either for each set A ∈(

[n]qi

)there is a color that is used for zi vertices on

(Ak

), or there is a set A ∈

([n]qi

)such that |XA| > sqi. This is obvious in case zi ≤ ti. If zi > ti (which istypically the case), then the negations of both statements imply that in XA

there are less than sqi colors, each of cardinality at most zi − 1. On theother hand, the colors from XA are used for at least 1

2

(qik

)k-sets in A, but

12

(qik

)> (zi − 1)sqi. This is a contradiction.

If for each qi-element set A there is a color in XA used at least zi times,then, arguing as in the proof of Theorem 4.10, we conclude that (i) from thelemma takes place.

If not, then x the largest index i, for which there exist a qi-element setA with |XA| > sqi, and choose such A. Clearly, i ≤ u − 1, otherwise wehave more than d colors in XA. Put Y := [n] \ A and consider the majoritycoloring κ′ of the qi+1-element subsets of Y . We denote by KGr

Y,qi+1the

Kneser hypergraph induced on(Yqi+1

).

We claim that at least one of the two holds. Either A κ′ is not proper, andwe again conclude that (i) holds (remark that for any qi+1-set B the set XB

has a color class of size zi+1 by the choice of i), or B in KGrY,qi+1

there is acolor from XA used for an (r−1)-tuple of pairwise disjoint qi+1-element sets.If the second possibility takes place, then we obtain (ii) from the lemma.Thus, we are left to show that one of the two possibilities claimed in thisparagraph takes place.

Assume that neither of the two options A, B takes place. In the caser = 2, this simply means that KGY,qi+1

is properly colored in less than d−sqicolors (the negation of B implies that the colors from XA are not used in κ′).Remark that β = 1 for r = 2 and thus qi+1 = 2qi for each i = 0, . . . , u− 1.

176

Therefore, n−2k−2l+1−sqi = d−sqi > χ(KGY,qi+1) = (n−qi)−2qi+1 +2,

which is equivalent to (s − 5)qi + 2(k + l) + 1 < 0. But, by the denition(4.21), s = 5 when r = 2. We arrive at a contradiction, which concludes theproof for r = 2.

In the case r ≥ 3 the negation of A, B means that KGrY,qi+1

is properlycolored, and that for each color α ∈ XA the collection of sets colored in αcontains at most r − 2 pairwise disjoint sets.

If r = 3, then construct a new coloring of KG3Y,qi+1

by arbitrarily groupingthe colors from XA into pairs and replacing each pair by a single new color.Remark that the new coloring uses at most d− sqi

2 colors and is still proper,since in any newly formed color there are no more than two pairwise disjointsets. Therefore, χ(KG3

n,k+l)− 1− sqi2 ≥ χ(KG3

Y,qi+1).

Recall that qi+1 ≤ 2lβ−1qi. For any r ≥ 2, we have

χ(KGrn,k+l)− χ(KGr

Y,qi+1) =⌈

n− (k + l − 1)r

r − 1

⌉−

⌈n− qi − (qi+1 − 1)r

r − 1

⌉<rqi+1 + qir − 1

≤ (2r + 1)lβ−1qir − 1

. (4.22)

Thus, for r = 3 we conclude that sqi2 < (2r+1)lβ−1qi

2 , which contradicts (4.21).Finally, consider the case r > 3. We again construct a new coloring of

KGrY,qi+1

that uses fewer colors than κ′, using the following procedure. First,split the colors from XA into groups of size k(r− 2) + 1 (with one remaininggroup of potentially smaller size). In each group choose one color α andsplit vertices (sets) from KGY,qi+1

of color α into k(r− 2) groups of pairwiseintersecting sets. To obtain such a split, take the largest family of pairwisedisjoint sets in KGY,qi+1

colored in α. It has size most r − 2, and thus itcovers the set U ⊂ Y of cardinality at most (r − 2)k. Each other set ofcolor α in Y must intersect U . We then split all sets of color α into familiesKj, j = 1, . . . , |U | of sets containing the j-th element of U .

Next, we adjoin each of the families Kj to one of the remaining k(r − 2)colors in the group. We get a proper coloring since none of the newly formedcolors contain more than r − 1 pairwise disjoint sets. The number of colorsused in the new coloring is less than d − b sqi

k(r−2)+1c ≤ d − b sqik(r−1)c. Thus,

comparing the inequality χ(KGrn,k+l)− 1− b sqi

k(r−1)c ≥ χ(KGrY,qi+1

) with theinequality (4.22), we get⌊ sqi

k(r − 1)

⌋+ 1 <

(2r + 1)lβ−1qir − 1

,

which contradicts the denition (4.21).

177

Lemma 4.13 tells us that if there exists a proper d-coloring of KGrn,k(p),

then there are subsets A1, . . . , Ar as in the lemma, such that the inducedsubgraph of KGr

n,k(p) on these subsets has no edge. In what follows wecalculate the probabilities of these events.

Assume that the rst possibility from Lemma 4.13 for a certain i takesplace. The probability of the corresponding event for the random hypergraphis at most (

n

qi

)r((qik

)zi

)r(1− p)zri . (4.23)

If the second possibility from Lemma 4.13 for a certain i takes place, thenthe probability of the corresponding event for the random hypergraph is atmost(

n

qi

)(n

qi+1

)r−1((qik

)ti

)((qi+1

k

)zi+1

)r−1

(1− p)tizr−1i+1 ≤ nrqi+1

((qi+1

k

)zi+1

)r(1− p)tiz

r−1i+1 . (4.24)

Typically, the last expression in (4.24) is much bigger than that in (4.23).Note that the total number of events is 2u + 1 ≤ 2 log n. Therefore, if eachhas probability less than 1

n , say, then we a.a.s. have χ(KGrn,k(p)) ≥ d+1. We

remark that this condition on the probability of a single event is by no meansrestrictive, since we are manipulating with expressions of much higher orderof growth. The following analogue of Lemma 4.11 with general bounds onthe chromatic number of random Kneser hypergraphs is proven by analogouscalculations:

Lemma 4.14. For n ≥ (k + l)r we a.a.s. have χ(KGrn,k(p)) ≥ d + 1 :=

χ(KGrn,k+l) if for each i = 0, . . . , u we have

3r(qi log n+ zi log(2sqi)

)− pzri → −∞ and (4.25)

3r(qi+1 log n+ zi+1 log(2sqi+1)

)− ptizr−1

i+1 → −∞. (4.26)

In what follows, we assume that p > 0 is xed and that k, l ≥ 2. Wehave log qi = Ω(log(2sqi+1)) for any i ≥ 0. Then the inequalities (4.25) and(4.26) follow from

zr−1i log qi, zri qi log n, tiz

r−2i+1 log qi, tiz

r−1i+1 qi+1 log n. (4.27)

We have zi+m/zi = Ω(qi+m/qi) for any r ≥ 2, k ≥ 2 and m ∈ [u − i].Therefore, for r ≥ 3 it is sucient to verify the inequalities (4.27) for i = 0.For r = 2 it is sucient to verify the rst, second and fourth inequality from(4.27) for i = 0 and the inequality t0 ≥ log n.

178

For r = 2, 3 we have s = O(l1/2), which implies zi = O(

(qik)l1/2qi

) log qi.

Therefore, the rst inequality from (4.27) is satised for r = 2, 3, and (4.27)reduces to the following:

zr0 (k + l) log n, tizr−2i+1 log qi, t0z

r−11 (k + l) log n, (4.28)

where for r = 3 the second inequality is again automatically satised. Letr = 2. Looking at the denitions (4.21) and using the fact that

(k+lk

)=

Ω((k + l)2), it is clear that the second condition is the most restrictive. Thefollowing inequality is sucient to satisfy (4.28) and implies both (4.7) and(4.8): (

k + l

k

) n log n. (4.29)

For r = 3, replacing the t0 factor with 1, we conclude that (4.28) followsfrom(

k + l

k

)3

(k + l)4l3/2 log n,

(2(k + l3/2)

k

)2

(k + l)3l log n. (4.30)

For xed l the rst condition is clearly more restrictive, and we get that(4.30) holds for k log1/(3l−4) n. For xed k the rst inequality is morerestrictive again, and we get that it holds for l log1/(3k−11/2) n. This givesthe inequalities (4.9) and (4.10).

For r > 3 we have to assume l ≥ 3 in order to get any good lower boundon zi: for l = 2 we have z0 = 1. But for l ≥ 3 we again have zi log qi, so itis again enough to verify the rst and third inequalities in (4.28). Replacingt0 with 1, we get that (4.28) is implied by(

k + l

k

)r kr(k + l)r+1lr/(r−1) log n and (4.31)(

2(k + lr/(r−1))

k

)r−1

kr−1(k + l)(k + l

rr−1)r−1

l log n. (4.32)

Similarly to the case r = 3, for both xed l and xed k (4.31) is morerestrictive. For xed l we get that (4.31) holds for k log

1r(l−2)−1 n. For

xed k it holds for l log1

r(k−1)− 2r−1r−1 n. This implies (4.11) and (4.12).

4.3.3. Simple lower bounds

In this section we present simple upper bounds for χ(KGrn,k(p)) and compare

them with the results of Theorem 4.2. If there exist a set A ⊂ [n] of size rk+l,

179

such thatKGr(n, k)(p)|A is an empty graph, then, coloring A into color 0 andthe rest as in the standard coloring ofKGr(n, k), we get that χ(KGr

n,k(p)) ≤χ(KGr

n,k) − bl/(r − 1)c. To estimate the probability of having such A, wend n sets of size l + 2k in [n], which have pairwise intersections of size atmost 1, and, and calculate the probability that one of those becomes empty.Note that the events for dierent sets are independent. The probability is(

1− (1− p)∏ri=1 (l+ikk )

)n≤ e−n(1−p)

∏ri=1 (l+ikk )

.

Therefore, ifn(1− p)

∏ri=1 (l+ikk ) →∞, (4.33)

then a.a.s. there exists such a set. If p, r, k are xed, then this condition issatised if for suciently large constant we have ecl

rk

= o(n), which impliesthat we can take l = Ω(log

1rk n). This shows that bounds (4.10), (4.12) are

essentially tight: the dierence between the lower and the upper bounds arein the degree of the logarithm.

If p, r, and l are xed, then the situation is more interesting. The condition

(4.33) is satised if ecr2k

= o(n), which could be fullled for k = Ω(log log n).This is very dierent from the bounds (4.9), (4.11). Of course, in the graphcase (r = 2) the gap between the upper and lower bounds is even bigger.

180

Chapter 5.

Coverings and ε-nets

My results presented in this chapter (joint with N. Mustafa and J. Pach) arepublished in [168, 169].

5.1. Introduction

Let X be a nite set and let R be a system of subsets of an underlyingset containing X. In computational geometry, the pair (X,R) is usuallycalled a range space. A subset X ′ ⊆ X is called an ε-net for (X,R) ifX ′ ∩ R 6= ∅ for every R ∈ R with at least ε|X| elements. The use of small-sized ε-nets in geometrically dened range spaces has become a standardtechnique in discrete and computational geometry, with many combinatorialand algorithmic consequences. In most applications, ε-nets precisely andprovably capture the most important quantitative and qualitative propertiesthat one would expect from a random sample. Typical applications includethe existence of spanning trees and simplicial partitions with low crossingnumber, upper bounds for discrepancy of set systems, LP rounding, rangesearching, streaming algorithms; see [187, 201].

For any subset Y ⊆ X, dene the projection of R on Y to be the setsystem

R|Y :=Y ∩R : R ∈ R

.

The Vapnik-Chervonenkis dimension or, in short, the VC-dimension of therange space (X,R) is the minimum integer d such that |R|Y | < 2|R| for anysubset Y ⊆ X with |Y | > d. According to the SauerShelah lemma [222, 226](discovered earlier by Vapnik and Chervonenkis [231]), for any range space(X,R) whose VC-dimension is at most d and for any subset Y ⊆ X, we have|R|Y | = O(|Y |d).

A straightforward sampling argument shows that every range space (X,R)has an ε-net of size O(1

ε log |R|X |). The remarkable result of Haussler and

181

Welzl [126], based on the previous work of Vapnik and Chervonenkis [231],shows that much smaller ε-nets exist if we assume that our range spacehas small VC-dimension. Haussler and Welzl [126] showed that if the VC-dimension of a range space (X,R) is at most d, then by picking a ran-dom sample of size Θ(dε log d

ε ), we obtain an ε-net with positive probability.Actually, they only used the weaker assumption that |R|Y | = O(|Y |d) forevery Y ⊆ X. This bound was later improved to (1 + o(1))(dε log 1

ε ), asd, 1

ε → ∞ [155]. In the sequel, we will refer to this result as the ε-net theo-rem. The key feature of the ε-net theorem is that it guarantees the existenceof an ε-net whose size is independent of both |X| and |R|X |. Furthermore, ifone only requires the VC-dimension of (X,R) to be bounded by d, then thisbound cannot be improved. It was shown in [155] that given any ε > 0 andinteger d ≥ 2, there exist range spaces with VC-dimension at most d, andfor which any ε-net must have size at least

(1− 2

d + 1d(d+2) + o(1)

)dε log 1

ε .

The eectiveness of ε-net theory in geometry derives from the fact thatmost geometrically dened range spaces (X,R) arising in applications havebounded VC-dimension and, hence, satisfy the preconditions of the ε-nettheorem.

There are two important types of geometric set systems, both involvingpoints and geometric objects in Rd, that are used in such applications. LetRbe a family of possibly unbounded geometric objects in Rd, such as the familyof all half-spaces, all balls, all polytopes with a bounded number of facets,or all semialgebraic sets of bounded complexity, i.e., subsets of Rd denedby at most D polynomial equations or inequalities in the d variables, each ofdegree at most D. Given a nite set of points X ⊂ Rd, we dene the primalrange space (X,R) as the set system induced by containment in the objectsfrom R. Formally, it is a set system with the set of elements X and setsX ∩R : R ∈ R. The combinatorial properties of this range space dependon the projectionR|X . Using this terminology, Radon's theorem [187] impliesthat the primal range space on a ground set X, induced by containment inhalf-spaces in Rd, has VC-dimension at most d+ 1 [201]. Thus, by the ε-nettheorem, this range space has an ε-net of size O(dε log 1

ε ).In many applications, it is natural to consider the dual range space, in

which the roles of the points and ranges are swapped. As above, let R bea family of geometric objects (ranges) in Rd. Given a nite set of objectsS ⊆ R, the dual range space induced by them is dened as the set system(hypergraph) on the ground set S, consisting of the sets Sx := S ∈ S : x ∈S for all x ∈ Rd. It can be shown that if for any X ⊂ Rd the VC-dimension

182

of the range space (X,R) is less than d, then the VC-dimension of the dualrange space induced by any subset of R is less than 2d [187].

Recent progress. In many geometric scenarios, however, one can nd smallerε-nets than those whose existence is guaranteed by the ε-net theorem. It hasbeen known for a long time that this is the case, e.g., for primal set systemsinduced by containment in balls in R2 and half-spaces in R2 and R3. Over thepast two decades, a number of specialized techniques have been developedto show the existence of small-sized ε-nets for such set systems [15, 16, 45,46, 47, 50, 155, 186, 190, 198, 207, 232, 233]. Based on these successes, itwas generally believed that in most geometric scenarios one should be able tosubstantially strengthen the ε-net theorem, and obtain perhaps even a O

(1ε

)upper bound for the size of the smallest ε-nets. In this direction, there havebeen two signicant recent developments: one positive and one negative.

Upper bounds. Following the work of Clarkson and Varadarajan [50], ithas been gradually realized that if one replaces the condition that the rangespace (X,R) has bounded VC-dimension by a more rened combinatorialproperty, one can prove the existence of ε-nets of size o(1

ε log 1ε ). To formulate

this property, we need to introduce some terminology.

Given a function ϕ : N→ R+, we say that the primal range space (X,R)has shallow-cell complexity ϕ if there exists a constant c = c(R) > 0 suchthat, for every Y ⊆ X and for every positive integer l, the number of atmost l-element sets in R|Y is O

(|Y | · ϕ(|Y |) · lc

). Note that if the VC-

dimension of (X,R) is d, then for every Y ⊆ X, the number of elements ofthe projection of the set system R to Y satises |R|Y | = O(|Y |d). However,the condition that (X,R) has shallow-cell complexity ϕ for some functionϕ(n) = O(nd

′), 0 < d′ < d− 1 and some constant c = c(R), implies not only

that |R|Y | = O(|Y |1+d′+c), but it reveals some nontrivial ner details aboutthe distribution of the sizes of the smaller members of R|Y .

Several of the range spaces mentioned earlier turn out to have low shallow-cell complexity. For instance, the primal range spaces induced by contain-ment of points in disks in R2 or half-spaces in R3 have shallow-cell com-plexity ϕ(n) = O(1). In general, it is known [187] that the primal rangespace induced by containment of points by half-spaces in Rd has shallow-cellcomplexity ϕ(n) = O

(nbd/2c−1

).

Dene the union complexity of a family of objects R, as the maximumnumber of faces (boundary pieces) of all dimensions that the union of any nmembers of R can have; see [2]. Applying a simple probabilistic technique

183

developed by Clarkson and Shor [51], one can nd an interesting relationshipbetween the union complexity of a family of objects R and the shallow-cellcomplexities of the dual range spaces induced by subsets S ⊂ R. Supposethat the union complexity of a family R of objects in the plane is O

(nϕ(n)

),

for some well-behaved non-decreasing function ϕ. Then the number of atmost l-element subsets in the dual range space induced by any S ⊂ R isO(l2 · |S|l ϕ( |S|l )

)= O

(|S|ϕ(|S|)l

)[225]; i.e., the dual range space induced by

S has shallow-cell complexity O(ϕ(n)

). According to the above denitions,

this means that for any S ⊂ R and for any positive integer l, the number ofl-element subsets S ′ ⊆ S for which there is a point p′ ∈ R2 contained in allelements of S ′, but in none of the elements of S\S ′, is at most O

(|S|ϕ(|S|)l

).

Note that for small values of l, the points p′ are not heavily covered (l times).Thus, the corresponding cells

⋂S∈S ′ S \

⋃T∈S\S ′ T of the arrangement S are

shallow, and the number of these shallow cells is bounded from above. Thisexplains the use of the term shallow-cell complexity.

A series of elegant results [15, 46, 233] illustrate that if the shallow-cellcomplexity of a set system is ϕ(n) = o(n), then it permits smaller ε-nets thanwhat is guaranteed by the ε-net theorem. The following theorem representsthe current state of the art (see [197] for a simple proof).

Theorem 5.1. Let (X,R) be a range space with shallow-cell complexity ϕ,where ϕ(n) = O(nd) for some constant d. Then, for every ε > 0, it hasan ε-net of size O

(1ε logϕ(1

ε )), where the constant hidden in the O-notation

depends on d.

Proof sketch. The main result in [46] shows the existence of ε-nets of sizeO(

1ε logϕ(|X|)

)for any non-decreasing function ϕ1. To get a bound in-

dependent of |X|, rst compute a small (ε/2)-approximation A ⊆ X for(X,R) [187]. It is known that there is such an A with |A| = O

(dε2 log 1

ε

)=

O( 1ε3 ), and for any R ∈ R, we have |R∩A||A| ≥

|R||X| −

ε2 . In particular, any

R ∈ R with |R| ≥ ε|X| contains at least an ε2-fraction of the elements of

A. Therefore, an (ε/2)-net for (A,R|A) is an ε-net for (X,R). Computingan (ε/2)-net for (A,R|A) gives the required set of size O

(2ε logϕ(|A|)

)=

O(

1ε logϕ( 1

ε3 ))

= O(

1ε logϕ(1

ε )).

Lower bounds. It was conjectured for a long time [190] that most geomet-rically dened range spaces of bounded Vapnik-Chervonenkis dimension havelinear-sized ε-nets, i.e., ε-nets of size O

(1ε

). These hopes were shattered by

Alon [5], who established a superlinear (but barely superlinear!) lower bound1Their result is in fact for the more general problem of small weight ε-nets.

184

on the size of ε-nets for the primal range space induced by straight lines inthe plane. Shortly after, Pach and Tardos [202] managed to establish a tightlower bound, Ω(1

ε log 1ε ) for the size of ε-nets in primal range spaces induced

by half-spaces in R4, and in several other geometric scenarios.

Theorem 5.2. [202] Let F denote the family of half-spaces in R4. For anyε > 0, there exist point sets X ⊂ R4 such that in the (primal) range spaces(X,F), the size of every ε-net is at least 1

9ε log 1ε .

My contribution. The goal of our research was to complete both avenuesof research opened by Theorems 5.1 and 5.2. Our rst theorem, provedin Section 2, generalizes Theorem 5.2 to Rd, for d ≥ 4. It provides anasymptotically tight bound in terms of both ε and d, and hence completelysettles the ε-net problem for half-spaces.

Theorem 5.3 ([168, 169]). For any integer d ≥ 4, real ε > 0 and anysuciently large integer n ≥ n0(ε), there exist primal range spaces (X,F)induced by n-element point sets X and collections of half-spaces F in Rd

such that the size of every ε-net for (X,F) is at least bd/4c9ε log 1ε .

As was mentioned in the rst subsection, for any d ≥ 1, the VC-dimensionof any range space induced by points and half-spaces in Rd is at most d+ 1.Thus, Theorem 5.3 matches, up to a constant factor independent of d and ε,the upper bound implied by the ε-net theorem of Haussler and Welzl. NogaAlon pointed out to us that it is very easy to show that for a xed ε > 0,the lower bound for ε-nets in range spaces induced by half-spaces in Rd hasto grow at least linearly in d. To see this, suppose that we want to obtain a13-net, say, for the range space induced by open half-spaces on a set X of 3dpoints in general position in Rd. Notice that for this we need at least d + 1points. Indeed, any d points of X span a hyperplane, and one of the openhalf-spaces determined by this hyperplane contains at least |X|3 points.

The key element of the proof of Theorem 5.2 [202] was to construct aset B of (k + 3)2k−2 axis-parallel rectangles in the plane such that for anysubset of them there is a set Q of at most 2k−1 points that hit none of therectangles that belong to this subset and all the rectangles in its complement(the precise statement is given in Section 5.3). We generalize this statementto Rd by constructing roughly d

2 times more axis-parallel boxes2 than in the

2An axis-parallel box in Rd is the Cartesian product of d + 1 intervals. For simplicity, in the sequel,they will be called boxes.

185

planar case, but the size of the set Q remains the same size. In Section 5.3,we prove

Lemma 5.4. Let k, d ≥ 2 be integers. Then there exists a set B of bd2c(k +3)2k−2 axis-parallel boxes in Rd such that for any subset S ⊆ B, one can nda 2k−1-element set Q of points with the property that

(i) Q ∩B 6= ∅ for any B ∈ B \ S, and(ii) Q ∩B = ∅ for any B ∈ S.

In the next section we show how this lemma implies the bound of The-orem 5.3, which is bd4c times better than the bound in Theorem 5.2. Theproof of Lemma 5.4 will be given in Section 5.3.

In Section 5.4, we show that the bound in Theorem 5.1 cannot be im-proved.

Denition 5.5. A function ϕ : R+ → R+ is called submultiplicative if thereexists an α, 0 < α < 1 such that

1) ϕα(x) ≤ ϕ(xα) for all suciently large x ∈ R+, and

2) ϕ(xy) ≤ ϕ(x)ϕ(y) for all suciently large x, y ∈ R+.

Theorem 5.6 ([168, 169]). Let d be a xed positive integer and let ϕ :R+ → R+ be any submultiplicative function with ϕ(n) = O(nd). Then, forany ε > 0 there exist range spaces (X,F) that have

(i) shallow-cell complexity ϕ, and for which(ii) the size of any ε-net is at least Ω(1

ε logϕ(1ε )).

Note that if ϕ(n) = Ω(n), then this theorem yields a lower bound for thesize of the smallest ε-nets in the constructed range spaces which is somewhatweaker than the bound Ω(dε log 1

ε ), valid for the old constructions in [155].Indeed, the property that the VC-dimension is at most d, imposed on therange spaces constructed in [155], implies that the range space has shallow-cell complexity O(nd−1).

Theorem 5.6 becomes interesting when ϕ(n) = o(n) and the upper boundO(

1ε logϕ(1

ε ))in Theorem 5.1 improves on the general upper boundO

(1ε log 1

ε

)guaranteed by the ε-net theorem. Theorem 5.6 shows that, if ϕ(n) = o(n),even this improved bound is asymptotically tight.

The best upper and lower bounds for the size of small ε-nets in range spaceswith a given shallow-cell complexity ϕ are based on purely combinatorialarguments, and they imply directly or indirectly all known results on ε-netsin geometrically dened range spaces (see [199] for a detailed discussion).

186

This suggests that the introduction of the notion of shallow-cell complexityprovided the right framework for ε-net theory.

5.2. Proof of Theorem 5.3 using Lemma 5.4

Let B be a set of d-dimensional axis-parallel boxes in Rd. We recall that thedual range space induced by B is the set system (hypergraph) on the groundset B consisting of the sets Bp := B ∈ B : p ∈ B for all p ∈ Rd.

Lemma 5.7. Let d ≥ 1 be an integer, and consider the dual range spaceinduced by a set of axis-parallel boxes B in Rd. Then there exists a functionf : B → R2d such that for every point p ∈ Rd, there is a half-space H in R2d

with f(B) : B ∈ Bp = H ∩ f(B) : B ∈ B.

Proof. By translation, we can assume that all boxes in B lie in the positiveorthant of Rd.

Consider the function g : B → R2d mapping a box B = [xl1, xr1]× [xl2, x

r2]×

· · ·×[xld, xrd] to the point (xl1, 1/x

r1, x

l2, 1/x

r2, . . . , x

ld, 1/x

rd) lying in the positive

orthant of R2d. Furthermore, for any p = (a1, a2, . . . , ad) ∈ Rd in the positiveorthant, let Cp denote the box [0, a1]× [0, 1/a1]× [0, a2]× [0, 1/a2]× · · · ×[0, ad]× [0, 1/ad] in R2d. Clearly, a point p lies in a box B in Rd if and onlyif g(B) ∈ Cp in R2d. Thus, g maps the set of boxes in B to a set of points inR2d, such that for any point p in the positive orthant of Rd, the set of boxesBp ⊂ B that contain p are mapped to the set of points that belong to thebox Bp. (Note that Bp contains the origin.)

We complete the proof by applying the following simple transformation([202, Lemma 2.3]) to the set Q = g(B): to each point q ∈ Q in the positiveorthant of R2d, we can assign another point q′ in the positive orthant of R2d

such that for each box in R2d that contains the origin, there is a half-spacewith the property that q belongs to the box if and only if q′ belongs to thecorresponding half-space. The mapping f(B) = (g(B))′ for every B ∈ Bmeets the requirements of the lemma.

Lemma 5.8. Given any integer d ≥ 2, a real number ε > 0, and a su-ciently large integer n ≥ n0(ε), there exists a set B of n axis-parallel boxesin Rd such that the size of any ε-net for the dual set system induced by B is

at leastbd2c9ε log 1

ε .

Proof. Let ε = α2k−1

with k ∈ N, k ≥ 2, and 13 ≤ α ≤ 2

3 . Applying Lemma 5.4,we obtain a set B of bd2c(k + 3)2k−2 boxes in Rd. We claim that the dualrange space induced by these boxes does not admit an ε-net of size (1−α)|B|.

187

Assume for contradiction that there is an ε-net S ⊆ B with | S | ≤ (1 −α)|B|. According to Lemma 5.4, there exists a set Q of 2k−1 points in Rd withthe property that no box in S contains any point of Q, but every member ofB \ S does. By the pigeonhole principle, there is a point p ∈ Q contained inat least

|B \ S ||Q|

≥ α|B||Q|

=α|B|2k−1

= ε|B|

members of B \ S. Thus, none of the at least ε|B| members of B hit by pbelong to S, contradicting the assumption that S was an ε-net.

Hence, the size of any ε-net in the dual range space induced by B is atleast

(1−α)|B| = (1−α)bd2c(k+3)2k−2 =

(1− α)α

2·bd

2c·k + 3

ε≥ 1

9·bd

2c·1ε·log

1

ε.

The system of boxes constructed above has a xed number of elements,depending on the value of 1/ε. We can obtain arbitrarily large constructionsby replacing each box of B ∈ B with several slightly translated copies of B(we refer the reader to [202] for details).

Now we are in a position to establish Theorem 5.3. By Lemma 5.7, anylower bound for the size of ε-nets in the dual range space induced by theset B of boxes in Rd gives the same lower bound for the size of an ε-netin the (primal) range space on the set of points f(B) ⊂ R2d correspondingto these boxes, in which the ranges are half-spaces in R2d. For any integerd ≥ 4 and any real ε > 0, Lemma 5.8 guarantees the existence of a set Bof n axis-parallel boxes in Rbd/2c such that any ε-net for the dual set system

induced by B has size at least bbd/2c

2 c9ε log 1

ε = bd/4c9ε log 1

ε . This fact, togetherwith Lemma 5.7, implies the stated bound.

5.3. Proof of Lemma 5.4

The proof of Lemma 5.4 is based on the following key statement.

Lemma A ([202]). Let k ≥ 2 be an integer. Then there exists a set R of(k + 3)2k−2 axis-parallel rectangles in R2 such that for any S ⊆ R, thereexists a 2k−1-element set Q of points in R2 with the property that

(i) Q ∩R 6= ∅ for any R ∈ R \ S, and(ii) Q ∩R = ∅ for any R ∈ S.

Denote the x- and y-coordinates of a point p ∈ R2 by x(p) and y(p)respectively, and set m = bd2c. Let R = R1, . . . , Rt, t = (k + 3)2k−2, be a

188

set of rectangles satisfying the conditions of Lemma A. By scaling, one canassume that R ⊂ [0, 1]2 for every R ∈ R.

Given that a box in Rd is the product of d intervals, the idea of theconstruction is to `lift' the rectangles in Lemma A, i.e., the set R, to boxesin Rd. So a rectangle R ∈ R can be mapped to a box in Rd which is theproduct d intervals: the rst two being the intervals dening R, and theother d − 2 intervals in the product being the full interval [0, 1]. One canthen again lift the same set R in a `non-interfering' way by mapping R to abox whose 3-rd and 4-th intervals are the intervals of R and the remainingintervals are [0, 1]. In this way, by packing intervals of each R ∈ R intodisjoint coordinates, one can lift R m times to get a set of bd2c · |R| boxes inRd.

Formally, for i = 1 . . .m, dene the injective functions fi that map a pointin R2 to a product of d intervals in Rd, as follows.

fi(p) = [0, 1]× · · · × [0, 1]︸ ︷︷ ︸2i− 2 intervals

×x(p)× y(p)× [0, 1]× · · · × [0, 1]︸ ︷︷ ︸d− 2i intervals

, p ∈ R2.

This mapping lifts each rectangleR ∈ R to the box fi(R) = fi(p) : p ∈ R,and each set of rectangles R′ ⊆ R to the set of boxes fi(R′) = fi(R) :R ∈ R′.

We now show that B =⋃mi=1 fi(R) is the desired set of bd2c(k + 3)2k−2

boxes in Rd. Let S ⊆ B be a xed set of boxes. For any index i ∈ [1,m],set Ri ⊆ R to be the set of preimage rectangles under fi of the boxes inS ∩fi(R), i.e.,Ri satises S ∩fi(R) = fi(Ri). LetQi = qi1, . . . , qi2k−1 ⊂ R2

be the set of points hitting all rectangles in R \ Ri and no rectangle in Ri;such a set exists by Lemma A. Now we argue that the set

Q =

(x(q1

j ), y(q1j ), . . . , x(qmj ), y(qmj )

): j ∈ [1, 2k−1]

if d is even,(

x(q1j ), y(q1

j ), . . . , x(qmj ), y(qmj ), 1)

: j ∈ [1, 2k−1]

if d is odd,

of 2k−1 points in Rd is the required set for S; i.e., Q hits all the boxes in B\S,and none of the boxes in B ∈ S. Take any box B ∈ B \ S; then there existsan index i and a rectangle R ∈ R\Ri such that R is the preimage rectangleof B under fi. By Lemma A, R contains a point q ∈ Qi, and thus B = fi(R)contains the point q′ ∈ Q with x(q) and y(q) in its (2i − 1)-th and 2i-thcoordinates, as all the remaining intervals dening B are [0, 1] and so eachsuch interval contains the corresponding coordinate of q′. On the other hand,let B ∈ S be a box with the preimage rectangle R ∈ Ri. By Lemma A, R

189

is not hit by any point of Qi, and thus for any point q′ ∈ Q, the (2i− 1)-thand 2i-th coordinates cannot both be contained in the corresponding twointervals dening B. Therefore, q′ does not hit B.

5.4. Proof of Theorem 5.6

The goal of this section is to establish lower bounds on the sizes of ε-nets inrange spaces with given shallow-cell complexity ϕ. Theorem 5.6 is a conse-quence of the following more precise statement.

Theorem 5.9. Let ϕ : R+ → R+ be a monotonically increasing submulti-plicative function3, which tends to innity and is bounded from above by apolynomial of constant degree.

For any 0 < δ < 110 one can nd an ε0 > 0 with the following property: for

any 0 < ε < ε0, there exists a range space on a set of n =logϕ( 1

ε )

ε elementswith shallow-cell complexity ϕ, in which the size of every ε-net is at least(1−4δ)

ε logϕ(1ε ).

Proof. The parameters of the range space are as follows:

n =logϕ(1

ε )

ε, m = εn = logϕ

(1

ε

), p =

nϕ1−2δ(n)(nm

)Let d be the smallest integer such that ϕ(n) = O(nd). In fact, we will assumethat nd−1 ≤ ϕ(n) ≤ c1n

d, for a suitable constant c1 ≥ 1, provided that nis large enough. In the most interesting case, when ϕ(n) = o(n), we haved = 1. Using that n ≥ logϕ(1/ε)

ε , if ε < ε0, we have the following logarithmicupper bound on m.

m = logϕ(1

ε

)≤ log

(c1ε−d) ≤ d log

c1

ε≤ d log n (5.1)

Consider a range space ([n],F) with a ground set [n] and with a systemof m-element subsets F , where each m-element subset of [n] is added toF independently with probability p. The next claim follows by a routineapplication of the Cherno bound.

Claim 5.10. With high probability, |F| ≤ 2nϕ1−2δ(n).

Theorem 5.9 follows by combining the next two lemmas that show that,with high probability, the range space ([n],F)

3Compare with Denition 5.5.

190

(i) does not admit an ε-net of size less than (1−4δ)ε logϕ(1

ε ), and(ii) has shallow-cell complexity ϕ.For the proofs, we need to assume that n = n(δ, d, ϕ) is a suciently large

constant, or, equivalently, that ε0 = ε0(δ, d) is suciently small.

Lemma 5.11. With high probability, the range space ([n],F) has shallow-cell complexity ϕ.

Proof. It is enough to show that for all suciently large x ≥ x0, every X ⊆[n], |X| = x, the number of sets of size exactly l in F|X is O(xϕ(x)), as thisimplies that the number of sets in F|X of size at most l is O(xϕ(x)l). In thecomputations below, we will also assume that l ≥ d + 1 ≥ 2; otherwise ifl ≤ d, and assuming x ≥ x0 ≥ 2d, we have(

x

l

)≤(x

d

)≤ xd ≤ xϕ(x)

where the last inequality follows by the assumption on ϕ(x), provided thatx is suciently large. We distinguish two cases.

Case 1: x > nϕδ/d(x)

. In this case, we trivially upper-bound |F|X | by |F|.By Claim 5.10, with high probability, we have

|F| ≤ 2n · ϕ1−2δ(n) ≤ 2n ·(ϕ(x) · ϕ

(nx

))1−2δ

( by the submultiplicativity of ϕ )

≤ 2n ·(ϕ(x) · ϕ

(ϕδ/d(x)

))1−2δ (as n/x ≤ ϕδ/d(x)

)≤ 2n ·

(c1ϕ(x)ϕδ(x)

)1−2δ (using ϕ(t) ≤ c1t

d)

≤ 2c′1nϕ(x)1−δ ≤ 2c′1xϕ(x)1−δ+δ/d = O(xϕ(x)).

Case 2: x ≤ nϕδ/d(x)

. Denote the largest integer x that satises this inequal-ity by x1. It is clear that x1 = o(n) (recall that ϕ is monotonically increasingand tends to innity). We also denote the system of all l-element subsetsof F|X by F|lX and the set of all l-element subsets of X by

(Xl

). Let E

be the event that F does not have the required ϕ(·)-shallow-cell complexityproperty. Then Pr[E] ≤

∑ml=2 Pr[El], where El is the event that for some

X ⊂ [n], |X| = x, there are more than xϕ(x) elements in F|lX . Then, for

191

any xed l ≥ d+ 1 ≥ 2, we have

Pr[El] ≤x1∑

x=x0

Pr[∃X ⊆ [n], |X| = x, |F|lX | > xϕ(x)

]

≤x1∑

x=x0

(n

x

) (xl)∑s=dxϕ(x)e

Pr[For xed X, |X| = x, |S ∈ F|X , |S| = l| = s

]

≤x1∑

x=x0

(n

x

) (xl)∑s=dxϕ(x)e

((xl

)s

)Pr[For xed X, |X| = x,S ⊆

(X

l

),

|S| = s, we have F|lX = S]

≤x1∑

x=x0

(n

x

) (xl)∑s=dxϕ(x)e

((xl

)s

)(1− (1− p)(

n−xm−l))s

(1− p)(n−xm−l)((

xl)−s)

(5.2)

≤x1∑

x=x0

(xl)∑s=dxϕ(x)e

(enx

)x(e (exl )ls

)s (p

(n− xm− l

))s(5.3)

≤x1∑

x=x0

(xl)∑s=dxϕ(x)e

(enx

)x(el+1xl−1

llϕ(x)p

(n

m

)ml

(n− x−m)l

)s(5.4)

≤x1∑

x=x0

(xl)∑s=dxϕ(x)e

(enx

)x((emxn

)l−1 e2mϕ1−2δ(n)

ϕ(x)

)s(5.5)

In the transition to the expression (5.3), we used several times (i) the bound(ab

)≤(eab

)bfor any a, b ∈ N; (ii) the inequality (1 − p)b ≥ 1 − bp for any

integer b ≥ 1 and real 0 ≤ p ≤ 1; and (iii) we upper-bounded the last factorof (5.2) by 1.

In the transition from (5.3) to (5.4) we lower-bounded s by xϕ(x). Wealso used the estimate

(n−xm−l)≤(nm

)ml

(n−x−m)l, which can be veried as follows.(

n− xm− l

)=

(n− xm

) l−1∏i=0

m− in− x−m+ (i+ 1)

≤(n− xm

)(m

n− x−m

)l≤(n

m

)ml

(n− x−m)l.

192

Finally, to obtain (5.5), we substituted the formula for p and used the factthat

ll(n− x−m)l =(l · (n− x−m)

)l ≥ (l · n2

)l ≥ nl,

as x ≤ x1 = o(n), m = εn ≤ n/4 for ε < ε0 ≤ 1/4 and l ≥ 2.Denote x2 = dn1−δe. We split the expression (5.5) into two sums Σ1 and

Σ2. Let

Σ1 :=

x2−1∑x=x0

(xl)∑s=dxϕ(x)e

(enx

)x((emxn

)l−1 e2mϕ1−2δ(n)

ϕ(x)

)s

Σ2 :=

x1∑x=x2

(xl)∑s=dxϕ(x)e

(enx

)x((emxn

)l−1 e2mϕ1−2δ(n)

ϕ(x)

)sThese two sums will be bounded separately. We have

Σ1 ≤x2−1∑x=x0

(xl)∑s=dxϕ(x)e

(enx

)x((emxn

)l−1 c1−2δ1 e2mnd−2dδ

xd−2dδϕ2δ(x)

)s(5.6)

≤x2−1∑x=x0

(xl)∑s=dxϕ(x)e

(enx

)x((emxn

)l−1−d+2dδ

Cmd+1−2dδ

)s(for some C > 0)

≤x2−1∑x=x0

(xl)∑s=dxϕ(x)e

(enx

)x((n−δ/2

)l−1−d+2dδ

Cmd+1

)s(5.7)

≤x2−1∑x=x0

xl(enx

)x (n−

δ2 ·2dδn

δ2

2

)xϕ(x)

≤x2−1∑x=x0

xl(enx

)xn−

xϕ(x)dδ2

2 (5.8)

≤x2−1∑x=x0

n2x−xϕ(x)dδ2

2 ≤x2−1∑x=x0

n−2x ≤ n

n2x0= o( 1

m

). (5.9)

To obtain (5.6), we used the property that ϕ(n) ≤ ϕ(x)ϕ(n/x) ≤ c1ϕ(x)(n/x)d,provided that n, x, n/x are suciently large. To establish (5.7), we used thefact that x ≤ x2 = n1−δ and that em ≤ ed log n ≤ nδ/2 (this follows from(5.1). In the transition to (5.8), we needed that l ≥ d + 1, d ≥ 1 and thatCmd+1 ≤ C(d log n)d+1 = o(nδ

2/2), by (5.1). Then we lower-bounded s byxϕ(x). To arrive at (5.9), we used that l ≤ x. The last inequality followsfrom the facts that x0 is large enough, so that ϕ(x) ≥ ϕ(x0) ≥ 8/(dδ2) andthat m = o(n).

193

Next, we turn to bounding Σ2. First observe that

ϕ1−2δ(n) ≤ ϕ1−2δ1−δ (n1−δ) ≤ ϕ

1−2δ1−δ (x) ≤ ϕ1−δ(x),

where we used the submultiplicativity and monotonicity of the function ϕ(n)and the fact that x ≥ x2 = n1−δ. Substituting the bound for ϕ1−2δ(n) in Σ2

and putting C = e2m, we obtain

Σ2 ≤x1∑

x=x2

(xl)∑s=dxϕ(x)e

(enx

)x((emxn

)l−1

Cϕ−δ(x)

)s≤

x1∑x=x2

xl(enx

)x (emxn

Cϕ−δ(x))xϕ(x)

(5.10)

≤x1∑

x=x2

(nx

)x−xϕ(x) (e1+x/(xϕ(x))mxl/(xϕ(x))Cϕ−δ(x)

)xϕ(x)

≤x1∑

x=x2

(nx

)x−xϕ(x) (C ′ϕ−δ/2(x)

)xϕ(x)

(for some constant C ′ > 0)

(5.11)

≤n(n

x1

)x2−x2ϕ(x2) (Cϕ−δ/2(x2)

)x2ϕ(x2)

≤(n

x1

)x2−x2ϕ(x2)

(5.12)

=(x1

n

)x2ϕ(x2)−x2= o(1/m).

In the transition to (5.10), we used that emx ≤ em2 ≤ ed2 log2 n < nand l ≥ 2. To get (5.11), we used that for some constant c > 1 we havexl/(xϕ(x)) ≤ cm/ϕ(x) ≤ clogϕ(x)/ϕ(x) = O(1) and that m ≤ ϕδ/2(x) for x ≥ x0.To obtain (5.12), we noticed that n1/(x2ϕ(x2)) = O(1). At the last equation, weused that x1 = o(n), ne/x1 →∞ as n→∞ and x2ϕ(x2)− x2 = Ω(n1−δ/2).

We have shown that for every l = 2, . . . ,m, Pr[El] = o(1/m).We concludethat Pr[E] ≤

∑ml=2 Pr[El] = o(1) and, hence, with high probability, the range

space ([n],F) has shallow-cell complexity ϕ.

Now we are in a position to prove that with high probability, the rangespace ([n],F) does not admit a small ε-net.

Lemma 5.12. With high probability, the size of any ε-net of the range space([n],F) is at least (1−4δ)

ε logϕ(1ε ).

Proof. Assume without loss of generality that δ < 1/10. Denote by µ theprobability that the range space has an ε-net of size t = (1−4δ)1

ε logϕ(1ε ) =

194

(1− 4δ)n. Then

µ ≤∑X⊆[n]|X|=t

Pr[X is an ε-net for F

]≤(n

t

)(1− p)(

n−tm ) ≤

(n

t

)e−p(

n−tm )

(5.13)

≤(ent

)te−nϕ

δ(n) ≤5ne−nϕδ(n) = o(1). (5.14)

Here, the crucial transition from (5.13) to (5.14) uses the inequality below.Since 1− ax > e−bx for b > a, 0 < x < 1/a− 1/b, we obtain that

p

(n− tm

)≥ p

(n

m

)(n−m− tn− t

)t≥ nϕ1−2δ(n)

(1− m

n− t

)t≥ nϕ1−2δ(n)

(1− (1 + δ/2)m

n

)t≥ nϕ1−2δ(n)e−

(1+δ)mtn

≥ nϕ1−2δ(n)e−(1−3δ) logϕ( 1ε ) ≥ nϕ1−2δ(n)ϕ−1+3δ

(1

ε

)≥ nϕδ(n).

Thus, Lemma 5.11 and Lemma 5.12 imply that with high probability therange space ([n],F) has shallow-cell complexity ϕ and it admits no ε-net ofsize less than (1−4δ)1

ε logϕ(1ε ). This completes the proof of the theorem.

195

Chapter 6.

Applications to distance problems

My results presented in this chapter (some of them joint with N. Alon andA.M. Raigorodskii) are published in [10, 11, 161, 162, 170].

6.1. Introduction and statement of results

6.1.1. Borsuk's problem

First we discuss the classical Borsuk partition problem. In 1933 K. Borsuk(see [41]) posed the following question: is it true that any set Ω ⊂ Rd havingdiameter 1 can be divided into some parts Ω1, . . . ,Ωd+1 whose diameters arestrictly smaller than 1? Here by the diameter of a set Ω we mean the quantity

diam Ω := supx,y∈Ω

|x− y|,

where, in turn, |x − y| denotes the standard Euclidean distance betweenvectors.

Let us use some additional notation. In this chapter, by f(Ω) we denotethe value

f(Ω) := minf : Ω = Ω1 ∪ . . . ∪ Ωf , ∀ i diam Ωi < diam Ω.

Furthermore,f(d) := max

Ω⊂Rd, diam Ω=1f(Ω),

i.e., f(d) is the minimum number of parts of smaller diameter, into which anarbitrary set of diameter 1 in Rd can be divided. In these terms, Borsuk'squestion is as follows: is it true that always f(d) = d + 1? The positiveanswer on this question is usually called Borsuk's conjecture.

The history of Borsuk's conjecture is very interesting. For a while most ofthe specialists in the eld of combinatorial geometry believed that the con-jecture should be true. Multiple results supporting it have been proved. For

196

example, if Ω has smooth boundary, then one certainly obtains f(Ω) ≤ d+1.This result is due to H. Hadwiger (see [124]), and it seems to show the evi-dence of the conjecture. However, the FranklWilson theorem (Theorem 2.1)has shifted the opinion: its application to the problem of determining thechromatic number of the space seemed very relevant to Borsuk's problem andsuggested that the conjecture may be false. Finally, in 1993 J. Kahn and G.Kalai [134] published a breakthrough paper, where, using the FranklWilsontheorem, they constructed a nite set of points in a very high dimension dthat could not be decomposed into d+ 1 subsets of smaller diameter.

Now, Borsuk's conjecture is shown to be true for d ≤ 3 and false ford ≥ 64, after a recent breakthrough due to A. Bondarenko (see [37, 132]).Also, we know that

(1.2255...+ o(1))√d ≤ f(d) ≤ (1.224...+ o(1))d.

Here the lower bound was found in 1999 by A.M. Raigorodskii (see [208]),who used the analogue of the FranklWilson theorem for 0,±1-vectors (cf.also Chapter 2), and the upper estimate is due to O. Schramm (see [224]).

The colorful history of Borsuk's conjecture is exhibited in numerous booksand survey papers. We refer the reader to [213], [123], [36], [43], [214], [212].

A careful analysis of all the known counterexamples to Borsuk's conjectureshows that they are always nite sets of points in Rd lying on spheres whoseradii are close to 1√

2. This is quite natural, since, by Jung's theorem (see

[133]), any set in Rd having diameter 1 can be covered by a ball of radius√d

2d+2 ∼1√2, and the intuition is that in order to get a counterexample, we

have to take a set with as big covering ball as possible. The main result inthis section is the following theorem which completely breaks such intuition.

Theorem 6.1 ([170]). Let Sd−1r ⊂ Rd be the sphere of radius r with center

at the origin. For any r > 12, there exists a d0 = d0(r) such that for every

d ≥ d0, one can nd a set Ω ⊂ Sd−1r which has diameter 1 and does not

admit a partition into d+ 1 parts of smaller diameter.

Actually, we shall prove stronger results. For details, see Section 6.2.

6.1.2. Distance graphs with high girth

We say that G = (V,E) is an distance graph in Rn, if V is a subset of Rn

and

E ⊆ x,y : x,y ∈ V, |x− y| = 1.

197

Such graphs arise naturally in the context of the problem of nding thechromatic number of the space. This famous question was posed by Nelsonin 1950: what is the minimum number χ(R2) of colors needed to color allpoints of the plane so that no two points at distance one receive the samecolor? Although this question does not sound too dicult, we do not knowthe answer yet. The best known upper and lower bounds are 5 ≤ χ(R2) ≤ 7,where the (rst nontrivial) lower bound was obtained recently by A. de Grey[121]. One may ask the same question for higher-dimensional spaces. Hereis the formal denition (see [24]):

χ(Rn) := minm ∈ N : Rn = H1 ∪ . . . ∪Hm : ∀i,∀x,y ∈ Hi |x− y| 6= 1.

This quantity was studied quite extensively (see the surveys [209], [229]and also [160]), e.g. there are nontrivial lower bounds for the value of χ(Rn),n ≤ 24. We will be interested in the behavior of χ(Rn) as n → ∞. Thefollowing asymptotic lower and upper bounds are due to A. Raigorodskii[209] and D. Larman, C. Rogers [177] respectively:

(ζlow + o(1))n ≤ χ(Rn) ≤ (3 + o(1))n, where ζlow = 1.239 . . .

The chromatic number of the space Rn is naturally related to the chro-matic numbers distance graphs in Rn. On the one hand, it immediately fol-lows from the denitions that for any such distance graph G, χ(G) ≤ χ(Rn).On the other hand, N.G. de Bruijn and P. Erdos [44] showed, using a stan-dard compactness argument, that there exists a distance graph G′ in Rn withnite number of vertices such that χ(G′) = χ(Rn).

The motivation for the studies presented in this section comes from thefollowing classical question. Can we construct graphs with arbitrarily largechromatic number and arbitrary girth (the length of the shortest cycle)?The positive answer to this question was given by P. Erdos [68]. He provedthat such graphs exist, although the proof was probabilistic, so there wasno explicit construction. Later, L. Lovasz [180] managed to give an explicitconstruction of such graphs.

It is natural to ask how large can the chromatic number of a distancegraph be if we additionally require that the graph has no cliques (completesubgraphs) or cycles of xed size. The question, whether there is a distancegraph in the plane with chromatic number 4 and without triangles (which areboth cliques of size 3 and cycles of length 3) was asked by P. Erdos [71]. Itwas answered positively. Moreover, P. O'Donnell ([63], [64]), in an attemptto prove that χ(R2) ≥ 5, proved that for any l ∈ N there exists a distancegraph in the plane with chromatic number 4 and girth greater than l.

198

We consider the following three families of distance graphs in Rn: C(n, k)is the family of all distance graphs that do not contain complete subgraphsof size k; Godd(n, k) is the family of all distance graphs that do not containodd cycles of length ≤ k; G(n, k) is the family of all distance graphs that donot contain cycles of length ≤ k. We obviously have the following inclusion:G(n, l) ⊂ Godd(n, l) ⊂ C(n, k) ⊂ C(n, k′), where l ≥ 3 and k′ ≥ k ≥ 3.

Let us dene the following quantities:

ζk := lim infn→∞

maxG∈C(n,k)

(χ(G))1/n,

ξoddk := lim infn→∞

maxG∈Godd(n,k)

(χ(G))1/n,

ξk := lim infn→∞

maxG∈G(n,k)

(χ(G))1/n.

For example, the bound ζk ≥ 1.1 means that there exists a sequence ofdistance graphs Gn ⊂ Rn, such that χ(Gn) ≥ (1.1 + o(1))n and none of Gi

contains a clique of size k.The values ζk and ξoddk were considered in several papers (see, e.g., [210]).

The most accurate estimates on ζk are due to myself [162], although I do notstate these results here (see also [57], where both ξoddk and ζk were considered).

There are two approaches to lower bound the quantity ζk. The rst ap-proach is probabilistic, and thus it does not give an explicit graph. One ad-vantage of this approach is that one can see that ζk ≥ ck, where limk→∞ ck =ζlow. In [57] this method gave nontrivial bounds only for k ≥ 5. A renementof this method suggested in [162] works for k ≥ 3.

The second approach is of code-theoretic avour. It provides us withexplicit constructions of such graphs and it works for k ≥ 3. Moreover, itgives much better bounds for small k. But as k grows, this method becomesworse than the probabilistic method. In particular, the bounds tend to someconstant that is signicantly smaller than ζlow.

A way to obtain bounds on ξoddk , k ≥ 5, is also code-theoretic. In [57] itwas proved that for any xed k we have ξoddk > 1. Bounds on both ζk andξoddk that are slightly weaker than the code-theoretic ones can be derived fromsimple geometric observations (see [162]).

We also note that both approaches rely on extremal set theoretic con-structions and on FranklWilson type theorems in particular.

Thus, there were non-trivial lower bounds on ζk, ξoddk . However, both pre-viously known probabilistic and code-theoretic approaches failed to provide

199

any estimate for ξk, k ≥ 4. The main result of this section is the followingtheorem.

Theorem 6.2 ([161]). For any xed k ≥ 3 we have ξk ≥ 1 + δ, whereδ = δ(k) is a positive constant that depends only on k.

This theorem is proved in Section 6.3.We say that a graph H is a forest if it doesn't contain cycles. Consider a

nite family H = H1, . . . , Hm of graphs. Let G(n,H) be the family of alldistance graphs in Rn that do not contain any of Hi ∈ H as a subgraph. Wedene the quantity ξ(H) as above:

ξ(H) := lim infn→∞

maxG∈G(n,H)

(χ(G))1/n.

Theorem 6.2 provides us with the following appealing corollary:

Corollary 6.3. For any nite family H of graphs such that no Hi ∈ H isa forest we have ξ(H) ≥ 1 + δ, where δ = δ(H) is a positive constant thatdepends only on H.

Proof. The proof is immediate. Let li be the length of the shortest cycle inHi, where Hi ∈ H. Then ξ(H) ≥ maxi ξli ≥ 1 + δ, where δ depends only onH.

Unfortunately, Theorem 6.2 says nothing about the family of graphs, onwhich this bound can be attained. So it is natural to raise the followingproblem:

Problem 1. Prove Theorem 6.2 using an explicit construction.

The second disadvantage of the method used for the proof of Theorem 6.2is the following. The graphs that nally can be obtained are not necessarilycomplete distance graphs, i.e. in the denition of their set of edges we have thestrict inclusion (not all possible edges are drawn). Here is another question:

Problem 2. Prove Theorem 6.2 using complete distance graphs.

Unfortunately, all we can prove is the following proposition.

Proposition 6.4 ([11]). For any g ∈ N there exists a sequence of completedistance graphs in Rd, d = 1, 2, . . . , with girth greater than g such that the

chromatic number of the graphs in the sequence grows as Ω(

dlog d

).

200

6.2. Borsuk's problem

Let us introduce the quantity

fr(d) = maxΩ⊂Sd−1r , diam Ω=1

f(Ω). :

In these terms, Theorem 6.1 says that for any r > 12 , there exists a d0 = d0(r)

such that for every d ≥ d0, fr(d) > d+ 1. Moreover, one has

Theorem 6.5. For any r > 12, there exist numbers k = k(r) ∈ N, c =

c(r) > 1 and a function δ = δ(d) = o(1) such that

fr(d) ≥ (c+ δ)2k√d.

Theorem 6.5 means that if r is xed and exceeds 12 , then the order of

magnitude of the value fr(d) is at least eg(d), where g(d) is just a xed positivepower of d. So not only the quantity fr(d) is greater than d+1 starting fromsome d0, but also it is substantially greater than the conjectured value. Thisfact allows us to use some optimization and to prove eventually the followingtheorem.

Theorem 6.6. Let r = r(d) = 12 + ϕ(d), where ϕ = o(1) and ϕ(d) ≥ c ln ln d

ln d

for all d and a large enough c > 0. Then, there exists a d0 such that ford ≥ d0, fr(d)(d) > d+ 1.

In other words, we can disprove Borsuk's conjecture by constructing setsof diameter 1 that lie on spheres with radii tending to 1

2 with d→∞. Here,in the case of ϕ(d) = Θ

(ln ln dln d

), there is already no room to spare in the

bound fr(d) > d+ 1. So using our approach, one only may discuss the valueof a constant c in Theorem 6.6, which is of course not signicant.

Now, a natural question arises: perhaps fr(d) ≤ d + 1, provided r =r(d) = O

(ln ln dln d

)? Unfortunately, we only can prove

Theorem 6.7. Let r = r(d) = 12 +ϕ(d), where ϕ = O(1/d). Then, fr(d) ≤

d+ 1.

Thus, we get a gap between the two functions 1d and

ln ln dln d . To reduce this

gap would be of a great interest.

The structure of the rest of our paper will be as follows. In Section 6.2.1,we shall prove both Theorem 6.1 and Theorem 6.5. Section 6.2.2 will bedevoted to the proof of Theorem 6.6. In Section 6.2.3, we shall briey discussTheorem 6.7 which is in fact very simple. In Section 6.2.4, we shall showthat in some sense, the constructions from Sections 6.2.1 and 6.2.2 are bestpossible.

201

6.2.1. Proofs of Theorems 6.1 and 6.5

Since Theorem 6.1 is an immediate consequence of Theorem 6.5, we justprove the second result.

Let us x an arbitrary r > 12 . Without loss of generality, we also assume

that r < 1√2. Let d be large enough (during the proof, we will see what it

means). To make clear our further exposition, we subdivide it into four parts.So in Section 6.2.1.1, we construct a d′-dimensional set Ω′ with d′ < d; inSection 6.2.1.2, we show that Ω′ can be transformed into some Ω ⊂ Sd−1

r ; inSection 6.2.1.3, we prove that fr(d) ≥ f(Ω) ≥ f(Ω′) ≥ (c+ δ)

2k√d with some

appropriate c > 1 and δ = o(1); in Section 6.2.1.4, we prove a key lemmawhich is formulated in Section 6.2.1.3.

6.2.1.1. Construction

Take

k := min

k′ ∈ N : r2 >

2k′ + 1

8k′

.

This value is correctly dened, since r > 12 and 2k′+1

8k′ →14 as k′ →∞. Put

n := maxm : m ≡ 0 (mod 4), m2k < d

.

Clearly2k√d− 5 ≤ n <

2k√d. (6.1)

Consider the function

u(a0) :=1 + 2k

(a02

)2k−1

2 + 4k(a02

)2k−1+ (4k − 2)

(a02

)2k. (6.2)

Obviously,

u(0) =1

2, u(2) =

2k + 1

8k< r2 <

1

2,

and u is monotone decreasing on the interval (0, 2). Therefore, the equation

1 + 2k(a02

)2k−1

2 + 4k(a02

)2k−1+ (4k − 2)

(a02

)2k= r2

has a unique solution a0 ∈ (0, 2).Take

a := min

a′ : a′ ≥ a0n

2,a′

4+n

4is a prime number

.

202

It is known (see [18]) that between x and x+O(x0.525

), there is certainly a

prime number. Thus, the quantity

p =a

4+n

4(6.3)

is asymptotically equal to p0n, where

p0 :=a0

8+

1

4∈(

1

4,1

2

).

Consider the set

Σ := x = (x1, . . . , xn) : ∀ i xi ∈ −1, 1, x1 = 1, x1 + . . .+ xn = 0.

Let A := a1, . . . , aw be the set of all possible 2k-character words over thealphabetX := 1, . . . , n, w := n2k. Assume that ai = ν(i, 1), . . . , ν(i, 2k),i = 1, . . . , w. Fix an x = (x1, . . . , xn) ∈ Σ. Consider

x∗2k :=(xν(1,1) · . . . · xν(1,2k), . . . , xν(w,1) · . . . · xν(w,2k),

√2ka2k−1x1, . . . ,

√2ka2k−1xn

).

Clearly the number of coordinates in any vector x∗2k equals w + n. Put

Ω′ :=x∗2k : x ∈ Σ

.

One can readily see that Ω′ lies in Rd′ with d′ = w < d. The point is thatfor any i, xi = x2k−1

1 · xi, since x1 = 1. The construction is complete.

6.2.1.2. Transforming Ω′ into an Ω ⊂ Sd−1r

First, let us calculate the diameter of Ω′. For the scalar product of any twovectors x∗2k,y∗2k ∈ Ω′, we have the relation(x∗2k,y∗2k

)=

w∑i=1

xν(i,1) · . . . · xν(i,2k) · yν(i,1) · . . . · yν(i,2k) + 2ka2k−1(x,y) =

=n∑

i1=1

. . .n∑

i2k=1

xi1 · . . . · xi2k · yi1 · . . . · yi2k + 2ka2k−1(x,y) =

=

(n∑

i1=1

xi1yi1

)· . . . ·

(n∑

i2k=1

xi2kyi2k

)+ 2ka2k−1(x,y) =

= (x,y)2k + 2ka2k−1(x,y).

Obviously the minimum of the form(x∗2k,y∗2k

)is attained on those and

only those pairs of vectors x,y ∈ Σ whose scalar product equals −a. Such

203

pairs of vectors do really exist for large enough values of d. Indeed, byConstruction, a ∼ a0n

2 and a0 < 2. So for large n, −a > −n. Moreover,a ≡ 0 (mod 4), and it is easy to see that for every two vectors x,y ∈ Σ, onenecessarily has (x,y) ∈ (−n, n] and (x,y) ≡ 0 (mod 4).

Thus, we get

diam2 Ω′ = 2(x∗2k,x∗2k

)− 2(a2k − 2ka2k) = 2n2k + 4ka2k−1n+ (4k− 2)a2k.

At the same time, Ω′ lies of course on the sphere Sd′−1ρ , where

ρ2 =(x∗2k,x∗2k

)= n2k + 2ka2k−1n.

Compressing Ω′ so that a new set Ω′′ has diameter 1, we see that Ω′′ ⊂Sd′−1r′ with

(r′)2 =n2k + 2ka2k−1n

2n2k + 4ka2k−1n+ (4k − 2)a2k.

Since, again by Construction, a ≥ a0n2 and u dened in (6.2) is monotone

decreasing, we get the inequalities

(r′)2 =n2k + 2ka2k−1n

2n2k + 4ka2k−1n+ (4k − 2)a2k

≤n2k + 2k

(a0n2

)2k−1n

2n2k + 4k(a0n2

)2k−1n+ (4k − 2)

(a0n2

)2k= u(a0) = r2.

Now, it remains to interpret Sd′−1r′ as an intersection of the sphere Sd−1

r

and a plane of dimension d′. This can be done, since d′ < d and r′ ≤ r. LetΩ be an image of Ω′′ under such interpretation.

6.2.1.3. Lower bound for f(Ω)

It is clear thatfr(d) ≥ f(Ω) = f(Ω′′) = f(Ω′).

So it remains to show that f(Ω′) ≥ (c+ δ)2k√d for appropriate c and δ. First,

assume that

f(Ω′) < f =|Ω′|

p−1∑i=0

C in

.

Then, Ω′ can be represented as

Ω′ = Ω′1 ∪ . . . ∪ Ω′f , ∀ i diam Ω′i < diam Ω′. (6.4)

204

Obviously the correspondence x→ x∗2k is a bijection between Σ and Ω′. So

|Ω′| = |Σ| = Cn2−1n−1

and, moreover, the partition (6.4) induces a partition

Σ = Σ1 ∪ . . . ∪ Σf .

By the choice of f and by pigeon-hole principle, there exists a part Σi with

|Σi| >p−1∑i=0

C in. In the next Section, we shall prove the following lemma.

Lemma 6.8. If Q ⊂ Σ is such that |Q| >p−1∑i=0

C in, then there exist x,y ∈ Q

with (x,y) = −a.

Of course Lemma 6.8 is applied only for n 1. By this lemma, Σi

contains two dierent vectors x,y with scalar product equal to −a. It means(cf. Section 6.2.1.2) that

diam Ω′i = |x∗2k − y∗2k| = diam Ω′,

which contradicts the properties of partition (6.4).Thus, we have shown that

f(Ω′) ≥ |Ω′|p−1∑i=0

C in

=C

n2−1n−1

p−1∑i=0

C in

.

Let us recall that p ∼ p0n, p0 ∈(

14 ,

12

). In this case, standard analytical

tools like Stirling's formula and Cherno's inequality entail the followingasymptotic relations:

Cn2−1n−1 = (2 + o(1))n,

p−1∑i=0

C in = (c′ + o(1))n, c′ < 2.

Taking the necessary ratio, we get

f(Ω′) ≥ (c+ δ′)n, c > 1, δ′ = o(1).

Finally, by inequalities (6.1) we have

fr(d) ≥ f(Ω′) ≥ (c+ δ′)n ≥ (c+ δ′)2k√d−5 = (c+ δ)

2k√d, δ = o(1).

Theorem 6.5 is proved.

205

6.2.1.4. Proof of Lemma 6.8

We shall use the now classical linear algebra method in combinatorics (see[6], [17], [211]).Consider an arbitrary Q := x1, . . . ,xs ⊂ Σ such that for every two dier-

ent i, j, one has (xi,xj) 6= −a. We need to show that s ≤p−1∑i=0

C in.

To each vector x ∈ Σ we assign a polynomial Px ∈ Z/pZ[y1, . . . , yn].Namely,

Px(y) =

p−1∏i=0,i 6≡−a (mod p)

(i− (x,y)), y = (y1, . . . , yn).

Here the product is taken over all the smallest non-negative residues imodulop except for i ≡ −a (mod p). So the degree of any Px does not exceed p−1.The most important property of such polynomials is as follows.

Property. For every x,y ∈ Σ, the congruence (x,y) ≡ −a (mod p) isequivalent to the congruence Px(y) 6≡ 0 (mod p).

Property is evident, and we shall just use it. However, before doing so, wemake a transformation of any Px into some P ′x. More precisely, we representevery polynomial Px as a linear combination of monomials. Of course eachmonomial has the form

yα1

i1· . . . · yαqiq , q ≤ p− 1.

If αν is even, we remove yiν from the monomial. If it is odd, we replace it by1. Clearly the new monomial is just a product of some variables. However,the new polynomials P ′x, x ∈ Σ, still are subject to Property. The point isthat in Property, only variables whose values are ±1 are considered.

By the just-given construction,

dim(linear span P ′xx∈Σ

)≤

p−1∑i=0

C in.

If we succeed now in showing that the vectors x1, . . . ,xs from the set Qcorrespond to the set P ′x1

, . . . , P ′xs of linearly independent polynomials (overthe eld Z/pZ), then Lemma 6.8 is proved.

So let us assume that

c1P′x1

(y) + . . .+ csP′xs

(y) ≡ 0 (mod p) ∀ y ∈ Σ. (6.5)

206

Take y = xi with an arbitrary i. On the one hand, (xi,xi) = n. ByConstruction, n − 4p = −a (see (6.3)). Therefore, (xi,xi) ≡ −a (mod p),i.e., by Property, P ′xi(xi) 6≡ 0 (mod p). On the other hand, if j 6= i, then(xi,xj) < n and (xi,xj) 6= −a. Moreover, since

a ≡ 0 (mod 4), n ≡ 0 (mod 4), (x,y) ≡ 0 (mod 4) ∀ x,y ∈ Σ,

we see that

(xi,xj) 6∈ n− p, n− 2p, n− 3p, n− 5p, n− 6p, n− 7p.

Finally, −a < 0, and so n−8p < −n, which means that (xi,xj) 6≡ a (mod p)and that, by Property, P ′xj(xi) ≡ 0 (mod p).

By relation (6.5), we get ci ≡ 0 (mod p) (here the primality of p is es-sential). Since i was arbitrary, we obtain the linear independence of ourpolynomials, and Lemma 6.8 is proved.

6.2.2. Proof of Theorem 6.6

Take r := 12 +ϕ, where ϕ(d) := 6 ln ln d

ln d . We shall prove that for large enoughd, fr(d) > d+ 2. Then, the whole assertion of Theorem 6.6 will follow, sincefor any r′ > r, Sdr′ ⊃ Sd−1

r .Of course, further exposition will be very close to the one in Section 6.2.1.

However, there will be some technical subtleties. First of all, we rewriteSection 6.2.1.1.

Take k := d1/ϕe. Then, for large values of d,

2k + 1

8k=

1

4+

1

8k≤ 1

4+ϕ

8<

1

4+ ϕ+ ϕ2 = r2.

In other similar inequalities below, we shall not write for large values of danymore; we shall just assume that d is big enough. As in Section 6.2.1.1,put

n := maxm : m ≡ 0 (mod 4), m2k < d

.

Now, inequality (6.1) means that

n ∼ dϕ2 = e(ln d)· 3 ln ln d

ln d = ln3 d.

Let a0 = 2− ϕ2 . We want to show that u(a0) < r2, provided u is the same

function as in (6.2). A standard computation is below. First,

u(a0) =1 + 2k

(a02

)2k−12 + 4k

(a02

)2k−1+ (4k − 2)

(a02

)2k < 2k(a0/2)2k−1

(8k − 2)(a0/2)2k·

1 + 12k(a0/2)2k−1

1 + 2(8k−2)(a0/2)2k

.

207

Further,2k

8k − 2=

1

4· 1

1− 14k

.

Since 11−x ≤ 1 + 2x, we have

2k

8k − 2≤ 1

4·(

1 +ϕ

2

).

Now,(a0/2)2k−1

(a0/2)2k=a0

2= 1− ϕ

4.

Finally,

1 + 12k(a0/2)2k−1

1 + 2(8k−2)(a0/2)2k

≤(

1 +1

2k(a0/2)2k−1

)·(

1− 1

(8k − 2)(a0/2)2k

). (6.6)

Clearly (a0

2

)2k

∼(a0

2

)2k−1

∼(

1− ϕ

4

) 2ϕ ∼ e−1/2.

Consequently, the right-hand side of (6.6) can be bounded from above by(1 +

1.1ϕ

2e−1/2

)·(

1− 0.9ϕ

8e−1/2

)≤ 1 +

3.6ϕ

8e−1/2.

Thus,

u(a0) ≤1

4·(

1 +ϕ

2

)·(

1− ϕ

4

)·(

1 +3.6ϕ

8e−1/2

)≤ 1

4·(

1 +ϕ

4

)·(

1 +3.6ϕ

8e−1/2

)≤ 1

4+ϕ

4<

1

4+ ϕ+ ϕ2 = r2.

Again, as in Construction, we take

a := min

a′ : a′ ≥ a0n

2,a′

4+n

4is a prime number

.

By the already cited results from [18], a < n and

p =a

4+n

4≤ n

2− ϕn

16+O

(n0.525

).

On the one hand,ϕn = Θ

((ln ln d)(ln d)2

).

208

On the other hand,

n0.525 = (ln d)3·0.525 = (ln d)1.575 = o((ln ln d)(ln d)2

).

Hence,p ≤ n

2− ϕn

20.

The sets Σ and Ω′ are just the same as in Section 6.2.1.1. The only changein Section 6.2.1.2, is in replacing the last equality by the inequality, in theestimate of (r′)2 by r2.

Reproducing Section 6.2.1.3 word for word, we obtain the bound

fr(d) ≥C

n2−1n−1

p−1∑i=0

C in

=12C

n2n

p−1∑i=0

C in

.

It remains to establish the estimate

12C

n2n

p−1∑i=0

C in

> d+ 2.

Clearlyp−1∑i=0

C in < nCp

n ≤ nCn2−xn , x =

[ϕn20

].

We have

Cn2n

Cn2−xn

=

(n2 + 1

)· . . . ·

(n2 + x

)(n2

)·(n2 − 1

)· . . . ·

(n2 − x+ 1

) =

(1 + 2

n

)· . . . ·

(1 + 2x

n

)(1− 2

n

)· . . . ·

(1− 2(x−1)

n

) =

= ex(x+1)

2−x(x−1)

2+O

(x3

n2

)= e

2x2

n+O

(x3

n2

)= e(1+o(1))ϕ

2n200 = e

Θ

((ln ln d)2

(ln d)2·(ln d)3

)= eΘ((ln ln d)2·(ln d)).

Thus,12C

n2n

p−1∑i=0

C in

= dΘ((ln ln d)2) > d+ 2,

and the proof is complete.

209

6.2.3. Proof of Theorem 6.7

Take S = Sd−1r with an arbitrary r. Inscribe into S a d-dimensional regular

simplex ∆. Consider its faces ∆1, . . . ,∆d+1. If O is the origin, then denoteby Si the set

conv ∆i, O ∩ S.Of course

S = S1 ∪ . . . ∪ Sd+1.

Moreover, it is well-known that (see [36])

diamSi = 2r

(1−Θ

(1

d

)).

If r = r(d) ≤ 12 + c

d with an appropriate c > 0, then we get diamSi < 1, andwe are done.

In principle, it's possible to obtain rather good upper bounds for fr(d)in general case. The simplest way for doing that is to use old results ofC.A. Rogers (see [219]): any sphere of radius r > 1

2 in Rd can be coveredby (2r + o(1))d spherical caps of diameter 1. This result gives already theestimate fr(d) ≤ (2r + o(1))d when r is xed. More subtle estimates canbe discovered by using an approach which appeared in the paper [42]. For

example, Rogers tells us that for r >√

38 , we can only show that fr(d) ≤

(c + o(1))d with c >√

32 . However, J. Bourgain and J. Lindenstrauss, the

authors of [42], provide us with a universal bound fr(d) ≤(√

32 + o(1)

)d.

Their ideas may be carefully applied even in some cases when r → 12 .

6.2.4. Improving Construction from Section 6.2.1.1 is hard

Let us discuss some key properties of Construction from Section 6.2.1.1 anda possibility of improving it. Indeed, one of the most important steps inConstruction was in assigning, to each n-dimensional vector x ∈ Σ, a d′-dimensional vector x∗2k, so that eventually we got a set Ω′ ⊂ Rd′. Thecorrespondence between vectors x from Σ and their images x∗2k was orga-nized in such a way that(

x∗2k,y∗2k)

= (x,y)2k + 2ka2k−1(x,y).

In other words, the scalar product of any two elements of the set Ω′ wasa polynomial depending on the scalar product of the preimages of thoseelements. Denote this polynomial by h∗a. Thus, h

∗a(t) = t2k + 2ka2k−1t.

210

In order to prove Theorems 6.1 6.6, we essentially used the followingproperties of h∗a (see Section 6.2.1.2): rst of all, h∗a, considered as functionon the interval [−n, n], attains its minimum at point −a ∈ (−n, 0) and so

dh∗adt

∣∣∣∣t=−a

= 0;

second, the quantity

h∗a(n)

2h∗a(n)− 2h∗a(−a)=

n2k + 2ka2k−1n

2n2k + 4ka2k−1n+ (4k − 2)a2k

is a close approximation to the radius of a sphere, on which a counterexampleto Borsuk's conjecture lies. Therefore, the results of Theorems 6.1 6.6 wouldbe improved, provided we could replace h∗a by another polynomial ha havingthe same rst property and a smaller value of the expression

ha(n)

2ha(n)− 2ha(−a).

Moreover, we may also vary the value a ∈ (0, n) by taking a ∼ a′n witha′ ∈ (0, 1) as close to 1 as necessary.

Let ∆ma be the set of all polynomials ha(t) = umt

m + . . . + u1t + u0 ofdegree m with non-negative coecients and having the property that ha,considered as function on the interval [−n, n], attains its minimum at point−a ∈ [−n, 0) and

dhadt

∣∣∣∣t=−a

= 0

(the last condition is non-trivial only for a = n).

Proposition 6.9. If m is even, then

mina∈(0,n]

minha∈∆m

a

ha(n)

2ha(n)− 2ha(−a)= min

h∈∆mn

h(n)

2h(n)− 2h(−n)=

h∗(n)

2h∗(n)− 2h∗(−n),

where h∗(t) := tm +mnm−1t. If m is odd, then

minh∈∆m

n

h(n)

2h(n)− 2h(−n)≥ min

h∈∆m−1n

h(n)

2h(n)− 2h(−n).

Before proving Proposition 6.9, let us briey comment on it. Actually,Proposition 6.9 tells us that, in Section 6.2.1.1 and Section 6.2.1.2, everythingwas done in an optimum way: it's better to take polynomials of even degree,and, among them, h∗a is asymptotically best possible (when a → n, h∗a →

211

t2k + 2kn2k−1t). It is worth noting that, in principle, for any h ∈ ∆ma , one

can transform Σ into such an Ω′ that the scalar product of any two vectorsfrom Ω′ is equal to the value of h at the scalar product of the preimagesof those two vectors (here it is important to assume that any h ∈ ∆m

a hasnon-negative coecients). However, according to Proposition 6.9, this factis already not quite useful for our purposes.

Proof of Proposition 6.9. First, we note that the quantity ha(−a) should benon-positive in order to minimize the ratio

ha(n)

2ha(n)− 2ha(−a)=

1

2− 2ha(−a)ha(n)

, (6.7)

since ha(n) > 0 for any ha. Polynomials ha ∈ ∆ma such that ha(−a) ≤ 0

do really exist, so we may assume that ha(−a) ≤ 0. Under this assumption,our minimization is equivalent to the maximization of the expression |ha(−a)|

ha(n)

over the set ∆ma ⊂ ∆m

a containing those and only those polynomials ha, forwhich ha(−a) ≤ 0.

Furthermore, we may suppose that u0 = 0, since, for u0 > 0, ratio (6.7)is denitely greater.

Now, let us prove that

mina∈(0,n]

minha∈∆m

a

ha(n)

2ha(n)− 2ha(−a)= min

h∈∆mn

h(n)

2h(n)− 2h(−n)(6.8)

or, which is the same, that

maxa∈(0,n]

maxha∈∆m

a

|ha(−a)|ha(n)

= maxh∈∆m

n

|h(−n)|h(n)

. (6.9)

Indeed, all the coecients of ha are non-negative, and so ha(a) ≤ ha(n).Therefore, for any a ∈ (0, n], we have

maxha∈∆m

a

|ha(−a)|ha(n)

≤ maxha∈∆m

a

|ha(−a)|ha(a)

= maxh∈∆m

n

|h(−n)|h(n)

,

which completes the proof of (6.8) and (6.9).It remains to show that

maxh∈∆m

n

|h(−n)|h(n)

≤ |h∗(−n)|h∗(n)

=m− 1

m+ 1.

Take an arbitrary polynomial h ∈ ∆mn . We represent it in the form

h = h1 + h2 =∑i

cαitαi +

∑i

cβitβi, cαi > 0, cβi > 0,

212

where, in h1, only odd degrees of t are taken, and, in h2, only even degreesare present. In this notation, we have

|h(−n)|h(n)

=

∣∣∣∣∑i

cαinαi −

∑i

cβinβi

∣∣∣∣∑i

cαinαi +

∑i

cβinβi.

Hence, we are led to check the inequality∑i

cαinαi ≤ m

∑i

cβinβi. (6.10)

However, we know that h′(−n) = 0, i.e.,∑i

αicαinαi−1 =

∑i

βicβinβi−1.

Then, we get the following series of inequalities:∑i

cαinαi ≤ n

∑i

αicαinαi−1 = n

∑i

βicβinβi−1 ≤ m

∑i

cβinβi.

Thus, (6.10) is true, and the proof of the rst part of Proposition is complete.

The second part would follow from the inequality

maxh∈∆m

n

|h(−n)|h(n)

≤ m− 2

m,

which is tantamount to∑i

cαinαi ≤ (m− 1)

∑i

cβinβi. (6.11)

Eventually, (6.11) is provided by the following series of estimates:∑i

cαinαi ≤ n

∑i

αicαinαi−1 = n

∑i

βicβinβi−1 ≤ (m− 1)

∑i

cβinβi.

Here we get the factor m− 1 instead of m, since m is odd and so βi ≤ m− 1for any i.

6.3. Distance graphs of high girth

6.3.1. Preliminaries

As a basis of our construction we will take a family G := G4i : i ∈ N ofdistance graphs, where G4n = (V4n, E4n), and

V4n := x = (x1, . . . , x4n) : xi ∈ 0, 1, 〈x,x〉 = 2n,

213

E4n := x,y : 〈x,y〉 = n.Here, as usual, 〈·, ·〉 denotes the Euclidean scalar product. In the next sub-section we will prove that for any k ∈ N there exists a family of graphs H4i

such that for each i the graph H4i is a subgraph of G4i and H4i has girthgreater than k. Moreover, χ(H4i) = (c+ δ(4i))4i, where c > 1 and δ(i)→ 0as i → ∞. This is all we need to prove since in any dimension of the form4i + j, j = 1, 2, 3 we can consider a plane of codimension j and embed anisometric copy of H4i there. As a result we obtain a sequence of graphs withdesired properties in all dimensions.

It is easy to see that |V4n| =(

4n2n

)= (2 + o(1))4n and |E4n| =

(4n2n

)(2nn

)2=

(4 + o(1))4n. We will use the following result from the paper [111]:

Theorem 6.10. For any ε > 0 there exists δ > 0 such that for any subset Sof V4n, |S| ≥ (2 − δ)4n, the number of edges in S (the cardinality of E4n|S)is greater than (4− ε)4n.

Remark 3. We do not give any numerical bounds for ξk since they are verydicult to derive. The reason is that we use Theorem 6.10, in which thereis no explicit dependency between ε and δ.

We will also need Lovasz Local Lemma (see [13]):

Theorem 6.11. Let A1, . . . , Am be events in an arbitrary probability spaceand J(1), . . . , J(m) be subsets of 1, . . . ,m. Suppose there are real numbersγi such that 0 < γi < 1, i = 1, . . . ,m. Suppose the following conditions hold:

1) Ai is independent of algebra generated by Aj, j 6∈ J(i) ∪ i.

2) Pr[Ai] ≤ γi∏

j∈J(i)(1− γj).

Then Pr[⋂m

i=1Ai

]≥∏m

i=1(1− γi) > 0.

We will use the following version of local lemma (see [31]):

Lemma 6.12. Let A1, . . . , Am and J(1), . . . , J(m) be as in Theorem 6.11.Suppose there are real numbers δi such that 0 < δi Pr[Ai] < 0.69, i =1, . . . ,m. Suppose the following condition holds:

ln δi ≥∑j∈J(i)

2δj Pr[Aj]. (6.12)

Then Pr[⋂m

i=1Ai

]≥∏m

i=1(1− δi Pr[Ai]) > 0.

214

Proof. This form of local lemma is easy to derive from Theorem 6.11. Wejust need to verify that the inequality 2 from Theorem 6.11 follows from theinequality (6.12). Indeed, we have the following inequality:

ln δi ≥∑j∈J(i)

2δj Pr[Aj] ≥∑j∈J(i)

− ln(1− δj Pr[Aj]),

since ln(1 − t) ≥ −t − t2 ≥ −2t for 0 < t < 0.69 (see [31]). We take anexponent of both sides:

δi ≥∏j∈J(i)

(1− δj Pr[Aj])−1.

Finally, we substitute δi = γi/Pr[Ai].

6.3.2. Proof of Theorem 6.2

Fix natural numbers k ≥ 3 and n. Let γ ∈ (0, 1) be a constant that will bedened later, and set p = γ4n. Consider a random subgraph G of the graphG4n in which all edges are chosen independently and uniformly with theprobability of each edge to occur equal to p. Namely, we have the probabilityspace (Ω4n,B4n, P4n), where

Ω4n := G = (V4n, E), E ⊆ E4n, B4n := 2Ω4n, P4n(G) := p|E|(1−p)|E4n|−|E|

for G = (V4n, E).Denote N := |V4n|. We dene two families of events on Ω4n. Firstly, for

some l we enumerate all l-element subsets of V4n and introduce the events

Xi := ith l-element subset is independent, i = 1, . . . , C lN .

Secondly, for each s = 3, . . . , k we enumerate all (labeled) cycles of lengths in G4n and introduce the events

Y sj := jth s-tuple is an s-cycle, j = 1, . . . , cs(G4n),

where cs(G4n) is the number of labeled s-cycles in G4n.Take l = (2−δ)4n, where δ > 0 is again some constant that will be dened

later. The statement of Theorem 6.2 will follow from the inequality

Pr

ClN⋂i=1

Xi ∩k⋂s=3

cs(G4n)⋂j=1

Y sj

> 0. (6.13)

215

Indeed, we obtain from (6.13) that there exists a subgraph G′ in G4n suchthat it does not contain cycles of length ≤ k and at the same time α(G′) ≤ l.The above means that G′ ∈ G(4n, k) and

χ(G′) ≥ N

l=

(2

2− δ+ o(1)

)4n

= (1 + δ′ + o(1))4n,

where δ′ is a positive constant which will be seen to depend only on k.To prove (6.13) we shall use Lemma 6.12. But before we apply it we have

to estimate the probabilities of the events Xi, Ysi .

We start with Xi. Put ai = |E(G4n|Wi)|, where Wi is an i-th l-element

subset of V4n. In other words, ai is the number of edges in Wi in the graphG4n. Then

Pr[Xi] = (1− p)ai ≤ e−pai = e−γ4nai.

Fix ε = ε(k) > 0 and choose δ = δ(ε, k) as in Theorem 6.10. Thenobviously ai ≥ (4− ε)4n.

We go on to Y si . It is easy to see that for each s-tuple Qs

i we havePr[Y s

i ] = ps = γ4ns.

We also need to analyze the dependencies between the events.For each Xi let us estimate the number of Y s

j on which it may depend.Note that if Xi and Y s

j are dependent then the corresponding sets Wi andQsj must have a common edge. The number of ways to make an s-cycle out

of a xed edge is not bigger than 2(s−2)4n. Thus the number of Y sj on which

Xi depends does not exceed ai2(s−2)4n.For all i, s, each Y s

i and Xi depend on not more than C lN events Xj.

It remains to estimate for each Y s1i the number of the events Y s2

j onwhich it depends. If they are dependent, Qs1

i and Qs2j must have a common

edge. Then it is easy to see that the number of such events does not exceeds12

4n(s2−2) = 24n(s2−2)(1+o(1)).For each event E ∈ Xi, Y

si we split the set J(E) (see Lemma 6.12) into

parts. First one (Jx(E)) contains all events of the type Xj. The other parts(Jys (E)) consist of the events of the type Y s

j . We want to apply Lemma 6.12,so we rewrite the conditions (6.12) for our events:

(Xi) ln δxi ≥ 2∑

j∈Jx(Xi)δxj e−γ4naj + 2

∑ks=3

∑j∈Jys (Xi)

δyj (s)γ4ns,

(Y s1i ) ln δyi (s1) ≥ 2

∑j∈Jx(Y

s1i ) δ

xj e−γ4naj + 2

∑ks=3

∑j∈Jys (Y

s1i ) δ

yj (s)γ

4ns.(6.14)

Fix a constant f = f(ε, δ, k) > 0, which will be dened later, and putδyi (s) = e, δxi = eγ

4n(1+f)ai. It is easy to see that for suciently large n we have

216

0 < δxi Pr[Xi] < 0.69, 0 < δyi (s) Pr[Y si ] < 0.69 (see Lemma 6.12). Then for

any j

δxj e−γ4naj = eγ

4n(1+f)aj−γ4naj = e−(γ−o(1))4naj ≤ e−(γ−o(1))4n(4−ε)4n = e−(

(4−ε)γ−o(1))4n.

We also have

C lN ≤

(eN

l

)l≤(

2

2− δ+ o(1)

)4n(2−δ)4n

= e(2−δ+o(1))4n

.

Thus for any i∑j∈Jx(Xi)

δxj e−γ4naj ≤

∑j∈Jx(Xi)

e−(

(4−ε)γ−o(1))4n

≤ C lNe−(

(4−ε)γ−o(1))4n≤ e(2−δ+o(1))

4n−(

(4−ε)γ−o(1))4n

= o(1),

if

γ >2− δ4− ε

. (6.15)

Similarly, if (6.15) holds, then∑j∈Jx(Y si )

δxj e−γ4naj = o(1).

Thereby, if we suppose that (6.15) holds, then the inequalities (6.14) willfollow from the system

(Xi) γ4n(1+f)ai ≥ 2∑k

s=3 eai24n(s−2)(1+o(1))γ4ns,

(Y s1i ) 1 ≥ 2

∑ks=3 e2

4n(s−2)(1+o(1))γ4ns.(6.16)

One can see that since γ < 1, both inequalities of (6.16) are consequencesof the following. For any function g(n) = o(1), any s = 3, . . . , k and allsuciently big n should hold

24n(s−2)(1+g(n))γ4n(s−1−f) = o(1). (6.17)

In turn, to prove this it is enough to check the inequality

s− 2 + (s− 1− f) log2 γ < 0 (6.18)

for s = 3, . . . , k. Any γ,

217

0 < γ < 2−k−2

k−1−f , (6.19)

satises (6.18), and also the system (6.16). Therefore, the system (6.14) issatised if both (6.15) and (6.19) hold:

2− δ4− ε

< γ < 2−k−2

k−1−f . (6.20)

Lastly, we can choose the parameters. We choose ε = ε(k) so that 24−ε <

2−k−2k−1 . Then we choose f = f(ε, δ, k) small enough so that 2−δ

4−ε < 2−k−2

k−1−f .

Finally, we choose γ = γ(f, ε, δ, k) ∈(

2−δ4−ε , 2

− k−2k−1−f

).

We have veried all the conditions of Lemma 6.12, hence the inequality(6.13) holds and Theorem 6.2 is proved.

218

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