A Note on -Critical Linear Hypergraphs Benny Sudakov
Abstract
J.P. Roudne as?wellas R. Aharoni and R. Ziv conjectured that every -critical linear hypergraph has at most +1 2 edges. In this note we prove this conjecture for 5, and obtain a nontrivial upper bound for general .
1 Introduction A linear hypergraph H is an ordered pair H = (V; E ), where V = V (H ) is a nite set of vertices and E = E (H ) is a collection of subsets of V , called edges, such that any pair e1 ; e2 2 E satis es je1 \ e2j 1. A subset T V is called a cover of H if it intersects every edge of H . The minimum cardinality of a cover is denoted by (H ), and is called the covering number of H . A hypergraph is called -critical if omitting any one of its edges reduces its covering number. Several problems in Extremal Set Theory deal with the estimation of the size of -critical systems of given types (see [2], [3] for various examples). Erd}os, Hajnal and Moon [1] proved that any ? critical graph has at most +1 edges. As a generalization of this result J.P. Roudne [2], as well as R. 2 Aharoni and R.Ziv [3] conjectured that the same estimate holds for any -critical linear hypergraph. In this note we prove this conjecture for 5. We also show that for any 5 the number of edges in a -critical linear hypergraph is at most 2 ? 3 + 5. It is worth noting that an easy upper bound for the number of edges in a -critical linear hypergraph is 2 ? + 1, as it is not too dicult to see that the maximum degree in each such hypergraph is at most and there is a set of ? 1 vertices that covers all the edges except one. But the trivial upper bound proves the conjecture only for = 2. The usual framework for dealing with -critical hypergraphs is by the so called set-pair method (see [2], [3] for more details). We start with some de nitions. De nition: A collection S = f(Ai; Bi)j1 i ng is called an (a; b) system i it satis es the following conditions:
Ai \ Bi = ; for all 1 i n
Department of Mathematics, Raymond and Beverly Sackler Faculty of Exact Sciences, Tel Aviv University, Tel Aviv, Israel. Email:
[email protected]. This research forms part of a Ph.D. thesis, written by the author under the supervision of Professor Noga Alon. Mathematics Subject Classi cation (1991): 05D05, 05D15.
1
Ai \ Bj 6= ; for i 6= j jAij = a for 1 i n and jBi \ Bj j < b for i 6= j: We denote by f (a; b) the maximum possible size of an (a; b) system. From every -critical linear hypergraph one can construct a ( ? 1; 2) system of the same size as follows. Let H = (V; E ) be a -critical linear hypergraph. By de nition, for any edge e 2 E one can nd a subset Xe V such that jXej = ? 1 and Xe intersects all the edges of H except e. Therefore the family f(Xe ; e)g forms a ( ? 1; 2) system of size jE j. Thus, the conjecture of Roudne , Aharoni and Ziv would follow from the following.
Conjecture 1.1 For all a 0: f (a; 2) ?a . +2 2
The conjecture of Aharoni and Ziv was initially formulated in this form. They also conjectured that even more is true:
Conjecture 1.2 For all pairs a and b: f (a; b) ?a b b. +
Note that the bound in Conjecture 1.2 is tight for all pairs a and b as shown by the following example. Let Y be a set of size a + b, then S = f(A; Y n A)j jAj = a; A Y g forms an (a; b) system of size ?a+b b . Here we prove the following theorem.
Theorem 1.3 Conjecture 1.1 is true for all a 4. Moreover, for any a 4: f (a; 2) a ? a + 3. 2
The proof of Theorem 1.3 is given in Section 2. In Section 3 we show how to apply this result and derive a simple upper bound for f (a; b) in general.
2 The Proof of Theorem 1.3 In this section we prove our main result. Let S = f(Ai ; Bi )j1 i ng be an (a; 2) system. Put X = [ni=1Ai. We may assume that Bi X for all i. Otherwise replace each Bi by Bi \ X , and note that the resulting family remains an (a; 2) system. We claim that no family of a + 2 of the sets Bi share a common point. Assume this is false, and renumber these sets as B1; : : :; Ba+2. Suppose +2 Bi. By de nition Aa+2 \ Ba+2 = ;, thus y 62 Aa+2. Therefore Aa+2 \ (Bi n fyg) 6= ; fyg = \ai=1 for all 1 i a + 1. Since the sets Bi n fy g are pairwise disjoint for i a + 1, we conclude that jAa+2j a + 1. This contradiction proves the claim. The following lemma may be interesting in its own right. Lemma 2.1 Let S = f(Ai; Bi)j1 i ng be an (a; 2) system, which contains a family of a + 1 of the sets Bi with a nonempty intersection. Then
jS j a + 1 + f (a ? 1; 2): 2
Proof: Suppose B ; : : :; Ba have a nonempty intersection. Let fyg = \ai Bi. Consider the new 1
+1 =1
+1
system S 0 = f(Aj nfy g; Bj )ja +2 j ng. It is easy to see that y 2 Aj for j a +2, since otherwise Aj \ (Bi n fyg) 6= ; for 1 i a + 1. But the sets Bi n fyg are pairwise disjoint for 1 i a + 1. Hence Aj has size at least a + 1, contradiction. Also y 62 Bj for j a + 2, since any a + 2 of the sets Bj have an empty intersection. Thus S 0 is an (a ? 1; 2) system, implying that
jS j = a + 1 + jS 0 j a + 1 + f (a ? 1; 2): 2 Now we can prove Conjecture 1.1 for a 3. Proof: The assertion is trivial for a = 0. Assuming a > 0, let S = f(Ai; Bi)j1 i ng be an (a; 2) system and suppose a 3. If S contains a family of a +1 of the sets Bi with a nonempty intersection, then we proceed by induction using Lemma 2.1. Otherwise, since A1 \ Bi 6= ; for 2 i n there is a point in A1 , which is contained in at least n?a 1 of the sets Bi . Thus n?a 1 a. Therefore n a2 +1. This proves Conjecture 1.1 for a 3: 2 Before proving Theorem 1.3 we need the following additional lemma.
Lemma 2.2 Let S = f(Ai; Bi)j1 i ng be an (a; 2) system, such that all Ai; 1 i n are
pairwise disjoint. Then jS j 2a + 1.
Proof: Since all Ai are pairwise disjoint and Bi \ Aj 6= ; for i 6= j , it follows that jBij n ? 1 for all i. By de nition of an (a; 2) system, jBi \ Bj j 1. Therefore, since Bi [nj=1 Aj for all i,
j[
n i=1
Aij = an j [
n i=1
Bi j
n X i=1
jBij ?
X i i: (3) Let jFa j = a ? t. By our assumptions we have that jFi j a and Pa1 jFij = a2 ? a + 3: Therefore t 1 and at least t + 2 families, namely F1; : : :; Ft+2 have size a precisely. We give a detailed proof that A(1) 1 \ A1 = ;. The same reasoning can be used to show that all (i) other sets Aj do not intersect A1 . for i t + 2. Claim 1: xi 62 A(1) 1 (1) Proof: Since x1 2 B1(1), by de nition x1 62 A(1) for some 2 i t + 2. 1 . Assume, rst, that xi 2 A1 (1) We claim in this case that all xi , 2 i t + 2, belong to A1 . Without loss of generality assume intersects Bi(j ) for all i. but xj does not for some j > 2. By de nition A(1) that x2 belongs to A(1) 1 1 has also a nonempty intersection with Bi(j ) n fxj g for all i. Since any a + 1 of the sets Bi Then A(1) 1 are disjoint we have that x2 62 Bi(j ) . Note that \i Bi(j ) = fxj g, thus Bi(j ) n fxj g are pairwise disjoint. has size at least a + 1, contradiction. Therefore all xi , 2 i t + 2, belong to A(1) Therefore A(1) 1 . 1 (1) Since any a + 1 of the sets Bi are disjoint, xj 62 Bi for all i and 2 j t + 2. By de nition A(1) 1 (1) (1) intersects Bi for i 2. Therefore we conclude that jA1 j (t + 1) + (a ? 1) = a + t a + 1, contradiction. Hence no xi , i t + 2, belongs to A(1) 1 : 2 (1) Claim 2: xk 62 A1 for all k t + 3. for 1 i t + 2 we have that Proof: Since xi 62 A(1) 1 (i) A(1) 6 ;; for 1 j a and for 2 i t + 2: 1 \ Bj n fxi g = (1) belongs to t + 1 of the sets Bi , one from We also have that jA(1) 1 j = a. Therefore each point in A1 (1) each of the families F2; : : :; Ft+2. Suppose xk belongs to A1 for some k t + 3. Since Fk is disjoint from Fi for 1 i t + 2 we conclude that xk belongs to t + 1 + jFk j t + 1 + a ? t = a + 1 of the sets Bi , contradiction. 2 is disjoint from A1 , since A1 = fx1 ; : : :; xag. Claim 1 together with Claim 2 prove that A(1) 1 (i) Similarly, each Aj is disjoint from A1 . Therefore all the sets Ai are pairwise disjoint. Thus by Lemma 2.2 jS j 2a + 1 < a2 ? a + 4; contradiction. This completes the proof of Theorem 1.3. 2
3 Concluding remarks It is not clear how to adapt our combinatorial method to get a good upper bound for f (a; b) in general. Perhaps some algebraic tools will be useful here. Note however that the following simple 4
observation, together with Theorem 1.3 gives some upper bound for the size of general (a; b) systems.
Proposition 3.1 For all pairs a and b: f (a; b + 1) af (a; b) + 1. Proof: Let S = f(Ai; Bi)j1 i ng be an (a; b) system. Suppose that A = fx ; : : :; xag. For 1
1
every xi consider the family Fi = f(Aj ; Bj )g of pairs in S where fBj g are all the sets Bi which contain the point xi . It is easy to see that f(A(ji); Bj(i) ) nfxigg forms an (a; b ? 1) system. Therefore jFij f (a; b ? 1). Since S = [ai=1 Fi [ f(A1; B1)g then, (i)
jS j
a X i=1
(i)
(i)
jFij + 1 af (a; b ? 1) + 1: 2
Acknowledgement: I am grateful to Noga Alon for his suggestions and comments on an early version of this paper.
References [1] P. Erd}os, A. Hajnal and J. Moon, A problem in graph theory, Amer. Math. Monthly 71 (1964), 1107-1110. [2] Z. Furedi, Matchings and covers in hypergraphs, Graphs and Combinatorics 4 (1988), 115-206. [3] Z. Tuza, Applications of the set-pair method in extremal hypergraph theory, Extremal problems for nite sets, Bolyai Society Mathematical Studies 3, Budapest, 1994, 479-514.
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