Generating all subsets of a finite set with disjoint unions

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1 Generating all subsets of a finite set with disjoint unions David Ellis, Benny Sudakov May 011 Abstract If X is an n-element set, we call a family G PX a k-generator for X if every x X can be expressed as a union of at most k disjoint sets in G. Frein, Lévêque and Sebő [10] conjectured that for n > k, the smallest k-generators for X are obtained by taking a partition of X into classes of sizes as equal as possible, and taking the union of the power-sets of the classes. We prove this conjecture for all sufficiently large n when k =, and for n a sufficiently large multiple of k when k 3. Keywords: generator, disjoint unions. 000 MSC: 05Dxx. 1 Introduction Let X be an n-element set, and let PX denote the set of all subsets of X. We call a family G PX a k-generator for X if every x X can be expressed as a union of at most k disjoint sets in G. For example, let (V i ) k i=1 be a partition of X into k classes of sizes as equal as possible; then F n,k := k P(V i ) \ { } i=1 St John s College, Cambridge, CB 1TP, UK. Department of Mathematics, UCLA, Los Angeles, CA 90095, USA. Research supported in part by NSF CAREER award DMS and by a USA-Israeli BSF grant. 1

2 is a k-generator for X. We call a k-generator of this form canonical. If n = qk + r, where 0 r < k, then F n,k = (k r)( q 1) + r( q+1 1) = (k + r) q k. Frein, Lévêque and Sebő [10] conjectured that for any k n, this is the smallest possible size of a k-generator for X. Conjecture 1 (Frein, Lévêque, Sebő). If X is an n-element set, k n, and G PX is a k-generator for X, then G F n,k. If n > k, equality holds only if G is a canonical k-generator for X. They proved this for k n 3k, but their methods do not seem to work for larger n. For k =, Conjecture 1 is a weakening of a conjecture of Erdős. We call a family G PX a k-base for X if every x X can be expressed as a union of at most k (not necessarily disjoint) sets in G. Erdős (see [11]) made the following Conjecture (Erdős). If X is an n-element set, and G PX is a -base for X, then G F n,. In fact, Frein, Lévêque and Sebő [10] made the analogous conjecture for all k. Conjecture 3 (Frein, Lévêque, Sebő). If X is an n-element set, k n, and G PX is a k-base for X, then G F n,k. If n > k, equality holds only if G is a canonical k-generator for X. Again, they were able to prove this for k n 3k. In this paper, we study k-generators when n is large compared to k. Our main results are as follows. Theorem 4. If n is sufficiently large, X is an n-element set, and G PX is a -generator for X, then G F n,. Equality holds only if G is of the form F n,. Theorem 5. If k N, n is a sufficiently large multiple of k, X is an n- element set, and G is a k-generator for X, then G F n,k. Equality holds only if G is of the form F n,k. In other words, we prove Conjecture 1 for all sufficiently large n when k =, and for n a sufficiently large multiple of k when k 3. We use some ideas of Alon and Frankl [1], and also techniques of the first author from [5], in which asymptotic results were obtained.

3 As noted in [10], if G PX is a k-generator (or even a k-base) for X, then the number of ways of choosing at most k sets from G is clearly at least the number of subsets of X. Therefore G k n, which immediately gives G n/k. Moreover, if G = m, then k i=0 ( ) m n. (1) i Crudely, we have so Hence, if k is fixed, then so k i=0 k 1 i=0 ( ) m m k 1, i ( ) m i ( ) m + m k 1. k ( ) m (1 + O(1/m)) n, k G (k!) 1/k n/k (1 o(1)). () Observe that if n = qk + r, where 0 r < k, then F n,k = (k + r) q k < (k + r) q = k n/k (1 + r/k) r/k < c 0 k n/k, (3) where c 0 := 1/ log log = (to 3 d.p.). Now for some preliminaries. We use the following standard notation. For n N, [n] will denote the set {1,,..., n}. If x and y are disjoint sets, we will sometimes write their union as x y, rather than x y, to emphasize the fact that the sets are disjoint. If k N, and G is a graph, K k (G) will denote the number of k-cliques in G. Let T s (n) denote the s-partite Turán graph (the complete s-partite graph on n vertices with parts of sizes as equal as possible), and let t s (n) = e(t s (n)). For l N, C l will denote the cycle of length l. If F is a (labelled) graph on f vertices, with vertex-set {v 1,..., v f } say, and t = (t 1,..., t f ) N f, we define the t-blow-up of F, F t, to be the 3

4 graph obtained by replacing v i with an independent set V i of size t i, and joining each vertex of V i to each vertex of V j whenever v i v j is an edge of F. With slight abuse of notation, we will write F t for the symmetric blow-up F (t,..., t). If F and G are graphs, we write c F (G) for the number of injective graph homomorphisms from F to G, meaning injections from V (F ) to V (G) which take edges of F to edges of G. The density of F in G is defined to be d F (G) = c F (G) G ( G 1) ( G F + 1), i.e. the probability that a uniform random injective map from V (F ) to V (G) is a graph homomorphism from F to G. Hence, when F = K k, the density of K k s in an n-vertex graph G is simply K k (G)/ ( n k). Although we will be interested in the density d F (G), it will sometimes be more convenient to work with the following closely related quantity, which behaves very nicely when we take blow-ups. We write Hom F (G) for the number of homomorphisms from F to G, and we define the homomorphism density of F in G to be h F (G) = Hom F (G) G F, i.e. the probability that a uniform random map from V (F ) to V (G) is a graph homomorphism from F to G. Observe that if F is a graph on f vertices, and G is a graph on n vertices, then the number of homomorphisms from F to G which are not injections is clearly at most ( ) f n f 1. Hence, d G (F ) if f is fixed. In the other direction, d F (G) h G(F )n f ( ) f n f 1 n(n 1) (n f + 1) h G(F ) O(1/n), (4) n f n(n 1) (n f + 1) h F (G) (1 + O(1/n))h F (G) (5) if f is fixed. Hence, when working inside large graphs, we can pass freely between the density of a fixed graph F and its homomorphism density, with an error of only O(1/n). Finally, we will make frequent use of the AM/GM inequality: 4

5 Theorem 6. If x 1,..., x n 0, then ( n ) 1/n x i 1 n i=1 n x i. i=1 The case k n via extremal graph theory. For n a sufficiently large multiple of k, it turns out to be possible to prove Conjecture 1 using stability versions of Turán-type results. We will prove the following Theorem 5. If k N, n is a sufficiently large multiple of k, X is an n- element set, and G is a k-generator for X, then G F n,k. Equality holds only if G is of the form F n,k. We need a few more definitions. Let H denote the graph with vertex-set PX, where we join two subsets x, y X if they are disjoint. With slight abuse of terminology, we call H the Kneser graph on PX (although this usually means the analogous graph on X (r) ). If F, G PX, we say that G k-generates F if every set in F is a disjoint union of at most k sets in G. The main steps of the proof: First, we will show that for any A PX with A Ω( n/k ), the density of K k+1 s in the induced subgraph H[A] is o(1). Secondly, we will observe that if n is a sufficiently large multiple of k, and G PX has size close to F n,k and k-generates almost all subsets of X, then K k (H[G]) is very close to K k (T k ( G )), the number of K k s in the k-partite Turán graph on G vertices. We will then prove that if G is any graph with small K k+1 -density, and with K k (G) close to K k (T k ( G )), then G can be made k-partite by removing a small number of edges. This can be seen as a (strengthened) variant of the Simonovits Stability Theorem [9], which states that any K k+1 -free graph G with e(g) close to the maximum e(t k ( G )), can be made k-partite by removing a small number of edges. This will enable us to conclude that H[G] can be made k-partite by the removal of a small number of edges, and therefore the structure of H[G] is close to that of the Turán graph T k ( G ). This in turn will enable us to show that the structure of G is close to that of a canonical k-generator F n,k (Proposition 9). Finally, we will use a perturbation argument to show that if n is sufficiently large, and G F n,k, then G = F n,k, completing the proof. 5

6 In fact, we will first show that if A PX with A Ω( n/k ), then the homomorphism density of K k+1 t in H[A] is o(1), provided t is sufficiently large depending on k. Hence, we will need the following (relatively wellknown) lemma relating the homomorphism density of a graph to that of its blow-up. Lemma 7. Let F be a graph on f vertices, let t = (t 1, t,..., t f ) N f, and let F t denote the t-blow-up of F. If the homomorphism density of F in G is p, then the homomorphism density of F t in G is at least p t 1t t f. Proof. This is a simple convexity argument, essentially that of [9]. It will suffice to prove the statement of the lemma when t = (1,..., 1, r) for some r N. We think of F as a (labelled) graph on vertex set [f] = {1,,..., f}, and G as a (labelled) graph on vertex set [n]. Define the function χ : [n] f {0, 1} by { 1 if i vi is a homomorphism from F to G, χ(v 1,..., v f ) = 0 otherwise. Then we have h F (G) = 1 n f (v 1,...,v f ) [n] f χ(v 1,..., v f ) = p. The homomorphism density h F (1,...,1,r) (G) of F (1,..., 1, r) in G is: h F (1,...,1,r) (G) = 1 n f 1+r (v 1,...,v f 1,v (1) f r,v() f,...,v(r) f ) [n]f 1+r i=1 χ(v 1,..., v f 1, v (i) f ) r = 1 1 χ(v n f 1 1,..., v f 1, v f ) n (v 1,...,v f 1 ) [n] f 1 v f [n] r 1 1 χ(v n f 1 1,..., v f 1, v f ) n (v 1,...,v f 1 ) [n] f 1 v f [n] r = 1 χ(v n f 1,..., v f 1, v f ) (v 1,...,v f 1,v f ) [n] f = p r. Here, the inequality follows from applying Jensen s Inequality to the convex function x x r. This proves the lemma for t = (1,..., 1, r). By 6

7 symmetry, the statement of the lemma holds for all vectors of the form (1,..., 1, r, 1,..., 1). Clearly, we may obtain F t from F by a sequence of blow-ups by these vectors, proving the lemma. The following lemma (a rephrasing of Lemma 4. in Alon and Frankl [1]) gives an upper bound on the homomorphism density of K k+1 t in large induced subgraphs of the Kneser graph H. Lemma 8. If A PX with A = m = (δ+1/(k+1))n, then h Kk+1 t(h[a]) (k + 1) n(δt 1). Proof. We follow the proof of Alon and Frankl cited above. Choose (k + 1)t members of A uniformly at random with replacement, (A (j) i ) 1 i k+1, 1 j t. The homomorphism density of K k+1 t in H[A] is precisely the probability that the unions t U i = j=1 are pairwise disjoint. If this event occurs, then U i n/(k + 1) for some i. For each i [k], we have ( t ) Pr{ U i n/(k + 1)} = Pr {A (j) i S} = S n/(k+1) S n/(k+1) A (j) i S X: S n/(k+1) Pr ( t j=1 ( S /m) t n ( n/(k+1) /m) t = n(δt 1). j=1 {A (j) i S} ) Hence, ( k ) Pr { U i n/(k + 1)} i=1 k Pr{ U i n/(k + 1)} (k + 1) n(δt 1). i=1 Therefore, as required. h Kk+1 t(h[a]) (k + 1) n(δt 1), 7

8 From the trivial bound above, any k-generator G has G n/k, so δ 1/(k(k + 1)), and therefore, choosing t = t k := k(k + 1), we see that Hence, by Lemma 7, Therefore, by (5), h Kk+1 t k (H[G]) (k + 1) n. h Kk+1 (H[G]) O k ( n/t k k). ( ) d Kk+1 (H[G]) O k n/t k k a k n (6) provided n is sufficiently large depending on k, where a k > 0 depends only on k. Assume now that n is a multiple of k, so that F n,k = k n/k k. We will prove the following stability result. Proposition 9. Let k N be fixed. If n is a multiple of k, and G PX has G (1 + η) F n,k and k-generates at least (1 ɛ) n subsets of X, then there exists an equipartition (S i ) k i=1 of X such that G ( k i=1ps i ) (1 Ck ɛ 1/k D k η 1/k ξ kn ) F n,k, where C k, D k, ξ k > 0 depend only on k. We first collect some results used in the proof. We will need the following theorem of Erdős [7]. Theorem 10 (Erdős). If r k, and G is a K k+1 -free graph on n vertices, then K r (G) K r (T k (n)). We will also need the following well-known lemma, which states that a dense k-partite graph has an induced subgraph with high minimum degree. Lemma 11. Let G be an n-vertex, k-partite graph with e(g) (1 1/k δ)n /. Then there exists an induced subgraph G G with G = n (1 δ)n and minimum degree δ(g ) (1 1/k δ)(n 1). 8

9 Proof. We perform the following algorithm to produce G. Let G 1 = G. Suppose that at stage i, we have a graph G i on n i+1 vertices. If there is a vertex v of G i with d(v) < (1 1/k η)(n i), let G i+1 = G i v; otherwise, stop and set G = G i. Suppose the process terminates after j = αn steps. Then we have removed at most (1 1/k η) j (( ) ( )) n n j (n i) = (1 1/k η) i=1 edges, and the remaining graph has at most ( ) ( ) k n j = (1 α) (1 1/k)n / k edges. But our original graph had at least edges, and therefore (1 1/k δ)n / (1 1/k η)(1 (1 α) )n / + (1 α) (1 1/k)n / (1 1/k δ)n /, so Choosing η = δ, we obtain and therefore so η(1 α) η δ. η(1 α) η(1 η), (1 α) 1 η, α 1 (1 η) 1/ η. Hence, our induced subgraph G has order and minimum degree G = n (1 δ)n, δ(g ) (1 1/k δ)(n 1). We will also need Shearer s Entropy Lemma. 9

10 Lemma 1 (Shearer s Entropy Lemma, [4]). Let S be a finite set, and let A be an r-cover of S, meaning a collection of subsets of S such that every element of S is contained in at least r sets in A. Let F be a collection of subsets of S. For A S, let F A = {F A : F F} denote the projection of F onto the set A. Then F r A A F A. In addition, we require two stability versions of Turán-type results in extremal graph theory. The first states that a graph with a very small K k+1 - density cannot have K r -density much higher than the k-partite Turán graph on the same number of vertices, for any r k. Lemma 13. Let r k be integers. Then there exist C, D > 0 such that for any α 0, any n-vertex graph G with K k+1 -density at most α has K r -density at most k(k 1) (k r + 1) (1 + Cα 1/(k+) + D/n). k r Proof. We use a straightforward sampling argument. Let G be as in the statement of the lemma. Let ζ ( n l) be the number of l-subsets U V (G) such that G[U] contains a copy of K k+1, so that ζ is simply the probability that a uniform random l-subset of V (G) contains a K k+1. Simple counting (or the union bound) gives ( ) l ζ α. k + 1 By Theorem 10, each K k+1 -free G[U] contains at most ( ) ( ) r k l r k K r s. Therefore, the density of K r s in each such G[U] satisfies k(k 1) (k r + 1) l r d Kr (G[U]) k r l(l 1) (l r + 1) k(k 1) (k r + 1) (1 + O(1/l)). (7) k r Note that one can choose a random r-set in graph G by first choosing a random l-set U, and then choosing a random r-subset of U. The density of 10

11 K r s in G is simply the probability that a uniform random r-subset of V (G) induces a K r, and therefore d Kr (G) = E U [d Kk (G[U])], where the expectation is taken over a uniform random choice of U. If U is K k+1 -free, which happens with probability 1 ζ, we use the upper bound (7); if U contains a K k+1, which happens with probability ζ, we use the trivial bound d Kk (G[U]) 1. We see that the density of K r s in G satisfies: k(k 1) (k r + 1) d Kr (G) (1 ζ) (1 + O(1/l)) + ζ k r ( ) k(k 1) (k r + 1) l + O(1/l) + α k r k + 1 k(k 1) (k r + 1) + O(1/l) + l k+1 α. k r Choosing l = min{ α 1/(k+), n} proves the lemma. The second result states that an n-vertex graph with a small K k+1 -density, a K k -density not too much less than that of T k (n), and a K k 1 -density not too much more than that of T k (n), can be made into a k-partite graph by the removal of only a small number of edges. Theorem 14. Let G be an n-vertex graph with K k+1 -density at most α, K k 1 -density at most (1 + β) k! k, k 1 and K k -density at least (1 γ) k! k, k where γ 1/. Then G can be made into a k-partite graph G 0 by removing at most ( edges, which removes at most ( K k s. β + γ + 8kk+1 (k + 1) k! β + γ + 8kk+1 (k + 1) k! α + k/n ) ( n ) ) ( ) k n α + k/n )( k 11

12 Proof. If k N, and G is a graph, let K k (G) = {S V (G) (k) : G[S] is a clique} denote the set of all k-sets that induce a clique in G. If S V (G), let N(S) denote the set of vertices of G joined to all vertices in S, i.e. the intersection of the neighbourhoods of the vertices in S, and let d(s) = N(S). For S K k (G), let f G (S) = d(t ). T S, T =k 1 We begin by sketching the proof. The fact that the ratio between the K k - density of G and the K k 1 -density of G is very close to 1/k will imply that the average Ef G (S) over all sets S K k (G) is not too far below n. The fact that the K k+1 -density of G is small will mean that for most sets S K k (G), every (k 1)-subset T S has N(T ) spanning few edges of G, and any two distinct (k 1)-subsets T, T S have N(T ) N(T ) small. Hence, if we pick such a set S which has f G (S) not too far below the average, the sets {N(T ) : T S, T = k 1} will be almost pairwise disjoint, will cover most of the vertices of G, and will each span few edges of G. Small alterations will produce a k-partition of V (G) with few edges of G within each class, proving the theorem. We now proceed with the proof. Observe that Ef G = = S K k (G) T S, T =k 1 d(t ) K k (G) T K k 1 (G) d(t ) K k (G) ( ) T Kk 1(G) d(t ) K k 1 (G)K k (G) = (kk k(g)) K k 1 (G)K k (G) = k K k(g) K k 1 (G) k (1 γ) k! 1 k k 1 k k 1 + β k! = 1 γ (n k + 1). 1 + β ( n k) ( n ) k 1 (The first inequality follows from Cauchy-Schwarz, and the second from our assumptions on the K k -density and the K k 1 -density of G.) 1

13 ( We call a set T K k 1 (G) dangerous if it is contained in at least α n k+1 ) Kk+1 s. Let D denote the number of dangerous (k 1)-sets. Double-counting the number of times a (k 1)-set is contained in a K k+1, we obtain: D ( ) ( ) ( ) n k + 1 k + 1 n α α, k + 1 since there are at most α ( n k+1) Kk+1 s in G. Hence, D ( ) n α. k 1 Similarly, we call a set S K k (G) treacherous if it is contained in at least α(n k) Kk+1 s. Double-counting the number of times a k-set is contained in a K k+1, we see that there are at most α ( n k) treacherous k-sets. Call a set S K k (G) bad if it is treacherous, or contains at least one dangerous (k 1)-set; otherwise, call S good. Then the number of bad k-sets is at most α ( n k ) + (n k + 1) ( ) n α = (k + 1) α k 1 so the fraction of sets in K k (G) which are bad is at most ( n k ), Suppose that (k + 1) α = kk (k + 1) α. (1 γ) k! (1 γ)k! k k max{ f G (S) : S is good} < (1 ψ)(n k + 1). Observe that for any S K k (G), we have f G (S) k(n k + 1), since d(t ) n k + 1 for each T S (k 1). Hence, (( Ef G < 1 kk (k + 1) ) α (1 ψ) + kk (k + 1) ) α k (n k + 1) (1 γ)k! (1 γ)k! ( 1 ψ + kk+1 (k + 1) ) α (n k + 1), (1 γ)k! a contradiction if ψ = ψ 0 := 1 1 γ 1 + β + kk+1 (k + 1) α (1 γ)k! 13 γ + β + kk+1 (k + 1) α. k!

14 Let S K k (G) be a good k-set such that f G (S) (1 ψ 0 )(n k + 1). Write S = {v 1,..., v k }, let T i = S \ {v i } for each i, and let N i = N(T i ) for each i. Observe that N i N j = N(S) for each i j, and N(S) = d(s) α(n k). Let W i = N i \ N(S) for each i; observe that the W i s are pairwise disjoint. Let R = V (G) \ k i=1w i be the set of leftover vertices. Observe that k N i \ N(S) = f G (S) kn(s) (1 ψ)(n k + 1) k α(n k), i=1 and therefore the number of leftover vertices satisfies R < (ψ + k α)n + k. We now produce a k-partition (V i ) k i=1 of V (G) by extending the partition (W i ) k i=1 of V (G) \ R arbitrarily to R, i.e., we partition the leftover vertices arbitrarily. Now delete all edges of G within V i for each i. The number of edges within N i is precisely the number of K k+1 s containing T i, which is at most α ( ) n k+1. The number of edges incident with R is trivially at most (ψ + k α)n(n 1) + k(n 1). Hence, the number of edges deleted was at most (ψ + k α)n(n 1) + k(n 1) + k ( ) n k + 1 α ( ) ( β + γ + 8kk+1 (k + 1) n α + k/n k! Removing an edge removes at most ( n k ) Kk s, and therefore the total number of K k s removed is at most ( ) ( ) β + γ + 8kk+1 (k + 1) n n α + k/n k! )( k ( = completing the proof. β + γ + 8kk+1 (k + 1) k! ). α + k/n ) ( k )( n k ), Note that the two results above together imply the following 14

15 Corollary 15. For any k N, there exist constants A k, B k > 0 such that the following holds. For any α 0, if G is an n-vertex graph with K k+1 -density at most α, and K k -density at least (1 γ) k! k k, where γ 1/, then G can be made into a k-partite graph G 0 by removing at most ( γ + Ak α 1/(k+) + B k /n ) ( ) n edges, which removes at most ( γ + Ak α 1/(k+) + B k /n ) ( )( ) k n k K k s. Proof of Proposition 9. Suppose G PX has G = m (1 + η) F n,k, and k-generates at least (1 ɛ) n subsets of X. Our aim is to show that G is close to a canonical k-generator. We may assume that ɛ 1/C k k and η 1/Dk k, so by choosing C k and D k appropriately large, we may assume throughout that ɛ and η are small. By choosing ξ k appropriately small, we may assume that n n 0 (k), where n 0 (k) is any function of k. We first apply Lemma 13 and Theorem 14 with G = H[G], where H is the Kneser graph on PX, G PX with G = m (1 + η) F n,k, and G k-generates at least (1 ɛ) n subsets of X. By (6), we have d Kk+1 (H[G]) a kn, and therefore we may take α = akn. Applying Lemma 13 with r = k 1, we may take β = bkn for some b k > 0. We have G = m (1 + η)(k n/k k), so ( m k ) mk k! < (1 + η)k k k n. k! Notice that k 1 ( ) m km k 1 k((1 + η)k n/k ) k 1 < (1 + η) k 1 k k (1 1/k)n. i i=0 Since G k-generates at least (1 ɛ) n subsets of X, we have K k (H[G]) (1 ɛ) n (1 + η) k 1 k k (1 1/k)n. 15

16 Hence, d Kk (H[G]) = K k(h[g]) ( m ) k where the last inequality follows from (1 ɛ)n (1 + η) k 1 k k (1 1/k)n ) ( (1+η)k n/k 1 ɛ (1 + η)k 1 k k n/k (1 + η) k k! k k (1 ɛ kη k k n/k ) k! k k, 1 ɛ (1 + η) k (1 ɛ)(1 η)k (1 ɛ)(1 kη) 1 ɛ kη. Therefore, the K k -density of H[G] satisfies where Let d Kk (H[G]) (1 γ) k! k k, γ = ɛ + kη + k k n/k. ( ) ( ) ψ = β + γ + 8kk+1 (k + 1) k α + k/n. k! By Theorem 14, there exists a k-partite subgraph G 0 of H[G] with ( ) m K k (G 0 ) K k (H[G]) ψ k ( ) m (1 ɛ) n (1 + η) k 1 k k (1 1/k)n ψ k (1 ɛ (1 + ) η)k k k ψ (1 + η) k 1 k k k/n n. k! k Writing we have φ = ɛ + (1 + η)k k k ψ + (1 + η) k 1 k k k/n, k! K k (G 0 ) (1 φ) n. 16

17 Let V 1,..., V k be the vertex-classes of G 0. By the AM/GM inequality, K k (G 0 ) ( k k ) i=1 V i V k i = (m/k) k, k i=1 and therefore G = m k(k k (G 0 )) 1/k k(1 φ) 1/k n/k, (8) recovering the asymptotic result of [5]. Moreover, any k-partite graph G 0 satisfies ( ) k e(g 0 ) (K k (G 0 )) /k. To see this, simply apply Shearer s Entropy Lemma with S = V (G 0 ), F = K k (G 0 ), and A = {V i V j : i j}. Then A is a (k 1)-cover of V (G 0 ). Note that F Vi V j E G0 (V i, V j ), and therefore (K k (G 0 )) k 1 e G0 (V i, V j ). {i,j} [k] () Applying the AM/GM inequality gives: (K k (G 0 )) k 1 {i,j} e G0 (V i, V j ) ( {i,j} e G 0 (V i, V j ) ( k ) ) ( k ) ( = ) ( k ) e(g 0 ) ( k, ) and therefore e(g 0 ) ( ) k (K k (G 0 )) /k, as required. It follows that ( ) k e(g 0 ) (1 φ) /k n/k ( ) ( k m (1 φ) /k (1 + η)k ) (1 η) (1 φ) /k (1 1/k)m / (1 η φ /k )(1 1/k)m / = (1 δ)(1 1/k)m /, 17

18 where δ = η + φ /k. Hence, G 0 is a k-partite subgraph of H[G] with G 0 = G = m, and e(g 0 ) (1 δ 1/k)m /. Applying Lemma 11 to G 0, we see that there exists an induced subgraph H of G 0 with and H (1 δ) G, (9) δ(h ) (1 1/k δ)( H 1). Let Y 1,..., Y k be the vertex-classes of H ; note that these are families of subsets of X. Clearly, for each i [k], Hence, for each i [k], Y i H δ(h ) (1/k + δ) H + 1. (10) Y i H (k 1)((1/k+ δ) H +1) (1/k (k 1) δ) H k+1. (11) For each i [k], let S i = y Y i y be the union of all sets in Y i. We claim that the S i s are pairwise disjoint. Suppose for a contradiction that S 1 S. Then there exist y 1 Y 1 and y Y which both contain some element p X. Since δ(h ) (1 1/k δ)( H 1), at least (1 1/k δ)( H 1) sets in i 1 Y i do not contain p. By (10), i 1 Y i = i 1 Y i (1 1/k + (k 1) δ) H + k 1, and therefore the number of sets in i 1 Y i containing p is at most (1 1/k + (k 1) δ) H + k 1 (1 1/k δ)( H 1) k δ H + k. The same holds for the number of sets in i Y i containing p, so the total number of sets in H containing p is at most k δ H + k. Hence, the total number of sets in G containing p is at most (k + 1) δm + k. 18

19 But then the number of ways of choosing at most k disjoint sets in G with one containing p is at most (1 + m k 1 )((k + 1) δm + k) = O k ( δ) n + O k ( (1 1/k)n ) < n 1 ɛ n, contradicting the fact that G k-generates all but ɛ n of the sets containing p. Hence, we may conclude that the S i s are pairwise disjoint. By definition, Y i PS i, and therefore Y i S i. But from (11), Y i (1 k(k 1) δ) H /k k + 1 (1 k(k 1) δ)(1 δ) G /k k + 1 (1 k(k 1) δ)(1 δ)(1 φ) 1/k n/k k + 1 (1 (k(k 1) + 1) δ φ 1/k ) n/k k + 1 > (1 k δ φ 1/k ) n/k k > n/k 1, using (9) and (8) for the second and third inequalities respectively. Hence, we must have S i n/k for each i, and therefore S i = n/k for each i, i.e. (S i ) k i=1 is an equipartition of X. Putting everything together and recalling that δ = η + φ /k and φ = O k (ɛ + η + c kn ), we have G ( k i=1ps i ) k Y i i=1 (1 k δ φ 1/k )k n/k k (1 C k ɛ 1/k D k η 1/k ξ kn )k n/k (provided n is sufficiently large depending on k), where C k, D k, ξ k > 0 depend only on k. This proves Proposition 9. We now prove the following Proposition 16. Let ν(n) = o(1). If G is a k-generator for X with G F n,k, and G ( k i=1ps i ) (1 ν) Fn,k, where (S i ) k i=1 is a partition of X into k classes of sizes as equal as possible, then provided n is sufficiently large depending on k, we have G = F n,k and G = k i=1ps i \ { }. Note that n is no longer assumed to be a multiple of k; the case k = and n odd will be needed in Section 3. 19

20 Proof. Let G and (S i ) k i=1 be as in the statement of the proposition. For each i [k], let F i = (PS i \ { }) \ G be the collection of all nonempty subsets of S i which are not in G. By our assumption on G, we know that F i o( Si ) for each i [k]. Let k E = G \ P(S i ) i=1 be the collection of extra sets in G; let E = M. By relabeling the S i s, we may assume that F 1 F F k. By our assumption on G, M k F 1. Let R = {y 1 s s k : y 1 F 1, s i S i i }; observe that the sets y 1 s s k are all distinct, so R = F 1 n S1. By considering the number of sets in E needed for G to k-generate R, we will show that M > k F 1 unless F 1 =. (In fact, our argument would also show that M > p k F 1 unless F 1 =, for any p k > 0 depending only on k.) Let N be the number of sets in R which may be expressed as a disjoint union of two sets in E and at most k other sets in G. Then N ( M ) k i=0 ( m i ) 1 k F 1 (k 1) (c 0k n/k ) k (k )! ( ) 4c k 0 k k F1 F S 1 n S 1 1 = o(1) F 1 n S 1 = o( R ), (1) where we have used G F n,k c 0 k n/k (see (3)), S 1 n/k, and F 1 = o( S1 ) in the second, third and fourth lines respectively. Now fix x 1 F 1. For j 1, let A j (x 1 ) be the collection of (k 1)-tuples (s,..., s k ) PS PS k such that may be expressed as a disjoint union x 1 s s k y 1 y y k with y j E but y i S i i j. Let A (x 1 ) be the collection of (k 1)-tuples (s,..., s k ) PS PS k such that x 1 s s k 0

21 may be expressed as a disjoint union of two sets in E and at most k other sets in G. Now fix j 1. For each (s,..., s k ) A j (x 1 ), we may write x 1 s s k = s 1 s s j 1 y j s j+1 s k, where y j = s j (x 1 \ s 1) E. Since y j S j = s j, different s j s correspond to different y j s E, and so there are at most E = M choices for s j. Therefore, ( ) A j (x 1 ) n S 1 S j M n S 1 S j F1 k F 1 k n S1, S 1 the last inequality following from the fact that S j S 1 1. Hence, k ( ) F1 A j (x 1 ) k(k 1) n S1 = o(1) n S1. (13) j= Observe that for each x 1 F 1, A (x 1 ) S 1 k A j (x 1 ) = PS PS 3 PS k, j=1 and therefore k A (x 1 ) + A 1 (x 1 ) + A j (x 1 ) n S1, j= so by (13), A (x 1 ) + A 1 (x 1 ) (1 o(1)) n S 1. Call x 1 F 1 bad if A (x 1 ) (k+) n S 1 ; otherwise, call x 1 good. By (1), at most a o(1)-fraction of the sets in F 1 are bad, so at least a 1 o(1) fraction are good. For each good set x 1 F 1, notice that A 1 (x 1 ) (1 (k+) o(1)) n S 1. Now perform the following process. Choose any (s,..., s k ) A 1 (x 1 ); we may write x 1 s s k = z (1) s s k with (s,..., s k ) PS PS k, z (1) E, z (1) S 1 = x 1, and z (1) \S 1. Pick p 1 z (1) \ S 1. At most 1 n S 1 of the members of A 1 (x 1 ) have union containing p 1, so there are at least (1 1 (k+) o(1)) n S 1 1

22 remaining members of A 1 (x 1 ). Choose one of these, (t,..., t k ) say. By definition, we may write x 1 t t k = z () t t k with (t,..., t k ) PS PS k, z () E, z () S 1 = x 1, and z () \ S 1. Since p 1 / z (), we must have z () z (1). Pick p z () \ S 1, and repeat. At most 3 4 n S 1 of the members of A 1 (x 1 ) have union containing p 1 or p ; there are at least ( 1 4 (k+) o(1)) n S 1 members remaining. Choose one of these, (u,..., u k ) say. By definition, we may write x 1 u u k = z (3) u u k with (u,..., u k ) PS PS k, z (3) E, z (3) S 1 = x 1, and z (3) \S 1. Note that again z (3) is distinct from z (1), z (), since p 1, p z (3). Continuing this process for k + 1 steps, we end up with a collection of k + 1 distinct sets z (1),..., z (k+1) E such that z (l) S 1 = x 1 l [k +1]. Do this for each good set x 1 F 1 ; the collections produced are clearly pairwise disjoint. Therefore, E (k + 1)(1 o(1)) F 1. This is a contradiction, unless F 1 =. Hence, we must have F = = F k =, and therefore G = k i=1p(s i ) \ { }, proving Proposition 16, and completing the proof of Theorem 5. 3 The case k = via bipartite subgraphs of H. Our aim in this section is to prove the k = case of Conjecture 1 for all sufficiently large odd n, which together with the k = case of Theorem 5 will imply Theorem 4. If n is sufficiently large, X is an n-element set, and G PX is a -generator for X, then G F n,. Equality holds only if G is of the form F n,. Recall that F n, = { n/ if n is even; 3 (n 1)/ if n is odd.

23 Suppose that X is an n-element set, and G PX is a -generator for X with G = m F n,. The counting argument in the Introduction gives ( ) m 1 + m + n, which implies that G (1 o(1)) n/. For n odd, we wish to improve this bound by a factor of approximately 1.5. Our first aim is to prove that induced subgraphs of the Kneser graph H which have order Ω( n/ ) are o(1)-close to being bipartite (Proposition 18). Recall that a graph G = (V, E) is said to be ɛ-close to being bipartite if it can be made bipartite by the removal of at most ɛ V edges, and ɛ-far from being bipartite if it requires the removal of at least ɛ V edges to make it bipartite. Using Szemerédi s Regularity Lemma, Bollobás, Erdős, Simonovits and Szemerédi [3] proved the following. Theorem 17 (Bollobás, Erdős, Simonovits, Szemerédi). For any ɛ > 0, there exists g(ɛ) N depending on ɛ alone such that for any graph G which is ɛ-far from being bipartite, the probability that a uniform random induced subgraph of G of order g(ɛ) is non-bipartite is at least 1/. Building on methods of Goldreich, Goldwasser and Ron [1], Alon and Krivelevich [] proved without using the Regularity Lemma that in fact, one may take g(ɛ) (log(1/ɛ))b (14) ɛ where b > 0 is an absolute constant. As observed in [], this is tight up to the poly-logarithmic factor, since necessarily, g(ɛ) 1 6ɛ. We will first show that for any fixed c > 0 and l N, if A PX with A c n/, then the density of C l+1 s in H[A] is at most o(1). To prove this, we will show that for any l N, there exists t N such that for any fixed c > 0, if A PX with A c n/, then the homomorphism density of C l+1 t in H[A] is o(1). Using Lemma 7, we will deduce that the homomorphism density of C l+1 in H[A] is o(1), implying that the density of C l+1 s in H[A] is o(1). This will show that H[A] is o(1)-close to being bipartite (Proposition 18). To obtain a sharper estimate for the o(1) term 3

24 in Proposition 18, we will use (14), although to prove Theorem 4, any o(1) term would suffice, so one could in fact use Theorem 17 instead of (14). We are now ready to prove the following Proposition 18. Let c > 0. Then there exists b > 0 such that for any A PX with A c n/, the induced subgraph H[A] can be made bipartite by removing at most (log log n) b A log n edges. Proof. Fix c > 0; let A PX with A = m c n/. First, we show that for any fixed l N, there exists t N such that the homomorphism density of C l+1 t s in H[A] is at most o(1). The argument is a strengthening of that used by Alon and Frankl to prove Lemma 4. in [1]. Let t N to be chosen later. Choose (l +1)t members of A uniformly at random with replacement, (A (j) i ) 1 i l+1, 1 j t. The homomorphism density of C l+1 t in H[A] is precisely the probability that the unions U i = t j=1 satisfy U i U i+1 = for each i (where the addition is modulo l + 1). We claim that if this occurs, then U i < ( 1 η)n for some i, provided η < 1/(4l + ). Suppose for a contradiction that U i U i+1 = for each i, and U i ( 1 η)n for each i. Then U i+ \ U i n U i+1 U i ηn for each i [l 1]. Since U l+1 \ U 1 l j=1(u j+1 \ U j 1 ), we have U l+1 \ U 1 l j=1 U j+1 \ U j 1 lηn. It follows that U 1 U l+1 (1/ (l + 1)η)n > 0 if η < 1/(4l + ), a contradiction. We now show that the probability of this event is very small. Fix i [k]. A (j) i 4

25 Observe that Pr{ U i (1/ η)n} = Pr = S X: S (1/ η)n S (1/ η)n S (1/ η)n Pr n ( (1/ η)n c n/ = (ηt 1)n c t n c t, ( t j=1 ( S /m) t ) t ( t j=1 {A (j) {A (j) i S} i S} ) ) provided t /η. Hence, ( l+1 ) l+1 Pr { U i (1/ η)n} Pr{ U i (1/ η)n} (l + 1) n c t. i=1 Therefore, i=1 h Cl+1 t(h[a]) (l + 1) n c t. Choose η = 1 8l and t = /η = 16l. By Lemma 7, h Cl+1 (H[A]) ((l + 1) n c t ) 1/tl+1 = (l + 1) 1/(16l)l+1 n/(16l)l+1 c 1/(16l)l = O( n/(16l)l+1 ). Observe that the number of (s+1)-subsets of A containing an odd cycle of H is at most s ( ) m (l + 1) m l+1 h Cl+1 (H[A]). (s l) l=1 Hence, the probability that a uniform random (s + 1)-subset of A contains an odd cycle of H is at most s l=1 m l+1 m(m 1) (m l) (s + 1)(s) ((s l) + 1)h C l+1 (H[A]) s(s + 1)!O( n/(16s)s+1 ), 5

26 (provided s O( m)). This can be made < 1/ by choosing s = a log n/ log log n, for some suitable a > 0 depending only on c. By (14), it follows that H[A] is ((log log n) b / log n)-close to being bipartite, for some suitable b > 0 depending only on c, proving the proposition. Before proving Theorem 4 for n odd, we need some more definitions. Let X be a finite set. If A PX, and i X, we define A i A + i = {x A : i / x}, = {x \ {i} : x A, i x}; these are respectively called the lower and upper i-sections of A. If Y and Z are disjoint subsets of X, we write H[Y, Z] for the bipartite subgraph of the Kneser graph H consisting of all edges between Y and Z. If B is a bipartite subgraph of H with vertex-sets Y and Z, and F PX, we say that B -generates F if for every set x F, there exist y Y and z Z such that y z =, yz E(B), and y z = x, i.e. every set in F corresponds to an edge of B. Proof of Theorem 4 for n odd. Suppose that n = l is odd, X is an n-element set, and G PX is a -generator for X with G = m F n, = 3 l. Observe that e(h[g]) l+1 G 1 l+1 3 l + 1, and therefore H[G] has edge-density at least l+1 3 l + 1 ) ( G l+1 3 l (3 l )(3 l 3) > 4. 9 (Here, the last inequality rearranges to the statement l > 0.) By Proposition 18 applied to G, we can remove at most (log log n) b log n G < (log log n) b 9 l log n edges from H[G] to produce a bipartite graph B. Let Y, Z be the vertexclasses of B; we may assume that Y Z = G. Define ɛ > 0 by {y z : y Y, z Z, y z = } = (1 ɛ) l+1 ; 6

27 then clearly, we have Note that Let ɛ 9 (log log n) b + 3 (l+1) = O log n e(b) (1 ɛ) l+1. (15) α = Y / l, β = Z / l. ( ) (log log n) b = o(1). log n By assumption, α + β 3 (l 1) < 3. Since Y Z e(b) ( ɛ) l, we have αβ ɛ. This implies that 1 ɛ < α, β < + ɛ. (16) (To see this, simply observe that to maximize αβ subject to the conditions α 1 ɛ and α + β 3, it is best to take α = 1 ɛ and β = + ɛ, giving αβ = ɛ 4ɛ < ɛ, a contradiction. It follows that we must have α > 1 ɛ, so β < + ɛ; (16) follows by symmetry.) From now on, we think of X as the set [n] = {1,,..., n}. Let First, we prove the following Claim 1. W 1 W = [n]. W 1 = {i [n] : Y + i Y /3}, W = {i [n] : Z + i Z /3}. Proof. Suppose for a contradiction that W 1 W [n]. Without loss of generality, we may assume that n / W 1 W. Let θ = Y + n / Y, φ = Z + n / Z ; then we have θ, φ 1/3. Observe that the number e n of edges between Y and Z which generate a set containing n satisfies (1 ɛ) l e n (θα(1 φ)β + φβ(1 θ)α) l = (θ + φ θφ)αβ l. (17) (Here, the left-hand inequality comes from the fact that B -generates all but at most ɛ l+1 subsets of [n], and therefore B -generates at least (1 ɛ) l sets containing n.) Notice that the function f(θ, φ) = θ + φ θφ, 0 θ, φ 1/3 7

28 is a strictly increasing function of both θ and φ for 0 θ, φ 1/3, and therefore attains its maximum of 4/9 at θ = φ = 1/3. Therefore, since α + β 3, we have 1 ɛ 4 9 αβ; 3/ 3 ɛ/ α, β 3/ + 3 ɛ/. Moreover, by the AM/GM inequality, αβ 9/4, so and therefore 1 ɛ 9 f(θ, φ), (18) 4 1/3 8ɛ/3 θ, φ 1/3. Thus Y, Z = (3/ o(1)) l and θ, φ = 1/3 o(1). Therefore, we have Y + n = l 1 (1 o(1)), Z + n = l 1 (1 o(1)), Y n = l (1 + o(1)), Z n = l (1 + o(1)). Observe that Gn = Yn Zn must -generate all but at most o( l ) of the sets in P{1,,..., n 1} = P{1,,..., l}, and therefore, by Proposition 9 for k = and n even, there exists an equipartition S 1 S of {1,,..., l} such that Yn contains at least (1 o(1)) l members of PS 1, and Zn contains at least (1 o(1)) l members of PS. Define U = {y Y : y S = }, V = {z Z : z S 1 = }. Since Un = (1 o(1)) l and Vn = (1 o(1)) l, we must have Yn \ Un = o( l ), and Zn \ Vn = o( l ). Our aim is now to show that Y n + \ U n + = o( l ), and Z n + \ V n + = o( l ). Clearly, we have Un PS 1, and Vn PS, so Un l and Vn l. Moreover, each set x Y n + \ U n + contains an element of S, and therefore x {n} is disjoint from at most l 1 sets in Vn PS. Similarly, each set x Z n + \ V n + contains an element of S 1, and therefore x {n} is disjoint from at most l 1 sets in Un PS 1. It follows that e n U + n V n + Y + n \ U + n l 1 + V + n U n + Z + n \ V + n l 1 + Y n \ U n Z + n + Z n \ V n Y + n U + n l + Y + n \ U + n l 1 + V + n l + Z + n \ V + n l 1 + o( l ). 8

29 On the other hand, by (17), we have e n (1 o(1)) l. Since Y n + = l 1 (1 o(1)), and Z n + = l 1 (1 o(1)), we must have Y n + \ U n + = o( l ), and Z n + \ V n + = o( l ), as required. We may conclude that Y \ U = o( l ) and Z \ V = o( l ). Hence, there are at most o( l ) sets in Y Z = G that intersect both S 1 and S. On the other hand, since Y n + = (1 o(1)) l 1 and Z n + = (1 o(1)) l 1, there are at least (1 + o(1)) l 1 sets s 1 S 1 such that s 1 {n} / Y, and there are at least (1 + o(1)) l 1 sets s S such that s {n} / Z. Taking all pairs s 1, s gives at least (1 + o(1)) l sets of the form {n} s 1 s (s 1 S 1, s 1 {n} / Y, s S, s {n} / Z). (19) Each of these requires a set intersecting both S 1 and S to express it as a disjoint union of two sets from G. Since there are o( l ) members of G intersecting both S 1 and S, G generates at most ( G + 1)o( l ) = o( l ) sets of the form (19), a contradiction. This proves the claim. We now prove the following Claim. W 1 W =. Proof. Suppose for a contradiction that W 1 W. Without loss of generality, we may assume that n W 1 W. As before, let θ = Y + n / Y, φ = Z + n / Z ; this time, we have θ, φ 1/3. Observe that ( ɛ) l e(b) (1 θφ)αβ l. (0) Here, the left-hand inequality is (15), and the right-hand inequality comes from the fact that there are no edges between pairs of sets (y, z) Y Z such that n y z. Since 1 θφ 8/9, we have Since α + β 3, it follows that Since αβ 9/4, we have ɛ 8 9 αβ. 3 (1 ɛ) α, β 3(1 + ɛ). ɛ 9 (1 θφ), 4 9

30 and therefore Hence, we have 1/3 θ, φ 1/3 + 8ɛ/3. Y + n = l 1 (1 o(1)), Z + n = l 1 (1 o(1)), Y n = l (1 + o(1)), Z n = l (1 + o(1)), so exactly as in the proof of Claim 1, we obtain a contradiction. Claims 1 and together imply that W 1 W is a partition of {1,,..., n} = {1,,..., l + 1}. We will now show that at least a (/3 o(1))-fraction of the sets in Y are subsets of W 1, and similarly at least a (/3 o(1))-fraction of the sets in Z are subsets of W. Let σ = Y \ P(W 1), τ = Z \ P(W ). Y Z Let y Y \ PW 1, and choose i y W ; since at least Z /3 of the sets in Z contain i, y has at most Z /3 neighbours in Z. Hence, ( ɛ) l e(b) ( 3 σαβ +(1 σ)αβ)l = (1 σ/3)αβ l (1 σ/3) 9 4 l, (1) and therefore σ 1/3 + 8ɛ/3, so Y P(W 1 ) (/3 8ɛ/3) Y. () Similarly, τ 1/3 + 8ɛ/3, and therefore Z P(W ) (/3 8ɛ/3) Z. If W 1 l 1, then Y P(W 1 ) l 1, so Y l 1 l /3 8ɛ/3 = ɛ < (1 ɛ)l, contradicting (16). Hence, we must have W 1 l. Similarly, W l, so { W 1, W } = {l, l + 1}. Without loss of generality, we may assume that W 1 = l and W = l + 1. We now observe that Z (3/ 6ɛ) l (3) 30

31 To see this, suppose that Z = (3/ η) l. Since Z + Y < 3 l, we have Y (3/ + η) l. Recall that any y Y \ PW 1 has at most Z /3 neighbours in Z. Thus, we have ( ɛ) l e(b) Y PW 1 Z + Y \ PW 1 3 Z l ( 3 η)l + ( 1 + η)l 3 ( 3 η)l = ( 1 3 η 3 η ) l. Therefore η 6ɛ, i.e. Z (3/ 6ɛ) l, as claimed. Since Z + Y < 3 l, we have Y (3/ + 6ɛ) l. (4) We now prove the following Claim 3. (a) P(W 1 ) \ Y ɛ l ; (b) Z \ PW ( ɛ + ɛ) l. Proof. We prove this by constructing another bipartite subgraph B of H with the same number of vertices as B, and comparing e(b ) with e(b). First, let D = min{ P(W ) \ Z, Z \ PW }, add D new members of P(W ) \ Z to Z, and delete D members of Z \ PW, producing a new set Z and a new bipartite graph B 1 = H[Y, Z ]. Since Z = Z ( + ɛ) l, we have Z \ PW ɛ l+1, i.e. Z is almost contained within PW. Notice that every member z Z \ PW had at most Y /3 neighbours in Y, and every new member of Z has at least Y P(W 1 ) (/3 8ɛ/3) Y neighbours in Y, using (). Hence, and therefore Second, let e(b) e(b 1 ) 8ɛ 8ɛ Y D Y 16ɛ 9 Z l = 4ɛ l, e(b 1 ) e(b) ɛ l+1 (1 3ɛ) l+1. C = min{ PW 1 \ Y, Y \ PW 1 }, add C new members of P(W 1 ) \ Y to Y, and delete C members of Y \ PW 1, producing a new set Y and a new bipartite graph B = H[Y, Z ]. Since Y (1 ɛ) l, we have Y PW 1 (1 ɛ) l. Since every deleted member of Y contained an element of W, it had at most (1 + ɛ) l neighbours in Z. (Indeed, such member of Y intersects l sets in PW, so has at most l neighbours in Z PW ; there are Z \ PW ɛ l+1 other sets in Z.) On 31

32 the other hand, every new member of Y is joined to all of Z PW, which has size at least Z PW (3/ 8ɛ) l. It follows that e(b ) e(b 1 ) + C( 1 10ɛ)l (1 3ɛ) l+1 + C( 1 10ɛ)l. (5) We now show that e(b ) (1 + ɛ) l+1. If Y l, then write Y = (1 + φ) l where φ 0; Y contains all of PW 1, and φ l extra sets. We have Z ( φ) l, and therefore by (3), φ 1/ + 6ɛ < 1. Note that every extra set in Y \ PW 1 has at most l neighbors in PW, and therefore at most (1 + ɛ) l neighbours in Z. Hence, e(b ) l ( φ) l + φ l (1 + ɛ) l = (1 + φɛ) l+1 (1 + ɛ) l+1. If, on the other hand, Y l, then since Y + Z 3 l, we have e(b ) Y Z l+1. Hence, we always have Combining (5) and (6), we see that C e(b ) (1 + ɛ) l+1. (6) 8ɛ 1/ 10ɛ l 0ɛ l, provided ɛ 1/100. This implies (a). Indeed, if PW 1 \ Y C 0ɛ l, then we are done. Otherwise, by the definition of C, we have Y \ PW 1 0ɛ l. Recall that by (16), Y (1 ɛ) l, and therefore Hence, Y PW 1 = Y Y \ PW 1 (1 ɛ) l 0ɛ l = (1 ɛ) l. P(W 1 ) \ Y ɛ l, (7) proving (a). Since e(b) (1 ɛ) l+1, e(b ) (1 + ɛ) l+1, and e(b ) e(b 1 ), we have e(b 1 ) e(b) e(b ) e(b) (1 + ɛ) l+1 (1 ɛ) l+1 = ɛ l+ (8) We now use this to show that D = min{ P(W ) \ Z, Z \ PW } ɛ l. Suppose for a contradiction that D ɛ l ; then it is easy to see that there must exist z Z \ PW with at least Y /3 8 ɛ l 3

33 neighbours in Y. Indeed, suppose that every z Z \ PW has less than Y /3 8 ɛ l neighbors in Y. Recall that every new member of Z has at least (/3 8ɛ) Y neighbours in Y. Hence, e(b 1 ) e(b) > 8D( ɛ ɛ) Y 8 ɛ l ( ɛ ɛ)(1 ɛ) l ɛ l+1 since ɛ < 1/16, contradicting (8). Hence, we may choose z Z \ PW with at least Y /3 8 ɛ l neighbours in Y. Without loss of generality, we may assume that n z W 1 ; then none of these neighbours can contain n. Hence, Y contains at most Y /3 + 8 ɛ l sets containing n. But by (7), Y contains at least (1 44ɛ) l 1 of the subsets of W 1 that contain n, and therefore Y (3/ o(1)) l. By (3), it follows that Y = (3/ o(1)) l and Z = (3/+o(1)) l, so Y contains (1 o(1)) l 1 sets containing n. Hence, by (18), so does Z. As in the proof of Claim 1, we obtain a contradiction. This implies that D = min{ P(W ) \ Z, Z \ PW } ɛ l, as desired. This implies (b). Indeed, if Z \ PW ɛ l, then we are done. Otherwise, by the definition of D, P(W ) \ Z ɛ l, and therefore Since Z ( + ɛ) l, we have Z PW ( ɛ) l. Z \ PW = Z Z PW ( + ɛ) l ( ɛ) l = ( ɛ + ɛ) l, proving (b). We conclude by proving the following Claim 4. Proof. Let P(W ) \ Z 4 ɛ l. F = P(W ) \ Z be the collection of sets in PW which are missing from Z, and let E 1 = Y \ PW 1 33

34 be the set of extra members of Y. Since G is a -generator for X, we can express all F l sets of the form w 1 f (w 1 W 1, f F ) as a disjoint union of two sets in G. All but at most ɛ l+1 of these unions correspond to edges of B. Since Z \ PW ( ɛ + ɛ) l, there are at most ( ɛ + ɛ) l Y edges of B meeting sets in Z \ PW. Call these edges of B bad, and the rest of the edges of B good. Fix f F ; we can express all l sets of the form w 1 f (w 1 W 1 ) as a disjoint union of two sets in G. If w 1 f is represented by a good edge, then we may write w 1 f = y 1 w where y 1 E 1 with y 1 W 1 = w 1, and w W, so for every such w 1, there is a different y 1 E 1. By (4), Y (3/ + 6ɛ) l, and by (7), Y PW 1 (1 ɛ) l, so E 1 = Y P(W 1 ) Y (3/ + 6ɛ) l (1 ɛ) l = (1/ + 8ɛ) l. Thus, for any f F, at most (1/ + 8ɛ) l unions of the form w 1 f correspond to good edges of B. All the other unions are generated by bad edges of B or are not generated by B at all, so (1/ 8ɛ) l F (ɛ + ɛ) l Y + ɛ l+1. Since Y (3/ + 6ɛ) l and ɛ is small, F 4 ɛ l, as required. We now know that Y contains all but at most o( l ) of PW 1, and Z contains all but at most o( l ) of PW. Since Y + Z < 3 l, we may conclude that Y = (1 o(1)) l and Z = ( o(1)) l. It follows from Proposition 16 that provided n is sufficiently large, we must have G = P(W 1 ) P(W ) \ { }, completing the proof of Theorem 4. 4 Conclusion We have been unable to prove Conjecture 1 for k 3 and all sufficiently large n. Recall that if G is a k-generator for an n-element set X, then G n/k. 34

35 In view of Proposition 18, it is natural to ask whether for any fixed k, all induced subgraphs of the Kneser graph H with Ω( n/k ) vertices can be made k-partite by removing at most o( n/k ) edges. This is false for k = 3, however, as the following example shows. Let n be a multiple of 6, and take an equipartition of [n] into 6 sets T 1,..., T 6 of size n/6. Let A = (T i T j ); {i,j} [6] () then A = 15( n/3 ), and H[A] contains a n/3 -blow-up of the Kneser graph K(6, ), which has chromatic number 4. It is easy to see that H[A] requires the removal of at least n/3 edges to make it tripartite. Hence, a different argument to that in Section 3 will be required. We believe Conjecture 1 to be true for all n and k, but it would seem that different techniques will be required to prove this. References [1] Alon, N., Frankl, P., The maximum number of disjoint pairs in a family of subsets, Graphs and Combinatorics 1 (1985), pp [] Alon, N., Krivelevich, M., Testing k-colorability, SIAM J. Discrete Math. 15 (00), Issue, pp [3] Bollobás, B., Erdős, P., Simonovits, M., Szemerédi, E., Extremal graphs without large forbidden subgraphs, Annals of Discrete Mathematics 3 (1978) pp [4] Chung, F.K.R., Frankl, P., Graham, R.L., Shearer, J.B., Some intersection theorems for ordered sets and graphs, J. Combin. Theory, Ser. A 43 (1986), pp [5] Ellis, D., Note on generating all subsets of a finite set with disjoint unions, Electron. J. Combin. 16, (009) no 1., Note 16. [6] Erdős, P., On extremal problems of graphs and generalized graphs, Israel J. Math. (1964), pp [7] Erdős, P., On the number of complete subgraphs contained in certain graphs, Publ. Math. Inst. Hung. Acad. Sci., Ser. A 7 (196), pp

36 [8] Erdős, P., Frankl, P., Rődl, V., The asymptotic number of graphs not containing a fixed subgraph and a problem for hypergraphs having no exponent, Graphs and Combinatorics (1986), pp [9] Erdős, P., Simonovits, M., Supersaturated graphs and hypergraphs, Combinatorica 3 (1983), [10] Frein, Y., Lévêque, B., Sebő, A., Generating all sets with bounded unions, Combin. Probab. Comput. 17 (008) pp [11] Füredi, Z., Katona, G.O.H., -bases of quadruples, Combin. Probab. Comput. 15 (006) pp [1] Goldreich, O., Goldwasser, S., and Ron, D., Property testing and its connection to learning and approximation, Proc. 37th Annual IEEE FOCS (1996), pp David Ellis D.Ellis@dpmms.cam.ac.uk Benny Sudakov b.sudakov@math.ucla.edu 36