THE KATONA THEOREM FOR VECTOR SPACES

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1 THE KATONA THEOREM FOR ECTOR SPACES PETER FRANKL AND NORIHIDE TOKUSHIGE Dedicated to Gyla O H Katoa o his 70th birthday Abstract e preset a vector space versio of Katoa s t-itersectio theorem 12 Let be the -dimesioal vector space over a fiite field, ad let F be a family of sbspaces of Sppose that dim(f F t holds for all F, F F The we show that F k=d k for + t = 2d, ad F k=d+1 k + 1 d for + t = 2d + 1 e also cosider the case whe the coditio dim(f F t is replaced with dim(f F t 1 1 Itrodctio I 1964, Katoa pblished his t-itersectio theorem 12, which is oe of the most basic reslts i extremal set theory It has bee exteded i may ways, oe of them beig a reslt cocerig a set-system avoidig jst oe itersectio de to Frakl ad Füredi 6 I this article, we show vector space versios of these reslts sig the liear algebra method e begi by recallig the Katoa s origial theorem Let X = {1, 2,, } ad let ( X k deote the set of all k-elemet sbsets of X Let P(X = ( X k be the power set of X e say that a family of sbsets F P(X is t-itersectig if F F t holds for all F, F F Let s defie a t-itersectig family K(, t of sbsets as follows For + t = 2d, let K(, t = ( X k=d k For + t = 2d + 1, choose a ( 1-elemet sbset Y X, ad set K(, t = ( ( X ( k=d+1 k Y d The Katoa s t-itersectio theorem states the followig Theorem 1 (12 Let 1 t ad let F P(X be t-itersectig The F K(, t Moreover if t > 1 the eqality holds iff F is isomorphic to K(, t For a family of sbsets F of X ad 0 we defie the -th shadow (F of F by (F = {G ( X : G F for some F F} The followig reslt is a key tool for the origial proof of Theorem 1 Date: November 13, 2013, 02:24pm 2000 Mathematics Sbject Classificatio Primary: 05D05 Secodary: 05B20 Key words ad phrases Katoa s t-itersectio theorem; avoidig jst oe itersectio; iclsio matrix; liear algebra methods The secod athor was spported by JSPS KAKENHI ad

2 2 P FRANKL AND N TOKUSHIGE Theorem 2 (12 Let 1 t k ad let F ( X k be t-itersectig The, for k t k, we have ( 2k t (F F ( 2k t Now we preset vector space versios of the above theorems Fix the q-elemet field F q ad let be the -dimesioal vector space over this field Let k deote the set of all k-dimesioal sbspaces of, let k = k = k 1 q i 1 i=0, ad let q k i 1 L( = k be the lattice of sbspaces of with respect to iclsio e say that a family of sbspaces F L( is t-itersectig if dim(f F t holds for all F, F F For 0 we defie the -th shadow F of F by F = {G : G F for some F F} The the correspodig reslt to Theorem 2 is as follows Theorem 3 Let 1 t k ad let F k be t-itersectig The, for k t k, we have 2k t F F k 2k t Let s defie a t-itersectig family K, t of sbspaces as follows For + t = 2d, let K, t = k=d k For + t = 2d + 1, choose a ( 1-dimesioal sbspace, ad set K, t = ( k=d+1 k d Usig Theorem 3 we will obtai the followig vector space versio of the Katoa s theorem Theorem 4 Let 1 t ad let F L( be t-itersectig The F K, t Moreover if t > 1 the eqality holds iff F is isomorphic to K, t e say that a family of sbsets F P(X is (t 1-avoidig if F F = t 1 holds for all distict F, F F Notice that if F is t-itersectig the it is (t 1- avoidig I 1975, Erdős 4 asked what happes if i Theorem 1 we weake the coditio t-itersectig to (t 1-avoidig Defie a (t 1-avoidig family K (, t 1 of sbsets of X by K (, t 1 = K(, t ( X k<t 1 k I 5, Frakl cojectred that this costrctio gives the maximm possible size for > 0 (t, ad he proved this for the case t = 2 (1-avoidig families for all This cojectre was solved by Frakl ad Füredi i sig the so-called liear algebra method e preset the correspodig vector space versio To state or reslt, we eed some defiitios e say that a family of sbspaces F L( is (t 1-avoidig if dim(f F t 1 holds for all distict F, F F Defie a (t 1-avoidig family K, t 1 of sbspaces of by K, t 1 = K, t k<t 1 k Theorem 5 Let t 1, > 0 (t, ad let F L( be (t 1-avoidig The F K, t 1 Moreover if t > 1 the eqality holds iff F is isomorphic to K, t 1 k

3 THE KATONA THEOREM FOR ECTOR SPACES 3 Sice a t-itersectig family is always a (t 1-avoidig family, the followig reslt is a obvios extesio of Theorem 3 (for the case k 2t 1, which will be sed to prove Theorem 5 Theorem 6 Let t 1, k 2t 1, ad let F k be (t 1-avoidig The, for k t k, we have 2k t F F 2k t I 6 the correspodig set-system versio of Theorem 6 is cojectred to be tre bt it is proved oly der the assmptio of k > k 0 (t This is becase the proof relies o a reslt of Frakl ad Sighi 10 statig that every k-iform, (t 1- avoidig family of sbsets is (k t-idepedet, provided k > k 0 (t (e will defie (k t-idepedece i Sectio 2 This proof, i tr, ses a divisibility property of itegers which reqires k > k 0 (t O the other had, we will se some basic properties of the cyclotomic polyomials to show that every k-iform, (t 1- avoidig family of sbspaces is (k t-idepedet provided k 2t 1 (Lemma 5 I this sese, Theorem 6 is a example where a vector space versio of a theorem has a stroger reslt tha a set-system versio, with a simpler proof Fially we metio the maximm size of k-iform, (t 1-avoidig families As for the case k 2t 1, we oly have the followig weaker bod, which is stated i 8 withot a proof (I 8 they claimed that Theorem 7 follows from their Theorem 11, bt this is tre oly for t-itersectig families Theorem 7 Let t 1, k 2t 1, ad let F k be (t 1-avoidig The F k t Frakl ad Graham 8 cojectre that if k 2t the the pper bod ca be improved to t k t (Theorem 7 for the case k = 2t 1 is almost sharp as described ( below O the other had, Frakl ad Füredi 7 obtaied the sharp pper bod t k t for the correspodig set-system versio, provided k 2t ad > 0 (k The proof techiqe sed i 7 is more combiatorial, ad differet from that i 6 For the case k 2t 1 we will derive the followig reslt from Theorem 7 Theorem 8 Let t 1, 2t 1 k > t 1, k, ad let F k be (t 1-avoidig The F 2k t t 1 k 2k t t 1 Theorem 8 is asymptotically tight as for fixed t, k e show the tightess (Theorem 9 i Sectio 4 sig a packig reslt of Rödl 14 e will se the liear algebra method to prove or reslts The proofs are similar to those i 6, bt we will follow the formlatio i the Babai Frakl book 2 The key idea is a idepedece of row vectors of the iclsio matrix This idea was already sed by Frakl ad Graham i 8, ad we cold se their reslts bt we choose to give direct ad elemetary proofs for self-completeess This paper is orgaized as follows I Sectio 2 we prepare some basic tools for the liear algebra method, ad prove Theorem 3 ad Theorem 4 (the Katoa theorem for vector spaces The i Sectio 3 we cosider families avoidig jst oe itersectio, k

4 4 P FRANKL AND N TOKUSHIGE ad prove Theorem 5 ad Theorem 6 I Sectio 4 we focs o iform families ad prove Theorem 7 ad Theorem 8 2 The Katoa theorem for vector spaces I this sectio, we prepare some basic tools for the liear algebra method, ad prove Theorem 3 ad Theorem 4 Let be the -dimesioal vector space over F q For 0 i k, F k, ad G i, defie the iclsio matrix M(F, G as follows This is a F G matrix whose (F, G-etry m(f, G, where F F ad G G, is defied by { 1 if F G, m(f, G = 0 if F G For F k ad 0 j i k, simple cotig yields M(F, i M( i, j = k j i j M(F, j (1 I fact, the (F, J-etry of (1, where F F ad J j, cots #{I i : J I F } I particlar, (1 shows the followig Lemma 1 Let 0 j i k ad F k The colsp M(F, j is cotaied i colsp M(F, i, where colsp M deotes the colm space of M over Q e say that F k is s-idepedet if the rows of M(F, s are liearly idepedet over Q, that is, the iclsio matrix has fll row-rak I this case, F s immediately follows Lemma 2 (8 Let 0 s k ad let F k be s-idepedet The k + s k + s F F (2 k Proof Let A B = deote the direct sm, that is, A B = {0} ad spa{a, B} = For each lie x 1 choose = x 1 so that x = Let F x = {G k 1 : x G F} k 1 Claim 1 F x k 1 is s-idepedet, that is, rak M(Fx, s = Fx e postpoe the proof of Claim 1, ad we first prove the lemma by idctio o k assmig Claim 1 Ieqality (2 trivially holds for the followig three cases: s = 0, = s, ad = k So let 1 s < < k ad assme that (2 is tre for k 1 By Claim 1 we ca apply the idctio hypothesis to F x k 1, ad we get 1 F x F x (k 1+s 1 (k 1+s (k 1 (3

5 THE KATONA THEOREM FOR ECTOR SPACES 5 By cotig #{(x, F 1 F : x F } i two ways, amely, by cotig the mber of edges i the correspodig bipartite graph from each side, we have F x = k 1 F (4 x 1 Similarly by cotig #{(x, G 1 F : x G}, we have 1 F x = 1 F (5 x 1 Usig (5, (3, ad (4, we get F (5 = 1 1 F x (3 1 1 x 1 x (4 = 1 1 k k 1+s F 1 = F 1 k 1+s k 1 F x k+s k+s k k 1+s 1 k 1+s k 1 This shows that (2 is tre for k as well, ad completes the idctio So all we eed is to prove Claim 1 Fix x 1 ad let 1 be sch that x = Divide s ito two parts s = C D, where C is the set of s-dimesioal sbspaces of ot cotaiig x, ad the remaiig part is D = {x T : T s 1 } (The C = q s 1 s ad D = 1 s 1 Let F x = {F F : x F } k e divide the colms of M(F x, s ito two blocks: M(F x, s = ( M(F x, C M(F x, D (6 A sbspace S s ca be represeted by a s matrix i redced echelo form with o zero rows (see, eg, 3, ad let ref(s deote the matrix e ca associate D with matrices for which leadig 1 i the last row occrs i the last colm The there is a atral bijectio from D to s 1 by takig the (s 1 ( 1 pricipal mior of ref(s Ths we may assme that M(F x, D = M(F x, s 1 This together with Lemma 1 gives colsp M(F x, D = colsp M(F x, s 1 colsp M(F x, s (7 If S C the the s ( 1 pricipal mior of ref(s determies a sbspace i s This gives a map φ : C s, ad for each S s we have φ 1 (S = q s becase φ(s = φ(s iff ref(s ad ref(s differ oly i the last colm Ths colms correspodig to S ad S i M(F x, C are the same iff φ(s = φ(s, ad M(F x, C ca be viewed as q s copies of M(F x, s Hece we have colsp M(F x, C = colsp M(F x, s (8

6 6 P FRANKL AND N TOKUSHIGE By (7 ad (8 with (6, it follows colsp M(F x, s colsp M(F x, s, ad rak M(F x, s rak M(F x, s (9 The opposite ieqality is trivial, ths we have eqality i (9 O the other had, sice F is s-idepedet, F x is also s-idepedet ad rak M(F x, s = F x Ths eqality i (9 yields that rak M(F x, s = F x Fially, otig that F x = F x ad M(F x, s = M(Fx, s, we have rak M(Fx, s = Fx as eeded This completes the proof of Claim 1 ad Lemma 2 Lemma 3 (8 Let 1 t k ad let F k be t-itersectig The F is (k t-idepedet Proof Let f(x = t i<k x i = 1 t i<k q x i 1 q 1 (10 By settig y = q x, we ca rewrite f(x as a polyomial g(y of degree k t i Qy, that is, f(x = g(y = q i y 1 q 1 Let ϕ s (y = s 1 i=0 t i<k q i y 1 q s i 1 The ϕ 0(y,, ϕ k t (y form a basis of the vector space (over Q of polyomials of degree k t with variable y Ths we ca determie a 0, a 1,, a k t Q iqely so that k t g(y = a s ϕ s (y s=0 I other words, otig that ϕ s (y = s 1 q x i 1 i=0 = x q s i 1 s, we ca determie a0,, a k t so that k t f(x = a x s s (11 Now defie a F F matrix A by s=0 For F, F F, the (F, F -etry of A is s=0 s=0 k t A = a s M(F, s M(F, s T k t a s #{ s : F F } = k t s=0 a dim(f F s s This eqals f(dim(f F by (11 Moreover, sig the t-itersectig property with (10, we have { f(dim(f F 0 if F F, = f(k 0 if F = F

7 THE KATONA THEOREM FOR ECTOR SPACES 7 Ths A is a diagoal matrix with o zero diagoal etries, ad rak A = F O the other had, it follows from Lemma 1 that the colsp M(F, s is cotaied i colsp M(F, k t for 0 s < k t, ad so colsp A is cotaied i colsp M(F, k t This gives rak M(F, k t rak A = F Ths M(F, k t has fll row-rak, amely, F is (k t-idepedet Proof of Theorem 3 By Lemma 3, F is (k t-idepedet So lettig s = k t i Lemma 2, we get the desired ieqality Proof of Theorem 4 e start with the followig simple cotig fact Claim 2 Let A a The #{B a : A B = } = q a( a Proof e may assme that ref(a = ( O I a The A B = gives ref(b = ( I a, ad there are q a( a ways for choosig the part Let G = G(a, a be a bipartite graph with the vertex partitio (G = a a ad the edge set E(G = {(A, B : A B = } The, by Claim 2, this is a q a( a -reglar graph For a vertex sbset A a, let N G (A = {B a : (A, B E(G for some A A} deote the eighborhood of A e cot the mber of edges betwee A ad N G (A i two ways The this mber is exactly q a( a A o oe had, ad at most q a( a N G (A o the other had Namely, we have the Hall coditio: Lemma 4 There is a bijectio ψ : a A N G (A Ths the bipartite graph has a perfect matchig, which ca be stated as follows a sch that A ψ(a = holds for all A a e will se ψ(a as a complemet of A here, ad also i the proof of Theorem 8 later (Notice that the orthogoal space A does ot ecessarily satisfy A A = The athors thak oe of the referees for otifyig this fact Let F k = F k, fk = F k, d = ( + t2, ad a = k t + 1 Claim 3 a F k + F a a for t k d Proof Let G a F k ad G F F k Usig Lemma 4 let H = ψ(g a Sice = G H F + H we have = dim(f + H ad dim(f H = dim F + dim H dim(f + H = k + ( a = t 1 The it follows from the t-itersectig property of F that H = ψ(g F a This gives ψ( a F k F a =, ad a F k + F a = ψ( a F k + F a a as desired

8 8 P FRANKL AND N TOKUSHIGE e otice for later se i the proof of Theorem 5 i Sectio 3 that the proof above did ot se the fll t-itersectig property, bt oly the (t 1-avoidig property Let t k < d Applyig Theorem 3 with = a = k t + 1, we have 2k t a F k F k a 2k t k = qk 1 q a 1 f k Sice qk 1 1 iff a k, that is, t 1, it follows that q a 1 af k f k with eqality holdig iff F k = or t = 1 The we ca ifer from Claim 3 that f k + f a a F k + F a (12 a (This is tre for k = d as well bt we will ot se this case Moreover if t > 1 the f k + f a = a iff fk = 0 ad f a = a where t k < d First cosider the case + t = 2d Applyig (12 for k = t, t + 1,, d 1 we have F = 1 f k = f + k=t f k = f + ( (f t + f 1 + (f t+1 + f (f d 1 + f d ( = K, t 1 2 d If t > 1 the eqality holds iff f k = 0 for 0 k < d ad f k = k for d k, amely, F is isomorphic to K, t Next cosider the case + t 1 = 2d Sice F d d is a t-itersectig family with 2d t < < 2d, we ca se a reslt i 11 to get 2d t 1 f d = F d = d d (13 Moreover if t > 1 the eqality holds iff F d = d for some ( 1-dimesioal sbspace Usig (12 for k = t, t + 1,, d 1 ad sig (13 for k = d, we have F = 1 f k = f + f k + f d k=t = f + ( (f t + f 1 + (f t+1 + f (f d 1 + f d+1 + f d ( = K, t 1 2 d + 1 d If t > 1 the eqality holds iff f k = 0 for 0 k < d, f k = k for d + 1 k, ad F d = d for some ( 1-dimesioal sbspace, amely, F is isomorphic to K, t This completes the proof of Theorem 4

9 THE KATONA THEOREM FOR ECTOR SPACES 9 3 Avoidig jst oe itersectio I this sectio we prove Theorem 5 ad Theorem 6 Lemma 5 Let t 1, k 2t 1, ad let F k be (t 1-avoidig The F is (k t-idepedet Proof e proceed as i Lemma 3 bt sig a differet f(x, that is, f(x = (qk 1 q x (q k 2 q x (q t q x (q k t 1(q k t 1 1 (q 1 (14 As we did i the proof of Lemma 3 we ca write f(x = k t s=0 a s x s for some a 0,, a k t Q Defie a F F matrix A by k t A = a s M(F, s M(F, s T The, for F, F F, the (F, F -etry of A is k t s=0 a s s By (14, we have f(x = 0 for x = t, t + 1,, k 1, ad f(k = k t j=1 s=0 q k j q k q j 1 = k t j=1 q k j (1 q j q j 1 dim(f F = f(dim(f F = ( 1 k t q (k 1+(k 2+ +t = ( 1 k t q (k t(k t+12 (15 For the remaiig vales except t 1, amely, for x = 0, 1,, t 2, we have f(x = q x(k t k x 1 k t ad k x 1 k x 1 = = (qk x 1 1 (q k t+1 1 Zq (16 k t t x 1 (q t x 1 1 (q 1 Recall that q 1 = j Φ j(q, where Φ j (q Zq is the j-th cyclotomic polyomial Let s look at the RHS of (16 The merator cotais Φ k t+1 (q as a factor comig from q k t+1 1 O the other had, j = t x 1 is the maximm j sch that Φ j (q appears i the deomiator as a factor Usig x 0 ad k 2t 1 we have t x 1 t 1 < k t + 1 So Φ k t+1 (q does ot appear i the deomiator Sice cyclotomic polyomials are pairwise relatively prime, it follows from (16 that Φ k t+1 (q divides k x 1 k t, amely, Φ k t+1 (q f(x for x = 0, 1,, t 2 Bt Φ k t+1 (q does ot divide f(k i Zq by (15 Note also that f(t 1 ever appears i A becase of the (t 1-avoidig property Coseqetly it follows that f(dim(f F Zq ad { Φ k t+1 (q f(dim(f F if F F, Φ k t+1 (q f(dim(f F if F = F

10 10 P FRANKL AND N TOKUSHIGE This meas that A is a diagoal matrix with o zero diagoal etries i the reside rig Zq(Φ k t+1 (q, ad ths rak A = F O the other had, it follows from (1 ad the defiitio of A that colsp A colsp M(F, k t Therefore we have F = rak A rak M(F, k t F Ths rak M(F, k t = F, amely, F is (k t-idepedet Proof of Theorem 6 This follows from Lemma 2 ad Lemma 5 Proof of Theorem 5 Let F k = F k, fk = F k, d = ( + t2, ad a = k t + 1 Let t k < d By Theorem 6, we have 2k t a F k F k a 2k t k = qk 1 q a 1 f k, ad so a F k f k with eqality holdig iff F k = or t = 1 The we ca ifer from Claim 3 (see the otice right after the proof of Claim 3 that f k + f a qk 1 q a 1 f k + f a (17 a Moreover if t > 1 the f k +f a = a iff fk = 0 ad f a = a where t k < d For k < t we will se a trivial pper bod f k k Case 1 + t = 2d e have 1 t 1 F = f k = f + f k + k=t f k = f + ( (f t + f 1 + (f t+1 + f (f d 1 + f d t 1 + f k (18 First sppose that f t 1 = 0 The applyig (17 for k = t, t + 1,, d 1 we have ( t 2 F = K, t d k If t > 1 the eqality holds iff f k = k for 0 k < t 1, fk = 0 for t 1 k < d ad f k = k for d k, amely, F is isomorphic to K, t 1 Next sppose that f t 1 0, that is, there is a F 0 F t 1 Sice F is (t 1- avoidig, o sbspace cotaiig F 0 ca be a member of F, which implies that (t 1 f k k k (t 1 for k t I particlar we have f d N M, where N = d, M = (t 1 d (t 1 Settig k = d 1 i (17 we have αf d 1 + f d N,

11 THE KATONA THEOREM FOR ECTOR SPACES 11 where α = qd 1 1 q d t 1 1 So f d 1 1 α (N f d Ths we have Hece we have f d 1 + f d 1 α (N f d + f d = 1 α N + (1 1 α f d f t 1 + f d 1 + f d 1 α N + (1 1 α (N M = N (1 1 α M + t 1 qd 1 q d t d q d 1 1 (t 1 d (t 1 The RHS is less tha d for > 0 (t Usig this with (17 for k = t, t +1,, d 2 we ca ifer from (18 that F < K, t 1 Case 2 + t 1 = 2d If F, F d the dim(f F t 1 Sice F is (t 1-avoidig, F d d is actally t-itersectig So we ca se a reslt i 11 to get 1 f d = F d d (19 Moreover if t > 1 the eqality holds iff F d = d for some ( 1-dimesioal sbspace rite F as follows: F = 1 t 1 f k = f + f k + k=t f k = f + ( (f t + f 1 + (f t+1 + f (f d 1 + f d+1 t 2 + f d + f t 1 + f k First sppose that f t 1 = 0 e se (17 for k = t, t + 1,, d 1, (19 for k = d, ad f k k for the remaiig k I this way we get F K, t 1 Moreover if t > 1 the eqality holds iff f k = k for 0 k < t 1, fk = 0 for t 1 k < d, f k = k for d + 1 k, ad Fd = d for some ( 1-dimesioal sbspace, amely, F is isomorphic to K, t 1 Next sppose that f t 1 0 The we ca arge as i Case 1 to coclde that f t 1 + f d 1 + f d+1 < d+1 for > 0 (t, which gives F < K, t 1 This completes the proof of Theorem 5 If we chage the defiitio of a (t 1-avoidig family F so that dim(f F t 1 is reqired for all F, F F, the f t 1 = 0 follows from this ew defiitio I this case the above proof shows that Theorem 5 holds withot assmig > 0 (t 4 Uiform families I this sectio we prove Theorem 7 ad Theorem 8 The we will show that Theorem 8 is asymptotically sharp sig a packig reslt of Rödl Proof of Theorem 7 This is a direct coseqece of Lemma 5

12 12 P FRANKL AND N TOKUSHIGE I the rest of this sectio we follow the proof i 7 Recall the bijectio ψ from Lemma 4 Proof of Theorem 8 Let b = 2t k 1 For B b let The we have F(B := {C ψ(b k b : B C F} B b F(B = k b F (20 Let k = k b = 2k 2t+1 ad t 1 = (t 1 b = k t The F(B is a k-iform, ( t 1-avoidig family with k = 2 t 1 (I fact if there are C 1, C 2 F(B sch that dim(c 1 C 2 = t 1, the F i = B C i F (i = 1, 2 bt dim(f 1 F 2 = b + ( t 1 = t 1, a cotradictio Ths, by Theorem 7, we have F(B b k t Now it follows from (20 ad (21 that as eeded F b b k t k b = = b k t (21 2k t t 1 k 2k t t 1 The bod i Theorem 8 is asymptotically sharp Namely, we have the followig Theorem 9 Let t 1 ad k > t 1 be fixed The for every ϵ > 0 there is a 0 sch that for all > 0 ad = F q there is a (t 1-avoidig family F k with F > (1 ϵ 2k t t 1 k 2k t t 1 To prove Theorem 9 we eed the followig variat of the packig theorem of Rödl 14 Theorem 10 Let r ad s be fixed The for every ϵ > 0 there is a 0 sch that for all > 0 ad = F q there is a family H r which satisfies dim(h H < s for all H, H H ad H > (1 ϵ s r s Proof of Theorem 9 By Theorem 10 we ca take a family S 2k t with dim(s S < t 1 for all S, S S ad S t 1 2k t t 1 as Let F = k (S The F is (t 1-avoidig ad F = 2k t k S becase k > t 1 Ths F satisfies the desired properties Fially we remark that Theorem 10 is derived from the followig reslt statig that almost reglar hypergraphs have almost perfect matchigs This reslt was origially obtaied by Frakl ad Rödl 9 ad we se a stroger versio give by Pippeger (see 1 or 13 Theorem 11 (9, 1, 13 Let F ( X k satisfy the followig (1 There is D sch that #{F F : x F } = D for all x X (2 For all {x, y} ( X 2, #{F F : {x, y} F } = o(d as D The there exist pairwise disjoit F 1,, F m F with m X k (as D ad hece X

13 THE KATONA THEOREM FOR ECTOR SPACES 13 Proof of Theorem 10 Let X = s ad k = r s Defie F := { R s : R ( r } X k The F is D-reglar, where D = s r s Moreover, for a pair {x, y} X, we have #{F F : {x, y} F } s 1 r s 1 = o(d D = q r s 1 I fact if for fixed r ad s, the D ad s 1 0, r s 1 q s 1 amely, s 1 r s 1 = o(d Ths, by Theorem 11, we have a matchig F1,, F m F with m s r s For 1 i m we ca write Fi = R i s The H := {R1,, R m } satisfies the desired properties of Theorem 10 r Ackowledgemet The athors thak the referees for their carefl readig ad may helpfl commets Refereces 1 N Alo, J Specer The probabilistic method (secod editio A iley-itersciece pblicatio, L Babai, P Frakl Liear Algebra Methods i Combiatorics ith Applicatios to Geometry ad Compter Sciece Prelimiary ersio 2 (September P J Camero Combiatorics: topics, techiqes, algorithms Cambridge Uiversity Press, P Erdős Problems ad reslts i graph theory ad combiatorial aalysis Proceedigs of the Fifth British Combiatorial Coferece (Uiv Aberdee, Aberdee, 1975, pp Cogresss Nmeratim, No X, Utilitas Math, iipeg, Ma, P Frakl A itersectio problem for fiite sets Acta Math Acad Sci Hgar 30 ( P Frakl, Z Füredi O hypergraphs withot two edges itersectig i a give mber of vertices J Combi Ser A, 36 ( P Frakl, Z Füredi Forbiddig jst oe itersectio J Combi Ser A, 39 ( P Frakl, R L Graham Itersectio theorems for vector spaces Eropea J Combi 6 ( P Frakl, Rödl Near perfect coverigs i graphs ad hypergraphs Eropea J Combi 6 ( P Frakl, N M Sighi Liear depedecies amog sbsets of a fiite set Erop J Combi, 4 ( P Frakl, R M ilso The Erdős Ko Rado theorem for vector spaces J Combi Ser A, 43 ( G O H Katoa Itersectio theorems for systems of fiite sets Acta Math Acad Sci Hg 15 ( N Pippeger, J Specer Asymptotic behavior of the chromatic idex for hypergraphs J Combi Ser A, 51 ( Rödl O a packig ad coverig problem Eropea J Combi 6 ( Shibya, Shibya-k, Tokyo , Japa address: peterfrakl@gmailcom College of Edcatio, Ryky Uiversity, Nishihara, Okiawa , Japa address: hide@ed-rykyacjp