EKR Sets for Large n and r

Transcription

1 EKR Set fo Lage n and The MIT Faculty ha made thi aticle openly available. Pleae hae how thi acce benefit you. You toy matte. Citation A Publihed Publihe Bond, Benjamin. "EKR Set fo Lage n and." Gaph and Combinatoic (206) http//dx.doi.og/0.007/ x Spinge Japan Veion Autho' final manucipt Acceed Mon Ja EST 209 Citable Link Tem of Ue Detailed Tem http//hdl.handle.net/72./020 Aticle i made available in accodance with the publihe' policy and may be ubject to US copyight law. Pleae efe to the publihe' ite fo tem of ue.

2 EKR et fo lage n and Benjamin Bond Maachuett Intitute of Technology, 77 Ma. Ave. Cambidge, MA, 0239 Abtact. Let A be a compeed, inteecting family and let X. Let A(X) = fa 2 A A \ X 6= ;g and S n; = (fg). Motivated by the Ed}o-Ko-Rado theoem, Bog aked fo which X [2; n] do we have ja(x)j js n;(x)j fo all compeed, inteecting familie A? We call X that atify thi popety EKR. Bog claied EKR et X uch that jxj. Babe claied X, with jxj, uch that X i EKR fo uciently lage n, and aked how lage n mut be. We pove n i uciently lage when n gow quadatically in. In the cae whee A ha a maximal element, we hapen thi bound to n > ' 2 implie ja(x)j js n;(x)j. We conclude by giving a geneating function that peed up computation of ja(x)j in compaion with the nave method. Key wod. Ed}o-Ko-Rado Theoem, Inteecting Family, Compeed Family. Intoduction The main object of tudy in thi pape ae compeed, inteecting familie. We begin by dening thee tem. Let denote the et of element ubet of = f; ; ng. We label element of in inceaing ode i.e. fo B = fb ; ; b g 2 we have bi < b i+. A family A i a ubet A. Call A inteecting if B; C 2 A implie B \ C 6= ;. Notice that A i tivially inteecting if n < 2 by the pigeonhole pinciple. We dene a patial ode known a the compeion ode on, a follow. Fo A = fa ; ; a g, and B = fb ; ; b g, dene A B if a i b i fo all i. A family A i compeed if A 2 A implie B 2 A fo B A. We extend thi patial ode to 2 fo C = fc ; ; c k g, we ay A C if k and a i c i fo i k. Fo example, thee ae + n 3 element A 2 with A f; 3g, ince the t two element mut be f; 2g o f; 3g. Let A be a family and X. Dene A(X) = fa 2 A A \ X 6= ;g. An impotant example of uch a family i S n;, dened by S n; = (fg) = fa 2 2 Ag If n and ae clea, then S n; will be implied to S. It i eay to check that S i compeed and inteecting. The following theoem i one of the fundamental eult about inteecting familie. Theoem (Ed}o-Ko-Rado)[4] (See alo [6]) Let n 2 and let A Then jaj jsj. be an inteecting family. In [3], Bog conideed a vaiant of the Ed}o-Ko-Rado theoem. Bog aked which et X [2; n] = f2; 3; ; ng have the popety that ja(x)j js(x)j fo all compeed, inteecting familie A. Call X with thi popety EKR. We aume X [2; n], becaue if 2 X, then S(X) = S and X i tivially EKR by the Ed}o-Ko-Rado theoem. Thee ae many X which ae not EKR. Fo example, conide the Hilton-Milne family N = S([2; + ]) [ f[2; + ]g, ee [7]. Then fo X = [2; + ], we have jn (X)j = js(x)j +.

3 2 Benjamin Bond The motivation fo conideing compeed familie i twofold. Fitly, the quetion i uninteeting without the equiement that A be compeed, ince fo any x 2 X, we have A = (fxg) maximize ja(x)j fo inteecting familie A. Secondly, abitay et lack tuctue, and by impoing moe condition, we may gain moe infomation. In fact, compeed familie and the hifting technique (ee [5] fo a uvey) ae poweful technique in extemal et theoy and can be ued to give a imple poof of the Ed}o-Ko-Rado theoem. In [3], Bog claied X that ae EKR fo jxj and gave a patial olution in the cae jxj <. Babe continued with Bog' wok in [2] by conideing jxj. To decibe hi eult, we intoduce the notion of eventually EKR et, which ae nite et X Z2 uch that fo xed, the et X i EKR fo uciently lage n. Notice that X doe not vay with o n, but whethe o not X i EKR may depend on and n. Theoem 2 (Babe) Let 3, n 2 and X [2; n] with jxj. If X 6 [2; + ], then X i eventually EKR if and only if one of the following hold. jxj = 2. jxj = 2 and f2; 3g \ X = ; 3. jxj = 3 and f2; 3g 6 X 4. jxj 4 Babe aked which n ae uciently lage to imply X i EKR. Thi pape povide bound on n. Baed on numeical eult fo mall n and tated in [2], Babe peculated that n wa ucient to imply X i EKR. Howeve, a will be een in Sectio, thi bound doe not hold in geneal. We eplace the uggeted bound of n with the following conjectue, which i uppoted by compute evidence fo 5. Conjectue 3 Let 2 and X [2; n] be eventually EKR. Then n > ' 2 implie X i EKR, whee ' = +p 5 2. Notice that in the cae = 2, thee ae only two maximal compeed, inteecting familie, namely S and the family ff; 2g; f; 3g; f2; 3gg, o the eult i eay in thi cae, and we aume 3. In ode to decibe eult towad Conjectue 3, we need the notion of geneating et. Thee wee intoduced by Ahlwede and Khachatian in [], and Babe conideed a vaiant denition, which i moe ueful in the peent context. Let G 2 and dene F (n; ; G) = fa 2 A G fo ome G 2 Gg Obeve that uch familie ae natually compeed. We may now tate the main theoem of thi pape. Theoem 4 Let 3 and let X be eventually EKR with jxj. Fo each c > log 2, thee exit an c uch that c implie that X i EKR. Futhemoe, fo 6, X i EKR fo n > 2. In the cae of a ingle geneato, we have a hape bound, which povide evidence fo Conjectue 3. Theoem 5 Let 3, let A = F (n; ; G) be an inteecting family with jgj =, and let X be eventually EKR with jxj. Then n > ' 2 implie ja(x)j js(x)j. The outline of the pape i a follow. In Sectio we give an example of a family A and et X howing that the coecient ' 2 in Conjectue 3 cannot be made any malle. Section 3 give a neceay and ucient condition fo a compeed family to be inteecting. Thi will be ueful in the ection that follow. The poof of Theoem 5 and 4 occu in Section 4 and 5, epectively. Finally, in Section 6 we give a geneating function that geatly peed up numeical computation fo jf (n; ; G)(X)j in compaion with the nave method.

4 EKR et fo lage n and 3 2. A family fo which ja(x)j > js(x)j In thi ection, we exhibit a family that how tightne of the coecient ' 2 in the bound n > ' 2 of Conjectue 3. We emak that fo compeed, inteecting familie A that ae not ubfamilie of S, the quantity ja(x)j i uually maximized by a family of the fom F (n; ; G) with jgj = (cf. Theoem 5). Fo uch familie, the family and choice of X given in Popoition 6 appea to be the laget. Thee obevation motivate Conjectue 3. Popoition 6 Let X = f2; 4; + 2g, A = F (n; ; ff2; 3gg). Fo 4, n 2, if ja(x)j > js(x)j then n < 3++p Poof. We begin by computing ja(x)ns(x)j and js(x)na(x)j. Conide the inteection of an element of one of thee familie with the et f; 2; 3; 4; + 2g. Fo an element of S(X)nA(X), the inteection mut be f; 4g; f; + 2g, o f; 4; + 2g. Fo an element of A(X)nS(X), the inteection mut be f2; 3g; f2; 3; 4g; f2; 3; + 2g; o f2; 3; 4; + 2g. Given one of thee et I, the numbe of element of the family that have that inteection with f; 2; 3; 4; + 2g i Multiplying by if i.e. a deied. ja(x)ns(x)j js(x)na(x)j = 3 jij. Thu + 4 ()!(n )! ()! and expanding, we get that ja(x)ns(x)j js(x)na(x)j > 0 if and only + ( 3 )n < 0; n < p ; 2 Notice that 3++p = ' 2 + o(), o the bound given in Conjectue 3 i tight, up to lowe ode tem. 3. Condition fo a family to be inteecting Thi ection detemine neceay and ucient condition fo a family to be inteecting that will be ueful late in the pape. The pupoe of thi ection i to pove the following popoition. Popoition 7 A compeed family A i inteecting if and only if fo any A = fa ; ; a g, B = fb ; b g 2 A thee exit a pai i; j, with i; j uch that i + j > maxfa i ; b j g. Note that thi i mot ueful in the cae whee A = F (n; ; G). In thi cae aume that fo ome A; B 2 A, we have an i; j uch that i + j > maxfa i ; b j g. Fo any C = fc ; ; c g A and D = fd ; ; d g B, we have i + j > maxfc i ; d j g, thu it i ucient to nd uch a pai i; j fo each pai of geneato. Poof. Aume that thee exit i; j with i + j > maxfa i ; b j g. Without lo of geneality, aume a i b j. Then a ; ; a i ; b ; ; b j b j, o we have i + j > b j element that ae le than o equal to b j, o two mut be the ame by the pigeonhole pinciple, o A \ B 6= ;. The oppoite diection i a diect conequence of Popoition 8. of [5] in the cae of 2 familie. When thi popoition i ephaed uing the notation of thi pape, and the eult i eticted to the peent condition, it tate that if A ; A 2 ae compeed, inteecting familie, and F i 2 A i i xed, then thee exit an ` uch that jf \ [; `]j + jf 2 \ [; `]j > `. To apply thi eult take A = F (n; ; fag), A 2 = F (n; ; fbg), F = A and F 2 = B, then let i = ja \ [; `]j, and j = jb \ [; `]j. Thi ha the following coollay. Coollay 8 Let A = fa ; ; a g. Thee exit an uch that A [; 2 ] if and only if F (n; ; fag) i an inteecting family.

5 4 Benjamin Bond Poof. If A [; 2 ] and fb ; ; b g; fc ; ; c g A, let i = j =, and we have + > maxfa ; a g maxfb ; c g, o F (n; ; fag) i inteecting. Conveely, aume thee exit i and j uch that i+j > a i ; a j. Without lo of geneality i j, o 2i i + j > a i Thi i ucient to how A [i; 2i ], ince a k a k+. 4. Poof of Theoem 5 Let A n;; = F (n; ; f[; 2 ]g) fo 2. Notice that by Coollay 8, fo the poof of Theoem 5 it i ucient to conide familie of thi fom. In [3], Bog poved that fo a compeed family A, and X; X 0 with X 0 X we have ja(x)j ja(x 0 )j. Notice that f + 2g, f4; + 2g, f2; 4; + 2g and f2; ; jxj; + 2g, ae the minimal element in each of the cae of Theoem 2, thu it uce to aume X i one of thee et. Thi i impotant, becaue it geatly educe the numbe of cae we need to conide. The poof will beak into eveal cae depending on vaiou poibilitie fo X, which i uniquely detemined by t = jxj. We aume thoughout that t and 2. Fo t 2, we will how that ja n;; (X)j i deceaing in, o it i ucient to how that js(x)j ja n;; (X)j fo mall value of. The cae t = i lightly moe complicated, ince ja n;; (X)j i not deceaing in. Fo thi cae we may wite ja n;; (X)j = D(n; ; ) + E(n; ; ) whee D(n; ; ) i deceaing in, and E(n; ; )=js(x)j goe to 0 a!. By conideing the atio ja n;; (X)j=jS(X)j, we will how that it i le than fo lage, and then ue a compute to check the eult fo mall value of. We begin by conideing the cae t The cae t 2 Lemma 9 If t 2, ja n;; (X)j i deceaing in fo 3. Alo, if n > ' 2 and t 3 then ja n;;2 (X)j ja n;;3 (X)j. Poof. Fit adde the cae Let B n;; = A n;;+ (X)nA n;; (X). Obeve that if A = fa ; ; a g 2 B n;;, then a = 2 and a + = 2 +. Notice that fo an element of A n;;+ na n;; thee ae 2 poibilitie fo the t element and n (2 + ) poibilitie fo the lat (we may aume that n + +, o 6= 0). element, hence ja n;;+ na n;; j = 2 Similaly ince 2 + 2, thee ae 2 t an element of X (notice thi i 0 if t 2 ). Subtacting give that jb n;; j = 2 element of An;;+ na n;; that do not contain 2 t Now conide C n;; = A n;; (X)nA n;;+ (X). Notice that if A = fa ; ; a g 2 C n;;, then fa ; ; a g [; 2 ] and a + > 2 +. We ue the ame counting method a ued above, namely, we t count the numbe of element of A n;; na n;;+ by conideing poibilitie fo the t element of a et and the lat element to nd that thee ae 2 uch element (Note that thi die lightly fo the method ued to count B n;; ince in that cae we new the value of the element in poition and + ). Then we count the numbe of element of A n;; na n;;+ that do not contain an element of X by conideing poibilitie fo the t and lat element, and multiplying. We nd thee ae 2 t. We then ubtact to get 2 2 t jc n;; j = Thi ame method of counting B n;; and C n;; will be epeated eveal time below with only light vaiation, o we omit the explanation of thoe computation. 2 We wih to how that jc n;; j jb n;; j. Notice that 2 t 2 2 t ince (2 t). Alo, 2, and combining thee how that jcn;; j jb n;; j, hence ja n;; j i deceaing in whe + 2. Now conide the cae that f2; 2 + g. Uing the counting method decibed above, modied fo the fact that now fo an element of B n;;, we have + 2 = a o a +, we nd that

6 EKR et fo lage n and 5 Let = 2 inequality by 2 jb n;; j = 2 2 t jc n;; j = and = 2 t. In thi cae, to how jcn;; j jb n;; j, by dividing both ide of thi and eaanging, it i ucient to how that n + + Since the lat tem i negative, it i enough to how n +. Notice that i laget in the cae t = 2, and in thi cae it implie to = 3 fo 2. Thi give jc 2 n;;j jb n;; j fo n > 25. Now adde the cae that 2 + < + 2. Uing the counting method given above, we nd that if t 4 and 2, o t 2 f2; 3g and 3, then 2 2 t 2 jb n;; j = ; 2 2 t 2 jc n;; j = We wih to how that jc n;; j jb n;; j. By eaanging tem and uing the identitie a b, a b = b a a b, we nd that howing jcn;; j jb n;; j i equivalent to howing 2 n 2 t 2 t + n b a b+ a b = Notice that we may aume that > t, becaue othewie the 2 t in the expeion fo jcj i 0, in which cae it i eay to ee that jc n;; j jb n;; j. Fom the expeion above, we ee that to how jc n;; j jb n;; j, it i ucient to how that ( ), which i eay to how n ( t+)(n ) by imple eaangement of the inequality. Thu jc n;; j jb n;; j. 2 n 6 It emain to check the cae that t = 3; = 2. Notice that in thi cae jc n;; j = 3 3 = + n 6 3 and jbn;; j = 3 3. Obeve that n 6 3 = (n )(n 3) ()() 3 ; and that fo n > ' 2 we (n )(n 3) have that (thi can be hown by cleaing denominato, uing the quadatic fomula ()() in a = n, and then maximizing with epect to ). Combining thi inequality with give jc n;; j jb n;; j. We now pove Theoem 5 in the cae t 2. Poof. By Lemma 9, it i ucient to how that js(x)j ja n;; (X)j in the cae (t; ) = (2; 2); (2; 3); (3; 2) and the cae t 4, = 2. We t adde the cae t 4, = 2. Uing the ame counting method a in Lemma 9, we count js(x)na n;;2 (X)j = n 3 n t and jan;;2 (X)nS(X)j = n 3. Since js(x)na n;;2 (X)j i deceaing in t, we may aume that t = 4. Thu it i ucient to how n 3 equivalent to n 3 0. Multiply thi expeion by ( )!(n )! to get that thi inequality i ()! (n 3)(n 4)(n ) ( ) (n 3)(n )(n ) 0 By letting n = a and expanding, we get a quadatic expeion in a. By uing the quadatic fomula, and then maximizing with epect to, we nd that js(x)j ja n;; (X)j fo a > ' 2. In the cae t = 2; = 2, we have js(x)na n;;2 (X)j = n 3 n 3 and jan;;2 (X)nS(X)j =. We may ue the ame method a in the cae t 4 to get js(x)j jan;; (X)j fo n > 2.

7 6 Benjamin Bond The cae t = 3, = 2 wa addeed in Section 2, o it only emain to check the cae t = 2; = 3. In thi cae we have js(x)na n;;3 (X)j = n 6 3 and jan;;3 (X)nS(X)j = n n 6 4. We ue the ame method, a in the cae t 4, but thi time we get a polynomial that i cubic in a. Uing the cubic fomula and maximizing, we nd that js(x)j ja n;;2 (X)j fo a The cae t = Lemma 0 Fo t =, we have ja n;; (X)j = D(n; ; ) + E(n; ; ) whee min(+;2 ) X i n i D(n; ; ) = and E(n; ; ) = ( + n i= + P 2 i=+3 i 2 n i othewie. Poof. We count the numbe of element in A n;; (X) with the -th element being i. Fo i min( + ; 2 ), thee ae i poibilitie fo the t element, one of the element mut be + 2, and thee ae n i poibilitie fo the emaining element. If 2 < + 2, all element of A n;; have the -th element le than + 2, o we have counted all of A n;; (X). Othewie, uing imila logic we get the expeion fo E(n; ; ). We adde D(n; ; ) and E(n; ; ) epaately. We t conide D(n; ; ). Lemma D(n; ; ) i deceaing in. n i Poof. In the cae that min( +; 2 ) = +, notice that fo i + we have that 2( ) i, o by inceaing, each tem i deceae. Alo, inceaing deceae the total numbe of tem, o in thi cae it i eay to ee that D(n; ; ) i deceaing in. Othewie, conide D(n; ; ) D(n; ; + ). Let C = fa 2 F (n; ; f[; 2 ]g) A; + 2 i in poition + o geateg. By eviewing the poof of Lemma 0 it i eay to ee that D(n; ; ) count jc j. Thu we have D(n; ; ) D(n; ; + ) = jc nc + j jc + nc j. We t adde the cae that Notice that C nc + conit of element that have the t element le that 2, and the + -t element i at leat Alo, notice that C + nc conit of element that have -th element equal to 2, and the + -t element i 2 +. By conideing uch element, we ee that jc nc + j jc + nc j = 2 It i eay to ee that that thi i geate than 0. The only emaining cae i whe + = + 2, and in thi cae we have jc nc + j jc + nc j = 2. Thu we ee D(n; ; ) i deceaing in. We now conide E(n; ; ). Notice that when t =, we have js(x)j =, o we begin by howing that E(n; ; )= i negligibly mall. Lemma 2 Let n= = a, and 4. Let Q(n; ; ) = E(n; ; )=. Fo any xed a 22, maxfq(n; ; )g =2 goe to 0 exponentially in a!, fo any. Fo a ' 2, we have Q(a; ; ) 325 3=2 (954). Poof. We adde each tem in E(n; ; ) individually. We begin by uing Stiling' appoximation, and then ue a compute to nd the maximal tem. We t adde P 2 i=+3 i 2 n i 2 fo <. We will conide each tem in the um individually, o the ame computation apply to the t tem of E(n; ; ), and to the cae =, o we omit the analyi of thoe cae. Dene q(n; ; i; ) by q(n; ; i; ) = i 2 n i 2 = ( )n(n ) i(i )( ) i!(n i)!!(n )!!(i )!( )!(n i + )!n!

8 EKR et fo lage n and 7 Recall Stiling' appoximation n! p 2n (n=e) n. Although thi only hold fo lage n, in geneal we have p 2n (n=e) n n! e p n (n=e) n. Applying thi give ( )n(n )e4 q(n; ; i; ) i(i )( )(2) i(n i)(n ) 2 2(i )( )(n i + )n i i (n i) n i (n ) n (i ) i ( ) (n i + ) n i + n n Notice that many of the powe of e fom Stiling' appoximation have canceled. We ubtitute n = a. We wih to nd fo which a thi tem goe to 0 exponentially in. To do thi, divide numeato and denominato by a, and pull out an -th oot. We alo ubtitute I = i=, and S = =. We have alo ued the fact ( )=(i ) =i. Thi give q(n; ; i; ) T (a; ; i; ) = 2 a(a )e 4 i 2 ( )(2) i(a i)(a ) 2 2(i )( )(a i + )a I I (a I) a I (a ) a S S (I S) I S ( S) S (a I + S) a I +S a a () I I (a I) a I (a ) a We t adde B(I; S; a) =. Notice that we have S S (I S) I S ( S) S (a I + S) a I +S aa the bound =2 S and I 2S. Uing a compute we compute the maximum of B(I; S; a) fo =2 S, I 2S, and a 22. Thi can be done uing the function minimized contained in SAGE. The maximum occu at S =, I =, a = 22, fo a 22 we have B(I; S; a) B(; ; 22) = In the cae that we etict a ' 2, we get B(I; S; a) 954. Thu each tem in Q(n; ; ) goe to 0 exponentially in fo a 22. To nih the poof of the t pat of the popoition, notice that we have hown each tem in the um goe to 0 exponentially in. Since thee ae at mot 4 tem, the um till goe to 0 exponentially in. We now wih to evaluate how quickly thi tem goe to 0 when a ' 2. We t ue a compute to how that T (a; ; i; ) i deceaing in a. Thi can be done by computing the patial T (a; ; i; and then computing the maximum of the patial deivative, uing the ame command a above. We nd that the maximum i , which occu at (a; ; i; ) = (50; 9; 82; 8), o T (a; ; i; ) i deceaing in a. Maximizing the t tem and uing the value fo B(I; S; a) computed above, we nd T (' 2 ; ; i; ) 325 p (954). Since q(n; ; i; ) T (a; ; i; ) and thee ae le than tem, we get Q(a; ; ; t) 325 3=2 (954) fo a ' 2. We need one moe lemma. Lemma 3 Fo contant b; c 2 N, with b c and n = a, we have lim! Poof. We expand the binomial coecient. n b c = n b c = (a )b c a b 2 () ( c + ) (n ) (n b + c ) () (n b + ) = c 2 (n ) b c n b 2 c Y i=2 i Y b+c j=0 n j n b Y k=2 n n k A goe to innity, the poduct go to, hence a deied. lim! n b c = lim! c 2 (n ) b c n b 2 = (a )b c a b 2 ;

9 8 Benjamin Bond We may now nih the poof of Theoem 5 in the cae t =. Poof. Notice D(n; ; 2) = n n 4 3, o by ubtituting n = a, D(n; ; 2) = a + (a )() (a)(a 3) Taking the deivative with epect to a, it i poible to how fo any that thi i deceaing in a when a > ' 2. By Lemma and 2, we have that ja n;; (X)j js(x)j D(n; ; ) + E(n; ; ) D(n; ; 2) = =2 (954) ' 2 + 2()('2 ) (' 2 )(' 2 3) =2 (954) By uing calculu and a compute, we check that thi i le than fo 27. Thu ja n;; (X)j js(x)j fo n ' 2 when 27. Uing the fomula given in Lemma 0, we can check that n ' 2 implie ja n;; (X)j js(x)j fo 4 27, which complete the poof of Theoem Poof of Theoem 4 To pove Theoem 4, we contuct a lage compeed family B that contain mot inteecting familie. It i not inteecting, but it till atie jb(x)j js(x)j fo uciently lage n, which implie ja(x)j js(x)j fo thee value of n. Recall that we may aume that when t = jxj = we have X = f + 2g, when t = 2 we have X = f4; + 2g, when t = 3 we have X = f2; 4; + 2g, and when t 4 then X = f2; ; t; + 2g. Popoition 4 Let B = F (n; ; ff; + gg) [ F (n; ; ff2; 3; + 2gg) [ Let A be an inteecting family. If A 6 S and A 6 A n;;2, then A B. [ =3 A n;; Poof. Conide an element A = fa ; ; a g 2 A. We begin by auming a =, and we conide the maximal poible value fo a 2. Fo the ake of contadiction, aume that a Since a i > a i, we have a i + i fo i 2. Since A 6 S, thee exit ome B = fb ; ; b g 2 A uch that b 2. By Popoition 7, thee exit a pai i; j uch that i + j > maxfa i ; b j g. If i =, then + j > b j, which implie b j = j, ince fo any B 2, we have bj j. Thi implie b =, which i a contadiction. So i 2 and we have i + j > a i + i, hence j >, which i impoible. Thu we cannot have a 2 + 2, o all A 2 A with mallet element ae contained in F (n; ; ff; + gg). Now aume that A [; 2 ] fo ome 2. If 3, then A 2 A n;; B. Thu we may aume = 2. If a =, we know A 2 B by the peviou paagaph, o we may aume a = 2, which implie a 2 = 3. To how A 2 F (n; ; ff2; 3; + 2gg), we ague a in the peviou paagaph. If a 3 + 3, then a i + i fo i 3. Since A 62 A n;;2, thee exit ome B 2 A with B 6 [2; 3]. Thi implie b 2 4. By Popoition 7, thee exit a pai i; j with i + j > maxfa i ; b j g. A in the peviou paagaph, if i 3, then j > which i impoible, o i 2 f; 2g. We t how we cannot have i =. If i =, then we cannot have j = ince a 2 = 2. If j 2, then notice that ince i =, we have + j > b j, but we alway have b j j, o b j = j. Since b j < b j+, thi implie b 2 = 2, which contadict b 2 4. Thu i 6=. Now we how we cannot have i = 2. Aume that i = 2. If j =, then we have 3 > a 2 = 3 which i impoible, hence j 2. Howeve, if j 2, then we have 2 + j > b j j, o b j 2 fj; j + g. Since b j b j, thi implie b 2 2 f2; 3g, which i fale. Thu i 62 f; 2g. Thi i impoible if a 3 + 3, o we have a 3 + 2, o A 2 F (n; ; f2; 3; + 2g). We now pove Theoem 4.

10 EKR et fo lage n and 9 Poof. A in the poof of Theoem 5, we t conide the cae t 2, then the cae t =. We conide jb(x)ns(x)j and js(x)nb(x)j. Fo all t, we have js(x)nb(x)j = n, ince the mallet element of a et in SnB mut be and the econd laget mut be at leat + 2, and if it inteect X, then it mut be + 2. Thee ae n choice fo the emaining element. To count jb(x)ns(x)j, notice that F (n; ; ff; + gg)ns(x) = ;. Alo, F (n; ; ff2; 3; + 2gg) F (n; ; ff2; 3gg), and jf (n; ; ff2; 3gg)nS(X)j = n To count the emaining element of B(X)nS(X) it i poible to how that ja n;; (X)nS(X)j i deceaing in by uing the ame agument a in Lemma 9 (i.e. Conide B n;; = (A n;;+ (X)nS(X))n(A n;; (X)nS(X)) and C n;; = (A n;; (X)nS(X))n(A n;;+ (X)nS(X)), and count jc n;; j and jb n;; j uing the method of Lemma 9, and check that jc n;; j jb n;; j). Thi give jb(x)ns(x)j n jan;;3 (X)nS(X)j + ja n;;4 (X)nS(X)j n 6 4 n n n n 8 5 Divide by. Notice that by Lemma 3, fo lage the ight hand ide i appoximately a 3(a )2 a 3(a )2 (a )2 6(a )3 3(a ) a 2 a 3 a 3 a 4 a 5 a 5 a 6 Notice that if we take a = c, then thi goe to 0, hence jb(x)ns(x)j i abitaily mall fo lage. n Alo, it i eay to check that = i inceaing in n fo any xed. So it i ucient to nd an fo which the inequality hold. Notice that n = (n )(n 3) ( + ) ()(n 3) (n + ) Since n b n b = n b, fo b, we get that n n n Take n = c fo ome contant c. Then ome imple manipulation give that c = n c c 2c! =c A!, thi goe to e =c. Thu fo n = c, js(x)=b(x)j= doe not go to 0, but jb(x)=s(x)j! 0 a!, o fo any c, thee exit a value c uch that fo n > c, we have that js(x)j ja(x)j fo all compeed, inteecting familie A. In paticula, by uing calculu and a compute (i.e. by checking that jb(x)ns(x)j= and ( n ) ae deceaing in fo uciently lage, then nding when thei um i le than ) we nd that when c = we have = 36 and fo c = 2 we have = 6. The cae fo t = 3 and t 4 ae vey imila to the cae t = 2. We till have js(x)nb(x)j =, but the numbe of element in F (n; ; f2; 3; + 2g)(X) that aen't in F (n; ; f; + 2g)(X) i n, and uing the ame method a above, thi goe to e =c a!. A above, n now n 3 (ja n;;3 (X)nS(X)j+jA n;;4 (X)nS(X)j)=! 0, o we jut need to chooe c o that e =c > e =c. Thi i atied fo c >. Thu fo n > log 2 c2, we have that js(x)j ja(x)j fo all compeed, inteecting familie A. In paticula, when t 3, c = 2 we ue a compute to nd that it hold with = 4. The cae t = i imila to the peviou cae, though we alo have to adde the contibution fom the E(n; ; ) tem. Howeve, a hown in Lemma 2, the tem E(n; ; )= i negligibly mall. Uing a imila method to the one ued in Lemma 2, we nd E(2 ; ; ) 3e , which allow u to ue a compute to nd that = 4.

11 0 Benjamin Bond 6. A geneating function fo ja(x)j In thi ection, we intoduce a geneating function that can be ued to compute ja(x)j fo any compeed family A and X. Thi method of calculating ja(x)j i much fate in pactice than enumeating element of A and checking if each inteect X. In ou expeiment, the method given below wa appoximately 40 time a fat. Befoe beginning we note that fo any non-empty et of geneato G 2, we may obtain a et of uch that F (n; ; G) = F (n; ; G 0 ) and jg 0 j jgj. Indeed, conide G = fg ; ; g k g 2 geneato G 0 G with k. Notice that if g i n +i, then fo A = fa ; ; a g G the inequality a i g i i tivially atied. So we may eplace g i by n + i in G without changing the et of element geneated by G. Then if k < we eplace G by G [ [n ( k) + ; n] to get a et with ize. Let G 0 be the family of et obtained in thi way, along with the element of G with ize. Notice we have F (n; ; G) = F (n; ; G 0 ). Thu we may aume any et of geneato i a family. Fo a family A, conide the function f A dened by Notice that f A (; ; ) = jaj. Let f A (x ; ; x n ) = X B2A i;x = Y i2b ( i =2 X 0 i 2 X We dene f A (x ; x n )j X=0 = f A ( ;X x ; 2;X x 2 ; ; n;x x n ). Notice that f A (x ; ; x n ) f A (x ; x n )j X=0 = f A(X) (x ; ; x n ) Fo A of the fom A = F (n; ; ffa ; ; a gg) we denote f A by f a;;a. Popoition 5 give a ecuive method fo computing f a;;an. Popoition 5 Fo a compeed family A = F (n; ; ffa ; ; a gg), we have f a;;a (x ; ; x n ) = Xa i= x i x i f fminfaj;i+j gg j= (x ; ; x n ); whee fminfa j ; i + j gg j= denote the element et whee the j-th element i minfa j; i + j g. Poof. Conide B 2 A = F (n; ; ffa ; ; a gg) with laget element i. Notice we mut have i a, and b j a j becaue B fa ; ; a g. Since b j b j+, and the laget element of B i i, the j-th element i at mot i+j. Combining thee obevation, we have fb ; ; b g fminfa j ; i+j gg j=. Conveely, evey B uch that b = i and fb ; ; b g minfa j ; i+j g i in A ince A i compeed, which give the deied expeion. Remak 6 Obeve that B G = fg ; ; g g and B H = fh ; ; h g if and only if B fminfg i ; h i gg i=. Thu by the peceding popoition, we may obtain f A with A = F (n; ; G) fo any et of geneato G uing the pinciple of incluion-excluion. Fo example with G = fg; Hg, f F (n;;fg;hg) = f F (n;;fgg) + f F (n;;fhg) f F (n;;ffminfgi;h igg i= g) 7. Concluding Remak Theoem 4 and 5 each eve an impotant pupoe. Peviouly, thee wa no known elation between n and that guaantee that one of the eventually EKR et claied by Babe i EKR. Theoem 4 give uch a bound. Howeve, it i unlikely that the bound given in Theoem 4 i tight, and Theoem 5 give a uggetion fo the optimal bound. One poible method fo poving Conjectue 3 i by anweing the following quetion, a light vaiant of one poed in [2]. Quetion 7 Given X, i thee a hot lit of familie, one of which maximize ja(x)j?

12 EKR et fo lage n and A natual choice fo uch a lit i A = fa n;; g =, and thi lit would be epecially ueful becaue of Theoem 5. Howeve, thi doe not hold in geneal, fo example when X = f4; + 2g, = 5, and n =, we have but ja n;; (X)j = js(x)j = 40; ja n;;2 (X)j = 2; ja n;;3 (X)j = 36; ja n;;4 (X)j = 40; ja n;;5 (X)j = 05; jf (n; ; ff2; 3; 4g; f3; 4; 6; 7gg)(X)j = 42 One may till hope that A povide uch a lit fo cetain value of n and. Anothe poible diection fo eeach concen t-inteecting familie. We ay a family A i t- inteecting if fo any A; B 2 A, we have ja \ Bj t. Quetion 8 Can ou eult be genealized to t-inteecting familie? Fo t-inteecting familie, [2] ugget conideing A(; X) = fa 2 A ja \ Xj g and ak fo which X do we have ja(; X)j jsn;(; t X)j fo all compeed and t-inteecting A, whee Sn; t = fa 2 [t] Ag. We upect it i poible to ue imila technique to thoe ued in [2] and thi pape to obtain patial eult in thi moe geneal cae. Acknowledgement. Thi eeach wa done at the Univeity of Minneota Duluth math REU, which i un by Joe Gallian, to whom the autho i thankful. The REU wa uppoted by the National Science Foundation and the Depatment of Defene (gant numbe ) and the National Secuity Agency (gant numbe H ). The autho would like to thank pogam advio Eic Riedl, Davie Rolnick, and Adam Hetebeg fo many helpful dicuion, a well a fo thei comment on thi pape. The autho would alo like to thank Duluth viito Jonathan Wang fo comment on thi pape, and an anonymou efeee fo helpful uggetion, which implied the poof of Theoem 5 and 4. Refeence [] R. Ahlwede and L. Khachatian, The complete inteection theoem fo ytem of nite et. Euopean J. Combin. 8 (2) (997) 25{36. [2] B. Babe, Maximal hitting fo n uciently lage. Gaph and Combinatoic, to appea. axiv v [3] P. Bog, Maximum hitting of a et by compeed inteecting familie, Gaph and Combinatoic. 27 (6) (20) 785{797 [4] P. Ed}o, C. Ko, and R. Rado, Inteection theoem fo ytem of nite et, Quat. J. Math. Oxfod Se. 2 (2) (96) 33{320. [5] P. Fankl, The hifting technique in extemal et theoy, in C. Whitehead, edito, Suvey in Combinatoic, volume 23 of London Math. Soc. Lectue Note Seie, Cambidge Univeity Pe, Cambidge, 987, pp. 8{0. [6] P. Fankl and Z. Fuedi, A new hot poof of the EKR theoem., Jounal of Combinatoial Theoy, Seie A. 9 (6) (202) [7] A. Hilton and E. Milne, Some inteection theoem fo ytem of nite et, Quat. J. Math. 8 () (967) Received Januay 203