ON THE SCOPE OF AVERAGING FOR FRANKL S CONJECTURE

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1 ON THE SCOPE OF AVERAGING FOR FRANKL S CONJECTURE GÁBOR CZÉDLI, MIKLÓS MARÓTI, AND E. TAMÁS SCHMIDT Abstrct. Lt F b union-closd fmily of substs of n m-lmnt st A. Lt n = F 2. For b A lt w(b) dnot th numbr of sts in F contining b minus th numbr of sts in F not contining b. Frnkl s conjctur from 979, lso known s th union-closd sts conjctur, stts tht thr xists n lmnt b A with w(b) 0. Th prsnt ppr dls with th vrg of th w(b), computd ovr ll b A. F is sid to stisfy th vrgd Frnkl s proprty if this vrg is nonngtiv. Although this much strongr proprty dos not hold for ll unionclosd fmilis, th first uthor [7] vrifid th vrgd Frnkl s proprty whnvr n 2 m 2 m/2 nd m 3. Th min rsult of this ppr shows tht () w cnnot rplc 2 m/2 with th uppr intgr prt of 2 m /3, nd (2) if Frnkl s conjctur is tru (t lst for m-lmnt bs sts) nd n 2 m 2 m /3 thn th vrgd Frnkl s proprty holds (i.., 2 m/2 cn b rplcd with th lowr intgr prt of 2 m /3). Th proof combins lmntry fcts from combintorics nd lttic thory. Th ppr is slf-contind, nd th rdr is ssumd to b fmilir nithr with lttics nor with combintorics.. Introduction nd th min thorm Givn n m-lmnt finit st A = {,..., m },fmily (or, in othr words, st) F of substs of A, i.. F P (A), is clld union-closd fmily (ovr A) ifx Y Fwhnvr X, Y F. W lwys ssum tht A is finit with 3 m := A nd n := F 2. It ws Ptr Frnkl in 979 who formultd th following conjctur, now known s Frnkl s conjctur or th union-closd sts conjctur: if F is s bov thn thr xists n lmnt of A which is contind in t lst hlf of th mmbrs of F. In spit of t lst thr dozn pprs, this conjctur is still opn. Hnc it will b convnint to us th following trminology: w sy tht Frnkl s conjctur holds ovr m-lmnt bs sts, if for ny unionclosd fmily F of substs of n m-lmnt (quivlntly, t most m-lmnt) st A with F 2, thr xists n lmnt of A which is contind in t lst hlf of th mmbrs of F. Clrly, it is sufficint to considr only thos union-closd sts tht contin th mpty st. Hnc in th squl, whn th siz F of F will b importnt, w will lwys ssum tht F. Dt: Jnury 8, 2008, rvisd: Dcmbr, Ky words nd phrss. Union-closd sts, Frnkl s conjctur, lttic. This rsrch ws prtilly supportd by th NFSR of Hungry (OTKA), grnt no. T , T nd K 6048.

2 2 GÁBOR CZÉDLI, MIKLÓS MARÓTI, AND E. TAMÁS SCHMIDT Th known chivmnts on Frnkl s conjctur blong to two ctgoris. Th first ctgory is constitutd by thos (in fct, th mjority of) rsults tht blong to pur combintorics, with rspct to both th sttmnts nd thir proofs. Thr r svrl dirctions nd th titls of th listd rfrncs spk for thmslvs, so w mntion only fw rsults rlvnt to our invstigtions. Bošnjk nd Mrković [5] prov tht Frnkl s conjctur holds ovr lvn-lmnt bs sts, whil Robrts [30] sttls th cs n = F 40 nd n<4m. As n opposit to Robrts rsult on smll fmilis, Go nd Yu [3] vrify th conjctur for vry lrg fmilis, i.. for thos with ( 3 () n 2 m 2 2)[m/3] ( ) m m For othr chivmnts of combintoril ntur cf.,.g., Norton nd Srvt [2] nd Vughn [32]. On cn rd mor bout th problm t [37] or, of cours, in Frnkl [2]. On th othr hnd, Stnly [3] nd Poonn [22] stblish nic lttic thortic vrsion of Frnkl s conjctur. (For dtils on cn lso s [7] or Ab nd Nkno [3].) This inititd sris of lttic thorticl pprs givn by Ab nd Nkno [], [2], [3], [4], Hrrmnn nd Lngsdorf [4], nd Rinhold [24] (som of which contind rsults lrdy known in th folklor); ths r th rsults blonging to th scond ctgory. Howvr, thr wr no rl links btwn th combintoril nd th lttic thorticl pprochs bfor [7], xcpt of cours for th sttmnt of thir quivlnc. In prticulr, rsults tht look combintoril wr provd by combintoril mthods, nd th lttic thorticl rsults hv not hd significnt influnc on combintorists. This is vry surprising, for th lttic thortic pproch hs t lst on obvious dvntg: whil it is firly difficult to visuliz union-closd fmily with, sy, (m, n) =(5, 2), dpicting th Hss digrm of th corrsponding twlv lmnt lttic crts no problm t ll. Probbly, [7] is th first cs whn purly combintoril sttmnt is provd within lttic thory. Similrly, th prsnt ppr blongs to nithr of th bovmntiond two ctgoris. Our min rsult is purly combintoril without mntioning lttics. Its proof is mixtur of lttic thory nd combintorics. Howvr, only th rudimnts of lttic thory nd thos of combintorics r usd. So th ppr is intndd to b slf-contind for most of th rdrs. Lt F b union-closd fmily ovr A nd lt th nottions n = F 2, m = A = {,..., m } b fixd throughout. For A lt (2) w() = {X F: X} {X F: / X}. Thn Frnkl s conjctur clims th xistnc of n A with w() 0. With th nottion w(f) = w() A A lt us sy tht F stisfis th vrgd Frnkl s proprty if w(f) 0. Although this proprty clrly implis tht Frnkl s conjctur holds for th givn F, it blongs to th folklor tht mny union-closd fmilis fil to stisfy th vrgd Frnkl s proprty.

3 AVERAGING FOR FRANKL S CONJECTURE 3 For givn m = A, th mximum vlu of n is of cours 2 m. union-closd fmilis F including, it is provd in [7] tht For lrg (3) F 2 m 2 m/2 = w(f) 0. This sttmnt is much strongr thn Go nd Yu s () in two snss: (3) covrs mny mor instncs of F, nd Frnkl s is rplcd by vrgd Frnkl s. Evn if 2 m/2 is bttr, i.., lrgr, thn th corrsponding xprssion in (), lrdy [7] obsrvs tht 2 m/2 is not th optiml vlu. Th originl trgt of th prsnt ppr ws to rplc 2 m/2 in (3) with th bst possibl vlu, in th dditionl hop tht th improvd vrsion of (3) givs mor informtion on th originl Frnkl s conjctur s wll. Unfortuntly, this xtr hop is not fulfilld yt, for th min thorm blow ssums th vlidity of Frnkl s conjctur. As usul, th uppr rsp. lowr intgr prt of rl numbr x will b dnotd by x rsp. x ; for xmpl 32/3 =. Min Thorm. Lt m 3, nd lt A b n m-lmnt st. () Thr xists union-closd fmily F ovr A with Fnd F = 2 m+ /3 = 2 m+ /3 such tht F fils th vrgd Frnkl s proprty. (2) Assum tht Frnkl s conjctur holds ovr m-lmnt bs sts. Thn ch union-closd fmily F ovr A with Fnd (4) n := F 2 m+ /3 stisfis th vrgd Frnkl s proprty. Frnkl s conjctur hs bn intnsivly studid nd it is lmost thr dcds old. Hnc it is rsonbl to conjctur tht (4) in itslf implis w(f) 0 for union-closd fmilis F. Th rst of th ppr is dvotd to th proof of this thorm nd will run s follows. First w introduc som intgr squncs, nd study thir lmntry proprtis. This rquirs only lmntry rgumnts with induction. Thn w study ordr-idls nd (ordr-) smi-idls (to b dfind ltr) of finit Booln lttics. W r intrstd in how to mximiz th sum (quivlntly, th vrg) of hights of lmnts of n ordr-idl or smi-idl X whn X is fixd. Using th proprtis of our intgr squncs w will show th xpctd but nontrivil fct tht this mximum is vilbl vi th obvious grdy lgorithm. Ordr-idls do not crt n invincibl problm; howvr, our trtmnt for smi-idls nds th ssumption tht Frnkl s conjctur holds ovr m-lmnt bs sts. This lds to nw conjctur, formultd t th nd of th ppr. Onc th grdy lgorithm for smi-idls is provn to b pproprit, th Min Thorm follows immditly from its obvious rformultion for smi-idls. Although this rformultion trnslts th problm to Lttic Thory, this is n sir nd lss lgnt trnsltion thn th usul on by Stnly [3] nd Poonn [22]. 2. Som intgr squncs W r going to dfin nd study thr kinds of intgr squncs: α, β nd γ. Th α rsp. β squncs clcult th totl hight of grdy ordr-idls rsp. grdy smi-idls (to b dfind ltr). Th γ squncs r th stpping ston btwn

4 4 GÁBOR CZÉDLI, MIKLÓS MARÓTI, AND E. TAMÁS SCHMIDT ths two kinds of idls, nd th γ squncs r lso ndd to undrstnd th innr structur of β squncs. In this sction, m nd n will dnot rbitrry nturl numbrs. Givn two intgr squncs =(,..., m ) nd b =(b,...,b n ), thir conctntion will b dnotd by (,..., m ) (b,...,b n )=(,..., m,b,...,b n ). Th lngth of squnc is dnotd by lngth( ); it is lwys positiv intgr nd w hv lngth( ) + lngth( b) = lngth( b). Whn lngth( ) = lngth( b) thn + b nd b r undrstood componntwis,.g., (,..., k )+(b,...,b k )=( + b,..., k + b k ). Th constnt squnc (j,j,...,j) will b dnotd by j; w us this nottion only in connction with ddition, so thr will b no mbiguity wht th lngth of j is. For xmpl, (,..., k )+ is( +,..., k + ). Whn thr xists squnc c with c = b thn w sy tht is (propr) initil sgmnt of b. Now, vi induction, lt us dfin α (0) = (0), α (i+) = α (i) ( + α (i) ), γ (i) = + α (i) (i =0,, 2...), nd β () = γ 0, β (i+) = γ (i) β (i), which mns tht For xmpl, β (j) = γ (j ) γ (j 2) γ () γ (0) (j =, 2, 3,...). α (3) + α (3) {}}{{}}{ α (4) =( 0,,, 2,, 2, 2, 3,, 2, 2, 3, 2, 3, 3, 4) nd }{{}}{{} + α (2) α (2) β (4) =(, 2, 2, 3, 2, 3, 3, 4,, 2, 2, 3,, 2, }{{}}{{}}{{}}{{} ). γ (3) γ (2) γ () γ (0) Notic tht lngth( α (i) ) = lngth( γ (i) )=2 i (0 i) whil lngth( β (i) )=2 i ( i). Th first mmbr of squnc is lwys indxd by. Th ith mmbr of α (n) will, of cours, b dnotd by α (n) i, nd similr nottions pply for β (n) nd γ (n). For convninc, lt α ( ) rsp. γ ( ) dnot th infinit squnc whos initil sgmnt of lngth 2 k is α (k) rsp. γ (k) for ch k N = {, 2,...}. For 0 k n, th subsquncs of th form sgm(, 2 k,i):=( i2 k +, i2 k +2,..., (i+)2 k), i =0,,...,2 n k r clld 2 k -sgmnts of =(,..., 2 n). Conscutiv 2 k -sgmnts will ply n importnt rol in th forthcoming considrtions. Lt sgm(, 2 k,i),...,sgm(, 2 k,i+ l ) nd sgm(, 2 k,j),...,sgm(, 2 k,j+ l ) b two fmilis of l conscutiv 2 k -sgmnts, nd considr two substs X nd Y of th corrsponding indx sts, i.., lt X {t : i2 k + t (i + l)2 k } nd Y {t : j2 k + t (j + l)2 k }. W sy tht X nd Y r qully positiond in

5 AVERAGING FOR FRANKL S CONJECTURE 5 ths fmilis of 2 k -sgmnts if X Y, x x +(j i)2 k is bijction. Tht is, qully positiond hs th nturl mning. Givn squnc =(,..., l ); th sum of t conscutiv mmbrs of bginning t th ith position will b dnotd by σ(, i, t) = i + i+ + + i+t. This nottion ssums tht i, 0 t nd i + t l. Notic tht σ(, i, 0)=0 by convntion. Th forthcoming lmms will b formultd only for α (n), but othr thn Lmm 2 thy will b obviously vlid nd usd for γ (n) s wll. As usul, N 0 = N {0} dnots th st of nonngtiv intgrs. Lmm. Lt i N nd j N 0 with i + j 2 n. σ( α (n),i,j). Thn σ( α (n),,j) Proof. W cn ssum tht i>. W us induction on j. Sinc α (n) is th only occurrnc of 0 in α (n), th sttmnt is vidnt for j. So w ssum tht j> nd th lmm holds for,...,j. For brvity, lt x = σ( α (n),,j) nd y = σ( α (n),i,j). If i j (pictorilly: if x nd y ovrlp) thn (5) x = σ( α (n),,i ) + σ( α (n),i,j i + ) nd y = σ( α (n),i,j i +)+σ( α (n),j+,i ), nd x y follows from th induction hypothsis. Hnc w cn ssum tht i>j. Th pictoril illustrtion blow (vn if it dos not rflct th full gnrlity) will b usful for wht coms nxt: α (k) + α (k) + α (k) ū + α (k) v + α (k) w + α (k) α (n) {}}{{}}{{}}{{}}{{}}{{}}{ =(,...,,,...,,,...,,...,,...,,,...,,,...,,...) α (n) =(,...,,,...,,,...,,...,,...,,,...,,,...,,...). }{{}}{{} z y Considr th uniqu k N 0 such tht 2 k j<2 k+. Th ssumption tht x nd y do not ovrlp implis tht k n. Firstly, w ssum tht k<n. Thn w cn choos thr conscutiv 2 k - sgmnts, sy sgm( α (n), 2 k,q), sgm( α (n), 2 k,q+ ) nd sgm( α (n), 2 k,q+ 2), such tht th summnds of y blong to this fmily B of conscutiv 2 k -sgmnts. (Mor prcisly but lss pictorilly, such tht q2 k + i nd i + j (q + 3)2 k.) Considr lso th fmily A of th 2 k -sgmnts sgm( α (n), 2 k, 0), sgm( α (n), 2 k, ) nd sgm( α (n), 2 k, 2), tht is, th first thr 2 k -sgmnts. Lt z b th sum of j conscutiv mmbrs in th first thr 2 k -sgmnts such tht th summnds of z in A nd th summnds of y in B r qully positiond. (Mor formlly, i = q2 k + r nd z = σ( α (n),r,j).) Obsrv tht A consists of α (k), + α (k), + α (k) whil B consists of ū + α (k), v + α (k), w + α (k) for pproprit positiv intgrs u, v, w. Now, 0 < u, v nd w, whnc th lmnts of th first thr 2 k - sgmnts r lss thn or qul to th corrsponding lmnts of sgm( α (n), 2 k,q), sgm( α (n), 2 k,q+), nd sgm( α (n), 2 k,q+2). Hnc z y, for z nd y r qully positiond. Furthr, x z follows from th prviously considrd ovrlpping cs, nd w conclud x y.

6 6 GÁBOR CZÉDLI, MIKLÓS MARÓTI, AND E. TAMÁS SCHMIDT Scondly, w ssum tht k = n. Thn α (n) consists of two 2 k -sgmnts only. Sinc x nd y do not ovrlp, w obtin tht j =2 k nd i =2 k +. Using α (n) = α (k) ( + α (k) ), w conclud tht x = σ( α (k),, 2 k ) <σ( + α (k),, 2 k )=y. Dfin th invrs of squnc =(, 2,..., k )s inv(, 2,..., k )=( k, k,..., ). Th proof, trivil induction, of th following lmm is lft to th rdr. Lmm 2. α (n) + inv( α (n) )= n. Now w formult sttmnt on th sum of th lst j mmbrs of α (n). Lmm 3. If j 2 n nd i<2 n j thn σ( α (n),i,j) σ( α (n), 2 n j +,j). Proof. Considr th squnc n α (n). This squnc is th invrs of α (n), so th sum of th lst j mmbrs bcoms th sum of th first j mmbrs. Hnc th ssrtion follows from Lmm. Lmm 4. Lt i, j < 2 n nd dfin { i + j, if i + j 2 n u := nd v := i + j u. 2 n, if i + j>2 n Thn σ( α (n),,i)+σ( α (n),,j) σ( α (n),,u)+σ( α (n),,v). Proof. If i + j 2 n thn v = 0 givs σ( α (n),,v) = 0, nd th ssrtion is trivil consqunc of Lmm. Hnc w cn ssum tht i + j>2 n. Lt us comput; th ppliction of Lmm 3 will b dnotd by, nd w will us tht j v =2 n i nd (thrfor) j>v: σ( α (n),,i)+σ( α (n),,j)=σ( α (n),,i)+σ( α (n),,v)+σ( α (n),v+,j v) σ( α (n),,i)+σ( α (n),,v)+σ( α (n),i+, 2 n i) = σ( α (n),, 2 n )+σ( α (n),,v)=σ( α (n),,u)+σ( α (n),,v). Lmm 5. If i<2 n thn β (n) i γ (n) i. Proof. An sy induction on n. Ifi 2 n thn β (n) i = γ (n) i by dfinition. Othrwis β (n) i = β (n ) γ (n ) < ( + γ (n ) ) i 2 n i 2 n i 2 n = γ (n) i. Lmm 6. If 0 j < 2 n nd 0 i 2 n thn σ( β (n), 2 n +,j) σ( β (n),i,j). Proof. For brvity, lt x = σ( β (n), 2 n +,j) nd y = σ( β (n),i,j). First ssum tht i + j 2 n. This mns tht th summnds of y li ntirly in γ (n ) ; howvr, th following illustrtion is only prticulr cs (for y nd z my ovrlp): γ (n ) β (n ) β (n) {}}{{}}{ =(,...,,...,,...,,...,,,...,,,,,,..., ) }{{}}{{}}{{} z y x For z = σ( β (n),,j) = σ( γ (n ),,j) w obtin x z from Lmm 5. z σ( γ (n ),i,j)=y by Lmm, whnc x y. Thn

7 AVERAGING FOR FRANKL S CONJECTURE 7 Now w ssum tht 2 n < i + j, which mns tht x nd y ovrlp. Lt z = σ( β (n), 2 n +,i+ j 2 n ), th intrsction of x nd y, u = σ( β (n),i + j, 2 n + i), v = σ( β (n),i,2 n + i), nd, furthr, lt w = σ( β (n),i+ j 2 n, 2 n + i). Thn x = z + u nd y = z + v, cf. th illustrtion blow (notic tht w nd v my ovrlp). γ (n ) β (n ) β (n) {}}{{}}{ =(,,...,,,...,,...,,...,,,,...,,,...,,...,,..., ) }{{}}{{}}{{}}{{} w v z u Sinc u nd w r qully positiond, Lmm 5 givs u w. Thn Lmm 3 pplid to γ (n ) yilds w v. Finlly, u v implis x = z + u z + v = y. Lmm 7. If 0 i<2 n nd 0 j 2 n thn σ( β (n ),,i)+σ( γ (n ),,j) σ( β (n),,i+ j). Proof. Sinc β (n) = γ (n ) β (n ), w cn comput: σ( β (n ),,i)+σ( γ (n ),,j)=σ( β (n), 2 n +,i)+σ( β (n),,j) σ( β (n),j+,i)+σ( β (n),,j)=σ( β (n),,i+ j), whr stnds for n ppliction of Lmm 6. For squnc, lt E(,i,...,j) dnot E(,i,...,j)= σ(, i, j i +) j i + = i + i+ + + j j i +, th vrg of th lmnts in th sgmnt ( i, i+,..., j ). Rmmbr tht γ ( ) ws introducd for convninc right bfor th dfinition of sgmnts; of cours γ ( ) could b rplcd by γ (n) in th following lmm. Lmm 8. Lt 2 n N. () For k =, 2,..., 2 n /3, E( β (n),,...,k)=e( γ ( ),,...,k) n/2. (2) E( β (n),,..., 2 n /3 +)=E( γ ( ),,..., 2 n /3 +)>n/2. (3) For 2 n /3 <k 2 n, E( β (n),,...,k) >n/2. Notic tht th qutions in Prts () nd (2) r clr by dfinitions. Although Prt (3) implis Prt (2), w will prov only Prts () nd (2). Prt (3) will not b provd, for it will not b usd in th squl nd its proof is similr to but considrbly lngthir thn th proofs of Prts () nd (2). Proof. Dfin S n = {k N: E( γ ( ),,...,t) n/2 for t =, 2,...,k}. On cn sily s tht proving tht 2 n /3 is th lrgst mmbr of S n is quivlnt to proving Prts () nd (2) for n. W prov this vi induction on n. Th cs n = 2 is vidnt. Now suppos tht n 3 nd tht Prts () nd (2) hold for n.

8 8 GÁBOR CZÉDLI, MIKLÓS MARÓTI, AND E. TAMÁS SCHMIDT Th first fw mmbrs of γ ( ) r dpictd blow for n =5: γ (n ) {}}{ γ (n 3) + γ (n 3) {}}{{}}{, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, }{{}}{{} γ (n 2) + γ (n 2) + γ (n ) {}}{ 2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,... Sinc 2 n 3 2 n /3, th induction hypothsis givs (6) E( γ ( ),,...,k) (n )/2 for k 2 n 3. Hnc {, 2,...,2 n 3 } S n. It follows from Lmm 2 tht (7) E( γ ( ),,...,2 n 3 )=(n )/2. Rwriting (6) from th first 2 n 3 -sgmnt to th scond on, which is + γ (n 3), w obtin E( γ ( ), 2 n 3 +,...,2 n 3 + k) (n +)/2 for k 2 n 3. This implis {, 2,...,2 n 2 } S n. Now, (7) for th scond 2 n 3 -sgmnt givs E( γ ( ), 2 n 3 +,...,2 n 3 +2 n 3 )=(n +)/2, which combind with (7) yilds (8) E( γ ( ),,...,2 n 2 )=n/2. Th shift from th first 2 n 2 -sgmnt to th scond on, + γ (n 2), chngs (8) into E( γ ( ), 2 n 2 +,...,2 n )=+n/2 =(n +2)/2, which togthr with (8) givs E( γ ( ),,...,2 n )=(n +)/2. Hnc 2 n / S n, nd w conclud tht th lrgst mmbr of S n is 2 n 2 + k for som 0 k<2 n 2. It follows from (8) tht k is th lrgst numbr in {0,,...,2 n 2 } such tht E( γ ( ), 2 n 2 +,...,2 n 2 + k) n/2. Now going from th scond 2 n 2 -sgmnt + γ (n 2) to th first 2 n 2 -sgmnt γ (n 2), w s tht k is th lrgst numbr in {0,,...,2 n 2 } such tht E( γ ( ),,...,k) n/2 =(n 2)/2. Hnc k is th lrgst mmbr of S n 2, whnc th induction hypothsis givs k = 2 n 2 /3. Thrfor, th lrgst mmbr of S n is 2 n n 2 /3 = 2 n 2 +2 n 2 /3 = 2 n /3. 3. Smi-idls nd Thorm quivlnt to th Min Thorm By lttic (L; ) w mn prtilly ordrd st such tht for ny x, y L th suprmum nd infimum of {x, y} xist; thy r dnotd by x y nd x y, rspctivly. W dl only with finit lttics; thy ncssrily hv uniqu lst lmnt 0 nd uniqu lrgst lmnt. For b L th subst {x L : x b} is dnotd by [, b] nd it is clld n intrvl of L. Whn =0orb = thn prticulr nottion pplis: =[, ] nd b =[0, b]. Th covring rltion is dfind vi b iff b nd [, b] = 2. If 0 thn is clld n tom of th lttic. Lt B m dnot th Booln lttic of siz 2 m. Ech x B m hs uniqu complmnt x stisfying x x = nd x x = 0. Notic tht B m is isomorphic to th powr st lttic ovr n m-lmnt st; th singlton sts, th complmnts of substs, th mpty st nd th whol st corrsponding to th toms, th complmnts of lmnts, 0 nd of B m, rspctivly. Th hight h(x) of n lmnt x B m is th lngth of ny mximl chin in x. In th powrst modl, th hight is just th numbr of lmnts of th givn subst. If X is subst of B m, thn th totl

9 AVERAGING FOR FRANKL S CONJECTURE 9 hight h(x) ofx is dfind to b th sum X h(). If X is nonmpty subst X of B m, thn its vrg hight is dfind to b nd dnotd by h(x) =h(x)/ X. A nonmpty subst X of B m is clld n ordr-idl if for ny x X, x X. If nonmpty st X is th union of crtin intrvls [ i,b i ] such tht th i r (not ncssrily distinct) toms, thn X is sid to b smi-idl of B m. Th gol of th nxt sction is to show tht whnvr th siz X of n ordridl is fixd, thn strightforwrd grdy lgorithm producs n ordr-idl with mximum totl hight. W lso prov th nlogous sttmnt for smi-idls; howvr, w could not void ssuming Frnkl s conjctur in tht cs. Th importnc of smi-idls is rvld by th following thorm nd th lmm following it. Thorm. Lt m 3. () Thr xists smi-idl X of B m such tht h(x) > m/2 nd X = 2 m /3. (2) Assum tht Frnkl s conjctur holds ovr m-lmnt bs sts. Thn for ch smi-idl X of B m, X 2 m /3 implis h(x) m/2. By propr sublgbr of (B m,, 0) w mn -closd subst Y of B m such tht 0 Y B m. Th importnc of th bov thorm is du to Lmm 9. Lt m N. () For X B m, X is smi-idl iff B m \ X is propr sublgbr of (B m,, 0). (2) Th Min Thorm nd Thorm r quivlnt sttmnts. (3) Assum tht Frnkl s conjctur holds ovr m-lmnt bs sts. Thn for ch smi-idl X of B m othr thn B m \{0}, X X /2 for som tom B m. Proof. Lt X b smi-idl, nd lt u, v Y = B m \ X. By wy of contrdiction, suppos u v/ Y. Thn [p, u v] X for som tom p B m. From p u v w conclud p u or p v, sy,p u, whnc u [p, u v] X contrdicts u Y. This shows tht Y is sublgbr of (B m,, 0). It is propr, for X. Now lt Y = B m \ X b propr sublgbr of (B m,, 0). Thn X. Lt u X, w hv to find n tom p B m with [p, u] Y =. Lt y dnot th join of th st Y u. Sinc y Y, w hv y<und thrfor thr is n tom p u \ y, which dos th job. This provs Prt (). Now, kping formul (2) nd th powrst modl of B m in mind, for propr sublgbr F of (B m,, 0), w lt w(f )= m w() is n tom of B m whr w() = F F =2 F F F =2 F F.

10 0 GÁBOR CZÉDLI, MIKLÓS MARÓTI, AND E. TAMÁS SCHMIDT Hnc w(f ) 0iff 0 (2 F F ) =2 ( F /2+ ) F tom tom tom ( ) =2 m F /2+ (, b): b F, tom =2 ( m F /2+ ) b F b, tom =2 ( m F /2+ ) h(b) =2 ( m F /2+h(F ) ). b F Dividing by 2 F w obtin tht w(f ) 0iffh(F ) m/2. Now lt X = B m \ F, smi-idl by Prt (). Sinc h(b m )=m/2, h(f ) m/2 iffh(x) m/2, complting th proof of Prt (2). Finlly, th strightforwrd dduction of Prt (3) from th bov rgumnts is lft to th rdr. 4. Grdy ordr-idls nd grdy smi-idls Givn n intgr k {,...,2 m }, w r looking for k-lmnt ordr-idl X of th Booln lttic B m with mximl totl hight h(x). It is nturl to construct X in grdy wy by induction on k. For k 3, vry k-lmnt ordr-idl hs th sm totl hight. Suppos w know how to construct (k )-lmnt ordridl Y of mximl totl hight. Thn w lt X = Y {x} such tht x/ Y, X is n ordr-idl nd h(x) is s lrg s possibl. Now w hv roughly dscribd our grdy lgorithm. Th gol of this sction is to show tht it mximizs th totl hight, indd. First of ll, mor xct dscription of th grdy lgorithm is ncssry. Lt, 2,..., m b fixd numrtion of th toms of B m. Associtd with this numrtion w dfin grdy numrtion of B m vi induction s follows. (For x, y B m, x y will dnot tht x prcds y in th grdy numrtion.) For m, lt coincid with th strict lttic ordr (which is linr ordr, i.., chin, in this cs). Now ssum tht m>. Clrly, m is m, th complmnt of m. Th isomorphism thorm of intrvls yilds tht m,..., m m is n numrtion of toms of m, which is Booln lttic. Lt 2 dnot th grdy numrtion of m ssocitd with th mntiond numrtion of its toms. Lt dnot th grdy numrtion of m ssocitd with th numrtion,, m of its toms. Finlly, lt b th conctntion of nd 2 in th following sns: for x, y B m, x y iff x, y m with x y, orx, y m with x 2 y,orx m nd y m. Th lft-hnd lttic in Figur shows th grdy numrtion of B 3 ssocitd with th from lft to right numrtion (i.., = c 2, 2 = c 3, 3 = c 5 ) of its toms. (Th grdy numrtion is dfind by th indxing: c i c j iff i<j.) Now, by grdy ordr-idl of B m w mn nonmpty subst of th form {x B m : x b} whr is grdy numrtion of B m, x b mns x b or x = b, nd b B m. In othr words, th grdy ordr-idls r th nonmpty initil sgmnts of grdy numrtions. Th smllst grdy ordr-idl is {0} (tk b = 0) whil th lrgst is B m (tk b = ). Th first prt of th following lmm justifis our trminology.

11 AVERAGING FOR FRANKL S CONJECTURE c 4 c 2 c 3 c 8 c 6 c c 7 c ` Figur Lmm 0. Lt X b grdy ordr-idl of B m. Thn X is n ordr-idl of B m, nd its totl hight is h(x) =σ( α (m),, X ), whnc h(x) =E( α (m),,..., X ). Th trivil induction proving this lmm is lft to th rdr. Now, lt, 2,..., m still b fixd numrtion of th toms of B m.forj =,...,mw dfin th intrvl I j =[ j, j ]. In prticulr, I =[, ] = B m, I m =[ m, m ]=[ m, m ] = B 0, nd in gnrl, I j = Bm j. In th right-hnd lttic in Figur, if w strt with th =, 2 = 9, 3 = 3, 4 = 5 numrtion (i.., from lft to right numrtion) of toms, w hv I =[, 8 ], I 2 =[ 9, 2 ], I 3 =[ 3, 4 ] nd I 4 = { 5 }. Notic tht {I,...,I m } is prtition of B m \{0}. Th squnc I,I 2,...,I m is clld stndrd intrvl squnc of B m ; thr r m! such squncs, for thr r m! numrtions of th toms. Now, by grdy smi-idl of B m w mn nonmpty subst X of th form (9) X = I I k U whr U is grdy ordr-idl of I k. (Hnc U but U = I k is prmittd.) Th uniqu k {,...,m} in (9) is sid to b th rnk of th grdy smi-idl X, nd it will b dnotd by rnk(x). Lmm. Lt X B m b grdy smi-idl of B m. Thn it is indd smi-idl, nd its totl hight is h(x) =σ( β (m),, X ), whnc h(x) =E( β (m),,..., X ). It follows from Lmms 0 nd tht for ch n 2 m rsp. k<2 m, B m includs grdy ordr-idl rsp. grdy smi-idl of siz n rsp. k; this obsrvtion will b rlvnt in th following proofs. Th following lmm is prsntd not only for its own intrst; it will b usd in th nxt sction. Lmm 2. Lt X nd Y b ordr-idls of B m such tht X = Y nd Y is grdy. Thn h(x) h(y ), i.., h(x) h(y ). Proof. W prov th lmm vi induction on m. W cn ssum tht m 2. Lt X b n ordr-idl of B m. It suffics to construct grdy ordr-idl U such tht U = X nd h(x) h(u). W cn (nd oftn will) chng X during th proof so

12 2 GÁBOR CZÉDLI, MIKLÓS MARÓTI, AND E. TAMÁS SCHMIDT tht h(x) dos not dcrs nd X rmins th sm. Now, for ltr rfrnc, w formult four fcts; thy r vidnt consquncs of dfinitions. If, for som k, Y is grdy ordr-idl of B k nd Z is n ordr-idl of B k thn (Fct ) Y 2 k implis tht Y contins cotom of B k ; (Fct 2) Y 2 k implis Y c for som cotom c of B k ; (Fct 3) if Z contins cotom d B k nd Z d is grdy ordr-idl of d thn Z is grdy ordr-idl of B k (hr d = d is n tom); (Fct 4) j Y implis tht B k hs j-lmnt grdy ordr-idl which is subst of Y. Now, if thr is n tom B m \ X thn X is n ordr-idl in, Booln lttic with m toms, whnc th induction hypothsis yilds n pproprit U in, which is grdy ordr-idl in B m s wll. Hnc w cn ssum tht X contins ll th toms of B m. Lt B m b n tom. Sinc B m is disjoint union of nd, X is th disjoint union of G := X nd G := X. (Figur 2, whr G = G 2 p, dpicts prticulr but importnt cs.) By th isomorphism thorm of intrvls, B : m p b G 2 G 0 b Figur 2 ϕ:, x x nd ψ :, x x r rciprocl isomorphisms. W know G from th ltst ssumption. Th injctivity of ψ nd th fct tht ψ(x) x X givs ψ(x) G for ny x G show tht G G. W cn ssum tht G rsp. G is grdy ordr-idl of rsp. such tht ψ(g) G. (Thn G G is ncssrily n ordr-idl, so it is still dnotd by X.) Indd, if this is not th cs thn, instd of G nd G, w could us H nd H obtind s follows. First th induction hypothsis llows us to rplc G by grdy ordr-idl H of with G = H nd h(g ) h(h ). Clrly, ϕ(h )is grdy ordr-idl of, nd Fct 4 givs grdy ordr-idl H of such tht H = G nd H ϕ(h ). Sinc h(g) h(h) by th induction hypothsis, H nd H do th job. Thr r thr css. First w ssum tht 2 m 2 G or 2 m X. W hv 2 m X in both css, for G G. By Fct, thr is n tom q B m

13 AVERAGING FOR FRANKL S CONJECTURE 3 such tht q X. By th induction hypothsis, thr xists subst Y with Y = X q nd h(y ) h(x q) such tht ithr Y is mpty or it is grdy ordr-idl of q. W cn lt U = Y q, which dos th job by Fct 3. Th cs whn G 2 m 2 is vn simplr. Indd, G p for som cotom p of by Fct 2. Lt q = ϕ(p), which is cotom of B m. Clrly, X q. Applying th induction hypothsis to q w obtin grdy ordr-idl U in q with h(x) h(u) nd X = U, nd U dos th job in B m, too. So w r lft with th cs dpictd in Figur 2; nmly, lt G < 2 m 2 < G nd X < 2 m. By Fct w cn choos n tom b such tht its complmnt p tkn in blongs to G. Notic tht p = b = ψ(b ). W cn ssum tht G [, b ], for othrwis G cn b rplcd by G -lmnt grdy ordr-idl of [, b ]. Lt G 2 = G [b, ], i = G 2, j = G, nd, to prpr n ppliction of Lmm 4 for n = m 2, lt u := i + j nd v := 0. Sinc X < 2 m, u<2 m 2. Choos u-lmnt grdy ordr-idl H 2 in [b, ]. Thn, by Fct 3, U := H 2 p is grdy ordr-idl of, whnc U is grdy ordr-idl of B m. Finlly, Lmms 4 nd 0 yild h(x) =h( p)+h(g 2 )+h(g) =h( p)+σ( γ (m 2),,i)+ σ( γ (m 2),,j) h( p) +σ( γ (m 2),,u)+σ( γ (m 2),, 0) = h( p) +h(h 2)+0= h(u). Lmm 3. Assum tht Frnkl s conjctur holds ovr m-lmnt bs sts. Thn h(x) E( β (m),,..., X ) for ch smi-idl X of B m. Proof. W prov th lmm vi induction on m. Lt X b smi-idl of B m. According to Lmm, it suffics to find grdy smi-idl Y of B m such tht X = Y nd h(x) h(y ). Prt (3) of Lmm 9 llows us to fix n numrtion,..., m of th toms in B m such tht with th nottions X 0 = X m, X =(X m ) \{ m }, X 2 = X { m } w hv X + X 2 X 0. Not tht X = X 0 X X 2 nd X i X j = for 0 i<j 2. Clrly, X 0 is smi-idl of m = B m. To s tht X is smi-idl of m = Bm, lt x X. Thn thr is n tom p B m with [p, x] X. Ifp m thn p m is n tom of m nd [p m,x] X.Ifp = m thn for ch tom q of m,[q, x] X. Thrfor X is indd smi-idl of m. Now lt I,...,I m b th stndrd intrvl squnc of m dtrmind by th numrtion,..., m of its toms. Similrly, lt J,...,J m b th stndrd intrvl squnc of m dtrmind by th numrtion m,..., m m of its toms. Notic tht th mppings ϕ j : I j J j, x x m nd ψ j : J j I j, x x m r rciprocl lttic isomorphisms for ch j {,...,m }. In fct, thy r th rstrictions of th rciprocl isomorphisms Now lt ϕ: m m, x x m nd ψ : m m, x x m. Y 0 = I I k U b grdy smi-idl of m (of rnk k) such tht Y 0 = X 0. Hr U dnots grdy ordr-idl of I k. Sinc X X 0, w cn choos grdy smi-idl Y

14 4 GÁBOR CZÉDLI, MIKLÓS MARÓTI, AND E. TAMÁS SCHMIDT of m such tht Y = X nd ψ(y ) Y 0 (cf. Fct 4 from th prvious proof). Lt l b th rnk of Y. Thn l k nd Y is of th form Y = J J l V, whr V is grdy ordr-idl of J l. Th induction hypothsis givs h(x 0 ) h(y 0 ) nd h(x ) h(y ), whnc h(x) h(y 0 Y X 2 ). Notic tht K = I J,K 2 = I 2 J 2,..., K m = I m J m, K m = { m } is th stndrd intrvl squnc of B m ssocitd with th numrtion,..., m of its toms. First w considr th cs whn k = l; th sitution for k = l = 3 is outlind in Figur 3. Sinc U is (grdy) ordr-idl of I k, V is n ordr-idl of J k nd B : m m I 2 I J 2 V J U 3 2 m 0 Figur 3 ψ k (V ) U, w conclud tht U V is n ordr-idl of K k = I k J k. Lt W b grdy ordr-idl of K k such tht W = U + V = U V. W know from Lmm 2 tht h(u V ) h(w ). Clrly, Z := (I J ) (I k J k ) W = K K k W is grdy smi-idl of B m, nd w hv h(y 0 Y ) h(z). If X 2 =, i.. m / X, thn X = Z nd h(x) =h(x 0 )+h(x ) h(y 0 )+h(y ) h(z), s rqustd. In cs of m X w nd n sy furthr stp: lt Z + b grdy smi-idl of B m such tht Z Z + nd Z + = Z +. Thn h(x) =h(x 0 )+h(x )+h( m ) h(y )+ h(z)+ h(z ) togthr with X = Z + sttls th cs k = l. Th gnrl cs is whn rnk(y 0 )=k l = rnk(y ). This will b sttld by n induction on k l. Fork l = 0 th job hs lrdy bn don. Now ssum tht k l>0; th sitution for (k, l) =(4, 2) is outlind in Figur 4. W r going to dfin pir (T 0,T ) of grdy smi-idls with th sm proprtis s thos ssumd for (Y 0,Y ) such tht rnk(t 0 ) rnk(t ) <k l.

15 AVERAGING FOR FRANKL S CONJECTURE 5 B : m m I 4 U I I I 2 J 4 J J V J m Figur 4 As n intrmdit stp w dfin grdy smi-idl R 0 of m nd grdy smi-idl R of m. Lt i = U nd j = V. In ordr to hrmoniz with th nottions of Lmm 4, dfin u = i + j if i + j 2 m l = J l nd lt u = J l othrwis. Lt v = i + j u. Now lt R = J J l V + whr V + is u-lmnt grdy ordr-idl of J l. Lt U b v-lmnt subst of I k such tht U is grdy ordr-idl of I k whn v>0. Thn R 0 = I I k U. Finlly, if V + = J l nd U thn lt T 0 = R 0 \ U, nd dd v = U mny nw lmnts to R to obtin th grdy smi-idl T of m. (Ths nw lmnts will of cours go into J l+.) In th othr cs whn V + J l or U =, w simply lt (T 0,T )=(R 0,R ). Clrly, w hv Y 0 Y = T 0 T nd rnk(y 0 ) rnk(y ) > rnk(t 0 ) rnk(t ). So w r lft with th duty of showing h(y 0 )+h(y ) h(t 0 )+h(t ). It follows from Lmms 4 nd 0 tht h Ik (U)+h Jl (V ) h Ik (U )+h Jl (V + ). Thrfor, msuring th totl hight in B m rthr thn in I k nd J l, w conclud h(u)+ U + h(v )+2 V h(u )+ U + h(v + )+2 V +, which implis h(u) +h(v ) h(u )+h(v + ), for U + V + = U + V nd V V +. Thrfor h(y 0 )+h(y ) h(r 0 )+h(r ). Finlly, h(r 0 )+h(r ) h(t 0 )+h(t ) is vidnt. 5. Th nd of th proof nd two conjcturs Now Thorm follows from Lmms 8, nd 3. Finlly, Lmm 9 gurnts tht Thorm implis th Min Thorm. W conclud th ppr with two conjcturs. Although thy r formultd in trms of lttic thory, which gos wll with th prsnt ppr, thy will b trnsltd to pur combintoril lngug ftrwrds. For ch smi-idl X of B m, h(x) E( β (m),,..., X ).

16 6 GÁBOR CZÉDLI, MIKLÓS MARÓTI, AND E. TAMÁS SCHMIDT Thr is function f(m) such tht f(m)/2 m/2 tnds to nd h(x) m/2 for vry smi-idl X B m with X f(m). According to Lmm 3 nd Thorm, positiv solution of Frnkl s conjctur would solv both problms in th ffirmtiv. Howvr, ths conjcturs might b sir (to prov or rfut) thn Frnkl s on. Th ids of Lmm 9 ld sily to th following combintoril intrprttion of th first conjctur. Givn k 2 A, w wnt to find union-closd fmily F, F P (A), such tht w(f) b miniml. Th conjctur ssrts tht w cn obtin such n F by th obvious grdy lgorithm in 2 A k stps, strting from F 0 = P (A) nd dlting just on mmbr of F i in th ith stp. Not tht Bošnjk nd Mrković [5] sttl th first conjctur for m. A positiv solution of th scond conjctur would simply sy tht vn if w cnnot lv th ssumption Frnkl s conjctur holds ovr m-lmnt sts out of th Min Thorm, th vrgd Frnkl s proprty holds for union-closd sts which r ssntilly lrgr thn thos trtd in [7] (nd mntiond in th Introduction.) Rfrncs [] Ttsuy Ab: Excss of lttic, Grphs nd Combintorics 8 (2002), [2] Ttsuy Ab: Strong smimodulr lttics nd Frnkl s conjctur, Algbr Univrslis 44 (2000), [3] Ttsuy Ab nd Bumpi Nkno: Frnkl s conjctur is tru for modulr lttics, Grphs nd Combintorics 4 (998), [4] Ttsuy Ab nd Bumpi Nkno: Lowr smimodulr typs of lttics: Frnkl s conjctur holds for lowr qusi-smimodulr lttics, Grphs nd Combintorics 6 (2000), no., 6. [5] I. Bošnjk nd P. Mrković: Th -lmnt cs of Frnkl s conjctur, Elctron. J. Combin. 5 (2008), no., Rsrch Ppr 88, 7 pp. [6] E. Brown nd T. P. Vughn: Configurtions with subst rstrictions, J. Combin. Mth. Combin. Comput. 48 (2004), [7] G. Czédli: On vrging Frnkl s conjctur for lrg union-closd sts, Journl of Combintoril Thory - Sris A, to ppr. [8] K. Dohmn: A nw prspctiv on th union-closd sts conjctur, Ars. Combin. 58 (200), [9] D. Duffus nd B. Snds: An inqulity for sizs of prim filtrs of finit distributiv lttics, Discrt Mth. 20 (999), [0] M. El-Zhr: A grph-thortic vrsion of th union-closd sts conjctur, J. Grph Thory 26 (997), [] L. F. Fitin nd J-C. Rnud: On union-closd sts nd Conwy s squnc, Bull. Austrl. Mth. Soc. 47 (993), [2] P. Frnkl: Extrml st systms. Hndbook of combintorics, Vol., 2, , Elsvir, Amstrdm, 995. [3] Widong Go nd Hongqun Yu: Not on th union-closd sts conjctur, Ars Combin. 49 (998), [4] C. Hrrmnn nd R. Lngsdorf: Frnkl s conjctur for lowr smimodulr lttics, hrrmnn/rchrch/ [5] R. T. Johnson nd T. P. Vughn: On union-closd fmilis. I, J. Combin. Th. Sr. A 84 (998), [6] G. Lo Fro: Union-closd sts conjctur: improvd bounds, J. Combin. Mth. Combin. Comput. 6 (994), [7] G. Lo Fro: A not on th union-closd sts conjctur, J. Austrl Mth. Soc. Sr. A 57 (994), [8] P. Mrković: An ttmpt t Frnkl s conjctur, Publ. Inst. Mth. (Bogrd) (N.S.) 8(95) (2007),

17 AVERAGING FOR FRANKL S CONJECTURE 7 [9] R. Morris: FC-fmilis nd improvd bounds for Frnkl s conjctur, Europn J. Combin. 27 (2006), [20] T. Nishimur nd S. Tkhshi: Around Frnkl s conjctur, Sci. Rp. Yokohm Nt. Univ. Sct. Mth. Phys. Chm. 43 (996), [2] R. M. Norton nd D. G. Srvt: A not of th union-closd sts conjctur, J. Austrl. Mth. Soc. Sr. A 55 (993) [22] B. Poonn: Union-closd fmilis, J. Combintoril Thory A 59 (992), [23] D. Rimr: An vrg st siz thorm, Combin. Probb. nd Comput. 2 (2003), [24] J. Rinhold: Frnkl s conjctur is tru for lowr smimodulr lttics, Grphs nd Combintorics 6 (2000), 5 6. [25] J-C. Rnud: Is th union-closd sts conjctur th bst possibl?, J. Austrl Mth. Soc. Sr. A 5 (99), [26] J-C. Rnud: A scond pproximtion to th boundry function on union-closd collctions, Ars. Combin. 4 (995), [27] J-C. Rnud nd D. G. Srvt: On th union-closd sts conjctur, Ars Combin. 27 (989), [28] J-C. Rnud nd D. G. Srvt: Improvd bounds for th union-closd sts conjctur, Ars Combin. 29 (990), [29] I. Rivl (d): Grphs nd Ordr, NATO Advncd Sci. Inst. Sr. C: Mth. nd Phys. Scincs 47, D. Ridl Publ. Co. Dordrcht Boston (985), p [30] I. Robrts, Tch. Rp. No. 2/92, School Mth. Stt., Curtin Univ. Tch., Prth, 992. [3] R. P. Stnly: Enumrtiv Combintorics, Vol. I., Blmont, CA: Wdsworth nd Brooks/Cool, 986. [32] T. P. Vughn: Fmilis implying th Frnkl conjctur, Europn J. Combin. 23 (2002), [33] T. P. Vughn: A not on th union-closd sts conjctur, J. Combin. Mth. Combin. Comput. 45 (2003), [34] T. P. Vughn: Tr-sts in union-closd fmily, J. Combin. Mth. Combin. Comput. 49 (2004), [35] P. Winklr: Union-closd sts conjctur, Austrl. Mth. Soc. Gz. 4 (987), p. 99. [36] P. Wójcik: Dnsity of union-closd fmilis, Discrt Mth. 05 (992), [37] wst/opnp/unionclos.html Univrsity of Szgd, Bolyi Institut, Szgd, Ardi vértnúk tr, HUNGARY 6720 E-mil ddrss: czdli@mth.u-szgd.hu URL: czdli/ Univrsity of Szgd, Bolyi Institut, Szgd, Ardi vértnúk tr, HUNGARY 6720 E-mil ddrss: mmroti@mth.u-szgd.hu URL: mmroti/ Mthmticl Institut of th Budpst Univrsity of Tchnology nd Economics, Műgytm rkp. 3, H-52 Budpst, Hungry E-mil ddrss, E. T. Schmidt: schmidt@mth.bm.hu URL: schmidt/