A Deza Frankl type theorem for set partitions

Transcription

1 A Deza Frankl ype heorem for se parons Cheng Yeaw Ku Deparmen of Mahemacs Naonal Unversy of Sngapore Sngapore Kok Bn Wong Insue of Mahemacal Scences Unversy of Malaya Kuala Lumpur, Malaysa Submed: Jan 21, 2015; Acceped: Mar 17, 2015; Publshed: Mar 30, 2015 Mahemacs Subjec Classfcaons: 05D99 Absrac A se paron of [n] s a collecon of parwse dsjon nonempy subses (called blocks) of [n] whose unon s [n]. Le B(n) denoe he famly of all se parons of [n]. A famly A B(n) s sad o be m-nersecng f any wo of s members have a leas m blocks n common. For any se paron P B(n), le τ(p ) = x : x} P } denoe he unon of s sngleons. Also, le µ(p ) = [n] τ(p ) denoe he se of elemens ha do no appear as a sngleon n P. Le F 2 = P B(n) : µ(p ) } ; F 2+1 ( 0 ) = P B(n) : µ(p ) ([n] \ 0 }) }. In hs paper, we show ha for r 3, here exss a consan n 0 = n 0 (r) dependng on r such ha for all n n 0, f A B(n) s (n r)-nersecng, hen F 2, f r = 2; A F 2+1 (1), f r = Moreover, equaly holds f and only f F 2, f r = 2; A = F 2+1 ( 0 ), f r = 2 + 1, for some 0 [n]. Keywords: -nersecng famly, Erdős-Ko-Rado, se parons he elecronc journal of combnaorcs 22(1) (2015), #P1.84 1

2 1 Inroducon Le [n] = 1,..., n}, and le ( ) [n] k denoe he famly of all k-subses of [n]. A famly A of subses of [n] s -nersecng f A B for all A, B A. One of he mos beauful resuls n exremal combnaorcs s he Erdős-Ko-Rado heorem. Theorem 1 (Erdős, Ko, and Rado [13], Frankl [15], Wlson [46]). Suppose A ( ) [n] k s -nersecng and n > 2k. Then for n (k + 1)( + 1), we have ( ) n A. k Moreover, f n > (k + 1)( + 1) hen equaly holds f and only f A = A ( ) [n] k : T A} for some -se T. In he celebraed paper [1], Ahlswede and Khacharan exended he Erdős-Ko-Rado heorem by deermnng he srucure of all -nersecng se sysems of maxmum sze for all possble n (see also [3, 14, 16, 17, 20, 25, 31, 36, 40, 42, 43, 45] for some relaed resuls). There have been many recen resuls showng ha a verson of he Erdős-Ko-Rado heorem holds for combnaoral objecs oher han se sysems. For example, an analogue of he Erdős-Ko-Rado heorem for he Hammng scheme s proved n [41]. A complee soluon for he -nersecon problem n he Hammng space s gven n [2]. Inersecng famles of permuaons were naed by Deza and Frankl n [10]. Some recen work done on hs problem and s varans can be found n [5, 7, 8, 11, 12, 19, 26, 28, 35, 37, 38, 39, 44]. The nvesgaon of he Erdős-Ko-Rado propery for graphs sared n [23], and gave rse o [4, 6, 21, 22, 24, 47]. The Erdős-Ko-Rado ype resuls also appear n vecor spaces [9, 18], se parons [27, 29, 30] and weak composons [32, 33, 34]. Le S n denoe he se of permuaons of [n]. A famly A S n s sad o be m- nersecng f for any σ, δ A, here s an m-se T [n] such ha σ(j) = δ(j) for all j T. Gven any σ S n, se µ(σ) = j [n] : µ(j) j},.e., µ(σ) s he se of all elemens n [n] ha are no fxed by σ. Le σ S n : µ(σ) }, f r = 2; F r = σ S n : µ(σ) ([n] \ 1}) }, f r = I can be verfed easly ha F r s (n r)-nersecng. Furhermore, Deza and Frankl [10] proved he followng heorem. Theorem 2 (Deza-Frankl). For r 3, here exss an n 0 = n 0 (r) such ha for all n n 0, f A S n s (n r)-nersecng, hen A F r. A se paron of [n] s a collecon of parwse dsjon nonempy subses (called blocks) of [n] whose unon s [n]. Le B(n) denoe he famly of all se parons of [n]. he elecronc journal of combnaorcs 22(1) (2015), #P1.84 2

3 I s well-known ha he sze of B(n) s he n-h Bell number, denoed by B n. A block of sze one s also known as a sngleon. We denoe he number of all se parons of [n] whch are sngleon-free (.e. whou any sngleon) by B n. A famly A B(n) s sad o be m-nersecng f P Q m for all P, Q A,.e., any wo of s members have a leas m blocks n common. Le I(n, m) denoe he se of all m-nersecng famles of se parons of [n]. For any se paron P B(n), le τ(p ) = x : x} P } denoe he unon of s sngleons. Also, le µ(p ) = [n] τ(p ) denoe he se of elemens ha do no appear as a sngleon n P. For any wo parons P, Q, we make he followng smple observaons: P and Q canno nersec n any sngleon x} where x µ(p ) µ(q) (here he operaon denoes he symmerc dfference of wo ses). P and Q mus nersec n every sngleon x} where x [n] (µ(p ) µ(q)). Le F 2 = P B(n) : µ(p ) } ; F 2+1 ( 0 ) = P B(n) : µ(p ) ([n] \ 0 }) }. I can be readly verfed ha F 2 I(n, n 2) and F 2+1 ( 0 ) I(n, n 2 1). Moreover, ( ) n F 2 = B, (1) ( ) ( ) n F 2+1 ( 0 ) = B + B n (2) In hs paper, we wll prove he followng heorem. Theorem 3. For r 3, here exss an n 0 = n 0 (r) such ha for all n n 0, f A B(n) s (n r)-nersecng, hen F 2, f r = 2; A F 2+1 (1), f r = Moreover, equaly holds f and only f F 2, f r = 2; A = F 2+1 ( 0 ), f r = 2 + 1, for some 0 [n]. Noe ha Theorem 3 can be consdered as an analogue of Theorem 2 for se parons. Le A 0 = x} : x [n]} and A 1 = x} : x [n] \ 1, 2}} 1, 2}}. Then F 2 = A 0 } and A 0, A 1 } I(n, n 2). So, A 0, A 1 } = 2 > F 2 = 1. Ths explans why r 3 s requred n Theorem 3. he elecronc journal of combnaorcs 22(1) (2015), #P1.84 3

4 2 Splng operaon In hs secon, we summarze some mporan resuls regardng he splng operaon for nersecng famly of se parons. We refer he reader o [27] for proofs whch are omed here. Le, j [n], j, and P B(n). Denoe by P [] he block of P whch conans. We defne he (, j)-spl of P o be he followng se paron: P \ P[] } }, P s j (P ) = [] \ }} f j P [], P oherwse. For a famly A B(n), le s j (A) = s j (P ) : P A}. Any famly A of se parons can be decomposed wh respec o gven, j [n] as follows: A = (A \ A j ) A j, where A j = P A : s j (P ) A}. Defne he (, j)-splng of A o be he famly S j (A) = (A \ A j ) s j (A j ). Surprsngly, urns ou ha for any A I(n, m), splng operaons preserve he sze and he nersecng propery. Lemma 4 ([27], Proposon 3.2). Le A I(n, m). Then S j (A) I(n, m) and S j (A) = A. A. A famly A of se parons s compressed f for any, j [n], j, we have S j (A) = Lemma 5 ([27], Proposon 3.3). Gven a famly A I(n, ), by repeaedly applyng he splng operaons, we evenually oban a compressed famly A I(n, ) wh A = A. Lemma 6. Le a, b be posve negers wh a + b n. Le P, Q B(n) be such ha P Q n a. If τ(p ) \ µ(q) n a b, hen P and Q have a leas b blocks of sze a leas 2 n common and µ(p ) µ(q) 2b. Proof. Snce τ(p ) \ µ(q) n a b, P and Q have a mos n a b sngleons n common. Now, P Q n a means ha P and Q have a leas n a blocks n common. Therefore, P and Q mus have a leas b blocks of sze a leas 2 n common. Le W 1,..., W b P Q wh W 2 for all. Then b =1 W µ(p ) µ(q) and W W j = for j. Ths mples ha 2b b =1 W µ(p ) µ(q). Lemma 7. If A I(n, n r), hen max P A µ(p ) 2r. Proof. Suppose max P A µ(p ) = 2r + s where s 1. Le P 0 A wh µ(p 0 ) = 2r + s. Then τ(p 0 ) = n 2r s. Noe ha P 0 n r for A I(n, n r). By Lemma 6 (ake Q = P = P 0 wh a = r, b = r + s), we have 2r + s = µ(p 0 ) 2(r + s). Thus, we have s 0, a conradcon. Hence, he lemma follows. he elecronc journal of combnaorcs 22(1) (2015), #P1.84 4

5 The followng heorem says ha he famly F 2+1 ( 0 ) s preserved when undong he splng operaons. Theorem 8. If 1, n 5 + 3, A I(n, n 2 1) and S j (A) = F 2+1 ( 0 ), hen A = F 2+1 ( 0 ). Proof. Suppose A F 2+1 ( 0 ). Then max P A µ(p ) ([n] \ 0 }) = + s wh s 1. Le P 0 A wh µ(p 0 ) ([n] \ 0 }) = + s. Then µ(p 0 ) = s or + s, dependng on wheher 0 µ(p 0 ) or no. By Lemma 7, µ(p 0 ) Snce n 5 + 3, here s a -se T [n] \ (µ(p 0 ) 0 }). Le A 1 = x} : x [n] \ (T 0 })} T 0 }}. Then A 1 F 2+1 ( 0 ) = S j (A). Now, τ(p 0 )\(T 0 }) = n 2 1 s, T 0 } = +1 2 and (T 0 }) µ(p 0 ). The only block n A 1 ha has sze greaer han one s T 0 }. Snce (T 0 }) µ(p 0 ), T 0 } / P 0. So, P 0 and A 1 have sngleons n common only. Noe ha he number of sngleons ha P 0 and A 1 have n common s exacly τ(p 0 ) \ (T 0 }) = n 2 1 s. Thus, P 0 A 1 n 2 1 s n 2 2. Ths means ha A 1 / A and A 1 = s j (C 1 ) for some C 1 A. Now, here are wo possbles for C 1 dependng on wheher j s n T 0 } or no: () C 1 = x} : x [n] \ (T, 0 })} T, 0 }} and j T 0 }. () C 1 = x} : x [n] \ (T, 0, j})} (T 0 }),, j}}. If () holds, hen (T, 0 }) / P 0 snce (T 0 }) µ(p 0 ), and τ(p 0 )\(T, 0 }) τ(p 0 ) \ (T 0 }) = n 2 1 s. Thus, P 0 C 1 n 2 1 s n 2 2, a conradcon. Suppose () holds. The number of sngleons ha P 0 and C 1 have n common s a mos τ(p 0 ) \ (T, 0, j}) τ(p 0 ) \ (T 0 }) n 2 1 s. Recall ha T 0 } / P 0. If, j} / P 0, hen P 0 C 1 n 2 1 s n 2 2, a conradcon. If, j} P 0, hen P 0 C 1 n 2 s. Snce P 0 C 1 n 2 1, we mus have s = 1, P 0 C 1 = n 2 1, and C 1 = x} : x [n]\(t, 0, j})} (T 0 }),, j}}. Bu hen µ(c 1 ) ([n]\ 0 }) = T, j} = +2 > +1 = µ(p 0 ) ([n]\ 0 }), conradcng he choce of P 0. Thus, A F 2+1 ( 0 ). By Lemma 4, A = S j (A) = F 2+1 ( 0 ). Hence, A = F 2+1 ( 0 ). 3 Man resul Lemma 9. Le 1, A B(n) and W [n]. Suppose ha W q and µ(p ) \ W 1 for all P A. Then here exss an n 0 = n 0 (q, ) such ha for all n n 0, Proof. Noe ha for each P A, A < n 0.5. µ(p ) = C 1 C 2, he elecronc journal of combnaorcs 22(1) (2015), #P1.84 5

6 where C 1 [n] \ W, C 1 1 and C 2 W. The number of such C 1 s a mos 1 ( ) n W 1 ( ) n, and he number of such C 2 s a mos 2 W 2 q. Furhermore, µ(p ) = µ(p ) \ W + µ(p ) W 1 + q. Therefore he number of Q A wh µ(q) = µ(p ) s a mos B µ(p ) B 1+q, where B m s he number of sngleon-free se parons of [m]. Thus A B 1 ( ) n 1+q 2 q. If = 1, hen A B q 2 q < n 0.5 provded ha n > (B q 2 q ) 2. Suppose 2. Then A B 1 1+q 2 (1 q n 1 ( + 1 j ) )! n =1 j=1 ) < B 1 1+q 2 (1 q + n 1 =1 = B 1+q 2 q n 1 < n 0.5, provded ha n ( B 1+q 2 q ) 2. Ths complees he proof of he lemma. Lemma 10. For 2, here exss an n 0 = n 0 () such ha for all n n 0, f A I(n, n 2), hen A F 2. Moreover, equaly holds f and only f A = F 2. Proof. Suppose A F 2. Then max P A µ(p ) = + s wh s 1. Le P 0 A wh µ(p 0 ) = + s. By Lemma 7, max P A µ(p ) 4. Clam. µ(p ) \ µ(p 0 ) 1 for all P A. Suppose here s a Q A wh µ(q) \ µ(p 0 ). Then τ(p 0 ) \ µ(q) n 2 s. Snce P 0 Q n 2, by Lemma 6, µ(p 0 ) µ(q) 2s. Therefore µ(q) = µ(q) \ µ(p 0 ) + µ(p 0 ) µ(q) + 2s. On he oher hand, µ(q) µ(p 0 ) = + s by he choce of P 0. Ths mples ha s 0, a conradcon. Hence, he clam follows. By Clam and Lemma 9 (ake W = µ(p 0 ) and q = 4), A < n 0.5. Noe ha B B 2 = 1 for 2. So, by equaon (1), F 2 = ( ) ( ) n B B n 1 1 (n j) n!!2 > 1 n 0.5, ( ) provded ha n max (!2 1 ) 2, 2 2. Thus, A < F 2. Suppose A F 2. Then A F 2 and equaly holds f and only f A = F 2. j=0 he elecronc journal of combnaorcs 22(1) (2015), #P1.84 6

7 Lemma 11. For 1, here exss an n 0 I(n, n 2 1) and A s compressed, hen = n 0 () such ha for all n n 0, f A A F 2+1 (1). Moreover, equaly holds f and only f A = F 2+1 ( 0 ) for some 0 [n]. Proof. Snce 1, B+1 B 2 = 1. Therefore, by equaon (2), for all a [n], ( ) ( ) n F 2+1 (a) = B + B n 1 +1 ( ) ( ) B n n 1 + ( ) = B n (n 1 j)! B ( n j=0 ) + n!2, (3) provded ha n 2. Suppose max P A µ(p ). Then µ(p ) ([n] \ 1}) for all P A. Hence, A F 2+1 (1) and he lemma follows. Suppose max P A µ(p ) = + s wh s 1. Le P 0 A wh µ(p 0 ) = + s. By Lemma 7, max P A µ(p ) Clam. If s 2, hen µ(p ) \ µ(p 0 ) 1 for all P A. Suppose here s a Q A wh µ(q) \ µ(p 0 ). Then τ(p 0 ) \ µ(q) n 2 s = n 2 1 (s 1). Snce P 0 Q n 2 1, by Lemma 6, P 0 and Q have a leas (s 1) blocks of sze a leas 2 n common and µ(p 0 ) µ(q) 2(s 1). Therefore µ(q) = µ(q) \ µ(p 0 ) + µ(p 0 ) µ(q) + 2(s 1). On he oher hand, µ(q) µ(p 0 ) = + s by he choce of P 0. Ths mples ha s 2. Snce s 2, we mus have s = 2, µ(q) = µ(p 0 ) = + 2 and P 0 and Q have exacly one block of sze 2 n common, say, j}. Snce A s compressed, s j (Q) A. Noe ha µ(s j (Q)) = µ(q) \ µ(p 0 ). So, µ(s j (Q)) \ µ(p 0 ) = and τ(p 0 ) \ µ(s j (Q)) = n 2 2 = n Snce P 0 s j (Q) n 2 1, by Lemma 6, µ(p 0 ) µ(s j (Q)) 2. Ths conradcs ha µ(s j (Q)) = µ(q) \ µ(p 0 ). Hence, he clam follows. Suppose s 2. By Clam and Lemma 9 (ake W = µ(p 0 ) and q = 4 + 2), A < n 0.5 for suffcenly large n. I hen follows from equaon (3) ha A < n 0.5 < n!2 F 2+1(1), f n (!2 ) 2. Suppose s = 1. Then µ(p ) + 1 for all P A. Le P 0, P 1,..., P m A be such ha for all 1 m, we have he elecronc journal of combnaorcs 22(1) (2015), #P1.84 7

8 () µ(p ) = + 1, and () µ(p ) \ ( 1 j=0 µ(p j)) =. We may assume ha m s he larges neger n he sense ha here s no R A wh µ(r) = + 1 and µ(r) \ ( m j=0 µ(p j)) =. If here s a Q A wh µ(q) \ ( m j=0 µ(p j)) + 1, hen µ(q) = + 1 and µ(q) µ(p 0 ) =. So, P 0 Q = τ(p 0 )\µ(q) = n 2 2 < n 2 1, a conradcon. Thus, µ(p ) \ ( m j=0 µ(p j)) for all P A, and A = A 1 A 2 where ( m } A 1 = P A : µ(p ) \ µ(p j )) 1, A 2 = P A : j=0 ( m } µ(p ) \ µ(p j )) = and µ(p ) =. j=0 Suppose m. Then m j=0 µ(p j) m j=0 µ(p j) ( + 1) 2. By Lemma 9 (ake W = m j=0 µ(p j) and q = ( + 1) 2 ), A 1 < n 0.5 for suffcenly large n. Noe ha he number of µ(r) wh R A and µ(r) = s a mos ( n ) and he number of Q A wh µ(q) = µ(r) s a mos B. Thus, ( ) A 2 B n. I hen follows from equaon (3) ha A A 1 + A 2 < n B ( n f n (!2 ) 2. Suppose m + 1. ) ( ) < n!2 + B n F 2+1 (1), Clam. There s a 0 [n] wh 0 P for = 0, 1, 2,..., + 1. Noe ha f µ(p ) µ(p j ) = for j, hen P P j = τ(p ) \ µ(p j ) = n 2 2 < n 2 1, a conradcon. So, µ(p ) µ(p j ) for j. By properes () and (), we may conclude ha µ(p ) µ(p j ) = 1 for all, j wh j. Le µ(p 1 ) µ(p 0 ) = 0 }, µ(p ) µ(p 0 ) = j 1 } and µ(p ) µ(p 1 ) = j 2 } where Snce µ(p ) = + 1 and µ(p ) \ ( 1 j=0 µ(p j)) =, j 1 = j 2 µ(p 1 ) µ(p 0 ) = 0 }. Thus, 0 P for = 0, 1, 2,..., + 1. Ths complees he proof of he clam. By Clam, µ(p ) = W 0 }, he elecronc journal of combnaorcs 22(1) (2015), #P1.84 8

9 for = 0, 1,..., + 1 and W W j = for j. Suppose A F 2+1 ( 0 ). Then here s a Q A wh µ(q) ([n] \ 0 }) = + 1,.e., µ(q) = + 1 and 0 / µ(q). Noe ha µ(q) µ(p ) for all, for oherwse, Q P = τ(q)\µ(p ) = n 2 2 < n 2 1. Therefore µ(q) W. Snce W W j = for j, µ(q) wll have a leas + 2 elemens, a conradcon. Hence, A F 2+1 ( 0 ), A F 2+1 ( 0 ) and equaly holds f and only f A = F 2+1 ( 0 ). Ths complees he proof of he lemma. Proof of Theorem 3. If r = 2, hen he heorem follows from Lemma 10. Suppose r = 2+1. By repeaedly applyng he splng operaons, we evenually oban a compressed famly A I(n, n 2 1) wh A = A (Lemma 5). I hen follows from Lemma 11 ha A = A F 2+1 (1) and equaly holds f and only f A = F 2+1 ( 0 ) for some 0 [n]. By Theorem 8, we may conclude ha A = F 2+1 ( 0 ) mples ha A = F 2+1 ( 0 ). Ths complees he proof of Theorem 3. Acknowledgmens We would lke o hank Peer Frankl for suggesng us o look a Theorem 2, and he anonymous referee for readng our paper. Ths projec was suppored by he Froner Scence Research Cluser, Unversy of Malaya (RG367-15AFR). References [1] R. Ahlswede and L. H. Khacharan, The complee nersecon heorem for sysems of fne ses, European J. Combn. 18 (1997), [2] R. Ahlswede and L.H. Khacharan, The damerc heorem n Hammng spaces Opmal ancodes, Adv. n Appl. Mah. 20 (1998), [3] C. Bey, On cross-nersecng famles of ses, Graphs Combn. 21 (2005), [4] P. Borg, Exremal -nersecng sub-famles of heredary famles, J. London Mah. Soc. 79 (2009), [5] P. Borg, On -nersecng famles of sgned ses and permuaons, Dscree Mah. 309 (2009), [6] P. Borg and F.C. Holroyd, The Erdos-Ko-Rado propery of varous graphs conanng sngleons, Dscree Mah. 309 (2009), [7] F. Brunk and S. Huczynska, Some Erdos-Ko-Rado heorems for njecons, European J. Combn. 31 (2010), [8] P. J. Cameron and C. Y. Ku, Inersecng famles of permuaons, European J. Combn. 24 (2003), [9] A. Chowdhury and B. Pakós, Shadows and nersecons n vecor spaces, J. Combn. Theory Ser. A 117 (2010), [10] M. Deza and P. Frankl, On he maxmum number of permuaons wh gven maxmal or mnmal dsance, J. Combn. Theory Ser. A 22 (1977), he elecronc journal of combnaorcs 22(1) (2015), #P1.84 9

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