aa r X i v : . [ m a t h . C O ] J a n Tilted Sperner Families
Imre Leader ∗ Eoin Long † June 8, 2018
Abstract
Let A be a family of subsets of an n -set such that A does not containdistinct sets A and B with | A \ B | = 2 | B \ A | . How large can A be? Our aimin this note is to determine the maximum size of such an A . This answersa question of Kalai. We also give some related results and conjectures. A set system
A ⊆ P [ n ] = P ( { , . . . , n } ) is said to be an antichain or Spernerfamily if A B for all distinct A, B ∈ A . Sperner’s theorem [5] says that anyantichain A has size at most (cid:0) n ⌊ n/ ⌋ (cid:1) . (See [2] for general background.)Kalai [3] noted that the antichain condition may be restated as: A does notcontain A and B such that, in the subcube of the n -cube spanned by A and B , they are the top and bottom points. He asked what happens if we ‘tilt’ thiscondition. For example, suppose that we instead forbid A , B such that A is1/3 of the way up the subcube spanned by A and B ? Equivalently, A cannotcontain two sets A and B with | A \ B | = 2 | B \ A | .An obvious example of such a system is any level set [ n ] ( i ) = { A ⊂ [ n ] : | A | = i } . Thus we may certainly achieve size (cid:0) n ⌊ n/ ⌋ (cid:1) . The system [ n ] ( ⌊ n/ ⌋ ) isnot maximal, as we may for example add to it all sets of size ⌊ n/ ⌋ − A must have size o (2 n ).Our aim in this note is to verify this. We show that the middle layer isasymptotically best, in the sense that the maximum size of such a family is(1 + o (1)) (cid:0) n ⌊ n/ ⌋ (cid:1) . We also find the exact extremal system, for n even andsufficiently large. We give similar results for any particular ‘forbidden ratio’ inthe subcube spanned.What happens if, instead of forbidding a particular ratio, we instead forbidan absolute distance from the bottom point? For example, for distance 1 thiswould correspond to the following: our set system A must not contain sets A and B with | A \ B | = 1. How large can A be? ∗ Department of Pure Mathematics and Mathematical Statistics, Centre for MathematicalSciences, Cambridge CB3 0WB, United Kingdom. E-mail: [email protected] † Department of Pure Mathematics and Mathematical Statistics, Centre for MathematicalSciences, Cambridge CB3 0WB, United Kingdom. E-mail: [email protected] of the second author is supported by a Benefactor Scholarship from St. John’sCollege, Cambridge. n (cid:0) n ⌊ n/ ⌋ (cid:1) , which is about (aconstant fraction of) 1 /n / of the whole cube. But we are not able to show thatthis is optimal: the best upper bound that we are able to give is 2 n /n . How-ever, if we strengthen the condition to A not having A and B with | A \ B | ≤ n (cid:0) n ⌊ n/ ⌋ (cid:1) , up to amultiplicative constant. In this section we consider the problem of finding the maximum size of a family A of subsets of [ n ] which satisfies p | A \ B | 6 = q | B \ A | for all A, B ∈ A where p : q is a fixed ratio. Initially we will focus on the first non-trivial case 1 : 2 (notethat 1 : 1 is trivial as then the condition just forbids two sets of the same sizein A ) and then at the end of the section we extend these results to any givenratio.As mentioned in the Introduction, for the ratio 1 : 2 we actually obtain theextremal family when n is even and sufficiently large. This family, which we willdenote by B , is a union of level sets: B = ∪ i ∈ I [ n ] ( i ) . Here the set I is definedas follows: I = { a i : i ≥ } ∪ { b i : i ≥ } , where a = b = n and a i and b i aredefined inductively by taking a i = ⌈ a i − ⌉ − b i = ⌊ b i − + n ⌋ + 1 for all i .For example, if n = 2 k then I = { k − } ∪ { i − ≤ i ≤ k − } ∪ { k − i + 1 :0 ≤ i ≤ k − } . Noting that for any sets A and B with either (i) | A | = l where l < n and | B | > l or (ii) | A | = l where l > n and | B | < l − n wehave | A \ B | 6 = 2 | B \ A | , we see that B satisfies the required condition. Our mainresult is the following. Theorem 1.
Suppose A is a set system on ground set [ n ] such that | A \ B | 6 =2 | B \ A | for all distinct A, B ∈ A . Then |A| ≤ (1 + o (1)) (cid:0) n ⌊ n/ ⌋ (cid:1) . Furthermore,if n is even and sufficiently large then |A| ≤ |B | , with equality if and only if A = B . The main step in the proof of Theorem 1 is given by the following lemma. Theproof is a Katona-type (see [4]) averaging argument.
Lemma 2.
Let A be a set system on [ n ] such that | A \ B | 6 = 2 | B \ A | for alldistinct A, B ∈ A . Then l X j = l |A j | (cid:0) nj (cid:1) ≤ for all l ≤ n and k X j =2 k − n |A j | (cid:0) nj (cid:1) ≤ for all k ≥ n , where A j = A ∩ [ n ] ( j ) .Proof. We only prove the first inequality, as the proof of the second is identical.Pick a random ordering of [ n ] which we denote by ( a , a , . . . , a ⌈ n ⌉ , b , . . . , b ⌊ n ⌋ ).2iven this ordering, let C i = { a j : j ∈ [2 i ] } ∪ { b k : k ∈ [ i + 1 , l ] } and let C = { C i : i ∈ [0 , l ] } . Consider the random variable X = |A ∩ C| . Since each set B ∈ [ n ] ( i ) is equally likely to be C i − l we have P [ B ∈ C ] = ( ni ) . Thus by linearityof expectation we have E ( X ) = l X i = l |A i | (cid:0) ni (cid:1) (1)On the other hand, given any C i , C j with i < j we have | C i \ C j | = 2 | C j \ C i | and so A can contain at most one of these sets. This gives E ( X ) ≤
1. Togetherwith (1) this gives the claimed inequality l X i = l | A i | (cid:0) ni (cid:1) ≤ Proof of Theorem 1.
We first show |A| ≤ (1 + o (1)) (cid:0) n ⌊ n/ ⌋ (cid:1) . By standard esti-mates (See e.g. Appendix A of [1]) we have | [ n ] ( ≤ αn ) ∪ [ n ] ( ≥ (1 − α ) n ) | = o ( (cid:0) n ⌊ n/ ⌋ (cid:1) )for any fixed α ∈ [0 , ), so it suffices to show that | S n i = n A i | ≤ (cid:0) n n (cid:1) . But thisfollows immediately from Lemma 2 by taking l = ⌊ n ⌋ .We now prove the extremal part of the claim in Theorem 1. We first showthat the maximum of f ( x ) = P ni =0 x i subject to the inequalities l X j = l x j (cid:0) nj (cid:1) ≤ , l ∈ { , , . . . , ⌊ n ⌋} (2)and k X j =2 k − n x j (cid:0) nj (cid:1) ≤ , k ∈ {⌈ n ⌉ , . . . , n } (3)from Lemma 2 occurs when x n/ = (cid:0) n n (cid:1) . Indeed, suppose otherwise. At leastone of these inequalities involving x n/ must occur with equality, as otherwisewe can increase x n/ slightly, increase the value of f ( x ) and still satisfy (2) and(3). Pick j > n as small as possible such that x j >
0. Let y n/ = x n/ + ǫ (cid:0) nn/ (cid:1) , y j = x j − ǫ (cid:0) nj (cid:1) and y i = x i for all other i . As f ( y ) > f ( x ) one of the (2) or(3) must fail. If ǫ is sufficiently small only the inequalities involving y n/ andnot y j can be violated. Choose k < n/ y k > y k doesnot occur in any inequality involving y j . Note that we must have j − k ≥ n .Decrease y k by ǫ (cid:0) nk (cid:1) . Since the only increased variable y n/ always occurs withone of y j or y k , it follows that y = ( y , . . . , y n ) satisfies (2) and (3).We claim that f ( y ) > f ( x ). Indeed, we must have either | j − n | ≥ n or | k − n | ≥ n . Without loss of generality assume that | k − n | ≥ n . Then since (cid:0) nn/ (cid:1) > (cid:0) n ( n/ (cid:1) + (cid:0) n n/ (cid:1) for sufficiently large n we have f ( y ) = f ( x )+ ǫ (cid:18) nn/ (cid:19) − ǫ (cid:18) nj (cid:19) − ǫ (cid:18) nk (cid:19) > f ( x )+ ǫ (cid:18) nn/ (cid:19) − ǫ (cid:18) n ( n/
2) + 1 (cid:19) − ǫ (cid:18) n n/ (cid:19) > f ( x ) . Therefore we must have x n/ = (cid:0) nn/ (cid:1) , as claimed.3ow, by the inequalities (2) and (3) we have x j = 0 for all n ≤ j ≤ n with j = n . From here it is easy to see by a weight transfer argument that f ( x ) hasa unique maximum when x i = (cid:0) ni (cid:1) for i ∈ I and x i = 0 otherwise. For a setsystem A these values of x i = |A i | can only be achieved if A = B , as claimed. (cid:3) We remark that the statement of Theorem 1 does not hold for all even n , ascan be seen for example by taking n = 4 and A = P [ n ] \ [ n ] (2) .We now extend Theorem 1 from the ratio 1 : 2 to any given ratio p : q .Let p : q be in its lowest terms and p < q . If A ∈ [ n ] ( i + a ) and B ∈ [ n ] ( i ) satisfy p | A \ B | = q | B \ A | then we have p ( a + b ) = q ( b ) where b = | B \ A | . Butthen pa = ( q − p ) b and since p and q are coprime we must have that ( q − p ) | a .Therefore any family A = S i ∈ I [ n ] ( i ) , where I is an interval of length q − p ,satisfies p | A \ B | 6 = q | B \ A | for all A, B ∈ A . Taking ⌊ n ⌋ ∈ I gives |A| =( q − p + o (1)) (cid:0) n ⌊ n/ ⌋ (cid:1) . Our next result shows that this is asymptotically bestpossible. Theorem 3.
Let p, q ∈ N be coprime with p < q . Let A be a set system onground set [ n ] such that p | A \ B | 6 = q | B \ A | for all distinct A, B ∈ A . Then |A| ≤ ( q − p + o (1)) (cid:0) n ⌊ n/ ⌋ (cid:1) . The following lemma performs an analogous role to that of Lemma 2 in theproof of Theorem 1.
Lemma 4.
Let A be a set system on [ n ] such that p | A \ B | 6 = q | B \ A | for alldistinct A, B ∈ A . Then X j ∈ J k |A j | (cid:0) nj (cid:1) ≤ where J k = { l : ⌈ pnp + q ⌉ ≤ l ≤ ⌊ qnp + q ⌋ , l ≡ k (mod ( q − p )) } for ≤ k ≤ q − p − .Proof. We only sketch the proof, as it is very similar to the proof of Lemma 2.For convenience we assume n = ( p + q ) m (this assumption is easily removed).Fix k ∈ [0 , q − p −
1] and let k ′ ≡ k − pm (mod ( q − p )) where k ′ ∈ [0 , q − p − n ] which we denote by ( a , a , . . . , a qm , b , . . . , b pm ).Given this ordering let C i = { a j : j ∈ [ qi + k ′ ] } ∪ { b j : j ∈ [ pi + 1 , pm ] } and let C = { C i : i ∈ [0 , m − } . (Here if k ′ = 0 we additionally adjoin C m to C .) Bychoice of k ′ , we have | C i | ∈ J k for all i ∈ [0 , m − C i and C j with i < j we have q | C i \ C j | = p | C j \ C i | , whichimplies that A contains at most one element of C . Using this the rest of theproof is as in Lemma 2.The proof of Theorem 3 is now identical to the proof of Theorem 1 takingLemma 4 in place of Lemma 2.For simplicity we have given in Lemma 4 only the inequalities that we neededin order to prove Theorem 3. Further inequalities involving smaller level setsanalogous to those in Lemma 2 can also be obtained in a similar fashion. Whilewe have not done so here, we note that it is possible to use these inequalities toagain find an exact extremal family for any given ratio p : q as in Theorem 1,provided q − p and n have the opposite parity and n is sufficiently large.4 Forbidding a fixed distance
In this final section we consider how large a family A can be if for all A, B ∈ A we do not allow A to have a constant distance from the bottom of the sub-cube formed with B . For ‘distance exactly 1’ this would mean that we ex-clude | A \ B | = 1 for A, B ∈ A . Here the following family A ∗ provides a lowerbound: let A ∗ consist of all sets A of size ⌊ n/ ⌋ such that P i ∈ A i ≡ r (mod n ),where r ∈ { , . . . , n − } is chosen to maximise |A ∗ | . Such a choice of r gives |A ∗ | ≥ n (cid:0) n ⌊ n/ ⌋ (cid:1) . Note that if we had | A \ B | = 1 for some A, B ∈ A ∗ then, since | A | = | B | , we would also have | B \ A | = 1. Letting A \ B = { i } and B \ A = { j } we then have i − j ≡ n ), giving i = j , a contradiction.We suspect that this bound is best. Conjecture 5.
Let
A ⊂ P [ n ] be a family which satisfies | A \ B | 6 = 1 for all A, B ∈ A . Then |A| ≤ (1 + o (1)) n (cid:0) n ⌊ n/ ⌋ (cid:1) . The following gives an upper bound that is a factor n / larger than this. Theorem 6.
Let
A ⊂ P [ n ] be a family such that | A \ B | 6 = 1 for all A, B ∈ A .Then there exists a constant C independent of n such that |A| ≤ Cn n .Proof. An easy estimate gives that the number of subsets of A in [ n ] ( ≤ n/ S [ n ] ( ≥ n/ is at most 4 (cid:0) nn/ (cid:1) = o ( n n ). Therefore it suffices to show that |A i | ≤ Cn (cid:0) ni (cid:1) for all i ∈ [ n , n ].To see this, note that since | A \ A ′ | 6 = 1 for all A, A ′ ∈ A , each B ∈ [ n ] ( i +1) contains at most one A ∈ A i . Double counting, we have n |A i | ≤ ( n − i ) |A i | = |{ ( A, B ) : A ∈ A i , B ∈ [ n ] ( i +1) , A ⊂ B }|≤ (cid:18) ni + 1 (cid:19) ≤ (cid:18) ni (cid:19) as required.Our final result gives an upper bound on the size of a family A in whichwe forbid ‘distance at most 1’ instead of ‘distance exactly 1’, i.e. where wehave | A \ B | > A, B ∈ A . Again, the family A ∗ constructed abovegives a lower bound for this problem. In general, if we forbid ‘distance at most k ’ then it is easily seen that the following family A ∗ k gives a lower bound of n k (cid:0) n ⌊ n/ ⌋ (cid:1) : supposing n is prime, let A ∗ k consist of all sets A of ⌊ n/ ⌋ whichsatisfy P i ∈ A i d ≡ n ) for all 1 ≤ d ≤ k .Our last result provides a upper bound which matches this up to a mul-tiplicative constant. The proof is again a Katona-type argument. Here thecondition | A \ B | > k rather than | A \ B | 6 = k seems to be crucial. Theorem 7.
Let k ∈ N . Suppose A is a set system on [ n ] such that | A \ B | > k for all distinct A, B ∈ A . Then |A| ≤ (2 k − o (1)) n k (cid:0) n ⌊ n/ ⌋ (cid:1) .Proof. Consider the family ∂ ( k ) A , the k -shadow of A , where ∂ ( k ) A = { B ∈ P [ n ] : B = A \ C for some A ∈ A and C ⊂ A with | C | = k } . A does not contain A, B with | A \ B | ≤ k , every element of ∂ ( k ) A iscontained in at most one element of A . Therefore we have | ∂ ( k ) A| = n X i =0 ( i ) k |A i | (4)where i k = i ( i − · · · ( i − k + 1). Now, since A does not contain A, B with | A \ B | ≤ k , it follows that ∂ ( k ) A is an antichain, and so by Sperner’s theoremwe have | ∂ ( k ) A| ≤ (cid:18) n ⌊ n/ ⌋ (cid:19) (5)Finally, an estimate of the sum of binomial coefficients (Appendix A of [1]) gives n − n / X i =0 |A i | ≤ n − n / X i =0 (cid:18) ni (cid:19) ≤ e − n / n . (6)Combining (4), (5) and (6) we obtain (cid:18) n ⌊ n/ ⌋ (cid:19) ≥ n − n / X i =0 ( i ) k |A i | + n X i = n − n / ( i ) k |A i |≥ n − n / X i =0 ( n − n / ) k |A i | − ( n − n / ) k e − n / n + n X i = n − n / ( n − n / ) k |A i | = ( n − o ( n )) k |A| − o ( (cid:18) n ⌊ n/ ⌋ (cid:19) )which gives the desired result.Taking k = 1 in Theorem 7 we obtain an upper bound which differs by afactor of 2 from the lower bound given by the family A ∗ . It would be interestingto close this gap. References [1] N. Alon and J. Spencer,
The Probabilistic Method , Wiley, 3rd ed., 2008.[2] B. Bollob´as,
Combinatorics: Set Systems, Hypergraphs, Families of Vectorsand Combinatorial Probability , Cambridge University Press, 1st ed, 1986.[3] G. Kalai, Personal communication (2010).[4] G.O.H. Katona, Two applications of Sperner type theorems,
Period. Math.Hungar. (1973), 19-26.[5] E. Sperner, Ein Satz ¨uber Untermengen einer endlichen Menge. Math. Z. ,27