An upper bound on the size of diamond-free families of sets

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by  Dániel Grósz

An upper bound on the size of diamond-free families of sets
Dániel Grósz∗ Abhishek Methuku† Casey Tompkins‡
Abstract
Let La(n,P) be the maximum size of a family of subsets of [n] = {1,2,...,n} not containing P
asa(weak)subposet. Thediamondposet,denotedB ,isdefinedonfourelementsx,y,z,w withthe
2
6
relations x < y,z and y,z < w. La(n,P) has been studied for many posets; one of the major open
1
problems is determining La(n,B ).
0 2
Studying the average number of sets from a family of subsets of [n] on a maximal chain in the
2
Boolean lattice 2[n] has been a fruitful method. We use a partitioning of the maximal chains and

n introduce an induction method to show that La(n,B2) ≤ (2.20711+o(1))(cid:0)⌊nn⌋(cid:1), improving on the
Ja earlier bound of (2.25+o(1))(cid:0)⌊nn⌋(cid:1)by Kramer, Martin and Young. 2
2
4
2
1 Introduction

]
O
Let[n]= 1,2,...,n . TheBooleanlattice2[n]isdefinedasthefamilyofallsubsetsof[n]= 1,2,...,n ,
C { } { }
and the ith level of 2[n] refers to the collection of all sets of size i. In 1928, Sperner proved the following
.
h well-known theorem.
t
a
Theorem1.1(Sperner[23]). If isafamilyofsubsetsof[n]suchthatnosetcontainsanother(A,B
m implies A B), then n F. Moreover, equality occurs if and only if is a level of maximum∈sizFe
[ 6⊂ |F|≤ ⌊n2⌋ F
in 2[n]. (cid:0) (cid:1)
1
v Definition 1.2. Let P be afinite poset, and be a family ofsubsets of[n]. We saythat P is contained
2 F
in as a (weak) subposet if there is an injection ϕ : P satisfying x < x ϕ(x ) ϕ(x )
3 F → F 1 p 2 ⇒ 1 ⊂ 2
for every x ,x P. is called P-free if P is not contained in as a weak subposet. We define the
3 1 2
∈ F F
6 corresponding extremal function as La(n,P):=max : is P-free .
{|F| F }
0
. A k-chain, denoted by P , is defined to be the poset on the set x ,x ,...,x with the relations
1 k 1 2 k
60 x1 ≤ x2 ≤ ··· ≤ xk. Using the above notation, Sperner’s theorem ca{n be stated a}s La(n,P2) = ⌊nn2⌋ .
LetΣ(n,k) denote the sumof the k largestbinomialcoefficients ofordern. An importantgeneralization
(cid:0) (cid:1)
1
ofSperner’stheoremduetoErdős [8]statesthatLa(n,P )=Σ(n,k). Moreover,equalityoccursifand
: k+1
v only if is the union of k of the largest levels in 2[n].
i F
X
Definition 1.3. The diamond poset, denoted (or or Q ), is a poset on four elements x,y,z,w ,
2 2 2
r B D { }
with the relations x < y,z and y,z < w. That is, is a subposet of a family of sets if there are
a 2
B A
different sets A,B,C,D with A B,C and B,C D. (Note that B and C are not necessarily
∈ A ⊂ ⊂
unrelated.)
TheVposetis aposeton x,y,z withthe relationsx y,z;the Λposetisdefinedon x,y,z withthe
{ } ≤ { }
relationsx,y z. Thatis,the Λisasubposetofafamily ofsets iftherearedifferentsetsB,C,D
≤ A ∈A
with B,C D.
⊂
The general study of forbidden poset problems was initiated in the paper of Katona and Tarján [14] in
1983. They determined the size of the largest family of sets containing neither a V nor a Λ. They also
gave an estimate on the maximum size of V-free families: 1+ 1 +o 1 La(n,V) 1+ 2 n .
n n ≤ ≤ n ⌊n2⌋
This result was later generalized by De Bonis and Katona [6] who obtained bounds for the r-fork poset,
(cid:0) (cid:0) (cid:1)(cid:1) (cid:0) (cid:1)(cid:0) (cid:1)
V defined by the relations x y ,y ,...,y . Other posets for which La(n,P) has been studied include
r 1 2 r
≤
∗DepartmentofMathematics,UniversityofPisa. e-mail: [email protected]
†DepartmentofMathematics,Central EuropeanUniversity, Budapest,Hungary. e-mail: [email protected]
‡AlfrédRényiInstituteofMathematics,HungarianAcademyofSciences. e-mail: [email protected]
1
complete two level posets, batons [24], crowns O (cycle of length 2k on two levels, asymptotically
2k
solved except for k 3,5 [12, 16]), butterfly [7], skew-butterfly [21], the N poset [9], harp posets
∈ { }
(l ,l ,...,l ),definedbykchainsoflengthl betweentwofixedelements[11],andrecentlythecomplete
1 2 k i
H
3 level poset K [22] among others. (See [10] for a nice survey by Griggs and Li.)
r,s,t
One of the first general results is due to Bukh [3] who determined the asymptotic value of La(n,P) for
all posets whose Hasse diagram is a tree: If T is a finite poset whose Hasse diagram is a tree of height
h(T) 2, then La(n,T)=(h(T) 1) n 1+O 1 .
≥ − n/2 n
⌊ ⌋
Using more general structures instea(cid:0)d of c(cid:1)h(cid:0)ains fo(cid:0)r d(cid:1)o(cid:1)uble counting, Burcsi and Nagy [4] obtained a
weaker version of this theorem for general posets showing that La(n,P) |P|+h(P) 1 n . Later
≤ 2 − n/2
⌊ ⌋
this was generalized by Chen and Li [5] and recently this general bound wa(cid:16)s improved by(cid:17)t(cid:0)he au(cid:1)thors of
the present article [13].
The most investigated poset for which even the asymptotic value of La(n,P) has yet to be determined
is the diamond which is the topic of our paper. The two middle levels of the Boolean lattice do not
2
containadiamoBnd, soLa(n, ) (2 o(1)) n . Using a simple andelegantargument,Griggs,Liand
Lu[11]showedthatLa(n,D2B)2<≥2.296− nn .(cid:0)S⌊on2m⌋(cid:1)etimeaftertheyhadannouncedthisbound,Axenovich,
Manske,and Martin [18] improvedthe(cid:0)⌊u2p⌋p(cid:1)er bound to 2.283 nn . This bound was further improvedto
2.273 ⌊nn2⌋ by Griggs, Li and Lu [11]. The best known uppe(cid:0)r⌊b2o⌋(cid:1)und on La(n,B2) is (2.25+o(1)) ⌊nn2⌋
due to Kramer, Martin and Young [15].
(cid:0) (cid:1) (cid:0) (cid:1)
Definition 1.4. A maximal chain or,for the restof this article, simply a chain ofthe Booleanlattice is
a sequence of sets , x , x ,x , x ,x ,x ,...,[n] with x ,x ,x ... [n]. We refer to x ,...,x
1 1 2 1 2 3 1 2 3 1 i
∅ { } { } { } ∈ { }
as the ith set on the chain. In particular, we refer to x as the first set on the chain, or just say that
1
{ }
the chain starts with the element x (as a singleton). We refer to x as the ith element added to form
1 i
the chain.
Definition 1.5. The Lubell function of a family of sets 2[n] is defined as
F ⊆
1
l(n, )= .
F n
F F
X∈F | |
(cid:0) (cid:1)
The notation is shortened to just l( ) when there is no ambiguity as to the Boolean lattice.
F
Observation 1.6. The Lubell function of a family is the average number of sets from on a chain,
F F
taken over all n! chains. In particular, the Lubell function of a level is 1, and the Lubell function of
an antichain is the number of chains containing a set from divided by n!. The Lubell function is
additive acrosFs a disjoint union of families of sets. FurthermoreF, l( ) n ([17]).
|F|≤ F ⌊n2⌋
(cid:0) (cid:1)
The Lubell function was derived from the celebrated BLYM inequality which was independently discov-
ered by Yamamoto, Meshalkin, Bollobás and Lubell. Using the Lubell function terminology, it states
that
BLYM inequality (Yamamoto,Meshalkin,Bollobás,Lubell[25,20,2,17]). If 2[n] is an antichain,
F ⊆
then l( ) 1.
F ≤
ForaposetP,letl(n,P)be themaximumofl(n, )overallfamilies 2[n] whicharebothP-freeand
F F ⊆
contain the empty set. Let l(P)=limsup l(n,P). Griggs, Li and Lu proved that
n
→∞
Lemma 1.7 (Griggs, Li and Lu [11]).
n
La(n, ) l( )+o(1) .
B2 ≤ B2 n
(cid:18) 2 (cid:19)
(cid:0) (cid:1)
(cid:4) (cid:5)
Kramer,Martin and Young used flag algebras to prove that
Lemma 1.8 (Kramer, Martin and Young [15]). l( )=2.25
2
B
thereby proving
2
Theorem 1.9 (Kramer, Martin and Young [15]).
n
La(n, ) (2.25+o(1)) .
B2 ≤ n
(cid:18) 2 (cid:19)
(cid:4) (cid:5)
The following construction shows that Lemma 1.8 is sharp.
Example 1.10. Let 2[n] consist of all the sets of the following forms: , e , e,o , o ,o where
1 2
F ⊆ ∅ { } { } { }
e denotes any even number in [n], and o, o and o denote any odd numbers in [n]. This family is
1 2
diamond-free, and l( )=2.25 o(1).
F ±
Example 1.11. This construction is a generalization of the previous one. Let A [n] with A = an.
⊆ | |
Let 2[n] consist of all the sets of the following forms: , e , e,o , o ,o where now e denotes
1 2
F ⊆ ∅ { } { } { }
any element of A, while o, o and o denote any elements of [n] A. This family is diamond-free, and
1 2
\
l( )=2+a a2 o(1). This family contains all size 2 sets that do not form a diamond with and the
F − ± ∅
singletons, so all maximal diamond-free families on levels 0, 1 and 2 that contain are of this form.
∅
The following restriction of the problem of diamond-free families has been investigated: How big can a
diamond-free family be if it can only contain sets from the middle three levels of 2[n] (denoted (n,3))?
B
Better bounds are known with this restriction. Axenovich, Manske and Martin showed that
Theorem 1.12 (Axenovich, Manske and Martin [18]). If (n,3) is diamond-free, then
2.20711+o(1) n . F ⊆ B |F| ≤
n
⌊2⌋
(cid:0) (cid:1)(cid:0) (cid:1)
Later, Manske and Shen improved it to 2.1547 n in [19] and recently, Balogh, Hu, Lidický and Liu
n
gave the best known bound of 2.15121 nn in (cid:0)[1⌊]2u⌋s(cid:1)ing flag algebras.
⌊2⌋
(cid:0) (cid:1)
Definition1.13. Wecallachainmaximal–non-maximal (MNM) withrespectto(w.r.t.) ifitcontains
F
a set from , and the biggest set contained in on the chain is not maximal in (i.e., there are other
F F F
sets from containing it on some other chains).
F
Itiseasytoseethatan -freefamilyisΛ-freeifandonlyifthefamilywegetbyadding isdiamond-free;
∅ ∅
adding increases the Lubell function by 1. In Section 2 of this paper, we prove the following lemma:
∅
Lemma 1.14. Let 2[n] be a Λ-free family that does not contain the empty set, nor any set larger
than n n for someFn⊆ N (that can be chosen independently of n). Assume that there are cn! MNM
′ ′
− ∈
chains. Then l( ) 1 min c+ 1,1 + min c+ 1,1 + 3.
F ≤ − n′ 4 n′ 4 n′
q
(cid:0) (cid:1) (cid:0) (cid:1)
It is easy to see that in Example 1.11 the number of MNM-chains is approximately a2n! (so a √c):
≈
these are the chains whose second set is e ,e with e ,e A. Thus this lemma is (asymptotically)
1 2 1 2
{ } ∈
sharp, and states that for a given number of MNM chains, Example 1.11 cannot be beaten (with some
restrictiononthesizesofthesets). Barringtherequirementthatthetopmostn levelsbe empty,Lemma
′
1.14isageneralizationofLemma1.8. TheproofofLemma1.7in[15]actuallyworkswiththe restriction
of Lemma 1.14 on the topmost sets with n′ = n2 −n32, immediately giving a new proof of Theorem 1.9.
Our proof of Lemma 1.14 includes an intricate induction step and a (non-combinatorial) lemma about
functions involving a lot of elementary algebra and calculus; but it does not require flag algebras, and it
does not use details of the structure of above the second level (except inside the induction).
F
Section 3 of this paper uses Lemma 1.14 to prove our main theorem:
Theorem 1.15. La(n, ) √2+3 +o(1) n < 2.20711+o(1) n .
B2 ≤ 2 ⌊n2⌋ ⌊n2⌋
(cid:16) (cid:17)(cid:0) (cid:1) (cid:0) (cid:1)(cid:0) (cid:1)
The proof is inspired by the proof of Theorem 1.12 (the same bound when restricted to 3 levels) as in
[18],using the idea ofgroupingchainsby the smallestsetcontainedin onachain(asdevelopedin[11]
F
and [15]).
It is conceivable that the upper bound can be improved, further, to 2.1547 n by using ideas similar
n
⌊2⌋
to those introduced in this paper and based on [19]; we continue working on it.
(cid:0) (cid:1)
3
1.25
1.25
1
1 0.75
0.5
0.75
0.25
0.5
0
0 0.25 0.5 0 0.25 0.5 0.75 1
Values of f(x,c) plotted in x, for c = Valuesoff(x,c)plottedinc,forx=0,0.05,0.01,...,1.
0,0.0125,0.025,...,0.25. Note that the 1−x part
oftheplotscoincide.
2 Λ-free families – Proof of Lemma 1.14
Definition 2.1. We define the following functions:
For x [0,1],c [0, ),
• ∈ ∈ ∞
1 x+ 1 1 c if x 1 and c<4 x x2 2
− 4(x x2) − ≤ 2 −
f(x,c)=xx22−−22xx(cid:16)++11.−2−5c+√c(cid:17) iiff xx≤≤ 1221 aanndd 414(cid:0)≤x−c(cid:0)x2(cid:1)2 ≤(cid:1)c≤ 14
1 x if 1 x.
 − 2 ≤
For x [0,1),c [0, ),a [0,1 x),a˜ 0,min a, c ,
• ∈ ∈ ∞ ∈ − ∈ x+a
h (cid:16) (cid:17)i
c a˜(x+a)
g(x,c,a,a˜)=a+(1 x a)f x+a, − +2a˜(1 x a).
− − 1 x a − −
(cid:18) − − (cid:19)
For x [0,1),c [0, ),a [0,1 x),a˜ 0,min a, c x ,
• ∈ ∈ ∞ ∈ − ∈ x+a −
h (cid:16) (cid:17)i
c (x+a˜)(x+a)
h(x,c,a,a˜)=a+(1 x a)f x+a, − +2a˜(1 x a)+x 3x(x+a).
− − 1 x a − − −
(cid:18) − − (cid:19)
Lemma 2.2. The functions above satisfy the following conditions:
1. For c [0, ), if c˜=min c,1 , then f(0,c)=1 c˜+√c˜.
∀ ∈ ∞ 4 −
2. f(x,c) is concave and mon(cid:0)otono(cid:1)usly increasing in c, and monotonously decreasing in x.
4
3. For x [0,1],c [0, ),a [0,1 x),a˜ 0,min a, c :g(x,c,a,a˜) f(x,c).
∀ ∈ ∈ ∞ ∈ − ∈ x+a ≤
h (cid:16) (cid:17)i
4. For x [0,1],c [0, ),a [0,1 x),a˜ 0,min a, c x :h(x,c,a,a˜) f(x,c).
∀ ∈ ∈ ∞ ∈ − ∈ x+a − ≤
h (cid:16) (cid:17)i
5. For c [0, ):1 x f(x,c).
∀ ∈ ∞ − ≤
We prove Lemma 2.2 in Appendix A.
We use induction on n to prove the following lemma. Lemma 1.14 is a special case of this lemma with
X = = .
X ∅
Lemma 2.3. Let 2[n] be a Λ-free family which does not contain , nor any set larger than n n
′
F ⊆ ∅ −
for some n′ ∈ N. Let us assume that we are given a “forbidden” set X ⊆ [n], with x = |Xn|. Also, let
be a “forbidden” antichain in which each set contains exactly one element of X (and may or may not
X
be a singleton). Let us assume that the sets in are disjoint from X, and unrelated to every set in .
F X
Let α = l( ), and let µn! be the number of chains which start with an element of X as a singleton,
X
but do not contain any set in . Assume, furthermore, that there are cn! MNM chains w.r.t. . Then
X F
l( ) f x,c+µ+ 1 (α µ x)+ 3.
F ≤ n′ − − − n′
Proposi(cid:0)tion 2.4. Lem(cid:1)ma 2.3 holds for n n.
′
≤
Proof. = . is an antichain, so, by the BLYM inequality, α 1. By Lemma 2.2 2. and 5.,
F ∅ X ≤
f x,c+µ+ 1 (α µ x)+ 3 f(x,c+µ) (α µ x) 1 x (α x) 0=l( ).
n′ − − − n′ ≥ − − − ≥ − − − ≥ F
(cid:0) (cid:1)
From now on we assume n n 1.
′
≤ −
Notation. Let A [n] X be the set of elements of [n] that appear as singletons in , and let a= |A|.
⊆ \ F n
Let be the family of those sets in which contain at least one element of A, but which are not
B F
singletons. Let β be the Lubell function . Let be the family of those sets in which only contain
B C F
elements of [n] X A.
\ \
Let A˜= e∈ A:(∃B ∈B : e ∈ B) , and let a˜ = |An˜|. Let c0n! be the number of chains that start with
e as a singleton for some e A˜, but do not contain any set from . Let νn! be the number of chains
t{ha}t start(cid:8)with an element of∈A˜ as(cid:9)a singleton, continue with an elBement of [n] X A as the second
\ \
element added to form the chain, yet do not contain any set from .
B
Let 1 = n > 1 and 1 = (x+a)n−1 1. These correction factors will account for the difference from
n 1 (x+a)(n 1) ≤
− −
the asymptotic behavior. (They are both typically close to 1. If x+a=0, let 1=1; it is irrelevant as it
will always be multiplied by x+a.)
Proposition 2.5. Every D is of the form d,o ,...,o with d X,o ,...,o [n] X A (where
1 k 1 k
∈X { } ∈ ∈ \ \
k may be 0).
Proof. D containsexactlyoneelementofX bydefinition. Ife A,thene / D,otherwiseD and e
∈ ∈ { }∈F
would be related.
Proposition 2.6. Sets in only contain one element of A. is an antichain, and the sets in are
B B B
also unrelated to every set in .
C
Proof. If e B with e A, and B was related to another set S , then e , B and S would
1 1 1
∈ ∈ B ∈ ∈ F { }
form a Λ. This applies to any S , as well as S = e for any e =e A.
2 1 2
∈B∪C { } 6 ∈
Proposition 2.7. a˜(x+a)1 a˜(x+a)1+ν =c c, and thus a˜ c .
≤ 0 ≤ ≤ (x+a)1
Proof. c c because a chain whose first set is e for some e A˜, but does not contain any set from
0
≤ { } ∈ B
is an MNM chain.
Any chain on which the singleton is e and the second set is e,d with e A˜ and d X A is always
{ } { } ∈ ∈ ∪
an MNM chain: e,d and any set that contains it is forbidden from being in either because it is not
{ } B
disjointfromX (whend X),orbecauseitwouldcontaintwoelementsofA(whend A). Thenumber
∈ ∈
of such chains is a˜n (an+xn 1) (n 2)!=a˜(x+a)n!1. And out of the chains which start with e ,
· − · − { }
and whose second set is e,o with some o [n] X A, νn! do not contain any set from .
{ } ∈ \ \ B
5
For a family of sets 2[n], let m( )n! be the number of chains which start with an element of X as a
A⊆ A
singleton and do not contain any set from . (For example, (l m)( )=α µ.) For a fixed d X, let
A − X − ∈
m ( )n! be the number of chains on which the singleton is d , and do not contain any element of .
d
A { } A
Proposition 2.8. For any d X, let = D :d D . We can assume without loss of generality
d
∈ X { ∈X ∈ }
that for any d ,d X, D d :D = D d :D . That is, if does not satisfy this
condition, we1sho2w∈a fam{ily\{ˆ1w}hich∈doXesd,1}and{also\{sa2ti}sfies ∈theXdc2o}nditions of LXemma 2.3’s statement
X
(each set contains exactly one element of X, the sets are unrelated to each other and to every set in ),
F
and for which f x,c+m( ˆ)+ 1 l( ˆ) m( ˆ) x f x,c+µ+ 1 (α µ x).
X n′ − X − X − ≤ n′ − − −
(cid:16) (cid:17) (cid:0) (cid:1) (cid:0) (cid:1)
Proof. Let d X be such that
0
∈
1
f x,c+ X m ( )+ X l( ) X m ( ) x
| | d0 Xd0 n − | | Xd0 −| | d0 Xd0 −
(cid:18) ′(cid:19)
1 (cid:0) (cid:1)
=min f x,c+ X m ( )+ X l( ) X m ( ) x .
d d d d d
d∈X(cid:20) (cid:18) | | X n′(cid:19)− | | X −| | X − (cid:21)
(cid:0) (cid:1)
Let ˆ = D d d :d X,D .
X \{ 0}∪{ } ∈ ∈Xd0
= , so α = l( ). It immediately follows from the definition of that if a chain
X d(cid:8)XXd d X Xd (cid:9) Xd
has d ∈as a singleton, and∈does not contain any set from , then it does not contain any set from
d
. {SoF}µ = m ( P). Similarly, l( ˆ) = X l( ) anXd m( ˆ) = X m ( ). Since f(x,c) is
X d X d Xd X | | Xd0 X | | d0 Xd0
monotonously in∈creasing and concave in c, using Jensen’s inequality
P
1
f x,c+ X m ( )+ X l( ) X m ( ) x
| | d0 Xd0 n − | | Xd0 −| | d0 Xd0 −
(cid:18) ′(cid:19)
(cid:0) (cid:1)
1 1
f x,c+ X m ( )+ X l( ) X m ( ) X x
d d d d d
≤ |X|"d X (cid:18) | | X n′(cid:19)− | |d X X −| |d X X −| | #
X∈ (cid:0) X∈ X∈ (cid:1)
1
f x,c+µ+ (α µ x).
≤ n − − −
(cid:18) ′(cid:19)
Sets in ˆ containexactly one element of X, andform an antichain. They are also unrelatedto every set
S :XS cannot contain any element of X, so it could only be related to a set in ˆ by being its subset.
∈F X
ButS mustalsobeunrelatedtoeveryD ,soitcannotbeasubsetofD d d either.
∈Xd0 ⊆X \{ 0}∪{ }
Corollary 2.9. With the assumption of Proposition 2.8,
either = d :d X (we refer to it as the singletons case),
• X { } ∈
or does n(cid:8)ot contain an(cid:9)y singleton (referred to as the no singleton case).
• X
Remark. Inthe singletonscase,the requirementforeverysetin tobe unrelatedtoeverysetin does
F X
notpose any restrictionon beyondother conditions of Lemma 2.3 (sets in are already disjoint from
F F
X). In the no singleton case, creates two restrictions:
X
The union of singletons in , A [n] .
• F ⊆ \ X
Setsin mustalsobeunrelatedtosetsiSn . Sucharestrictionexistsonlyif containssetsbigger
• C X X
than 2: sets in are necessarily unrelated to any two-element set that contains some d X since
C ∈
they may not contain d.
Example 2.10. LetC [n] X,andlet = d,o :d X,o C . Thenα=l( )= 2·xn·|C|·(n−2)! =
⊆ \ X { } ∈ ∈ X n!
2x|Cn|1, and µ = xn·(xn+ann−! 1)·(n−2)! = x(x+a(cid:8))1. The only restrict(cid:9)ion on F that this X creates is that
the union of singletons A [n] X C.
⊆ \ \
In other words,let us assume that α=l( )=2xγ1 for γ R (without assuming that is of the above
X ∈ X
form). Then it is possible that a = |A| can be as big as 1 x γ with not creating any restrictions
n − − X
on (depending on the actual structure of , namely, if it is made up of sets of size 2 as above; then
C X
α=2x(1 x a)1 and µ=x(x+a)1). But if a>1 x γ, this can only be possible at the same time
− − − −
as α=2xγ1 if contains sets bigger than 2, and thus creates restrictions on .
X C
6
Now some calculations follow in preparationfor applying induction.
Proposition 2.11. In the no singleton case, there are at least α x(1 x a)1 n! chains start with
− − −
o for some o [n] X A, and which contain a set from .
{ } ∈ \ \ X (cid:0) (cid:1)
Proof. A total of αn! chains contain a set D . By Proposition 2.5, the singleton on such a chain is
∈ X
either from X or [n] X A. Under our assumptions, D 2, it contains only one element of X, and
\ \ | | ≥
no element of A. So on a chain through D which has d X as its singleton, the second set must be
{ } ⊆
d,o with o [n] X A. The number of chains which contain a set from , and which start with an
{ } ∈ \ \ X
element of d X as their singleton, is at most xn (1 x a)n (n 2)!=x(1 x a)1n!. On the rest,
∈ · − − · − − −
the singleton is from [n] X A.
\ \
Proposition2.12. β a˜(1 x a)1+ν n! chains startwith o forsome o [n] X A, andcontain
− − − { } ∈ \ \
a set from .
B (cid:0) (cid:1)
Proof. A total of βn! chains contain a set from . A set in is of the form e,o ,...,o with e
1 k
A˜,o ,...,o [n] X A,k 1. A chain that coBntains a B B , and does not{start with o} for som∈e
1 k
o [n] X ∈A, m\ust\start w≥ith an element of A˜, and contin∈uBe with an element of [n] X{ }A as the
∈ \ \ \ \
second element added to form the chain. There are a˜n (1 x a)n (n 2)! = a˜(1 x a)1n! such
· − − · − − −
chains, out of which νn! do not contain any set from . So a˜(1 x a)1 ν n! chains contain a set
B − − −
from and start with an element of A˜. The rest start with o for some o [n] X A.
B {(cid:0)} ∈ (cid:1)\ \
Proposition 2.13. In the no singleton case, the number of chains of the form , d , d,o ,... with
∅ { } { }
d X,o [n] X A, which do not contain any set from , is µ x(x+a)1 n!.
∈ ∈ \ \ X −
(cid:0) (cid:1)
Proof. A total of µn! chains start with an element of X and do not contain any set from . The chains
X
ofthe form , d , d ,d ,... with d X,d X A never containa set from when containsno
1 1 2 1 2
∅ { } { } ∈ ∈ ∪ X X
singleton. The number of these chains is xn (xn+an 1) (n 2)!= x(x+a)1 n!. For the rest, the
· − · −
second element added to form the chain is from [n] X A.
\ \ (cid:0) (cid:1)
Notation. LetX =X A. Let = d,o :d X,o [n] X A ,andlet = e,o :e A A˜,o
′
∪ Y { } ∈ ∈ \ \ Z { } ∈ \ ∈
[n] X A . In the singletons case, let = . In the no singleton case, let = .
\ \ (cid:8) X′ Y ⊔B⊔Z (cid:9) (cid:8) X′ X ⊔B⊔Z
Propositio(cid:9)n(cid:9) 2.14. The three families which make up are indeed disjoint in both cases, and their
′
X
union forms an antichain.
Proof. is an antichain by Proposition 2.6; is an antichain by definition; and and are an-
B X Y Z
tichains because both consist of size 2 sets only. Let D = d,o ,...,o , Y = d,o ,
1 k
B = e ,p ,...,p and Z = e ,q with d X, e {A˜, e A˜ }A,∈oX,o,p ,q {[n] }X∈ AY,
1 1 l 2 1 2 i i
{ } ∈ B { } ∈ Z ∈ ∈ ∈ \ ∈ \ \
and l 1. B is unrelated to D by definition, and to Y because d / B and B 2. Z is unrelated to D
≥ ∈ | |≥
and Y because d / Z and e / D,Y; Z is unrelated to B because e / Z and e / B.
2 1 2
∈ ∈ ∈ ∈
Proposition 2.15. Sets in are disjoint from X , and they are unrelated to every set in (in both
′ ′
C X
cases).
Proof. ForeveryC ,C [n] X Aanditisunrelatedtoeverysetin bydefinition. C isunrelated
∈C ⊆ \ \ X
to every set in by Proposition 2.6. It also cannot be a superset of a Y or a Z , since those
B ∈ Y ∈ Z
contain an element of X or A; neither a proper subset of Y or Z because Y = Z =2 C .
| | | | ≤| |
Proposition 2.16. The number of chains that start with an element of [n] X and contain a set from
′
\
is
′
X
at least x(1 x a)1+ β a˜(1 x a)1+ν +(1 x a)(a a˜)1 n! in the singletons case,
• − − − − − − − −
and
(cid:2) (cid:0) (cid:1) (cid:3)
at least α x(1 x a)1 + β a˜(1 x a)1+ν +(1 x a)(a a˜)1 n! in the no single-
• − − − − − − − − −
ton case.
(cid:2)(cid:0) (cid:1) (cid:0) (cid:1) (cid:3)
Proof. Thenumberofchainsonwhichthe singletonis o witho [n] X =[n] X A,andthe second
′
{ } ∈ \ \ \
setis o,d withd X,isxn (1 x a)n (n 2)! =x(1 x a)1n!. Thenumberofchainsonwhich
the sin{glet}on∈iYs o , and∈the secon·ds−et is− o,e· − with e −A −A˜, is (1 x a)n (a a˜)n (n 2)!=
{ } { }∈Z ∈ \ − − · − · −
(1 x a)(a a˜)1n!. The rest follows from Proposition 2.11 and Proposition 2.12.
− − −
7
Proposition 2.17. The number of chains on which the singleton is o with o [n] X , the second set
′
{ } ∈ \
is o,d with d X =X A, and which do not contain any set from , is
′ ′
{ } ∈ ∪ X
νn! in the singletons case, and
•
µ x(x+a)1+ν n! in the no singleton case.
• −
Proof.(cid:0)Let = ,A ,A(cid:1),...,A ,[n] be a chain with A A ... A [n]. Let ϕ( )
1 2 n 1 1 2 n 1
A ∅ − ∅ ⊂ ⊂ ⊂ ⊂ − ⊂ A
be the chain ,A A ,A ,A ,...,A ,[n]. (In other words, in the order in which elements of [n] are
2 1 2 3 n 1
∅ \ −
added to form the chain, the first two are swapped.) ϕ is a bijection.
It is easy to check that does not contain singletons. ϕ is a bijection between chains of the form
′
X
, o , o,d ,... containing no set from , and chains of the form , d , o,d ,... containing no set
′
∅ { } { } X ∅ { } { }
from , with o [n] X and d X A. Classifying the chains , d , o,d ,... by d:
′ ′
X ∈ \ ∈ ∪ ∅ { } { }
For d X, o,d in the singletons case. In the no singleton case, µ x(x+a)1 n! chains of
• ∈ { }∈Y −
the form , d , o,d ,... contain no set from by Proposition 2.13; these chains also contain no
∅ { } { } X (cid:0) (cid:1)
set from or , since sets from those do not contain any element of X.
B Z
For d A˜, the number of chains of this form which contain no set from is νn!; these chains also
• contai∈n no set from , or , since sets from those contain no elementBof A˜.
X Y Z
For d A A˜, o,d .
• ∈ \ { }∈Z
Summing these cases, we get the statement of the proposition.
Let[n] X A= o ,o ,...,o . Let o ,[n] = S [n]:o S ; andfor a family of sets , let
1 2 (1 x a)n i i
\ \ { − − } { ⊆ ∈ } A
A−oi ={S\{oi}:S ∈A}. Let Ci′ = C∩ o(cid:2)i,[n] (cid:3)−oi, and Xi′ = X′∩ oi,[n] −oi. Ci′ ⊆2[n]\{oi} is
a Λ-free family which does not contain (since o / A, so o / ), nor any set larger than n 1 n.
i i ′
(cid:0)∅ (cid:2) (cid:3)∈(cid:1) { }∈F (cid:0) (cid:2) (cid:3)(cid:1) − −
Sets in are disjoint from X , and are unrelated to sets in by Proposition 2.15. Moreover,every set
Ci′ ′ Xi′
in contains exactly one element of X . Therefore, the conditions of Lemma 2.3 are satisfied for the
Xi′ ′
family Ci′ ⊆2[n]\{oi} where the corresponding“forbidden” set is X′ ⊆[n]\{oi}, with |nX′1| =(x+a)1 and
the corresponding “forbidden” antichain is . −
Xi′
|nX′1| =(x+a)1. Let α′i =l(n−1,Xi′). (Here the Lubell function on the Booleanlattice 2[n]\{oi} of order
n−−1 is used.) Since Xi′ is an antichain, α′i(n−1)! is the number of chains in 2[n]\{oi} that contain a set
from Xi′. Chains of 2[n]\{oi} correspondto chains of 2[n] that start with {oi}. So by Proposition 2.16, in
the singletons case
(1 x a)n
− −
α′i ≥ x(1−x−a)1+ β−a˜(1−x−a)1+ν +(1−x−a)(a−a˜)1 n,
i=1
X (cid:2) (cid:0) (cid:1) (cid:3)
and in the no singleton case
(1 x a)n
− −
α α x(1 x a)1 + β a˜(1 x a)1+ν +(1 x a)(a a˜)1 n.
′i ≥ − − − − − − − − −
i=1
X (cid:2)(cid:0) (cid:1) (cid:0) (cid:1) (cid:3)
Let µ′i(n−1)! be the number of chains in the Booleanlattice 2[n]\{oi} which start with an element of X′
as a singleton, but do not contain any set from . By Proposition 2.17, in the singletons case
Xi′
(1 x a)n
− −
µ =νn,
′i
i=1
X
and in the no singleton case
(1 x a)n
− −
µ′i = µ−x(x+a)1+ν n.
i=1
X (cid:0) (cid:1)
Letc′i(n−1)!bethenumberofMNMchainsw.r.t. Ci′ in2[n]\{oi}. Thecorresponding2[n]-chains,starting
with o , areMNM chains w.r.t. . The total number ofMNM chains w.r.t. is cn!, outof whichc n!
i 0
{ } F F
start with an element of A as a singleton. By Proposition 2.7,
(1 x a)n
− −
c =(c c )n=(c a˜(x+a)1 ν)n.
′i − 0 − −
i=1
X
8
Example 2.18. Let X = = and = e,o : e A,o [n] A . Then A˜ = A, β = 2a(1 a)1,
X ∅ B { } ∈ ∈ \ −
and ν = 0. X′ = A, and Xi′ = {e} : e ∈ A(cid:8) . i(=1−1a)nα′i = a(1−a(cid:9))1n, α′i = a1 = |nX′1| and µ′i = 0.
i(=1−1x−a)nc′i = c−a2 n and, o(cid:8)n average, c′i(cid:9)= Pc1−aa2. −
−
PExample 2.19.(cid:0)Let X(cid:1)= = and ˆ := e,o ,o : e A,o ,o [n] A . Then X = A,
1 2 1 2 ′
X ∅ B ⊆ B { } ∈ ∈ \
and e,o : e A,o [n] A o . Chains on 2[n] of the form , e , e ,o , e ,o,e ,...
Xi′ ⊆ { } ∈ ∈ \ \{ i} (cid:8) ∅ { 1} { 1(cid:9) } { 1 2}
do not inte(cid:8)rsect B. So i(=1−1a)nµ′i = ν ≥ a(cid:9)2(1−a)121a1(an(−n1)2−)1n (greater if B $ Bˆ), and, on average,
µ′i ≥a2121a1(an(−n1)2−)1 =x′P2xx′(′((nn−11))−11) where x′ = |nX′1|. In the−case of B =Bˆ, the size of the sets in C is at
− − − −
least 3, and the size of those in is at least 2.
Ci′
Proposition 2.20.
(1 x a)n (1 x a)n
1 − − 1 − −
l( )= l(n 1, ) and l( )=a+β+l( )=a+β+ l(n 1, ).
C n − Ci′ F C n − Ci′
i=1 i=1
X X
(Still understanding the one parameter version l( ) as l(n, ) for a family 2[n].)
F F F ⊆
Proof. Every chain in the Boolean lattice 2[n] that intersects has an o as a singleton, and thus
i
C { }
corresponds to a chain in the Boolean lattice [o ,[n]] o that intersects .
i − i Ci′
(1 x a)n (1 x a)n
1 1 − − 1 − −
l( )= = = l(n 1, ).
C n! |H∩C| n! |H∩Ci′| n − Ci′
HisacXhainin2[n] Xi=1 HisachainXin[oi,[n]]−oi Xi=1
Let = e : e A . Then = . So l( ) = l( )+l( )+l( ) with l( ) = |A| = a and
A { } ∈ F A⊔B⊔C F A B C A n
l( )=β.
B (cid:8) (cid:9)
We now prove Lemma 2.3 (and thus Lemma 1.14) using induction on n. According to Proposition 2.4,
Lemma 2.3 holds for n n. By induction and Lemma 2.2 2.,
′
≤
1 3
l(n 1, ) f (x+a)1,c +µ + α µ (x+a)1 +
− Ci′ ≤ ′i ′i n − ′i− ′i− n
(cid:18) ′(cid:19) ′
1 (cid:0) 3 (cid:1)
≤f x+a,c′i+µ′i+ n − α′i−µ′i−(x+a)1 + n .
(cid:18) ′(cid:19) ′
(cid:0) (cid:1)
So, by Proposition 2.20, we have
(1 x a)n (1 x a)n
1 − − 1 − − 1
l( )= l(n 1, ) f x+a,c +µ +
C n − Ci′ ≤ n ′i ′i n
i=1 i=1 (cid:18) ′(cid:19)
X X
(1 x a)n (1 x a)n (1 x a)n
1 − − − − − − 1 3(1 x a)n
− n α′i− µ′i− (x+a)1+ n · −n− .
i=1 i=1 i=1 ′
X X X
 
We handle the caseof1 x a=0separately. If1 x a=0,A=[n] X and,since anynon-singleton
− − − − \
e ,e ,... wouldformaΛwiththesingletons e , e , = andl( )=a=1 x. Thisis
1 2 1 2
{ }∈F { } { }∈F F A F −
onlypossibleinthesingletonscase,sinceanon-singletonin wouldhavetocontainelementsof[n] X A.
X \ \
In the singletons case α=x and µ=0, so l( )=1 x f(x,c) f x,c+µ+ 1 (α µ x)+ 3
F − ≤ ≤ n′ − − − n′
by Lemma 2.2 5. From now on, we assume that 1 x a>0.
− − (cid:0) (cid:1)
Since f is concave in c, by Jensen’s inequality, and since f is monotonously decreasing in x,
l( ) (1 x a)f x+a, i(=1−1x−a)nc′i+ i(=1−1x−a)nµ′i + 1
C ≤ − − P (1−x−Pa)n n′!
(1 x a)n (1 x a)n
1 − − 1 − − 3(1 x a)
−n α′i− n µ′i−(1−x−a)(x+a)1+ −n − .
i=1 i=1 ′
X X
 
9
Correction term calculations that we will use later (assuming n n 1):
′
≤ −
1 x a (x+a˜)(1 x a) 1 x a
(1 1)(x+a˜)(x+a)+ − − = − − + − −
− n n 1 n
′ − ′ (1)
(1+x+a˜)(1 x a) 1 (x+a)2 1
− − − .
≤ n ≤ n ≤ n
′ ′ ′
1 x a 1 x a 1
(1 1)a˜(x+a)+ − − (1 1)(x+a˜)(x+a)+ − − . (2)
− n ≤ − n ≤ n
′ ′ ′
3(1 x a)
2a˜(1 x a)(1 1)+2(1 1)x 2(1 1)+(1 1) x(x+a)+ − −
− − − − − − − n
′ (3)
2a+3x (cid:0) 3(1 x a) 3(cid:1)
+ − − .
≤ n 1 n ≤ n
− ′ ′
3(1 x a) 2a 3(1 x a) 3
2a˜(1 x a)(1 1)+ − − + − − . (4)
− − − n ≤ n 1 n ≤ n
′ − ′ ′
In the singletons case:
(c a˜(x+a)1 ν)+ν 1
l( ) a+β+(1 x a)f x+a, − − +
F ≤ − − 1 x a n
(cid:18) − − ′(cid:19)
x(1 x a)1+ β a˜(1 x a)1+ν +(1 x a)(a a˜)1
− − − − − − − − −
(cid:16)(cid:2) (cid:0) 3(cid:1)(1 x a) (cid:3)
ν (1 x a)(x+a)1 + − −
− − − − n
′
c a˜(x+a)1 1(cid:17) 3(1 x a)
=a+(1 x a)f x+a, − + +2a˜(1 x a)1+ − −
− − 1 x a n − − n
(cid:18) − − ′(cid:19) ′
c a˜(x+a)+(1 1)a˜(x+a)+ 1 x a
=a+(1 x a)f x+a, − − −n′− +2a˜(1 x a)
− − 1 x a − −
(cid:18) − − (cid:19)
3(1 x a)
+2a˜(1 x a)(1 1)+ − − .
− − − n
′
ByLemma2.22.and3.,and(2)and(4)intheCorrectiontermcalculations(notethatinthiscaseα=x
and µ=0),
1 3 1 3 1 3
l( ) g x,c+ ,a,a˜ + f x,c+ + =f x,c+µ+ (α µ x)+ .
F ≤ n n ≤ n n n − − − n
(cid:18) ′ (cid:19) ′ (cid:18) ′(cid:19) ′ (cid:18) ′(cid:19) ′
(Note that a˜ c , so 0 c−a˜(x+a)1 c+n1′−a˜(x+a), and a˜ c+n1′.)
≤ (x+a)1 ≤ 1 x a ≤ 1 x a ≤ x+a
− − − −
In the no singleton case:
(c a˜(x+a)1 ν)+µ x(x+a)1+ν 1
l( ) a+β+(1 x a)f x+a, − − − +
F ≤ − − 1 x a n
(cid:18) − − ′(cid:19)
α x(1 x a)1 + β a˜(1 x a)1+ν +(1 x a)(a a˜)1
− − − − − − − − − −
(cid:16)(cid:2)(cid:0) (cid:1) (cid:0) (cid:1) 3(1 x a) (cid:3)
[µ x(x+a)1+ν] (1 x a)(x+a)1 + − −
− − − − − n
′
c+µ (x+a˜)(x+a)1(cid:17) 1
=a+(1 x a)f x+a, − + (α µ x)
− − 1 x a n − − −
(cid:18) − − ′(cid:19)
3(1 x a)
+2a˜(1 x a)1+(2 1 1)x (2 1+1)x(x+a)+ − −
− − · − − · n
′
c+µ (x+a˜)(x+a)+(1 1)(x+a˜)(x+a)+ 1 x a
=a+(1 x a)f x+a, − − −n′−
− − 1 x a
(cid:18) − − (cid:19)
+2a˜(1 x a)+x 3x(x+a) (α µ x)
− − − − − −
3(1 x a)
+2a˜(1 x a)(1 1)+2(1 1)x 2(1 1)+(1 1) x(x+a)+ − − .
− − − − − − − n
′
(cid:0) (cid:1)
10