Partially Ordered Sets
Corollary 3.33 We have the inequality
3.10 The Sauer–Shelah Theorem
If a collection of setsCis not a VC class (that is, if the Vapnik-Chervonenkis dimen-sion of Cis infinite), thenΨC[m] = 2m for all m ∪ N. However, we shall prove that ifVCD(C) = d, thenΨC[m]is bounded asymptotically by a polynomial of degree d.
Definition 3.59 LetCbe a collection of subsets of a finite set S, let s ∪ S, and let C(s)be the collectionCS−{s}.
Chas a pair(A,B)at s if A,B∪C, B∧A, and A−B = {s}. Note that if(A,B)is a pair ofCat s, then B= A− {s}.
Define the subcollectionsP⊃(C,s)andP⊃⊃(C,s)ofCas
P⊃(C,s)= {A∪C | (A,B)is a pair at s for some B∪C}
= {A∪C | s∪ A,A− {s} ∪C},
P⊃⊃(C,s)= {B∪C | (A,B)is a pair at s for some A∪C}
= {B∪C | s∅∪B,B⊆ {s} ∪C}.
Lemma 3.60 LetCbe a collection of subsets of a finite set S, let s ∪S, and letC(s)
be the collectionCS−{s}.
The following statements hold:
(i) if(A1,B1),(A1,B2),(A2,B1),(A2,B2)are pairs at s ofC, then A1=A2and B1=B2;
(ii) we have|P⊃(C,s)| = |P⊃⊃(C,s)|; (iii) |C| − |C(s)| = |P⊃⊃(C,s)|;
(iv) C(s)= {C∪C | s∅∪C} ⊆ {C− {s} | C∪Cand s∪C}; (v) |C(s)| = |C| − |P⊃⊃(C,s)|.
Proof For part(i), Definition 3.59 implies
B1∧ A1,B2∧A1,B1∧A2,B2∧ A2,
and A1=B1⊆ {s}, A1=B2⊆ {s}, A2=B1⊆ {s}, and A2=B2⊆ {s}. Therefore, A1=B1⊆ {s} = A2, which implies B1=A1− {s} = A2− {s} = B2.
For part(ii), let f :P⊃(C,s)−∈P⊃⊃(C,s)be the function f(A)= A− {s}. It is easy to verify that f is a bijection and this implies|P⊃(C,s)| = |P⊃⊃(C,s)|.
To prove part(iii)let B∪P⊃⊃(C,s). By the definition ofP⊃⊃(C,s)we have B ∪C, s∅∪B, and B⊆ {s} ∪C.
Define the mappingg : P⊃⊃(C,s) −∈ C−C(s) asg(B) = B⊆ {s}. Note that B⊆ {s} ∪ C−CS−{s}, sog is a well-defined function. Moreover,g is one-to one because B1⊆ {s} = B2⊆ {s}, s ∅∪ B1, and s ∅∪ B2imply B1 = B2. Also,gis a surjection because if D∪C−CS−{s}, then s∪ D and D− {s} ∪P⊃⊃(C,s). Thus,g is a bijection, which implies|C| − |C(s)| = |C−C(s)| = |P⊃⊃(C,s).
To prove Part(iv)let C ∪C(s). If s∅∪C, then C belongs to the first collection of the union; otherwise, that is if s ∪C, C− {s}belongs to the second collection and the equality follows.
Finally, for Part(v), by applying Part(iv)we can write
|C(s)| = |{C∪C | s∅∪C}| + |{C− {s} | C ∪Cand s∪C}|
−|{C∪C | s∅∪C} ∩ {C− {s} | C∪Cand s∪C}|
130 3 Combinatorics
= |{C∪C | s∅∪C}| + |{C− {s} | C ∪Cand s∪C}| − |P⊃⊃(C,s)|
= |C| − |P⊃⊃(C,s)|.
Lemma 3.61 LetCbe a collection of sets of a non-empty finite set S and let s0be an element of S. IfP⊃⊃(C,s0)shatters a subset T of S− {s0}, thenCshatters T⊆ {s0}. Proof SinceP⊃⊃(C,s0)shatters T , for every subset U of T there is B ∪ P⊃⊃(C,s0) such that U =T∩B. Let W be a subset of T⊆ {s0}. If s0∅∪W , then W ∧T and by the previous asumption, there exists B ∪P⊃⊃(C,s0)such that W =(T ⊆ {s0})∩B.
If s0 ∪W , then there exists B1∪P⊃⊃(C,s0)such that for W1=W − {s0}we have W1=T ∩B1. By the definition ofP⊃⊃(C,s0), B1⊆ {s0} ∪Cand
(T ⊆ {s0})∩(B1⊆ {s0})=(T ∩B)⊆ {s0} =W1⊆ {s0} =W. Thus,Cshatters T ⊆ {s0}.
We saw that we haveVCD(C)∞log2|C|. For collections of subsets of finite sets we have a stronger result.
Theorem 3.62 Let C be a collection of sets of a non-empty finite set S with VCD(C)=d. We have
2d∞ |C| ∞(|S| +1)d.
Proof The first inequality, reproduced here for completeness, was discussed earlier.
For the second inequality the argument is by induction on|S|. The basis case,
|S| =1 is immediate.
Suppose that the inequality holds for collections of subsets with no more than n elements and let S be a set containing n+1 elements. Let s0be an arbitrary but fixed element of S. By Part (iv) of Lemma 3.60 we have|C| = |Cs0| + |P⊃⊃(C,s0)|.
The collectionCs0 consists of subsets of S− {s0}. SinceVCD(C)=d, it is clear thatVCD(Cs0)∞d and, by inductive hypothesis|Cs0| ∞(|S− {s0}| +1)d.
We claim thatVCD(P⊃⊃(C,s0))∞d−1. Suppose that P⊃⊃(C,s0)= {B ∪C | s0∅∪B,B⊆ {s0} ∪C}
shatters a set T , where T ∧ S and|T| d. Then, by Lemma 3.61, C would shatter T ⊆ {s0}; since |T ⊆ {s0}| ⊕ d +1, this would lead to a contradiction.
Therefore, we have VCD(P⊃⊃(C,s0)) ∞ d −1 and, by the inductive hypothesis,
|P⊃⊃(C,s0))| ∞(|S− {s0}| +1)d−1. These inequalities imply
|C| = |Cs0| + |P⊃⊃(C,s0)|
∞(|S− {s0}| +1)d+(|S− {s0}| +1)d−1
=(|S− {s0}| +1)d−1(|S− {s0}| +2)
∞(|S| +1)d,
which completes the proof.
Lemma 3.63 LetCbe a collection of subsets of a finite set S, let s ∪S, and letC(s)
be the collectionCS−{s}. IfVCD(P⊃⊃(C,s))=n−1 in S− {s}, thenVCD(C)=n.
Proof SinceVCD(P⊃⊃(C,s))=n−1 in S− {s}, there exists a set T ∧S− {s}with
|T| =n−1 that is shattered byP⊃⊃(C,s). LetGbe the subcollection ofCdefined by:
G=P⊃(C,s)⊆P⊃⊃(C,s)
= {A∪C | A− {s} ∪C} ⊆ {B ∪C | B⊆ {s} ∪C}.
We claim thatGshatters T⊆ {s}. Two cases may occur for a subset W of T ⊆ {s}: (i) If s ∅∪W , then W ∧T and, since T is shattered by byP⊃⊃(C,s)it follows that
there exists G∪Gsuch that W =G∩(T ⊆ {s}).
(ii) If s ∪ W , let G⊃ ∪P⊃⊃(C,s)be set such that W− {s} = T ∩G⊃, which exists by the previous argument. Then, we have G⊃∪P⊃⊃(C,s), so G⊃⊆ {s} ∪P⊃(C,s) and(G⊃∩ {s})∩(T ⊆ {s})=W .
Thus,Gshatters T⊆ {s}and so doesC.
For n,k∪Nand 0∞k∞n define the number⎟n
∞k
as n
∞k
⎜
= k i=0
n i
⎜ . Clearly,⎟n
∞0
=1 and⎟n
n
=2n.
Theorem 3.64 Letν:N2−∈Nbe the function defined by ν(d,m)=
1 if m=0 or d=0
ν(d,m−1)+ν(d−1,m−1) otherwise. We have
ν(d,m)= m
∞d
⎜
for d,m∪N.
Proof The argument is by strong induction on s = i+m. The base case, s = 0, implies m =0 and d =0, and the equality is immediate. Suppose that the equality holds forν(d⊃,m⊃), where d⊃+m⊃<d+m. We have
132 3 Combinatorics
Theorem 3.65 (Sauer–Shelah Theorem) LetCbe a collection of sets such that VCD(C)=d. If m∪Nis a number such that d∞m, thenΨC[m] ∞ν(d,m). Proof The argument is by strong induction on s =d+m. For the base case, s=0 we have d =m=0 and this means that the collectionCshatters only the empty set.
Thus,ΨC[0] = |C∅| =1, and this impliesΨC[0] =1=ν(0,0).
Proof The argument is by induction on d. In the basis step, d=2 both members are equal to 2.
Suppose the inequality holds for d. We have (d+1)d+1
because
Suppose that the inequality holds for m>d 1. We have ν(d,m+1)=ν(d,m)+ν(d−1,m)
It is easy to see that the inequality 2 md−1
and, therefore, is valid. This yields immediately the inequality of the lemma.
The next theorem discusses the asymptotic behavior of the functionν:
Theorem 3.68 The functionνsatisfies the inequality:
ν(d,m) < em
134 3 Combinatorics The argument is by induction on d1. The basis case, d =1 is immediate. Suppose that 2⎟d
The last inequality holds because the sequence ⎟
1+1dd Proof The statement is a direct consequence of Theorem 3.68.
Let u : B2k −∈ B2 be a Boolean function of k arguments and let C1, . . . ,Ck
Proof Let S be a subset of U that consists of m elements. The collection(Ci)Sis not larger thanν(d,m). For a set in the collection W ∪ u(C1, . . . ,Ck)S we can write W =S∩u(C1, . . . ,Ck), or, equivalently, 1W =1S·u(1C1, . . . ,1Ck). By Exercise 25 of Chap.11, there exists a Boolean functiongSsuch that
1S·u(1C1, . . . ,1Ck)=gS(1S·1C1, . . . ,1S·1Ck)=gS(1S∩C1, . . . ,1S∩Ck).
Since there are at mostν(d,m)distinct sets of the form S∩Cifor every i , 1∞i ∞k, it follows that there are at most(ν(d,m))kdistinct sets W , hence u(C1, . . . ,Ck)[m] ∞ (ν(d,m))k.
Theorem 3.68 implies
u(C1, . . . ,Ck)[m] ∞ em d
kd
.
We observed that ifΨC[m]<2m, thenVCD(C) <m. Therefore, to limit the Vapnik-Chervonenkis dimension of the collection u(C1, . . . ,Ck)it suffices to require that
⎟em
d
kd
<2m.
Let a = md. The last inequality can be written as(ea)kd <2ad; equivalently, we have(ea)k <2a, which yields k < loga(ea). Ifδ(k)is the least integer a such that k< loga(ea), then m∞δ(k)d, which gives our conclusion.
Example 3.72 If k=2, the least integer a such thatloga(ea) >2 is k=10, as it can be seen by graphing this function; thus, ifC1,C2are two collection of concepts with VCD(C1)=VCD(C2)=d, the Vapnik-Chervonenkis dimension of the collections C1≡C2orC1∗C2is not larger than 10d.
Lemma 3.73 Let S,T be two sets and let f : S −∈ T be a function. IfDis a collection of subsets of T , U is a finite subset of S andC= f−1(D)is the collection {f−1(D) | D∪D}, then|CU| ∞ |Df(U)|.
Proof Let V = f(U)and denote f U byg. For D,D⊃∪Dwe have (U∩ f−1(D))⊥(U∩ f−1(D⊃))
=U∩(f−1(D)⊥ f−1(D⊃))=U∩(f−1(D⊥D⊃))
=g−1(V ∩(D⊥D⊃))=g−1(V∩D)⊥g−1(V ⊥D⊃).
Thus, C =U∩f−1(D)and C⊃=U∩f−1(D⊃)are two distinct members ofCU, then V∩D and V∩D⊃are two distinct members ofDf(U). This implies|CU| ∞ |Df(U)|. Theorem 3.74 Let S,T be two sets and let f : S −∈T be a function. IfDis a collection of subsets of T andC= f−1(D)is the collection{f−1(D) | D ∪D}, thenVCD(C)∞VCD(D). Moreover, if f is a surjection, thenVCD(C)=VCD(D). Proof Suppose thatCshatters an n-element subset K = {x1, . . . ,xn}of S, so|CK| = 2n By Lemma 3.73 we have|CK| ∞ |Df(U)|, so |Df(U)| 2n, which implies
136 3 Combinatorics
|f(U)| =n and|Df(U)| = 2n, because f(U)cannot have more than n elements.
Thus,Dshatters f(U), soVCD(C)∞VCD(C).
Suppose now that f is surjective and H = {t1, . . . ,tm}is an m element set that is shattered byD. Consider the set L = {u1, . . . ,um}such that ui ∪ f−1(ti)for 1∞i ∞m. Let U be a subset of L. Since H is shattered byD, there is a set D∪D such that f(U)=H∩D, which implies U =L∩ f−1(D). Thus, L is shattered by Cand this means thatVCD(C)=VCD(D).
Definition 3.75 The density ofCis the number
dens(C)=inf{s∪R>0 | ΨC[m] ∞c·ms for every m∪N}, for some positive constant c.
Theorem 3.76 Let S,T be two sets and let f : S −∈T be a function. IfDis a collection of subsets of T andC= f−1(D)is the collection{f−1(D) | D ∪D}, thendens(C)∞dens(D). Moreover, if f is a surjection, thendens(C)=dens(D). Proof Let L be a subset of S such that|L| =m. Then,|CL| ∞ |Df(L)|. In general, we have|f(L)| ∞m, so|Df(L)| ∞D[m] ∞cms. Therefore, by Lemma 3.73, we have|CL| ∞ |Df(L)| ∞D[m] ∞cms, which impliesdens(C)∞dens(D).
If f is a surjection, then, for every finite subset M of T such that|M| =m there is a subset L of S such that|L| = |M|and f(L)=M. Therefore,D[m] ∞ΨC[m] and this impliesdens(C)=dens(D).
IfC,Dare two collections of sets such thatC∧D, thenVCD(C)∞ VCD(D) anddens(C)∞dens(D).
Theorem 3.77 LetCbe a collection of subsets of a set S and letC⊃ = {S−C | C ∪C}. Then, for every K ∪P(S)we have|CK| = |C⊃K|.
Proof We prove the statement by showing the existence of a bijection f :CK −∈
C⊃K. If U ∪ CK, then U =K ∩C, where C ∪C. Then S−C ∪ C⊃and we define f(U)= K∩(S−C)= K−C ∪C⊃K. The function f is well-defined because if K ∩C1=K∩C2, then K−C1=K−(K∩C1)=K −(K∩C2)=K−C2.
It is clear that if f(U)= f(V)for U,V ∪CK, U =K∩C1, and V =K∩C2, then K−C1=K−C2, so K∩C1=K∩C2and this means that U =V . Thus, f is injective. If W ∪C⊃K, then W =K∩C⊃for some C⊃∪C. Since C⊃=S−C for some C ∪C, it follows that W=K−C, so W = f(U), where U =K ∩C.
Corollary 3.78 LetCbe a collection of subsets of a set S and letC⊃ = {S−C | C ∪C}. We havedens(C)=dens(C⊃)andVCD(C)=VCD(C⊃).
Proof This statement follows immediately from Theorem 3.77.
Theorem 3.79 For every collection of sets we havedens(C)∞VCD(C). Further-more, ifdens(C)is finite, thenCis a VC-class.
Proof IfCis not a VC-class the inequalitydens(C)∞VCD(C)is clearly satisfied.
Suppose now that Cis a VC-class andVCD(C) = d. By Sauer–Shelah Theorem (Theorem 3.65) we have ΨC[m] ∞ ν(d,m); then, by Theorem 3.68, we obtain ΨC[m] ∞⎟em
d
d
, sodens(C)∞d.
Suppose now thatdens(C)is finite. SinceΨC[m] ∞cms ∞2mfor m sufficiently large, it follows thatVCD(C)is finite, soCis a VC-class.
LetD be a finite collection of subsets of a set S. In Supplement 6 of Chap.1 the partitionρDwas defined as consisting of the nonempty sets of the form{D1a1∩ D2a2∩ · · · ∩Drar, where(a1,a2, . . . ,ar)∪ {0,1}r.
Definition 3.80 A collectionD= {D1, . . . ,Dr}of subsets of a set S is independent if the partition ρDhas the maximum numbers of blocks, that is, it consists of 2r blocks.
IfDis independent, then the Boolean subalgebra generated byDin the Boolean algebra (P(S),{∩,⊆, ¯,∅,S})contains 22r sets, because this subalgebra has 2r atoms. Thus, ifDshatters a subset T with|T| = p, then the collectionDT contains 2psets, which implies 2p∞22r, or p∞2r.
LetCbe a collection of subsets of a set S. The independence number ofC, I(C) is:
I(C)=sup{r | {C1, . . . ,Cr}
is independent for some finite{C1, . . . ,Cr} ∧C}.
The next theorem is an analog of Theorem 3.74 for the independence number of a collection.
Theorem 3.81 Let S,T be two sets and let f : S −∈T be a function. IfDis a collection of subsets of T andC= f−1(D)is the collection{f−1(D) | D ∪D}, then I(C)∞I(D). Moreover, if f is a surjection, then I(C)=I(D).
Proof Let E = {D1, . . . ,Dp}be an independent finite subcollection of D. The partitionρEcontains 2rblocks. By Supplement 30 of Chap.11, the number of atoms of the subalgebra generated by {f−1(D1), . . . ,f−1(Dp)}is not greater than 2r. Therefore, I(C)∞I(D); from the same supplement it follows that if f is surjective, then I(C)=I(D).
Theorem 3.82 IfCis a collection of subsets of a set S such thatVCD(C) 2n, then I(C)n.
Proof Suppose thatVCD(C)2n, that is, there exists a subset T of S that is shattered byCand has at least 2nelements. Then, the collectionCT contains at least 22n sets, which means that the Boolean subalgebra ofP(T)generated byTCcontains at least 2natoms. This implies that the subalgebra ofP(S)generated byCcontains at least this number of atoms, so I(C)n.
138 3 Combinatorics
4. Prove that the number of antisymmetric relations on a finite set that has n elements is 2n·3n(n−1)2 . by the pidgeonhole principle, that there are two numbers p,q ∪S that have the same odd part. The smaller such number divides the larger one.
9. Starting from Newton’s binomial formula, prove the following identities:
n
m+n+p
17. LetCbe a collection of subsets of a finite set S.
(a) Prove that if C ∩ D ∅= ∅ for every pair(C,D)of members of C, then
140 3 Combinatorics (b) By applying the inclusion-exclusion principle, prove that the number of
derangements is
(c) Use a combinatorial argument to prove that
n! = intersect T , which implies T = ∅. Thus, every cover contributes only to the term (−1)0|d(C,∅)|k. If
Ci = V ⊂ S, then the Ci contribute to every term corresponding to T = S −V , so the total contribution of C1, . . . ,Ck
is
T∧S−V(−1)T, and this sum equals 0 because every nonempty set has an equal number of even and odd-sized subsets. Thus, only the collection of sets that cover the entire set S contributes to ck.
21. Let P be a subset of{1, . . . ,2n}such that|P| =n+1. Prove that there exists a pair of numbers in P×P whose components are relatively prime numbers.
22. Prove thatR(m,p,q)∞ n contains a subsequence that is either strictly increasing, strictly decreasing, or constant. Extend this result to countable, totally ordered sets.
Hint: Consider the complete graph on the set{n0,n1, . . .}, and color each edge (ni,nj)with i < j with red if ni <nj, blue if ni >nj, and white if ni =nj. 24. Let a = (a1, . . . ,an) be a sequence in Seqn(N) such that r ∞ min ai and
let p = maxai. If p = (p, . . . ,p) ∪ Seqn(N), prove thatRamsey(p,r) Ramsey(a,r).
25. The left shift and the right shift on PERMn are the mappings lshift,rshift : PERMn−∈PERMndefined by
lshift(a1,a2, . . . ,an)=(a2, . . . ,an,a1), rshift(a1,a2, . . . ,an)=(an,a1. . . ,an−1), for every(a1, . . . ,an)∪PERMn, respectively.
(a) Prove that lshift and rshift are inverse to each other.
(b) Two permutations f,g∪ PERMnare equivalent, f ◦g, if there exists an integer k such that lshift(k)(f)=g. Here h(k)denotes the kthiteration of h.
Prove that “◦” is an equivalence on PERMn.
26. Prove that if 0 ∞ p ∞ (n +1)! −1, then there exists a unique sequence a = (a1, . . . ,an)∪Seq(nn)such that 0∞ai ∞i and p=a1·1!+a2·2!+· · ·+an·n!. 27. Consider the polynomial[x]n = x(x−1)· · ·(x−n +1)called the factorial
power of x with exponent n. The coefficients of this polynomial
[x]n=s(n,n)xn+s(n,n−1)xn−1+ · · · +s(n,i)xi + · · · +s(n,0) are known as the Stirling numbers of the first kind. Prove that
[x]n=(−1)n[n−x−1]n; s(n,0)=0; s(n,n)=1;
s(n+1,k)=s(n,k−1)−ns(n,k).
Hint: To prove the last equality, observe that[x]n+1= [x]n(x−n)and seek the coefficient of xkin both sides.
28. Prove that for p,n∪Nwe have [p]n =
0 if 0∞ p∞n−1,
p!
(p−n)! if pn. Furthermore, prove that
e=√
p=0
[p]n
p! for any non-negative integer n.
29. Let S,U be two finite sets with|U| =n. Prove that the number of functions of the form f :S −∈U that have the same kernel partitionρ= {H1, . . . ,Hk} ∪ PART(S)with k∞n equals the number of injective functions from a set with k elements into U , and therefore this number is[n]k=n(n−1)· · ·(n−k+1). 30. Let α(ρ)the number of blocks of a partitionρ ∪ PART(S), where|S| = m.
Prove that for every u∪Nwe have
{[u]α(ρ) | ρ∪PART(S)} =um.
Solution: Let U be a finite set such that|U| =u. There are umfunctions of the form f : S −∈U and each such function has a kernel partition with k blocks with k∞n. As observed before, there exist[u]ksuch parttions, which yields the above equality.
31. Prove that every polynomial xncan be written as
142 3 Combinatorics
32. Prove that the Stirling numbers of the second kind satisfy the equalities:
(a) S(n,2)=2n−1−1 and S(n,n−1)=⎟n
2
; (b) S(n,i)=i S(n−1,i)+S(n−1,i−1).
Solution: We discuss only the equality (b). A partition of a set A = {a1, . . . ,an}into i blocks can be obtained from a partition of A− {an} into i blocks and adding anto one of these blocks or by placing anin a block by itself. The first term of the second member of the equality counts parti-tions obtained by the first construction, while the second term corresponds to the second construction.
33. The total number of partitions of a set having n elements is denoted by Bnand is known as the nthBell number. Clearly, Bn=n (b) the number of solutions in positive integers is⎟n−1
p−1
.
Solution: Let S = {a,b}be a two-element set and let a = (a, . . . ,a) ∪ Seqn(S)be a sequence of length n. Let a⊃ ∪ Seqn+p−1be the sequence obtained from a by inserting p−1 elements b. The number of a symbols between any two consecutive bs yields a solution in natural numbers if adjacent b symbols are allowed and there are⎟n+p−1
p−1
configurations of a⊃, each corresponding to a choice of p−1 positions out of a total of n+p−1 possible places. Further, any solution of the equation can be obtained in this manner. The second part follows immediately from the first part.
36. Prove that|IPn(ν)|equals the number of solutions of the equation y1+· · ·+yν= n−νsuch that y1y2· · ·yν0. Infer that|IPn(ν)| =ν
k=1|IPn−ν(k)|. 37. Let f : {1, . . . ,n} −∈ {1, . . . ,n}be a function and Df be the set Df =
{f(i+1)− f(i) | 1∞i ∞n−1} ⊆ {f(1)−1,n− f(n)}.
Prove that f is monotonic if and only if Df is a non-negative solution of the
equation x1+x2+ · · · +xn+xn+1=n. Infer the number of monotonic trans-formations of the set{1, . . . ,n}.
38. Prove that the number of Ferrers diagrams that can be placed in an m×n rectangle is⎟m+n
n
.
39. The definition of binomial coefficients can be extended to pairs of the form(x,k), where x∪Cand k∪Zby writing
where[z]k is the polynomial introduced in Exercise 27. Show that the equality
⎟n
k
=⎟ n
n−k
fails when n<0; however, the addition identity (3.6) z holds for every complex number z and every integer k.
40. A partition k∪IPnis self-conjugate if k⊃=k. Using Ferrers diagrams prove that the number of self-conjugate partitions in IPnequals the number of partitions of n into distinct odd numbers.
41. Let S be a set having n elements.
(a) A collectionAof subsets of S has the intersecting property if A,B ∪ A implies A ∩B ∅= ∅. Prove that if Ahas the intersecting property, then
|A| ∞2n−1.
(b) Prove that there are collections of subsets of S that have the intersecting property and contain 2n−1subsets.
Solution: For Part 1, note that, for any subset B, at most one of the sets B, S−B may belong toA. Therefore,Amay not contain more than half of the members ofP(S), so|A| ∞2n−1. A collection of sets that answers Part (b) is the set of subsets of S that contain a fixed element a of S. Clearly, there are 2n−1such sets.
42. Prove Sperner’s theorem using Supplement 8 of Chap.1.
Solution: LetC be a Sperner system on a finite set S such that|S| = n and Sperner system with the same number of sets asC. The process is repeated until a Sperner systemC⊃that contains no sets with fewer thann−21elements is obtained andC⊃= |C|. Thus each set inC⊃has at leastn+21 elements. Now the process is reversed using theβC⊃by replacing every set ofC⊃of size n+21 by a set of size
144 3 Combinatorics
43. Let CandD be two collections of subsets of a set S. In Definition 1.18, we introduced the collectionD−CasC−D= {U−V | U ∪C,V ∪D}. Prove that|C−C||C|.
Solution: LetDbe a collection of subsets of a set S and letD⊃= {S−D | D∪ D}. Observe that|D| = |D⊃|.
IfCandDare two collections of subsets of S, we have|C≡D⊃| = |(C≡D⊃)⊃| =
|C⊃∗D|. By the previous observation and by Corollary 3.47, we can write
|C| · |D| = |C| · |D⊃| ∞ |C≡D⊃| · |C∗D⊃|
= |C⊃∗D| · |C∗D⊃| = |D−C| · |C−D|.
If we now chooseD=C, the previous inequality yields|C|2∞ |C−C|2, which gives the desired inequality.
44. Prove that ifCandDare hereditary or dually hereditary families of subsets of a finite set S, then|C| · |D| ∞2n· |C∩D|.
1 if n is a product of an even number of distinct primes,
−1 if n is a product of an odd number of distinct primes,
Solution: For n = 1, the equality is immediate. Suppose that n > 1. Only numbers m that are products of distinct prime numbers contribute to the
sum
50. LetC,Dbe two collections of subsets of a set S. Prove that for every m∪Nwe haveΨC⊆D[m] =max{ΨC[m], ΨD[m]}.
51. Let S be a nonempty set and letC= {{x} | x∪S}. Prove thatVCD(C)=1.
52. Let S be a nonempty set. Prove that ifCis a collection of subsets of S such that
|C|2, thenVCD(C)1.
53. Let U be a finite set and letCbe a collection of subsets of U such that|C|2.
Prove thatVCD(C) > 1+lnln|C||U|.
Solution: Observe thatΨC[|U|] = |C|. Therefore, by Sauer–Shelah Theorem (Theorem 3.65) and by Theorem 3.68, we have
|C| ∞ e|U|
d
⎜d
,
where d is the VC dimension of the collectionC. The last inequality implies ln|C| ∞d(1+ln|U| −ln d),
so ln|C| ∞d(1+ln|U|), which gives the desired inequality.
54. Prove that ifCis a chain of subsets of a set S, thenVCD(C)=1.
55. Let C be a collection of subsets of S. Prove that if T is a subset of S, then VCD(CT)∞VCD(C).
56. LetCbe a collection of sets such that C,C⊃∪Cand C∅=C⊃implies C∩C⊃= ∅. Prove thatVCD(C)=1.
57. Let S be a set and letC1, . . . ,Cnbe n chains in the poset(P(S),∧). Define the collectionCasC= {⎭n
i=1Ci | Ci ∪Ci,1∞i∞n}. Prove thatVCD(C)∞n.
Solution: Let T be a subset of S such that|T| = n+1. Clearly, T has n+1 subsets that have n elements.
For each i at most one n-element subset of T is the intersection of the form T ∩C, where C∪ Ci. Indeed, if we would have two distinct n-element sets of the form T∩C⊃and T ∩C⊃⊃, where C⊃,C⊃⊃∪Cthis would imply the existence of x⊃ ∪(T ∩C⊃)−(T ∩C⊃⊃)and of x⊃⊃ ∪(T ∩C⊃⊃)−(T ∩C⊃), which would mean that x⊃ ∪C⊃−C⊃⊃and x⊃⊃∪C⊃⊃−C⊃, thus contradicting theCi is a chain of sets. Let Ui be this n-element when it exists.
Let W be an n-element subset of T such that W = T ∩C for some C =
⎭n
i=1Ci ∪ C. Then, either Cj ∩T = W or Ci ∩T = T for 1∞ j ∞ n and Ci ∩T = W for at least one i , 1 ∞ i ∞ n. Therefore, W = Ui for some i , 1∞i ∞n, which shows that at most n subsets of T that contain n elements can be obtained as intersections of T with the elements ofC. Thus, T is not shattered byCandVCD(C)∞n.
58. For 1∞i ∞n and a ∪Rlet Ci,a = {x∪Rn | x=(x1, . . . ,xn),xi ∞a}. The chain of setsCi is defined by{Ci,a | a ∪ R}for 1∞ i ∞ n. Prove thatC =
⎭n
i=1Ci shatters the set B = {e1, . . . ,en}, where ei = (0, . . . ,0,1,0, . . . ,0) has 1 as its i th component for 1∞i∞n, soVCD(C)=n.
146 3 Combinatorics 59. The statement included here is a generalization of Example 3.58. Prove that the Vapnik-Chervonenkis dimension of the collection of rectangular subsets ofRn given by
C= n
⎪
i=1
[ai,bi] | ai,bi ∪ ˆR,ai ∞bi, for 1∞i ∞n
⎦
is 2n. If ai = −√for all ai, 1∞i∞n, thenVCD(C)=n.
60. Let S be a set that contains at least two elements and let Cbe a collection of subsets S. Suppose that for every two-element subset of S, T = {t1,t2}, there exist U,V ∪Csuch that T ∧U and T∩V = ∅. ThenVCD(C)=1 if and only ifCis a chain.
Solution: Suppose thatVCD(C)=1 butCis not a chain. Then,Ccontains two sets C⊃,C⊃⊃ such that neither C⊃ ∧ C⊃⊃ nor C⊃⊃ ∧ C⊃. Let c⊃ ∪ C⊃−C⊃⊃ and c⊃⊃ ∪C⊃⊃−C⊃. Then, the two element set T = {c⊃,c⊃⊃}is shattered byC, which impliesVCD(C)2. The reverse implication follows from Supplement 54.
61. Prove that ifCis a collection of subsets of a set S such that{∅,S} ∧ C, then VCD(C)=1 if and only ifCis a chain.
62. Let S be a finite set and letCbe a collection of subsets of S. Prove that|C| =
|{K ∪P(S) | Cshatters K}|.
Solution: Let x be an element of S and letνx :C−∈Pbe the injective mapping introduced in Supplement 26 of Chap.1. We claim that ifνx(C) = {νx(C) | C ∪ C}shatters K , thenCshatters K . If x ∅∪ K , thenCK = νx(C)K, so the statement obviously holds. If x∪ K and L∧K− {x}, then there is F ∪νx(C) such that F∩K =L⊆ {x}and T =νx(C)for some C ∪C. Since x ∪F , both F and F− {x}belong toC, soCshatters K .
Definew(C)=
{|C| | C∪C}. LetC⊃be a collection of sets obtained fromC by applying transforms of the formνx, such thatw(C⊃)is minimal. For C ∪C⊃ and x∪ K we must have C− {x} ∪C⊃because otherwisew(νx(C⊃)) < w(C⊃), contradicting the minimality ofC⊃. Thus,C⊃is hereditary, so it shatters any set it contains. Since|C⊃| = |C|(by Supplement 26 of Chap.1, andCshatters at least as many sets asCwe obtain the desired equality.
63. Let(P,∞)and(Q,∞)be two locally finite posets having the Möbius function μPandμQ, respectively. In Exercise 30 of Chap.1we saw that(P×Q,∞)is a locally finite poset. Prove that its Möbius functionμP×P is given by
μP×P((x,y), (x⊃,y⊃))=μP(x,x⊃)μ(y,y⊃), where x∞x⊃and y∞ y⊃.
Bibliographical Comments
Supplement 20 was obtained in [11]. The inequality from Supplement 43 was
Supplement 20 was obtained in [11]. The inequality from Supplement 43 was