Partially Ordered Sets
Definition 2.46 A well-founded poset is a partially ordered set where every nonempty subset has a minimal element
2.8 Posets and Zorn’s Lemma
A statement equivalent to a fundamental principle of set theory known as the Axiom of Choice is Zorn’s lemma stated below.
Zorn’s Lemma: If every chain of a poset(S,)has an upper bound, then S has a maximal element.
Theorem 2.81 The following three statements are equivalent for a poset(S,): (i) If every chain of(S,)has an upper bound, then S has a maximal element
(Zorn’s Lemma).
(ii) If every chain of(S,)has a least upper bound, then S has a maximal element.
(iii) S contains a chain that is maximal with respect to set inclusion (Hausdorff maximality principle).
Proof (i) implies (ii)is immediate.
(ii) implies (iii):Let(CHAINS(S),∈)be the poset of chains of S ordered by set inclusion. By Theorem 2.41, every chain{Ui | i∈ I}of the poset(CHAINS(S),∈) has a least upper bound {Ui | i ∈ I}in the poset(CHAINS(S),∈). Therefore, by (ii), (CHAINS(S),∈)has a maximal element that is a chain of(S,)that is maximal with respect to set inclusion.
(iii) implies (i):Suppose that S contains a chain W that is maximal with respect to set inclusion and that every chain of(S,)has an upper bound. Letwbe an upper bound of W .
Ifw∈ W , thenwis a maximal element of S. Indeed, if this were not the case, then S would contain an element t such thatw < t and W∅ {t}would be a chain that would strictly include W .
Ifw ∩∈ W , then W∅ {w}would be a chain strictly including W , which, again, would contradict the maximality of W . Thus,wis a maximal element of(S,).
Denote byPORD(S)the collection of partial order relations on the set S.
Definition 2.82 Letρ,ρ∪∈ PORD(S). The partial orderρ∪is an extension ofρif (x,y)∈ρimplies(x,y)∈ρ∪. Equivalently, we shall say thatρ∪extendsρ.
An important consequence of Zorn’s lemma is the next statement, which shows that any partial order defined on a set can be extended to a total order on the same set.
Theorem 2.83 (Szpilrajn’s Theorem) Let(S,)be a poset. There is a total order ∪on S that is an extension of.
Proof LetPORD(S,)be the set of partial order relations that can be defined on the set S and contain the relation “”; clearly, the relation “” itself is a member of PORD(S,). We will apply Zorn’s lemma to the poset(PORD(S,),∈).
LetR= {ρi | i ∈ I}be a chain of(PORD(S,),∈); that is, a chain of partial ordersρirelative to set inclusion such that xy implies(x,y)∈ρifor every i∈ I and all x,y∈S. We claim that the relationρ= Ris a partial order on S.
Indeed, sinceιS ∈∈ ρi for i∈ I we haveιS∈ρ, soρis a reflexive relation.
To prove thatρis antisymmetric let x,y∈ S be two elements such that(x,y)∈ρ and(y,x)∈ρ. By the definition ofρ, there exist i,j ∈ I such that(x,y)∈ρi and (y,x)∈ρj. SinceRis a chain, we have eitherρi ∈ρjorρj ∈ρi. In the first case, both(x,y)and(y,x)belong toρj, so x =y because of the antisymmetry ofρj; in the second case, the same conclusion follows because(x,y)and(y,x)belong toρi. Thus,ρis indeed antisymmetric.
We leave it to the reader to prove the transitivity ofρ. Thus,ρis a partial order that includes “”, and the arbitrary chainRhas an upper bound. By Zorn’s lemma
2.8 Posets and Zorn’s Lemma 91
Fig. 2.7 Hasse diagrams of three total orders on the set {0,x,y,z,1}
(a) (b) (c)
the poset(PORD(S,),∈)has a maximal element∪. We now prove that∪is a total order.
Suppose that(u, v)and(v,u)are two distinct ordered pairs of elements of S such that u∩∪vandv∩∪u. We show that this supposition leads to a contradiction.
Let1be the relation on S given by
1= {(x,y)∈S×S | x∪ y} ∅ {(u, v)}
∅{(z, v)∈S× {v} | z∪v} ∅ {(u,t)∈ {u} ×S | u ∪t}.
SinceιS ∈∪∈1, it follows that1is reflexive.
To prove the antisymmetry of1, suppose that p1q and q1 p. Sincev∩∪u, it follows that(p,q)∩=(u, v). Thus, the following cases may occur:
(i) If p∪q and q∪ p, then p=q by the antisymmetry of∪.
(ii) If p =u, we have u 1 q and q 1 u. By the definition of1, this implies u∪q and q∪u, respectively, so q=u= p.
(iii) If q = v, we have p 1 v andv 1 p, which imply p ∪ v andv ∪ p, respectively. Thus, p=v=q.
We leave the proof of transitivity for “1” to the reader.
Note that ∪ is strictly included in 1 because u ∩∪ v. This contradicts the maximality of the partial order∪, so∪must be a total order.
Example 2.84 Consider the poset(N5,)introduced in Example 2.55. The posets (N5,i), where 1⊕i 3 whose Hasse diagrams are shown in Fig.2.7a–c are such that≡iandiis a total order for 1i 3. Also, it is easy to see that we have actually=1∞3.
Corollary 2.85 Let(S,)be a poset and let x and y be two incomparable elements in(S,). There exists a total order∪on S that extendssuch that x ∪ y and a total order∪∪that extendssuch that y∪∪x.
Proof This statement follows immediately from Szpilrajn’s theorem.
Exercises and Supplements
1. Define the relationon the setNnby(p1, . . . ,pn)⊕(q1, . . . ,qn)if pi qi
for 1i n. Prove that(Nn,)is a partially ordered set.
2. Prove that acyclicity is a hereditary property; this means that if a relationπ∈ S×S is acyclic andθ∈π, thenθis also acyclic.
3. Let f :R−⊃R>0andg:R>0−⊃Rbe the functions defined by f(x)=ex for x ∈Randg(x)=ln x for X ∈R>0. Prove that f andgare mutually inverse isomorphisms between the posets(R,)and(R>0,).
4. Let S and T be two sets and letbe the relation on S T defined by f g if Dom(f)∈Dom(g)and f(s)=g(s)for every s∈Dom(f). Prove thatis a partial order on ST .
5. Prove that a binary relationρon a set S is a strict partial order on S if and only if it is irreflexive, transitive, and antisymmetric.
6. Let(S,)be a poset. An order ideal is a subset I of S such that x ∈ I and yx implies y∈ I . IfI(S,)is the collection of order ideals of(S,), prove thatK∈I(S,)implies
K∈I(S,). Further, argue that S ∈I(S,). 7. Let(S,)be a poset. An order filter is a subset F of S such that x ∈ F and
yx implies y∈ F . IfF(S,)is the collection of order filters of(S,), prove thatK∈F(S,)implies
K∈F(S,). Further, show that S∈I(S,). 8. Let(S,)be a finite poset. Prove that S contains at least one maximal and at
least one minimal element.
9. Let(S,)be a finite poset, where S= {x1, . . . ,xn}. Construct the sequence of posets((S1,1), (S2,2), . . .)as follows. Let(S1,1)=(S,). For 1i n, choose xpi to be the first element of Siin the sequence s=(x1, . . . ,xn)that is minimal in(Si,). Define Si+1=Si− {xpi}andi+1=i ∞(Si+1×Si+1). Prove that the sequence(xp1, . . . ,xpn)is a total order on S that extends the partial order.
10. Let S be an infinite set and let(C,∈)be the partially ordered set of its cofinite sets. Prove that for every U,V ∈Cboth sup{U,V}and inf{U,V}exist.
11. Does the poset of partial functions(ST,)introduced in Exercise 4 have a least element?
12. Let(S,)be a poset and let U and V be two subsets of S such that U ∈ V . Prove that if both sup U and sup V exist, then sup Usup V . Prove that if both inf U and inf V exist, then inf V inf U .
13. Prove that the Completeness Axiom ofRimplies that for any positive real num-bers x,y there exists n∈Nsuch that nx >y (Archimedes’ property ofR).
14. Suppose that S and T are subsets ofRthat are bounded above. Prove that S∅T is bounded above and sup S∅T =max{sup S,sup T}.
15. Letλandπbe two partitions of a finite set S. Prove that|λ| + |π||λ√π| +
|λ⊇π|.
16. Prove that ifλis a partition of a set S and|λ| =k, then there arek
2
partitions that coverλ.
Exercises and Supplements 93
Fig. 2.8 The Hasse diagram of the standard example
Fig. 2.9 Hasse diagram of the poset Tm,p,q
17. Let(S,)be a poset. Prove that if a chain in S has at most p elements and an antichain has at most q elements, then|S| ⊕pq.
18. Let(S,)be a poset. Prove that(S,)is a chain if and only if for every subset T of S both sup T and inf T exist and{sup T,inf T} ∈T .
Let(S,)be a poset. A realizer of(S,)is a family of total orders on S,R= {i
| i ∈ I}such that
=⎛
{i| i ∈ I}.
If(S,)is a finite poset, the dimension of(S,)is the smallest size d of a realizer of(S,). The dimension of a finite poset(S,)is denoted bydim(S,).
19. Let S= {x1, . . . ,xn}be a finite set. Prove that the discrete partial orderιSon S has dimension 2.
Solution: Consider the total order1= TO(x1, . . . ,xn)and its dual 2= TO(xn, . . . ,x2,x1). Note that(x,x∪)∈1 ∞2if and only if x =x∪; that is, if and only if(x,x∪)∈ιS.
20. Let(Sn,)be the poset whose Hasse diagram is given in Fig.2.8, where Sn= {x1, . . . ,xn,y1, . . . ,xn}. This poset was introduced in [1] and is known as the standard example. Prove thatdim(Sn,)=n.
21. Consider the poset(Tp,m,q,), whose Hasse diagram is given in Fig.2.9. The set Tp,m,qconsists of three sets of pairwise incomparable elements{z1, . . . ,zp}, {u1, . . . ,um}, and{w1, . . . , wq}such that zi <uj < wkfor every 1i p,
1 j m, and 1kq. Prove that if at least one of the numbers p,m,q is greater than 1, thendim(Tp,m,q,)=2.
22. Prove that the set of partial order relations on a set S is a closure system on the set S×S.
23. Prove that the transitive closure of an acyclic relation is a strict partial order.
24. Prove that if{(Si,i) | 1i n}is a family of totally ordered posets, then the lexicographic product(S1× · · · ×Sn,→)is a total order.
25. Let(S1,1)and(S2,2)be two posets and let f :S1−⊃S2be a monotonic mapping. Prove that if S2has a least element 0, then f−1(0)is an order filter of S1, and if S2has a greatest element 1, then f−1(1)is an order ideal of S1. 26. Let(S,)be a poset. Define the mapping f:S −⊃P(S)by f(x)= {y∈
S | x<y}.
(a) Prove that fis an antimonotonic mapping between the posets(S,)and (P(S),∈).
(b) If C is a chain in(S,), prove that f(C)is a chain in(P(S),∈). (c) Let (S,) and (S,∪) be two posets defined on the set S. Prove that
f∞∪(x)= f(x)∞ f∪(x)for every x∈S.
27. In the proof of Szpilrajn’s theorem, we introduced the set of partial orders that extend the partial order “”. The inclusion between relations defines a partial order onPORD(S,). We saw that the maximal elements ofPORD(S,)are total orders on S and that the least element ofPORD(S,)is the relation itself.
Let(S,)be a poset. Prove that there exists a collection of total orders{i| i∈ I}on S such that=
i∈I i.
Solution: Ifis itself a total order, then the desired collection of total orders consists ofitself. Suppose therefore thatis not total, and letINC(S,)be the set of all pairs of incomparable elements of(S,).
For each pair(x,y)∈INC(S,), consider the total orders∪x y and∪∪x ythat extendsuch that x ∪x y y and y∪x y x. Clearly,
∈ ⎛
{∪x y ∞∪∪x y| (x,y)∈INC(S,)}.
Suppose that
{∪x y ∞∪∪x y| (x,y)∈INC(S,)}contains a pair of elements (r,s)INC(S,). Then, we have both r ∪r s s and r ∪∪r s s. Since s ∪∪r s r , this would imply r =s by the antisymmetry of∪∪r s. This, however, contradicts the incomparability of(r,s)in(S,). Thus, for any pair(u, v)∈
{∪x y ∞∪∪x y| (x,y)∈INC(S,)}, we have uvorvu, which shows that
= ⎛
{∪x y ∞∪∪x y| (x,y)∈INC(S,)}.
A poset(S,)is locally finite if every interval[x,y]of S is a finite set.
28. Prove that the poset(N,)is locally finite.
Exercises and Supplements 95 29. Let S be a finite set. Prove that the poset(Seq(S),i n f), wherei n f is the
partial order introduced in Example 2.6, is locally finite.
Let(P,), and(Q,)be two posets. Their product is the poset(P×Q,)where (x,y)(x∪,y∪)if x x∪and yy∪.
30. Let(P,), and(Q,)be two posets. Prove that(P×Q,)is locally finite if and only if both(P,)and(Q,)are locally finite.
31. Prove that if(P,)and(Q,)are graded posets by the grading functions h andg, respectively, then(P ×Q,)is graded by the function f defined by
f(p,q)=h(p)g(q)for(p,q)∈ P×Q.
32. Letψ :S×S −⊃Rbe the Riemann function of a locally finite poset(S,), and letψkbe the productψ∧ψ∧ · · · ∧ψ, which contains kψfactors, where k∈N. Prove that:
(a) ψ2(x,y)= |[x,y]|if xy.
(b) ψk(x,y)gives the number of multichains of length k that can be interpolated between x and y.
Bibliographical Comments
There is a vast body of literature dealing with posets and their applications and a substantial number of references that focus on combinatorial study of posets. Among these we mention [2–5].
Two very useful referrences are [6] and [7].
References
1. B. Dushnik, E.W. Miller, Partially ordered sets. Am. J. Math. 63, 600–610 (1941)
2. W.M. Trotter, Combinatorics and Partially Ordered Sets (The Johns Hopkins University Press, Baltimore, 1992)
3. W.T. Trotter, Partially ordered sets, in Handbook of Combinatorics, ed. by R.L. Graham, M.
Grötschel, L. Lovász (The MIT Press, Cambridge, 1995), pp. 433–480
4. R.P. Stanley, Enumerative Combinatorics, vol. 1 (Cambridge University Press, Cambridge, 1997)
5. J.H. van Lint, R.M. Wilson, A Course in Combinatorics (Cambridge University Press, Cam-bridge, second edition, 2002)
6. M. Barbut, B. Montjardet, Ordre ar Classification - Algèbre et Combinatoire (Hachette Uni-versité, Paris, 1970)
7. N. Caspard, B. Leclerc, B. Montjardet, Finite Ordered Sets (Cambridge University Press, Cambridge, 2012)
Chapter 3
Combinatorics
3.1 Introduction
Combinatorics is the area of mathematics concerned with counting collections of mathematical objects. We begin by discussing several elementary combinatorial issues such as permutations, the power set of a finite sets, the inclusion-exclusion principle, and continue with more involved combinatorial techniques that are relevant for data mining, such as the combinatorics of locally finite posets, Ramsey’s Theorem, various combinatorial properties of collection of sets. The chapter concludes with two sections dedicated to the Vapnik-Chervonenkis dimension of a collection and to the Sauer–Shelah theorem.
3.2 Permutations
Definition 3.1 A permutation of a set S is a bijection f :S−∈S.
A permutation f of a finite set S= {s0, . . . ,sn−1}is completely described by the sequence(f(s0), . . . ,f(sn−1)). No two distinct components of such a sequence may be equal because of the injectivity of f , and all elements of the set S appear in this sequence because f is surjective. Therefore, the number of permutations equals the number of such sequences, which allows us to conclude that there are n(n−1)· · ·2·1 permutations of a finite set S with|S| =n.
The number n(n −1)· · ·2·1 is denoted by n!. This notation is extended by defining 0! =1 to capture the fact that there exists exactly one bijection of∅, namely the empty mapping.
The set of permutations of the set S = {1, . . . ,n}is denoted by PERMn. If f ∪PERMnis such a permutation, we write
f :
1 · · · i · · · n a1· · ·ai · · ·an
⎜ ,
D. A. Simovici and C. Djeraba, Mathematical Tools for Data Mining, 97 Advanced Information and Knowledge Processing, DOI: 10.1007/978-1-4471-6407-4_3,
© Springer-Verlag London 2014
98 3 Combinatorics where ai = f(i)for 1∞i ∞n. To simplify the notation, we specify f just by the sequence(a1, . . . ,ai, . . . ,an).
Since{1, . . . ,n}is a finite set, for every x ∪ {1, . . . ,n}and f ∪ PERMnthere exists k∪Nsuch that x = fk(x). If k is the least number with this property, the set {x, f(x), . . . , fk−1(x)}is the cycle of x and is denoted by Cf,x. The number|Cf,x| is the length of the cycle.
Cycles of length 1 are said to be trivial.
Note that each pair of elements fi(x) and fj(x) of Cf,x are distinct for 0∞i,j ∞ |Cf,x| −1.
If z ∪ Cf,x and|Cf,x| = k, then z= fj(x)for some j , 0 ∞ j ∞ k−1. Since x= fk(x), it follows that x = fk−j(z), which shows that x∪Cf,z. Consequently, Cf,x =Cf,z.
Thus, the cycles of a permutation f ∪PERMnform a partitionρf of{1, . . . ,n}. Definition 3.2 A k-cyclic permutation of{1, . . . ,n}is a permutation such thatρf
consists of a cycle of length k,(j1, . . . ,jk)and a number of n−k cycles of length 1.
A transposition of{1, . . . ,n}is a 2-cyclic permutation.
Note that if f is a transposition of{1, . . . ,n}, then f2=1S.
Theorem 3.3 Let f be a permutation in PERMn, andρf = {Cf,x1, . . . ,Cf,xm}be the cycle partition associated to f . Define the cyclic permutations g1, . . . ,gm of {1, . . . ,n}as
gp(t)=
f(t) if t∪Cf,xp, t otherwise.
Then,gpgq=gqgpfor every p,q such that 1∞ p,q ∞m.
Proof Observe first that u∪Cf,x if and only if f(u)∪Cf,xfor any cycle Cf,x. We can assume that p ∅= q. Then, the cycles Cf,xp and Cf,xq are disjoint. If u ∅∪ Cf,xp ⊆Cf,xq, then we can writegp(gq(u)) = gp(u) =u andgq(gp(u)) = gq(u)=u.
Suppose now that u ∪ Cf,xp −Cf,xq. We havegp(gq(u)) = gp(u) = f(u). On the other hand,gq(gp(u)) = gq(f(u)) = f(u)because f(u) ∅∪ Cf,xq. Thus, gp(gq(u))=gq(gp(u)). The case where u∪Cf,xq−Cf,xpis treated similarly. Also, note that Cf,xp ∩Cf,xq = ∅, so, in all cases, we havegp(gq(x))=gq(gp(u)).
The set of cycles{g1, . . . ,gm}is the cyclic decomposition of the permutation f . Definition 3.4 A standard transposition is a transposition that changes the places of two adjacent elements.
Example 3.5 The permutation f ∪PERM5given by f :
1 2 3 4 5 1 3 2 4 5
⎜
is a standard transposition of the set{1,2,3,4,5}. On the other hand, the permutation
g:
1 2 3 4 5 1 5 3 4 2
⎜
is a transposition but not a standard transposition of the same set because the pair of elements involved is not consecutive.
If f ∪ PERMnis specified by the sequence(a1, . . . ,an), we refer to each pair (ai,aj)such that i < j and ai >aj as an inversion of the permutation f . The set of all such inversions is denoted by INV(f). The number of elements of INV(f)is denoted by inv(f).
A descent of a permutation f ∪PERMnis a number j such that 1∞ j ∞n−1 and aj >aj+1. The set of descents of f is denoted by D(f).
Example 3.6 Let f ∪PERM6be:
f :
1 2 3 4 5 6 4 2 5 1 6 3
⎜ .
We have INV(f) = {(4,2), (4,1), (4,3), (2,1), (5,1), (5,3), (6,3)} and inv(f)
=7. Furthermore, D(f)= {1,3,5}.
It is easy to see that the following conditions are equivalent for a permutation f ∪PERMn:
(i) f =1S; (ii) inv(f)=0;
(iii) D(f)= ∅.
Theorem 3.7 Every permutation f ∪ PERMncan be written as a composition of transpositions.
Proof If D(f)= ∅, then f =1Sand the statement is vacuous. Suppose therefore that D(f) ∅= ∅, and let j ∪ D(f), which means that(aj,aj+1)is an inversion f . Letg be the standard transposition that exchanges aj and aj+1. It is clear that inv(gf)=inv(f)−1. Thus, ifgiare the transpositions that correspond to all standard inversions of f for 1∞i ∞ p=inv(f), it follows thatgp· · ·g1f has 0 inversions and, as observed above,gp· · ·g1f =1S. Sinceg2 =1S for every transpositiong, we have f =gp· · ·g1, which gives the desired conclusion.
100 3 Combinatorics Theorem 3.8 If f ∪ PERMn, then inv(f)equals the least number of standard transpositions, and the number of standard transpositions involved in any other factorization of f as a product of standard transposition differs from inv(f)by an even number.
Proof Let f = hq· · ·h1 be a factorization of f as a product of standard trans-positions. Then, h1· · ·hqf =1S and we can define the sequence of permutations fl =hl· · ·h1f for 1 ∞l ∞q. Since each hi is a standard transposition, we have inv(fl+1)−inv(fl)=1 or inv(fl+1)−inv(fl)= −1. If
|{l | 1∞l∞q−1 and inv(fl+1)−inv(fl)=1}| =r,
then|{l | 1∞l∞q−1 and inv(fl+1)−inv(fl)= −1}| =q−r , so inv(f)+r− (q−r)=0, which means that q =inv(f)+2r . This implies the desired conclusion.
An important characteristic of permutations is their parity. Namely, the permuta-tion parity is defined as the parity of the number of their inversions: a permutapermuta-tion
f ∪PERMnis even (odd) if inv(f)is an even (odd) number.
Theorem 3.8 implies that any factorization of a permutation as a product m stan-dard transpositions determines whether the permutation is odd or even.
Note that any transposition is an odd permutation. Indeed, if f ∪ PERMnis a transposition of i and j , where i< j we have
f =(1,2, . . . ,i−1,j,i+1, . . . ,j−1,i,j+1, . . . ,n).
The number j generates j−i inversions, and each of the numbers i+1, . . . ,j−1 generates one inversion because they are followed by i . Thus, the total number of inversions is j−i+(j−i−1)=2(j−i)−1, which is obviously an odd number.
Theorem 3.9 A cyclic permutation f of length k is the composition of k−1 trans-positions.
Proof Let(j1, . . . ,jk)be the cycle of length k of f . It is immediate that f is the product of the k−1 transpositions(j1,j2), (j2,j3), . . . , (jk−1,j1).
Thus, the parity of a cyclic permutation of even length is odd.
Corollary 3.10 Let f ∪PERMnbe a permutation that has cνcycles of lengthνfor ν⊕1. The parity of f is the parity of the number c2+c4+ · · ·; in other words, the parity of a permutation is given by the parity of the number of its even cycles.
Proof By Theorem 3.9 a cyclic transposition of length ν is the composition of ν−1 transpositions. Thus, if f has cν cycles of length ν, then f is a product
of
ν1cν(ν−1)transpositions. It is clear that t he parity of this sum is deter-mined by those terms whereν−1 is impar. Thus the parity of f is given by the parity of c2+3c4+5c6+ · · · and this equals the parity of c2+c4+c6+ · · ·.