The set of tolerances on A is denoted by TOL(S).
Example 1.106 Let a be a nonnegative number and letρa⊆R×Rbe the relation defined by
ρa= {(x,y)∈S×S | |x−y|a}.
It is clear thatρais reflexive and symmetric; however,ρais not transitive in general.
For example, we have(3,5) ∈ ρ2and(5,6)∈ ρ2, but(3,6)∈ ρ2. Thus,ρ2is a tolerance but is not an equivalence.
1.3.7 Partitions and Covers
Next, we introduce the notion of partition of a set, a special collection of subsets of a set.
Definition 1.107 Let S be a nonempty set. A partition of S is a nonempty collection π = {Bi | i ∈ I}of nonempty subsets of S, such that ⎜
{Bi | i ∈ I} = S, and Bi∩Bj = ∅for every i,j∈I such that i=j.
Each set Biofπis a block of the partitionπ.
The set of partitions of a set S is denoted by PART(S). The partition of S that consists of all singletons of the form {s}with s ∈ S will be denoted by αS; the partition that consists of the set S itself will be denoted byωS.
Example 1.108 For the two-element set S = {a,b}, there are two partitions: the partitionαS= {{a},{b}}and the partitionωS= {{a,b}}.
For the one-element set T = {c}, there exists only one partition,αT =ωT = {{t}}. Example 1.109 A complete list of partitions of a set S = {a,b,c}consists of
π0= {{a},{b},{c}},π1= {{a,b},{c}}, π2= {{a},{b,c}}, π3= {{a,c},{b}}, π4= {{a,b,c}}.
Clearly,π0=αSandπ4=ωS.
Definition 1.110 Let S be a set and letπ,σ ∈ PART(S). The partition πis finer than the partitionσif every block C ofσis a union of blocks ofπ. This is denoted byπσ.
Theorem 1.111 Letπ= {Bi | i∈I}andσ= {Cj | j ∈J}be two partitions of a set S.
Forπ,σ∈PART(S), we haveπσif and only if for every block Bi ∈πthere exists a block Cj∈σsuch that Bi⊆Cj.
1.3 Relations and Functions 29 Proof Ifπσ, then it is clear for every block Bi ∈πthere exists a block Cj ∈ σ such that Bi⊆Cj.
Conversely, suppose that for every block Bi∈πthere exists a block Cj∈σsuch that Bi⊆Cj. Since two distinct blocks ofσare disjoint, it follows that for any block Bi ofπ, the block Cj ofσthat contains Biis unique. Therefore, if a block B ofπ intersects a block C ofσ, then B⊆C.
Let Q = ⎜
{Bi ∈ π | Bi ⊆ Cj}. Clearly, Q ⊆ Cj. Suppose that there exists x∈Cj−Q. Then, there is a block Bβ∈πsuch that x∈Bβ∩Cj, which implies that Bβ ⊆Cj. This means that x∈Bβ ⊆C, which contradicts the assumption we made about x. Consequently, Cj=Q, which concludes the argument.
Note thatαSπωSfor everyπ∈PART(S).
Two equivalence classes either coincide or are disjoint. Therefore, starting from an equivalenceρ∈EQ(U), we can build a partition of the set U.
Definition 1.112 The quotient set of the set U with respect to the equivalenceρis the partition U/ρ, where
U/ρ= {[u]ρ | u∈U}.
An alternative notation for the partition U/ρisπρ.
Moreover, we can prove that any partition defines an equivalence.
Theorem 1.113 Letπ= {Bi | i∈I}be a partition of the set U. Define the relation ρπby(x,y)∈ρπ if there is a set Bi ∈πsuch that{x,y} ⊆Bi. The relationρπis an equivalence.
Proof Let Bi be the block of the partition that contains u. Since{u} ⊆Bi, we have (u,u)∈ρπ for any u∈U, which shows thatρπis reflexive.
The relation ρπ is clearly symmetric. To prove the transitivity ofρπ, consider (u, v), (v,w)∈ρπ. We have the blocks Biand Bjsuch that{u, v} ⊆Biand{v,w} ⊆ Bj. Sincev ∈ Bi ∩Bj, we obtain Bi = Bj by the definition of partitions; hence, (u,w)∈ρπ.
Corollary 1.114 For any equivalence ρ ∈ EQ(U), we have ρ = ρπρ. For any partitionπ∈PART(U), we haveπ=πρπ.
Proof The argument is left to the reader.
The previous corollary amounts to the fact that there is a bijectionφ:EQ(U)−→
PART(U), where φ(ρ) = πρ. The inverse of this mapping,Ψ : PART(U) −→
EQ(U), is given byψ(π)=ρπ.
Also, note that, forπ,π⇒∈PART(S), we haveππ⇒if and only ifρπ ⊆ρπ⇒. We say that a subset T of a set S isπ-saturated if it is aρπ-saturated set.
Theorem 1.115 For any mapping f : U −→ V , there is a bijection h:U/ker(f)−→f(U).
Proof Consider the ker(f)class[u]of an element u∈U, and define h([x])=f(x). The mapping h is well-defined for if u⇒ ∈ [u], then(u,u⇒) ∈ ker(f), which gives f(u)=f(u⇒).
Further, h is onto since if y∈f(U), then there is u∈U such that f(u)=y, and this gives y=h([u]).
To prove the injectivity of h, assume that h([u])=h([v]). This means that f(u)= f(v); hence,(u, v)∈ker(f), which means, of course, that[u] = [v].
An important consequence of the previous proposition is the following decompo-sition theorem for mappings.
Theorem 1.116 Every mapping f :U−→V can be decomposed as a composition of three mappings: a surjectiong:U−→U/ker(f), a bijection h:U/ker(f)−→
f(U), and an injection k:f(U)−→V .
Proof The mappingg : U −→ U/ker(f)is defined by g(u) = [u]for u ∈ U, while k:f(A)−→B is the inclusion mapping given by k(v)=vfor allv∈f(U). Therefore, k(h(g(u)))=k(h([u]))=k(f(u))=f(u)for all u∈U.
A generalization of the notion of partition is introduced next.
Definition 1.117 Let S be a set. A cover of S is a nonempty collectionCof nonempty subsets of S,C= {Bi | i∈I}, such that⎜
{Bi | i∈I} =S.
The set of covers of a set S is denoted by COVERS(S).
Example 1.118 Let S be a set. The collectionPk(S)of subsets of S that contain k elements is a cover of S for every k 1. For k =1,P1(S)is actually the partition αS.
The notion of collection refinement introduced in Definition 1.12 is clearly applicable to covers and will be used in Sect.15.5.
Definition 1.119 A Sperner collection of subsets of a set S is a collectionCsuch thatC⊆P(S)and for any C,C⇒∈C, C =C⇒implies that C⊆C⇒and C⇒⊆C.
The set of Sperner collections on S is denoted bySPER(S).
LetCandDbe two Sperner covers of S. DefineCDif for every C∈Cthere exists D∈Dsuch that C⊆D.
The relationis a partial order on the collectionSPER(S). Indeed, the relation is clearly reflexive and transitive. So we need to verify only that it is antisymmetric.
Suppose thatC,D∈SPER(S),CD,andDC. If C∈C, there exists D∈D such that C ⊆ D. On the other hand, sinceD C, there exists C⇒ ∈ Csuch that D⊆C⇒, so C⊆C⇒. SinceCis a Sperner collection this is possible only if C =C⇒, so D=C, which impliesC⊆D. Applying a similar argument to an arbitrary D∈D yields the conclusion thatD⊆C, soC=D, which allows us to conclude that “” is antisymmetric and, therefore, a partial order onSPER(S).
1.4 Countable Sets 31
1.4 Countable Sets
A set is called countable if it is either empty or the range of a sequence. A set that is not countable is called uncountable.
Note that if S is a countable set and f : S −→T is a surjection, then T is also countable.
Example 1.120 Every finite set is countable. Let S be a finite set. If S = ∅, then S is countable. Otherwise, suppose that S= {a0, . . . ,an−1}, where n1. Define the sequence s as
s(i)=
ai if 0in−1, an−1 otherwise. It is immediate thatRan(s)=S.
Example 1.121 The setNis countable becauseN=Rans, where s is the sequence s(n) = n for n ∈ N. A similar argument can be used to show that the set Z is countable. Indeed, let t be the sequence defined by
t(n)= n−1
2 if n is odd
−n2 if n is even.
Let m be an integer. If m >0, then m=t(2m−1); otherwise (that is, if m 0), m=t(−2m), so z=Ran(t).
Example 1.122 We prove now that the set N×N is countable. To this end, consider the representation of pairs of natural numbers shown in Fig.1.2. The pairs of the setN×Nare scanned in the order suggested by the dotted arrows. The 0th pair is (0,0), followed by(0,1),(1,0),(0,2),(1,1),(2,0), etc. We define the bijection β :N×N−→Nasβ(p,q)=n, where n is the place occupied by the pair(p,q)in the previous list. Thus,β(0,0)=0,β(0,1)=1,β(2,0)=5, and so on.
In general, bijections of the form h : N×N −→ Nare referred to as pairing functions, soβis an example of a pairing function.
The existence of the inverse bijectionβ−1:N−→N×Nshows thatN×Nis indeed a countable set becauseN×N=Ran(β−1).
Another example of a bijection betweenN×NandPcan be found in Exercise 22.
Starting from countable sets, it is possible to construct uncountable sets, as we see in the next example.
Example 1.123 Let F be the set of all functions of the form f :N−→ {0,1}. We claim that F is not countable.
If F were countable, we could write F = {f0,f1, . . . ,fn, . . .}. Define the function g : N −→ {0,1}byg(n)= fn(n)for n∈ N, where0¯ =1 and1¯ = 0. Note that g=fnfor every fnin F becauseg(n)=fn(n)=fn(n), that is,gis different from fn
Fig. 1.2 Representation ofN×N.
at least on n for every n∈N. Butgis a function defined onNwith values in{0,1}, so it must equal some function fm from F. This contradiction implies that F is not countable.