Chapter 2 ELEMENTARY SET THEORY

Chapter 2 ELEMENTARY SET THEORY 2.1

Introduction

We adopt the “naive” (as opposed to axiomatic) point of view for set theory and regard the notions of a set as primitive and well-understood without formal definitions. We just assume that a set A is a collection of objects characterized by some defining property that allows us to think of the objects as a whole entity. The defining property has to be such that it must be clear whether a given object belongs to the set or not. The objects possessing the property are called elements or members of the set. We denote sets by common capital letters A, B, C, etc. and elements or objects of the sets by lower case letters a, b, c,etc. For example, we write A = {a, b, c} to indicate that and if

x

A

is a collection of elements

belongs to

A

we write

2.1.1 Description of a Set

x ∈ A.

If

x

a, b, c.

If

x

A is a x∈ / A.

is any element, if

doesn’t belong to

A

we write

set,

To describe a particular set, we have to indicate the property that characterizes its elements. To start, lets consider a set which has no members. Since a set is determined by its elements, there is only one such set which is called the

set, which is denoted by ∅. Any be non-empty or non-void.

set

A,

empty set,

or the

null

consisting of one or more elements is said to

A non-void set A is said to be finite if A contains n distinct elements, where n is some positive integer. Such a set A is said to be of order n. The null set is defined to be finite with order zero. A set consisting of exactly one element, say A = {a}, is called a singleton . If a set A is not finite, then we say that A is Definition 34

infinite.

To describe a finite set, one way to proceed is by listing or enumerating members. For example,

,

integers 1 3 and 5.

S = {1, 3, 5} means that S

Note that we use braces to denote the set and that the order

of the elements plays no role in describing the set. Thus the sets

S




=

{5, 3, 1}

its

is the set whose elements are the

are to be viewed as exactly the same set.

S

=

{1, 3, 5}

and

Elementary Set Theory

20

But in general, — and especially for infinite sets —, we do not describe a set by listing all its members. A convenient method of characterizing sets is as follows. Suppose that for each elements x of a set A there is a propositional function P (x) which is either true or false. We may then define a set C which consists of all elements x ∈ A such that P (x) is true, and we may write C

= {x ∈ A : P (x) is true}.

For brevity of notation, we will omit the words ”is true” and, when it is clear which set x belongs to, we will also omit the reference set A. Thus we will sometimes write {x : P (x)} . From a given set, we can form new sets, called subsets of the given set. In general we say that a set A is a subset of a set belongs also to B. More formally

B

whenever every element of

A

Definition 35 Given two sets A and B, we say that A is a subset of B , or that A is contained in B, and denote it A ⊆ B (or B ⊇ A), iff ∀x : x ∈ A ⇒ x ∈ B. We say that A is a proper subset of B, or that A is strictly contained in B , and denote it A ⊂ B, iff [A ⊆ B ∧ ˜ (B ⊆ A)] is true. Definition 36 We say that a set A is equal to B and write A = B iff x ∈ A ⇐⇒ x ∈ B, A B A = B ⇔ [A ⊆ B ∧ (B ⊆ A)] A different B A = B A B x y x = y. x y x = y. Exercise 37 Show that if A is any set, then A ⊆ A and ∅ ⊆ A. Exercise 38 Show that A = B ⇐⇒ [A ⊆ B ∧ (B ⊆ A)] . Definition 39 Let X be a set and let A ⊆ X.The complement of subset A with respect to X is the set of elements of X which do not belong to A. We denote the complement of A with respect to X by ACX = {x ∈ X : x ∈/ A}. i.e.,

.

and

We say that

equal. If write

and If

have the same elements. Equivalently is

from

, and we write

, if

and

are not

denote the same element of a set we say that they are equal and we

and

denote distinct elements of a set, we write

When it is clear that the complement is with respect to X , we simply say the and write AC = {x ∈ X : x ∈/ A}. In every discussion involving sets, we will always have a given fixed set in mind from which we take elements and subsets. We will call this set the universal set, and we will usually denote it by X. complement of A,

Throughout the remainder of the present chapter, X will denote always an arbitrary non-empty fixed set.

Exercise 40 Prove the following statements. Let A,B, C be subsets of X. Then 1. if A ⊆ B and B ⊆ C then A ⊆ C.

Operation on sets

21

2. XC = ∅ 3. ∅C = X
 4. AC C = A 5. A ⊆ B ⇔ B C ⊆ AC 6. A = B ⇔ AC = BC

Next we need to consider sets whose elements are sets themselves. For example, if A,B and C are subsets of X, then the collection A = {A,B,C } is a set, whose elements are A,B and C. We usually call a set whose elements are subsets of X a

X or a collection of subsets of X. In terms of notation, this concludes our hierarchical system, where lowercase letters refer to elements of X, upper case letter to subsets of X and script letters to families of subsets of X. Lets note that the empty set ∅ is a subset of X. It is possible then to form family of subsets of

a nonempty set whose only element is the empty set, i.e. {∅}. In this case, {∅} is a singleton. We see that ∅ ⊆ {∅} and ∅ ∈ {∅}. There is a special family of subsets of to which we have given a special name.

X

power set

by

P (A) .

A

Let A be any subset in X. We define the power class of or the of to be the family of all subsets of . We denote the power set of A Specifically

Definition 41

A

A

P (A) = {B : B ⊆ A} The power class of the empty set is P (∅) = {∅},i.e. the singleton of ∅. The power class of a singleton is P ({a}) = {∅,{a}}.Notice that the power set of A always contains A and ∅. In general, if A is a finite set with n elements, then P (A) contains 2n elements. This is why sometimes the power set of A is (improperly) denoted as 2A.

Exercise 42 Prove that if A is a finite set with n elements, then P (A) contains 2n elements.

2.2 Operation on sets

2.2.1 Union and Intersection Definition 43 Let A and B be two subsets of X.We define the union of sets A and B , denoted by A ∪ B, as the set of all elements that are in A or B; i.e.

A ∪ B = {x ∈ X : [(x ∈ A)

∨ (x ∈ B )]}.

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22

We also define the of sets A and B , denoted by A ∩ B, as the set of all elements that are in both A and B; i.e. intersection

A ∩ B = {x ∈ X : [(x ∈ A) ∧ (x ∈ B)]}.

Both definitions can be extended to finite or infinite collection of sets. Definition 44 Two sets are said to be disjoint if they don’t have any element in common, i. e.. if A ∩ B = ∅.

Exercise 45 Prove the following properties 1. A ∩ AC = ∅; A ∪ AC = X. 2. Commutative laws:

A ∪ B = B ∪ A; A ∩ B = B ∩ A.

• •

3. Associative law:

A ∪ (B ∪ C ) = (A ∪ B) ∪ C = A ∪ B ∪ C ; A ∩ (B ∩ C ) = (A ∩ B) ∩ C = A ∩ B ∩ C.

• •

4. Distributive law:

A ∪ (B ∩ C ) = (A ∪ B) ∩ (A ∪ C ); A ∩ (B ∪ C ) = (A ∩ B) ∪ (A ∩ C ). 5. (A ⊆ B ) ⇔ [(A ∪ B ) = B ] ∧ [(A ∩ B ) = A] • •

6. De Morgan’s laws (A ∩ B )C = AC ∪ B C (A ∪ B )C = AC ∩ B C

• •

From the associative law established in the previous exercise, there is no ambiguity in writing A ∪ B ∪ C. Extending this concept, let n be any positive integer and let A1 , A2 ..., An denote subsets of X. The set A1 ∪ A2 ∪ ... ∪ An is defined to be the set of all x ∈ X which belong to at least one of the subsets Ai , and we write


A = A1 ∪ A ∪ ... ∪ A n

=1

n

2

i

{x ∈ X : x ∈ Ai for some i = 1, ...,n}

=

i


A = A1 ∩ A ∩ ... ∩ A

Using a similar argument we define the intersection of

n

=1

i

i

2

n

=

n subsets of X as

{x ∈ X : x ∈ Ai for all i = 1, ...,n}

Relations

23

Theorem 46

Let A1 , A2 ..., An


n A C n AC i=1 i i=1 i
n A
A

•

[

•

[

]

i=1

be subsets of

X. Then

=

C = i]

n

C i

=1

i

Proof. Is left as an exercise. Use your results from the previous exercise.

2.2.2

Relative Difference

Definition 47 Let A and B be two subsets of X. The relative difference of B and

B − A) ,— also known as the relative complement of A in B — is defined as the set of elements in B that do not belong to A. Hence B − A = {x ∈ X : x ∈ B ∧ x ∈/ A}. A , denoted as (

Exercise 48 Prove the following statements. Let A and B be two subsets of X. Then 1. A − B = A ∩ B C . 2. 3. 4. 5.

A − B )C = AC ∪ B. A = B ⇔ A − B = B − A. B − (B − A) = A ⇔ A ⊆ B. B−A=B ⇔B∩A=∅

(

Exercise 49 Let B and A1, A2..., An be subsets of X. Then



B−A) =1
 n n 2. B − [ i=1 Ai i=1 B − A )

1. B − [ ni=1 A ] = i

] =

n i

(

i

(

i

2.3 Relations

2.3.1 Ordered pairs

So far the order of the elements has not been relevant. Thus the set {a, b} has been considered equivalent to the set {b, a} . There are times however where the order is important. When we wish to indicate that a set of two elements a and b is ordered, we enclose the elements in parentheses: (a, b) . Then a is called the first element and b the second element. A set of such two elements is called an ordered pair, and is defined in the following way.

Definition 50 That is

An ordered pair (a, b)

(a, b) = {{a}, {a, b}}.

is a set whose members are

{a}

and

From the definition, we can derive a main property of ordered pairs.

{a, b}.

Elementary Set Theory

24

Theorem 51 (x, y) = (a, b) ⇐⇒ x = a and y = b Proof. Lets first prove the converse. If x = a and y = b then (a, b) = {{a} , {a, b}} = {{x} , {x, y}} = (x, y) Now lets assume (x, y ) = (a, b) . Then by definition we have {{x} , {x, y }} = {{a} , {a, b}} . Consider two cases, depending on whether x = y or x = y. If x = y then {x} = {x, y} ,so (x, y) = {{x}} . Since (x, y) = (a, b) then we have {{x}} = {{a} , {a, b}} .

{}

The set on the left hand side has only one member x .Thus the set on the right must also have one member, so a = a, b , concluding that a = b. Then x = a , so x = a and x = a. Thus x = y = a = b. If x = y, then it must be a = b — otherwise we would be arriving to a contradiction —. From the assumption we know that (x, y) = (a, b) , so

{} {} 

{} { } 

{{ }} {{ }}

{x} ∈ {{a} , {a, b}} implying that {x} = {a} or {x} = {a, b} . In either case we have a ∈ {x} , and so

a = x. By the same argument we have

{x, y} ∈ {{a} , {a, b}} and given that x = y it has to be {x, y } = {a, b} . Now x = a, x = y and y ∈ {a, b} implies y

= b.

Exercise 52 Extend the definition of an ordered pair to a finite set of n elements. Call it n-dimensional vector or n-tuple.

2.3.2 Cartesian Product Definition 53 The Cartesian product of two sets A and B, written A × B, is the set of all ordered pairs (a, b) such that a ∈ A and b ∈ B. That is A×B

= {(a, b) : a ∈ A,

Notation We will write A2 = A × A.

and

b ∈ B}

2.3.3 Properties It is possible to derive some important properties, which we enumerate below.

A ∩ B ) × (C ∩ D) = (A × C ) ∩ (B × D). 2. (A ∪ B ) × C = (A × C ) ∪ (B × C ). Exercise 54 Prove the two properties stated above. 1. (

Relations

25

2.3.4 Binary Relations

A binary relation between two objects a and b is a condition involving a and b that

is either true or false. For example, the relation “less than” is a relation between positive integers. When considering a relation between two objects, in some cases it is necessary to know which object comes first. For instance, if “Mark is taller than Paul” is true then “Paul is taller than Mark” is false. Thus, it is natural for the formal definition of a relation to depend on the concept of an ordered pair.

Definition 55 Let A and B be sets. A relation between A and B R of A

×

is any subset

a ∈ A and b ∈ B are related by R if (a, b) ∈ R, and we aRb”. If B = A, then we speak of a relation R ⊆ A × A being

B. We say that

denote this by saying “ a

relation on A.

Definition 56 Let R be a relation between two sets A and B. The domain of R, D(R) {a ∈ A : ∃ b ∈ B  (a, b) ∈ R} denoted

, is the set of all first elements of members of R. Formally,

D(R) =

R, denoted R(R), is the set of all second elements of R. Formally, R(R) = {b ∈ B : ∃a ∈ A  (a,b) ∈ R}.Clearly from the definition we obtain D (R) ⊆ A, and R(R ) ⊆ B. .

The

range

of

Definition 57 A binary relation R on a set A can have the following properties: Completeness: ∀a, b ∈ A, aRb ∨ bRa. Reflexive property: ∀a ∈ A, aRa. Irreflexive property: ∀a ∈ A, ˜(aRa). Symmetric property: ∀a, b ∈ A, aRb ⇒ bRa. Antisymmetric property: ∀a, b ∈ A and a = b, aRb ⇒ ˜ (bRa) . Transitive property: ∀a, b, c ∈ A, (aRb ∧ bRc) ⇒ aRc.

1. The relation “≤” defined on S = {1, 2, 3, 4, 5} is complete, reflexive and antisymmetric.

Example 58

2. Let S be the set of all people who live in the US, and suppose that two people x and y are related by R if x lives within a mile of y. Then R is not complete, is reflexive and symmetric, but not transitive.

Exercise 59 Let R be the relation “to be taller and younger than” on the set S of all the Econ 897 students.

What properties does R have?

We are specifically interested in two types of relations which come up frequently in economics and mathematics, namely equivalence and order relations.

Elementary Set Theory

26

2.3.5

Equivalence relations A relation R on a set S is an equivalence relation if it is reflexive, symmetric and transitive. An equivalence relation is often denoted by ”=” or ”˜”.

Definition 60

Note that completeness is not a requirement for equivalence. This type of relation is singled out because it possess the properties naturally associated with the idea of equality.

Example 61 Let S be the set of all people who live in the US, and suppose that two people x and y are related by R if “x has the same age as y”, then R is reflexive, transitive and symmetric. Hence R is an equivalence relation.

Exercise 62 Let S be the set of all lines in a plane. Determine which of the following are equivalence relations

• x R y if “x is parallel to y” • x R y if “x is perpendicular to y”

Given an equivalence relation R on a set S, it is natural to group together all the elements that are related to a particular element. More precisely

Definition 63 Let R

an element of

A.

{y ∈ A : yRx} .

The

A2

be an equivalence relation on a set A, and let x be equivalence class of x with respect to R is the set Ex =

⊆

Exercise 64 Prove the following statements:

1) “Let S be an equivalence relation defined on a set A. Then every two different equivalence classes must be disjoint”. 2) “Let S be an equivalence relation defined on a set A. Then every element of A belongs to at most one equivalence class”.

Thus we see that an equivalence relation R on a set S breaks S into disjoint subsets in a natural way. This operation on a set is known as a partition. 2.3.6 Partition Definition 65 Let A be a subset of X. A partition of A is a collection nonempty subsets of

x ∈ A belongs to some subset S ∈ A. For all S, T ∈ A, if S = T, then S ∩ T = ∅.

1. Each 2.

A such that

A

of

Relations

27

Example 66 Let S be the set of all students in a particular university. For x and y in S, define xRy iff x and y were born in the same calendar year. Then R is an equivalence relation, and a typical equivalence class is the set of all students who were born “in the same year as student z”.

Exercise 67 Consider the previous example. If y ∈ Ez , does this mean that y and z

are the same age?. Would it have been the same to define the equivalence relation as the set of all students who were born in a particular year, say “in 1980”?

Not only does an equivalence relation on a set S determine a partition of S, but the partition can be used to determine the relation. We formalize this as a theorem Let R be an equivalence relation on a set S. Then A = {Ex : x ∈ S } is a partition of S . The relation “belongs to the same piece as” is the same as R. Conversely, if P is a partition of S, let R be defined by xRy iff x and y are in the same piece of the partition. Then R is an equivalence relation and the corresponding partition into equivalence classes is the same as P .

Theorem 68

Proof. Since R is reflexive, each element x ∈ S belongs to some equivalence Ex ∈ A. Moreover, from a previous exercise, we know that two equivalence

class classes must be either equal or disjoint. Hence all the different equivalence classes have empty intersection, making A a partition on S . Now suppose that P is the relation “belongs to the same piece (equivalence class) as ”. Then

xPy ⇔ x, y ∈ Ez for some z ∈ S ⇔ xR zand yRz for some z ∈ S ⇔ xRy Thus P and R are the same. Conversely, suppose that P is a partition of S and let R be defined by xRy iff x and y are in the same piece of the partition. Clearly, R is reflexive and symmetric. To see that R is transitive, suppose that xRy and yRz . We know that y ∈ A for some A ∈ P , therefore x ∈ A , z ∈ A and hence xRz. Finally, the equivalence classes of R correspond to the pieces of

P

because of the way

R

was defined.

2.3.7 Order relations Definition 69 A binary relation is called a quasi-order or preorder if it is reflexive and transitive.

Example 70

Let S be the set of all students in a particular university. Let R be the relation “is at least as tall as and as old as”. Then R is not complete, but is reflexive and transitive. Then it is a preorder.

Elementary Set Theory

28

Definition 71 A binary relation is called an order if it is, reflexive, transitive, and antisymmetric. It is called a strict order if it is irreflexive, transitive and antisymmetric. Exercise 72 Is the preorder of the previous example an order? Justify. Definition 73 An order that is also complete is called a complete or total order.∗ Notation To stress that some orders may not be complete, we sometimes call them partial orders. An order is often denoted by ”
” or ” ”. The notation

a < b indicates a strict order where a
b y  (>)x instead of x
( 3 2n n!

⇒ ≤ ⇒ ≤

2.5 Appendix: Binomial Coefficients

Binomial coefficients are used in many sciences for counting events and computing probabilities, and they constitute a useful mathematical tool in different areas in Economics.

Definition 85 The factorial of a natural number n, n!, is computed as n! = n (n −

1) (n − 2)....2 · 1. The factorial of 0 is defined to be 0! = 1.

Factorial numbers are useful for computing permutations of distinguishable elements. For example, suppose that you have to compute the number of possible

arrangements of 10 individuals in a line. For the first position you have ten possible choices. Once the first position is taken, you have 9 remaining options for the second place. So for the first two positions you have 10 · 9 = 90 possibilities. For the third position there are only 8 possible guys to pick and so on. So for a line of 10 guys we have 10 · 9 · 8...2 · 1 = 10! options. A very used property of permutations is that n! = n (n − 1)!. This is trivial to show by construction from the definition and is therefore left as an exercise. The intuition is the same as before. To compute the possible rearrangements of n elements we have n options for the first place, and for each of those n choices we have all the permutations of the remaining (n − 1) elements. The permutation of elements is the basis of binomial coefficients, also called

combinations .

D 86 The combination of n elements, taken in groups of k is denoted by
nefinition 
n n k

,

and is computed as

k

=

!

n−k )! k! .

(

The rationale for this definition is the following. Arrange the n elements in a line. We will say that we obtain the same combination of k elements whenever the same k elements occupy the first k positions in the line. This is very important, when we consider combinations, the ordering of the elements is immaterial. We already know that the total number of permutations of the n elements is n! . However, from all those n! arrangements, many of them correspond to the same combination, meaning that the same k objects occupy the first k positions. If each combination is counted N times by computing n!, we then know that the true number of combinations will be nN! . The remaining point is to find N . To do this, just take a given arrangement and compute all the permutations that give the same combination of k elements. We have that for a given position of the last (n k ) elements, we can rearrange the first

−

Appendix: Binomial Coefficients

33

k elements in k ! ways; and in a similar fashion, for a given position of the first k elements, we can rearrange the last (n − k ) elements in (n − k )! ways. Therefore the total number of rearrangements that
 give the same combination is N = (n − k )! k !.

n This leads to the final result that k

=

n! n−k)! k! .

(

Exercise 87 Computing the factorial of a number may be cumbersome if n is large

( try 1000! for example), but show that when computing combinatorial numbers this difficulty may be overcome by calculating


n k

if

k

= n (n − 1) ... (n −kk! + 2) (n − k + 1)

is not too large. Just to convince yourself, compute


1000
4

.

Example 88

Suppose you pick four cards from the top of a 52 card pile . What is the chance of your getting a poker of aces? The probability is computed as the number 
of favorable events divided by the total number of events. The latter is simply 524 =

The former is simply 1, since there is only one combination that comprises a poker of aces. Therefore the chance is (521 ) = 2701,725 ˜ 3.7 10−6 52! 48! 4! .

4

Our first use of binomial coefficients will be to provide a guess for the binomial formula, that we have proven by induction above. In general, this is a limitation of the proofs by induction. We have to come up with the result before the proof begins. To find a guess of the binomial formula we know that (a + b)n = (
a + b) (a + b) (a + b) ... (a + b) n times

by the distributive law of multiplication, we know that each term of the binomial sum is calculated as the multiplication of n elements, some of them being a and being the is therefore obvious that the binomial sum will have the form nrest b. It n k n −k (a + b) = k=0 αk a b . The only problem is to find the αk . But that is easy if

we use combinatorial numbers. We can think of the term ak bn−k as a list of n ordered elements, with k a’s and (n − k) b’s. For example if k = 1, then we know that there is only one a, either in position 1,2, 3,..., n

k

− 1 , n and the rest are b s. In general, for a fixed particular


all we have to do is to find in how many different ways can we rearrange the

a’s k

and b’s. This is the same as computing in how many different positions can we find

a’s

and (n

− k) b s, which is the
nnumber of combinations of the n available positions


taken in groups of

k.

So

αk

= k

, and the binomial formula is computed as n

(a + b) =

n    n

k k=0

ak bn−k

34

Elementary Set Theory

2.6 References S.R. Lay, Analysis with an Introduction to Proof . Chapter 2. Third Edition. Prentice Hall. A. Matozzi, Lecture Notes Econ 897 University of Pennsylvania Summer 2001. A. N. Mitchell and C. J. Herget, Applied Algebra and Functional Analysis. Dover Publications. H.L. Royden, Real Analysis . Third Edition. Prentice Hall.