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 )]}.
Elementary Set Theory
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 (<)y.
and a = b . It is equivalent to write
An example of partial order is set inclusion; for each set X, the relation ⊆
is a partial order in the power set P(X ). The reason for qualifying “partial” is that
sometimes the order between two elements is not defined, i.e., there can be M, N ∈
P(X ) such that neither M ⊆ N nor N ⊆ M.
Definition 74
An ordered set is a set S in which an order is defined.
Again, the order can be either partial or complete. For the case of a complete
ordered set ,the following statement can be shown
Exercise 75 Let S be a complete ordered set. Then ∀x, y ∈ S, one and only one of
the statements x < y, x = y, y < x is true.
2.4 The Set of Natural Numbers. Induction
Lets assume the existence of a set N
consisting of the positive integers, also called
natural numbers. Thus
N = {1, 2, 3, 4...}
This set can be characterized from the following set of Axioms, called the
Peano axioms in honor of the Italian mathematician Giuseppe Peano, who developed
this approach in the late 19th century. We suppose that there exist a set P whose
elements are called natural numbers and a relation on P called successor with
the following properties:
P1.
There exists a natural number, denoted by 1, that is not the successor of any
other natural number.
Note that any complete relation is also reflexive. This makes our definition of complete ordering
somehow redundant, given that one requirement is implied by another. Nevertheless this way of
defining an order is customary (at least in economics) and we decided to stick with the widely
adopted definition.
∗
29
The Set of Natural Numbers. Induction
P2. Every natural number has a unique successor. If m ∈ P, then we let m
the successor of
m.
denote
P3. Every natural number except 1 is the successor of exactly one natural number.
P4. If M is a set of natural numbers such that
1. 1 ∈ M and
2. ∀ k ∈ P, if k ∈ M , then k ∈ M,then M = P.
Axioms P1 to P3 express the intuitive notion that 1 is the first natural number
and that we can progress through the natural numbers in succession one at a time.
Axiom P4 is the equivalent of what we will call later the principle of mathematical
induction.
Exercise 76 Consider the relation “successor” on N. What are its properties?
Using the previous axioms, we can define what addition and multiplication
means. Basically
D1. ∀n ∈ P, define n + 1 = n .
D2. Let n, m ∈ P. If m = k and n + k is defined, then define n + m = n + k
(n + k ) .
=
D3. ∀n ∈ P, define n · 1 = n.
D4. Let n, m ∈ P. If m = k and n · k is defined, then define n · m = n · k
= (n · k )+ n.
Note that in D2 and D4 the existence of k is assured by axiom P3.
Based on the definition of sum, we can now introduce a strict order “<” and
an order “≤” relations on N
Definition 77 Let N be the set of natural numbers. For x, y ∈ N
iff ∃ n ∈ N : x + n = y. We also say that x ≤ y
iff x < y
Exercise 78 Verify that “<” is indeed a strict order on N
[(
)
.
∨
we say that x < y
x = y )]
(
is true.
Elementary Set Theory
30
2.4.1 The Well Ordering Property and the Principle of Mathematical Induction
There is an additional property of N that expresses in a precise way the idea that
each nonempty subset of N must have a least (first) element.
Theorem 79 (well-ordering property of N ) If S
there exists an element
m∈S
such that
m≤k
is a nonempty subset of
k ∈ S.
for all
N, then
Proof. Lets consider two cases: if 1 ∈ S, and if 1 ∈/ S. In the first case,
we know from axiom P1 that 1 is not the successor of any element of N. Then for
every element k of S it must be 1 ≤ k. So m exists and is 1. If 1 ∈
/ S the proof goes
by contradiction. Suppose that there is no such least element m. We will construct
the set M = {n ∈ N : n ∈
/ S ∧ {1, 2, ..., n − 1, n} ∩ S = ∅} . This is the largest set of
consecutive natural numbers starting at 1 that do not intersect with S. Clearly 1 ∈ M
by assumption. Now if l ∈ M, then l = l + 1 must also belong to M. For if not then
l
∈S
and then it would be the least element of Therefore from axiom P4
implying that
S = ∅, contradicting
S.
S
the assumption that
M
= N,
is nonempty.
One important tool to be used when proving theorems about the natural
numbers comes from axiom P4 and is known as the principle of mathematical
induction.
It enables us to conclude that a given statement about natural numbers
is true for all natural numbers without having to verify it for each number one at a
time.
Theorem 80 (Principle of Mathematical Induction) Let P (n) be a statement that
is either true or false for each n ∈ N. Then P (n) is true for all n ∈ N provided that
1. P(1) is true, and
2. For each k ∈ N, if P (k) is true then P (k + 1) is also true.
Proof. The proof is trivial if we rely on axiom P4. Lets call S to the set
S = {k ∈ N : P (k )
∈ S and
S = N, implying
From the assumptions we know that a) 1
k + 1) ∈ S .
n.
(
From axiom P4 we have that
}
is true
b)
N, k ∈ S ⇒
P (n) holds for all
∀k ∈
that
Exercise 81 Take the well ordering property as an axiom and prove the Principle
of Mathematical Induction without relying on axiom P4. (Hint: the proof goes by
contradiction. Construct an auxiliary set S = {n ∈ N : P (n) is false} and use the
well ordering property to guarantee the existence of a least element
m ∈ S.)
It is customary to refer to the verification of part 1 of the theorem as the
basis
for induction and part 2 as the induction step.The assumption that P (k ) is true
in verifying part 2 is known as the induction hypothesis. It is essential that both
parts be verified to have a valid proof using mathematical induction.
The Set of Natural Numbers. Induction
31
Exercise 82 Prove the following theorem: Let
≥
that is either true or false for each n
m. Then
that
1. P m is true and
2. For each k
,if P k is true then P
( )
2.4.2
∈N
m ∈ N and let P (n) be a statement
()
P (n) is true for all n ≥ m provided
(k + 1) is true
Applications
An useful and important theorem is proven byinduction.
Write x! for x(x − 1)(x −
n
2) · · · 3 · 2 · 1 for
x ≥ 1, set 0! = 1 —, and write x for the ratio x nn−x for all x n.
n n Observe that k + k−1 = n+1
— for further details on combinatory numbers and
k
!
!(
their use, check the appendix of this chapter —.
)!
Theorem 83 The Binomial Theorem (a + b)n = nk=0 nk an−k bk .
Proof. The formula is trivially valid for n = 1. (Basis of induction) Suppose
:
the formula of the binomial theorem holds for n (inductive step). Show then that it
also holds for n + 1.
n+1
(a + b)
n
= (a + b)(a + b) = (a + b)
a
k
n
n
=
a
b +
a
k
k
n
n
=
b +
a
n
n
n
n−k+1 k
k=0
n
=
n−k+1 k
n
k =0
n+1
n −k k
b
k=0
by assumption
n−k k+1
b
an− +1b
chg of var l=k+1
k
l−1
l=1
k=0
n
n n+1 n
n
n− +1
0 a + l=1 l + l − 1 a b
l
l
l
l
+
n
n
bn+1
n + 1
n + 1
n n + 1 n− +1
n+1
n+1
=
0 a + l=1 l a b + n + 1 b
n+1 n + 1 (n+1)−k k
=
a
b
k
l
l
k=0
which is indeed the corresponding expression for n + 1.
Exercise 84 Prove or give a counterexample of the following statements
1. For everypositive integer n, n2 + 3n + 8 is an even number.
n
2. ∀n ∈ N, ni=1 i = (n+1)
2
3. ∀n ∈ N, ni=1 i2 = (2n+1)6(n+1) n
2 n2
4. ∀n ∈ N, ni=1 i3 = (n+1)
= (ni=1 i)2
4
5. Assuming x = 1 , 1 + x + x2 + ... + xn = 1−1x−nx+1 , ∀n ∈ N
Elementary Set Theory
32
6. ∀n ∈ N , n > 3 ⇒ n2 ≤ n !
7. ∀n ∈ N, n > 3
n2 2n
8.∀n ∈ N, n > 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 thennumber 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.