Chapter 4

FUNCTIONS

Until now we have learnt how to prove statements, we have introduced the basics of

set theory and used those concepts to present and characterize the main features of

the sets of interest in this course: N ,Z,Q and R. In this chapter we will refine the

notion of a relation to define a function, and use functions to study the cardinality

of sets.

4.1 Functions

Definition 128 Let A and B be sets. A function between A and B is a nonempty

relation f AB such that if (a, b) f and (a, b) f , then b = b. The domainof f, denoted by dom f , is the set of all first elements of members of f. The range of

f, denoted by ran f , is the set of all second elements of members of f. Symbolically

dom f = {a A : b B (a, b) f}

ran f = {b B : a A (a, b) f}

The set B is referred to as the codomain of f. If it happens that the domain of f

is equal to all A, then we say that f is a function from A into B and we write

f : A B.

If in the previous definition we allow the binary relation f to be such that(a, b) f , (a, b) f and b = b , then we are talking of a different object, calleda correspondence. When a function consists of just a few ordered pairs, it can beidentified by listing them. But usually they are too many to list, so the function isdefined by specifying the domain and giving a rule for determining the second elementin the ordered pair that corresponds to a particular first element. For example

f ={(x, y) : y = x2 and x R

}

The domain of the function would be obtained either from the context or by

stating it explicitly. In this part of the course we deal mainly with functions from R

into R. Therefore,unless told otherwise, when a function is given by a formula, the

domain is taken to be the largest subset of R for which the formula will always yield

a real number.

48 Functions

4.1.1 Properties of Functions

Definition 129 A function f : A B is called surjective (or is said to map Aonto B) if B = ran f. A surjective function is also referred to as a surjection.

The question on whether or not a function is surjective depends on the choice ofthe codomain. A function can always be made surjective by restricting the codomainto being equal to the range, but sometimes this is not convenient. For a function fgiven by the formula f (n) =

n, there is no simple description for the range. It is

just easier to write f : N R and not be more precise.If it happens that no member of the codomain appears more than once as a

second element of the members of f, then we have another important type of function

Definition 130 A function f : A B is called injective (or one-to-one ) if, forall a and a in A, f (a) = f (a) implies that a = a. An injective function is alsoreferred to as an injection.

If a function is both surjective and injective, then it is particularly well be-haved.

Definition 131 A function f : A B is called a bijection or a bijective function

if it is both surjective and injective.

Example 132 The relation f = {(x, y) R2 : y = x2} is a function but it is neithersurjective nor injective. If we define it as f : R+ R, it becomes injective. If we

define it as f : R+ R+, it becomes bijective.

Exercise 133 Consider the following function: Let A be a nonempty set and let S be

a subset of A. We may define a function S: A {0, 1} by

S(a) =

{1, if a S0, if a / S

This function is called the characteristic function (or indicator function) andis widely used in probability and statistics. Under what conditions is

S: a) injective,

b) surjective and c) bijective?

Exercise 134 Prove that a function is one to one if and only if x1 = x2 = f(x1) =f(x2).

Functions 49

4.1.2 Functions acting on sets

When thinking of a function as a transforming its domain into its range, we may

wish to consider what happens to certain subsets of the domain. Or we may wish to

identify the set of all points in the domain that are mapped into a particular subset

of the range. To do this we use the following notation:

Notation Suppose that f : A B. If C A, we let f (C) represent the subset{f (x) : x C} of B. The set f (C) is called the image of C in B. If D B,we let f1 (D) represent the subset {x A : f (x) D} of A. The set f1 (D)is called the pre-image of D in A or f inverse of D.

Remark 135 The symbol f1 is not to be thought of as an inverse function applied

to points in the range of f. In particular, given a point y in B it makes no sense

to talk about f1 (y) as a point in A. We shall see later how this idea can be mademeaningful in some cases, but for now we can apply f1 only to a subset of B, and

by doing so we obtain a subset of A.

Exercise 136 Suppose that f : A B .Let C, C1,and C2 be subsets of A and let D,

D1, and D2 be subsets of B. Then show that the following hold:

(a)C f1 [f (C)](b) f [f1 (D)] D(c) f (C1 C2) f (C1) f (C2)(d) f (C1 C2) = f (C1) f (C2)(e) f1 (D1 D2) = f

1 (D1) f1 (D2)

(f) f1 (D1 D2) = f1 (D1) f1 (D2)(g) f1 (B D) = A f1 (D)

Exercise 137 Suppose that f : A B. Let C, C1and C2 be subsets of A and let Dbe a subset of B. Prove that the following hold:

(a) If f is injective then f1 [f (C)] = C(b) If f is surjective, then f [f1 (D)] = D(c) If f is injective, then f (C1 C2) = f (C1) f (C2)

4.1.3 Composition of Functions

Given two functions f : A B and g : B C, then for any a A, f (a) B. ButB is in the domain of g, so g can be applied to f (a). This yields g(f (a)) an elementof C. This correspondence is called the composition function of g and f and isdenoted g f ( read g of f ). Thus

(g f) (a) = g (f (a))

In terms of ordered pairs we have

g f = {(a, c) A C : b B (a, b) f and (b, c) g}

50 Functions

Notice that g f is in general different from f g. In fact, f g may be mean-

ingless unless ran g domf. However, the composition of functions is associative

and it also preserves the properties of being surjective or injective.

Example 138 Let g(x) = ex+1 and f(x) = x 3 be functions defined on the realline. It follows that g(f (x)) = ef(x)+1 = ex2 and f(g(x)) = g(x) 3 = ex+1 3.

Exercise 139 Let f : A B, g : B C, and h : C D.Prove that h (g f) =(h g) f.

Theorem 140 Let f : A B and g : B C. Then(a) If f and g are surjective, then g f is surjective.(b) If f and g are injective, then g f is injective.(c) If f and g are bijective then g f is bijective.Proof. (a) Since g is surjective, then ran g = C. That is, for every c C ,

there exists b B such that (b, c) g. Now, since f is also surjective, ran f =B, andfor such b there exists a A such that (a, b) f. But then for every c C , thereexists a A such that (a, c) (g f ) , so g f is surjective.

(b) is left as an exercise(c) Results trivially from (a) and (b)

4.1.4 Inverse Functions

Given a function f : A B, we have seen how f determines a relationship betweensubsets of B and subsets of A. That is, given D B, we have the pre-image f1 (D)in A. We would like to extend this idea so that f1 can be applied to a point in B toobtain a point in A. That is, suppose D = {y} , where y B. There are two thingsthat can prevent f1 (D) from being a point in A : It may be that f1 (D) is empty,and it may be that f1 (D) contains several points instead of just one.

Exercise 141 Given f : A B and y B, under what conditions on f can weassert that there exists an x in A such that f(x) = y ?

Exercise 142 Given f : A B and y B, under what conditions on f can weassert that there exists a unique x in A such that f (x) = y ?

Given a bijection f : A B, we see that each y in B corresponds to exactlyone x in A, the unique x such that f (x) = y. This correspondence defines a functionfrom B into A called the inverse of f and denoted f1.

Definition 143 Let f : A B be bijective. The inverse function of f is the

function f1 given by

f1 = {(y, x) B A : (x, y) f}

Functions 51

If f : A B is bijective, then it follows that f1 : B A is also bijective.Indeed, since dom f = A and ran f = B , we have that dom f1 = B and ran f1 =A. Thus f1 is a mapping from B onto A. Since f is a function, a given x in A cancorrespond to only one y in B. This means that f1 is injective, and hence bijective.

When f is followed by f1, the effect is to map x in A onto f (x) in B andthen back to x in A. That is, (f1 f) (x) = x, for every x A. A function definedon a set A that maps each element in A onto itself is called the identity function onA, and is denoted by iA.Thus we can say that f

1f = iA. Furthermore, if f (x) = y,

then x = f1 (y) , so that

(f f1 (y)

)= f

(f1 (y)

)= f (x) = y

Thus f f1 = iB . Let us summarize these results in the following theorem

Theorem 144 Let f : A B be bijective.Then(a) f1 : B A is bijective,(b) f1 f = iA and f f1 = iB.

Exercise 145 Given f : A B, and the identity functions iA : A A andiB : B B. Show that

(a) iB f = f(b) f iA = f

The definition provided above for the inverse function requires the function f

to be bijective and proves as a theorem that f1 f = iA and f f1 = iB. It is

possible however, to define the inverse function using this property and finding that

being a bijection is a necessary and sufficient condition for invertibility as a theorem

Definition 146 A function f : A B is invertible if there exists a function

f1 : B A such that f1 f = iA and f f1= iB. The function f

1 is called the

inverse of f

Theorem 147 A function f : A B is invertible if and only if it is one to one andonto. (i.e. bijective)

Proof. Sufficiency follows by contradiction. Suppose f is invertible but notbijective. Then either f is not one to one or not onto. Invertibility implies that thereexists a function f1 : B A such that f1 f = iA and f f1 = iB .

If f is not one to one, there must be x1, x2 A,x1 = x2 such that f(x1) =f(x2) = y. But then f1 (y) = f1 (f (x1)) = x1 and f1 (y) = f1 (f (x2)) = x2,which contradicts the fact that f1 is a function.

If f is not onto, then there must be a b B such that b / ran f. By assumptionon f1, f1 (b) exists in A and is a = f1 (b) . But then f (a) ran f, implyingf f1 (b) = iB (b) which contradicts the assumption that f is invertible.

52 Functions

Necessity follows by construction. Suppose f : A B is bijective. By as-sumption, for every a A there exists a b B, such that (a, b) f. Lets construct acandidate relation for the inverse as

g = {(b, a) B A : (a, b) f}

To prove that g = f1, we have to show that a) g is a function, b) it is such thatg : B A and c) g f = iA and f g = iB.

a) Since f is a function, then f is a nonempty relation, then g must be alsononempty. Now, assume that there exist (b, a) and (b, a) g, with a = a. If thisis true, then (a, b) f and (a, b) f , contradicting the assumption that f is one toone. Then g is a function.

b) Suppose that b B : a A (b, a) g. This is equivalent to sayingthat b B : a A (a, b) f, contradicting the assumption that f is onto. Theng : B A.

c) Assume g f = iA. By definition of composition we have

g f = {(a, a) AA : b B (a, b) f and (b, a) g}

Since g f = iA, there must be a pair (a, a) g f with a = a. But this implies

that for some b B, (a, b) f and (a, b) f, contradicting the assumption that f isinjective. By the same argument, if f g = iB, then there must be a pair (b, b) f gwith b = b. This implies that for some a A, (b, a) g and (a, b) f. But if(b, a) g then by construction (a, b) f. This and the fact that (a, b) f,with b = b,contradicts the assumption that f is a function.

There is one last result relating inverse functions and composition. It will beof particular interest when studying infinite sets.

Theorem 148 Let f : A B and g : B C be bijective. Then the compositiong f : A C is bijective and (g f)1 = f1 g1.

Proof. We already know from the properties of composition that g f isbijective. Thus g f is invertible. Now

g f = {(a, c) : b B (a, b) f and (b, c) g} ,

so that

(g f )1 = {(c, a) : b B (a, b) f and (b, c) g}

={(c, a) : b B (b, a) f1 and (c, b) g1

}

= f1 g1

Exercise 149 Instead of the set approach, use the associative property of the com-

position of functions to prove this last theorem.

Cardinality 53

4.2 Cardinality

The concept of cardinality emerges from the desire to compare the relative sizes of

sets. We shall begin by defining what it means for two sets to be the same size and

then approach the question of comparing size. Certainly, it is reasonable to say that

two sets S and T are the same size if there is a bijective function f : S T, for this

function will set up a one to one correspondence between the elements of each set.

Definition 150 Two sets S and T are called equinumerous, and we write S T,

if there exists a bijective function f : S T .

Observe that this is just another example of an equivalence relation. In fact,

satisfies reflexivity, symmetry, and transitivity.

Exercise 151 If F is a family of sets, then the concept of being equinumerous is a

relation on F. Show that is an equivalence relation.

Since is an equivalence class, it partitions any family of sets into disjoint

equivalence classes. With each equivalence class we associate a cardinal number, that

we think of as giving the size of the set.

Definition 152 A set S is called finite if S = or if there is n N and a bijection

f : {1, 2, ..., n} S.A set which is not finite is called infinite.

It will be convenient to abbreviate the set {1, 2, ..., n} by In. Thus we can say

that S is finite iff S = or if S is equinumerous with In for some n N.

Definition 153 The cardinal number of In is n, and if InS, we say that S has

n elements. The cardinal number of is taken to be 0. If a cardinal number is not

finite, it is called transfinite.

Definition 154 A set S is denumerable if there exists a bijection f : N S. If

a set is finite or denumerable, it is called countable. If a set is not countable, it is

uncountable. The cardinal number of a denumerable set is denoted by 0.

Remark 155 Note that in Rudin, a countable set is the same as a denumerable set.

Sets that we call countable in these notes, Rudin calls at most countable.

This symbol is read aleph nought. Aleph is the first letter of the Hebrew alphabet.

54 Functions

4.2.1 Countable Sets

Two examples of countable sets are the set of natural numbers N and the set of even

natural numbers E. It should seem at first glance that N should be bigger than E.Indeed, E is a proper subset of N and, in fact, it contains only half of N. But whatis half of 0? Actually, both sets are equinumerous. Just consider the bijectionf : N E, given by f (n) = 2n. So E also has cardinality 0.

Exercise 156 Find a bijection f : N Z, thereby showing that the set Z of all

integers is also denumerable.

A sequence is a function f : N B,B R. The values of f , denoted by thesymbol {xn}, are called the terms of the sequence. The symbol {xn} is a short cutfor x1, x2, x3, . . . xn.

If a nonempty set S is finite, then there exist n N and a bijection f :In S. Using the function f, we can count off the members of S as follows:f (1) , f (2) , f (3) , ..., f (n) . Letting f (k) = sk for 1 k n, we obtain the morefamiliar notation S = {s1, s2, ..., sn} . The same kind of counting is possible for a de-numerable set, and this is why both kinds of sets are called countable. For example,

if T is denumerable, then there exists a bijection g : N T , and we may writeT = {g (1) , g (2) , g (3) , ...} or T = {t1, t2, t3, ...} ,where g (n) = tn.

Theorem 157 Let S be a countable set and let T S. Then T is countable.

Proof. If T is finite, we are done. Thus lets assume that T S is an infiniteset. Since T is infinite, so must be S, so S is denumerable (since it is countable).Therefore there exists a bijection f : N S and we can write S as a list of distinctmembers S = {s1, s2, s3, ...} , where f (n) = sn.Now let A = {n N : sn T} . SinceA is a nonempty subset of N, it has a least member, say a1. Similarly, the set A{a1}has a least member, say a2. In general, having chosen a1, a2, ..., ak, let ak+1 be the least

member of A{a1, a2, ..., ak} . Essentially, if we select from our listing of S those termsthat are in T and keep them in the same order, then an is the subscript of the nth

term in this new list.

Now define a function g : N N by g (n) = an. Since T is infinite, g isdefined for every n N . Since a

n+1 / {a1, ...., an} , g must be injective. Thus thecomposition f g is also injective. Since each element of T is somewhere in the listingof S, g (N) includes all the subscripts of terms in T. Thus f g is a bijection from Nonto T and T is also denumerable.

Using the previous theorem, we can derive two useful criteria for determiningwhen a set is countable.

Theorem 158 Let S be a nonempty set. The following three conditions are equiva-lent:

(a) S is countable

Cardinality 55

(b) There exists an injection f : S N.(c) There exists a surjection g : N SProof. Suppose that S is countable. Then there exists a bijection h : J S,

where J = In for some n N if S is finite and J = N if S is denumerable. In eithercase, h1 is a bijection from S onto J and hence an injection (at least) from S to N.Thus (a) implies (b).

Now suppose that there exists an injection f : S N. Then f is a bijectionfrom S to f(S), so f1 is a bijection from f(S) back to S. We use f1 to obtaina function g from all N onto S as follows: Let p be any fixed number of S. Defineg : N S by

g (n) =

{f1 (n) if n f (S)p if n / f (S)

Then g [f (S)] = f1 [f (S)] = S and g [N f (S)] = {p} , so that g is a surjectionfrom N onto S. Thus (b) implies (c).

Finally, suppose that there exists a surjection g : N S. Define h : S N by

h (s) is the smallest n N such that g (n) = s

Then h is an injection from S to N, and hence a bijection from S onto the subseth (S) of N. Since N is countable, the previous theorem implies that h (S) is countable.Since S and h (S) are equinumerous, S is also countable.

So far we have shown that the set of integers is countable. Now we willestablish a couple of lemmas to show a surprising result. The set of rational numbers

is also countable.

Lemma 159 Let S and T be nonempty countable sets. The union S T is also

countable.

Proof. Since S and T are countable, there exist surjections f : N S and

g : N T. Define h : N S T by

h (n) =

{f(n+1

2

)if n is odd

g(n

2

)if n is even

Then h is surjective, so S T is countable.

Recalling that every natural number can be written as a product of primes,

and that this representation is unique except for the order of the factors, we can show

that the Cartesian product of two countable sets is also countable.

Lemma 160 Let S and T be nonempty countable sets. The Cartesian product ST

is also countable.

56 Functions

Proof. Since S and T are countable, there exist injections f : S N andg : T N. Define h : S T N by

h (s, t) = 2f(s) 3g(t), where s S and t T.

Then h is injective, for if h (s, t) = h (u, v) , then

2f(s) 3g(t) = 2f(u) 3g(v).

Since the prime factored form of a number is unique, we have f (s) = f (u) andg (t) = g (v) . Finally, since both f and g are one to one, this implies that s = u andt = v. Thus h is also injective and then S T is also countable.

Exercise 161 Let S and T be nonempty countable sets. Show that the Cartesian

product ST is also countable by finding an explicit surjective function h : NST.

(Hint: Look at theorem 167)

Theorem 162 The set of rational numbers is countable.

Proof. Let Q+ and Q be the set of positive rationals and negative rationals

respectively. We first show that Q+ is countable. Any member of Q+ can be written

uniquely as m/n, where m,n N and m and n having no common factors except form = n = 1 . Define f : Q+ N by

f (m,n) = 2m 3n.

Then f is injective as in the previous lemma and so Q+ is countable. The mapping

g : Q+ Q given by g (r) = r is clearly bijective, so Q+ and Q are equinumerous.That is, Q is countable. Since Q = Q+ Q {0} , by applying the first lemmatwice we obtain that Q is countable.

By generalizing the first lemma, we can show that a countable family of count-

able sets is also countable. But to do this we need first some notation to handle

families of infinitely many sets

Definition 163 If for each j in a nonempty set J there corresponds a set Aj, then

A = {Aj : j J}

is called an indexed family of sets with J as the index set. The union of all sets in

A is given by{Aj : j J} = {x : x Aj for some j J} .

The intersection of all sets in A is given by

{Aj : j J} = {x : x Aj for all j J} .

Cardinality 57

Notation Other notations for{Aj : j J} include

jJ

Aj or

A

If J = {1, 2, ..., n} , we may write

n

j=1

Aj or

nj=1

Aj

and if J = N , the common notation is

j=1

Aj or

j=1Aj

Similar variations of the notation apply to intersections.

There are some situations where a family of sets has not been indexed but we

still want to take the union or the intersection of all the sets. If B is a nonempty

collection of sets, then we let

BB

B = {x : B B x B}

and

BB

B = {x : B B , x B }

Exercise 164 FindBB

B andBB

B for each collection B.(a) B =

{[1, 1 +

1

n

]: n N

}

(b) B ={(1, 1 +

1

n

): n N

}

(c) B ={(

1

n,1

n

): n N

}

(d) B = {[2, x] : x R and x > 2}(e) B = {[0, 3] , (1, 5) , [2, 4)}

Exercise 165 Let {Aj : j J} be an indexed family of sets, and let B be a set. Provethe following identities

(a) B

[jJ

Aj

]=jJ

(B Aj)

(b)B

[jJ

Aj

]=jJ

(B Aj)

(c)B

[jJ

Aj

]=jJ

(B Aj)

58 Functions

The De Morgans Laws which we saw in chapter 2 are valid for any number

of collection of sets.

Theorem 166 Let A ={Aj : j J} be a (finite or infinite) collection of indexed setsAj. Then

(jJAj

)C=

jJACj

(jJAj

)C=

jJACj

Proof. We will prove only the first equality and leave the proof of the second

as a useful exercise. For convenience let B =(

jJAj

)Cand C =

jJACj . If

x B then x / BC which means that x / Aj for any j J. This means thatx ACj for all j J. Hence x C. This proves that B C. In order to show thatB = C, we need to prove that C B as well. So take x C. Hence x ACj for

all j J = x / Aj for any j J. Thus x /

jJAj = x

(jJ

Aj

)C. This

completes the proof.

Now we have all we need to prove the desired theorem

Theorem 167 Let A ={Aj : j J} be a countable family of countable sets. Thenthe union S =

jJAj is also a countable set.

Proof. Since empty sets contribute nothing to the union, we can assume

without loss of generality that all the sets are nonempty. Since the family is countable,

we can replace the index set J by N and consider A ={An: n N}. If the original

family had only a finite number of sets, A1, ..., Ak, let An = A1 for every n > k.

Now for each set An there is a surjection fn : N An, so we can write An =

{an1

, an2

, an3

, ....} , where fn(j) = a

nj. We now arrange the elements of

n=1An

in

the following way:

A1 : a11 a12 a13 a14 . . .

A2 : a21 a22 a23 a24 . . .

A3 : a31 a32 a33 a34 . . .

A4 : a41 a42 a43 a44 . . .

...

Hint: Remember that (AC)C = A.

Cardinality 59

By moving along each diagonal of the array in the manner indicated, we obtain

a listing of all the elements in

n=1An

:

a11, a12, a21, a31, a22, a13, a14, ....

This listing defines a surjection f : N

n=1An, so the union is countable.

Exercise 168 Use the previous theorem, to show that the set Q is countable.

Now that we know that Q is countable, we may wonder if R is countable too.

This is not so by the following theorem

Theorem 169 The set of all real numbers R is uncountable.

Proof. Since any subset of a countable set is countable, it suffices to show that

the interval J = (0, 1) is uncountable. If J were countable, we could list its membersand have

J = {x1, x2, x3, ...} = {xn : n N}

We shall show that leads to a contradiction by constructing a real number that is in J

but is not included in list of xns. Each element of J has a infinite decimal expansion,

so we can write

x1 = 0.a11a12a13...,

x2 = 0.a21a22a23...,

x3 = 0.a31a32a33...,

...

where each aij {0, 1,2, ..., 9} . We now construct a real number y = 0.b1b2b3... bydefining

bn=

{2 if ann = 2

3 if ann = 2

Since each digit in the decimal expansion of y is either 2 or 3, y J. But y is not one

of the numbers xnsince it differs from x

nin the nth decimal place. This contradicts

the assumption that J is countable, so J must be uncountable.

Exercise 170 Suppose you are teaching this theorem and a student asks this ques-

tion: We have shown previously that the set of rational numbers is countable. Using

this proof we have shown that the reals are not countable. Why cant I use this same

proof to prove that the rationals constitute a countable set? Where does the proof fail

to be correct?

60 Functions

We sometimes denote the cardinal number of R as . Since Q is countable

and R is not, we say that > 0 ( they are unequal transfinite cardinals) meaningthat you cannot find a 1-1 correspondence between R and Q.

Exercise 171 Let A={{sn} : the elements of s

nare 0s and 1s} . Is A countable?

Prove it.

Exercise 172 Show that the following pairs of sets are equinumerous(a) S = [0, 1] and T = [1, 3](b) S = [0,1] and T = [0, 1)(c) S = [0,1) and T = (0, 1)(d) S = (0,1) and T = (0,)(e) S = (0, 1) and T = R

4.3 References

S.R. Lay, Analysis with an Introduction to Proof .Chapter 2. Third Edition. Pren-

tice Hall.

A. Matozzi, Lecture Notes Econ 897 University of Pennsylvania Summer

2001.

H.L. Royden, Real Analysis . Third Edition. Prentice Hall.

Some authors prefer to use c ( for continuum ) instead of .