Analysis with an introduction to proof 5th by steven lay ch02

Chapter 2
Sets and Functions

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Section 2.3, Slide 1

1-1


Section 2.3
Functions

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Section 2.3, Slide 2

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Definition 2.3.1

Let A and B be sets. A function from A to B is a nonempty relation f ⊆ A × B that

satisfies the following two conditions:
1. Existence: For all a in A, there exists a b in B such that (a, b) ∈ f.
2. Uniqueness: If (a, b) ∈ f

and (a, c) ∈

f , then b = c.

That is, given any element a in A, there is one and only one element b in B such that (a, b) ∈ f.
Set A is called the domain of f and is denoted by dom f.
Set B is referred to as the codomain of
We write f : A → B to indicate

f

f.

has domain A and codomain B.

The range of

f, denoted rng  f, is the set of all second elements of members of f . That is,

rng f = {b ∈ B : ∃ a ∈ A

(a, b) ∈ f }.

If (x, y) ∈ f , we often say that f maps x onto y and use the notation y = f (x).
Domain A

x

Codomain B

•

•

f (x)

rng  f
f:A→B

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Section 2.3, Slide 3

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*Similar to Practice 2.3.2 in the text.

Practice 2.3.2*
Let

A = {1, 2, 3} and B = {2, 4, 6, 8}. Which of the following relations are functions

from A to B?
B
A

R = {(1, 6), (2, 4)}

2
1
4
2

6

This is not a function. It violates the existence condition.
The number 3 in set A is not related to anything in set B.

3
8

B
R = {(1, 2), (2, 8), (3, 6), (1, 4)}

A
2
1
4
2
6

This is not a function. It violates the uniqueness condition.
The number 1 in set A is related to two numbers in set B.

3
8

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Section 2.3, Slide 4

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*Similar to Practice 2.3.2 in the text.

Practice 2.3.2*
Let

A = {1, 2, 3} and B = {2, 4, 6, 8}. Which of the following relations are functions

from A to B?
B
A

R = {(1, 6), (2, 8), (3, 2)}

2
1
4
2
6

This is a function from A to B.
It is OK that nothing in set A is related to number 4 in set B.

3
8

A
1
2
3

B

R = {(1, 4), (2, 6), (3, 4)}

2
This is a function from A to B.
4
6

It is OK that both 1 and 3 in set A are related to 4 in set B.

8
In both of these functions, there is at least one element in B that is not related to
anything in A.
When there are no extra elements in B, the function is given a special name.

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Section 2.3, Slide 5

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Properties of Functions
Definition 2.3.4
A function

f : A → B is called surjective (or is said to map A onto B) if B = rng

f.

A surjective function is also referred to as a surjection.
The question of whether or not a function is surjective depends on the choice of codomain.
A function can always be made surjective by restricting the codomain to being equal to
the range, but sometimes this is not convenient.

If it happens that no member of the codomain appears more than once as a second
element in one of the ordered pairs, then we have another important type of function.

Definition 2.3.5
A function

f : A → B is called injective (or one-to-one) if, for all a and a′ in A,

f (a) = f (a′) implies that a = a′. An injective function is also referred to as an injection.
If a function is both surjective and injective, then it is particularly well behaved.

Definition 2.3.6
A function

f : A → B is called bijective or a bijection if it is both surjective and injective.

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

Section 2.3, Slide 6

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Example 2.3.7
Consider the function given by the formula

f (x) = x2.

y
4

y = h (x)

If we take
y = f (x)

for both the domain and codomain

so that f :    → 

, then f is not surjective

y = g (x)

because there is no real number that maps onto –1.

2

If we limit the codomain to be the set [0, ∞), then the
function

–2

g:

→ [0, ∞) such that g(x) = x2 is surjective.

x

2

y = –1

Since

g(−1) = g(1), we see that g is not injective when defined on all of

But restricting

.

g to be defined on only [0, ∞), it becomes injective.

Thus h: [0, ∞) → [0, ∞) such that h(x) = x2 is bijective.

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Section 2.3, Slide 7

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Practice 2.3.12

Consider the four functions pictured below. For each function, the domain and codomain
are sets consisting of two or three points as indicated. Classify each function as being
surjective, injective, bijective, or none of these.

•
• • •

• • •

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

This function is neither.

•

•

• • •

This function is surjective.

•

•

•

• • •

•

•

This function is injective.

This function is bijective.

Section 2.3, Slide 8

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Functions Acting on Sets
When thinking of a function as transforming its domain into its range, we may wish to
Or we may wish to identify the
consider what happens to certain subsets of the domain.
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 2.3.13
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.

f
A

B

f

–1

f

(D)

f (A)

–1

D
f (C)

C

f

If D ⊆ B, we let f – 1(D) represent the subset {x ∈ A: f (x) ∈ D} of A.
called the pre-image of D

Note: At this time, f

–1

in A or “ inverse of D.
 f

The set

–
f   1(D) is

 ”

is only applied to sets, not points.

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Section 2.3, Slide 9

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Example 2.3.13

Let f :   → 

be given by

y

f (x) = x2. Then the following hold.

2
y=x

4

If C = [0, 2], then
1

f (C1)

f (C1) = [0, 4].

2

–2

2

C

x

1

–2

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Section 2.3, Slide 10

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Example 2.3.13

Let f :   → 
y

be given by

f (x) = x2. Then the following hold.

2
y=x

4

If C = [0, 2], then
1
If C = [−1, 2], then
2

f (C2)

f (C1) = [0, 4].

f (C2) = [0, 4].

2

–2

2

C

x

2

–2

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Section 2.3, Slide 11

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Example 2.3.13

Let f :   → 
y

be given by

f (x) = x2. Then the following hold.

2
y=x

4

If C = [0, 2], then

1
If C = [−1, 2], then
2

f (C3)

f (C1) = [0, 4].

f (C2) = [0, 4].

2
If C = [−2, −1] ∪ [0, 1], then
3

–2

2

C

f (C3) = [0, 4].

x

3
–2

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Section 2.3, Slide 12

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Example 2.3.13

Let f :   → 
y

be given by

f (x) = x2. Then the following hold.

2
y=x

4

f (C1) = [0, 4].

If C = [0, 2], then
1
If C = [−1, 2], then
2

D

f (C2) = [0, 4].

2

1

If C = [−2, −1] ∪ [0, 1], then
3

If D = [0, 4], then
1
–2

2

f

–1

f  –

 1

f (C3) = [0, 4].
(D ) = [−2, 2].
1

x

(D )
1

–2

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Section 2.3, Slide 13

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Example 2.3.13

Let f :   → 
y

be given by

f (x) = x2. Then the following hold.

2
y=x

4

f (C1) = [0, 4].

If C = [0, 2], then
1
If C = [−1, 2], then
2

D

f (C2) = [0, 4].

2

2

If C = [−2, −1] ∪ [0, 1], then
3

If D = [0, 4], then
1
–2

2

x
If D = [−1, 4], then
2

f

–1

f –

 1

f  –

 1

f (C3) = [0, 4].
(D ) = [−2, 2].
1

(D ) = [−2, 2].
2

(D )
2

–2

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Section 2.3, Slide 14

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Example 2.3.13

Let f :   → 

be given by

y

f (x) = x2. Then the following hold.

2
y=x

4

f (C1) = [0, 4].

If C = [0, 2], then
1
If C = [−1, 2], then
2

D

f (C2) = [0, 4].

2

3

If C = [−2, −1] ∪ [0, 1], then
3

If D = [0, 4], then
1
–2

2

x
If D = [−1, 4], then
2

f

–1

(D )
3

If D = [1, 4], then
3

f –

 1

f 

–

f –

 1

f (C3) = [0, 4].
(D ) = [−2, 2].
1

(D ) = [−2, 2].
2

 1
(D ) = [−2, −1] ∪ [1, 2].
3

–2

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Section 2.3, Slide 15

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Given a function

f : A → B, there are many relationships that hold between the images

and pre-images of subsets of A and B.

Several of these are included in the next theorem. The proofs are left to the practice
problems and the exercises.

Theorem 2.3.16
Suppose that

f : A → B. Let C, C , and C be subsets of A and let D, D , and D be
1

2
1
2

subsets of B. Then the following hold:

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– 1

(a)

C

(b)

f [ f  – 1(D)] ⊆ D

(c)

f (C ∩ C )

(d)

f (C ∪ C ) = f (C ) ∪ f (C )
1
2
1
2

(e)

 f  – 1(D1 ∩ D2) =  f  – 1(D1) ∩  f – 1(D2)

(f )

 f 

(g)

 f 

⊆ f 

1

– 1
– 1

[ f

 (C)]

⊆ f (C1) ∩

2

(D ∪ D ) =  f
1
2

(B \D) = A\ f 

– 1

– 1

(D ) ∪  f
1

– 1

f (C )
2

(D )
2

(D)

Section 2.3, Slide 16

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To see that equality may not hold in Theorem 2.3.16 (a) and (c), consider the
following situation:
Given

f : A → B, suppose C1 and C2 are nonempty subsets of A such that

C1 ∩ C2 = ∅ and f (C1) = f (C2).

A

B
f

C1

f (C1)

C2
f

We see that f 

–

f (C2)

 (C1)] = C1 ∪ C2, and this is larger than C1.

 1
[ f

This shows that equality may not hold in part (a): C

Also, since C1 ∩ C2 = ∅, we have f (C1 ∩ C2 ) = ∅.

⊆ f 

– 1

[ f

 (C)].

But f (C1) ∩ f (C2) = f (C1) ≠ ∅,

This shows that equality may not hold in part (c): f (C1 ∩ C2) ⊆ f (C1) ∩

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

f (C ).
2

Section 2.3, Slide 17

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While Theorem 2.3.16 states the strongest results that hold in general, if we apply
certain restrictions on the functions involved, then the containment symbols in
parts (a), (b), and (c) may be replaced by equality.

Theorem 2.3.18
Suppose that f : A → B. Let C, C1, and C2 be subsets of A and let D be a subset of B.
Then the following hold:

(a)

If f is injective, then f 

(b)

If f is surjective, then f [ f

(c)

If f is injective, then f (C

 

1 ∩ C2 )

–

 1

[ f (C)]

= C.

–

 1

=

(D)] = D.

f

Proof of (c): We only have to show that f (C1) ∩

 (C1) ∩

f

 (C2).

f (C ) ⊆ f (C ∩ C ), since the
2
1
2

converse inclusion is Theorem 2.3.16(c).

Then y ∈ f (C1) and

To this end, let y ∈ f (C ) ∩ f (C ).
1
2

y ∈  f (C2).

It follows that there exists a point x in C such that f (x ) = y.
1
1
1

Similarly, there exists a point x in C such that f (x ) = y.
2
2
2
Since

f is injective and f (x ) = y = f (x ), we must have x = x .
1
2
1
2

That is, x ∈ C  ∩ C .
1
1
2

Copyright © 2013, 2005, 2001 Pearson Education, Inc.

But then y = f (x1) ∈

f (C ∩ C ). ♦
1
2

Section 2.3, Slide 18

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Composition of Functions

then for any a ∈ A, f (a) ∈ B.

If f and g are functions with f : A → B and g : B → C,
But B is the domain of g, so g can be applied to

f (a). This yields g  f (a)), an element of C.
 (

B

A

g

f

C

g  f (a))
 (

f (a)

•

•

a

•

( g ○ f )(a)
 

g ○ f 

This establishes a correspondence between a in A and g ( f (a)) in C.
This correspondence is called the composition function of

by g

  (read “g of f ”).

It defines a function g

○f

( g ○ f )(a)

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f and g and is denoted
 

○f

: A → C given by

= g ( f (a)) for all a ∈ A.

 

Section 2.3, Slide 19

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Our next theorem tells us that composition of functions preserves the properties of
being surjective or injective. The proof is in the book and the exercises.

Theorem 2.3.20
Let f : A → B and g : B → C.

Then

(a) If f and g are surjective, then

g ○ f  is

(b) If f and g are injective, then

injective.

g ○ f  is

surjective.

(c) If f and g are bijective, then

g ○ f  is

bijective.

Inverse Functions
Given a function f : A → B, we have seen how

f determines a relationship between

That is, given D ⊆ B, we have the pre-image

subsets of B and subsets of A.

We would like to be able to extend this idea so that
to obtain a point in A.

That is, suppose

There are two things that can prevent
It may be that

f 

D = { y},

– 1

can be applied to a point in B

where y ∈ B.

– 1

(D) from being a point in A:

–
f   1(D) is empty,

and it may be that

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f 

–
f   1(D) in A.

f 

– 1

(D) contains several points instead of just one.

Section 2.3, Slide 20

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Practice 2.3.21

Given f : A → B and y ∈ B, under what conditions on

assert that there exists an x in A such that

Practice 2.3.22

f can we

Ans: f must be surjective.

f (x) = y?

Given f : A → B and y ∈ B, under what conditions on

f can we

Ans: f must be bijective.

f (x) = y?

assert that there exists a unique x in A such that

Given a bijection f : A → B, we see that each y in B corresponds to exactly one x in A,
the unique x such that f (x) = y.

This correspondence defines a function from B into A

called the inverse of f and denoted f  –  . Thus x = f  –
 1

A

 1

( y).

B
f 
x = f 

–1
 

y = f  ( x)

( y)

.  

•

•
f 

–1
 

Definition 2.3.23
Let f

: A → B be bijective. The inverse function of

f is the function given by
f  –   = {(
 1

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 y, x) ∈ B × A

: (x, y) ∈ f }.

Section 2.3, Slide 21

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If f : A → B is bijective, then it follows that

f 

– 1 

: B → A is also bijective.

–
–
Indeed, since dom f = A and rng f = B, we have dom f   = B and rng f    = A.
 1

Thus f

– 1

 1

is a mapping from B onto A.

Since f is a function, a given x in A can correspond to only one y in B.
This means that f 

–

 1

is injective, and hence bijective.

When f is followed by f 
That is, ( f 

–

 1

 ○

–

 1, the effect is to map x in A onto

f (x) in B and then back to x in A.

 f )(x) = x, for every x ∈ A.

A function defined on a set A that maps each element in A onto itself is called the
identity function on A, and is denoted by iA.
Furthermore, if f (x) =
Thus f ○

f 

– 1

= iB.

y, then x = f

So, f 
– 1

–

 1

 ○

 f  = iA.

–
f   1 )( y) = f ( f – 1 ( y)) = f (x) =

( y), so that ( f ○

y.

We summarize these results in the following theorem.

Theorem 2.3.24
Let f : A → B be bijective. Then
(a)

f 

– 1

 : B → A is bijective

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(b)

f 

– 1

○

A and f

f =i

○ f 

– 1

=i

B.

Section 2.3, Slide 22

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Our final theorem relates inverse functions and composition.

Theorem 2.3.28
Let f : A → B and
and ( g ○ f ) − 1 =

f 

– 1

○ g 

− 1

g : B → C be bijective. Then the composition g ○ f : A → C is bijective

.

Proof: We know from Theorem 2.3.20 that

g ○ f is bijective, so g ○ f has an inverse

denoted by ( g ○ f ) − 1, and this inverse maps C onto A.

g  f
○

f

g

A

B
f 

–1

C
–1
g 

( g  f )
○

 

–1

–1 –1
= f    g 
○

We are asked to verify the equality of the two functions (

 

− 1
– 1
− 1
 g ○ f ) 
and f 
○ g  ,

as sets of ordered pairs.

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Section 2.3, Slide 23

1-23


To show (

 g ○ f ) 

− 1

 

– 1
⊆ f 
○ g − 1, we suppose (c, a) ∈ ( g ○ f ) − 1.

By the definition of an inverse function, this means (a, c) ∈ g 

○ f .

The definition of composition implies that
∃ b ∈ B such that (a, b) ∈
Since f and

g are bijective, this means that

(b, a) ∈ f 

That is, f 

–

f and (b, c) ∈ g.

– 1

and (c, b) ∈ g 

But then,

– 1
 1
(b) = a and g  (c) = b.

( f 
so that (c, a) ∈ ( f 

– 1
.

– 1

○

–

 1

○ g 

(

− 1
– 1
− 1
– 1
)(c) = f 
g 
(c)) = f  (b) = a,

− 1

g 

− 1
) and ( g  ○  f ) − 1 ⊆ ( f – 1 ○ g  ).

The reverse set inclusion is Exercise 31. ♦

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Section 2.3, Slide 24

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