Analysis with an introduction to proof 5th by steven lay ch02a

Chapter 2
Sets and Functions

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

Section 2.2, Slide 1

1-1


Section 2.2
Relations

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

Section 2.2, Slide 2

1-2


In listing the elements of a set, the order is not important.

So, {1, 3} = {3, 1}.

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 is called the second.

The important property of ordered pairs is that
(a, b) = (c, d )

iff

a = c and b = d.

Ordered pairs can be defined using basic set theory in a clever way. Essentially, we
identify the two elements in the ordered pair and specify which one comes first.

Definition 2.2.1
The ordered pair (a, b) is the set whose members are {a, b} and {a}.
In symbols
we have (a, b) =

{{a}, {a, b}}, where we have written the singleton set first.

The acceptability of this definition depends on the ordered pairs actually having the
property expected of them.

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

This we prove in the following theorem.

Section 2.2, Slide 3

1-3


Definition:

Theorem 2.2.2:

Proof:

(a, b) =

{{a}, {a, b}} and (c, d) = {{c}, {c, d}}

(a, b) = (c, d) iff a = c and b = d.

If a = c and b = d, then
(a, b) =

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

Conversely, suppose that (a, b) = (c, d).
We wish to conclude that

Then we have
a = c and b = d.

Consider two cases: when a = b and when a ≠ b.

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

If a = b, then {a} = {a, b}, so

{{a}, {a, b}} = {{c}, {c, d}}.

Since (a, b) = (c, d), we then have

{{a}} = {{c}, {c, d}}.
The set on the left has only one member, {a}. Thus the set on the right can have only one member, so
{c} = {c, d} and c = d.
{a} = {c} and a = c.
On the other hand,

But then {{a}}

= {{c}}, so

Thus, a = b = c = d.
if a ≠ b …

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

Section 2.2, Slide 4

1-4


Definition:

Theorem 2.2.2:

(a, b) =

{{a}, {a, b}} and (c, d) = {{c}, {c, d}}

(a, b) = (c, d) iff a = c and b = d.

On the other hand, if a ≠ b,

then from the preceding argument it follows that c ≠ d.

Since (a, b) = (c, d), we must have
{a} ∈ {{c},

{c, d}},

which means that {a} = {c} or {a} = {c, d}.
In either case, we have

Again, since (a, b) = (c, d), we must have

a = c.

c ∈ {a}, so

{a, b} ∈ {{c},

{c, d}}.

Thus {a, b} = {c} or {a, b} = {c, d}. But {a, b} has two distinct members and {c} has only one, so we must have
{a, b} = {c, d}.
Now a = c, a ≠ b and b ∈ {c, d}, which implies that

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

b = d. ♦

Section 2.2, Slide 5

1-5


Definition 2.2.4

If A and B are sets, then the Cartesian product (or cross product) of A and B,
written A × B, is the set of all ordered pairs (a, b) such that a ∈ A and b ∈ B.
In symbols, A × B = {(a, b) : a ∈ A and b ∈ B}.

Example 2.2.5
If A and B are intervals of real numbers, then in the Cartesian coordinate system with
A on the horizontal axis and B on the vertical axis, A × B is represented by a rectangle.
For example, if

and B is the interval (2,
then
 4], A × B is the

A is the interval [1, 4)

rectangle shown below:

y

Note: The solid lines indicate the
left and top edges are included.

[

4
)

The dashed lines indicate the right

A×B

B

and bottom edges are not included.
2

[
1

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

A

)
4

x

Section 2.2, Slide 6

1-6

A relation between two objects a and b is a condition involving a and b that is either
true or false.

When it is true, we say a is related to b; otherwise, a is not related to b.

For example, “less than” is a relation between positive integers.
We have

1 < 3 is true,

2 < 7 is true,

5 < 4 is false.

When considering a relation between two objects, it is necessary to know which object
comes For
first.instance, 1 < 3 is true but 3 < 1 is false.

So it is natural for the formal

definition of a relation to depend on the concept of an ordered pair.

Definition 2.2.7

Let A and B be sets. A relation between

A and B is any subset R of A × B.

We say that an element a in A is related by R to an element b in B if (a, b) ∈ R,
and we often denote this by writing “a R b.”

The first set A is referred to as the domain of the relation and denoted by dom R.
If B = A, then we speak of a relation R ⊆ A × A being a relation on A.

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

Section 2.2, Slide 7

1-7


Example 2.2.8

Returning to Example 2.2.5 where A = [1, 4) and B = (2, 4], the relation a R b given by
“a < b” is graphed as the portion of A × B that lies to the left of the line x = y.

y

[

4

x=y

A×B

)

R

B

2

[

A

)

1

4

x=1

x

x=3

For example, we can see that 1 in A is related to all b in (2, 4].
But 3 in A is related only to those b that are in (3, 4].

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

Section 2.2, Slide 8

1-8

Certain relations are singled out because they possess the properties naturally associated
with the idea of equality.

Definition 2.2.9
A relation R on a set S is an equivalence relation if it has the following properties
for all x, y, z in S:
(a) x R x

(reflexive property)

(b) If x R y, then y R x.

(symmetric property)

(c) If x R y and y R z, then x R z.

(transitive property)

Example 2.2.10

Determine which properties apply to each relation.

(a) Define a relation R on

by x R y if x ≤ y.

It is reflexive and transitive, but not symmetric.
(b) Let S be the set of all lines in the plane and let R be the relation “is parallel to.”
It is reflexive (if we agree that a line is parallel to itself), symmetric, and transitive.
(c) Let S be the set of all people who live in Chicago, and suppose that two people x and y

are related by R if x lives within a mile of y.
It is reflexive and symmetric, but not transitive.

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

Section 2.2, Slide 9

1-9


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, we define the equivalence class (with respect to R

  ) of x ∈ S to be the set

E

x

={

y∈S

 : y R x}.

Since R is reflexive, each element of S is in some equivalence class.
Furthermore, two different equivalence classes must be disjoint.

That is, if two

equivalence classes overlap, they must be equal.

Ex
•

•

x

y

and, by symmetry, x R w.
This implies x′ ∈ Ey and Ex ⊆ Ey.

x

y

Ey

x
•

We claim that E = E .

To see this, suppose that w ∈ E ∩ E .

w

•

y

x′

For any x′ ∈ E we have x′ R x.
x
Also, w ∈ E , so w R y.

y

But w ∈ E , so w R x
x
Using transitivity twice, we have x′ R y.

The reverse inclusion follows in a similar manner.

Thus we see that an equivalence relation R on a set S breaks S into disjoint pieces in a natural way.
These pieces are an example of a partition of the set.

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

Section 2.2, Slide 10

1-10


Definition 2.2.12
A partition of a set S is a collection P

of nonempty subsets of S such that

(a)

Each x ∈ S belongs to some subset A ∈ P .

(b)

For all A, B ∈ P , if A ≠ B, then A ∩ B = ∅.

A member of P is called a piece of the partition.
Example 2.2.13
Let S = {1, 2, 3}.

Then the collection P =

The the collection P = {{1, 2}, {3}}
S

is also a partition of S.
S

1

2

{{1}, {2}, {3}} is a partition of S.

3

P = {{1}, {2}, {3}}
The collection P = {{1, 2}, {2, 3}}

1
2

3

P = {{1, 2}, {3}}

is not a partition of S because {1, 2} and {2, 3}

are not disjoint.

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

Section 2.2, Slide 11

1-11


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.

Theorem 2.2.17
Let R be an equivalence relation on a set S. Then { x : x ∈ S} is a partition of S.
E

The relation “belongs to the same piece as” is the same as R.

Conversely, if P

is a partition of S, let P be defined by x P y iff x and y are in the same

piece of the partition. Then P is an equivalence relation and the corresponding partition
into equivalence classes is the same as P .

Proof: Let R be an equivalence relation on S. We have already shown that {Ex : x ∈ S}

is a partition ofNow
S. suppose that P is the relation “belongs to the same piece
Then
(equivalence class) as.”
xPy

iff

x, y ∈ E

z for some z ∈ S.

iff

x R z and

y R z for some z ∈ S.

iff

x R z and

z R y for some z ∈ S.

iff

xRy

Thus, P and R are the same relation.

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

Section 2.2, Slide 12

1-12


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.

Theorem 2.2.17
Let R be an equivalence relation on a set S. Then {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 P be defined by x P y iff x and y are in the same

piece of the partition. Then P is an equivalence relation and the corresponding partition
into equivalence classes is the same as P .

Proof:
Conversely, suppose that P is a partition of S and let P be defined by x
are in the same piece of the partition.

To see that P is transitive, suppose that x P y and y P z.

 P y

iff x and y

Clearly, P is reflexive and symmetric.
Then y ∈ E ∩ E .

x

z

But this implies that E = E by the contrapositive of 2.2.12(b), so x P z.

x

z

Finally, the equivalence classes of P correspond to the pieces of P because of the way
P was defined. ♦

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

Section 2.2, Slide 13

1-13