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

So it is natural for the formal definition of a relation to depend on the concept of an ordered pair.. We say that an element a in A is related by R to an element b in B if a, b  R, an

Chapter 2

Sets and Functions

Section 2.2 Relations

In symbols

we have (a, b) = {{a}, {a, b}}, where we have written the singleton set first

Essentially, we

identify the two elements in the ordered pair and specify which one comes first

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

Definition 2.2.1

The acceptability of this definition depends on the ordered pairs actually having the

property expected of them This we prove in the following theorem

Thus the set on the right can have only

one member, so

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

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

Conversely, suppose that (a, b) = (c, d) Then we have {{a}, {a, b}} = {{c}, {c, d}}

We wish to conclude that a = c and b = d Consider two cases: when a = b and when a 

b.

But then {{a}} = {{c}}, so

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

{{a}} = {{c}, {c, d}}

The set on the left has only one member, {a}.

{c} = {c, d} and c = d.

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

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

On the other hand, if a  b 

…

But {a, b} has two distinct members and

{c} has only one, so we must have

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 c  {a}, so a = c Again, since (a, b) = (c, d), we must have

{a, b}  {{c}, {c, d}}

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

{a, b} = {c, d}.

Now a = c, a  b and b  {c, d}, which implies

that

b = d 

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

then A  B is the

rectangle shown below:

2 4

x y

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 A is the interval [1,4)

[ )

A  B

A

B

and B is the interval (2,4],

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

The dashed lines indicate the right and bottom edges are not included

So it is natural for the formal definition of a relation to depend on the concept of an ordered pair

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 first For instance, 1 < 3 is true but 3 < 1 is false

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 “aR  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

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.

2 4

x

y

[ )

A

B

A  B x = y

R

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]

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

(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

Determine which properties apply to each relation

That is, if two

equivalence classes overlap, they must be equal

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

x







x

y

w

E y

E x

To see this, suppose that w  E x  E y

We claim that E x = E y For any x  E x we have xR  x But w  E x , so w  R  x

and, by symmetry, x  R  w Also, w  E y , so w  R  y Using transitivity twice, we have x R y

This implies x  E y and E x  E 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

Definition 2.2.12

(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   = {{1},{2},{3}} is a partition of S

1

S

The the collection P   = {{1,2},{3}} is also a partition of S

P   = {{1},{2},{3}} P   = {{1,2},{3}}

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 { E x : 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: Let R be an equivalence relation on S We have already shown that {E x : x  S}

is a partition of S.

Now suppose that P is the relation “belongs to the same piece

(equivalence class) as.” Then

xPy iff x, y  E z    for some z  S.

iff xRz and yRz    for some z  S.

iff xRy

Thus, P and R are the same relation

iff xRz and zRy    for some z  S.

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 {E x : 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  P  y iff x and y

are in the same piece of the partition Clearly, P is reflexive and symmetric

To see that P is transitive, suppose that x  P  y and y  P  z Then y  E x  E z

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

Finally, the equivalence classes of P correspond to the pieces of P because of the way

P was defined 