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Chapter  2 Sets and Functions Chapter  2 Sets and Functions

Chapter 2 Sets and Functions - PowerPoint Presentation

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Chapter 2 Sets and Functions - PPT Presentation

Section 22 Relations In symbols we have a b a ID: 750694

partition relation equivalence set relation partition set equivalence iff definition ordered related piece suppose reflexive property true classes elements

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Slide1

Chapter 2

Sets and FunctionsSlide2

Section 2.2

RelationsSlide3

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 ordered pair (a, b) is the set whose members are {a, b} and {a}.

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

This we prove in the following theorem.Slide4

Thus the set on the right can have only one member, so

Definition: (

a

,

b) = {{a}, {a

, b}} and (c,

d) = {{c

}, {c, d}

}Theorem 2.2.2: (a, b) = (c, d) iff a =

c and b = d. Proof:

If a = c and b = d, then(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 …Slide5

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. 

Theorem 2.2.2: (a,

b) = (c

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

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

d) = {{c

}, {c, d}}Slide6

then

 

B is the rectangle shown below:

1

4

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. In symbols, A 

B = {(a

, b) : a

 A and b  B}.

Example 2.2.5If 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  BA

Band B is the interval (2, 4],

Note: The solid lines indicate the

left and top edges are included.The dashed lines indicate the rightand bottom edges are not included. Slide7

So it is natural for the formal

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

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 first.

For instance, 1 < 3 is true but 3 < 1 is false.Definition 2.2.7Let 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.Slide8

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

.

1

4

2

4

x

y

[

)

[ )A

B

A

 B

x = yRFor 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].

x = 1

x = 3Slide9

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

are related by

R

if

x

lives within a mile of

y

.Determine which properties apply to each relation.It is

reflexive and symmetric, but not transitive. Slide10

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

Ex = { 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

xy

w

E

y

E

x

To see this, suppose that

w

Ex  E

y.

We claim that Ex =

Ey. For any x  

Ex we have x R x.

But w  E

x, so

w R x

and, by symmetry, x R w.

Also, w

 Ey, so w R

 y. Using transitivity twice, we have

x R y. This implies

x  Ey and Ex

 Ey

. 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. Slide11

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 = {{1},

{2}, {3}} is a partition of

S

.

1

2

3

S

1

2

3

S

The

the

collection

P

=

{{1,

2}, {3}

} is also a partition of S. P

= {{1}, {2},

{3}}P

= {{1, 2},

{3}}

The collection P =

{{1,

2},

{2,

3}

}

is not a partition of

S

because {1,

2} and {2,

3}

are not disjoint.Slide12

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 2.2.17Let 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: 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

x

P

y

iff

x

, y  E

z for some

z  S.

iff x R z and y R

z for some z  S.

iff x R yThus,

P and R

are the same relation.iff x R

z and z R y for some z  S.Slide13

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 2.2.17Let 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 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  Ez. But this implies that E

x = Ez 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. 