Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically,...

145
Order Relations and Functions

Transcript of Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically,...

Page 1: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Order Relationsand

Functions

Page 2: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Problem Session Tonight

7:00PM – 7:50PM380-380X

Optional, but highly recommended!

Page 3: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Recap from Last Time

Page 4: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Relations

● A binary relation is a property that describes whether two objects are related in some way.

● Examples:● Less-than: x < y● Divisibility: x divides y evenly● Friendship: x is a friend of y● Tastiness: x is tastier than y

● Given binary relation R, we write aRb iff a is related to b by relation R.

Page 5: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Order Relations

Page 6: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

“x is larger than y”

“x is faster than y”

“x is tastier than y”

“x is a subset of y”

“x divides y”

“x is a part of y”

Page 7: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Informally

An order relation is a relation that ranks elements against one another.

Do not use this definition in proofs!It's just an intuition!

Page 8: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Properties of Order Relations

x ≤ y

Page 9: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Properties of Order Relations

x ≤ y

1 ≤ 5 5 ≤ 8and

Page 10: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Properties of Order Relations

x ≤ y

1 ≤ 5 5 ≤ 8and

1 ≤ 8

Page 11: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Properties of Order Relations

x ≤ y

42 ≤ 99 99 ≤ 137and

Page 12: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Properties of Order Relations

x ≤ y

42 ≤ 99 99 ≤ 137and

42 ≤ 137

Page 13: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Properties of Order Relations

x ≤ y

x ≤ y y ≤ zand

Page 14: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Properties of Order Relations

x ≤ y

x ≤ y y ≤ zand

x ≤ z

Page 15: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Properties of Order Relations

x ≤ y

x ≤ y y ≤ zand

x ≤ z

Transitivity

Page 16: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Properties of Order Relations

x ≤ y

Page 17: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Properties of Order Relations

x ≤ y

1 ≤ 1

Page 18: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Properties of Order Relations

x ≤ y

42 ≤ 42

Page 19: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Properties of Order Relations

x ≤ y

137 ≤ 137

Page 20: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Properties of Order Relations

x ≤ y

x ≤ x

Page 21: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Properties of Order Relations

x ≤ y

x ≤ x

Reflexivity

Page 22: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Properties of Order Relations

x ≤ y

Page 23: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Properties of Order Relations

x ≤ y

19 ≤ 21

Page 24: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Properties of Order Relations

x ≤ y

19 ≤ 2121 ≤ 19?

Page 25: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Properties of Order Relations

x ≤ y

19 ≤ 2121 ≤ 19?

Page 26: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Properties of Order Relations

x ≤ y

42 ≤ 137

Page 27: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Properties of Order Relations

x ≤ y

42 ≤ 137137 ≤ 42?

Page 28: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Properties of Order Relations

x ≤ y

42 ≤ 137137 ≤ 42?

Page 29: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Properties of Order Relations

x ≤ y

137 ≤ 137

Page 30: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Properties of Order Relations

x ≤ y

137 ≤ 137137 ≤ 137?

Page 31: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Properties of Order Relations

x ≤ y

137 ≤ 137137 ≤ 137

Page 32: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Antisymmetry

A binary relation R over a set A is called antisymmetric iff

For any x ∈ A and y ∈ A,If xRy and y ≠ x, then yRx.

Equivalently:

For any x ∈ A and y ∈ A,if xRy and yRx, then x = y.

Page 33: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

An Intuition for Antisymmetry

Self-loops allowed

Only one edge between nodes

For any x ∈ A and y ∈ A,If xRy and y ≠ x, then yRx.

Page 34: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Partial Orders

● A binary relation R is a partial order over a set A iff it is

● reflexive,● antisymmetric, and● transitive.

● A pair (A, R), where R is a partial order over A, is called a partially ordered set or poset.

Page 35: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Partial Orders

Why “partial”?

● A binary relation R is a partial order over a set A iff it is

● reflexive,● antisymmetric, and● transitive.

● A pair (A, R), where R is a partial order over A, is called a partially ordered set or poset.

Page 36: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Gold Silver Bronze

46 29 29

38 27 23

29 17 19

24 26 32

14 15 17

Total

104

88

65

82

4613 8 7 28

2012 Summer Olympics

Inspired by http://tartarus.org/simon/2008-olympics-hasse/Data from http://www.london2012.com/medals/medal-count/

14 15 17 4611 19 14 44

14 15 17 4611 11 12 34

Page 37: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Gold Silver Bronze

46 29 29

38 27 23

29 17 19

24 26 32

14 15 17

Total

104

88

65

82

4613 8 7 28

2012 Summer Olympics

Inspired by http://tartarus.org/simon/2008-olympics-hasse/Data from http://www.london2012.com/medals/medal-count/

14 15 17 4611 19 14 44

14 15 17 4611 11 12 34

Page 38: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Define the relationship

(gold0, total0)R(gold1, total1)

to be true when

gold0 ≤ gold1 and total0 ≤ total1

Page 39: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

10446

8838

29 8224

2813

4411

3411

65

Page 40: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

10446

8838

29 8224

2813

4411

3411

65

Page 41: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

10446

8838

29 8224

2813

4411

3411

65

Page 42: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

10446

8838

29 8224

2813

4411

3411

65

Page 43: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

10446

8838

29 8224

2813

4411

3411

65

Page 44: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

10446

8838

29 8224

2813

4411

3411

65

Page 45: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

10446

8838

29 8224

2813

4411

3411

65

Page 46: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

10446

8838

6529 8224

2813

4411

3411

Page 47: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

10446

8838

6529 8224

2813

4411

3411

Page 48: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

10446

8838

6529 8224

2813

4411

3411

Page 49: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

10446

8838

6529 8224

2813

4411

3411

Page 50: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

10446

8838

6529 8224

2813

4411

3411

Page 51: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

10446

8838

6529 8224

2813

4411

3411

Page 52: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

10446

8838

6529 8224

2813

4411

3411

Page 53: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

10446

8838

6529 8224

2813

4411

3411

Page 54: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Partial and Total Orders

● A binary relation R over a set A is called total iff for any x ∈ A and y ∈ A, that xRy or yRx.● It's possible for both to be true.

● A binary relation R over a set A is called a total order iff it is a partial order and it is total.

● Examples:● Integers ordered by ≤.● Strings ordered alphabetically.

Page 55: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

10446

8838

6529 8224

2813

4411

3411

Page 56: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

10446

8838

6529 8224

2813

4411

3411

Page 57: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

10446

8838

6529 8224

2813

4411

3411

Page 58: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

10446

8838

6529 8224

2813

4411

3411

Page 59: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

10446

8838

6529 8224

2813

4411

3411

Page 60: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

10446

8838

6529 8224

2813

4411

3411

Page 61: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

10446

8838

6529 8224

2813

4411

3411

More Medals

Fewer Medals

Page 62: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

10446

8838

6529 8224

2813

4411

3411

More Medals

Fewer Medals

Page 63: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Hasse Diagrams

● A Hasse diagram is a graphical representation of a partial order.

● No self-loops: by reflexivity, we can always add them back in.

● Higher elements are bigger than lower elements: by antisymmetry, the edges can only go in one direction.

● No redundant edges: by transitivity, we can infer the missing edges.

Page 64: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Indian Mediterranean

Mexican

Chinese Italian

American

Tasty

Not Tasty

Dorm

Page 65: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

0

3

1

2

4

...Larger

Smaller

Page 66: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Hasse Artichokes

Page 67: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Hasse Artichokes

Page 68: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Summary of Order Relations

● A partial order is a relation that is reflexive, antisymmetric, and transitive.

● A Hasse diagram is a drawing of a partial order that has no self-loops, arrowheads, or redundant edges.

● A total order is a partial order in which any pair of elements are comparable.

Page 69: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

For More on the Olympics:

http://www.nytimes.com/interactive/2012/08/07/sports/olympics/the-best-and-worst-countries-in-the-medal-count.html

Page 70: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Functions

Page 71: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

A function is a means of associating each object in one set with an object in some

other set.

Page 72: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

DikdikNubian

IbexSloth

Page 73: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain
Page 74: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Black and White

Page 75: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Terminology

● A function f is a mapping such that every value in A is associated with a unique value in B.● For every a ∈ A, there exists some b ∈ B with f(a) = b.

● If f(a) = b0 and f(a) = b1, then b0 = b1.

● If f is a function from A to B, we sometimes say that f is a mapping from A to B.● We call A the domain of f.● We call B the codomain of f.

– We'll discuss “range” in a few minutes.

● We denote that f is a function from A to B by writing

f : A → B

Page 76: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Is This a Function from A to B?

A B

Page 77: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Is This a Function from A to B?

A B

Page 78: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Is This a Function from A to B?

Each object in the domain has to be associated with exactly one object in the

codomain!

Each object in the domain has to be associated with exactly one object in the

codomain!

A B

Page 79: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Is This a Function from A to B?

California

New York

Delaware

Washington DC

Sacramento

Dover

Albany

A B

Each object in the domain has to be associated with exactly one object in the

codomain!

Each object in the domain has to be associated with exactly one object in the

codomain!

Page 80: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Is This a Function from A to B?

Love-a-Lot

Tenderheart

Wish

Funshine

Friend

It's fine that nothing is associated with Friend; functions do not need

to use the entire codomain.

It's fine that nothing is associated with Friend; functions do not need

to use the entire codomain.

A B

Page 81: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Defining Functions

● Typically, we specify a function by describing a rule that maps every element of the domain to some element of the codomain.

● Examples:● f(n) = n + 1, where f : ℤ → ℤ● f(x) = sin x, where f : ℝ → ℝ● f(x) = ⌈x⌉, where f : ℝ → ℤ

● When defining a function it is always a good idea to verify that● The function is uniquely defined for all elements in the

domain, and● The function's output is always in the codomain.

Page 82: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Defining Functions

Typically, we specify a function by describing a rule that maps every element of the domain to some element of the codomain.

Examples:

f(n) = n + 1, where f : ℤ → ℤ

f(x) = sin x, where f : ℝ → ℝ● f(x) = ⌈x⌉, where f : ℝ → ℤ

When defining a function it is always a good idea to verify that

The function is uniquely defined for all elements in the domain, and

The function's output is always in the codomain.

This is the ceiling function – the smallest integer greater than or equal to x. For

example, 1 = 1, 1.37 = 2, ⌈ ⌉ ⌈ ⌉

and = 4.⌈π⌉

This is the ceiling function – the smallest integer greater than or equal to x. For

example, 1 = 1, 1.37 = 2, ⌈ ⌉ ⌈ ⌉

and = 4.⌈π⌉

Page 83: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Defining Functions

● Typically, we specify a function by describing a rule that maps every element of the domain to some element of the codomain.

● Examples:● f(n) = n + 1, where f : ℤ → ℤ● f(x) = sin x, where f : ℝ → ℝ● f(x) = ⌈x⌉, where f : ℝ → ℤ

● When defining a function it is always a good idea to verify that● The function is uniquely defined for all elements in the

domain, and● The function's output is always in the codomain.

Page 84: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Piecewise Functions

● Functions may be specified piecewise, with different rules applying to different elements.

● Example:

● When defining a function piecewise, it's up to you to confirm that it defines a legal function!

f (n)={ −n /2 if n is even(n+1)/2 otherwise

Page 85: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

♀☿

♂♃♄♅♆

Page 86: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

♀☿

♂♃♄♅♆

Page 87: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Mercury

Venus

Earth

Mars

Jupiter

Saturn

Uranus

Neptune

Pluto

♀☿

♂♃♄♅♆

Page 88: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Mercury

Venus

Earth

Mars

Jupiter

Saturn

Uranus

Neptune

Pluto

♀☿

♂♃♄♅♆

Page 89: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Mercury

Venus

Earth

Mars

Jupiter

Saturn

Uranus

Neptune

♀☿

♂♃♄♅♆

Page 90: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Mercury

Venus

Earth

Mars

Jupiter

Saturn

Uranus

Neptune

♀☿

♂♃♄♅♆

Page 91: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Injective Functions

● A function f : A → B is called injective (or one-to-one) if each element of the codomain has at most one element of the domain associated with it.● A function with this property is called an

injection.

● Formally:

If f(x0) = f(x1), then x0 = x1

● An intuition: injective functions label the objects from A using names from B.

Page 92: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain
Page 93: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain
Page 94: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Front Door

BalconyWindow

BedroomWindow

Page 95: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Front Door

BalconyWindow

BedroomWindow

Page 96: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Surjective Functions

● A function f : A → B is called surjective (or onto) if each element of the codomain has at least one element of the domain associated with it.● A function with this property is called a

surjection.● Formally:

For any b ∈ B, there exists atleast one a ∈ A such that f(a) = b.

● An intuition: surjective functions cover every element of B with at least one element of A.

Page 97: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Injections and Surjections

● An injective function associates at most one element of the domain with each element of the codomain.

● A surjective function associates at least one element of the domain with each element of the codomain.

● What about functions that associate exactly one element of the domain with each element of the codomain?

Page 98: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Katniss Everdeen

Merida

Hermione Granger

Page 99: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Katniss Everdeen

Merida

Hermione Granger

Page 100: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Bijections

● A function that associates each element of the codomain with a unique element of the domain is called bijective.● Such a function is a bijection.

● Formally, a bijection is a function that is both injective and surjective.

● A bijection is a one-to-one correspondence between two sets.

Page 101: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Compositions

Page 102: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

www.microsoft.com www.apple.com www.google.com

Page 103: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

www.microsoft.com www.apple.com www.google.com

Page 104: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

www.microsoft.com www.apple.com www.google.com

Page 105: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

www.microsoft.com www.apple.com www.google.com

Page 106: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

www.microsoft.com www.apple.com www.google.com

Page 107: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

www.microsoft.com www.apple.com www.google.com

Page 108: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

www.microsoft.com www.apple.com www.google.com

Page 109: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Function Composition

● Let f : A → B and g : B → C.● The composition of f and g (denoted

g ∘ f) is the function g ∘ f : A → C defined as

(g ∘ f)(x) = g(f(x))● Note that f is applied first, but f is on the

right side!● Function composition is associative:

h ∘ (g ∘ f) = (h ∘ g) ∘ f

Page 110: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Function Composition

● Suppose f : A → A and g : A → A.● Then both g ∘ f and f ∘ g are defined.● Does g ∘ f = f ∘ g?● In general, no:

● Let f(x) = 2x● Let g(x) = x + 1● (g ∘ f)(x) = g(f(x)) = g(2x) = 2x + 1● (f ∘ g)(x) = f(g(x)) = f(x + 1) = 2x + 2

Page 111: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Cardinality Revisited

Page 112: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Cardinality

● Recall (from lecture one!) that the cardinality of a set is the number of elements it contains.● Denoted |S|.

● For finite sets, cardinalities are natural numbers:● |{1, 2, 3}| = 3● |{100, 200, 300}| = 3

● For infinite sets, we introduce infinite cardinals to denote the size of sets:● |ℕ| = ℵ0

Page 113: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Comparing Cardinalities

● The relationships between set cardinalities are defined in terms of functions between those sets.

● |S| = |T| is defined using bijections.

|S| = |T| iff there is a bijection f : S → T

, , ,

, ,,

Page 114: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Theorem: If |R| = |S| and |S| = |T|, then |R| = |T|.

Proof: We will exhibit a bijection f : R → T. Since |R| = |S|, there is abijection g : R → S. Since |S| = |T|, there is a bijection h : S → T.

Let f = h ∘ g; this means that f : R → T. We prove that f is abijection by showing that it is injective and surjective.

To see that f is injective, suppose that f(r0) = f(r1). We will showthat r0 = r1. Since f(r0) = t and f(r1) = t, we know (h ∘ g)(r0) = tand (h ∘ g)(r1) = t. By definition of composition, we haveh(g(r0)) = h(g(r1)). Since h is a bijection, h is injective, we havethat g(r0) = g(r1). Since g is a bijection, g is injective, we havethat r0 = r1. Therefore, f is injective.

To see that f is surjective, consider any t ∈ T. We will show thatthere is some r ∈ R such that f(r) = t. Since h is a bijection fromS to T, h is surjective, so there is some s ∈ S such that h(s) = t.Since g is a bijection from R to S, g is surjective, so there is somer ∈ R such that g(r) = s. Thus f(r) = (h ∘ g)(r) = h(g(r)) = h(s) = tas required, so f is surjective.

Since f is injective and surjective, it is bijective. Thus there is abijection from R to T, so |R| = |T|. ■

Page 115: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Theorem: If |R| = |S| and |S| = |T|, then |R| = |T|.

Proof: We will exhibit a bijection f : R → T. Since |R| = |S|, there is abijection g : R → S. Since |S| = |T|, there is a bijection h : S → T.

Let f = h ∘ g; this means that f : R → T. We prove that f is abijection by showing that it is injective and surjective.

To see that f is injective, suppose that f(r0) = f(r1). We will showthat r0 = r1. Since f(r0) = t and f(r1) = t, we know (h ∘ g)(r0) = tand (h ∘ g)(r1) = t. By definition of composition, we haveh(g(r0)) = h(g(r1)). Since h is a bijection, h is injective, we havethat g(r0) = g(r1). Since g is a bijection, g is injective, we havethat r0 = r1. Therefore, f is injective.

To see that f is surjective, consider any t ∈ T. We will show thatthere is some r ∈ R such that f(r) = t. Since h is a bijection fromS to T, h is surjective, so there is some s ∈ S such that h(s) = t.Since g is a bijection from R to S, g is surjective, so there is somer ∈ R such that g(r) = s. Thus f(r) = (h ∘ g)(r) = h(g(r)) = h(s) = tas required, so f is surjective.

Since f is injective and surjective, it is bijective. Thus there is abijection from R to T, so |R| = |T|. ■

Page 116: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Theorem: If |R| = |S| and |S| = |T|, then |R| = |T|.

Proof: We will exhibit a bijection f : R → T. Since |R| = |S|, there is abijection g : R → S. Since |S| = |T|, there is a bijection h : S → T.

Let f = h ∘ g; this means that f : R → T. We prove that f is abijection by showing that it is injective and surjective.

To see that f is injective, suppose that f(r0) = f(r1). We will showthat r0 = r1. Since f(r0) = t and f(r1) = t, we know (h ∘ g)(r0) = tand (h ∘ g)(r1) = t. By definition of composition, we haveh(g(r0)) = h(g(r1)). Since h is a bijection, h is injective, we havethat g(r0) = g(r1). Since g is a bijection, g is injective, we havethat r0 = r1. Therefore, f is injective.

To see that f is surjective, consider any t ∈ T. We will show thatthere is some r ∈ R such that f(r) = t. Since h is a bijection fromS to T, h is surjective, so there is some s ∈ S such that h(s) = t.Since g is a bijection from R to S, g is surjective, so there is somer ∈ R such that g(r) = s. Thus f(r) = (h ∘ g)(r) = h(g(r)) = h(s) = tas required, so f is surjective.

Since f is injective and surjective, it is bijective. Thus there is abijection from R to T, so |R| = |T|. ■

Page 117: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Theorem: If |R| = |S| and |S| = |T|, then |R| = |T|.

Proof: We will exhibit a bijection f : R → T. Since |R| = |S|, there is abijection g : R → S. Since |S| = |T|, there is a bijection h : S → T.

Let f = h ∘ g; this means that f : R → T. We prove that f is abijection by showing that it is injective and surjective.

To see that f is injective, suppose that f(r0) = f(r1). We will showthat r0 = r1. Since f(r0) = t and f(r1) = t, we know (h ∘ g)(r0) = tand (h ∘ g)(r1) = t. By definition of composition, we haveh(g(r0)) = h(g(r1)). Since h is a bijection, h is injective, we havethat g(r0) = g(r1). Since g is a bijection, g is injective, we havethat r0 = r1. Therefore, f is injective.

To see that f is surjective, consider any t ∈ T. We will show thatthere is some r ∈ R such that f(r) = t. Since h is a bijection fromS to T, h is surjective, so there is some s ∈ S such that h(s) = t.Since g is a bijection from R to S, g is surjective, so there is somer ∈ R such that g(r) = s. Thus f(r) = (h ∘ g)(r) = h(g(r)) = h(s) = tas required, so f is surjective.

Since f is injective and surjective, it is bijective. Thus there is abijection from R to T, so |R| = |T|. ■

Page 118: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Theorem: If |R| = |S| and |S| = |T|, then |R| = |T|.

Proof: We will exhibit a bijection f : R → T. Since |R| = |S|, there is abijection g : R → S. Since |S| = |T|, there is a bijection h : S → T.

Let f = h ∘ g; this means that f : R → T. We prove that f is abijection by showing that it is injective and surjective.

To see that f is injective, suppose that f(r0) = f(r1). We will showthat r0 = r1. Since f(r0) = t and f(r1) = t, we know (h ∘ g)(r0) = tand (h ∘ g)(r1) = t. By definition of composition, we haveh(g(r0)) = h(g(r1)). Since h is a bijection, h is injective, we havethat g(r0) = g(r1). Since g is a bijection, g is injective, we havethat r0 = r1. Therefore, f is injective.

To see that f is surjective, consider any t ∈ T. We will show thatthere is some r ∈ R such that f(r) = t. Since h is a bijection fromS to T, h is surjective, so there is some s ∈ S such that h(s) = t.Since g is a bijection from R to S, g is surjective, so there is somer ∈ R such that g(r) = s. Thus f(r) = (h ∘ g)(r) = h(g(r)) = h(s) = tas required, so f is surjective.

Since f is injective and surjective, it is bijective. Thus there is abijection from R to T, so |R| = |T|. ■

R S T

Page 119: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Theorem: If |R| = |S| and |S| = |T|, then |R| = |T|.

Proof: We will exhibit a bijection f : R → T. Since |R| = |S|, there is abijection g : R → S. Since |S| = |T|, there is a bijection h : S → T.

Let f = h ∘ g; this means that f : R → T. We prove that f is abijection by showing that it is injective and surjective.

To see that f is injective, suppose that f(r0) = f(r1). We will showthat r0 = r1. Since f(r0) = t and f(r1) = t, we know (h ∘ g)(r0) = tand (h ∘ g)(r1) = t. By definition of composition, we haveh(g(r0)) = h(g(r1)). Since h is a bijection, h is injective, we havethat g(r0) = g(r1). Since g is a bijection, g is injective, we havethat r0 = r1. Therefore, f is injective.

To see that f is surjective, consider any t ∈ T. We will show thatthere is some r ∈ R such that f(r) = t. Since h is a bijection fromS to T, h is surjective, so there is some s ∈ S such that h(s) = t.Since g is a bijection from R to S, g is surjective, so there is somer ∈ R such that g(r) = s. Thus f(r) = (h ∘ g)(r) = h(g(r)) = h(s) = tas required, so f is surjective.

Since f is injective and surjective, it is bijective. Thus there is abijection from R to T, so |R| = |T|. ■

R S T

g

g

g

Page 120: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Theorem: If |R| = |S| and |S| = |T|, then |R| = |T|.

Proof: We will exhibit a bijection f : R → T. Since |R| = |S|, there is abijection g : R → S. Since |S| = |T|, there is a bijection h : S → T.

Let f = h ∘ g; this means that f : R → T. We prove that f is abijection by showing that it is injective and surjective.

To see that f is injective, suppose that f(r0) = f(r1). We will showthat r0 = r1. Since f(r0) = t and f(r1) = t, we know (h ∘ g)(r0) = tand (h ∘ g)(r1) = t. By definition of composition, we haveh(g(r0)) = h(g(r1)). Since h is a bijection, h is injective, we havethat g(r0) = g(r1). Since g is a bijection, g is injective, we havethat r0 = r1. Therefore, f is injective.

To see that f is surjective, consider any t ∈ T. We will show thatthere is some r ∈ R such that f(r) = t. Since h is a bijection fromS to T, h is surjective, so there is some s ∈ S such that h(s) = t.Since g is a bijection from R to S, g is surjective, so there is somer ∈ R such that g(r) = s. Thus f(r) = (h ∘ g)(r) = h(g(r)) = h(s) = tas required, so f is surjective.

Since f is injective and surjective, it is bijective. Thus there is abijection from R to T, so |R| = |T|. ■

R S T

g

g

g

h

h

h

Page 121: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Theorem: If |R| = |S| and |S| = |T|, then |R| = |T|.

Proof: We will exhibit a bijection f : R → T. Since |R| = |S|, there is abijection g : R → S. Since |S| = |T|, there is a bijection h : S → T.

Let f = h ∘ g; this means that f : R → T. We prove that f is abijection by showing that it is injective and surjective.

To see that f is injective, suppose that f(r0) = f(r1). We will showthat r0 = r1. Since f(r0) = t and f(r1) = t, we know (h ∘ g)(r0) = tand (h ∘ g)(r1) = t. By definition of composition, we haveh(g(r0)) = h(g(r1)). Since h is a bijection, h is injective, we havethat g(r0) = g(r1). Since g is a bijection, g is injective, we havethat r0 = r1. Therefore, f is injective.

To see that f is surjective, consider any t ∈ T. We will show thatthere is some r ∈ R such that f(r) = t. Since h is a bijection fromS to T, h is surjective, so there is some s ∈ S such that h(s) = t.Since g is a bijection from R to S, g is surjective, so there is somer ∈ R such that g(r) = s. Thus f(r) = (h ∘ g)(r) = h(g(r)) = h(s) = tas required, so f is surjective.

Since f is injective and surjective, it is bijective. Thus there is abijection from R to T, so |R| = |T|. ■

R S T

g

g

g

h

h

h

h ∘ g

h ∘ g

h ∘ g

Page 122: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Theorem: If |R| = |S| and |S| = |T|, then |R| = |T|.

Proof: We will exhibit a bijection f : R → T. Since |R| = |S|, there is abijection g : R → S. Since |S| = |T|, there is a bijection h : S → T.

Let f = h ∘ g; this means that f : R → T. We prove that f is abijection by showing that it is injective and surjective.

To see that f is injective, suppose that f(r0) = f(r1). We will showthat r0 = r1. Since f(r0) = t and f(r1) = t, we know (h ∘ g)(r0) = tand (h ∘ g)(r1) = t. By definition of composition, we haveh(g(r0)) = h(g(r1)). Since h is a bijection, h is injective, we havethat g(r0) = g(r1). Since g is a bijection, g is injective, we havethat r0 = r1. Therefore, f is injective.

To see that f is surjective, consider any t ∈ T. We will show thatthere is some r ∈ R such that f(r) = t. Since h is a bijection fromS to T, h is surjective, so there is some s ∈ S such that h(s) = t.Since g is a bijection from R to S, g is surjective, so there is somer ∈ R such that g(r) = s. Thus f(r) = (h ∘ g)(r) = h(g(r)) = h(s) = tas required, so f is surjective.

Since f is injective and surjective, it is bijective. Thus there is abijection from R to T, so |R| = |T|. ■

Page 123: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Theorem: If |R| = |S| and |S| = |T|, then |R| = |T|.

Proof: We will exhibit a bijection f : R → T. Since |R| = |S|, there is abijection g : R → S. Since |S| = |T|, there is a bijection h : S → T.

Let f = h ∘ g; this means that f : R → T. We prove that f is abijection by showing that it is injective and surjective.

To see that f is injective, suppose that f(r0) = f(r1). We will showthat r0 = r1. Since f(r0) = t and f(r1) = t, we know (h ∘ g)(r0) = tand (h ∘ g)(r1) = t. By definition of composition, we haveh(g(r0)) = h(g(r1)). Since h is a bijection, h is injective, we havethat g(r0) = g(r1). Since g is a bijection, g is injective, we havethat r0 = r1. Therefore, f is injective.

To see that f is surjective, consider any t ∈ T. We will show thatthere is some r ∈ R such that f(r) = t. Since h is a bijection fromS to T, h is surjective, so there is some s ∈ S such that h(s) = t.Since g is a bijection from R to S, g is surjective, so there is somer ∈ R such that g(r) = s. Thus f(r) = (h ∘ g)(r) = h(g(r)) = h(s) = tas required, so f is surjective.

Since f is injective and surjective, it is bijective. Thus there is abijection from R to T, so |R| = |T|. ■

Page 124: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Theorem: If |R| = |S| and |S| = |T|, then |R| = |T|.

Proof: We will exhibit a bijection f : R → T. Since |R| = |S|, there is abijection g : R → S. Since |S| = |T|, there is a bijection h : S → T.

Let f = h ∘ g; this means that f : R → T. We prove that f is abijection by showing that it is injective and surjective.

To see that f is injective, suppose that f(r0) = f(r1). We will showthat r0 = r1. Since f(r0) = f(r1), we know (h ∘ g)(r0) = (h ∘ g)(r1).By definition of composition, we have h(g(r0)) = h(g(r1)). Since his a bijection, h is injective. Thus since h(g(r0)) = h(g(r1)), we havethat g(r0) = g(r1). Since g is a bijection, g is injective, so becauseg(r0) = g(r1) we have that r0 = r1. Therefore, f is injective.

To see that f is surjective, consider any t ∈ T. We will show thatthere is some r ∈ R such that f(r) = t. Since h is a bijection fromS to T, h is surjective, so there is some s ∈ S such that h(s) = t.Since g is a bijection from R to S, g is surjective, so there is somer ∈ R such that g(r) = s. Thus f(r) = (h ∘ g)(r) = h(g(r)) = h(s) = tas required, so f is surjective.

Since f is injective and surjective, it is bijective. Thus there is abijection from R to T, so |R| = |T|. ■

Page 125: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Theorem: If |R| = |S| and |S| = |T|, then |R| = |T|.

Proof: We will exhibit a bijection f : R → T. Since |R| = |S|, there is abijection g : R → S. Since |S| = |T|, there is a bijection h : S → T.

Let f = h ∘ g; this means that f : R → T. We prove that f is abijection by showing that it is injective and surjective.

To see that f is injective, suppose that f(r0) = f(r1). We will showthat r0 = r1. Since f(r0) = f(r1), we know (h ∘ g)(r0) = (h ∘ g)(r1).By definition of composition, we have h(g(r0)) = h(g(r1)). Since his a bijection, h is injective. Thus since h(g(r0)) = h(g(r1)), we havethat g(r0) = g(r1). Since g is a bijection, g is injective, so becauseg(r0) = g(r1) we have that r0 = r1. Therefore, f is injective.

To see that f is surjective, consider any t ∈ T. We will show thatthere is some r ∈ R such that f(r) = t. Since h is a bijection fromS to T, h is surjective, so there is some s ∈ S such that h(s) = t.Since g is a bijection from R to S, g is surjective, so there is somer ∈ R such that g(r) = s. Thus f(r) = (h ∘ g)(r) = h(g(r)) = h(s) = tas required, so f is surjective.

Since f is injective and surjective, it is bijective. Thus there is abijection from R to T, so |R| = |T|. ■

Page 126: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Theorem: If |R| = |S| and |S| = |T|, then |R| = |T|.

Proof: We will exhibit a bijection f : R → T. Since |R| = |S|, there is abijection g : R → S. Since |S| = |T|, there is a bijection h : S → T.

Let f = h ∘ g; this means that f : R → T. We prove that f is abijection by showing that it is injective and surjective.

To see that f is injective, suppose that f(r0) = f(r1). We will showthat r0 = r1. Since f(r0) = f(r1), we know (h ∘ g)(r0) = (h ∘ g)(r1).By definition of composition, we have h(g(r0)) = h(g(r1)). Since his a bijection, h is injective. Thus since h(g(r0)) = h(g(r1)), we havethat g(r0) = g(r1). Since g is a bijection, g is injective, so becauseg(r0) = g(r1) we have that r0 = r1. Therefore, f is injective.

To see that f is surjective, consider any t ∈ T. We will show thatthere is some r ∈ R such that f(r) = t. Since h is a bijection fromS to T, h is surjective, so there is some s ∈ S such that h(s) = t.Since g is a bijection from R to S, g is surjective, so there is somer ∈ R such that g(r) = s. Thus f(r) = (h ∘ g)(r) = h(g(r)) = h(s) = tas required, so f is surjective.

Since f is injective and surjective, it is bijective. Thus there is abijection from R to T, so |R| = |T|. ■

Page 127: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Theorem: If |R| = |S| and |S| = |T|, then |R| = |T|.

Proof: We will exhibit a bijection f : R → T. Since |R| = |S|, there is abijection g : R → S. Since |S| = |T|, there is a bijection h : S → T.

Let f = h ∘ g; this means that f : R → T. We prove that f is abijection by showing that it is injective and surjective.

To see that f is injective, suppose that f(r0) = f(r1). We will showthat r0 = r1. Since f(r0) = f(r1), we know (h ∘ g)(r0) = (h ∘ g)(r1).By definition of composition, we have h(g(r0)) = h(g(r1)). Since his a bijection, h is injective. Thus since h(g(r0)) = h(g(r1)), we havethat g(r0) = g(r1). Since g is a bijection, g is injective, so becauseg(r0) = g(r1) we have that r0 = r1. Therefore, f is injective.

To see that f is surjective, consider any t ∈ T. We will show thatthere is some r ∈ R such that f(r) = t. Since h is a bijection fromS to T, h is surjective, so there is some s ∈ S such that h(s) = t.Since g is a bijection from R to S, g is surjective, so there is somer ∈ R such that g(r) = s. Thus f(r) = (h ∘ g)(r) = h(g(r)) = h(s) = tas required, so f is surjective.

Since f is injective and surjective, it is bijective. Thus there is abijection from R to T, so |R| = |T|. ■

Page 128: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Theorem: If |R| = |S| and |S| = |T|, then |R| = |T|.

Proof: We will exhibit a bijection f : R → T. Since |R| = |S|, there is abijection g : R → S. Since |S| = |T|, there is a bijection h : S → T.

Let f = h ∘ g; this means that f : R → T. We prove that f is abijection by showing that it is injective and surjective.

To see that f is injective, suppose that f(r0) = f(r1). We will showthat r0 = r1. Since f(r0) = f(r1), we know (h ∘ g)(r0) = (h ∘ g)(r1).By definition of composition, we have h(g(r0)) = h(g(r1)). Since his a bijection, h is injective. Thus since h(g(r0)) = h(g(r1)), we havethat g(r0) = g(r1). Since g is a bijection, g is injective, so becauseg(r0) = g(r1) we have that r0 = r1. Therefore, f is injective.

To see that f is surjective, consider any t ∈ T. We will show thatthere is some r ∈ R such that f(r) = t. Since h is a bijection fromS to T, h is surjective, so there is some s ∈ S such that h(s) = t.Since g is a bijection from R to S, g is surjective, so there is somer ∈ R such that g(r) = s. Thus f(r) = (h ∘ g)(r) = h(g(r)) = h(s) = tas required, so f is surjective.

Since f is injective and surjective, it is bijective. Thus there is abijection from R to T, so |R| = |T|. ■

Page 129: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Theorem: If |R| = |S| and |S| = |T|, then |R| = |T|.

Proof: We will exhibit a bijection f : R → T. Since |R| = |S|, there is abijection g : R → S. Since |S| = |T|, there is a bijection h : S → T.

Let f = h ∘ g; this means that f : R → T. We prove that f is abijection by showing that it is injective and surjective.

To see that f is injective, suppose that f(r0) = f(r1). We will showthat r0 = r1. Since f(r0) = f(r1), we know (h ∘ g)(r0) = (h ∘ g)(r1).By definition of composition, we have h(g(r0)) = h(g(r1)). Since his a bijection, h is injective. Thus since h(g(r0)) = h(g(r1)), we havethat g(r0) = g(r1). Since g is a bijection, g is injective, so becauseg(r0) = g(r1) we have that r0 = r1. Therefore, f is injective.

To see that f is surjective, consider any t ∈ T. We will show thatthere is some r ∈ R such that f(r) = t. Since h is a bijection fromS to T, h is surjective, so there is some s ∈ S such that h(s) = t.Since g is a bijection from R to S, g is surjective, so there is somer ∈ R such that g(r) = s. Thus f(r) = (h ∘ g)(r) = h(g(r)) = h(s) = tas required, so f is surjective.

Since f is injective and surjective, it is bijective. Thus there is abijection from R to T, so |R| = |T|. ■

Page 130: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Theorem: If |R| = |S| and |S| = |T|, then |R| = |T|.

Proof: We will exhibit a bijection f : R → T. Since |R| = |S|, there is abijection g : R → S. Since |S| = |T|, there is a bijection h : S → T.

Let f = h ∘ g; this means that f : R → T. We prove that f is abijection by showing that it is injective and surjective.

To see that f is injective, suppose that f(r0) = f(r1). We will showthat r0 = r1. Since f(r0) = f(r1), we know (h ∘ g)(r0) = (h ∘ g)(r1).By definition of composition, we have h(g(r0)) = h(g(r1)). Since his a bijection, h is injective. Thus since h(g(r0)) = h(g(r1)), we havethat g(r0) = g(r1). Since g is a bijection, g is injective, so becauseg(r0) = g(r1) we have that r0 = r1. Therefore, f is injective.

To see that f is surjective, consider any t ∈ T. We will show thatthere is some r ∈ R such that f(r) = t. Since h is a bijection fromS to T, h is surjective, so there is some s ∈ S such that h(s) = t.Since g is a bijection from R to S, g is surjective, so there is somer ∈ R such that g(r) = s. Thus f(r) = (h ∘ g)(r) = h(g(r)) = h(s) = tas required, so f is surjective.

Since f is injective and surjective, it is bijective. Thus there is abijection from R to T, so |R| = |T|. ■

Page 131: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Theorem: If |R| = |S| and |S| = |T|, then |R| = |T|.

Proof: We will exhibit a bijection f : R → T. Since |R| = |S|, there is abijection g : R → S. Since |S| = |T|, there is a bijection h : S → T.

Let f = h ∘ g; this means that f : R → T. We prove that f is abijection by showing that it is injective and surjective.

To see that f is injective, suppose that f(r0) = f(r1). We will showthat r0 = r1. Since f(r0) = f(r1), we know (h ∘ g)(r0) = (h ∘ g)(r1).By definition of composition, we have h(g(r0)) = h(g(r1)). Since his a bijection, h is injective. Thus since h(g(r0)) = h(g(r1)), we havethat g(r0) = g(r1). Since g is a bijection, g is injective, so becauseg(r0) = g(r1) we have that r0 = r1. Therefore, f is injective.

To see that f is surjective, consider any t ∈ T. We will show thatthere is some r ∈ R such that f(r) = t. Since h is a bijection fromS to T, h is surjective, so there is some s ∈ S such that h(s) = t.Since g is a bijection from R to S, g is surjective, so there is somer ∈ R such that g(r) = s. Thus f(r) = (h ∘ g)(r) = h(g(r)) = h(s) = tas required, so f is surjective.

Since f is injective and surjective, it is bijective. Thus there is abijection from R to T, so |R| = |T|. ■

Page 132: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Theorem: If |R| = |S| and |S| = |T|, then |R| = |T|.

Proof: We will exhibit a bijection f : R → T. Since |R| = |S|, there is abijection g : R → S. Since |S| = |T|, there is a bijection h : S → T.

Let f = h ∘ g; this means that f : R → T. We prove that f is abijection by showing that it is injective and surjective.

To see that f is injective, suppose that f(r0) = f(r1). We will showthat r0 = r1. Since f(r0) = f(r1), we know (h ∘ g)(r0) = (h ∘ g)(r1).By definition of composition, we have h(g(r0)) = h(g(r1)). Since his a bijection, h is injective. Thus since h(g(r0)) = h(g(r1)), we havethat g(r0) = g(r1). Since g is a bijection, g is injective, so becauseg(r0) = g(r1) we have that r0 = r1. Therefore, f is injective.

To see that f is surjective, consider any t ∈ T. We will show thatthere is some r ∈ R such that f(r) = t. Since h is a bijection fromS to T, h is surjective, so there is some s ∈ S such that h(s) = t.Since g is a bijection from R to S, g is surjective, so there is somer ∈ R such that g(r) = s. Thus f(r) = (h ∘ g)(r) = h(g(r)) = h(s) = tas required, so f is surjective.

Since f is injective and surjective, it is bijective. Thus there is abijection from R to T, so |R| = |T|. ■

Page 133: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Theorem: If |R| = |S| and |S| = |T|, then |R| = |T|.

Proof: We will exhibit a bijection f : R → T. Since |R| = |S|, there is abijection g : R → S. Since |S| = |T|, there is a bijection h : S → T.

Let f = h ∘ g; this means that f : R → T. We prove that f is abijection by showing that it is injective and surjective.

To see that f is injective, suppose that f(r0) = f(r1). We will showthat r0 = r1. Since f(r0) = f(r1), we know (h ∘ g)(r0) = (h ∘ g)(r1).By definition of composition, we have h(g(r0)) = h(g(r1)). Since his a bijection, h is injective. Thus since h(g(r0)) = h(g(r1)), we havethat g(r0) = g(r1). Since g is a bijection, g is injective, so becauseg(r0) = g(r1) we have that r0 = r1. Therefore, f is injective.

To see that f is surjective, consider any t ∈ T. We will show thatthere is some r ∈ R such that f(r) = t. Since h is a bijection fromS to T, h is surjective, so there is some s ∈ S such that h(s) = t.Since g is a bijection from R to S, g is surjective, so there is somer ∈ R such that g(r) = s. Thus f(r) = (h ∘ g)(r) = h(g(r)) = h(s) = tas required, so f is surjective.

Since f is injective and surjective, it is bijective. Thus there is abijection from R to T, so |R| = |T|. ■

Page 134: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Theorem: If |R| = |S| and |S| = |T|, then |R| = |T|.

Proof: We will exhibit a bijection f : R → T. Since |R| = |S|, there is abijection g : R → S. Since |S| = |T|, there is a bijection h : S → T.

Let f = h ∘ g; this means that f : R → T. We prove that f is abijection by showing that it is injective and surjective.

To see that f is injective, suppose that f(r0) = f(r1). We will showthat r0 = r1. Since f(r0) = f(r1), we know (h ∘ g)(r0) = (h ∘ g)(r1).By definition of composition, we have h(g(r0)) = h(g(r1)). Since his a bijection, h is injective. Thus since h(g(r0)) = h(g(r1)), we havethat g(r0) = g(r1). Since g is a bijection, g is injective, so becauseg(r0) = g(r1) we have that r0 = r1. Therefore, f is injective.

To see that f is surjective, consider any t ∈ T. We will show thatthere is some r ∈ R such that f(r) = t. Since h is a bijection fromS to T, h is surjective, so there is some s ∈ S such that h(s) = t.Since g is a bijection from R to S, g is surjective, so there is somer ∈ R such that g(r) = s. Thus f(r) = (h ∘ g)(r) = h(g(r)) = h(s) = tas required, so f is surjective.

Since f is injective and surjective, it is bijective. Thus there is abijection from R to T, so |R| = |T|. ■

Page 135: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Theorem: If |R| = |S| and |S| = |T|, then |R| = |T|.

Proof: We will exhibit a bijection f : R → T. Since |R| = |S|, there is abijection g : R → S. Since |S| = |T|, there is a bijection h : S → T.

Let f = h ∘ g; this means that f : R → T. We prove that f is abijection by showing that it is injective and surjective.

To see that f is injective, suppose that f(r0) = f(r1). We will showthat r0 = r1. Since f(r0) = f(r1), we know (h ∘ g)(r0) = (h ∘ g)(r1).By definition of composition, we have h(g(r0)) = h(g(r1)). Since his a bijection, h is injective. Thus since h(g(r0)) = h(g(r1)), we havethat g(r0) = g(r1). Since g is a bijection, g is injective, so becauseg(r0) = g(r1) we have that r0 = r1. Therefore, f is injective.

To see that f is surjective, consider any t ∈ T. We will show thatthere is some r ∈ R such that f(r) = t. Since h is a bijection fromS to T, h is surjective, so there is some s ∈ S such that h(s) = t.Since g is a bijection from R to S, g is surjective, so there is somer ∈ R such that g(r) = s. Thus f(r) = (h ∘ g)(r) = h(g(r)) = h(s) = tas required, so f is surjective.

Since f is injective and surjective, it is bijective. Thus there is abijection from R to T, so |R| = |T|. ■

Page 136: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Theorem: If |R| = |S| and |S| = |T|, then |R| = |T|.

Proof: We will exhibit a bijection f : R → T. Since |R| = |S|, there is abijection g : R → S. Since |S| = |T|, there is a bijection h : S → T.

Let f = h ∘ g; this means that f : R → T. We prove that f is abijection by showing that it is injective and surjective.

To see that f is injective, suppose that f(r0) = f(r1). We will showthat r0 = r1. Since f(r0) = f(r1), we know (h ∘ g)(r0) = (h ∘ g)(r1).By definition of composition, we have h(g(r0)) = h(g(r1)). Since his a bijection, h is injective. Thus since h(g(r0)) = h(g(r1)), we havethat g(r0) = g(r1). Since g is a bijection, g is injective, so becauseg(r0) = g(r1) we have that r0 = r1. Therefore, f is injective.

To see that f is surjective, consider any t ∈ T. We will show thatthere is some r ∈ R such that f(r) = t. Since h is a bijection fromS to T, h is surjective, so there is some s ∈ S such that h(s) = t.Since g is a bijection from R to S, g is surjective, so there is somer ∈ R such that g(r) = s. Thus f(r) = (h ∘ g)(r) = h(g(r)) = h(s) = tas required, so f is surjective.

Since f is injective and surjective, it is bijective. Thus there is abijection from R to T, so |R| = |T|. ■

Page 137: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Theorem: If |R| = |S| and |S| = |T|, then |R| = |T|.

Proof: We will exhibit a bijection f : R → T. Since |R| = |S|, there is abijection g : R → S. Since |S| = |T|, there is a bijection h : S → T.

Let f = h ∘ g; this means that f : R → T. We prove that f is abijection by showing that it is injective and surjective.

To see that f is injective, suppose that f(r0) = f(r1). We will showthat r0 = r1. Since f(r0) = f(r1), we know (h ∘ g)(r0) = (h ∘ g)(r1).By definition of composition, we have h(g(r0)) = h(g(r1)). Since his a bijection, h is injective. Thus since h(g(r0)) = h(g(r1)), we havethat g(r0) = g(r1). Since g is a bijection, g is injective, so becauseg(r0) = g(r1) we have that r0 = r1. Therefore, f is injective.

To see that f is surjective, consider any t ∈ T. We will show thatthere is some r ∈ R such that f(r) = t. Since h is a bijection fromS to T, h is surjective, so there is some s ∈ S such that h(s) = t.Since g is a bijection from R to S, g is surjective, so there is somer ∈ R such that g(r) = s. Thus f(r) = (h ∘ g)(r) = h(g(r)) = h(s) = tas required, so f is surjective.

Since f is injective and surjective, it is bijective. Thus there is abijection from R to T, so |R| = |T|. ■

Page 138: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Theorem: If |R| = |S| and |S| = |T|, then |R| = |T|.

Proof: We will exhibit a bijection f : R → T. Since |R| = |S|, there is abijection g : R → S. Since |S| = |T|, there is a bijection h : S → T.

Let f = h ∘ g; this means that f : R → T. We prove that f is abijection by showing that it is injective and surjective.

To see that f is injective, suppose that f(r0) = f(r1). We will showthat r0 = r1. Since f(r0) = f(r1), we know (h ∘ g)(r0) = (h ∘ g)(r1).By definition of composition, we have h(g(r0)) = h(g(r1)). Since his a bijection, h is injective. Thus since h(g(r0)) = h(g(r1)), we havethat g(r0) = g(r1). Since g is a bijection, g is injective, so becauseg(r0) = g(r1) we have that r0 = r1. Therefore, f is injective.

To see that f is surjective, consider any t ∈ T. We will show thatthere is some r ∈ R such that f(r) = t. Since h is a bijection fromS to T, h is surjective, so there is some s ∈ S such that h(s) = t.Since g is a bijection from R to S, g is surjective, so there is somer ∈ R such that g(r) = s. Thus f(r) = (h ∘ g)(r) = h(g(r)) = h(s) = tas required, so f is surjective.

Since f is injective and surjective, it is bijective. Thus there is abijection from R to T, so |R| = |T|. ■

Page 139: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Theorem: If |R| = |S| and |S| = |T|, then |R| = |T|.

Proof: We will exhibit a bijection f : R → T. Since |R| = |S|, there is abijection g : R → S. Since |S| = |T|, there is a bijection h : S → T.

Let f = h ∘ g; this means that f : R → T. We prove that f is abijection by showing that it is injective and surjective.

To see that f is injective, suppose that f(r0) = f(r1). We will showthat r0 = r1. Since f(r0) = f(r1), we know (h ∘ g)(r0) = (h ∘ g)(r1).By definition of composition, we have h(g(r0)) = h(g(r1)). Since his a bijection, h is injective. Thus since h(g(r0)) = h(g(r1)), we havethat g(r0) = g(r1). Since g is a bijection, g is injective, so becauseg(r0) = g(r1) we have that r0 = r1. Therefore, f is injective.

To see that f is surjective, consider any t ∈ T. We will show thatthere is some r ∈ R such that f(r) = t. Since h is a bijection fromS to T, h is surjective, so there is some s ∈ S such that h(s) = t.Since g is a bijection from R to S, g is surjective, so there is somer ∈ R such that g(r) = s. Thus f(r) = (h ∘ g)(r) = h(g(r)) = h(s) = tas required, so f is surjective.

Since f is injective and surjective, it is bijective. Thus there is abijection from R to T, so |R| = |T|. ■

Page 140: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Properties of Cardinality

● Equality of cardinality is an equivalence relation. For any sets R, S, and T:● |S| = |S|. (reflexivity)● If |S| = |T|, then |T| = |S|. (symmetry)● If |R| = |S| and |S| = |T|, then |R| = |T|.

(transitivity)

Page 141: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Comparing Cardinalities

● We define |S| ≤ |T| as follows:

|S| ≤ |T| iff there is an injection f : S → T

The ≤ relation over set cardinalities is a total order. For any sets R, S, and T:

|S| ≤ |S|. (reflexivity)

If |R| ≤ |S| and |S| ≤ |T|, then |R| ≤ |T|. (transitivity)

If |S| ≤ |T| and |T| ≤ |S|, then |S| = |T|. (antisymmetry)

Either |S| ≤ |T| or |T| ≤ |S|. (totality)

These last two proofs are extremely hard.

The antisymmetry result is the Cantor-Bernstein-Schroeder Theorem; a fascinating read, but beyond the scope of this course.

Totality requires the axiom of choice. Take Math 161 for more details.

Page 142: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Comparing Cardinalities

● We define |S| ≤ |T| as follows:

|S| ≤ |T| iff there is an injection f : S → T

The ≤ relation over set cardinalities is a total order. For any sets R, S, and T:

|S| ≤ |S|. (reflexivity)

If |R| ≤ |S| and |S| ≤ |T|, then |R| ≤ |T|. (transitivity)

If |S| ≤ |T| and |T| ≤ |S|, then |S| = |T|. (antisymmetry)

Either |S| ≤ |T| or |T| ≤ |S|. (totality)

These last two proofs are extremely hard.

The antisymmetry result is the Cantor-Bernstein-Schroeder Theorem; a fascinating read, but beyond the scope of this course.

Totality requires the axiom of choice. Take Math 161 for more details.

Page 143: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Comparing Cardinalities

● We define |S| ≤ |T| as follows:

|S| ≤ |T| iff there is an injection f : S → T

The ≤ relation over set cardinalities is a total order. For any sets R, S, and T:

|S| ≤ |S|. (reflexivity)

If |R| ≤ |S| and |S| ≤ |T|, then |R| ≤ |T|. (transitivity)

If |S| ≤ |T| and |T| ≤ |S|, then |S| = |T|. (antisymmetry)

Either |S| ≤ |T| or |T| ≤ |S|. (totality)

These last two proofs are extremely hard.

The antisymmetry result is the Cantor-Bernstein-Schroeder Theorem; a fascinating read, but beyond the scope of this course.

Totality requires the axiom of choice. Take Math 161 for more details.

Page 144: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Comparing Cardinalities

● We define |S| ≤ |T| as follows:

|S| ≤ |T| iff there is an injection f : S → T

The ≤ relation over set cardinalities is a total order. For any sets R, S, and T:

|S| ≤ |S|. (reflexivity)

If |R| ≤ |S| and |S| ≤ |T|, then |R| ≤ |T|. (transitivity)

If |S| ≤ |T| and |T| ≤ |S|, then |S| = |T|. (antisymmetry)

Either |S| ≤ |T| or |T| ≤ |S|. (totality)

These last two proofs are extremely hard.

The antisymmetry result is the Cantor-Bernstein-Schroeder Theorem; a fascinating read, but beyond the scope of this course.

Totality requires the axiom of choice. Take Math 161 for more details.

Page 145: Order Relations and Functions - Stanford University · 2012-10-09 · Defining Functions Typically, we specify a function by describing a rule that maps every element of the domain

Comparing Cardinalities

● We define |S| ≤ |T| as follows:

|S| ≤ |T| iff there is an injection f : S → T● The ≤ relation over set cardinalities is a total order. For any

sets R, S, and T:● |S| ≤ |S|. (reflexivity)● If |R| ≤ |S| and |S| ≤ |T|, then |R| ≤ |T|. (transitivity)● If |S| ≤ |T| and |T| ≤ |S|, then |S| = |T|. (antisymmetry)● Either |S| ≤ |T| or |T| ≤ |S|. (totality)

● These last two proofs are extremely hard.● The antisymmetry result is the Cantor-Bernstein-Schroeder

Theorem; a fascinating read, but beyond the scope of this course.● Totality requires the axiom of choice. Take Math 161 for more

details.