Relations. Functions.Bijection and counting.

Cartesian product

Functions

Bijection

Inclusion-Exclusion

p. 2

Cartesian products

Given two setsA= {1, 2,3}B = {1, 2,3, 4}

Their Cartesian product

A× B = {(1,1), (2,1), (3,1),(1,2), (2,2), (3,2),(1,3), (2,3), (3,3),(1,4), (2,4), (3,4)}

Question: What is the cartesian product of Z×Z?

(Z is the set of all integers)

Cartesian product

Functions

Bijection

Inclusion-Exclusion

p. 3

Cartesian product Z×Z= Z2

x

y

Origin (0, 0)

(1, 2)

(−3, 0)

(−1,−3)

All pairs ofintegers are in Z2,for exmaple(1, 2) ∈ Z2

Cartesian product

Functions

Bijection

Inclusion-Exclusion

p. 4

Is it a Cartesian product?

x

y

Cartesian product

Functions

Bijection

Inclusion-Exclusion

p. 5

Cartesian product of Z×N

x

y

Cartesian product

Functions

Bijection

Inclusion-Exclusion

p. 6

Is it a Cartesian product?

x

y

Cartesian product

Functions

Bijection

Inclusion-Exclusion

p. 7

Cartesian product of N×N= N2

x

y

Cartesian product

Functions

Bijection

Inclusion-Exclusion

p. 8

Is it a Cartesian product?

x

y

(2,0)(−2,0)

Cartesian product

Functions

Bijection

Inclusion-Exclusion

p. 9

Is it a Cartesian product? No

x

y

(2,0)(−2,0)

R=�

(x , y) ∈ Z2�

� max(|x |, |y|) = 2

⊆ Z2

Cartesian product

Functions

Bijection

Inclusion-Exclusion

p. 10

Is it a Cartesian product? No

x

y

(2,0)(−2,0)

A subset R ⊆ Z×Z of a Cartesianproduct is called a relation.

Cartesian product

Functions

Bijection

Inclusion-Exclusion

p. 11

Relations

Given any two sets

A= {♠,♣,♥,♦} and B = {1,2, 3, . . .}

Def. A subset R of the Cartesian product A× B is called a relationfrom the set A to the set B.

R=�

(♠, 99), (♥, 15), (♣, 105), (♣, 1), (♣, 15)

⊆ A× B

Cartesian product

Functions

Bijection

Inclusion-Exclusion

p. 12

Relations

Example of a relation:

S = set of students

C = set of classes

R= {(s, c) | student s takes class c}

Alice

Bob

Eve

Mark

Zack

CS 150

CS 235

Math 254

Hist 152

Cartesian product

Functions

Bijection

Inclusion-Exclusion

p. 13

Airline routemap

We connect two cities if the airline operates a flight from one tothe other. Is it a relation?

Cartesian product

Functions

Bijection

Inclusion-Exclusion

p. 14

Airline routemap

Routes ⊆ Cities× Cities

Cartesian product

Functions

Bijection

Inclusion-Exclusion

p. 15

Airline routemap

Cartesian product

Functions

Bijection

Inclusion-Exclusion

p. 16

Any subset of A× B is a relation

x

y

R ⊆ Z×ZR=�

(−1,2), (1, 2), (−1, 1),(1, 1), (−2,−1), (−1,−2),(0,−2), (1,−2), (2,−1)

Cartesian product

Functions

Bijection

Inclusion-Exclusion

p. 17

Any subset of A× B is a relation

x

y

(−1, 1)

R ⊆ R×RR=�

(x , y) ∈ R×R | (x+1)2+(y−1)2 ≤ 2

Cartesian product

Functions

Bijection

Inclusion-Exclusion

p. 18

A function is a relation too!

x

y

R ⊆ R×RR=�

(x , y) ∈ R×R | y = f (x)

Cartesian product

Functions

Bijection

Inclusion-Exclusion

p. 19

Relation {(x , y) | y = |x |}

x

y

Z→ Z

3

2

1

0

−1

−2

−3

3

2

1

0

−1

−2

−3

Cartesian product

Functions

Bijection

Inclusion-Exclusion

p. 20

Relation {(x , y) | y = x rem 3}

x

y

Z→ Z

3

2

1

0

−1

−2

−3

3

2

1

0

−1

−2

−3

Cartesian product

Functions

Bijection

Inclusion-Exclusion

p. 21

Functions

Def. A relation R ⊆ A× B is a function (a functional relation) if forevery a ∈ A, there is at most one b ∈ B so that (a, b) ∈ R.

A= {1, 2,3}, B = {1,2, 3,4}

R1 = {(1, 1), (2,2), (3,3)}R2 = {(1, 2), (1,3), (1,4), (2,3), (2,4), (3,4)} ← Not a functionR3 = {(1, 2), (2,3), (3,4)}

1

2

3

1

2

3

4

1

2

3

1

2

3

4

1

2

3

1

2

3

4

Cartesian product

Functions

Bijection

Inclusion-Exclusion

p. 22

Functions

Functional relation R ⊆ A× B defines a unique way to map eachelement from the set A to an element from the set B.

There is a well-known and convenient notation for functions:

f (a) = b where a ∈ A and b ∈ B

It maps elements from A to B:

f : A→ B

Af−→ B

Cartesian product

Functions

Bijection

Inclusion-Exclusion

p. 23

Functions

Def. For the function f : A→ B, set A is called domain, and set Bis called codomain.

x

y

z

1

2

3

4f : A→ B

domain( f ) = A= {x , y, z}codomain( f ) = B = {1, 2,3, 4}image( f ) = f (A) = {2, 3}

Def. f (a) is the image of a point a ∈ A.

Def. The image of a function f , denoted by f (A), is the set of allimages of all points a ∈ A

f (A) = {x | ∃a ∈ A ( f (a) = x)}.

The image of a function is also called range.

Cartesian product

Functions

Bijection

Inclusion-Exclusion

p. 24

Onto

Def. A function f : A → B is called onto if and only if for everyelement b ∈ B there is an element a ∈ A with f (a) = b.

In other words, the image f (A) is the whole codomain B.

f : A→ B is onto

v

w

x

y

z

1

2

3

4

g : A→ B is not onto

v

w

x

y

z

1

2

3

4

Cartesian product

Functions

Bijection

Inclusion-Exclusion

p. 25

One-to-one

Def. A function f : A → B is said to be one-to-one if and only iff (x) = f (y) implies that x = y for all x , y ∈ A.

f : A→ B is one-to-one

w

x

y

z

0

1

2

3

4

g : B→ C is not one-to-one

0

1

2

3

4

1/5

1/6

1/7

Cartesian product

Functions

Bijection

Inclusion-Exclusion

p. 26

Bijection

Def. The function f is a bijection (also called one-to-one correspon-dence) if and only if it is both one-to-one and onto.

f : A→ B is a bijection

v

w

x

y

z

0

1

2

3

4

g : A→ B is not a bijection

v

w

x

y

z

0

1

2

3

4

Cartesian product

Functions

Bijection

Inclusion-Exclusion

p. 27

Bijection

f : A→ B is one-to-one, butnot onto

w

x

y

z

0

1

2

3

4

g : C → D is onto, but notone-to-one

v

w

x

y

z

0

1

2

3

So, both functions are not bijections.

Cartesian product

Functions

Bijection

Inclusion-Exclusion

p. 28

Bijection. Observation

Bijection Rule. Given two sets A and B, if there exists a bijection

f : A→ B, then |A|= |B|.

v

w

x

y

z

0

1

2

3

4

We can count the size of the set A, instead of the size of B!

Cartesian product

Functions

Bijection

Inclusion-Exclusion

p. 29

Bijection Rule.

Consider two similar problems:

(a) How many bit strings contain exactly three 1s and two 0s?

11010

(b) How many strings can be composed of three ‘A’s and five ‘b’s sothat an ‘A’ is always followed by a ‘b’?

AbAbbAbb

We show that this two problems are equivalent by constructing abijection.

Cartesian product

Functions

Bijection

Inclusion-Exclusion

p. 30

Bijection Rule.

Let X be the set of bit strings

X = {11010, . . .}

and Y be the set of ‘A’ and ‘b’ strings

Y = {AbAbbAbb, . . .}

We can construct a bijection f : X → Y :

1 gets replaced by Ab0 gets replaced by b

Cartesian product

Functions

Bijection

Inclusion-Exclusion

p. 31

Bijection Rule.

f : 11100 7→ Ab Ab Ab b b11010 7→ Ab Ab b Ab b11001 7→ Ab Ab b b Ab10110 7→ Ab b Ab Ab b10101 7→ Ab b Ab b Ab10011 7→ Ab b b Ab Ab01110 7→ b Ab Ab Ab b01101 7→ b Ab Ab b Ab01011 7→ b Ab b Ab Ab00111 7→ b b Ab Ab Ab

Function f is one-to-one and onto, so it is a bijection. Therefore,the cardinalities of two sets are equal: |X |= |Y |=

�53

�

= 10.

Cartesian product

Functions

Bijection

Inclusion-Exclusion

p. 32

Bijection. Observation

Observation For every bijection f : A→ B, exists an inverse function

f −1 : B→ A

f : A→ B1

v

w

x

y

z

0

1

2

3

4

The inverse function is a bijection too.

Cartesian product

Functions

Bijection

Inclusion-Exclusion

p. 33

Bijection. Observation

Observation For every bijection f : A→ B, exists an inverse function

f −1 : B→ A

f : A→ B1

v

w

x

y

z

0

1

2

3

4

f −1 : B→ A

0

1

2

3

4

v

w

x

y

z

The inverse function is a bijection too.

Cartesian product

Functions

Bijection

Inclusion-Exclusion

p. 34

Bijection. Observation

Given two bijections f : A→ B and g : B→ C .Consider their composition

h(x) = g( f (x))

f : A→ B

w

x

y

z

1

2

3

4

g : B→ C

1

2

3

4

1/5

2/5

3/5

4/5

h : A→ C is a bijection, and therefore |A|= |C |.

Cartesian product

Functions

Bijection

Inclusion-Exclusion

p. 35

Bijection. Observation

Given two bijections f : A→ B and g : B→ C .Consider their composition

h(x) = g( f (x))

f : A→ B

w

x

y

z

1

2

3

4

g : B→ C

1

2

3

4

1/5

2/5

3/5

4/5

h : A→ C

w

x

y

z

1/5

2/5

3/5

4/5

h : A→ C is a bijection, and therefore |A|= |C |.

Cartesian product

Functions

Bijection

Inclusion-Exclusion

p. 36

Bijection. Counting subsets

Please, count the number of subsets of a set

A= {a, b, c, d, e}

by reducing the problem to counting bit strings.

Let’s find a bijection between the power set

P (A) = {∅, {a}, {b}, . . .}

and the set of bit stings of length 5:

{0, 1}5 = {00000, 00001,00010, 00011, . . .}

Cartesian product

Functions

Bijection

Inclusion-Exclusion

p. 37

Bijection. Counting subsets

Please, count the number of subsets of a set

A= {a, b, c, d, e}

by reducing the problem to counting bit strings.

Let’s find a bijection f between the power set

P (A) = {∅, {a}, {b}, . . .}

and the set of bit stings of length 5:

{0, 1}5 = {00000, 00001,00010, 00011, . . .}

Cartesian product

Functions

Bijection

Inclusion-Exclusion

p. 38

Bijection. Counting subsets

f :P ({a, b, c, d, e})→ {0,1}5

0s and 1s encode the membership of the five elements of{a, b, c, d, e}

f :∅ 7→ 00000{a} 7→ 10000{b} 7→ 01000{a, b} 7→ 11000{c} 7→ 00100{a, c} 7→ 10100{b, c} 7→ 01100{a, b, c} 7→ 11100

. . .skipping um.. 23 subsets

{a, b, c, d, e} 7→ 11111

The cardinality�

�{0, 1}5�

�= 25 = 32

Therefore, by the bijectionrule,�

�P (A)�

�= 25

Cartesian product

Functions

Bijection

Inclusion-Exclusion

p. 39

Inclusion-Exclusion principle

We remember the subtraction rule for the union of two sets

|A∪ B|= |A|+ |B| − |A∩ B|

Can it be generalized for a union of n sets

|A1 ∪ . . .∪ An|= |A1|+ . . .+ |An| − 〈something〉?

Of course, it can!

Cartesian product

Functions

Bijection

Inclusion-Exclusion

p. 40

Inclusion-Exclusion principle

We remember the subtraction rule for the union of two sets

|A∪ B|= |A|+ |B| − |A∩ B|

Can it be generalized for a union of n sets

|A1 ∪ . . .∪ An|= |A1|+ . . .+ |An| − 〈something〉?

Of course, it can!

Cartesian product

Functions

Bijection

Inclusion-Exclusion

p. 41

Inclusion-Exclusion principle

Union of three sets

|A1 ∪ A2 ∪ A3|= |A1|+ |A2|+ |A3|− |A1 ∩ A2| − |A2 ∩ A3| − |A1 ∩ A3|+ |A1 ∩ A2 ∩ A3|

|{1,2, 3} ∪ {2, 3,4} ∪ {3, 4,1}|= 3+ 3+ 3− 2− 2− 2+ 1= 4

Cartesian product

Functions

Bijection

Inclusion-Exclusion

p. 42

Inclusion-Exclusion principle

Union of n sets

|A1 ∪ . . .∪ An|= the sum of the sizes of the individual setsminus the sizes of all two-way intersections

plus the sizes of all three-way intersections

minus the sizes of all four-way intersections

plus the sizes of all five-way intersections

etc.