Cartesian product cross product A and B A B A B f a b j a ...stacho/macm101.pdf · De nition 1 The...

Definition 1

The Cartesian product (or cross product) of A and B , denoted byA× B , is the set

A× B = {(a, b) | a ∈ A and b ∈ B}

1. the elements (a, b) of A× B are ordered pairs

2. for pairs (a, b), (c , d) we have

(a, b) = (c , d) ⇐⇒ a = c and b = d

Definition 2

The n-fold product of sets A1, A2, . . . , An is the set of n-tuples

A1 × A2 × . . .× An = {(a1, a2, . . . , an) | ai ∈ Ai for all 1 ≤ i ≤ n}

() October 30, 2007 1 / 12

Definition 1


A× B = {(a, b) | a ∈ A and b ∈ B}



(a, b) = (c , d) ⇐⇒ a = c and b = d

Definition 2



() October 30, 2007 1 / 12

Definition 1


A× B = {(a, b) | a ∈ A and b ∈ B}



(a, b) = (c , d) ⇐⇒ a = c and b = d

Definition 2



() October 30, 2007 1 / 12

Definition 1

A× B = {(a, b) | a ∈ A and b ∈ B}

A = {2, 3, 4}B = {4, 5}

a) A×B = {(2, 4), (2, 5), (3, 4), (3, 5), (4, 4), (4, 5)}b) B×A = {(4, 2), (4, 3), (4, 4), (5, 2), (5, 3), (5, 4)}OO

//

2 3 4

4

5

◦◦◦◦◦◦

A× B

A

B

() October 30, 2007 2 / 12

Definition 1

A× B = {(a, b) | a ∈ A and b ∈ B}

A = {2, 3, 4}B = {4, 5}

a) A×B = {(2, 4), (2, 5), (3, 4), (3, 5), (4, 4), (4, 5)}b) B×A = {(4, 2), (4, 3), (4, 4), (5, 2), (5, 3), (5, 4)}OO

//

2 3 4

4

5

◦◦◦◦◦◦

A× B

A

B

() October 30, 2007 2 / 12

Definition 1

A× B = {(a, b) | a ∈ A and b ∈ B}

A = {2, 3, 4}B = {4, 5}

a) A×B = {(2, 4), (2, 5), (3, 4), (3, 5), (4, 4), (4, 5)}b) B×A = {(4, 2), (4, 3), (4, 4), (5, 2), (5, 3), (5, 4)}OO

//

2 3 4

4

5

◦◦◦◦◦◦

A× B

A

BOO

//

2

3

4

4 5

◦ ◦◦ ◦◦ ◦

B × A

B

A

() October 30, 2007 2 / 12

B = {4, 5}a) B2 = B × B = {(4, 4), (4, 5), (5, 4), (5, 5)}b) B3 = B × B × B = {(4, 4, 4), (4, 4, 5), (4, 5, 4), (4, 5, 5),

(5, 4, 4), (5, 4, 5), (5, 5, 4), (5, 5, 5)}OO

//

4

5

4 5

◦◦◦◦

B × B

B

B

() October 30, 2007 3 / 12

B = {4, 5}a) B2 = B × B = {(4, 4), (4, 5), (5, 4), (5, 5)}b) B3 = B × B × B = {(4, 4, 4), (4, 4, 5), (4, 5, 4), (4, 5, 5),

(5, 4, 4), (5, 4, 5), (5, 5, 4), (5, 5, 5)}OO

//

4

5

4 5

◦◦◦◦

B × B

B

B

() October 30, 2007 3 / 12

B = {4, 5}a) B2 = B × B = {(4, 4), (4, 5), (5, 4), (5, 5)}b) B3 = B × B × B = {(4, 4, 4), (4, 4, 5), (4, 5, 4), (4, 5, 5),

(5, 4, 4), (5, 4, 5), (5, 5, 4), (5, 5, 5)}

OO

//

::tttttttttttttttttttttttttttttt

4444

4

5

4 5

45

◦◦◦◦◦◦◦◦

B × B × B

B

B

B

() October 30, 2007 3 / 12

{4, �} × {x , y} × {♥,♠,♣}Tree Diagram

()

(4)

(�)

(4, x)

(4, y)

(�, x)

(�, y)

(4, x ,♥)

(4, x ,♠)

(4, x ,♣)

(4, y ,♥)

(4, y ,♠)

(4, y ,♣)

(�, x ,♥)

(�, x ,♠)

(�, x ,♣)

(�, y ,♥)

(�, y ,♠)

(�, y ,♣)

��

????????????

oooooo

OOOOOO

oooooo

OOOOOO

gggggg

WWWWWW

gggggg

WWWWWW

gggggg

WWWWWW

gggggg

WWWWWW

() October 30, 2007 4 / 12

Definition 3

A (binary) relation from A to B is a subset of A× B .A (binary) relation on A is a subset of A× A.

A = {2, 3, 4} and B = {4, 5}

a) R1 = {(2, 4), (3, 5)}b) R2 = {(2, 4), (3, 4), (4, 4)}c) R3 = {(2, 4), (2, 5), (4, 4), (4, 5)}d) R4 = ∅

OO

//

2 3 4

4

5

◦◦

R1

A

B

() October 30, 2007 5 / 12

Definition 3


A = {2, 3, 4} and B = {4, 5}

a) R1 = {(2, 4), (3, 5)}b) R2 = {(2, 4), (3, 4), (4, 4)}c) R3 = {(2, 4), (2, 5), (4, 4), (4, 5)}d) R4 = ∅

OO

//

2 3 4

4

5

◦◦

R1

A

B

() October 30, 2007 5 / 12

Definition 3


A = {2, 3, 4} and B = {4, 5}

a) R1 = {(2, 4), (3, 5)}b) R2 = {(2, 4), (3, 4), (4, 4)}c) R3 = {(2, 4), (2, 5), (4, 4), (4, 5)}d) R4 = ∅

OO

//

2 3 4

4

5

◦ ◦ ◦

R2

A

B

() October 30, 2007 5 / 12

Definition 3


A = {2, 3, 4} and B = {4, 5}

a) R1 = {(2, 4), (3, 5)}b) R2 = {(2, 4), (3, 4), (4, 4)}c) R3 = {(2, 4), (2, 5), (4, 4), (4, 5)}d) R4 = ∅

OO

//

2 3 4

4

5

◦◦

◦◦

R3

A

B

() October 30, 2007 5 / 12

Definition 3


A = {2, 3, 4} and B = {4, 5}

a) R1 = {(2, 4), (3, 5)}b) R2 = {(2, 4), (3, 4), (4, 4)}c) R3 = {(2, 4), (2, 5), (4, 4), (4, 5)}d) R4 = ∅

OO

//

2 3 4

4

5

R4

A

B

() October 30, 2007 5 / 12

Relation R = {(x , y) ∈ Z× Z | 1 ≤ x ≤ y ≤ 4} == {(1, 1), (1, 2), (1, 3), (1, 4), (2, 2), (2, 3), (2, 4), (3, 3), (3, 4), (4, 4)}

OO

//

1 2 3 4

1

2

3

4

◦◦◦◦

◦◦◦◦◦ ◦

R Notation

(x , y) ∈ R

m

xRy

(think of R as ≤)

() October 30, 2007 6 / 12


OO

//

1 2 3 4

1

2

3

4

◦◦◦◦

◦◦◦◦◦ ◦

R Notation

(x , y) ∈ R

m

xRy

(think of R as ≤)

() October 30, 2007 6 / 12


OO

//

1 2 3 4

1

2

3

4

◦◦◦◦

◦◦◦◦◦ ◦

R Notation

(x , y) ∈ R

m

xRy

(think of R as ≤)

() October 30, 2007 6 / 12

{Justin, Joey , Kevin, Nick} × {Britney , Christina, Jessica, Kelly , Sarah}

Justin

Joey

Kevin

Nick

Britney

Christina

Jessica

Kelly

Sarah

ddddddddddddddddd

ZZZZZZZZZZZZZZZZZ

OOOOOOOOOOOOOOOOOOO

DDDDDDDDDDDDDDDDDDDDDD

::::::::::::::::::::::::::

ooooooooooooooooooddddddddddddddddd

ZZZZZZZZZZZZZZZZZ

OOOOOOOOOOOOOOOOOO


zzzzzzzzzzzzzzzzzzzzzz

oooooooooooooooooooddddddddddddddddd

ZZZZZZZZZZZZZZZZZ

OOOOOOOOOOOOOOOOOOO

��


oooooooooooooooooooddddddddddddddddd

ZZZZZZZZZZZZZZZZZ

() October 30, 2007 7 / 12

Who dated whom? {(Ju, Br), (Ju, Je), (Jo, Ke), (Jo, Sa),(Ke, Br), (Ke, Ch), (Ni , Ch), (Ni , Ke), (Ni , Sa)}

Justin

Joey

Kevin

Nick

Britney

Christina

Jessica

Kelly

Sarah

dddddddddddddddddddddddddddddddddd

OOOOOOOOOOOOOOOOOOO

OOOOOOOOOOOOOOOOOOO

OOOOOOOOOOOOOOOOOO

OOOOOOOOOOOOOOOOOO





ooooooooooooooooooo

ooooooooooooooooooo


zzzzzzzzzzzzzzzzzzzzzzdddddddddddddddddddddddddddddddddd

ZZZZZZZZZZZZZZZZZ

ZZZZZZZZZZZZZZZZZ

() October 30, 2007 8 / 12

Who is dating whom? {(Ju, Je), (Ke, Br), (Ni , Ch), (Ni , Ke)}

Justin

Joey

Kevin

Nick

Britney

Christina

Jessica

Kelly

Sarah

() October 30, 2007 9 / 12

Theorem 4

For any set A, we have A× ∅ = ∅ ( and ∅ × A = ∅)

Proof. If (a, b) ∈ A× ∅, then a ∈ A and b ∈ ∅, impossible. �

Theorem 5

For any sets A, B , C

a) A× (B ∩ C ) = (A× B) ∩ (A× C )

b) A× (B ∪ C ) = (A× B) ∪ (A× C )

c) (A ∩ B)× C = (A× C ) ∩ (B × C )

d) (A ∪ B)× C = (A× C ) ∪ (B × C )

Proof. a) (a, b) ∈ A× (B ∩ C ) ⇐⇒ a ∈ A and b ∈ B ∩ C ⇐⇒a ∈ A and b ∈ B and b ∈ C ⇐⇒ (a, b) ∈ A× B and(a, b) ∈ A× C ⇐⇒ (a, b) ∈ (A× B) ∩ (A× C ) �

() October 30, 2007 10 / 12

Theorem 4

For any set A, we have A× ∅ = ∅ ( and ∅ × A = ∅)

Proof. If (a, b) ∈ A× ∅, then a ∈ A and b ∈ ∅, impossible. �

Theorem 5

For any sets A, B , C

a) A× (B ∩ C ) = (A× B) ∩ (A× C )

b) A× (B ∪ C ) = (A× B) ∪ (A× C )

c) (A ∩ B)× C = (A× C ) ∩ (B × C )

d) (A ∪ B)× C = (A× C ) ∪ (B × C )

Proof. a) (a, b) ∈ A× (B ∩ C ) ⇐⇒ a ∈ A and b ∈ B ∩ C ⇐⇒a ∈ A and b ∈ B and b ∈ C ⇐⇒ (a, b) ∈ A× B and(a, b) ∈ A× C ⇐⇒ (a, b) ∈ (A× B) ∩ (A× C ) �

() October 30, 2007 10 / 12

Observation 6

For any two sets A, B, the number of elements in A× B is

|A× B | = |A| · |B |

Hence there are exactly |P(A× B)| = 2|A×B| = 2|A|·|B| differentrelations from A to B.

() October 30, 2007 11 / 12

Observation 6

For any two sets A, B, the number of elements in A× B is

|A× B | = |A| · |B |

Hence there are exactly |P(A× B)| = 2|A×B| = 2|A|·|B| differentrelations from A to B.

() October 30, 2007 11 / 12

Exercises:

5.1.7 - If A = {1, 2, 3, 4, 5} and B = {w , x , y , z}, how manyelements are there in P(A× B). Answer: 220 = 1, 048, 576

5.1.3 - For A = {1, 2, 3} and B = {2, 4, 5}a) |A× B | =? Answer: 9

b) # of relations from A to B ? Answer: 29 = 512

c) # of relations on A ? Answer: 29 = 512

d) # of relations from A to Bthat contain (1, 2) and (1, 5) ? Answer: 27 = 128

e) # of relations from A to Bthat contain exactly five ordered pairs ? Answer:

(95

)= 126

f) # of relations on A thatcontain at least seven elements ? Answer:

(97

)+

(98

)+

(99

)= 121

() October 30, 2007 12 / 12

Exercises:







(95

)= 126


(97

)+

(98

)+

(99

)= 121

() October 30, 2007 12 / 12

Exercises:







(95

)= 126


(97

)+

(98

)+

(99

)= 121

() October 30, 2007 12 / 12

Exercises:







(95

)= 126


(97

)+

(98

)+

(99

)= 121

() October 30, 2007 12 / 12

Exercises:







(95

)= 126


(97

)+

(98

)+

(99

)= 121

() October 30, 2007 12 / 12

Exercises:







(95

)= 126


(97

)+

(98

)+

(99

)= 121

() October 30, 2007 12 / 12

Exercises:







(95

)= 126


(97

)+

(98

)+

(99

)= 121

() October 30, 2007 12 / 12

Exercises:







(95

)= 126


(97

)+

(98

)+

(99

)= 121

() October 30, 2007 12 / 12

Exercises:







(95

)= 126


(97

)+

(98

)+

(99

)= 121

() October 30, 2007 12 / 12

Exercises:







(95

)= 126


(97

)+

(98

)+

(99

)= 121

() October 30, 2007 12 / 12

Exercises:







(95

)= 126


(97

)+

(98

)+

(99

)= 121

() October 30, 2007 12 / 12

Exercises:







(95

)= 126


(97

)+

(98

)+

(99

)= 121

() October 30, 2007 12 / 12

Exercises:







(95

)= 126


(97

)+

(98

)+

(99

)= 121

() October 30, 2007 12 / 12

Exercises:







(95

)= 126


(97

)+

(98

)+

(99

)= 121

() October 30, 2007 12 / 12

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Cartesian product cross product A and B A B A B f a b j a ...stacho/macm101.pdf · De nition 1 The...

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