Cardinal Numbers and the Continuum Hypothesis Bernd Schr oder - - PowerPoint PPT Presentation

▶
cardinal numbers and the continuum hypothesis
SMART_READER_LITE
LIVE PREVIEW

Cardinal Numbers and the Continuum Hypothesis Bernd Schr oder - - PowerPoint PPT Presentation

Feb 15, 2023 7 likes •1.41k views

Finite Sizes Infinite Sizes Cardinal Arithmetic Cardinal Numbers and the Continuum Hypothesis Bernd Schr oder logo1 Bernd Schr oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum

slide-1
SLIDE 1

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

Cardinal Numbers and the Continuum Hypothesis

Bernd Schr¨

  • der

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-2
SLIDE 2

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

Introduction

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-3
SLIDE 3

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

Introduction

  • 1. We want a standard “size” for each set, just like the

number of elements (which is a natural number) is the standard size for finite sets.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-4
SLIDE 4

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

Introduction

  • 1. We want a standard “size” for each set, just like the

number of elements (which is a natural number) is the standard size for finite sets.

  • 2. Ordinal numbers will not quite work because different
  • rdinal numbers can have the same size.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-5
SLIDE 5

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

Introduction

  • 1. We want a standard “size” for each set, just like the

number of elements (which is a natural number) is the standard size for finite sets.

  • 2. Ordinal numbers will not quite work because different
  • rdinal numbers can have the same size.
  • 3. Plus, once we are past that, we want to do arithmetic.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-6
SLIDE 6

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

Introduction

  • 1. We want a standard “size” for each set, just like the

number of elements (which is a natural number) is the standard size for finite sets.

  • 2. Ordinal numbers will not quite work because different
  • rdinal numbers can have the same size.
  • 3. Plus, once we are past that, we want to do arithmetic.
  • 4. To start, consider the arithmetic of finite set sizes.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-7
SLIDE 7

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

Theorem.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-8
SLIDE 8

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Theorem. Let A,B and C be finite sets.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-9
SLIDE 9

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Theorem. Let A,B and C be finite sets. Then the following

hold.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-10
SLIDE 10

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Theorem. Let A,B and C be finite sets. Then the following

hold.

  • 1. If A∩B = /

0, then |A∪B| = |A|+|B|.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-11
SLIDE 11

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Theorem. Let A,B and C be finite sets. Then the following

hold.

  • 1. If A∩B = /

0, then |A∪B| = |A|+|B|.

  • 2. |A∪B| = |A|+|B|−|A∩B|.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-12
SLIDE 12

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Theorem. Let A,B and C be finite sets. Then the following

hold.

  • 1. If A∩B = /

0, then |A∪B| = |A|+|B|.

  • 2. |A∪B| = |A|+|B|−|A∩B|.

3. |A∪B∪C|=|A|+|B|+|C|−|B∩C|−|A∩B|−|A∩C|+|A∩B∩C|.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-13
SLIDE 13

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Theorem. Let A,B and C be finite sets. Then the following

hold.

  • 1. If A∩B = /

0, then |A∪B| = |A|+|B|.

  • 2. |A∪B| = |A|+|B|−|A∩B|.

3. |A∪B∪C|=|A|+|B|+|C|−|B∩C|−|A∩B|−|A∩C|+|A∩B∩C|.

  • 4. |A×B| = |A|·|B|.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-14
SLIDE 14

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Theorem. Let A,B and C be finite sets. Then the following

hold.

  • 1. If A∩B = /

0, then |A∪B| = |A|+|B|.

  • 2. |A∪B| = |A|+|B|−|A∩B|.

3. |A∪B∪C|=|A|+|B|+|C|−|B∩C|−|A∩B|−|A∩C|+|A∩B∩C|.

  • 4. |A×B| = |A|·|B|.
  • 5. With AB denoting the set of all functions from B to A, we

have

  • AB

= |A||B|.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-15
SLIDE 15

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

Proof (parts 2 and 3 only).

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-16
SLIDE 16

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

Proof (parts 2 and 3 only). |A∪B|+|A∩B|

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-17
SLIDE 17

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

Proof (parts 2 and 3 only). |A∪B|+|A∩B| =

  • A∪(B\A)
  • +|A∩B|

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-18
SLIDE 18

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

Proof (parts 2 and 3 only). |A∪B|+|A∩B| =

  • A∪(B\A)
  • +|A∩B|

= |A|+|B\A|+|A∩B|

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-19
SLIDE 19

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

Proof (parts 2 and 3 only). |A∪B|+|A∩B| =

  • A∪(B\A)
  • +|A∩B|

= |A|+|B\A|+|A∩B| = |A|+

  • B\(A∩B)
  • +|A∩B|

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-20
SLIDE 20

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

Proof (parts 2 and 3 only). |A∪B|+|A∩B| =

  • A∪(B\A)
  • +|A∩B|

= |A|+|B\A|+|A∩B| = |A|+

  • B\(A∩B)
  • +|A∩B|

= |A|+|B|

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-21
SLIDE 21

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

Proof (parts 2 and 3 only). |A∪B|+|A∩B| =

  • A∪(B\A)
  • +|A∩B|

= |A|+|B\A|+|A∩B| = |A|+

  • B\(A∩B)
  • +|A∩B|

= |A|+|B| |A∪B∪C|

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-22
SLIDE 22

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

Proof (parts 2 and 3 only). |A∪B|+|A∩B| =

  • A∪(B\A)
  • +|A∩B|

= |A|+|B\A|+|A∩B| = |A|+

  • B\(A∩B)
  • +|A∩B|

= |A|+|B| |A∪B∪C| =

  • A∪(B∪C)
  • Bernd Schr¨
  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-23
SLIDE 23

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

Proof (parts 2 and 3 only). |A∪B|+|A∩B| =

  • A∪(B\A)
  • +|A∩B|

= |A|+|B\A|+|A∩B| = |A|+

  • B\(A∩B)
  • +|A∩B|

= |A|+|B| |A∪B∪C| =

  • A∪(B∪C)
  • =

|A|+|B∪C|−

  • A∩(B∪C)
  • Bernd Schr¨
  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-24
SLIDE 24

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

Proof (parts 2 and 3 only). |A∪B|+|A∩B| =

  • A∪(B\A)
  • +|A∩B|

= |A|+|B\A|+|A∩B| = |A|+

  • B\(A∩B)
  • +|A∩B|

= |A|+|B| |A∪B∪C| =

  • A∪(B∪C)
  • =

|A|+|B∪C|−

  • A∩(B∪C)
  • =

|A|+|B|+|C|−|B∩C|−

  • (A∩B)∪(A∩C)
  • Bernd Schr¨
  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-25
SLIDE 25

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

Proof (parts 2 and 3 only). |A∪B|+|A∩B| =

  • A∪(B\A)
  • +|A∩B|

= |A|+|B\A|+|A∩B| = |A|+

  • B\(A∩B)
  • +|A∩B|

= |A|+|B| |A∪B∪C| =

  • A∪(B∪C)
  • =

|A|+|B∪C|−

  • A∩(B∪C)
  • =

|A|+|B|+|C|−|B∩C|−

  • (A∩B)∪(A∩C)
  • =

|A|+|B|+|C|−|B∩C|−

  • |A∩B|+|A∩C|−
  • (A∩B)∩(A∩C)
  • Bernd Schr¨
  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-26
SLIDE 26

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

Proof (parts 2 and 3 only). |A∪B|+|A∩B| =

  • A∪(B\A)
  • +|A∩B|

= |A|+|B\A|+|A∩B| = |A|+

  • B\(A∩B)
  • +|A∩B|

= |A|+|B| |A∪B∪C| =

  • A∪(B∪C)
  • =

|A|+|B∪C|−

  • A∩(B∪C)
  • =

|A|+|B|+|C|−|B∩C|−

  • (A∩B)∪(A∩C)
  • =

|A|+|B|+|C|−|B∩C|−

  • |A∩B|+|A∩C|−
  • (A∩B)∩(A∩C)
  • =

|A|+|B|+|C|−|B∩C|−|A∩B|−|A∩C|+|A∩B∩C|

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-27
SLIDE 27

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

Proof (parts 2 and 3 only). |A∪B|+|A∩B| =

  • A∪(B\A)
  • +|A∩B|

= |A|+|B\A|+|A∩B| = |A|+

  • B\(A∩B)
  • +|A∩B|

= |A|+|B| |A∪B∪C| =

  • A∪(B∪C)
  • =

|A|+|B∪C|−

  • A∩(B∪C)
  • =

|A|+|B|+|C|−|B∩C|−

  • (A∩B)∪(A∩C)
  • =

|A|+|B|+|C|−|B∩C|−

  • |A∩B|+|A∩C|−
  • (A∩B)∩(A∩C)
  • =

|A|+|B|+|C|−|B∩C|−|A∩B|−|A∩C|+|A∩B∩C|

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-28
SLIDE 28

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

Example.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-29
SLIDE 29

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Example. In the restaurant “Zum Adler” 41 people are dining.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-30
SLIDE 30

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Example. In the restaurant “Zum Adler” 41 people are dining.

25 people have ordered bone marrow dumpling soup as an appetizer.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-31
SLIDE 31

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Example. In the restaurant “Zum Adler” 41 people are dining.

25 people have ordered bone marrow dumpling soup as an

  • appetizer. 32 people have ordered blood sausage as the main

course.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-32
SLIDE 32

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Example. In the restaurant “Zum Adler” 41 people are dining.

25 people have ordered bone marrow dumpling soup as an

  • appetizer. 32 people have ordered blood sausage as the main
  • course. 18 people have ordered black forest cake for desert.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-33
SLIDE 33

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Example. In the restaurant “Zum Adler” 41 people are dining.

25 people have ordered bone marrow dumpling soup as an

  • appetizer. 32 people have ordered blood sausage as the main
  • course. 18 people have ordered black forest cake for desert. 12

people will have bone marrow dumpling soup and blood sausage.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-34
SLIDE 34

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Example. In the restaurant “Zum Adler” 41 people are dining.

25 people have ordered bone marrow dumpling soup as an

  • appetizer. 32 people have ordered blood sausage as the main
  • course. 18 people have ordered black forest cake for desert. 12

people will have bone marrow dumpling soup and blood

  • sausage. 9 people will have bone marrow dumpling soup and

black forest cake.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-35
SLIDE 35

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Example. In the restaurant “Zum Adler” 41 people are dining.

25 people have ordered bone marrow dumpling soup as an

  • appetizer. 32 people have ordered blood sausage as the main
  • course. 18 people have ordered black forest cake for desert. 12

people will have bone marrow dumpling soup and blood

  • sausage. 9 people will have bone marrow dumpling soup and

black forest cake. 15 people will have blood sausage and black forest cake.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-36
SLIDE 36

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Example. In the restaurant “Zum Adler” 41 people are dining.

25 people have ordered bone marrow dumpling soup as an

  • appetizer. 32 people have ordered blood sausage as the main
  • course. 18 people have ordered black forest cake for desert. 12

people will have bone marrow dumpling soup and blood

  • sausage. 9 people will have bone marrow dumpling soup and

black forest cake. 15 people will have blood sausage and black forest cake. How many people will have all three dishes?

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-37
SLIDE 37

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Example. In the restaurant “Zum Adler” 41 people are dining.

25 people have ordered bone marrow dumpling soup as an

  • appetizer. 32 people have ordered blood sausage as the main
  • course. 18 people have ordered black forest cake for desert. 12

people will have bone marrow dumpling soup and blood

  • sausage. 9 people will have bone marrow dumpling soup and

black forest cake. 15 people will have blood sausage and black forest cake. How many people will have all three dishes? M :=set of all people who will have bone marrow dumpling soup.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-38
SLIDE 38

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Example. In the restaurant “Zum Adler” 41 people are dining.

25 people have ordered bone marrow dumpling soup as an

  • appetizer. 32 people have ordered blood sausage as the main
  • course. 18 people have ordered black forest cake for desert. 12

people will have bone marrow dumpling soup and blood

  • sausage. 9 people will have bone marrow dumpling soup and

black forest cake. 15 people will have blood sausage and black forest cake. How many people will have all three dishes? M :=set of all people who will have bone marrow dumpling

  • soup. S :=set of all people who will have blood sausage.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-39
SLIDE 39

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Example. In the restaurant “Zum Adler” 41 people are dining.

25 people have ordered bone marrow dumpling soup as an

  • appetizer. 32 people have ordered blood sausage as the main
  • course. 18 people have ordered black forest cake for desert. 12

people will have bone marrow dumpling soup and blood

  • sausage. 9 people will have bone marrow dumpling soup and

black forest cake. 15 people will have blood sausage and black forest cake. How many people will have all three dishes? M :=set of all people who will have bone marrow dumpling

  • soup. S :=set of all people who will have blood sausage.

C :=set of all people who will have black forest cake.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-40
SLIDE 40

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Example. In the restaurant “Zum Adler” 41 people are dining.

25 people have ordered bone marrow dumpling soup as an

  • appetizer. 32 people have ordered blood sausage as the main
  • course. 18 people have ordered black forest cake for desert. 12

people will have bone marrow dumpling soup and blood

  • sausage. 9 people will have bone marrow dumpling soup and

black forest cake. 15 people will have blood sausage and black forest cake. How many people will have all three dishes? M :=set of all people who will have bone marrow dumpling

  • soup. S :=set of all people who will have blood sausage.

C :=set of all people who will have black forest cake. |M ∩S∩C|

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-41
SLIDE 41

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Example. In the restaurant “Zum Adler” 41 people are dining.

25 people have ordered bone marrow dumpling soup as an

  • appetizer. 32 people have ordered blood sausage as the main
  • course. 18 people have ordered black forest cake for desert. 12

people will have bone marrow dumpling soup and blood

  • sausage. 9 people will have bone marrow dumpling soup and

black forest cake. 15 people will have blood sausage and black forest cake. How many people will have all three dishes? M :=set of all people who will have bone marrow dumpling

  • soup. S :=set of all people who will have blood sausage.

C :=set of all people who will have black forest cake. |M ∩S∩C| = |M∪S∪C|−|M|−|S|−|C|+|M∩S|+|M∩C|+|S∩C|

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-42
SLIDE 42

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Example. In the restaurant “Zum Adler” 41 people are dining.

25 people have ordered bone marrow dumpling soup as an

  • appetizer. 32 people have ordered blood sausage as the main
  • course. 18 people have ordered black forest cake for desert. 12

people will have bone marrow dumpling soup and blood

  • sausage. 9 people will have bone marrow dumpling soup and

black forest cake. 15 people will have blood sausage and black forest cake. How many people will have all three dishes? M :=set of all people who will have bone marrow dumpling

  • soup. S :=set of all people who will have blood sausage.

C :=set of all people who will have black forest cake. |M ∩S∩C| = |M∪S∪C|−|M|−|S|−|C|+|M∩S|+|M∩C|+|S∩C| = 41−25−32−18+12+9+15

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-43
SLIDE 43

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Example. In the restaurant “Zum Adler” 41 people are dining.

25 people have ordered bone marrow dumpling soup as an

  • appetizer. 32 people have ordered blood sausage as the main
  • course. 18 people have ordered black forest cake for desert. 12

people will have bone marrow dumpling soup and blood

  • sausage. 9 people will have bone marrow dumpling soup and

black forest cake. 15 people will have blood sausage and black forest cake. How many people will have all three dishes? M :=set of all people who will have bone marrow dumpling

  • soup. S :=set of all people who will have blood sausage.

C :=set of all people who will have black forest cake. |M ∩S∩C| = |M∪S∪C|−|M|−|S|−|C|+|M∩S|+|M∩C|+|S∩C| = 41−25−32−18+12+9+15 = 2

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-44
SLIDE 44

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Example. In the restaurant “Zum Adler” 41 people are dining.

25 people have ordered bone marrow dumpling soup as an

  • appetizer. 32 people have ordered blood sausage as the main
  • course. 18 people have ordered black forest cake for desert. 12

people will have bone marrow dumpling soup and blood

  • sausage. 9 people will have bone marrow dumpling soup and

black forest cake. 15 people will have blood sausage and black forest cake. How many people will have all three dishes? M :=set of all people who will have bone marrow dumpling

  • soup. S :=set of all people who will have blood sausage.

C :=set of all people who will have black forest cake. |M ∩S∩C| = |M∪S∪C|−|M|−|S|−|C|+|M∩S|+|M∩C|+|S∩C| = 41−25−32−18+12+9+15 = 2

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-45
SLIDE 45

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

Definition.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-46
SLIDE 46

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Definition. A cardinal number is an ordinal number α so that

for all ordinal numbers β that are equivalent to α we have α ⊆ β.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-47
SLIDE 47

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Definition. A cardinal number is an ordinal number α so that

for all ordinal numbers β that are equivalent to α we have α ⊆ β. Definition.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-48
SLIDE 48

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Definition. A cardinal number is an ordinal number α so that

for all ordinal numbers β that are equivalent to α we have α ⊆ β.

  • Definition. For every infinite set S we define the cardinality |S|
  • f S to be the unique cardinal number α that is equivalent to S.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-49
SLIDE 49

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

Theorem.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-50
SLIDE 50

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Theorem. Cantor-Schr¨
  • der-Bernstein Theorem.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-51
SLIDE 51

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Theorem. Cantor-Schr¨
  • der-Bernstein Theorem. Let A and B

be sets so that there is an injective function f : A → B and an injective function g : B → A.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-52
SLIDE 52

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Theorem. Cantor-Schr¨
  • der-Bernstein Theorem. Let A and B

be sets so that there is an injective function f : A → B and an injective function g : B → A. Then there is a bijective function h : A → B.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-53
SLIDE 53

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Theorem. Cantor-Schr¨
  • der-Bernstein Theorem. Let A and B

be sets so that there is an injective function f : A → B and an injective function g : B → A. Then there is a bijective function h : A → B. Proof.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-54
SLIDE 54

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Theorem. Cantor-Schr¨
  • der-Bernstein Theorem. Let A and B

be sets so that there is an injective function f : A → B and an injective function g : B → A. Then there is a bijective function h : A → B.

  • Proof. Define F(X) := A\g
  • B\f[X]
  • for all X ⊆ A.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-55
SLIDE 55

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Theorem. Cantor-Schr¨
  • der-Bernstein Theorem. Let A and B

be sets so that there is an injective function f : A → B and an injective function g : B → A. Then there is a bijective function h : A → B.

  • Proof. Define F(X) := A\g
  • B\f[X]
  • for all X ⊆ A. Then

X ⊆ Y implies f[X] ⊆ f[Y]

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-56
SLIDE 56

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Theorem. Cantor-Schr¨
  • der-Bernstein Theorem. Let A and B

be sets so that there is an injective function f : A → B and an injective function g : B → A. Then there is a bijective function h : A → B.

  • Proof. Define F(X) := A\g
  • B\f[X]
  • for all X ⊆ A. Then

X ⊆ Y implies f[X] ⊆ f[Y], B\f[X] ⊇ B\f[Y]

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-57
SLIDE 57

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Theorem. Cantor-Schr¨
  • der-Bernstein Theorem. Let A and B

be sets so that there is an injective function f : A → B and an injective function g : B → A. Then there is a bijective function h : A → B.

  • Proof. Define F(X) := A\g
  • B\f[X]
  • for all X ⊆ A. Then

X ⊆ Y implies f[X] ⊆ f[Y], B\f[X] ⊇ B\f[Y], g

  • B\f[X]
  • ⊇ g
  • B\f[Y]
  • Bernd Schr¨
  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-58
SLIDE 58

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Theorem. Cantor-Schr¨
  • der-Bernstein Theorem. Let A and B

be sets so that there is an injective function f : A → B and an injective function g : B → A. Then there is a bijective function h : A → B.

  • Proof. Define F(X) := A\g
  • B\f[X]
  • for all X ⊆ A. Then

X ⊆ Y implies f[X] ⊆ f[Y], B\f[X] ⊇ B\f[Y], g

  • B\f[X]
  • ⊇ g
  • B\f[Y]
  • ,

F(X)

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-59
SLIDE 59

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Theorem. Cantor-Schr¨
  • der-Bernstein Theorem. Let A and B

be sets so that there is an injective function f : A → B and an injective function g : B → A. Then there is a bijective function h : A → B.

  • Proof. Define F(X) := A\g
  • B\f[X]
  • for all X ⊆ A. Then

X ⊆ Y implies f[X] ⊆ f[Y], B\f[X] ⊇ B\f[Y], g

  • B\f[X]
  • ⊇ g
  • B\f[Y]
  • ,

F(X) = A\g

  • B\f[X]
  • Bernd Schr¨
  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-60
SLIDE 60

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Theorem. Cantor-Schr¨
  • der-Bernstein Theorem. Let A and B

be sets so that there is an injective function f : A → B and an injective function g : B → A. Then there is a bijective function h : A → B.

  • Proof. Define F(X) := A\g
  • B\f[X]
  • for all X ⊆ A. Then

X ⊆ Y implies f[X] ⊆ f[Y], B\f[X] ⊇ B\f[Y], g

  • B\f[X]
  • ⊇ g
  • B\f[Y]
  • ,

F(X) = A\g

  • B\f[X]
  • ⊆ A\g
  • B\f[Y]
  • Bernd Schr¨
  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-61
SLIDE 61

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Theorem. Cantor-Schr¨
  • der-Bernstein Theorem. Let A and B

be sets so that there is an injective function f : A → B and an injective function g : B → A. Then there is a bijective function h : A → B.

  • Proof. Define F(X) := A\g
  • B\f[X]
  • for all X ⊆ A. Then

X ⊆ Y implies f[X] ⊆ f[Y], B\f[X] ⊇ B\f[Y], g

  • B\f[X]
  • ⊇ g
  • B\f[Y]
  • ,

F(X) = A\g

  • B\f[X]
  • ⊆ A\g
  • B\f[Y]
  • = F(Y).

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-62
SLIDE 62

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Theorem. Cantor-Schr¨
  • der-Bernstein Theorem. Let A and B

be sets so that there is an injective function f : A → B and an injective function g : B → A. Then there is a bijective function h : A → B.

  • Proof. Define F(X) := A\g
  • B\f[X]
  • for all X ⊆ A. Then

X ⊆ Y implies f[X] ⊆ f[Y], B\f[X] ⊇ B\f[Y], g

  • B\f[X]
  • ⊇ g
  • B\f[Y]
  • ,

F(X) = A\g

  • B\f[X]
  • ⊆ A\g
  • B\f[Y]
  • = F(Y).

Let C :=

  • H ∈ P(A) : H ⊆ F(H)
  • Bernd Schr¨
  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-63
SLIDE 63

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Theorem. Cantor-Schr¨
  • der-Bernstein Theorem. Let A and B

be sets so that there is an injective function f : A → B and an injective function g : B → A. Then there is a bijective function h : A → B.

  • Proof. Define F(X) := A\g
  • B\f[X]
  • for all X ⊆ A. Then

X ⊆ Y implies f[X] ⊆ f[Y], B\f[X] ⊇ B\f[Y], g

  • B\f[X]
  • ⊇ g
  • B\f[Y]
  • ,

F(X) = A\g

  • B\f[X]
  • ⊆ A\g
  • B\f[Y]
  • = F(Y).

Let C :=

  • H ∈ P(A) : H ⊆ F(H)
  • and let c ∈ C.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-64
SLIDE 64

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Theorem. Cantor-Schr¨
  • der-Bernstein Theorem. Let A and B

be sets so that there is an injective function f : A → B and an injective function g : B → A. Then there is a bijective function h : A → B.

  • Proof. Define F(X) := A\g
  • B\f[X]
  • for all X ⊆ A. Then

X ⊆ Y implies f[X] ⊆ f[Y], B\f[X] ⊇ B\f[Y], g

  • B\f[X]
  • ⊇ g
  • B\f[Y]
  • ,

F(X) = A\g

  • B\f[X]
  • ⊆ A\g
  • B\f[Y]
  • = F(Y).

Let C :=

  • H ∈ P(A) : H ⊆ F(H)
  • and let c ∈ C. Then there

is an H ∈ P(A) with c ∈ H ⊆ F(H) ⊆ F(C).

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-65
SLIDE 65

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Theorem. Cantor-Schr¨
  • der-Bernstein Theorem. Let A and B

be sets so that there is an injective function f : A → B and an injective function g : B → A. Then there is a bijective function h : A → B.

  • Proof. Define F(X) := A\g
  • B\f[X]
  • for all X ⊆ A. Then

X ⊆ Y implies f[X] ⊆ f[Y], B\f[X] ⊇ B\f[Y], g

  • B\f[X]
  • ⊇ g
  • B\f[Y]
  • ,

F(X) = A\g

  • B\f[X]
  • ⊆ A\g
  • B\f[Y]
  • = F(Y).

Let C :=

  • H ∈ P(A) : H ⊆ F(H)
  • and let c ∈ C. Then there

is an H ∈ P(A) with c ∈ H ⊆ F(H) ⊆ F(C). Hence C ⊆ F(C).

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-66
SLIDE 66

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Theorem. Cantor-Schr¨
  • der-Bernstein Theorem. Let A and B

be sets so that there is an injective function f : A → B and an injective function g : B → A. Then there is a bijective function h : A → B.

  • Proof. Define F(X) := A\g
  • B\f[X]
  • for all X ⊆ A. Then

X ⊆ Y implies f[X] ⊆ f[Y], B\f[X] ⊇ B\f[Y], g

  • B\f[X]
  • ⊇ g
  • B\f[Y]
  • ,

F(X) = A\g

  • B\f[X]
  • ⊆ A\g
  • B\f[Y]
  • = F(Y).

Let C :=

  • H ∈ P(A) : H ⊆ F(H)
  • and let c ∈ C. Then there

is an H ∈ P(A) with c ∈ H ⊆ F(H) ⊆ F(C). Hence C ⊆ F(C). Then F(C) ⊆ F

  • F(C)
  • .

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-67
SLIDE 67

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Theorem. Cantor-Schr¨
  • der-Bernstein Theorem. Let A and B

be sets so that there is an injective function f : A → B and an injective function g : B → A. Then there is a bijective function h : A → B.

  • Proof. Define F(X) := A\g
  • B\f[X]
  • for all X ⊆ A. Then

X ⊆ Y implies f[X] ⊆ f[Y], B\f[X] ⊇ B\f[Y], g

  • B\f[X]
  • ⊇ g
  • B\f[Y]
  • ,

F(X) = A\g

  • B\f[X]
  • ⊆ A\g
  • B\f[Y]
  • = F(Y).

Let C :=

  • H ∈ P(A) : H ⊆ F(H)
  • and let c ∈ C. Then there

is an H ∈ P(A) with c ∈ H ⊆ F(H) ⊆ F(C). Hence C ⊆ F(C). Then F(C) ⊆ F

  • F(C)
  • . By definition of C, F(C) ⊆ C.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-68
SLIDE 68

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Theorem. Cantor-Schr¨
  • der-Bernstein Theorem. Let A and B

be sets so that there is an injective function f : A → B and an injective function g : B → A. Then there is a bijective function h : A → B.

  • Proof. Define F(X) := A\g
  • B\f[X]
  • for all X ⊆ A. Then

X ⊆ Y implies f[X] ⊆ f[Y], B\f[X] ⊇ B\f[Y], g

  • B\f[X]
  • ⊇ g
  • B\f[Y]
  • ,

F(X) = A\g

  • B\f[X]
  • ⊆ A\g
  • B\f[Y]
  • = F(Y).

Let C :=

  • H ∈ P(A) : H ⊆ F(H)
  • and let c ∈ C. Then there

is an H ∈ P(A) with c ∈ H ⊆ F(H) ⊆ F(C). Hence C ⊆ F(C). Then F(C) ⊆ F

  • F(C)
  • . By definition of C, F(C) ⊆ C. Thus

C = F(C).

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-69
SLIDE 69

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

Proof (concl.).

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-70
SLIDE 70

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

Proof (concl.). C

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-71
SLIDE 71

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

Proof (concl.). C = F(C)

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-72
SLIDE 72

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

Proof (concl.). C = F(C) = A\g

  • B\f[C]
  • Bernd Schr¨
  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-73
SLIDE 73

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

Proof (concl.). C = F(C) = A\g

  • B\f[C]
  • implies

g

  • B\f[C]
  • = A\C

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-74
SLIDE 74

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

Proof (concl.). C = F(C) = A\g

  • B\f[C]
  • implies

g

  • B\f[C]
  • = A\C and then B\f[C] = g−1[A\C].

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-75
SLIDE 75

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

Proof (concl.). C = F(C) = A\g

  • B\f[C]
  • implies

g

  • B\f[C]
  • = A\C and then B\f[C] = g−1[A\C]. Hence

g−1

  • A\C is bijective from A\C onto B\f[C].

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-76
SLIDE 76

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

Proof (concl.). C = F(C) = A\g

  • B\f[C]
  • implies

g

  • B\f[C]
  • = A\C and then B\f[C] = g−1[A\C]. Hence

g−1

  • A\C is bijective from A\C onto B\f[C]. Define h : A → B

by h|C := f|C and h|A\C := g−1

  • A\C.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-77
SLIDE 77

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

Proof (concl.). C = F(C) = A\g

  • B\f[C]
  • implies

g

  • B\f[C]
  • = A\C and then B\f[C] = g−1[A\C]. Hence

g−1

  • A\C is bijective from A\C onto B\f[C]. Define h : A → B

by h|C := f|C and h|A\C := g−1

  • A\C. Then h|C : C → f[C] and

h|A\C : A\C → B\f[C] are bijective.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-78
SLIDE 78

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

Proof (concl.). C = F(C) = A\g

  • B\f[C]
  • implies

g

  • B\f[C]
  • = A\C and then B\f[C] = g−1[A\C]. Hence

g−1

  • A\C is bijective from A\C onto B\f[C]. Define h : A → B

by h|C := f|C and h|A\C := g−1

  • A\C. Then h|C : C → f[C] and

h|A\C : A\C → B\f[C] are bijective. So h is surjective.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-79
SLIDE 79

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

Proof (concl.). C = F(C) = A\g

  • B\f[C]
  • implies

g

  • B\f[C]
  • = A\C and then B\f[C] = g−1[A\C]. Hence

g−1

  • A\C is bijective from A\C onto B\f[C]. Define h : A → B

by h|C := f|C and h|A\C := g−1

  • A\C. Then h|C : C → f[C] and

h|A\C : A\C → B\f[C] are bijective. So h is surjective. To prove that h is injective, let x,y ∈ A be so that x = y.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-80
SLIDE 80

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

Proof (concl.). C = F(C) = A\g

  • B\f[C]
  • implies

g

  • B\f[C]
  • = A\C and then B\f[C] = g−1[A\C]. Hence

g−1

  • A\C is bijective from A\C onto B\f[C]. Define h : A → B

by h|C := f|C and h|A\C := g−1

  • A\C. Then h|C : C → f[C] and

h|A\C : A\C → B\f[C] are bijective. So h is surjective. To prove that h is injective, let x,y ∈ A be so that x = y. If x,y ∈ C, then h(x)

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-81
SLIDE 81

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

Proof (concl.). C = F(C) = A\g

  • B\f[C]
  • implies

g

  • B\f[C]
  • = A\C and then B\f[C] = g−1[A\C]. Hence

g−1

  • A\C is bijective from A\C onto B\f[C]. Define h : A → B

by h|C := f|C and h|A\C := g−1

  • A\C. Then h|C : C → f[C] and

h|A\C : A\C → B\f[C] are bijective. So h is surjective. To prove that h is injective, let x,y ∈ A be so that x = y. If x,y ∈ C, then h(x) = f(x)

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-82
SLIDE 82

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

Proof (concl.). C = F(C) = A\g

  • B\f[C]
  • implies

g

  • B\f[C]
  • = A\C and then B\f[C] = g−1[A\C]. Hence

g−1

  • A\C is bijective from A\C onto B\f[C]. Define h : A → B

by h|C := f|C and h|A\C := g−1

  • A\C. Then h|C : C → f[C] and

h|A\C : A\C → B\f[C] are bijective. So h is surjective. To prove that h is injective, let x,y ∈ A be so that x = y. If x,y ∈ C, then h(x) = f(x) =

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-83
SLIDE 83

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

Proof (concl.). C = F(C) = A\g

  • B\f[C]
  • implies

g

  • B\f[C]
  • = A\C and then B\f[C] = g−1[A\C]. Hence

g−1

  • A\C is bijective from A\C onto B\f[C]. Define h : A → B

by h|C := f|C and h|A\C := g−1

  • A\C. Then h|C : C → f[C] and

h|A\C : A\C → B\f[C] are bijective. So h is surjective. To prove that h is injective, let x,y ∈ A be so that x = y. If x,y ∈ C, then h(x) = f(x) = f(y)

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-84
SLIDE 84

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

Proof (concl.). C = F(C) = A\g

  • B\f[C]
  • implies

g

  • B\f[C]
  • = A\C and then B\f[C] = g−1[A\C]. Hence

g−1

  • A\C is bijective from A\C onto B\f[C]. Define h : A → B

by h|C := f|C and h|A\C := g−1

  • A\C. Then h|C : C → f[C] and

h|A\C : A\C → B\f[C] are bijective. So h is surjective. To prove that h is injective, let x,y ∈ A be so that x = y. If x,y ∈ C, then h(x) = f(x) = f(y) = h(y).

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-85
SLIDE 85

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

Proof (concl.). C = F(C) = A\g

  • B\f[C]
  • implies

g

  • B\f[C]
  • = A\C and then B\f[C] = g−1[A\C]. Hence

g−1

  • A\C is bijective from A\C onto B\f[C]. Define h : A → B

by h|C := f|C and h|A\C := g−1

  • A\C. Then h|C : C → f[C] and

h|A\C : A\C → B\f[C] are bijective. So h is surjective. To prove that h is injective, let x,y ∈ A be so that x = y. If x,y ∈ C, then h(x) = f(x) = f(y) = h(y). If x,y ∈ C, then h(x)

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-86
SLIDE 86

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

Proof (concl.). C = F(C) = A\g

  • B\f[C]
  • implies

g

  • B\f[C]
  • = A\C and then B\f[C] = g−1[A\C]. Hence

g−1

  • A\C is bijective from A\C onto B\f[C]. Define h : A → B

by h|C := f|C and h|A\C := g−1

  • A\C. Then h|C : C → f[C] and

h|A\C : A\C → B\f[C] are bijective. So h is surjective. To prove that h is injective, let x,y ∈ A be so that x = y. If x,y ∈ C, then h(x) = f(x) = f(y) = h(y). If x,y ∈ C, then h(x) = g−1(x)

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-87
SLIDE 87

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

Proof (concl.). C = F(C) = A\g

  • B\f[C]
  • implies

g

  • B\f[C]
  • = A\C and then B\f[C] = g−1[A\C]. Hence

g−1

  • A\C is bijective from A\C onto B\f[C]. Define h : A → B

by h|C := f|C and h|A\C := g−1

  • A\C. Then h|C : C → f[C] and

h|A\C : A\C → B\f[C] are bijective. So h is surjective. To prove that h is injective, let x,y ∈ A be so that x = y. If x,y ∈ C, then h(x) = f(x) = f(y) = h(y). If x,y ∈ C, then h(x) = g−1(x) = g−1(y)

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-88
SLIDE 88

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

Proof (concl.). C = F(C) = A\g

  • B\f[C]
  • implies

g

  • B\f[C]
  • = A\C and then B\f[C] = g−1[A\C]. Hence

g−1

  • A\C is bijective from A\C onto B\f[C]. Define h : A → B

by h|C := f|C and h|A\C := g−1

  • A\C. Then h|C : C → f[C] and

h|A\C : A\C → B\f[C] are bijective. So h is surjective. To prove that h is injective, let x,y ∈ A be so that x = y. If x,y ∈ C, then h(x) = f(x) = f(y) = h(y). If x,y ∈ C, then h(x) = g−1(x) = g−1(y) = h(y).

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-89
SLIDE 89

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

Proof (concl.). C = F(C) = A\g

  • B\f[C]
  • implies

g

  • B\f[C]
  • = A\C and then B\f[C] = g−1[A\C]. Hence

g−1

  • A\C is bijective from A\C onto B\f[C]. Define h : A → B

by h|C := f|C and h|A\C := g−1

  • A\C. Then h|C : C → f[C] and

h|A\C : A\C → B\f[C] are bijective. So h is surjective. To prove that h is injective, let x,y ∈ A be so that x = y. If x,y ∈ C, then h(x) = f(x) = f(y) = h(y). If x,y ∈ C, then h(x) = g−1(x) = g−1(y) = h(y). Otherwise, WLOG x ∈ C and y ∈ C.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-90
SLIDE 90

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

Proof (concl.). C = F(C) = A\g

  • B\f[C]
  • implies

g

  • B\f[C]
  • = A\C and then B\f[C] = g−1[A\C]. Hence

g−1

  • A\C is bijective from A\C onto B\f[C]. Define h : A → B

by h|C := f|C and h|A\C := g−1

  • A\C. Then h|C : C → f[C] and

h|A\C : A\C → B\f[C] are bijective. So h is surjective. To prove that h is injective, let x,y ∈ A be so that x = y. If x,y ∈ C, then h(x) = f(x) = f(y) = h(y). If x,y ∈ C, then h(x) = g−1(x) = g−1(y) = h(y). Otherwise, WLOG x ∈ C and y ∈ C. Then h(x) ∈ f[C] and h(y) ∈ B\f[C].

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-91
SLIDE 91

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

Proof (concl.). C = F(C) = A\g

  • B\f[C]
  • implies

g

  • B\f[C]
  • = A\C and then B\f[C] = g−1[A\C]. Hence

g−1

  • A\C is bijective from A\C onto B\f[C]. Define h : A → B

by h|C := f|C and h|A\C := g−1

  • A\C. Then h|C : C → f[C] and

h|A\C : A\C → B\f[C] are bijective. So h is surjective. To prove that h is injective, let x,y ∈ A be so that x = y. If x,y ∈ C, then h(x) = f(x) = f(y) = h(y). If x,y ∈ C, then h(x) = g−1(x) = g−1(y) = h(y). Otherwise, WLOG x ∈ C and y ∈ C. Then h(x) ∈ f[C] and h(y) ∈ B\f[C]. So h(x) = h(y).

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-92
SLIDE 92

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

Proof (concl.). C = F(C) = A\g

  • B\f[C]
  • implies

g

  • B\f[C]
  • = A\C and then B\f[C] = g−1[A\C]. Hence

g−1

  • A\C is bijective from A\C onto B\f[C]. Define h : A → B

by h|C := f|C and h|A\C := g−1

  • A\C. Then h|C : C → f[C] and

h|A\C : A\C → B\f[C] are bijective. So h is surjective. To prove that h is injective, let x,y ∈ A be so that x = y. If x,y ∈ C, then h(x) = f(x) = f(y) = h(y). If x,y ∈ C, then h(x) = g−1(x) = g−1(y) = h(y). Otherwise, WLOG x ∈ C and y ∈ C. Then h(x) ∈ f[C] and h(y) ∈ B\f[C]. So h(x) = h(y).

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-93
SLIDE 93

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

Theorem.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-94
SLIDE 94

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Theorem. Let A be an infinite set.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-95
SLIDE 95

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Theorem. Let A be an infinite set. Then A×A is equivalent to

A.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-96
SLIDE 96

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Theorem. Let A be an infinite set. Then A×A is equivalent to

A. Sketch of proof.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-97
SLIDE 97

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Theorem. Let A be an infinite set. Then A×A is equivalent to

A. Sketch of proof. Apply Zorn’s Lemma to the set F of all bijective functions f : X ×X → X, where X ⊆ A.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-98
SLIDE 98

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Theorem. Let A be an infinite set. Then A×A is equivalent to

A. Sketch of proof. Apply Zorn’s Lemma to the set F of all bijective functions f : X ×X → X, where X ⊆ A. To prove that a maximal element f : Y ×Y → Y of F must be a bijective function from A×A to A

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-99
SLIDE 99

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Theorem. Let A be an infinite set. Then A×A is equivalent to

A. Sketch of proof. Apply Zorn’s Lemma to the set F of all bijective functions f : X ×X → X, where X ⊆ A. To prove that a maximal element f : Y ×Y → Y of F must be a bijective function from A×A to A, assume that Y is not equivalent to A.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-100
SLIDE 100

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Theorem. Let A be an infinite set. Then A×A is equivalent to

A. Sketch of proof. Apply Zorn’s Lemma to the set F of all bijective functions f : X ×X → X, where X ⊆ A. To prove that a maximal element f : Y ×Y → Y of F must be a bijective function from A×A to A, assume that Y is not equivalent to A. There must be an injective function from Y to A\Y.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-101
SLIDE 101

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Theorem. Let A be an infinite set. Then A×A is equivalent to

A. Sketch of proof. Apply Zorn’s Lemma to the set F of all bijective functions f : X ×X → X, where X ⊆ A. To prove that a maximal element f : Y ×Y → Y of F must be a bijective function from A×A to A, assume that Y is not equivalent to A. There must be an injective function from Y to A\Y. Let Z ⊆ A\Y be equivalent to Y and expand f to a function from (Y ∪Z)×(Y ∪Z) to Y ∪Z

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-102
SLIDE 102

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Theorem. Let A be an infinite set. Then A×A is equivalent to

A. Sketch of proof. Apply Zorn’s Lemma to the set F of all bijective functions f : X ×X → X, where X ⊆ A. To prove that a maximal element f : Y ×Y → Y of F must be a bijective function from A×A to A, assume that Y is not equivalent to A. There must be an injective function from Y to A\Y. Let Z ⊆ A\Y be equivalent to Y and expand f to a function from (Y ∪Z)×(Y ∪Z) to Y ∪Z, using that Y ×Z ∪Z ×Z ∪Z ×Y is equivalent to Z.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-103
SLIDE 103

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Theorem. Let A be an infinite set. Then A×A is equivalent to

A. Sketch of proof. Apply Zorn’s Lemma to the set F of all bijective functions f : X ×X → X, where X ⊆ A. To prove that a maximal element f : Y ×Y → Y of F must be a bijective function from A×A to A, assume that Y is not equivalent to A. There must be an injective function from Y to A\Y. Let Z ⊆ A\Y be equivalent to Y and expand f to a function from (Y ∪Z)×(Y ∪Z) to Y ∪Z, using that Y ×Z ∪Z ×Z ∪Z ×Y is equivalent to Z.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-104
SLIDE 104

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

Definition.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-105
SLIDE 105

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Definition. Cardinal arithmetic.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-106
SLIDE 106

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Definition. Cardinal arithmetic. Let α and β be cardinal

numbers and let A and B be sets with |A| = α and |B| = β.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-107
SLIDE 107

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Definition. Cardinal arithmetic. Let α and β be cardinal

numbers and let A and B be sets with |A| = α and |B| = β.

  • 1. α +β :=
  • A×{0}∪B×{1}
  • Bernd Schr¨
  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-108
SLIDE 108

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Definition. Cardinal arithmetic. Let α and β be cardinal

numbers and let A and B be sets with |A| = α and |B| = β.

  • 1. α +β :=
  • A×{0}∪B×{1}
  • 2. αβ := |A×B|

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-109
SLIDE 109

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Definition. Cardinal arithmetic. Let α and β be cardinal

numbers and let A and B be sets with |A| = α and |B| = β.

  • 1. α +β :=
  • A×{0}∪B×{1}
  • 2. αβ := |A×B|
  • 3. αβ :=
  • AB

, where AB := {f : f is a function from A to B}.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-110
SLIDE 110

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Definition. Cardinal arithmetic. Let α and β be cardinal

numbers and let A and B be sets with |A| = α and |B| = β.

  • 1. α +β :=
  • A×{0}∪B×{1}
  • 2. αβ := |A×B|
  • 3. αβ :=
  • AB

, where AB := {f : f is a function from A to B}. Theorem.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-111
SLIDE 111

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Definition. Cardinal arithmetic. Let α and β be cardinal

numbers and let A and B be sets with |A| = α and |B| = β.

  • 1. α +β :=
  • A×{0}∪B×{1}
  • 2. αβ := |A×B|
  • 3. αβ :=
  • AB

, where AB := {f : f is a function from A to B}.

  • Theorem. Let α,β be cardinal numbers so that one of α and β

is infinite.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-112
SLIDE 112

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Definition. Cardinal arithmetic. Let α and β be cardinal

numbers and let A and B be sets with |A| = α and |B| = β.

  • 1. α +β :=
  • A×{0}∪B×{1}
  • 2. αβ := |A×B|
  • 3. αβ :=
  • AB

, where AB := {f : f is a function from A to B}.

  • Theorem. Let α,β be cardinal numbers so that one of α and β

is infinite. Then α +β = max{α,β}

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-113
SLIDE 113

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Definition. Cardinal arithmetic. Let α and β be cardinal

numbers and let A and B be sets with |A| = α and |B| = β.

  • 1. α +β :=
  • A×{0}∪B×{1}
  • 2. αβ := |A×B|
  • 3. αβ :=
  • AB

, where AB := {f : f is a function from A to B}.

  • Theorem. Let α,β be cardinal numbers so that one of α and β

is infinite. Then α +β = max{α,β} and αβ = max{α,β}.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-114
SLIDE 114

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Definition. Cardinal arithmetic. Let α and β be cardinal

numbers and let A and B be sets with |A| = α and |B| = β.

  • 1. α +β :=
  • A×{0}∪B×{1}
  • 2. αβ := |A×B|
  • 3. αβ :=
  • AB

, where AB := {f : f is a function from A to B}.

  • Theorem. Let α,β be cardinal numbers so that one of α and β

is infinite. Then α +β = max{α,β} and αβ = max{α,β}. Proof.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-115
SLIDE 115

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Definition. Cardinal arithmetic. Let α and β be cardinal

numbers and let A and B be sets with |A| = α and |B| = β.

  • 1. α +β :=
  • A×{0}∪B×{1}
  • 2. αβ := |A×B|
  • 3. αβ :=
  • AB

, where AB := {f : f is a function from A to B}.

  • Theorem. Let α,β be cardinal numbers so that one of α and β

is infinite. Then α +β = max{α,β} and αβ = max{α,β}.

  • Proof. Good exercise.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-116
SLIDE 116

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Definition. Cardinal arithmetic. Let α and β be cardinal

numbers and let A and B be sets with |A| = α and |B| = β.

  • 1. α +β :=
  • A×{0}∪B×{1}
  • 2. αβ := |A×B|
  • 3. αβ :=
  • AB

, where AB := {f : f is a function from A to B}.

  • Theorem. Let α,β be cardinal numbers so that one of α and β

is infinite. Then α +β = max{α,β} and αβ = max{α,β}.

  • Proof. Good exercise.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-117
SLIDE 117

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

Theorem.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-118
SLIDE 118

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Theorem. Let α,β and γ be cardinal numbers.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-119
SLIDE 119

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Theorem. Let α,β and γ be cardinal numbers. Then
  • 1. αβαγ = αβ+γ

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-120
SLIDE 120

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Theorem. Let α,β and γ be cardinal numbers. Then
  • 1. αβαγ = αβ+γ
  • 2. αγβ γ = (αβ)γ

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-121
SLIDE 121

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Theorem. Let α,β and γ be cardinal numbers. Then
  • 1. αβαγ = αβ+γ
  • 2. αγβ γ = (αβ)γ

3.

  • αβγ

= αβγ

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-122
SLIDE 122

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Theorem. Let α,β and γ be cardinal numbers. Then
  • 1. αβαγ = αβ+γ
  • 2. αγβ γ = (αβ)γ

3.

  • αβγ

= αβγ Proof (part 1 only).

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-123
SLIDE 123

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Theorem. Let α,β and γ be cardinal numbers. Then
  • 1. αβαγ = αβ+γ
  • 2. αγβ γ = (αβ)γ

3.

  • αβγ

= αβγ Proof (part 1 only). Let A,B,C be disjoint sets with |A| = α, |B| = β and |C| = γ.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-124
SLIDE 124

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Theorem. Let α,β and γ be cardinal numbers. Then
  • 1. αβαγ = αβ+γ
  • 2. αγβ γ = (αβ)γ

3.

  • αβγ

= αβγ Proof (part 1 only). Let A,B,C be disjoint sets with |A| = α, |B| = β and |C| = γ. Then AB∪C is equivalent to the set AB ×AC

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-125
SLIDE 125

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Theorem. Let α,β and γ be cardinal numbers. Then
  • 1. αβαγ = αβ+γ
  • 2. αγβ γ = (αβ)γ

3.

  • αβγ

= αβγ Proof (part 1 only). Let A,B,C be disjoint sets with |A| = α, |B| = β and |C| = γ. Then AB∪C is equivalent to the set AB ×AC via f ∈ AB∪C

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-126
SLIDE 126

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Theorem. Let α,β and γ be cardinal numbers. Then
  • 1. αβαγ = αβ+γ
  • 2. αγβ γ = (αβ)γ

3.

  • αβγ

= αβγ Proof (part 1 only). Let A,B,C be disjoint sets with |A| = α, |B| = β and |C| = γ. Then AB∪C is equivalent to the set AB ×AC via f ∈ AB∪C → (f|B,f|C).

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-127
SLIDE 127

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Theorem. Let α,β and γ be cardinal numbers. Then
  • 1. αβαγ = αβ+γ
  • 2. αγβ γ = (αβ)γ

3.

  • αβγ

= αβγ Proof (part 1 only). Let A,B,C be disjoint sets with |A| = α, |B| = β and |C| = γ. Then AB∪C is equivalent to the set AB ×AC via f ∈ AB∪C → (f|B,f|C). Hence αβ+γ

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-128
SLIDE 128

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Theorem. Let α,β and γ be cardinal numbers. Then
  • 1. αβαγ = αβ+γ
  • 2. αγβ γ = (αβ)γ

3.

  • αβγ

= αβγ Proof (part 1 only). Let A,B,C be disjoint sets with |A| = α, |B| = β and |C| = γ. Then AB∪C is equivalent to the set AB ×AC via f ∈ AB∪C → (f|B,f|C). Hence αβ+γ =

  • AB∪C
  • Bernd Schr¨
  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-129
SLIDE 129

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Theorem. Let α,β and γ be cardinal numbers. Then
  • 1. αβαγ = αβ+γ
  • 2. αγβ γ = (αβ)γ

3.

  • αβγ

= αβγ Proof (part 1 only). Let A,B,C be disjoint sets with |A| = α, |B| = β and |C| = γ. Then AB∪C is equivalent to the set AB ×AC via f ∈ AB∪C → (f|B,f|C). Hence αβ+γ =

  • AB∪C

=

  • AB ×AC
  • Bernd Schr¨
  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-130
SLIDE 130

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Theorem. Let α,β and γ be cardinal numbers. Then
  • 1. αβαγ = αβ+γ
  • 2. αγβ γ = (αβ)γ

3.

  • αβγ

= αβγ Proof (part 1 only). Let A,B,C be disjoint sets with |A| = α, |B| = β and |C| = γ. Then AB∪C is equivalent to the set AB ×AC via f ∈ AB∪C → (f|B,f|C). Hence αβ+γ =

  • AB∪C

=

  • AB ×AC

= αβαγ.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-131
SLIDE 131

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

  • Theorem. Let α,β and γ be cardinal numbers. Then
  • 1. αβαγ = αβ+γ
  • 2. αγβ γ = (αβ)γ

3.

  • αβγ

= αβγ Proof (part 1 only). Let A,B,C be disjoint sets with |A| = α, |B| = β and |C| = γ. Then AB∪C is equivalent to the set AB ×AC via f ∈ AB∪C → (f|B,f|C). Hence αβ+γ =

  • AB∪C

=

  • AB ×AC

= αβαγ.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-132
SLIDE 132

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

A Mysterious Gap

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-133
SLIDE 133

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

A Mysterious Gap

ℵ0 denotes the first infinite cardinal number.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-134
SLIDE 134

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

A Mysterious Gap

ℵ0 denotes the first infinite cardinal number. That is, ℵ0 = ω = N0.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-135
SLIDE 135

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

A Mysterious Gap

ℵ0 denotes the first infinite cardinal number. That is, ℵ0 = ω = N0. We know that 2ℵ0 is not countable.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-136
SLIDE 136

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

A Mysterious Gap

ℵ0 denotes the first infinite cardinal number. That is, ℵ0 = ω = N0. We know that 2ℵ0 is not countable. But is it equal to the first uncountable ordinal ℵ1?

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-137
SLIDE 137

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

A Mysterious Gap

ℵ0 denotes the first infinite cardinal number. That is, ℵ0 = ω = N0. We know that 2ℵ0 is not countable. But is it equal to the first uncountable ordinal ℵ1? Axiom.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-138
SLIDE 138

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

A Mysterious Gap

ℵ0 denotes the first infinite cardinal number. That is, ℵ0 = ω = N0. We know that 2ℵ0 is not countable. But is it equal to the first uncountable ordinal ℵ1?

  • Axiom. The Continuum Hypothesis.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-139
SLIDE 139

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

A Mysterious Gap

ℵ0 denotes the first infinite cardinal number. That is, ℵ0 = ω = N0. We know that 2ℵ0 is not countable. But is it equal to the first uncountable ordinal ℵ1?

  • Axiom. The Continuum Hypothesis. 2ℵ0 = ℵ1.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

slide-140
SLIDE 140

logo1 Finite Sizes Infinite Sizes Cardinal Arithmetic

A Mysterious Gap

ℵ0 denotes the first infinite cardinal number. That is, ℵ0 = ω = N0. We know that 2ℵ0 is not countable. But is it equal to the first uncountable ordinal ℵ1?

  • Axiom. The Continuum Hypothesis. 2ℵ0 = ℵ1.

Bernd Schr¨

  • der

Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis