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

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 ) � logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

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 | logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

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 | logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

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

Finite Sizes Infinite Sizes Cardinal Arithmetic Example. In the restaurant “Zum Adler” 41 people are dining. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

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. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

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. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

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. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

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. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

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. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

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. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

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? logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

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. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

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. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

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. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

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 | logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

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 | logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

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 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

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 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

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 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

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

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 . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

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. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

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 | of S to be the unique cardinal number α that is equivalent to S. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

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

Finite Sizes Infinite Sizes Cardinal Arithmetic Theorem. Cantor-Schr¨ oder-Bernstein Theorem . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

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

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

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

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

Finite Sizes Infinite Sizes Cardinal Arithmetic Theorem. Cantor-Schr¨ oder-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 ] logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

Finite Sizes Infinite Sizes Cardinal Arithmetic Theorem. Cantor-Schr¨ oder-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 ] logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

Finite Sizes Infinite Sizes Cardinal Arithmetic Theorem. Cantor-Schr¨ oder-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 ] , � � � � B \ f [ X ] ⊇ g B \ f [ Y ] g logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

Finite Sizes Infinite Sizes Cardinal Arithmetic Theorem. Cantor-Schr¨ oder-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 ] , � � � � B \ f [ X ] ⊇ g B \ f [ Y ] g , F ( X ) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

Finite Sizes Infinite Sizes Cardinal Arithmetic Theorem. Cantor-Schr¨ oder-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 ] , � � � � B \ f [ X ] ⊇ g B \ f [ Y ] g , � � F ( X ) = A \ g B \ f [ X ] logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

Finite Sizes Infinite Sizes Cardinal Arithmetic Theorem. Cantor-Schr¨ oder-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 ] , � � � � B \ f [ X ] ⊇ g B \ f [ Y ] g , � � � � F ( X ) = A \ g B \ f [ X ] ⊆ A \ g B \ f [ Y ] logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

Finite Sizes Infinite Sizes Cardinal Arithmetic Theorem. Cantor-Schr¨ oder-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 ] , � � � � B \ f [ X ] ⊇ g B \ f [ Y ] g , � � � � F ( X ) = A \ g B \ f [ X ] ⊆ A \ g B \ f [ Y ] = F ( Y ) . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

Finite Sizes Infinite Sizes Cardinal Arithmetic Theorem. Cantor-Schr¨ oder-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 ] , � � � � B \ f [ X ] ⊇ g B \ f [ Y ] g , � � � � F ( X ) = A \ g B \ f [ X ] ⊆ A \ g B \ f [ Y ] = F ( Y ) . � � � Let C : = H ∈ P ( A ) : H ⊆ F ( H ) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

Finite Sizes Infinite Sizes Cardinal Arithmetic Theorem. Cantor-Schr¨ oder-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 ] , � � � � B \ f [ X ] ⊇ g B \ f [ Y ] g , � � � � 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 . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

Finite Sizes Infinite Sizes Cardinal Arithmetic Theorem. Cantor-Schr¨ oder-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 ] , � � � � B \ f [ X ] ⊇ g B \ f [ Y ] g , � � � � 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 ) . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

Finite Sizes Infinite Sizes Cardinal Arithmetic Theorem. Cantor-Schr¨ oder-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 ] , � � � � B \ f [ X ] ⊇ g B \ f [ Y ] g , � � � � 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 ) . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

Finite Sizes Infinite Sizes Cardinal Arithmetic Theorem. Cantor-Schr¨ oder-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 ] , � � � � B \ f [ X ] ⊇ g B \ f [ Y ] g , � � � � 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 ) . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

Finite Sizes Infinite Sizes Cardinal Arithmetic Theorem. Cantor-Schr¨ oder-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 ] , � � � � B \ f [ X ] ⊇ g B \ f [ Y ] g , � � � � 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 . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

Finite Sizes Infinite Sizes Cardinal Arithmetic Theorem. Cantor-Schr¨ oder-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 ] , � � � � B \ f [ X ] ⊇ g B \ f [ Y ] g , � � � � 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 ) . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

Finite Sizes Infinite Sizes Cardinal Arithmetic Proof (concl.). logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

Finite Sizes Infinite Sizes Cardinal Arithmetic Proof (concl.). C logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

Finite Sizes Infinite Sizes Cardinal Arithmetic Proof (concl.). C = F ( C ) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

Finite Sizes Infinite Sizes Cardinal Arithmetic � � Proof (concl.). C = F ( C ) = A \ g B \ f [ C ] logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

Finite Sizes Infinite Sizes Cardinal Arithmetic � � Proof (concl.). C = F ( C ) = A \ g B \ f [ C ] implies � � B \ f [ C ] = A \ C g logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

Finite Sizes Infinite Sizes Cardinal Arithmetic � � Proof (concl.). C = F ( C ) = A \ g B \ f [ C ] implies = A \ C and then B \ f [ C ] = g − 1 [ A \ C ] . � � B \ f [ C ] g logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

Finite Sizes Infinite Sizes Cardinal Arithmetic � � Proof (concl.). C = F ( C ) = A \ g B \ f [ C ] implies = A \ C and then B \ f [ C ] = g − 1 [ A \ C ] . Hence � � B \ f [ C ] g g − 1 � A \ C is bijective from A \ C onto B \ f [ C ] . � logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

Finite Sizes Infinite Sizes Cardinal Arithmetic � � Proof (concl.). C = F ( C ) = A \ g B \ f [ C ] implies = A \ C and then B \ f [ C ] = g − 1 [ A \ C ] . Hence � � B \ f [ C ] g 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 . � logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

Finite Sizes Infinite Sizes Cardinal Arithmetic � � Proof (concl.). C = F ( C ) = A \ g B \ f [ C ] implies = A \ C and then B \ f [ C ] = g − 1 [ A \ C ] . Hence � � B \ f [ C ] g 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. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

Finite Sizes Infinite Sizes Cardinal Arithmetic � � Proof (concl.). C = F ( C ) = A \ g B \ f [ C ] implies = A \ C and then B \ f [ C ] = g − 1 [ A \ C ] . Hence � � B \ f [ C ] g 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. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

Finite Sizes Infinite Sizes Cardinal Arithmetic � � Proof (concl.). C = F ( C ) = A \ g B \ f [ C ] implies = A \ C and then B \ f [ C ] = g − 1 [ A \ C ] . Hence � � B \ f [ C ] g 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 . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

Finite Sizes Infinite Sizes Cardinal Arithmetic � � Proof (concl.). C = F ( C ) = A \ g B \ f [ C ] implies = A \ C and then B \ f [ C ] = g − 1 [ A \ C ] . Hence � � B \ f [ C ] g 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 ) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

Finite Sizes Infinite Sizes Cardinal Arithmetic � � Proof (concl.). C = F ( C ) = A \ g B \ f [ C ] implies = A \ C and then B \ f [ C ] = g − 1 [ A \ C ] . Hence � � B \ f [ C ] g 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 ) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

Finite Sizes Infinite Sizes Cardinal Arithmetic � � Proof (concl.). C = F ( C ) = A \ g B \ f [ C ] implies = A \ C and then B \ f [ C ] = g − 1 [ A \ C ] . Hence � � B \ f [ C ] g 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 ) � = logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

Finite Sizes Infinite Sizes Cardinal Arithmetic � � Proof (concl.). C = F ( C ) = A \ g B \ f [ C ] implies = A \ C and then B \ f [ C ] = g − 1 [ A \ C ] . Hence � � B \ f [ C ] g 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 ) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

Finite Sizes Infinite Sizes Cardinal Arithmetic � � Proof (concl.). C = F ( C ) = A \ g B \ f [ C ] implies = A \ C and then B \ f [ C ] = g − 1 [ A \ C ] . Hence � � B \ f [ C ] g 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 ) . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

Finite Sizes Infinite Sizes Cardinal Arithmetic � � Proof (concl.). C = F ( C ) = A \ g B \ f [ C ] implies = A \ C and then B \ f [ C ] = g − 1 [ A \ C ] . Hence � � B \ f [ C ] g 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 ) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

Finite Sizes Infinite Sizes Cardinal Arithmetic � � Proof (concl.). C = F ( C ) = A \ g B \ f [ C ] implies = A \ C and then B \ f [ C ] = g − 1 [ A \ C ] . Hence � � B \ f [ C ] g 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 ) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

Finite Sizes Infinite Sizes Cardinal Arithmetic � � Proof (concl.). C = F ( C ) = A \ g B \ f [ C ] implies = A \ C and then B \ f [ C ] = g − 1 [ A \ C ] . Hence � � B \ f [ C ] g 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 ) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

Finite Sizes Infinite Sizes Cardinal Arithmetic � � Proof (concl.). C = F ( C ) = A \ g B \ f [ C ] implies = A \ C and then B \ f [ C ] = g − 1 [ A \ C ] . Hence � � B \ f [ C ] g 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 ) . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

Finite Sizes Infinite Sizes Cardinal Arithmetic � � Proof (concl.). C = F ( C ) = A \ g B \ f [ C ] implies = A \ C and then B \ f [ C ] = g − 1 [ A \ C ] . Hence � � B \ f [ C ] g 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 . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

Finite Sizes Infinite Sizes Cardinal Arithmetic � � Proof (concl.). C = F ( C ) = A \ g B \ f [ C ] implies = A \ C and then B \ f [ C ] = g − 1 [ A \ C ] . Hence � � B \ f [ C ] g 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 ] . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

Finite Sizes Infinite Sizes Cardinal Arithmetic � � Proof (concl.). C = F ( C ) = A \ g B \ f [ C ] implies = A \ C and then B \ f [ C ] = g − 1 [ A \ C ] . Hence � � B \ f [ C ] g 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 ) . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

Finite Sizes Infinite Sizes Cardinal Arithmetic � � Proof (concl.). C = F ( C ) = A \ g B \ f [ C ] implies = A \ C and then B \ f [ C ] = g − 1 [ A \ C ] . Hence � � B \ f [ C ] g 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 ) . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

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

Finite Sizes Infinite Sizes Cardinal Arithmetic Theorem. Let A be an infinite set. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

Finite Sizes Infinite Sizes Cardinal Arithmetic Theorem. Let A be an infinite set. Then A × A is equivalent to A. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

Finite Sizes Infinite Sizes Cardinal Arithmetic Theorem. Let A be an infinite set. Then A × A is equivalent to A. Sketch of proof. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

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 . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

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 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

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 . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis

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 . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Cardinal Numbers and the Continuum Hypothesis