Countable Sets Bernd Schr oder logo1 Bernd Schr oder Louisiana - PowerPoint PPT Presentation

Finite vs. Infinite Countability Examples Theorem. N is equivalent to Z . � n − 1 2 ; if n is odd , Proof. The function f ( n ) : = if n is even , is bijective. − n 2 ; (Good exercise.) Definition. A set C is called countably infinite iff there is a bijective function f : N → C. A set C is called countable iff C is finite or countably infinite. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Theorem. If C is countable and S ⊆ C, then S is countable. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Theorem. If C is countable and S ⊆ C, then S is countable. Proof. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Theorem. If C is countable and S ⊆ C, then S is countable. Proof. WLOG let S be infinite. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Theorem. If C is countable and S ⊆ C, then S is countable. Proof. WLOG let S be infinite. Let f : N → C be bijective and let n 1 : = min f − 1 [ S ] . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Theorem. If C is countable and S ⊆ C, then S is countable. Proof. WLOG let S be infinite. Let f : N → C be bijective and let n 1 : = min f − 1 [ S ] . For k ∈ N , define n k + 1 recursively. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Theorem. If C is countable and S ⊆ C, then S is countable. Proof. WLOG let S be infinite. Let f : N → C be bijective and let n 1 : = min f − 1 [ S ] . For k ∈ N , define n k + 1 recursively. Once � � f − 1 [ S ] \{ n 1 ,..., n k } n 1 ,..., n k are chosen, let n k + 1 : = min . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Theorem. If C is countable and S ⊆ C, then S is countable. Proof. WLOG let S be infinite. Let f : N → C be bijective and let n 1 : = min f − 1 [ S ] . For k ∈ N , define n k + 1 recursively. Once � � f − 1 [ S ] \{ n 1 ,..., n k } n 1 ,..., n k are chosen, let n k + 1 : = min . Define g : N → S by g ( k ) : = f ( n k ) . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Theorem. If C is countable and S ⊆ C, then S is countable. Proof. WLOG let S be infinite. Let f : N → C be bijective and let n 1 : = min f − 1 [ S ] . For k ∈ N , define n k + 1 recursively. Once � � f − 1 [ S ] \{ n 1 ,..., n k } n 1 ,..., n k are chosen, let n k + 1 : = min . Define g : N → S by g ( k ) : = f ( n k ) . Then g is injective logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Theorem. If C is countable and S ⊆ C, then S is countable. Proof. WLOG let S be infinite. Let f : N → C be bijective and let n 1 : = min f − 1 [ S ] . For k ∈ N , define n k + 1 recursively. Once � � f − 1 [ S ] \{ n 1 ,..., n k } n 1 ,..., n k are chosen, let n k + 1 : = min . Define g : N → S by g ( k ) : = f ( n k ) . Then g is injective (and it really maps into S ). logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Theorem. If C is countable and S ⊆ C, then S is countable. Proof. WLOG let S be infinite. Let f : N → C be bijective and let n 1 : = min f − 1 [ S ] . For k ∈ N , define n k + 1 recursively. Once � � f − 1 [ S ] \{ n 1 ,..., n k } n 1 ,..., n k are chosen, let n k + 1 : = min . Define g : N → S by g ( k ) : = f ( n k ) . Then g is injective (and it really maps into S ). Suppose for a contradiction that g is not surjective. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Theorem. If C is countable and S ⊆ C, then S is countable. Proof. WLOG let S be infinite. Let f : N → C be bijective and let n 1 : = min f − 1 [ S ] . For k ∈ N , define n k + 1 recursively. Once � � f − 1 [ S ] \{ n 1 ,..., n k } n 1 ,..., n k are chosen, let n k + 1 : = min . Define g : N → S by g ( k ) : = f ( n k ) . Then g is injective (and it really maps into S ). Suppose for a contradiction that g is not surjective. Let b : = min f − 1 � � S \ g [ N ] logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Theorem. If C is countable and S ⊆ C, then S is countable. Proof. WLOG let S be infinite. Let f : N → C be bijective and let n 1 : = min f − 1 [ S ] . For k ∈ N , define n k + 1 recursively. Once � � f − 1 [ S ] \{ n 1 ,..., n k } n 1 ,..., n k are chosen, let n k + 1 : = min . Define g : N → S by g ( k ) : = f ( n k ) . Then g is injective (and it really maps into S ). Suppose for a contradiction that g is not surjective. Let � � b : = min f − 1 � � f − 1 [ S ] ∩{ 1 ,..., b − 1 } � S \ g [ N ] and let k : = � . � � logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Theorem. If C is countable and S ⊆ C, then S is countable. Proof. WLOG let S be infinite. Let f : N → C be bijective and let n 1 : = min f − 1 [ S ] . For k ∈ N , define n k + 1 recursively. Once � � f − 1 [ S ] \{ n 1 ,..., n k } n 1 ,..., n k are chosen, let n k + 1 : = min . Define g : N → S by g ( k ) : = f ( n k ) . Then g is injective (and it really maps into S ). Suppose for a contradiction that g is not surjective. Let � � b : = min f − 1 � � f − 1 [ S ] ∩{ 1 ,..., b − 1 } � S \ g [ N ] and let k : = � . � � Then f − 1 [ S ] ∩{ 1 ,..., b − 1 } = { n 1 ,..., n k } logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Theorem. If C is countable and S ⊆ C, then S is countable. Proof. WLOG let S be infinite. Let f : N → C be bijective and let n 1 : = min f − 1 [ S ] . For k ∈ N , define n k + 1 recursively. Once � � f − 1 [ S ] \{ n 1 ,..., n k } n 1 ,..., n k are chosen, let n k + 1 : = min . Define g : N → S by g ( k ) : = f ( n k ) . Then g is injective (and it really maps into S ). Suppose for a contradiction that g is not surjective. Let � � b : = min f − 1 � � f − 1 [ S ] ∩{ 1 ,..., b − 1 } � S \ g [ N ] and let k : = � . � � Then f − 1 [ S ] ∩{ 1 ,..., b − 1 } = { n 1 ,..., n k } and � � f − 1 [ S ] \{ n 1 ,..., n k } b = min . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Theorem. If C is countable and S ⊆ C, then S is countable. Proof. WLOG let S be infinite. Let f : N → C be bijective and let n 1 : = min f − 1 [ S ] . For k ∈ N , define n k + 1 recursively. Once � � f − 1 [ S ] \{ n 1 ,..., n k } n 1 ,..., n k are chosen, let n k + 1 : = min . Define g : N → S by g ( k ) : = f ( n k ) . Then g is injective (and it really maps into S ). Suppose for a contradiction that g is not surjective. Let � � b : = min f − 1 � � f − 1 [ S ] ∩{ 1 ,..., b − 1 } � S \ g [ N ] and let k : = � . � � Then f − 1 [ S ] ∩{ 1 ,..., b − 1 } = { n 1 ,..., n k } and � � f − 1 [ S ] \{ n 1 ,..., n k } b = min . But then b = n k + 1 , contradiction. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Theorem. If C is countable and S ⊆ C, then S is countable. Proof. WLOG let S be infinite. Let f : N → C be bijective and let n 1 : = min f − 1 [ S ] . For k ∈ N , define n k + 1 recursively. Once � � f − 1 [ S ] \{ n 1 ,..., n k } n 1 ,..., n k are chosen, let n k + 1 : = min . Define g : N → S by g ( k ) : = f ( n k ) . Then g is injective (and it really maps into S ). Suppose for a contradiction that g is not surjective. Let � � b : = min f − 1 � � f − 1 [ S ] ∩{ 1 ,..., b − 1 } � S \ g [ N ] and let k : = � . � � Then f − 1 [ S ] ∩{ 1 ,..., b − 1 } = { n 1 ,..., n k } and � � f − 1 [ S ] \{ n 1 ,..., n k } b = min . But then b = n k + 1 , contradiction. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Theorem. The set N × N is countable logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Theorem. The set N × N is countable n ✻ ✲ m logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Theorem. The set N × N is countable n ✻ ✲ m 1 2 3 4 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Theorem. The set N × N is countable n ✻ 4 3 2 1 ✲ m 1 2 3 4 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Theorem. The set N × N is countable n ✻ 4 r r r r 3 r r r r 2 r r r r 1 r r r r ✲ m 1 2 3 4 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Theorem. The set N × N is countable n ✻ ( 1 , 4 ) ( 2 , 4 ) ( 3 , 4 ) ( 4 , 4 ) 4 r r r r ( 1 , 3 ) ( 2 , 3 ) ( 3 , 3 ) ( 4 , 3 ) 3 r r r r ( 1 , 2 ) ( 2 , 2 ) ( 3 , 2 ) ( 4 , 2 ) 2 r r r r ( 1 , 1 ) ( 2 , 1 ) ( 3 , 1 ) ( 4 , 1 ) 1 r r r r ✲ m 1 2 3 4 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Theorem. The set N × N is countable n ✻ . ··· . . ( 1 , 4 ) ( 2 , 4 ) ( 3 , 4 ) ( 4 , 4 ) 4 r r r r ( 1 , 3 ) ( 2 , 3 ) ( 3 , 3 ) ( 4 , 3 ) 3 r r r r ··· ( 1 , 2 ) ( 2 , 2 ) ( 3 , 2 ) ( 4 , 2 ) 2 r r r r ( 1 , 1 ) ( 2 , 1 ) ( 3 , 1 ) ( 4 , 1 ) 1 r r r r ✲ m 1 2 3 4 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Theorem. The set N × N is countable n ✻ . ··· . . ( 1 , 4 ) ( 2 , 4 ) ( 3 , 4 ) ( 4 , 4 ) 4 r r r r ( 1 , 3 ) ( 2 , 3 ) ( 3 , 3 ) ( 4 , 3 ) 3 r r r r ··· first one ■ ( 1 , 2 ) ( 2 , 2 ) ( 3 , 2 ) ( 4 , 2 ) 2 r r r r ( 1 , 1 ) ( 2 , 1 ) ( 3 , 1 ) ( 4 , 1 ) 1 r r r r ✲ m 1 2 3 4 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Theorem. The set N × N is countable n ✻ . ··· . . ( 1 , 4 ) ( 2 , 4 ) ( 3 , 4 ) ( 4 , 4 ) 4 r r r r next two ■ ( 1 , 3 ) ( 2 , 3 ) ( 3 , 3 ) ( 4 , 3 ) 3 r r r r ··· first one ■ ( 1 , 2 ) ( 2 , 2 ) ( 3 , 2 ) ( 4 , 2 ) 2 r r r r ( 1 , 1 ) ( 2 , 1 ) ( 3 , 1 ) ( 4 , 1 ) 1 r r r r ✲ m 1 2 3 4 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Theorem. The set N × N is countable n ✻ . ··· . . next three ■ ( 1 , 4 ) ( 2 , 4 ) ( 3 , 4 ) ( 4 , 4 ) 4 r r r r next two ■ ( 1 , 3 ) ( 2 , 3 ) ( 3 , 3 ) ( 4 , 3 ) 3 r r r r ··· first one ■ ( 1 , 2 ) ( 2 , 2 ) ( 3 , 2 ) ( 4 , 2 ) 2 r r r r ( 1 , 1 ) ( 2 , 1 ) ( 3 , 1 ) ( 4 , 1 ) 1 r r r r ✲ m 1 2 3 4 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Theorem. The set N × N is countable n next four ✻ ■ . ··· . . next three ■ ( 1 , 4 ) ( 2 , 4 ) ( 3 , 4 ) ( 4 , 4 ) 4 r r r r next two ■ ( 1 , 3 ) ( 2 , 3 ) ( 3 , 3 ) ( 4 , 3 ) 3 r r r r ··· first one ■ ( 1 , 2 ) ( 2 , 2 ) ( 3 , 2 ) ( 4 , 2 ) 2 r r r r ( 1 , 1 ) ( 2 , 1 ) ( 3 , 1 ) ( 4 , 1 ) 1 r r r r ✲ m 1 2 3 4 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Theorem. The set N × N is countable f ( m , n ) : = 1 2 ( m + n − 1 )( m + n − 2 )+ n n next four ✻ ■ . ··· . . next three ■ ( 1 , 4 ) ( 2 , 4 ) ( 3 , 4 ) ( 4 , 4 ) 4 r r r r next two ■ ( 1 , 3 ) ( 2 , 3 ) ( 3 , 3 ) ( 4 , 3 ) 3 r r r r ··· first one ■ ( 1 , 2 ) ( 2 , 2 ) ( 3 , 2 ) ( 4 , 2 ) 2 r r r r ( 1 , 1 ) ( 2 , 1 ) ( 3 , 1 ) ( 4 , 1 ) 1 r r r r ✲ m 1 2 3 4 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Proof. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Proof. We claim that the function f : N × N → N defined by f ( m , n ) : = 1 2 ( m + n − 1 )( m + n − 2 )+ n is bijective. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Proof. We claim that the function f : N × N → N defined by f ( m , n ) : = 1 2 ( m + n − 1 )( m + n − 2 )+ n is bijective. First, an auxiliary equation. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Proof. We claim that the function f : N × N → N defined by f ( m , n ) : = 1 2 ( m + n − 1 )( m + n − 2 )+ n is bijective. First, an auxiliary equation. f ( m − 1 , n + 1 ) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Proof. We claim that the function f : N × N → N defined by f ( m , n ) : = 1 2 ( m + n − 1 )( m + n − 2 )+ n is bijective. First, an auxiliary equation. f ( m − 1 , n + 1 ) 1 � �� � = ( m − 1 )+( n + 1 ) − 1 ( m − 1 )+( n + 1 ) − 2 + n + 1 2 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Proof. We claim that the function f : N × N → N defined by f ( m , n ) : = 1 2 ( m + n − 1 )( m + n − 2 )+ n is bijective. First, an auxiliary equation. f ( m − 1 , n + 1 ) 1 � �� � = ( m − 1 )+( n + 1 ) − 1 ( m − 1 )+( n + 1 ) − 2 + n + 1 2 1 = 2 ( m + n − 1 )( m + n − 2 )+ n + 1 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Proof. We claim that the function f : N × N → N defined by f ( m , n ) : = 1 2 ( m + n − 1 )( m + n − 2 )+ n is bijective. First, an auxiliary equation. f ( m − 1 , n + 1 ) 1 � �� � = ( m − 1 )+( n + 1 ) − 1 ( m − 1 )+( n + 1 ) − 2 + n + 1 2 1 = 2 ( m + n − 1 )( m + n − 2 )+ n + 1 = f ( m , n )+ 1 . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Proof (injectivity). logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Proof (injectivity). We will prove by induction on m that f ( m , n ) = f ( a , b ) implies ( m , n ) = ( a , b ) . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Proof (injectivity). We will prove by induction on m that f ( m , n ) = f ( a , b ) implies ( m , n ) = ( a , b ) . Base step, m = 1: logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Proof (injectivity). We will prove by induction on m that f ( m , n ) = f ( a , b ) implies ( m , n ) = ( a , b ) . Base step, m = 1: Let f ( 1 , n ) = f ( a , b ) . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Proof (injectivity). We will prove by induction on m that f ( m , n ) = f ( a , b ) implies ( m , n ) = ( a , b ) . Base step, m = 1: Let f ( 1 , n ) = f ( a , b ) . Then f ( 1 , n ) = f ( a , b ) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Proof (injectivity). We will prove by induction on m that f ( m , n ) = f ( a , b ) implies ( m , n ) = ( a , b ) . Base step, m = 1: Let f ( 1 , n ) = f ( a , b ) . Then f ( 1 , n ) = f ( a , b ) 1 1 2 ( 1 + n − 1 )( 1 + n − 2 )+ n = 2 ( a + b − 1 )( a + b − 2 )+ b logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Proof (injectivity). We will prove by induction on m that f ( m , n ) = f ( a , b ) implies ( m , n ) = ( a , b ) . Base step, m = 1: Let f ( 1 , n ) = f ( a , b ) . Then f ( 1 , n ) = f ( a , b ) 1 1 2 ( 1 + n − 1 )( 1 + n − 2 )+ n = 2 ( a + b − 1 )( a + b − 2 )+ b n ( n − 1 )+ 2 n = ( a + b − 1 )( a + b − 2 )+ 2 b logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Proof (injectivity). We will prove by induction on m that f ( m , n ) = f ( a , b ) implies ( m , n ) = ( a , b ) . Base step, m = 1: Let f ( 1 , n ) = f ( a , b ) . Then f ( 1 , n ) = f ( a , b ) 1 1 2 ( 1 + n − 1 )( 1 + n − 2 )+ n = 2 ( a + b − 1 )( a + b − 2 )+ b n ( n − 1 )+ 2 n = ( a + b − 1 )( a + b − 2 )+ 2 b n 2 + n ( a + b − 1 ) 2 − ( a + b − 1 )+ 2 b = logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Proof (injectivity). We will prove by induction on m that f ( m , n ) = f ( a , b ) implies ( m , n ) = ( a , b ) . Base step, m = 1: Let f ( 1 , n ) = f ( a , b ) . Then f ( 1 , n ) = f ( a , b ) 1 1 2 ( 1 + n − 1 )( 1 + n − 2 )+ n = 2 ( a + b − 1 )( a + b − 2 )+ b n ( n − 1 )+ 2 n = ( a + b − 1 )( a + b − 2 )+ 2 b n 2 + n ( a + b − 1 ) 2 − ( a + b − 1 )+ 2 b = n 2 + n ( a + b − 1 ) 2 +( b − a + 1 ) = logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Proof (injectivity, cont.). logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Proof (injectivity, cont.). n 2 + n = ( a + b − 1 ) 2 +( b − a + 1 ) logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Proof (injectivity, cont.). n 2 + n = ( a + b − 1 ) 2 +( b − a + 1 ) Suppose for a contradiction that a + b − 1 < n . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Proof (injectivity, cont.). n 2 + n = ( a + b − 1 ) 2 +( b − a + 1 ) Suppose for a contradiction that a + b − 1 < n . Then b − a + 1 ≤ b < n and ( a + b − 1 ) 2 +( b − a + 1 ) < n 2 + n logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Proof (injectivity, cont.). n 2 + n = ( a + b − 1 ) 2 +( b − a + 1 ) Suppose for a contradiction that a + b − 1 < n . Then b − a + 1 ≤ b < n and ( a + b − 1 ) 2 +( b − a + 1 ) < n 2 + n , contradiction. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Proof (injectivity, cont.). n 2 + n = ( a + b − 1 ) 2 +( b − a + 1 ) Suppose for a contradiction that a + b − 1 < n . Then b − a + 1 ≤ b < n and ( a + b − 1 ) 2 +( b − a + 1 ) < n 2 + n , contradiction. Thus a + b − 1 ≥ n . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Proof (injectivity, cont.). n 2 + n = ( a + b − 1 ) 2 +( b − a + 1 ) Suppose for a contradiction that a + b − 1 < n . Then b − a + 1 ≤ b < n and ( a + b − 1 ) 2 +( b − a + 1 ) < n 2 + n , contradiction. Thus a + b − 1 ≥ n . Suppose for a contradiction that a + b − 1 = n + k for some k ∈ N . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Proof (injectivity, cont.). n 2 + n = ( a + b − 1 ) 2 +( b − a + 1 ) Suppose for a contradiction that a + b − 1 < n . Then b − a + 1 ≤ b < n and ( a + b − 1 ) 2 +( b − a + 1 ) < n 2 + n , contradiction. Thus a + b − 1 ≥ n . Suppose for a contradiction that a + b − 1 = n + k for some k ∈ N . Then ( a + b − 1 ) 2 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Proof (injectivity, cont.). n 2 + n = ( a + b − 1 ) 2 +( b − a + 1 ) Suppose for a contradiction that a + b − 1 < n . Then b − a + 1 ≤ b < n and ( a + b − 1 ) 2 +( b − a + 1 ) < n 2 + n , contradiction. Thus a + b − 1 ≥ n . Suppose for a contradiction that a + b − 1 = n + k for some k ∈ N . Then ( a + b − 1 ) 2 = ( n + k ) 2 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Proof (injectivity, cont.). n 2 + n = ( a + b − 1 ) 2 +( b − a + 1 ) Suppose for a contradiction that a + b − 1 < n . Then b − a + 1 ≤ b < n and ( a + b − 1 ) 2 +( b − a + 1 ) < n 2 + n , contradiction. Thus a + b − 1 ≥ n . Suppose for a contradiction that a + b − 1 = n + k for some k ∈ N . Then ( a + b − 1 ) 2 = ( n + k ) 2 = n 2 + 2 kn + k 2 , so that logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Proof (injectivity, cont.). n 2 + n = ( a + b − 1 ) 2 +( b − a + 1 ) Suppose for a contradiction that a + b − 1 < n . Then b − a + 1 ≤ b < n and ( a + b − 1 ) 2 +( b − a + 1 ) < n 2 + n , contradiction. Thus a + b − 1 ≥ n . Suppose for a contradiction that a + b − 1 = n + k for some k ∈ N . Then ( a + b − 1 ) 2 = ( n + k ) 2 = n 2 + 2 kn + k 2 , so that b − a + 1 = − ( 2 k − 1 ) n − k 2 . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Proof (injectivity, cont.). n 2 + n = ( a + b − 1 ) 2 +( b − a + 1 ) Suppose for a contradiction that a + b − 1 < n . Then b − a + 1 ≤ b < n and ( a + b − 1 ) 2 +( b − a + 1 ) < n 2 + n , contradiction. Thus a + b − 1 ≥ n . Suppose for a contradiction that a + b − 1 = n + k for some k ∈ N . Then ( a + b − 1 ) 2 = ( n + k ) 2 = n 2 + 2 kn + k 2 , so that b − a + 1 = − ( 2 k − 1 ) n − k 2 . Hence n + k − b + 1 = a logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Proof (injectivity, cont.). n 2 + n = ( a + b − 1 ) 2 +( b − a + 1 ) Suppose for a contradiction that a + b − 1 < n . Then b − a + 1 ≤ b < n and ( a + b − 1 ) 2 +( b − a + 1 ) < n 2 + n , contradiction. Thus a + b − 1 ≥ n . Suppose for a contradiction that a + b − 1 = n + k for some k ∈ N . Then ( a + b − 1 ) 2 = ( n + k ) 2 = n 2 + 2 kn + k 2 , so that b − a + 1 = − ( 2 k − 1 ) n − k 2 . Hence n + k − b + 1 = a = ( 2 k − 1 ) n + k 2 + b + 1 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Proof (injectivity, cont.). n 2 + n = ( a + b − 1 ) 2 +( b − a + 1 ) Suppose for a contradiction that a + b − 1 < n . Then b − a + 1 ≤ b < n and ( a + b − 1 ) 2 +( b − a + 1 ) < n 2 + n , contradiction. Thus a + b − 1 ≥ n . Suppose for a contradiction that a + b − 1 = n + k for some k ∈ N . Then ( a + b − 1 ) 2 = ( n + k ) 2 = n 2 + 2 kn + k 2 , so that b − a + 1 = − ( 2 k − 1 ) n − k 2 . Hence n + k − b + 1 = a = ( 2 k − 1 ) n + k 2 + b + 1, which implies that 2 b = ( − 2 k + 2 ) n + k − k 2 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Proof (injectivity, cont.). n 2 + n = ( a + b − 1 ) 2 +( b − a + 1 ) Suppose for a contradiction that a + b − 1 < n . Then b − a + 1 ≤ b < n and ( a + b − 1 ) 2 +( b − a + 1 ) < n 2 + n , contradiction. Thus a + b − 1 ≥ n . Suppose for a contradiction that a + b − 1 = n + k for some k ∈ N . Then ( a + b − 1 ) 2 = ( n + k ) 2 = n 2 + 2 kn + k 2 , so that b − a + 1 = − ( 2 k − 1 ) n − k 2 . Hence n + k − b + 1 = a = ( 2 k − 1 ) n + k 2 + b + 1, which implies that 2 b = ( − 2 k + 2 ) n + k − k 2 ≤ 0 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Proof (injectivity, cont.). n 2 + n = ( a + b − 1 ) 2 +( b − a + 1 ) Suppose for a contradiction that a + b − 1 < n . Then b − a + 1 ≤ b < n and ( a + b − 1 ) 2 +( b − a + 1 ) < n 2 + n , contradiction. Thus a + b − 1 ≥ n . Suppose for a contradiction that a + b − 1 = n + k for some k ∈ N . Then ( a + b − 1 ) 2 = ( n + k ) 2 = n 2 + 2 kn + k 2 , so that b − a + 1 = − ( 2 k − 1 ) n − k 2 . Hence n + k − b + 1 = a = ( 2 k − 1 ) n + k 2 + b + 1, which implies that 2 b = ( − 2 k + 2 ) n + k − k 2 ≤ 0, contradiction. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Proof (injectivity, cont.). n 2 + n = ( a + b − 1 ) 2 +( b − a + 1 ) Suppose for a contradiction that a + b − 1 < n . Then b − a + 1 ≤ b < n and ( a + b − 1 ) 2 +( b − a + 1 ) < n 2 + n , contradiction. Thus a + b − 1 ≥ n . Suppose for a contradiction that a + b − 1 = n + k for some k ∈ N . Then ( a + b − 1 ) 2 = ( n + k ) 2 = n 2 + 2 kn + k 2 , so that b − a + 1 = − ( 2 k − 1 ) n − k 2 . Hence n + k − b + 1 = a = ( 2 k − 1 ) n + k 2 + b + 1, which implies that 2 b = ( − 2 k + 2 ) n + k − k 2 ≤ 0, contradiction. Thus a + b − 1 = n . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Proof (injectivity, cont.). n 2 + n = ( a + b − 1 ) 2 +( b − a + 1 ) Suppose for a contradiction that a + b − 1 < n . Then b − a + 1 ≤ b < n and ( a + b − 1 ) 2 +( b − a + 1 ) < n 2 + n , contradiction. Thus a + b − 1 ≥ n . Suppose for a contradiction that a + b − 1 = n + k for some k ∈ N . Then ( a + b − 1 ) 2 = ( n + k ) 2 = n 2 + 2 kn + k 2 , so that b − a + 1 = − ( 2 k − 1 ) n − k 2 . Hence n + k − b + 1 = a = ( 2 k − 1 ) n + k 2 + b + 1, which implies that 2 b = ( − 2 k + 2 ) n + k − k 2 ≤ 0, contradiction. Thus a + b − 1 = n . Hence b − a + 1 = n logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Proof (injectivity, cont.). n 2 + n = ( a + b − 1 ) 2 +( b − a + 1 ) Suppose for a contradiction that a + b − 1 < n . Then b − a + 1 ≤ b < n and ( a + b − 1 ) 2 +( b − a + 1 ) < n 2 + n , contradiction. Thus a + b − 1 ≥ n . Suppose for a contradiction that a + b − 1 = n + k for some k ∈ N . Then ( a + b − 1 ) 2 = ( n + k ) 2 = n 2 + 2 kn + k 2 , so that b − a + 1 = − ( 2 k − 1 ) n − k 2 . Hence n + k − b + 1 = a = ( 2 k − 1 ) n + k 2 + b + 1, which implies that 2 b = ( − 2 k + 2 ) n + k − k 2 ≤ 0, contradiction. Thus a + b − 1 = n . Hence b − a + 1 = n , and then 2 b = 2 n and a = 1 logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Proof (injectivity, cont.). n 2 + n = ( a + b − 1 ) 2 +( b − a + 1 ) Suppose for a contradiction that a + b − 1 < n . Then b − a + 1 ≤ b < n and ( a + b − 1 ) 2 +( b − a + 1 ) < n 2 + n , contradiction. Thus a + b − 1 ≥ n . Suppose for a contradiction that a + b − 1 = n + k for some k ∈ N . Then ( a + b − 1 ) 2 = ( n + k ) 2 = n 2 + 2 kn + k 2 , so that b − a + 1 = − ( 2 k − 1 ) n − k 2 . Hence n + k − b + 1 = a = ( 2 k − 1 ) n + k 2 + b + 1, which implies that 2 b = ( − 2 k + 2 ) n + k − k 2 ≤ 0, contradiction. Thus a + b − 1 = n . Hence b − a + 1 = n , and then 2 b = 2 n and a = 1, as was to be proved. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Proof (injectivity, cont.). n 2 + n = ( a + b − 1 ) 2 +( b − a + 1 ) Suppose for a contradiction that a + b − 1 < n . Then b − a + 1 ≤ b < n and ( a + b − 1 ) 2 +( b − a + 1 ) < n 2 + n , contradiction. Thus a + b − 1 ≥ n . Suppose for a contradiction that a + b − 1 = n + k for some k ∈ N . Then ( a + b − 1 ) 2 = ( n + k ) 2 = n 2 + 2 kn + k 2 , so that b − a + 1 = − ( 2 k − 1 ) n − k 2 . Hence n + k − b + 1 = a = ( 2 k − 1 ) n + k 2 + b + 1, which implies that 2 b = ( − 2 k + 2 ) n + k − k 2 ≤ 0, contradiction. Thus a + b − 1 = n . Hence b − a + 1 = n , and then 2 b = 2 n and a = 1, as was to be proved. Hence f ( 1 , n ) = f ( a , b ) implies ( 1 , n ) = ( a , b ) . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Proof (injectivity, concl.). logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Proof (injectivity, concl.). Induction step ( m − 1 ) → m : logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Proof (injectivity, concl.). Induction step ( m − 1 ) → m : Let f ( m , n ) = f ( a , b ) . logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets

Finite vs. Infinite Countability Examples Proof (injectivity, concl.). Induction step ( m − 1 ) → m : Let f ( m , n ) = f ( a , b ) . WLOG m , a � = 1. logo1 Bernd Schr¨ oder Louisiana Tech University, College of Engineering and Science Countable Sets