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

  1. 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

  2. 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

  3. 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

  4. 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

  5. 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

  6. 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

  7. 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

  8. 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

  9. 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

  10. 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

  11. 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

  12. 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

  13. 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

  14. 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

  15. 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

  16. 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

  17. 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

  18. 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

  19. 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

  20. 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

  21. 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

  22. 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

  23. 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

  24. 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

  25. 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

  26. 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

  27. 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

  28. 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

  29. 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

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

  31. 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

  32. 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

  33. 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

  34. 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

  35. 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

  36. 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

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

  38. 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

  39. 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

  40. 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

  41. 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

  42. 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

  43. 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

  44. 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

  45. 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

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

  47. 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

  48. 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

  49. 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

  50. 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

  51. 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

  52. 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

  53. 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

  54. 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

  55. 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

  56. 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

  57. 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

  58. 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

  59. 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

  60. 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

  61. 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

  62. 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

  63. 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

  64. 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

  65. 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

  66. 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

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

  68. 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

  69. 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

  70. 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

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