Lecture 24 Enoch Cheung November 18 2013 1 Let A B be countably infinite sets By definition there are bijections f A N and g B N Therefore there is a bijection h A B N N a b 7 f a g b To see that h is an injection consider a b a0 b0 A B and suppose h a b h a0 b0 then by definition f a g b f a0 g b0 By injectivity of f and g this implies that a a0 and b b0 so a b a0 b0 To see that h is surjective consider n m N then by surjectivity of f and g there are a A such that f a n and b B such that g b B Therefore a b A B such that h a b n m Therefore h A B N N is a bijection Recall from last time that s N N N defined s m n 2m 1 2n 1 is a bijection We claim that s h A B N is a bijection To see that s h is injective consider a b a0 b0 A B then if s h a b s h a0 b0 then by injectivity of s h a b h a0 b0 and by injectivity of h we have a b a0 b0 To see that s h is surjective for all n N since s is surjective there is some n0 m0 N such that s n0 m0 n and by surjectivity of h there is some a b A B such that h a b n0 m0 Therefore s h a b s n0 m0 n Therefore s h A B N is a bijection which shows that A B is countable 2 Diagonalization Show that there does not exists an injection f P A A for any set A Suppose such injection f P A A exists Consider X f Y Y A f Y Y Note that X Y so let a f X To see that this is a contradiction note that there are two cases a X or a X Suppose a X then a f X X so by definition f X X which is a contradiction On the other hand if a X then there is some Y A and f Y Y such that f Y a X However since f is injective f Y a f X implies that X Y which is a contradiction because a f X X but f Y Y 3 In class we proved that the open interval 0 1 was uncountable a Prove that 0 1 0 1 by showing that the function f 0 1 0 1 below is injective x 21 0 1 1 f x 2n x 2n 1 for some n N x otherwise Injectivity Suppose x y 0 1 and f x f y If f x f y 0 then note that the only possibility is x y 21 If f x f y 21n for some n N then note that the only possibility 1 is x y 2n 1 Otherwise f x f y does not take the forms above so f x f y x y Surjectivity Consider any y 0 1 If y 0 then f 21 0 If y 21n for some n N then 1 f 2n 21n Otherwise y 6 0 and y 0 1 so y 0 1 so f y y 1 b Prove that 0 1 R by showing that the function f x ln 1 x x is a bijection f 0 1 R We will provide an inverse to f Note that for any a 6 0 and b 6 1 1 1 1 1 a b 1 b b 1 a a a a b 1 Therefore let g R 0 1 be defined g y 1 ey 1 to see that the fraction is well defined note that for any y R ey 0 This also shows that ey 1 1 so ey1 1 1 and ey 1 0 so 0 ey1 1 1 so g has appropriate domain and codomain To check that g is the inverse of f note that for any x 0 1 g f x 1 1 x eln x 1 1 1 1 x 1 x 1 x x and for any y R f g y ln 1 ey 1 1 ey 1 1 ln 1 1 ey 1 thus g f 1 is the inverse of f so f is a bijection 2 1 ln ey 1 1 y