Uncountable Cantors Theorem shows how to keep finding Sets bigger - PowerPoint PPT Presentation

Mathematics for Computer Science Infinite Sizes MIT 6.042J/18.062J Are all sets the same size? NO! Uncountable Cantor’s Theorem shows how to keep finding Sets bigger infinities. Albert R Meyer, March 4, 2015 Albert R Meyer, March 4, 2015 Cantor.1 Cantor.2 Countable Sets Diagonal Arguments ω { } Suppose s 0 ,s 1 ,s 2 , …∈ 0,1 A is countable iff can list it: a 0 ,a 1 ,a 2 ,…. example: 0 1 2 3 . . . n n+1 . . . s 0 0 0 1 0 . . . 0 0 . . . * { } ::= {finite bit strings} 0,1 0 1 1 0 . . . 0 1 . . . s 1 1 0 0 0 . . . 1 0 . . . s 2 ω { } Claim: ::= {∞-bit strings} 0,1 1 0 1 1 . . . 1 1 . . . s 3 . . . 1 is uncountable. . . . 1 . . 0 Albert R Meyer, March 4, 2013 Cantor.4 Albert R Meyer, March 4, 2015 Cantor.3 1

Diagonal Arguments Diagonal Arguments ω ω { } { } Suppose s 0 ,s 1 ,s 2 , …∈ 0,1 Suppose s 0 ,s 1 ,s 2 , …∈ 0,1 0 1 2 3 . . . n n+1 . . . 1 0 0 1 0 . . . 0 0 . . . � s 0 1 …differs from every row! 0 0 1 1 0 . . . 0 1 . . . s 1 0 ω 1 { } 1 0 1 0 0 . . . 1 0 . . . s 2 So cannot be listed: 0,1 1 0 1 0 1 . . . 1 1 . . . s 3 0 this diagonal sequence 0 . . . 1 0 will be missing 0 . . . 1 0 1 . . 0 1 Albert R Meyer, March 4, 2013 Cantor.5 Albert R Meyer, March 4, 2013 Cantor.6 ω { } Strictly Smaller 0,1 is uncountable ω ⎞ ⎛ { } ⎟ ⎜ So surj 0,1 ⎟ NOT ⎜ ⎜ N ⎟ and ⎜ ⎟ ⎝ ⎠ A strict B ::= NOT (A surj B) ω { } surj N 0,1 obviously A is “strictly smaller” than B ω { } N "strictly smaller" than 0,1 ω { } So N strict 0,1 Albert R Meyer, March 4, 2013 Cantor.8 Albert R Meyer, March 4, 2015 Cantor.9 2

Cantor’s Theorem Diagonal Arguments Suppose A = {a,b,s,t, … ,d,e, … } A strict pow(A) pow(A) = {f(a),f(b),f(s), … ,f(d), … } for every set, A a b s t . . . d e . . . f(a) . (finite or infinite) f(b) . f(s) . f(t) . . Albert R Meyer, March 4, 2013 Cantor.11 Albert R Meyer, March 4, 2015 Cantor.10 Diagonal Arguments Diagonal Arguments Suppose A Suppose A = {a,b,s,t, … ,d,e, … } = {a,b,s,t, … ,d,e, … } pow(A) = {f(a),f(b),f(s), … ,f(d), … } pow(A) = {f(a),f(b),f(s), … ,f(d), … } a b s t c . . d e . . . a b s t c . . d e . . . f(a) a s t e . f(a) a a s t e . f(b) a b c d . f(b) a b b c d . f(s) b t . f(s) b t . s f(t) s t c d . f(t) s t t c d . f(c) b s d e f(c) b s d e c . . . . . . Albert R Meyer, March 4, 2013 Albert R Meyer, March 4, 2013 Cantor.12 Cantor.13 3

A strict Pow(A) Diagonal Arguments Suppose A = {a,b,s,t, … ,d,e, … } Pf: say have fcn f:A � pow(A). pow(A) = {f(a),f(b),f(s), … ,f(d), … } Define a subset of A that is not in D a b s t c . . d e . . . the range of f: namely f(a) s s t t e . f(b) a a c c d . D::= {a ∈ A | a ∉ f(a)} f(s) b b s s t t . f(t) s s c c d . ∉ range(f) D Now since it differs f(c) b b s s c c d e from set f(a) at element a! . . . . . Albert R Meyer, March 4, 2013 Cantor.14 Albert R Meyer, March 4, 2015 Cantor.15 str ict pow( N ) N A strict Pow(A) So no f-arrow into D. That is, f is not a surjection. pow( N ) is uncountable QED Albert R Meyer, March 4, 2015 Albert R Meyer, March 4, 2015 Cantor.21 Cantor.22 4

{0,1} ω again Proving Uncountability Lemma: If A is uncountable We know {0,1} ω bij pow( N ) and C surj A then C is uncountable pow( N ) and uncountable by Cantor, Albert R Meyer, March 4, 2015 Albert R Meyer, March 4, 2015 Cantor.24 Cantor.26 {0,1} ω again Real Numbers Uncountable ω { } R surj 0,1 We know {0,1} ω bij pow( N ) map ± r to binary rep 3 1/3 = 111.010101… pow( N ) and uncountable by maps to 111010101… Cantor, so {0,1} ω uncountable. Albert R Meyer, March 4, 2015 Albert R Meyer, March 4, 2015 Cantor.27 Cantor.28 5

Real Numbers Uncountable ω { } R surj 0,1 map ± r to binary rep 1/2 = .100000… 1/2 maps to 100000… � = .0111111… -1/2 maps to 0111111… � Albert R Meyer, March 4, 2015 Cantor.29 6

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