Discrrete mathematics for computer science 11uncountable

Uncountable Sets 2/22/12 Countably Infinite There are as many natural numbers as integers 8… 0, -1, 1, -2, 2, -3, 3, -4, … f(n) = n/2 if n is even, -(n+1)/2 if n is odd is a bijection from Natural Numbers → Integers 2/13/12 Infinite Sizes Are all infinite sets the same size? NO! Cantor’s Theorem shows how to keep finding bigger infinities 2/22/12 P(N) • How many sets of natural numbers? • The same as there are natural numbers? • Or more? 2/22/12 Countably Infinite Sets ::= {finite bit strings} … is countably infinite Proof: List strings shortest to longest, and alphabetically within strings of the same length 2/22/12 Countably infinite Sets = {e, 0, 1, 00, 01, 10, 11, …} = { f(0), = { e, 0, 1, f(1), f(2), 00, 01, 10, f(3), f(4), …} 11, 000, 2/22/12 … } Uncountably Infinite Sets What about infinitely long bit strings? Like infinite decimal fractions but with bits Claim: ::= {∞-bit strings} is uncountable 2/22/12 Diagonal Arguments Suppose n n+1 s0 0 0 s1 1 s2 0 s3 1 1 2/22/12 Diagonal Arguments • Suppose n n+1 s0 1 0 s1 1 s2 10 s3 1 1 2/22/12 01 10 Diagonal Arguments Suppose ⋯ …differs from every row! So cannot be listed: this diagonal sequence will be missing 2/22/12 10 Cantor’s Theorem For every set, A (finite or infinite), there is no bijection A↔P(A) 2/22/12 11 There is no bijection A↔P(A) Pf by by contradiction: contradiction: suppose Pf f:A↔P(A) is a bijection Let W::= {a  A|a f(a)}, so for any a, a W iff a f(a) f is a bijection, so W=f(a0), for some a0 A (∀a) a f(a0) iff a f(a ) 2/22/12 12 There is no bijection A↔P(A) Pf by by contradiction: contradiction: suppose Pf f:A↔P(A) is a bijection Let W::= {a  A|a f(a)}, so for any a, a W iff a f(a) f is a bijection, so W=f(a0), for some a0 A 2/22/12 a 0contradiction f(a0) iff a 0f(a ) 13 So P(N) is uncountable P(N) = set of subsets of N ↔ {0,1}ω ↔ infinite “binary decimals” representing reals in the range 2/22/12 14 ... contradiction: suppose Pf f:A↔P(A) is a bijection Let W::= {a  A|a f(a)}, so for any a, a W iff a f(a) f is a bijection, so W=f(a0), for some a0 A (∀a) a f(a0) iff a f(a ) 2/22/12 12 There is no bijection... contradiction: suppose Pf f:A↔P(A) is a bijection Let W::= {a  A|a f(a)}, so for any a, a W iff a f(a) f is a bijection, so W=f(a0), for some a0 A 2/22/12 a 0contradiction f(a0) iff a 0f(a ) 13 So... every row! So cannot be listed: this diagonal sequence will be missing 2/22/12 10 Cantor’s Theorem For every set, A (finite or infinite), there is no bijection A↔P(A) 2/22/12 11 There is no bijection