Uncountable Sets 2/22/121. Countably Infinite 2/13/122 There are as many natural numbers as integers 0 1 2 3 4 5 6 7 8 … 0, -1, 1, -2, 2, -3, 3, -4, 4.

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Uncountable Sets 2/22/121

Countably Infinite 2/13/122 There are as many natural numbers as integers 0 1 2 3 4 5 6 7 8 … 0, -1, 1, -2, 2, -3, 3, -4, 4 … f(n) = n/2 if n is even, -(n+1)/2 if n is odd is a bijection from Natural Numbers → Integers

Infinite Sizes Are all infinite sets the same size? NO! Cantor’s Theorem shows how to keep finding bigger infinities. 2/22/123

P(N) How many sets of natural numbers? The same as there are natural numbers? Or more? 2/22/124

Countably Infinite Sets ::= {finite bit strings} … is countably infinite 2/22/125 Proof: List strings shortest to longest, and alphabetically within strings of the same length

Countably infinite Sets = {e, 0, 1, 00, 01, 10, 11, …} 2/22/126 = {e, 0, 1, 00, 01, 10, 11, 000, …} = {f(0), f(1), f(2), f(3), f(4), …}

Uncountably Infinite Sets Claim: ::= {∞-bit strings} is uncountable. 2/22/127 What about infinitely long bit strings? Like infinite decimal fractions but with bits

Diagonal Arguments Suppose 0123...nn+1... s0s0 0010...00... s1s1 0110...01... s2s2 1000...10... s3s3 1011...11......1...1..0 2/22/128

Diagonal Arguments Suppose 0123...nn+1... s0s0 0010...00... s1s1 0110...01... s2s2 1000...10... s3s3 1011...11......1...1..0 0 0 0 0 1 1 1 2/22/129

So cannot be listed: this diagonal sequence will be missing …differs from every row! Diagonal Arguments Suppose 0 0 0 0 1 1 1 ⋯ 2/22/1210

Cantor’s Theorem For every set, A (finite or infinite), there is no bijection A↔P(A) 2/22/1211

There is no bijection A↔P(A) W::= {a ∈ A | a ∉ f(a)}, so for any a, a ∈ W iff a ∉ f(a). f is a bijection, so W=f(a 0 ), for some a 0 ∈ A. ( ∀ a) a ∈ f(a 0 ) iff a ∉ f(a ). Pf by contradiction: suppose f:A ↔ P(A) is a bijection. Let Pf by contradiction: 2/22/1212

There is no bijection A↔P(A) W::= {a ∈ A | a ∉ f(a)}, so for any a, a ∈ W iff a ∉ f(a). f is a bijection, so W=f(a 0 ), for some a 0 ∈ A. a ∈ f(a 0 ) iff a ∉ f(a ). Pf by contradiction: suppose f:A ↔ P(A) is a bijection. Let Pf by contradiction: 0 0 0 contradiction 2/22/1213

So P(N) is uncountable P(N) = set of subsets of N ↔ {0,1} ω ↔ infinite “binary decimals” representing reals in the range 0..1 2/22/1214

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