Presentation is loading. Please wait.

Presentation is loading. Please wait.

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.

Similar presentations


Presentation on theme: "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."— Presentation transcript:

1 Uncountable Sets 2/22/121

2 Countably Infinite 2/13/122 There are as many natural numbers as integers … 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

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

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

5 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

6 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), …}

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

8 Diagonal Arguments Suppose nn+1... s0s s1s s2s s3s /22/128

9 Diagonal Arguments Suppose nn+1... s0s s1s s2s s3s /22/129

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

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

12 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

13 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: contradiction 2/22/1213

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


Download ppt "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."

Similar presentations


Ads by Google