Download presentation

Presentation is loading. Please wait.

Published byClay Cordy Modified over 3 years ago

1
Uncountable Sets 2/22/121

2
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

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

9
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

10
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

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: 0 0 0 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 0..1 2/22/1214

Similar presentations

OK

CS 310 – Fall 2006 Pacific University CS310 The Halting Problem Section 4.2 November 15, 2006.

CS 310 – Fall 2006 Pacific University CS310 The Halting Problem Section 4.2 November 15, 2006.

© 2018 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google

Ppt on effective communication skills Ppt on my school days Make a ppt on how to control noise pollution Ppt on launching a new product in india Ppt on condition monitoring of transformers Ppt on accounting standard 17 Ppt on shipping industry in india Ppt on bluetooth security system Ppt on world diabetes day images Download free ppt on active and passive voice