Counting ∞ Review: Bijections Counting Infinite Sets A function f : A → B is: ◮ one-to-one (an injection ) if f ( x ) = f ( y ) implies x = y . Or, x � = y implies f ( x ) � = f ( y ) . Distinct inputs, distinct outputs. ◮ onto (a surjection ) if for each y ∈ B , there is an x ∈ A with Are there more blue dots or red dots? How did we know that there were the same number of dots of f ( x ) = y . Every element in B is hit. each color, without counting ? Did you count all of the dots? Then, f is a bijection if it is both an injection and a surjection. You found a bijection between the blue dots and red dots! How did you know the answer? A bijection “rearranges” the elements of A to form B . To count infinities, we will take the definition of “same size” to Today: We count to ∞ and beyond. be “there exists a bijection between the sets”. Countability Hilbert’s Hotel I Hilbert’s Hotel II What does it mean for us to “count” the elements of a set? Now suppose that a new bus of passengers arrives. There is a Consider an infinite hotel, one room for each n ∈ N . The rooms Our model for counting: N = { 0 , 1 , 2 ,... } . new guest n for each positive integer n . are all filled by guests. A set A is called countable if there exists a bijection between A Can we still accommodate the guests? A new guest arrives. Can we accommodate the new guest? and a subset of N . ◮ Any finite set is countable. Consider the set For each n ∈ N , move guest in room n to room 2 n . Put the i th For each n ∈ N , move the guest in room n to n + 1. Then place Odin’s notable children = { Hela , Thor , Loki } . new guest into the i th odd-numbered room. the new guest in room 0. f ( Hela ) = 0 , f ( Thor ) = 1 , f ( Loki ) = 2 . We found a bijection f : Z → N . In other words, we found a bijection f : N ∪{− 1 } → N . Then, f : Odin’s notable children → { 0 , 1 , 2 } is a bijection. f ( n ) = 2 n for n ∈ N , f ( − n ) = 2 n − 1 for positive n . f ( − 1 ) = 0 , f ( n ) = n + 1 for n ∈ N . ◮ N itself is countable. Adding a countably infinite number of elements to N does not ◮ If A is countable and infinite, then we say it is countably Adding one more element to N does not change its size. change its size. infinite . ◮ What else is countable?
Proving the Bijection Formally Countably Infinite Sets Be Careful Recall: If A and B are finite and have the same size, then if Here are some countably infinite sets. f : A → B is injective or surjective, then it is both. Is the following a listing of Z ? ◮ N . N ∪{− 1 } . Z . ◮ The set of even numbers. The set of odd numbers. 0 , 1 , 2 , 3 ,..., − 1 , − 2 , − 3 ,... This is not true for infinite sets, so we must check both injectivity and surjectivity. ◮ The set of prime numbers. Where does the element − 1 show up in the list? Why is the set of prime numbers countably infinite? It is infinite f ( n ) = 2 n for n ∈ N , f ( − n ) = 2 n − 1 for positive n . (we proved this). But we can list them. To give a listing of a set A , every element of A must show up at Proof that f is bijective . some finite index in the list. 2 , 3 , 5 , 7 , 11 ,... ◮ One-to-one: Assume f ( x ) = f ( y ) . Prove x = y . ◮ In the example above, we never “reach” the element − 1. ◮ If f ( x ) = f ( y ) are odd, then − 2 x − 1 = − 2 y − 1. So, x = y . The list is exhaustive . Every prime number shows up in the list. Here is a valid listing of Z : ◮ If f ( x ) = f ( y ) are even, then 2 x = 2 y . So, x = y . An exhaustive list is equivalent to a bijection. ◮ Onto: Consider any n ∈ N . Either n is even or odd. 0 , 1 , − 1 , 2 , − 2 , 3 , − 3 ,... ◮ If n is even, then n = 2 k for some k ∈ N . Then, f ( k ) = n . f ( 2 ) = 0 , f ( 3 ) = 1 , f ( 5 ) = 2 , f ( 7 ) = 3 , f ( 11 ) = 4 ,... Be careful with “. . . ” in the middle of your listing. ◮ If n is odd, then n = 2 k − 1 for some positive k . Then A set whose elements can be listed is countable. f ( − k ) = n . Hilbert’s Hotel III The Formal Injection Bijections Compose We found an injection Now a countably infinite number of buses arrive, each bus Fact : If f : A → B and g : B → C are bijections, then so is g ◦ f . containing a countably infinite number of passengers. f : { prime numbers }×{ 1 , 2 , 3 ,... } → { odd numbers } Proof . Can we accomodate the guests? given by f ( p , i ) = p i -th odd number. ◮ If g ( f ( x )) = g ( f ( y )) , then g is one-to-one so f ( x ) = f ( y ) . First, “make room for ∞ ” (send guest n to room 2 n as before). ◮ Since f is one-to-one, then x = y . So g ◦ f is one-to-one. Since { prime numbers } , { 1 , 2 , 3 ,... } , and { odd numbers } all ◮ If c ∈ C , then there is a b ∈ B such that g ( b ) = c (since g is have the same size as N , this is the same as finding an injection Label each bus with a prime number p . Label each person in onto). the bus with a positive integer. g : N × N → N . ◮ There is an a ∈ A such that f ( a ) = b (since f is onto). Send the i th person in bus p to the p i -th odd numbered room. ◮ So, g ( f ( a )) = g ( b ) = c . So g ◦ f is onto. Why? There are bijections ◮ Bus 2’s passengers get sent to: 2 · 2 1 − 1, 2 · 2 2 − 1, 2 · 2 3 − 1, . . . f 1 : N → { prime numbers } , Bijections compose. ◮ Bus 3’s passengers get sent to: 2 · 3 1 − 1, 2 · 3 2 − 1, f 2 : N → { 1 , 2 , 3 ,... } , Exercise : If there are bijections f : A → A ′ and g : B → B ′ , then 2 · 3 3 − 1, . . . f 3 : N → { odd numbers } , h ( a , b ) = ( f ( a ) , g ( b )) is a bijection A × B → A ′ × B ′ . Adding a countably infinite number of countable infinities to N so we get an injection g ( m , n ) = f − 1 does not change its size. 3 ( f ( f 1 ( m ) , f 2 ( n ))) .
Bijections Compose Is Q Countable? Interleaving Argument Suppose that A is a countable alphabet. Consider the set of all finite strings whose symbols come from A . Fact : If f : A → B is a bijection, and there are bijections Is Q countable? f 1 : A → A ′ and f 2 : B → B ′ , then there is a bijection g : A ′ → B ′ . ◮ We found an injection N × N → N . So, N × N is countable. A is countable. ◮ Since Z has the same size as N , then Z × Z is countable. Proof . Proof . ◮ Every rational number q ∈ Q can be written as q = a / b , f A B ◮ List the alphabet A = { a 1 , a 2 , a 3 ,... } . where a , b ∈ Z , b > 0, and a / b is in lowest terms. ◮ Step 0: List the empty string. f 1 f 2 ◮ This defines an injection Q → Z × Z . g ◮ Step 1: List all strings of length ≤ 1 using symbols from A ′ B ′ ◮ An injection implies that Q is “smaller” than N × N , so Q is { a 1 } . a 1 . countable. Define g = f 2 ◦ f ◦ f − 1 1 . The composition of bijections is a ◮ Step 2: List all strings of length ≤ 2 using symbols from ◮ On the other hand, Q is infinite, so Q is countably infinite. bijection. { a 1 , a 2 } . a 1 , a 2 , a 1 a 1 , a 1 a 2 , a 2 a 1 , a 2 a 2 . ◮ Step 3: List all strings of length ≤ 3 using symbols from Principle: To show that a set A is countable, we only need to To show that A has the same size as N , we can show that A find an injection from A into a countable set. { a 1 , a 2 , a 3 } . has the same size as A ′ , where A ′ has the same size as N . ◮ Continue forever. This exhaustively lists the members of the set. Polynomials with Rational Coefficients Is R Countable? Cantor’s Diagonalization Argument Is R countable? First, let us study the closed unit interval [ 0 , 1 ] . ◮ Assume we could list all numbers in [ 0 , 1 ] . Each element of [ 0 , 1 ] can be represented as a infinite-length Consider the set of polynomials with rational coefficients. Is this ◮ Form a new number in [ 0 , 1 ] by changing each number in decimal string. set countable? the diagonal. ◮ For example, take the element 0 . 37. This can also be ◮ This number cannot be the i th element of the list because For a polynomial, e.g., P ( x ) = ( 2 / 3 ) x 4 − 2 x 2 +( 1 / 10 ) x + 9, represented as 0 . 36999 ... . it differs in the i th digit. think of it as a string: ( 2 / 3 , 0 , − 2 , 1 / 10 , 9 ) . Suppose we had a list of all numbers in [ 0 , 1 ] . ◮ We found an element not in our original list! 0 . 9 9 1 ... The alphabet is Q , countably infinite. ◮ So, [ 0 , 1 ] is uncountable . 0 . 0 2 3 ... Each polynomial is a finite-length string from the alphabet. 0 . 2 8 9 ... What happens when we try to apply the diagonalization . . . . . ... . . . . . . . . . . argument to N ? The polynomials with rational coefficients are countable. ◮ We get a number with infinitely many digits. If we change the numbers on the diagonal, 0.929. . . , we get a ◮ This is not a natural number! Not a contradiction. number which is not in the list. ◮ Change all 8s to 1s and change all other numbers to 8s.
Recommend
More recommend
