Analysis with an introduction to proof 5th by steven lay ch02b

Chapter Sets and Functions Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 2.4, Slide 1-1 Section 2.4 Cardinality Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 2.4, Slide 1-2 How can we compare the sizes of two sets? If S = {x ∈   : x = 9}, then S = {–3, 3} and we say that S has two elements If T = {1, 7, 11}, then T has three elements and we think of T as being “larger” than S These intuitive ideas are fine for small (finite) sets, but how can we compare the size of (infinite) sets like or ? Certainly, it is reasonable to say that two sets S and T are the same size if there is a bijective function f  : S → T, for this function will set up a one-to-one correspondence between the elements of each set Definition 2.4.1 Two sets S and T are called equinumerous, and we write S ~ T, if there exists a bijective function from S onto T It is easy to see that “~” is an equivalence relation on any family of sets, and it partitions the family into disjoint equivalence classes With each equivalence class we associate a cardinal number that we think of as giving the size of the set In dealing with finite sets, it will be convenient to abbreviate the set {1,  2, …, n} by In Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 2.4, Slide 1-3 Definition 2.4.3 A set S is said to be finite if S = ∅ or if there exists and a bijection f : In → S If a set is not finite, it is said to be infinite Definition 2.4.4 The cardinal number of In is n, and if S ~ In, we say that S has n elements The cardinal number of ∅ is taken to be If a cardinal number is not finite, it is called transfinite Definition 2.4.6 A set S is said to be denumerable if there exists a bijection f :  → S If a set is finite or denumerable, it is called countable If a set is not countable, it is uncountable The cardinal number of a denumerable set is denoted by ℵ0 “Aleph” is the first letter of the Hebrew alphabet Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 2.4, Slide 1-4 The relationships between the various “sizes” of sets is shown below countable sets infinite sets finite denumerable uncountable sets sets sets There are two kinds of countable sets: finite sets and denumerable sets The denumerable sets are infinite sets The infinite sets that are not denumerable sets are uncountable sets We have not yet shown that there exist any infinite sets that are not denumerable Before doing so, we look at some properties of countable sets Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 2.4, Slide 1-5 Countable Sets Example 2.4.7 It would seem at first glance that the set of natural numbers should be “bigger” than the set E of even natural numbers Indeed, E is a proper subset of it contains only “half” of and, in fact, But what is “half ” of ℵ0? Our experience with finite sets is a poor guide here, for : and E are actually equinumerous! 10 12 14 f (n) = 2n E: The function f (n) = 2n is a bijection from Copyright © 2013, 2005, 2001 Pearson Education, Inc onto E, so E also has cardinality ℵ0 Section 2.4, Slide 1-6 If a nonempty set S is finite, then there exists n ∈  and a bijection f : In → S Using the function f , we can count off the members of S as follows: f  (1), f (2), f (3),  …, f (n) Letting f (k) = sk for ≤ k ≤ n, we obtain the more familiar notation S = {s1, s2, …, sn} The same kind of counting process is possible for a denumerable set, and this is why both kinds of sets are called countable For example, if T is denumerable, then there exists a bijection and we may write T = {g  (1), g (2), g (3),…} or T = {t 1, t2, t 3,  …}, where g :  → T, g (n) = tn This ability to list the members of a set as a first, second, third and so on, characterizes countable sets If the list terminates, then the set is finite If the list does not terminate, then the set is denumerable Theorem 2.4.9 Let S be a countable set and let T ⊆ S Then T is countable The proof of this “obvious” result is not difficult, but it uses the Well-Ordering Property of that we won’t encounter until Chapter Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 2.4, Slide 1-7 The following theorem follows readily from Theorem 2.4.9 The details of the proof are in the text Theorem 2.4.10 Let S be a nonempty set The following three conditions are equivalent (a) S is countable (b) There exists an injection f : S → (c) There exists a surjection g :   → S Example 2.4.11(a) Suppose S and T are countable sets Then S ∪ T is countable Theorem 2.4.10 implies that there exist surjections f :  Define h: → S ∪ T by   n +1  f  ÷,    h( n) =  g  n ,   ÷  Then h is surjective, so S ∪ T is countable Copyright © 2013, 2005, 2001 Pearson Education, Inc  → S and g :   → T if n is odd if n is even We can see this as follows: Section 2.4, Slide 1-8 Given surjections f : → S and g : → T, and the even integers with g to count T let h use the odd integers with f to count S h f S • f (1) = h(1) • f (2) = h(3) • f (3) = h(5) • g T • g (1) = h(2) • g (2) = h(4) • g (3) = h(6) • h Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 2.4, Slide 1-9 We can adapt this process to show that the set of rational numbers is countable We begin by constructing a rectangular array of fractions positive integers The first row contains all the The second row contains all the fractions with denominator equal to The third row contains all the fractions with denominator equal to 3, etc Moving along each diagonal of the array in the manner indicated, we obtain a listing of all the 2 3 3 elements in the positive rationals 4 5 4 Copyright © 2013, 2005, 2001 Pearson Education, Inc 3, 4, This listing defines a surjection f : + 1, 1, 2, 1, 2, so But, + : , , → + , is countable = + ∪ {0} ∪ , – so is countable, too In the text this is generalized to show that any union of a countable family of countable sets is countable Section 2.4, Slide 10 1-10 Theorem 2.4.12 The set of real numbers is uncountable Proof: Since any subset of a countable set is countable (Theorem 2.4.9), it suffices to If J were countable, we could list its show that the interval J = (0, 1) is uncountable members and have J = {x1, x2, x3, …} = {xn : n ∈ } We shall show that this leads to a contradiction by constructing a real number that is in J Each element of J but is not included in the list of xn’s has an infinite decimal x1 = a11 a12 a13 ×××, expansion, so we can write x2 = a21 a22 a23 ×××where , each a x3 = a31 a32 a33 ×××, × × × We now construct a real number y = b1 b2 b3 ⋅ ⋅ ⋅ by defining Since each digit in the decimal expansion of i j ∈ {0,  …, 9}  1, 2, if ann ≠ 2, bn =  3, if ann = y is either or 3, y ∈ J But y is not one of the numbers xn, since it differs from xn in the nth decimal place This contradicts our assumption that J is countable, so J must be uncountable ♦ Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 2.4, Slide 11 1-11 Ordering of Cardinal Numbers Our approach to comparing the size of two sets is built on the definition of equinumerous and our understanding of functions no larger than T Intuitively, if f : S → T is injective, then S can be Not only is this true for finite sets, but we have observed that it also holds for countable sets (Theorem 2.4.10) Since we think of cardinal numbers as representing the size of a set, we shall use them when comparing sizes Definition 2.4.14 We denote the cardinal number of a set S by | S |, so that we have | S | = | T | iff S and T are equinumerous That is, | S | = | T | iff there exists a bijection We define | S | ≤ | T | to mean that there exists an injection f : S → T f : S → T As usual, | S | < | T | means that | S | ≤ | T | and | S | ≠ | T | The basic properties of our ordering of cardinals are included in Theorem 2.4.15 The proofs are all straightforward and are left to the exercises Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 2.4, Slide 12 1-12 Theorem 2.4.15 Let S, T, and U be sets (a) If S ⊆ T, then | S | ≤ | T | (b) | S | ≤ | S | (c) If | S | ≤ | T | (d) If m, n ∈ (e) If S is finite, then | and | T | ≤ | U |, then | S | ≤ | U | and m ≤ n, then |{1, 2,  …,  m}| ≤ |{1, 2,  …,  n}|  S | < ℵ0 Notes: (a) This corresponds to our intuitive feeling about the relative sizes of subsets (b) This is the reflexive property (c) This is the transitive property (d) This means that the order of m and n as integers is the same as the order for the finite cardinals m and n (e) Recall that ℵ0 is the transfinite cardinal number of Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 2.4, Slide 13 1-13 It is customary to denote the cardinal number of ⊆ Since by c, for continuum , we have ℵ0 ≤ c In fact, since is countable and is uncountable, we have ℵ0 < c Thus Theorem 2.4.15(e) implies that ℵ0 and c are unequal transfinite cardinals Are there any others? The answer is an emphatic yes, as we see in our next theorem Definition 2.4.16 Given any set S, let P (S) denote the collection of all the subsets of S The set P  (S) is called the power set of S Theorem 2.4.18 For any set S, we have | S | < |P  (S) Recall: | S | < | T | | means there exists an injection from S to T, but no bijection from S to T Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 2.4, Slide 14 1-14 Theorem 2.4.18 For any set S, we have | S | < |P  (S) | Proof: The function g  S →  P  (S) given by g  : To prove that | S | ≠ |P  (S)  |, we show that no function from S to P  (s) = {s} is clearly injective, so | S | ≤ |P  (S) |  (S) can be surjective Then for each x ∈ S, f (x) is a subset of S Suppose that f : S → P  (S) Now for some x in S it may be that x is in the subset Let f (x) and for others it may not be T = {x ∈ S : x ∉ f (x)} We have T ⊆ S, so T ∈ P  (S) If f were surjective, then T = f ( y) for some y ∈ S Now either y ∈ T or y ∉ T, but both possibilities lead to contradictions: If y ∈ T, then y ∉ f ( y) by the definition of T But f ( y) = T, so y ∉ f ( y) implies y ∉ T On the other hand, if y ∉ T, then since f ( y) = T, we have y ∉ f ( y) But then y ∈ T, by the definition of T Thus we conclude that no function from S to P Copyright © 2013, 2005, 2001 Pearson Education, Inc (S) can be surjective, so | S | ≠ |P  (S) | ♦ Section 2.4, Slide 15 1-15 By applying Theorem 2.4.18 again and again, we obtain an infinite sequence of transfinite cardinals each larger than the one preceding: ℵ0 = | | < |P ( Does the cardinal c = | )| < |P (P ( ))| < |P (P (P ( )))| < ⋅⋅⋅ | fit into this sequence? In Exercise 24 we sketch the proof that |P  ( ) | = c In Exercise 11 we show that every infinite set has a denumerable subset Since (by Theorem 2.4.9) every infinite subset of a denumerable set is denumerable, we see that ℵ0 is the smallest transfinite cardinal What is the first cardinal greater than ℵ0? We know that c > ℵ0, but is there any cardinal number λ such that ℵ0 < λ < c ? That is, is there any subset of with size “in between” and ? The conjecture that there is no such set was first made by Cantor and is known as the continuum hypothesis In 1900 it was included as the first of Hilbert’s famous 23 unsolved problems Whether it is true or false is still an unanswered— perhaps unanswerable—question Copyright © 2013, 2005, 2001 Pearson Education, Inc Section 2.4, Slide 16 1-16 ... contains all the fractions with denominator equal to The third row contains all the fractions with denominator equal to 3, etc Moving along each diagonal of the array in the manner indicated, we obtain... Education, Inc  → S and g :   → T if n is odd if n is even We can see this as follows: Section 2.4, Slide 1-8 Given surjections f : → S and g : → T, and the even integers with g to count T let h... function from S onto T It is easy to see that “~” is an equivalence relation on any family of sets, and it partitions the family into disjoint equivalence classes With each equivalence class we