1-1 Copyright © 2013, 2005, 2001 Pearson Education, Inc. Section 2.4, Slide 1 Chapter 2 Sets and Functions

1-2 Copyright © 2013, 2005, 2001 Pearson Education, Inc. Section 2.4, Slide 2 Section 2.4 Cardinality

1-3 Copyright © 2013, 2005, 2001 Pearson Education, Inc. Section 2.4, Slide 3 How can we compare the sizes of two sets? If S = {x  : x 2 = 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 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 I n.

1-4 Copyright © 2013, 2005, 2001 Pearson Education, Inc. Section 2.4, Slide 4 Definition A set S is said to be finite if S =  or if there exists and a bijection f : I n  S. If a set is not finite, it is said to be infinite. Definition The cardinal number of I n is n, and if S ~ I n, we say that S has n elements. The cardinal number of  is taken to be 0. If a cardinal number is not finite, it is called transfinite. Definition 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  0. “Aleph” is the first letter of the Hebrew alphabet.

1-5 Copyright © 2013, 2005, 2001 Pearson Education, Inc. Section 2.4, Slide 5 The relationships between the various “sizes” of sets is shown below. countable setsinfinite sets denumerable sets finite sets uncountable 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.

1-6 Copyright © 2013, 2005, 2001 Pearson Education, Inc. Section 2.4, Slide 6 Countable Sets 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 and, in fact, it contains only “half” of. But what is “half ” of  0 ? Example Our experience with finite sets is a poor guide here, for and E are actually equinumerous! E : : The function f (n) = 2n is a bijection from onto E, so E also has cardinality  0. f (n) = 2n

1-7 Copyright © 2013, 2005, 2001 Pearson Education, Inc. Section 2.4, Slide 7 If a nonempty set S is finite, then there exists n  and a bijection f : I n  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) = s k for 1  k  n, we obtain the more familiar notation S = {s 1, s 2, …, s n }. 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 g :  T, and we may write T = {g (1), g (2), g (3),…} or T = {t 1, t 2, t 3, …}, where g (n) = t n. 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 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 3.

1-8 Copyright © 2013, 2005, 2001 Pearson Education, Inc. Section 2.4, Slide 8 Theorem 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. The following theorem follows readily from Theorem The details of the proof are in the text. Example (a) Suppose S and T are countable sets. Then S  T is countable. Theorem implies that there exist surjections f :  S and g :  T. Define h:  S  T by Then h is surjective, so S  T is countable. We can see this as follows:

1-9 Copyright © 2013, 2005, 2001 Pearson Education, Inc. Section 2.4, Slide ...... S ...... T f g h h f (1) = h(1) f (2) = h(3) f (3) = h(5) g (1) = h(2) g (2) = h(4) g (3) = h(6) Given surjections f :  S and g :  T, let h use the odd integers with f to count S and the even integers with g to count T.

1-10 Copyright © 2013, 2005, 2001 Pearson Education, Inc. Section 2.4, Slide 10 The first row contains all the positive integers. We can adapt this process to show that the set of rational numbers is countable. We begin by constructing a rectangular array of fractions. The second row contains all the fractions with denominator equal to 2. 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 elements in the positive rationals : + This listing defines a surjection f : , so is countable. + + But, =  {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.

1-11 Copyright © 2013, 2005, 2001 Pearson Education, Inc. Section 2.4, Slide 11 Theorem The set of real numbers is uncountable. Proof: Since any subset of a countable set is countable (Theorem 2.4.9), it suffices to show that the interval J = (0, 1) is uncountable. If J were countable, we could list its members and have J = {x 1, x 2, x 3, …} = {x n : n  }. We shall show that this leads to a contradiction by constructing a real number that is in J but is not included in the list of x n ’s. Each element of J has an infinite decimal expansion, so we can write where each a i j  {0, 1, …, 9}. We now construct a real number y = 0. b 1 b 2 b 3    by defining Since each digit in the decimal expansion of y is either 2 or 3, y  J. But y is not one of the numbers x n, since it differs from x n in the nth decimal place. This contradicts our assumption that J is countable, so J must be uncountable. 

1-12 Copyright © 2013, 2005, 2001 Pearson Education, Inc. Section 2.4, Slide 12 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. Intuitively, if f : S  T is injective, then S can be no larger than T. Not only is this true for finite sets, but we have observed that it also holds for countable sets (Theorem ). Since we think of cardinal numbers as representing the size of a set, we shall use them when comparing sizes. Definition 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 f : S  T. We define | S |  | T | to mean that there exists an injection 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 The proofs are all straightforward and are left to the exercises.

1-13 Copyright © 2013, 2005, 2001 Pearson Education, Inc. Section 2.4, Slide 13 Theorem Let S, T, and U be sets. (a)If S  T, then | S |  | T |. (b)| S |  | S |. (c)If | S |  | T | and | T |  | U |, then | S |  | U |. (d)If m, n  and m  n, then | {1, 2, …, m} |  | {1, 2, …, n} |. (e)If S is finite, then | S | <  0. 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  0 is the transfinite cardinal number of.

1-14 Copyright © 2013, 2005, 2001 Pearson Education, Inc. Section 2.4, Slide 14 It is customary to denote the cardinal number of by c, for continuum. Since, we have  0  c.  In fact, since is countable and is uncountable, we have  0 < c. Thus Theorem (e) implies that  0 and c are unequal transfinite cardinals. Are there any others? The answer is an emphatic yes, as we see in our next theorem. Definition 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 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.

1-15 Copyright © 2013, 2005, 2001 Pearson Education, Inc. Section 2.4, Slide 15 Theorem For any set S, we have | S | < | P (S) |. Proof: The function g : S  P (S) given by g (s) = {s} is clearly injective, so | S |  | P (S) |. To prove that | S |  | P (S) |, we show that no function from S to P (S) can be surjective. Suppose that f : S  P (S).Then for each x  S, f (x) is a subset of S. Now for some x in S it may be that x is in the subset f (x) and for others it may not be. Let 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 (S) can be surjective, so | S |  | P (S) |. 

1-16 Copyright © 2013, 2005, 2001 Pearson Education, Inc. Section 2.4, Slide 16 By applying Theorem again and again, we obtain an infinite sequence of transfinite cardinals each larger than the one preceding:  0 = | | <    < | P ( )|< | P ( P ( ))|< | P ( P ( P ( )))| Does the cardinal c = | | 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  0 is the smallest transfinite cardinal. What is the first cardinal greater than  0 ? We know that c >  0, but is there any cardinal number such that  0 < < 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.