A Primer on Intuitive Set Theory

Transcription

1 A Primer on Intuitive Set Theory Sam Smith January 2005 No one shall expel us from the paradise which Cantor has created for us. David Hilbert Introduction. In these notes, we give highlights of the theory of sets as developed by Georg Cantor ( ). Cantor s results, most of which he obtained in the 1870s, represent a singular achievement in the history of mathematics. They were at once completely novel and unexpected, and, at the same time, compellingly elegant and natural. Virtually from scratch, Cantor developed a thorough system of measure and comparsion for infinite sets which is now part of the basic vocabulary of mathematics. Cantor s theory was also, from its inception, extremely controversial. Fundamental questions raised by Cantor s methodology and and his stiking results became central issues for both philosophers and mathematicians. Cantor struggled with the controversies himself until his death. We give a brief indication of two important aspects of controversy regarding Cantor s methods in this introduction. As noted by many important thinkers of his day, there was a serious price of admission to Cantor s paradise : To enter, one must take sides on the controversial issue of the existence of actual or completed mathematical infinity. For recall that Aristotle and his adherents argued that to speak of 1

2 an actual infinite totality is incoherent: infinitude is a process, something that is never completed. Thus listing the counting or natural numbers is sensible, but writing the totality 1, 2, 3,... {1, 2, 3,...} of all natural numbers is not. Cantor s first move was to reject this prohibition and to allow for infinite totalities. In fact, Cantor went substantially further. Not only was he willing to consider the possible existence of infinite totalities, in pursuing his revolutionary program Cantor proposed making a detailed mathematical study of their properties. This particular controversy deepens when we discuss other number systems, specifically the so-called real numbers. While a rational number may be succinctly defined as a ratio of two integers, how do we make sense of numbers such as 2 or π or 3 π? A rigorous construction of the real numbers from the rationals was given by a friend and contemporary of Cantor s Richard Dedekind in the 1860 s. In this construction, real numbers appear as certain (infinite!) subsets of the rational numbers called cuts. For our purposes, we will understand real numbers using an idea which goes back to grade school: decimal expansion. We have 2 = , π = , and 3 π = Note that we must conscience a completed infinity just to write down a single real number. A second controversy related to Cantor s work stems from an observation made by Betrand Russell at the turn of the twentieth century. In what became a fundamental issue for both mathematicians and philosophers, Russell showed that the most obvious notion of what a set is leads directly to paradox. Russell s paradox demonstrated the need for careful and precise axiomatic proofs in set theory and logic. Cantor s set theory was ultimately redeveloped axiomatically using the Zermelo-Frankel axioms (Appendix D). Moreover, twentieth century set theorists focused as much on the meaning of a theory of sets, a perspective known as meta-theory, as on proving the 2

3 theorems within this theory. In these notes, we will focus primarily on the latter proving theorems and we will largely ignore the issues raised by Russell s paradox. The non-axiomatic approach we take is referred to as naive or intuitive set theory for this reason. 1. The Basics. In this section we establish some basic terminology and recall some elementary facts about sets. We begin with 1.1 Definitions. A set S is any collection of objects. The members of S are called the elements. We write x S to indicate that the object x is an element of S. If A is a set and every element of A is an element of S we say A is a subset of S and write A S. The set with no elements is called the empty set and denoted by φ. 1.2 Examples. Some important sets which will figure prominently below are 1. The set N of natural or counting numbers; N = {1, 2, 3,...}. For each element m N we have the finite subset N m = {1, 2, 3,..., m} 2. The integers Z = {..., 2, 1, 0, 1, 2...}. 3. The rational numbers Q which is the set of all ratios of integers with nonzero denominator; Q = {p/q p, q Z, q 0}. 4. The real numbers R. As mentioned in the introduction, the representation of elements of R is itself a significant issue. We will express a positive real number x R as an infinite decimal x = b n b n 1 b 1 b 0.a 1 a 2 a 3 where each b j and a i is a digit i.e. a number between 0 and 9 inclusive. This representation is not unique since, e.g., and represent the same number. It turns out that if we make a commitment to uniformly choose the former or the latter in every case where it occurs then our decimal expansion will be unique and well-defined. We elect to always choose the repeating 0 s over 9 s in what follows Now notice that integers and rational numbers are also real numbers. What do their decimal expansions look like? For integers, the answer is immediate. With our convention 1 = etc. As far as fractions go, we have 1/3 =.33333, 1/12 = and 1/7 =

4 Note that in each case, after some preliminary digits, we have a repeating pattern. To be precise, say a decimal expansion x = b n b n 1 b 1 b 0.a 1 a 2 a 3 is repeating if there exists natural numbers k and j such that a i = a i+j for all i k. Note that 1/3 (k = 1, j = 1) and 1/12 (k = 3, j = 1) and 1/7 (k = 1, j = 6) are all repeating. We have Theorem 1.3 A real number x is rational if and only if x has a repeating decimal expansion. Proof. Suppose x has a repeating decimal expansion. We show x is a ratio of two integers. It clearly suffices to assume x is a pure decimal, i.e. has no integer part. In this case, with k and j as in the definition, we may write x =.a 1 a 2 a 3 a k a k+1 a k+j where we are using the usual repeating decimal notation. Notice that multiplying x by powers of 10 serves to shift the decimals of x to the left of the decimal point. In particular, and 10 k x = a 1 a k.a k+1 a k+j 10 k+j x = a 1 a k a k+1 a k+j.a k+1 a k+j. Subtracting, cancels the decimal part and gives (10 k+j 10 k )x = (a 1 a k a k+1 a k+j ) (a 1 a k ). The point is that the right hand side is an integer. Let s call it m. We then have x = m/(10 k+j 10 k ) is a rational number. Conversely, suppose x = p/q is rational for integers p and q. To find the decimal expansion of x we use long division to divide q into p. For example, to find the expansion of 1/7 we divide 7 into : 7 divides into 10 once with remainder 3, our first digit is 1. The remainder 3 dictates the next step: 7 divides into 30 four times with remainder 2, our second digit is 4, 7 divides into 20 twice with remainder 6 our third digit is 2, 7 divides into 60 eight times etc. Now observe that the remainders that we get at each stage are between 0 and q (or in our example 7). Thus the remainder eventually 4

5 repeats. But once the remainder repeats the digits repeat, as needed. We have now introduced the most important numeric sets that we will encounter. It is important to realize, however, that any object can be an element of a set. In particular, sets can themselves be elements of other sets. For instance, we must distinguish between the set A = { 2, N} with two elements and the set B = { 2, 1, 2, 3,...} with infinitely many elements. We assume the reader is familiar with the basic set operations of union, intersection and difference. The definitions are A B = {x x A or x B} A B = {x x A and x B} A B = {x x A and x B}. These operations are related by De Morgan s Laws: Theorem 1.4 For any sets A, B and C we have i) A (B C) = (A B) (A C) and ii) A (B C) = (A B) (A C). Proof. We prove i) and leave ii) as an exercise. To prove two sets S and T are equal, we must show 1) that every element of S is an element of T ; i.e., that S T, and 2) that every element of T is an element of S: T S. So let x A (B C) be any element. Then, by definition, x A and x B C. The latter means that x cannot be in B nor in C. Thus x A and x B and x C which implies x (A B) (A C). Conversely, suppose y (A B) (A C) is any element. Then y A B and y A C. In either case, y A. Moreover, y B and y C and so y B C. Thus y A (B C), as needed. Exercise 1. Prove Theorem 1.4, part ii). We will also consider the cartesian product A B of two sets: A B = {(x, y) : x A and y B}. 5

6 For example, N 2 N 3 = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3)}. In general, if A had m elements and B has n elements then A B has mn elements. Thus cartesian product is like a multiplication for sets. The power set P (A) of a given set A is defined to be the set of all subsets of A. For example P (N 2 ) = {φ, {1}, {2}, N 2 }. Of course, if A is an infinite set then so is P (A). For finite sets, the power set behaves like an exponential. Specifically, we have the following theorem which shows why P (A) is sometimes denoted 2 A. Theorem 1.5 Suppose A has m elements. Then P (A) has 2 m elements. Proof. If A = φ is the empty set then P (A) = {φ} has 1 = 2 0 elements. If A has 1 element, say A = {a}, then P (A) = {φ, A} has 2 elements and the theorem holds for m = 1, as well. But to prove this theorem, we must consider sets of size m, for each possible m N! This seems, at first glance, an impossibility. However, a natural approach is the following: we have proven the theorem true for m = 0 and m = 1. We will now assume the theorem true for some fixed m and argue that, with this assumption, the theorem is true for the next number m + 1. This approach logically establishes the theorem for all m; it is called the principle of induction. So fix m and assume P (A) has 2 m elements for all sets with m elements. Let B be any set with m + 1 elements. We need to show that P (B) has 2 m+1 or twice as many elements as P (A). Now, since B has m+1 elements, we may write B = A {b} where A has m elements and b A. By our assumption, the theorem is true for A. That is, there are exactly 2 m subsets of A. Now notice that there are exactly two kinds of subsets of B: those which have b in them and those which do not. The first kind of subsets are precisely the subsets of A. Thus there are 2 m of them. Let us call them A 1,..., A 2 m. Then notice that there are also exactly 2 m subsets of B of the second type namely, A 1 {b},..., A 2 m {b}. Thus P (B) has twice as many elements as P (A), as needed. Exercise 2. Let A = {1, 2, 3} and B = {3, 5, 7}. Write down the elements 6

7 of the sets a) P (A B) b) A P (A B) c) P (A) P (B). 2. One-to-One Correspondences and Cardinality of Sets. Given two sets A and B, a natural question to ask is whether they have the same size. When A and B are finite we can just count elements and see. When A and B are infinite, however, this is a more interesting question. How do we compare set sizes when both are infinite? An answer comes from examining the meaning of counting. When we count a set A, we decide which element of A is number 1, which is 2 and which is 3 etc. That is, we define a function from N m to A where m is the size of A. Morevoer, this function assigns exactly one element of N m to each element of A. This is called a one-to-one correspondence. We make this notion precise with 2.1 Definitions. A function f : A B between two sets A and B is a rule which assigns to each element x A a uniquely determined element y = f(x) B. A function f : A B is one-to-one if, for every pair x 1, x 2 A, f(x 1 ) = f(x 2 ) in B implies x 1 = x 2. A function f : A B is onto if, for every element y B, there exists x A such that f(x) = y. A function f : A B which is both one-to-one and onto is called a one-to-one correspondence between A and B. 2.2 Examples. 1. The function f : Z Z given by f(x) = 2x is one-to-one. For suppose f(x 1 ) = f(x 2 ) for some integers x 1, x 2. Then 2x 1 = 2x 2 which implies x 1 = x 2. Note that f is not onto since, e.g., 3 is not of the form 2x for any integer x. In fact, f defines a one-to-one correspondence between the set of all integers and the set of all even integers. This is an example of Galileo s paradox: an infinite set can be in one-to-one correspondence with a proper subset of itself! 7

8 2. The function f : (0, 1) (1, + ) 1 defined by f(x) = 1/x is a oneto-one correspondence. For suppose f(x 1 ) = f(x 2 ) for some x 1, x 2 (0, 1). Then 1/x 1 = 1/x 2 which clearly implies x 1 = x 2. For onto, note that if y (1, + ) then y > 1 which implies 0 < 1/y < 1. Thus take x = 1/y and we have f(x) = y. 3. There exists a one-to-one correspondence between the interval ( 2, 2) and R. Here we must define a function g : ( 2, 2) R ourselves. With Example 2 in mind, we set g(x) = 1/x for x ( 1, 0) (0, 1). Note that g is one-to-one and maps onto (, 1) (1, + ). Next we make g(0) = 0. Now all that s left to do is find a one-to-one correspondence between ( 2, 1] [1, 2) and [ 1, 0) (0, 1]. We accomplish this using linear functions: for x ( 2, 1] let g(x) = 2 x and for x [1, 2) let g(x) = 2 x. It is then an easy exercise (cf. Exercises 3 and 4 below) to show that g restricted to ( 2, 1] [1, 2) is one-to-one and onto. Thus g : ( 2, 2) R is a one-to-one correspondence. Our definiton of g can be summarized as g(x) = 2 x 2 < x 1 1/x 1 < x < 0 0 x = 0 1/x 0 < x < 1 2 x 1 x < 2. It may help to graph g to see how it works. Exercise 3. Define f : (0, 1) ( 1, 3) by f(x) = 4x 1. Prove that f is a one-to-one correspondence. Exercise 4. Prove that, given any two points a, b there is a one-to-one correspondence between the interval (a, b) and ( 2, 2). (Try a linear function!) One-to-one correspondence allows us to decide when two sets have the same size, regardless of the meaning of the elements. Let us write A B if there exists a one-to-one correspondence f : A B. Notice that A B is either true or false for any given ordered pair of sets A, B. That is, there either is or is not a one-to-one correspondence f. We say that is a relation 1 Given two real numbers a < b, (a, b) = {x R a < x < b}. We allow a = or b = + to indicate unbounded intervals to the left or right, respectively. Square brackets indicate the endpoint (when not infinite) is included. 8

9 on any set of sets. In fact, is called an equivalence relation. The formal definitions are as follows. 2.3 Definitions. Let S be any set. A relation on S is a function R : S S {T, F }. Here R(x, y) = T means x is related to y written x y while R(x, y) = F means x is not related to y or, notationally, x y. A relation on a set S is an equivalence relation if the following three properties hold: i) (reflexivity) x x for all x S, ii) (symmetry) x y implies y x for all x, y S, and, iii) (transitivity) x y and y z imply x z for all x, y, z S. Equivalence relations behave like equality, =, which is the prototype. Note that there are many relations which are not equivalence relations. For example, inequality and strict inequality < on the reals and rationals. These relations are prototypes for a class of relations called order relations. We will study these in 6. The following fundamental result is proved using inverses and compositions of functions. We recall these notions and give the proof in Appendix A. Theorem 2.4 Let S be any set of sets. 2 The relation of one-to-one correspondence is an equivalence relation on S. Ultimately, in 6, we will assign to each set A a number A called the cardinal number or cardinality of A which captures the size of A. When A is finite, A will equal the number of elements of A. When A is infinite the definition of the cardinal number of A is trickier to make. For the moment, we will content ourselves with defining three relations: equality A = B, inequality A B and strict inequality A < B of cardinals numbers: 2.5 Definitions. We say the cardinality of A is equal to the cardinality of B and write A = B if there exists a one-to-one correspondence f : A B; i.e., if A B. We say the cardinality of A is less than or equal to the cardinality of B 2 We would like S to be the set of all sets. Unfortunately, the existence of such a universal set leads directly to paradoxes! 9

10 and write A B if there exists a one-to-one function f : A B. If A B but there is no one-to-one correspondence between A and B we say the cardinality of A is strictly less than that of B and write A < B. In our new notation, Examples 1,3 and Exercise 4 now say Z = {even integers} ( 2, 2) = R and (a, b) = ( 2, 2) for any a < b. Notice how natural and necessary the reflexive, symmetric and transitive laws are. In particular, we can write (a, b) = ( 2, 2) = R. This sentence has meaning: it says that there are as many points in any interval of R as there are in all of R! It is important to realize that just because we call our relation the equality of cardinal numbers does not mean that it will behave like equality. We still have to prove the desired properties (Theorem 2.4) in Appendix A. As a case in point, consider our notion of inequality. Note that, for any sets A, B if A B then A B. The needed one-to-one function i : A B is called the inclusion function. It is defined by i(a) = a for all a A. (Of course, i is onto if and only if A = B.) Since the closed interval [ 2, 2] is a subset of R, we have [ 2, 2] R. By the same token, the open interval ( 2, 2) is a subset of the closed interval [ 2, 2] and so ( 2, 2) [ 2, 2]. By Example 3 above, R ( 2, 2). Taking the composition (see Theorem A.4) we obtain R [ 2, 2]. Thus both [ 2, 2] R and R [ 2, 2] are true. Can we conclude that [ 2, 2] = R? What is the one-to-one correpondence g : [ 2, 2] R? The function f : ( 2, 2) R defined in Example 3 does not extend to [ 2, 2]. We made crucial use of the fact that the 2 and 2 were not included when we defined f. Besides f is already onto; where would we send the extra points 2 and 2? We must construct an altogether new one-to-one correpondence and it is not at all clear how! In general, the question we are asking is whether, for any sets A and B, if A B and B A can we conclude that A = B? Such a theorem would require producing, out of thin air, a one-to-one correspondence from two unrelated one-to-one functions. That this can be done is the substance of a deep theorem conjectured by Cantor and proved by Schroeder and Bernstein. We give the proof in Appendix B. 10

11 Theorem 2.6 (Schroeder-Bernstein) Suppose f : A B and g : B A are one-to-one functions. Then there exists a one-to-one correspondence between A and B. 3. Countable and Uncountable Sets. Definition 3.1. We say a set is A is countable if A N. If A = N then we say A is countably infinite. If A is neither countable nor countably infinite we say A is uncountable. Note that all finite sets are countable. Of course, the natural numbers N themselves are countably infinite. Suppose A is countably infinite. Let f : N A be a one-to-one correspondence. Let a 1 = f(1) A, a 2 = f(2) A, etc. Then, since f is one-to-one, each of the a i are distinct elements of A. Since f is onto A = {a 1, a 2, a 3,...}. Conversely, if A = {a 1, a 2, a 3,...} we can define a one-to-one correspondence between N and A directly: namely, send 1 to a 1, 2 to a 2, etc. Thus the countably infinite sets are precisely those whose elements can be listed or counted in an exhaustive way. This observation gives a method for proving a set A is countable; namely, we show that there is an exhaustive way to list the elements of A in the form above. For example, Theorem 3.2 The integers Z are countably infinite. Proof. Note that Z = {0, 1, 1, 2, 2, 3, 3,...}. This is clearly an exhaustive list. Somewhat more interesting is Theorem 3.3 The rationals Q are countably infinite. Proof. It suffices to prove that the postive rationals, written Q +, are countably infinite. For we can then use the trick for dealing with negatives and zero above. To each postive rational p/q we associate the natural number p + q, the sum of numerator and denominator. Of course, there are many rationals for 11

12 each sum. For example, 1/4, 2/3, 3/2 and 4/1 all have sum 5. In fact, there are clearly n 1 different rationals for each sum n > 1. We will order the rationals first by the size of this sum and then, for each sum n, in increasing order of numerator. What we obtain is the following list of the positive rationals: Q + = {1/1, 1/2, 2/1, 1/3, 2/2, 3/1, 1/4, 2/3, 3/2, 4/1,...}. Note that this is an exhaustive list: a given rational p/q appears in the pth position among the rationals with sum p + q. Unfortunately, there are many repetitions on this list, e.g., 1/1, 2/2,.... However, if we delete all representations of a positive rational except the first, what results is the needed exhaustive list. The countability of the rationals is a surprising result since there appear to be many more rationals then integers. For example, what is the smallest rational number greater than zero? A moment s thought reveals that there is none. In fact, between any two real numbers r and s there always lies a rational number. For this reason, the rationals are said to be dense in R. Nonetheless, we managed to find a new way of ordering the rationals which exhausted them all. In perhaps his most famous result, Cantor showed that such a trick could not be played on the real numbers. The technique of the proof is known as Cantor s Diagonal Argument. Theorem 3.4 The real numbers R are uncountable. Proof. It suffices to prove that the interval (0, 1) is not countably infinite. We will show that in any list of real numbers between 0 to 1 there will always be at least one missing number between 0 and 1. The result follows. So let x 1, x 2, x 3, x 4,... be any list of real numbers in (0, 1). We may write these number in their decimal expansion: x 1 =.a 11 a 12 a 13 a x 2 =.a 21 a 22 a 23 a x 3 =.a 31 a 32 a 33 a x 4 =.a 41 a 42 a 43 a

13 Here a ij is the jth decimal digit of x i. Define a number z digit by digit as follows: Let z 1 be any integer from 0 to 9 other than a 11. Let z 2 be any integer from 0 to 9 other than a 22. Continue in this way to obtain z 3, z 4,... with the property that z i a ii. 3 Let z =.z 1 z 2 z 3 z 4... Then z (0, 1) since it is a pure decimal. Moreover, z x i for all i because z and x i differ in the ith digit. With our cardinality notation, Theorems can be summarized in one line: N = Z = Q < R. A natural question now arises: Is there any set A whose cardinality lies strictly between that of the natural numbers and the reals? We know A cannot be an interval (a, b) since (a, b) = R. On the other hand, A must be larger than the rationals. Based on this evidence, Cantor made the following conjecture: 3.5 The Continuum Hypothesis There is no set A with N < A < R. Proving or refuting Cantor s Hypothesis and its generalization (see 5) were listed among the most important open problems by David Hilbert in his famous address to the International Congress of Mathematicians at the turn of the twentieth century. Significantly, the ultimate resolution of these problems were not theorems in set theory but rather meta-theoretic results to the effect that neither a proof nor a disproof of these hypotheses is possible within the context of set theory. 4. Countability of Unions and Cartesian Products. In this section, we consider the extent to which countability is preserved under the set operations unions and cartesian products. We settle the question for unions and leave cartesian products as exercises. Our first result 3 To be more precise, we should say z i = 2 if a ii = 1 and z i = 1 if a ii 1. 13

14 asserts that a countable union of countable sets is still countable. Theorem 4.1 Let A 1, A 2, A 3,... be any collection of countable sets. Then the union A = A 1 A 2 A 3 is also a countable set. Proof. We may as well assume each set A i is countably infinite and that there are no intersections. As usual, the former means we can exhaustively list the elements of A i. Write A i = {a i1, a i2, a i3,...} so that a ij is the jth element of A i. We can then obtain an exhaustive listing of A by ordering the elements a ij in increasing order of the sum of the subscripts i + j. For each sum we list in increasing order of i. Specifically, is the needed exhaustive list of A. A = {a 11, a 12, a 21, a 22, a 13, a 22, a 31,...}, We can now deduce a remarkable consequence of the fact that Q < R. Let I = R Q, the irrational numbers. Irrational numbers are characterized as having infinite, nonrepeating decimal expansions. It is elementary to prove that 2 I. In fact, for all m N, m I unless m is a perfect square. Other famous irrational numbers include π and e. Corollary 4.2 The irrational numbers I are uncountable. Proof If I is countable then R = Q I is also, contradicting Theorem Exercise 5. Let S be any infinite set. Prove that S has a countably infinite subset A. Thus N is the smallest infinite cardinal. (Your argument will require the Axiom of Choice.) Exercise 6. Let S be any infinite set and B any countably infinite set. Prove that S B = S. Hint: Use Exercise 5. Exercise 7. Prove that a set A is infinite if and only if A is in one to one 14

15 correspondence with a proper subset of itself. Exercise 8. Let A and B be countable sets. Prove that the cartesian product A B is also countable. Exercise 9. Use induction and Exercise 5 to prove that, for any n 1, if A 1, A 2,..., A n are countable sets then so is A 1 A 2 A n. Exercise 10. Prove that the infinite cartesian product N 2 N 2 is an uncountable set. Hint: Note that a typical element of this set is an infinite sequence of 1s and 2s. Use Cantor s Diagonal Argument. Corollary 4.2 was, as you can imagine, anathema to the many nineteenth century mathematicians and philosophers who did not believe in the existence of irrational numbers. Cantor not only proved the existence of such numbers (without actually producing one example!), but he proved that there are fundamentally more of these strange numbers then there are the familiar rationals. Corollary 4.2 shows that, although the rationals are dense in R, they are actually very rare; most real numbers are irrational. An even more surprising result uses the full strength of Theorem 4.1. Define a real number a to be algebraic if a is a root of any polynomial P (x) = a n x n + a n 1 x n 1 + a 1 x + a 0 whose coefficients a 0, a 1,..., a n are all rational numbers. Let A be the set of all algebraic numbers. Of course, Q A since every rational satisfies a linear polynomial with rational coefficients. Note that m A for all rational m: the polynomial is P (x) = x 2 m. Thus many irrational numbers are algebraic numbers. Nonetheless, we have Theorem 4.3 The algebraic numbers A are countable. Proof. We will express A as a countable union of countable sets. The result then follows from Theorem 4.1. Let A 1 be the set of all real numbers which satisfy linear polynomials with rational coefficients. Then A 1 = Q and so is countable. Let A 2 be the set of all real numbers which satisfy quadratic polynomials with rational coefficients. A quadratic polynomial P (x) is determined by three rational numbers, a 0, a 1, a 2, the constant, linear and quadratic coeffficients. Thus the set of all quadratic polynomials with rational coefficients is 15

16 in one-to-one correspondence with the cartesian product Q Q Q. This set is countable by Exercies 7. Thus we can exhaustively list the quadratic polynomials with rational coefficients: P 1 (x), P 2 (x),.... Now each quadratic polynomial has at most two roots. Let a i1, a i2 be the roots for P i (x). (If P i (x) has duplicate roots we should delete one of these; if P i (x) has no roots we will just omit P i (x) altogether.) We can now exhaustively list the elements of A 2 ; namely, A 2 = {a 11, a 12, a 21, a 22, a 31, a 32,...}. In general, let A n be the set of all real numbers which satisfy nth degree polynomials with rational coefficients. The set of all nth degree polynomials with rational coefficients is countable, by Exercise 7 again, since it is in one-to-one correspondence with the cartesian product of n + 1 copies of the rationals. Thus we can enumerate these polynomials: P 1 (x), P 2 (x),... and let a i1,..., a in be the roots of P i (x) with the same caveats above. An exhaustive listing of A n is then given by A n = {a 11, a 12,..., a 1n, a 21, a 22,..., a 2n, a 31, a 32,... a 3n,...}. Finally, observe that A = A 1 A 2 A 3. Now let T = R A; T is called the set of transcendental numbers. These are real numbers which do not satisfy any polynomial with rational coefficients. It took two centuries for mathematicians to prove that π is transcendental. The proof, given in 1882 by C.L.F. Lindemann, was one of the most celebrated results of the nineteenth century. Nine years earlier, using difficult techniques of complex analysis Charles Hermite had proved that e was transcendental. Thus the following corollary to Theorem 4.3 is remarkable since it guarantees that most real numbers are transcendental despite the fact that we can only name a few! Corollary 4.4 The transcendental numbers T are uncountable. 5. Power Sets and Sizes of Infinity. Cantor s Diagonal Argument (Theorem 3.4) establishes that there are different sizes of infinity. Specifically, we know that there are countably infinite sets like N and Q and then there are the real numbers R which are 16

17 uncountable. The Continuum Hypothesis proposes that there are no set sizes in between. In this section we address the question whether there are any cardinal numbers larger then R. We begin, in the spirit of the previous section, by proving that countability is not preserved by the power set operation. Theorem 5.1 The power set of the natural numbers P (N) is uncountable. 4 Proof. We use Cantor s Diagonal Argument to argue that there can be no exhaustive list of all subsets of N. So suppose that A 1, A 2, A 3,... is any list of subsets of N. We construct a subset A N which is not on the list. If 1 A 1 we do not include 1 in A. If 1 A 1 then we do put 1 in A. Similarly, we include 2 in A if and only if 2 A 2 and so on for each i N. The precise definition of A is A = {i N i A i }. It is obvious that A does not coincide with any A i and so is not on the list. To find a set with cardinality larger than that of R we might look at the power set P (R). Since P (N) > N, perhaps the same is true for R. The following result, known as Cantor s Theorem, asserts remarkably that this not only works for N and R but for any set A! Theorem 5.2 For any set A, A < P (A). Proof. We must prove 1) that there is a one-to-one function f : A P (A), and 2) that there can not be any one-to-one function g : P (A) A. Part 1) is easy: Define f : A P (A) by f(x) = {x} for all x A. It is obvious that f is one-to-one since {x} = {y} as subsets of A if and only if x = y in A. To prove 2), suppose there is a one-to-one function g : P (A) A. We then define a function h : A P (A) as follows: Set h(a) = φ if a A is not in the image of g. Otherwise, a = g(x) for some unique subset X A (since g is one-to-one). Set h(a) = X. Observe that, by definition, h is onto. Consider the subset R of A defined by R = {x A x h(x)}. In words, R is the set of all elements of A which are not members of their associated subset h(x) in P (A). Since h is onto, there must be some element a A with h(a) = R. 4 Using binary numbers, it is easy to prove the stronger result that P (N) [0, 1]. 17

18 We now ask the following simple question: Is a R? Suppose the answer is yes; i.e. that a R. Then a R = h(a). Thus a does not satisfy the condition for membership of R. But this means a R. The answer yes is contradictory. Suppose, on the other hand, the answer is no; that a R. Then a R = h(a) and so a h(a). But by its very definition, a should be in R. Thus no will not do for an answer either. The only conclusion we can make in the face of this contradiction is that the original assumption of a one-to-one function g : A P (A) is impossible. Thus A < P (A). Cantor s Theorem has the amazing consequence that there are infinitely many sizes of infinity. Specifically we have the following list of sets whose cardinalities are in strictly increasing order: φ, N 1, N 2,..., N m,..., N, R, P (R), P (P (R)), P (P (P (R))),... Since R P (N) (see Footnote 4, above) the infinite sets in this sequence are all generated by the power set operation. The question occurs, of course, whether there are other sets which can be inserted in this sequence. Cantor s Generalized Continuum Hypothesis proposes that the answer is no: 5.3 The Generalized Continuum Hypothesis. Given any infinite set A, there is no set B with A < B < P (A). Exercise 11. Let A be any countable set and n any natural number. Prove that the set of all subsets of A with exactly n elements is a countable set. (Try induction.) Exercise 12. Let A be any countable set. Prove that the set F (A) consisting of all finite subsets of A is a countable set. (Use Exercise 11.) 6. Well-Orderings, Ordinal and Cardinal Numbers. The purpose of this section is to develop the theory of cardinal numbers of sets. Our goal is to assign to each set A a number A called the cardinal 18

19 number of A which will capture the size of A. Of course, we recognize that if A is an infinite set then A will not be a number in any ordinary sense of the word. What do we intend A to mean? What do we want a system of cardinal numbers to do for us? The second question is easier to answer. We would like our cardinal numbers to allow us to compare any pair of sets A and B: to say either A and B have the same cardinal number, or, that one has a larger cardinal number than the other. This goal is the basis for our construction. In 2, we defined the inequality of cardinal numbers A B for any two sets A, B to mean that there exists a one-to-one function f : A B. Recall that A B defines a relation on any set of sets. The Schroeder- Bernstein Theorem shows that the inequality of cardinal numbers behaves as expected with respect to the equivalence relation, one-to-one correspondence. Specifically, A B and B A implies A = B. So what s the problem? Well, very simply, given two sets A and B do we know that one of the following A B or B A must be true? Given two sets A and B must there either be a one-to-one function f : A B or, if not, a one-to-one function from g : B A? If you think about it for a moment, you ll agree that it seems highly plausible. Unfortunately, there also seems to be no direct way to prove it. The answer, as we will see, is yes. Happily, the constructions needed for the proof also furnish us with a natural way to define the cardinal numbers. The properties we want to be true for inequality of cardinal numbers are those of an order relation. The official definition is 6.1 Definition. Let S be a set and a relation on S. Then is called an order relation on S if i) (comparability) for all x, y either x y or y x, ii) (antisymmetry) for all x, y S x y and y x if and only if x = y and iii) (transitivity) for all x, y, z S if x y and y z then x z. If is an order on a set S we write x < y whenever x y but x y. It is important to notice that inequality of cardinal numbers does not define an order relation on most sets of sets. The problem is not i), which we will prove below, but ii) even with the Schroeder-Berstein Theorem. The point is that the equality referred to in ii) is genuine equality. If A B and B A then we know A = B. But this just means that there is a one-to-one correspondence between A and B, not that they are equal as sets. 19

20 Suppose, however, that we focus on a collection of sets no two of which are in one-to-one correspondence. Then A B and B A would imply A = B as sets! This observation becomes our strategy. We will prove the comparability of all sets. We will then construct a canoncial collection of sets C with the property that for any set A, A C for exactly one set C in our collection. The sets C will be called the cardinal numbers. We thus undertake to prove the comparability of all sets under the relation of inequality of cardinal numbers. Our method will be to consider the class of well-ordered sets, prove that these sets are all comparable, and then prove actually every set can be well-ordered. We begin with 6.2 Definition. Let be an order relation on a set S. We say that well-orders S if, for every nonempty subset A S, there exists an element a 0 A with the property that a 0 a for all a A. 6.3 Examples. 1) The natural numbers N are well-ordered by the usual inequality relation. However, none of the other number sets Z, Q, R are well-ordered by inequality since many subsets of these sets will not have a least element. 2) Given any set S of sets, we have the relation of set inclusion which clearly satisfies ii) and iii). Of course, will usually not be an order because most sets are not comparabe. A set C of sets which is ordered by is called a chain. For example, {φ, {1}, {1, 2}, {1, 2, 3},..., N} is a chain contained in P (N). 3) Since sets can themselves be elements of sets, is a relation on any set S of sets also. Note that, unless there is a set A S which is an element of itself, pairs of sets A, B S will not satisfy both A B and B A. Thus condition ii) above will be vacuously true. Condition iii) is obviously true. But, again, i) will be false for most S. Examples of sets S which are ordered by can be obtained by taking any set A and forming the set B = A {A} whose elements are those of A plus one more element, the set A itself. Of course, A B. Next let C = B {B}. Then S = {A, B, C} is well-ordered by. We could continue this process to obtain an infinite set well-ordered 20

21 by, namely S = {A, A {A}, A {A {A}}, A {A {A}} {A {A {A}},...}. Note that S N. Some important features of well-ordered sets follow directly from the definition. Let S be well-ordered by a relation. Taking S to be the subset of S in the definition, we see that S has a least element, say s 0. Now let S 1 = S s 0. If S 1 is nonempty, then it has a least element, say s 1. Continuing, we obtain an increasing list of elements s 0 < s 1 < s 2 < s 3 < of S. Does this mean that S is countable? Not necessarily. There could be many elements of S not obtainable in this way. While every non-maximal element in a well-ordered set S has an immediate successor (Exercise 13, below) many elements besides the minimal element s 0 may not have an immediate predecessor. For example, in the chain given above in P (N), the element N has no immediate predecessor. A given set S well-ordered by contains many well-ordered subsets. Given any element x S we form the initial segment determined by x by letting S(x) = {s S s < x}. Clearly, S(x) is itself well-ordered by. Note that x S(x). Exercise 13. Let S be a set well-ordered by. Suppose A S has the property that, for all s S, if s a for some a A then actually s A. Prove that, either A = S or A is an intial segment of S; i.e., that A = S(x) for some x S. (Hint: Consider the complement of A in S.) Exercise 14. Let S be a set well-ordered by. Let x S. Suppose X is not maximal in S; i.e., there exists an element z S with z > x. Prove that there then exists an element s(x) in S, called the immediate successor, of x such that i) x < s(x) and ii) if x < z then s(x) z, as well. (Hint: Consider the complement of the set S(x) {x} in S.) We now set out to prove that all well-ordered sets are comparable by oneto-one functions. That is, for all well-ordered sets S, T, either S T or T S. Our method is to single out a canonical class of well-ordered sets called ordinals and prove that every well-ordered set is in one-to-one correspondence with one of these. We then prove that the ordinals are themselves 21

22 all comparable. Our definition recalls Example 3 above. 6.4 Definition. We say a set O of sets is an ordinal if i) O is well-ordered by and ii) if A is any object and A S for some set S O then actually A O. Of course, the empty set φ is an ordinal. The sets S constructed in Example 3, however, are not likely to be ordinals. For if A has any elements then these must be included in S by ii), which they are not. In fact, the only way that the construction in Example 3 will work is if we take A = φ. We have then the following list of ordinals: φ, {φ}, {φ, {φ}}, {φ, {φ}, {φ, {φ}}}, {φ, {φ}, {φ, {φ}}, {φ, {φ}, {φ, {φ}}},.... In general, if O is an ordinal then so is the set O {O}. This operation, which produces a new ordinal from a given one, is called the successor operation. Let us write succ(o) = O {O}. Note that the sequence of ordinals above was generated by the successor operation. For convenience, we will give these ordinals new names. Let 0 be the empty set φ. Let 1 = succ( 0). In general, let n = succ(n 1) for n N. We have simply made an explicit correspondence between the nonempty sets on our list and the natural numbers. Notice that all the all the ordinals created so far form a chain; that is, they are ordered (in fact, well-ordered) by. Thus our next result gives a method for creating a new ordinal from these. Theorem 6.5 Let {O α } be any chain of ordinals. Then the union O = α O α is an ordinal, as well. Proof. We must first check the three conditions in Definition 6.4. Let A, B O. Then, since the O α form a chain, we can find O α large enough so that both A, B O α. But now O α is an ordinal and so, in particular, ordered by. Thus A and B are comparable with regard to and obey antisymmetry. Of course, the transitivity of holds for any set of sets. 22

23 To see that well-orders O, let A be any nonempty subset of O. Choose O α so that A O α φ. Let S A O n be the least element of this nonempty subset of the ordinal O α. Suppose there exist R A with R S. Then, by property ii) in Definition 6.4, R O α ; i.e., R A O α But this means S is not the least element of this subset and so there can be no such R. Thus S is the least element of A. Finally, suppose R S where S O. Again, find O α large enough so that S O α. Since O α is an ordinal R O α and so R O. The ordinal O in Theorem 6.5 is called the limit ordinal of the chain {O α }. We obtain our first infinite ordinal by taking the limit of the finite ordinals listed above. Specifically, let ω = Note that ω N. This construction of the natural numbers is due to John Von Neuman. Of course, now that we have ω we can use our successor operation to obtain another list of new ordinals. Given any ordinal O, let O + n denote the result of applying the successor function n-times to O. We then have the following list of infinite ordinals: ω, ω + 1, ω + 2, ω + 3,..., ω + n,.... Taking the limit of these ordinals, we obtain a new ordinal which we call ω + ω or ω 2. And, having produced ω 2, we can apply the successor function repeatedly to obtain ω 2 + n. Taking the limit ordinal yields ω 3 and so on. Moreover, we can take the limit of all the ordinals obtained by this process to get ω ω = ω 2. Exponentials are obtain similarly. To summarize, 23

24 we have the following list of ordinals in increasing order of : φ, 1, 2, 3,..., n,... ω, ω + 1, ω + 2, ω + 3,..., ω + n,... ω 2, ω 2 + 1, ω 2 + 2, ω 2 + 3,..., ω 2 + n,... ω 3, ω 3 + 1, ω 3 + 2, ω 3 + 3,..., ω 3 + n,..... ω n, ω n + 1, ω n + 2, ω n + 3,..., ω n + n,..... ω 2, ω 2 + 1, ω 2 + 2, ω 2 + 3,..., ω 2 + n,... ω 2 + ω, ω 2 + ω + 1, ω 2 + ω + 2, ω 2 + ω + 3,..., ω 2 + ω + n,..... ω 3, ω 3 + 1, ω 3 + 2, ω 3 + 3,..., ω 3 + n,..... ω n, ω n + 1, ω n + 2, ω n + 3,..., ω n + n,..... ω ω, ω ω + 1, ω ω + 2, ω ω + 3,..., ω ω + n,.... It will turn out that the ordinals obtainable by the process described above are actually all the ordinals. In fact, we will ultimately prove that, up to one-to-one correspondence, every set appears on this list somewhere! To begin, we prove that every well-ordered set appears on this list. Actually we prove a little more. Given sets S and T ordered by and, respectively, we say a function f : S T is order-preserving if x 1 x 2 in S implies f(x 1 ) f(x 2 ) in T. An order preserving one-to-one correspondence f : S T is called an isomorphism of ordered sets. In this case, we write S = T. Theorem 6.6 Let S be any set well-ordered by a relation. Then S is isomorphic to a unique ordinal O. In fact, O is obtained by repeated use of the successor function and limit process applied to the empty set, as described above. Proof. If S is empty there is nothing to prove. Suppose S has one element.. 24

25 Note that every one element set is trivially well-ordered. Thus we should prove that there is only one ordinal with one element, namely 1 = {φ}. Well suppose O = {A} is a one element ordinal. Suppose a A. Then, by the definition of an ordinal, a O which is false. Thus A = φ. Of course, we won t get far in this step-by-step manner. Thus we shift gears. Using Theorem 6.5, we show that if our current theorem is true for an intial segment S(y) of a well-ordered set S then it is true for S(y) {y}, as well. Specifically, suppose y S has the property that for all x < y there is an isomorphism of S(x) {x} onto a unique ordinal created by our process. Let us write O x for the unique ordinal from our list and f x : S(x) {x} O x for the isomorphism. Of course, the ordinals O x for x < y form a chain: any collection of ordinals created by our process forms a chain! Thus let O = O x. x<y By Theorem 6.5, O is an ordinal. We define an isomorphism f : S(y) O as follows. Given x S(y) since x < y we can use the isomorphism f x and set f(x) = f x (x). The point here is that the functions f x are all compatible. That is, t x implies f t = f x on S(t) {t}) since both are order preserving one-to-one correspondences. We can now extend f to an isomorphism g : S(y) {y} O {O} by setting g(x) = f(x) for x S(y) and g(y) = {O}. It is obvious that g is an isomorphism. Moreover, since each of the O x were unique ordinals it is clear the the ordinal O {O}, is also unique. We can now complete the poof with one trick. Define a subset R of S by R = {y S for all x y, S(x) {x} = O for some unique ordinal O obtainable by our process.} We wish to prove R = S. First observe that R φ. For, if s 0 is the least element in S then trivially s 0 R by the first paragraph of the proof. Now suppose, for a contradiction, that R S. Using Exercise 13, we see that, by its very definition, R is an initial segment of S. Write R = S(y 0 ) for some y 0 S. Recall that y 0 R. Using the second paragraph of the proof, we can find an isomorphism S(y 0 ) {y 0 } onto a unique ordinal O. But this means y 0 R = S(y 0 ), a contradiction. We conclude that R = S, as needed. 25

26 Corollary 6.7 Let O be any ordinal. Then O is obtained by repeated use of the successor function and limit process applied to the empty set. Proof. This is a direct consequence of the uniqueness in Theorem 6.6. Corollary 6.8 Let S and T be any two well-ordered sets. S T or T S. Then either Proof. By Theorem 6.5, S = O 1 and T = O 2 for some ordinals O 1, O 2. But the ordinals are clearly ordered by by our construction. Thus either O 1 O 2 or O 2 O 1. Since set inclusion determines a one-to-one function, the result follows from the transitivity of inequality of cardinal numbers. With Corollary 6.8, we are one step away from completing our goal. All we need to know now is that every set can be well-ordered by some order relation. This remarkable fact is known as the Well-Ordering Theorem. We give the proof in Appendix C. Theorem 6.9 Given any set S there exists an order relation such that S is well-ordered by. With the Well-Ordering Theorem, Theorem 6.6 and Corollary 6.8 now holds for all sets. Corollary 6.10 Given any set S there exists a unique ordinal O such that S O. Corollary 6.11 Given any pair of sets S, T either S T or T S. It is interesting to consider what a well-ordering of the reals R would look like. What would be the least element? How would the order be defined? Unfortunately, the proof of the Well-Ordering Theorem doesn t give much of a clue. A natural question arises from Corollary 6.10; namely, which ordinal on our list corresponds to R? Answering this question amounts to resolving the 26

27 Continuum Hypothesis. We conclude this section by defining the cardinal numbers. Remember that we would like to have exactly one cardinal number for each set size. Thus we define the cardinal numbers to be those ordinals which occur first on our ordered list for each given cardinality. To be precise, we make 6.12 Definition. A cardinal number is an ordinal O with the property that, if O is another ordinal with O O, then O < O. Thus the first cardinal number is 0 = φ, the second is 1 = {φ}. Each of the ordinals n are cardinal numbers; these are called the finite cardinal numbers. We may write N n = n. Of course, it is more natural to drop the overline notation and write n = 0, 1, 2,... for the finite cardinals. The next cardinal number is the limit ordinal ω. It is customary to denote this cardinal number by ℵ 0. From Sections 3 and 4, have N = Z = Q = A = ℵ 0. Now note that ω + n for all n N is still a countable set since the finite union of countable sets is countable. In fact, by Theorem 4.1, ω n is countable since it is a countable union of countable sets. Of course, we know that, at some point, we will encounter an uncountable ordinal on our list. Just when, is a very good question! Let ℵ 1 denote this first uncountable ordinal. We can restate Cantor s Continuum Hypothesis as: 6.13 The Continuum Hypothesis. R = ℵ 1. By Cantor s Theorem (Theorem 5.2), given any set A there exists a set, namely the power set, P (A) with strictly larger cardinality. Using Corollary 6.10, we see that to each ordinal O we can assign a cardinal number ℵ O so that the ℵ O are in strictly increasing order of cardinality. The list of cardinal numbers thus begins 0, 1, 2,..., ℵ 0, ℵ 1, ℵ 2,..., ℵ ω, ℵ ω+ 1, ℵ ω+ 2,.... The Generalized Continuum Hypothesis can be restated as 6.14 The Generalized Continuum Hypothesis. For any ordinal O, P (ℵ O ) = ℵ O