Peter Kahn. Spring 2007

Transcription

1 MATH 304: CONSTRUCTING THE REAL NUMBERS Peter Kahn Spring 2007 Contents 1 The Natural Numbers The Peano Axioms Proof by induction Comparing integral systems Definition by Induction Addition Ordering the natural numbers Variations on induction Well-ordering Multiplication The natural numbers and cardinality Except for illustrative examples, the following sections describe a systematic construction of the real numbers using only the basic concepts of set theory. They follow the standard progression of such constructions: natural numbers, integers, rational numbers, and real numbers. Exercises with asterisks are considered challenging. 1 The Natural Numbers The natural numbers N consist of the positive integers 1, 2, 3,..., together with the number 0. Whereas small positive integers have been used, with varied symbolic representation, since prehistoric times, the number 0 is a relatively late addition The construction of the real numbers described in Chapter 3 of Robert Stoll s 1963 book Set Theory and Logic, W.H. Freeman and Company, San Francisco, London, inspired the development in these notes. There will be occasional references to Stoll s book. c May 21,

2 (estimated at around 300 B.C.), hence slightly less natural than the others. Thus, some mathematicians exclude it from N. This doesn t change the theory in any significant way, just the form of certain definitions and proofs. It turns out to be slightly more convenient to include 0 in N, so this is what we (and most contemporary mathematicians) do. There are a number of different approaches to constructing the natural numbers. Our point of view will be that the entire set N = {0, 1, 2, 3,...} is given as a member of our universe, together with one crucial property of N. We codify this property in the form of an axiom, which we shall use to analyze N and to derive new properties. In this approach, we are reminded of the dictum of the distinguished German mathematician Leopold Kronecker ( ): God created the integers, all else is the work of man. In our derivations of new properties of N, we shall insist on using only the basic axiom, together with the rules of symbolic logic and introductory set theory as presented in the earlier chapters. Of course, once we have proved a result, we may use it as well. By deriving all the usual features of N in this way, not only do we place them on a solid logical foundation but we also validate our choice of initial assumptions. 1.1 The Peano Axioms To focus on key properties of N, we make the following definitions: Definition 1. An integral system < X, x 0, s > consists of a set X, a distinguished element x 0 X, called the initial element, and a function s : X X, called the successor function, satisfying: 2

3 a. s is an injection. b. x 0 R(s). c. Suppose that Y is a subset of X containing x 0 such that, for every x Y we must have s(x) Y. Then Y = X. Property (c) is known as the Principle of Induction (for < X, x 0, s >). The stated property of Y (namely, that if x Y, then s(x) Y ) is sometimes described by the statement Y is closed under s. It is also sometimes abbreviated by writing s(y ) Y, where s(y ) is shorthand for {z : z = s(x), for some x Y }. The Principle of Induction states that there is only one subset of X that contains x 0 and is closed under s: namely X itself. We ll see shortly how this translates into the usual method of proof by induction. Let us now define the function σ : N N by the rule σ(n) = n + 1. We shall assume The Axiom of Integrality: < N, 0, σ > is an integral system. The Axiom of Integrality is a fundamental assumption that we are making about the natural numbers; its three features, as given in Definition 1, are part of our concept of N. A nearly identical presentation of these features (written as five items) is known as the Peano Axioms, named after the Italian mathematician G. Peano, The basic axiom for the natural numbers who published them in In the definition of the successor function σ, we used the notation n + 1 because it is the common way that we write the successor of n. We do not mean by this that we have yet defined or have available a notion of addition of natural numbers, and 3

4 that n + 1 is the result of adding 1 to n. Addition is a secondary concept that we shall define later. To emphasize this, we shall for the time being use the notation σ(n) = n, avoiding n + 1 until later. 1.2 Proof by induction The student may recognize the Principle of Induction from other courses, perhaps notably beginning calculus. Anticipating the way we are going to be using this principle for proofs later in these notes, let us formulate the general idea here in the terminology and symbolism of the Symbolic Logic notes and then reformulate it in more familiar terms. Proofs by induction involve statements of the form ( n)p (n), where P is a predicate involving the single variable n, which is assumed to range over N. Let T (P ) be the truth set of P. By this we mean the set of all specializations a of the variable n for which P (a) is true. To prove the given statement, we must show that T (P ) = N. According to the Principle of Induction for < N, σ, 0 >, this can be accomplished (k by showing two things: namely, that (i) 0 T (P ), and that (ii) ( k N)( T (P ) ) ( k T (P ) )). Reformulating this, we get the familiar version of the induction principle: namely, to prove that ( n)p (n) is true, first prove that P (0) is true, and then prove ( k N) ( P (k) is true P (k ) is true ). Familiar version of induction The first part is often called the initial step of the induction proof, and the second is called the inductive step. The hypothesis in the inductive step is often called the 4

5 inductive hypothesis or induction hypothesis. We present one example of this method that should be familiar to someone who has studied calculus. The example is intended as an illustration, not as part of the logical development we are presenting. So, we shall be assuming the standard properties of the natural numbers as they are used in arithmetic, elementary algebra, and calculus courses. For any natural number n, let P (n) be the predicate (2n+1) = (n+1) 2. We use induction to prove ( n)p (n). P (0) is the assertion 1 = (0 + 1) 2, which is obviously true. We must now show that ( k) ( P (k) P (k + 1) ). So, choose any natural number k and assume P (k), i.e., assume (2k + 1) = (k + 1) 2. We must prove P (k+1). Begin by writing P (k+1): (2(k+1)+1) = ((k+1)+1) 2. The left-hand side of this, can be written as ( (2k + 1) ) + (2(k + 1) + 1). Since P (k) is true, by the induction assumption, we may replace the part in large parentheses by (k + 1) 2, obtaining (k + 1) 2 + 2(k + 1) + 1, which equals ( (k + 1) + 1 ) 2. This calculation shows that Familiar example of an induction proof (2(k + 1) + 1) = ( (k + 1) + 1 ) 2, which is just the assertion P (k + 1). Therefore induction step is complete. The principle of induction now implies that P (n) is true for all n, so the desired result is proved. We now return to the logical development and illustrate method of proof by induction by proving the following simple but useful statement: Proposition 1. Every natural number is either equal to 0 or is the successor of 5

6 another natural number. This proposition may seem obvious to us, but it is not contained directly in the definition we have given, so it needs to be proved. Proof. The predicate in the statement of the proposition is n {0} R(σ), so its truth set is obviously {0} R(σ). Let us call this X for short. We show that X satisfies the hypotheses of the induction principle. The first property, that 0 X, is true by definition. Now choose any k X. Since k = σ(k), we obviously have k R(σ). But R(σ) X, by definition, so k X. By the principle of induction, Another example of an induction proof X = N, as desired. We remark that the Axiom of Integrality asserts in part that σ is injective. Together with the foregoing proposition, this means that every non-zero natural number n can be written in the form m, with m uniquely determined by n. We may call m the predecessor of n. Exercise 1. Using the standard facts about natural numbers that you have learned in earlier courses, prove the following by induction on n: a (2n) = n(n + 1). b n 3 = ( n(n + 1)/2 ) Comparing integral systems We now wish to compare different integral systems, and for this, we use the following definition: 6

7 Definition 2. Let < X, x 0, s > and < Y, y 0, t > be two integral systems. An integral isomorphism of integral systems < X, x 0, s > < Y, y 0, t > is a function f : X Y such that a. f is a bijection. b. f(x 0 ) = y 0. c. ( x X), f(s(x)) = t(f(x)). If an integral isomorphism exists, we say that < X, x 0, s > is integrally isomorphic to < Y, y 0, t >. An integral isomorphism in this sense, simply establishes a bijection between elements of X and Y in such a way as to match initial elements and preserve the successor relation. Thus, isomorphic integral systems are just copies of one another. Exercise 2. Prove that this notion of integral isomorphism is an equivalence relation. Exercise 3. Prove that for every integral system < X, x 0, s >, there is at most one integral isomorphism < N, 0, σ > < X, x 0, s >. (Hint: Suppose that f and g are integral isomorphisms, and prove by induction that ( n) f(n) = g(n).) The existence of at least one isomorphism requires the notion of definition by induction, which is discussed below. Exercise 4. In this exercise you may use the standard properties of numbers (real, rational, natural, etc.). For each of the following examples, state which is an integral system and which isn t. In either case, justify your answer. 7

8 a. < Z, 0, σ >, where Z denotes the set of all (i.e., positive, negative and zero) integers, and 0 and σ are as before. b. < Q +, 0, σ >, where Q + is the set of non-negative rational numbers, and 0 and σ are as before. c. < N(10), 10, σ >, where N(10) is the set of natural numbers 10, and σ is as before. d. < N, 0, σ >, where N denotes the non-positive integers, and 0 and σ are as before. 1.4 Definition by Induction The principle of induction is used not only to prove statements of the sort already indicated, but it is also used to define functions that have N as their domain. For example, assuming for the moment that we have full use of the system of real numbers R, choose any real number a. We might try to define an exponential function f : N R as follows: f(0) = 1; f(n ) = af(n). Indeed, for the first few values of n, we get f(0) = 1, f(1) = af(0) = a, f(2) = a a = a 2, f(3) = a f(2) = a a 2 = a 3,..., etc. For those familiar with computer science, this is just the standard recursive definition of a n. The definition of f at the successor of n is defined in terms of its value at n. Two familiar examples of definition by induction As another example, we might try to define the function g : N R as follows: g(0) = 1; g(n ) = n g(n). The first few values of this function are 1, 1, 2, 6, 24, 120,.... Most students have seen this function before: it is just n!, the factorial function. Here, 8

9 the definition of g at the successor of n depends both on n itself and on the value of g at n. These definitions work fine if one is interested only in finitely many values of the function and this is all that comes into play in computer science or combinatorics. By themselves, these rules for defining (or, more precisely, attempting to define) the function, do not actually give a function with domain N. They give only a finite sequence of function values. This gap is filled by the following theorem, which shows that the examples just given do indeed determine functions with domain N. It also gives a general criterion for making valid definitions by induction. The theorem can be proved using the principle of induction, but the proof is fairly lengthy, so we shall omit it. A proof can be found in the first two chapters of Stoll s book. We shall use the result stated by the theorem, however. Theorem 1. Suppose that A is a given set with a fixed element a 0 selected in A. Suppose further that we have functions h n : A A, one for each n N. Then, there exists a unique function k : N A satisfying (i) k(0) = a 0, and (ii) k(n ) = h n (k(n)). Notice that the theorem requires us to have available the element a 0 and the stated functions h n. It then gives us the desired function k. The functions h n are either given as part of the background information or can be defined by some method other than induction. Let us illustrate this by showing how it applies to the two examples above. For the first example, let A be the set of real numbers R, and let a be a real number as before. For each n, let h n : R R be given by the formula h n (x) = ax. (In this case, all these functions h n are equal.) Notice that the definition of h n does not use 9

10 induction. Then, the theorem tells us that there exists a unique function k such that k(0) = a and k(n ) = h n (k(n)) = ak(n). This is precisely the exponential function f of the first example. For the second example, let A = N and now choose a = 1. Define h n by the formula h n (x) = n x. The reader should convince himeself/herself that this definition does not use induction. In this case, all the functions h n are different from each other. The theorem now tells us that there exists a unique function k : N N satisfying k(0) = 1 and k(n ) = n k(n). This is precisely the factorial function g of the second example. Exercise 5. Use the method of definition by induction to prove that for each integral system, < X, x 0, s >, there exists at least one integral isomorphism f :< N, 0, σ > < X, x 0, s >. Exercises 3 and 5 tell us that every integral system is a copy of < N, 0, σ > in a unique way, or, to put it another way, there is only one integral isomorphism class of integral systems, and every representative is isomorphic to < N, 0, σ > in a unique way. Because of this, we say that the properties in Definition 1 characterize (or determine, or define) < N, 0, σ > Henceforth, we write < N, 0, σ > more briefly as N We now use this method of definition by induction to define addition of natural Uniqueness of natural numbers numbers. 1.5 Addition We have started with the most primitive features that lie at the heart of our concept of the natural numbers, encoded as The Integrality Axiom (or Peano s Axioms). No- 10

11 tice that, except for the operation of succession, (i.e., counting), we are not assuming anything about operations with the natural numbers (such as addition and multiplication). In particular, we are in a pre-arithmetic mode with respect to N. We now remedy this by defining the operation of addition, +, for N. Induction can then be used to verify that this operation has all the familiar properties. The definition and derivation of the properties of multiplication are similar; we do this later. Choose any m N. We shall use induction to define a function a m : N N, and we shall denote the function value a m (n) by m + n. To define a m we make use of Theorem 1, and this requires us to define functions h n : N N. In this case, the definition is easy: we define each h n to be the successor function σ. We must also choose a fixed member of N, which in this case, we choose to equal m. Then Theorem 1 tells us that there exists a unique function N N, which we shall denote a m such that a. a m (0) = m. b. a m (n ) = (a m (n)). Definition 3. For any m and n in N, we denote a m (n) by m + n and call it the sum of m and n ( or the result of adding m to n, or m plus n). In terms of the + notation, the two defining features of a m may be written as follows: a. m + 0 = m. b. m + n = (m + n). 11

12 We are left with the important task of justifying this definition, which amounts to showing that it gives us the operation of addition with which we are all familiar, i.e., that it has all the familiar properties. Those that are not immediate consequences of the definition are all proved by induction. We sketch a few proofs by induction for illustration. Lemma 1. Choose any natural numbers l, m, and n. Then: (i) m+0 = m = 0+m; (ii) m + 1 = m = 1 + m; (iii) l + (m + n) = (l + m) + n. Proof. (i) m + 0 = a m (0) = m, by definition, proving the first equality. The second equality, namely, m = a 0 (m) must be proved by induction on m. By definition, a 0 (0) = 0, so the assertion is true for m = 0. Now assume it is true for m = k: that is, k = a 0 (k). Evaluate a 0 (k ) = (a 0 (k)) = k, the first equality following from property (b) of a 0 and the second from the induction hypothesis. This completes the proof of (i). (ii) Evaluate m + 1 = a m (1) = a m (0 ) = (a m (0)) = m. (The reader should fill in the reasons for each equality.) The second equality, namely, m = a 1 (m), is proved by induction. When m = 0, both sides reduce to 1, so the assertion is true for m = 0. Assume it for m = k, i.e., k = a 1 (k), and then evaluate a 1 (k ) = (a 1 (k)) = (k ). Switching the order of the equality, we obtain (k ) = a 1 (k ). But this is precisely the second equality in assertion (ii) of the lemma, with m = k, as desired. Therefore, the induction, hence the proof of (ii), is complete. (iii) Let P (n) be the predicate ( l)( m)(l + (m + n) = (l + m) + n). 12

13 We prove ( n)p (n) by induction on n. P (0) is the statement ( l)( m)(l + (m + 0) = (l + m) + 0). Both sides reduce to l + m, by what has already been proved, so P (0) is true. Now assume P (k) for some k. We evaluate the left-hand side of P (k ) for any selected l and m: l + (m + k ) = l + (m + k) = ( l + (m + k) ) = ( (l + m) + k ) = (l + m) + k. Again, we leave it to the reader to supply reasons for each step. The resulting equality is precisely P (k ), so the induction proof of (iii) is complete. The reader may have noticed that statement (i) of the lemma is the familiar additive identity rule for arithmetic and that the third statement of the lemma is the familiar associative law for addition. Usually, these are stated as axioms or properties that we assume. The lemma shows, however, that these can be proved from more rudimentary concepts. The associative law is needed because addition is defined to be a binary operation on natural numbers. Once more than two numbers are to be added, we must decide how to group them so that we can do the addition two numbers at a time. In the case of three numbers, there are two ways of grouping the numbers (without changing their order) or, as we may say, there are two ways of associating the numbers (into pairs). The associative law tells us that these two ways give equal sums. Exercise 6. Show that there are five ways of associating four numbers into pairs, 13

14 and then prove, using the associative law, that all five ways produce the same sum. How many ways of associating five numbers are there? List them. This exercise shows that the associative law can be extended to the case of four numbers, and similar arguments extend the law to five numbers, etc. Indeed, there is a general inductive argument (which we do not go into in these notes) that extends the result to any number of numbers. This means that, for addition, we do not need to use parentheses to express arbitrarily long sums (e.g., a 0 + a a n ), since any association into pairs yields the same result. So, from now on we feel free to omit parentheses in these situations. Incidentally, earlier, we expressed some discomfort in using the notation n + 1 to represent the successor of n, since we had not yet defined addition. However, now, in light of assertion (ii) of the lemma, we are able to use either n + 1 or 1 + n in place of n. Exercise 7. Use the definition of addition and the method of proof by induction to prove the following statements: a. ( m)( n)(m + n = n + m). the commutative law for addition b. ( k)( m)( n) ( (m + k = n + k) (m = n) ). the cancellation law for addition The statements in the earlier lemma and in the exercise give the most important features of the additive algebra of the natural numbers, so we summarize them in the following theorem. Further algebra will come into play when we discuss multiplication later. 14

15 Theorem 2. Let x, y, z be any natural numbers. Then the following hold: a. x + (y + z) = (x + y) + z. associativity of addition Basic properties of addition b. (x + z = y + z) x = y. cancellation for addition c. x + 0 = 0 + x = x identity law for addition d. x + y = y + x. commutativity of addition 1.6 Ordering the natural numbers Although we have spoken informally about a sequential aspect to the natural numbers, the concept of an integral system does not, by itself, contain any explicit information about an ordering of the elements, except for the reference to one element as the initial element. But the natural ordering of the counting numbers is one of their basic features, so we now introduce this into the picture with a precise definition. Definition 4. Given m, n N, we write m n (or, equivalently, n m) if and only if there exists a k N such that m + k = n. For later use, it is convenient to rephrase this definition in the following way: Let m and n be any natural numbers. The equation m + x = n has a solution in N if and only if m n. Exercise 8. Prove the following addendum to the foregoing: If the equation has a solution in N, then the solution is unique. 15

16 Exercise 9. Show that is a partial order on N. Exercise 10. Let < be the strict ordering corresponding to. Show that m < n if and only if there exists a k N such that k 0 and m + k = n (or, equivalently, k > 0 and m + k = n). Exercise 11. Let m, n, and p be any natural numbers. Prove that m n m + p n + p. Exercise 12. Suppose that m, n, p and q are natural numbers such that m+n = p+q. Prove that if m < p, then n > q. Exercise 13. * Let m be any natural number. Prove by induction on n, that no natural number n satisfies m < n < m + 1. Again, this is an example of one of those so-called obvious statements that is not so obvious to prove. Another way of stating this fact is the following: For all m and n, n < m + 1 if and only if n m. Convince yourself that this is, indeed, equivalent to the assertion in the preceding exercise. Proposition 2. is a linear order on N. Proof. (a) A linear order is a partial order for which any two elements are comparable (cf. Set Theory notes). So, we have to show that for any m and n in N, either m n or n m. We prove this by induction on m. So, let P (m) be the assertion ( n) ( ) (m n) (n m). Since, for any n, 0 + n = n, it follows that 0 n, and so P (0) is true. 16

17 Now assume P (m) is true for m = k. We prove P (k ) by first choosing any n. We must show k n or n k. By the induction hypothesis, there are two cases, k n or n k: (i) Suppose n k. Since k = k + 1, we have k k, so, by the transitivity of, n k. So P (k ) holds in this case. (ii) Now suppose that k n. Case (i) includes the case n = k, so we may as well assume k < n. We shall show that k n. Since k < n, it cannot be the case that n = 0. Therefore, n has a unique predecessor p, i.e., n = p + 1. By Exercise 13, the strict inequality k < n = p + 1 is equivalent to k p, i.e., p = k + l, for some l N. Therefore, n = p + 1 = k l, which implies k n, as desired. This completes the induction. Often the fact that is a linear order is phrased in terms of <, which we then also call a (strict) linear ordering. In terms of <, the condition is phrased as follows: For Strict linear ordering any natural numbers m and n, exactly one of the following holds: m < n, m = n, or n < m. Exercise 14. Prove the following, for all m, n, p N: a. m < n n < p m < p ; This is just the transitivity of <. Since the transitivity of has already been proved, you may use that for this proof. b. n = 0 n > 0. c. (n < n). d. m < n σ(m) = n σ(m) < n. 17

18 1.7 Variations on induction Using the linear ordering of N, it is now easy to formulate two variations of the usual method of induction. Variation 1: This method does not have a commonly used name since it amounts to little more than a re-labeling. We might call it shifted induction, however, when we refer to it. This kind of induction arises when the predicate one would like to verify is not usefully defined for some small values of n. Perhaps P (n) is defined only for n some fixed number n 0 which depends on the particular problem. We can still do induction in this case, except that our initial value is no longer 0 but is n 0. Shifted induction The situation we are facing then is that P (n 0 ) is known to be true and ( k) ( P (k) P (k + 1) ) is known to be true, provided k n 0. We would then like to be able to conclude that P (n) is true for all n n 0. Essentially, our entire frame of reference is shifted over by n 0 units. This variant of induction is easy to derive from the standard induction principle via the following re-labeling trick. Define a new predicate Q by the formula Q(n) = P (n 0 +n). Since, by definition, k n 0 if and only if there is a natural number m such that k = n 0 + m, k ranges over all natural number n 0 precisely when m ranges over all natural numbers. Therefore, the situation we confront can now be reformulated as follows: Q(0) is true, and ( m) ( Q(m) Q(m + 1) ) is true. So, ordinary induction implies that Q(n) is true for all natural numbers n, and, hence, P (n) is true for all natural numbers n n 0. Variation 2: This method is sometimes called strong induction or complete induction even though it is equivalent to ordinary induction. In this form of induction, we still Strong induction 18

19 have the same initial step. But the induction step is replaced by the following ( ( k) ( i) ( (i k) P (i) ) ) P (k + 1). That is, if, for all k, P (1), P (2),..., P (k) are all true, then P (k + 1) is true. The initial step plus the induction step then imply the usual conclusion: ( n)p (n) is true. Exercise 15. Given a predicate P = P (n), formulate a predicate Q such that by applying ordinary induction to Q, you get strong induction for P. Of course, Variation 1 can be combined with strong induction: i.e., you don t have to start at 0 in strong induction. Here is an example showing how strong induction, although logically equivalent to ordinary induction, is a convenient tool in proofs. We shall be referring to standard multiplicative properties of natural numbers, which have not yet been established in our logical development, but we are doing so only for illustration. Proposition 3. Every natural number n > 1 can be written as a product of primes. Proof. This proof is just an illustrative sketch. Since there are no natural numbers n satisfying 1 < n < 2, we may start the strong induction at n = 2. When n = 2, the proposition is immediate, since 2 itself is a prime, so it is a one-fold product of primes. Suppose the proposition is true for all natural numbers i k, and consider the natural number k + 1. If it is prime, we are done. If not, then by definition, it can be written as a product of two natural numbers k + 1 = ab, with neither a nor b equal to k + 1 or 1. Therefore, both a and b are < k + 1. Since there are no natural numbers strictly between k and k + 1, both a and b are k, hence the Example of proof by strong induction 19

20 strong induction hypothesis may be applied to them. They are thus both products of primes, and so, clearly their product is a product of primes. The point here is that since we are dealing with a multiplicative property, simply knowing the proposition for k may not be adequate for proving it for k + 1, since these two numbers are not related multiplicatively in a nice way. However, knowing the proposition for all natural numbers k gives us the flexibility we need for the proof. This concludes our discussion of strong induction. 1.8 Well-ordering Finally, here is another basic and useful fact about N, which plays a role in proofs later on. Proposition 4. Every non-empty subset S of N has a least element (i.e., an element that is minimal among all elements in S). Proof. Let S be any subset of N. The proof will use the contrapositive method. Assume that S has no minimal element, and let T be the set difference N \ S. Let P (n) be the predicate ( k) ( (k n) (k T ) ). This asserts that the set of all natural numbers less than or equal to n is a subset of T. Clearly P (0) is true, since it asserts simply that 0 T. If 0 were not in T, it would be in the complementary set S, and it would clearly be a minimal member of S, contradicting our assumption about S. Exercise 16. Prove the induction step ( m)p (m) P (m + 1). The exercise concludes the induction proof that ( n)p (n) is true. But this means that T = N, hence S =. 20

21 Thus, assuming that S has no minimal element, the above proof shows that it must be empty, which is precisely the contrapositive of the statement of the proposition. This property of N is known as the well-ordering principle. We sometimes say, simply, the natural numbers are well-ordered. Let us be more precise about the definition of well-ordering. We say a linear ordering on a set S (or a strict linear ordering on S) is a well-ordering provided that every The natural numbers are wellordered non-empty subset of S has a minimal element. So, to prove that a relation on S is a well-ordering, first, one has to prove that it is a linear ordering (or a strict linear ordering), and then one has to prove that it satisfies this extra minimality property. The following exercise shows that there are well-ordered sets that do not satisfy the Axiom of Integrality (i.e., induction does not hold). Exercise 17. * Consider the cartesian product N N = N 2. Using the ordering of N just defined, we now endow N 2 with the associated (strict) lexicographic order: namely, we write (a, b) < (c, d) if and only if (a < c) or ((a = c) (b < d)). a. Show that this defines a strict linear order on N 2, and then show that this linear order well-orders N 2. b. Define a function τ : N 2 N 2 as follows: For any (a, b) N 2, let T (a, b) be Example of a wellordered set for which induction fails the set of all (c, d) > (a, b). Since N 2 is well-ordered, T (a, b) has a minimal element, say (m, n). Define τ(a, b) = (m, n). Prove that τ is injective and that (0, 0) R(τ). From the definition, it should be clear that τ is a kind of successor function. (In fact, it is not hard to prove that τ(a, b) = (a, b + 1), but, unless you prove this, do not use this fact for this part of the exercise.) 21

22 c. Show that < N 2, τ, (0, 0) > fails to be an integral system. That is, show why the induction principle fails. (Here, you may want to prove and use the formula for τ given above.) 1.9 Multiplication We proceed similarly to the case of addition, using the method of induction to define multiplication. Choose any natural number m. We shall define a function b m : N N which is supposed to represent multiplication by m. To do this by the method of induction, we need to define functions h n : N N, which we define as follows: h n (x) = x + m, for each n. Note that we use addition in this definition, but that s okay because addition has already been defined. The method of induction also requires us to choose a particular element in N: our choice is the natural number 0. Then, Theorem 1 tells us that there exists a unique function b m : N N satisfying b m (0) = 0, and b m (n ) = b m (n) + m. The first few values of b m are: 22

23 b m (0) = 0 b m (1) = m b m (2) = m + m b m (3) = (m + m) + m and so on. Clearly, b m represents repeated addition of m, which coincides with our intuitive understanding of multiplication. We write b m (n) as mn and call it the product of m by n or m times n. (Occasionally, we interpose a dot, as in a b, if this makes the product more readable.) In terms of the product notation that we just introduced, the key defining properties of b m can be written as m 0 = 0, and m n = mn + m. We now list the basic properties of multiplication as a theorem. Theorem 3. For any natural numbers l, m, and n, the following are true: a. (lm)n = l(mn). the associative law for multiplication b. lm = ml. the commutative law for multiplication c. l 1 = l = 1 l. the multiplicative identity law Basic properties of multiplication 23

24 d. 0 m = 0 = m 0. e. Suppose that l 0. Then, we have (l m = l n) m = n. the cancellation law for multiplication f. Suppose that l 0. Then we have (l m l n) m n. g. l(m + n) = lm + ln and (l + m)n = ln + mn. the distributive laws Except for the second equality in item (d), which is true by definition, all of the statements in this theorem can be proved by the method of induction. The proofs are not very different from those already done for the facts about addition. However, the first equality in (d) can also be proved directly (i.e., without induction), provided one has first proved the second distributive law. The proof is a little trick involving the evaluation of 0+0 m as follows: 0+0 m = 0 m = (0+0)m = 0 m+0 m. Thus, we get m = 0 m + 0 m, which, by additive cancellation, yields 0 = 0 m. We give this proof because a nearly identical proof can be used in more general contexts, which we ll encounter later. Exercise 18. Prove the statements of Theorem The natural numbers and cardinality In this section we discuss cardinality issues related to N, making use of the material in Section 6 of the Set Theory notes. The student should refer to that section when needed for reminders about definitions. We begin by defining the initial segments of the set N. Definition. Let n be any natural number, and define the n th initial segment N n to be {k : (k N) (k < n)}. 24

25 Obviously, N 0 is the empty set. Moreover, since, for any natural number m, there are no natural numbers strictly between m and m + 1 (Exercise 13), we have N 1 = {0}, N 2 = {0, 1}, N 3 = {0, 1, 2}. More generally, for n > 0, we could write N n = {0, 1,..., m}, where here m is the predecessor of n. To promote ease of reading, it is useful to give this number its usual name, namely n 1. However, we need to remember that subtraction has not yet been defined, so at this point, the use of n 1 does not indicate that we are subtracting 1 from n, only that we are considering the immediate predecessor of n. With this notation understood, we may sometimes write N n as {0, 1, 2,..., n 1}. 25

26 Definition. A set S is called finite if and only if #(S) = #(N n ), for some natural number n. A set that is not finite is called infinite. The reader should have no trouble with the following exercise. Finite and infinite sets Exercise 19. Suppose that S is a set such that #(S) = #(N n ), for some natural number n, and suppose that x is an object not in S. Then #(S {x}) = #(N n+1 ) We now recall the definition of the relation defined in the chapter Set Theory. If S and T are two sets, we write #(S) #(T ) to indicate that there is an injection S T but no bijection S T. In Set Theory, we cite some theorems that imply that is a strict linear order. We wish to compare the strict order relation < on N with the relation on the collection of initial segments N n. In order to keep assumptions to a minimum, we shall use only the fact that is transitive and irreflexive in this context, not that it is a linear order. In particular, we do not assume the Schröder-Bernstein Theorem or the fact that any two cardinalities are comparable. Lemma 2. Let m and n be any natural numbers such that m < n. Then there is no injective map N n N m. Proof. The proof is by induction on m. When m = 0, the result is obvious, because N m is empty and N n is not, and there is no map whatsoever from a nonempty set to the empty set, much less an injective map. So, assume inductively that the result holds for m = k. We show that when n > k + 1, there is no injective map N n N k+1. This proof will proceed by contradiction. Accordingly, suppose that n > k + 1 and there is an injective map f : N n N k+1. Since n > k + 1 > k, we may use the inductive hypothesis to conclude that the range 26

27 of f cannot be contained in N k = {0, 1,..., k 1}. That is, there must be some a such that f(a) = k. If a = n 1, i.e., if f(n 1) = k, then f maps all the other elements, i.e., the elements of N n 1, injectively into N k. But, since n > k + 1, we have n 1 > k, so we may use the induction hypothesis to conclude that this injective mapping N n 1 N k cannot exist. Therefore, f cannot satisfy f(n 1) = k. This means that f(a) = k, for some a other than n 1; it also means, since f is injective, that f(n 1) = b, where b is some element of N k+1 other than k. We now define a new map f : N n N k+1 as follows: f (x) = f(x), as long as x is neither equal to a or to n 1; f (a) = b, and f (n 1) = k. That is, f is almost identical with f except that it switches the values of f at a and n 1. Therefore, f is an injective map from N n to N k+1 satisfying f (n 1) = k. But we have just shown in the preceding paragraph that such a map cannot exist. This is the desired contradiction. Corollary 3. For every pair of natural numbers m and n, m < n #(N m ) #(N n ). Comparing < and Proof. We first prove the implication in the = direction. Let any pair of natural numbers m and n be given such that m < n. Then N m N n, and so #(N m ) #(N n ). Lemma 2 implies that we cannot have #(N m ) = #(N n ), which proves that #(N m ) #(N n ). Now we prove the converse. Suppose #(N m ) #(N n ). Since this implies that there is an injective map N m N n, Lemma 2 implies that we cannot have m > n. But the assumption #(N m ) #(N n ) also implies that there is no bijection N m N n, so we 27

28 cannot have m = n. The only remaining possibility, then, is that m < n. Exercise 20. For all natural numbers m and n, m = n if and only if #(N m ) = #(N n ). We may conclude, therefore, that the set of natural numbers corresponds bijectively to the set of all finite cardinalities in such a way as to preserve the strict orderings defined on these sets. In particular, this shows that is a well-ordering on the set of all cardinalities N n, n ranging over N, indeed a copy of the strict linear ordering < on N. New notation: It is clearly convenient to have some more compact notation for the cardinality #(N n ), and, of course, we all know that the conventional terminology and notation fits the bill: namely, instead of #(N n ), we write simply n, and we refer to any set with that cardinality as a set with cardinality n. The results just proved provide logical justification for adopting this convention. That is, even though the concept of n as an element of the set N is very different from the concept of the cardinality of the set N n, since the two correspond bijectively in an order-preserving way, there is little cost and considerable convenience in conflating the two. In a similar spirit, we shall use < instead of when talking about finite cardinalities. That is, we shall use m < n to mean either that the natural number m precedes the natural number n or that the cardinality m is strictly less than the cardinality n. Proposition 5. Let n be any natural number, and let S be any subset of N n. Then, there exists a natural number m n such that #(S) = #(N m ). Exercise 21. * Prove the above proposition by induction on n. The initial step of the induction is almost obvious. We give a hint for the inductive step. Assume that 28

29 the result holds for n = k, and let S be a subset of N k+1 = {0, 1,..., k}. If S is already contained in N k = {0, 1,..., k 1}, then the induction hypothesis applies immediately. If not, this means that k must be a member of S. Let T = S \ {k}, show that the induction hypothesis can be applied to T, and then use this, together with Exercise 19 to get the desired result for S. This proposition gives us good information about subsets of N n, but it doesn t tell us about finite sets in general. To approach the general case, we make a definition: we call a subset S of N bounded if there exists a natural number n such that S N n. Bounded and unbounded subsets of N Otherwise, we say it is unbounded. Proposition 5 tells us that bounded sets are finite. Exercise 22. Prove that if f : N N is an injective map, then its range R(f) is unbounded. (Hint: Use Lemma 2.) Use this to conclude that N is infinite. Proposition 6. If S is a finite subset of N, then S is bounded. Proof. We prove the contrapositive: If S is unbounded, then S is infinite. So, suppose that S is unbounded. We begin by defining a function f : S S. Note that this is not an inductive definition. First observe that S, since is obviously bounded. Next, choose any n S. We shall define f(n). Let T n be the set {m : (m S) (m > n)}. If T n =, this would mean that every m S satisfied m n, i.e., S N n+1, contradicting the assumed unboundedness of S. So T n is non-empty. By the wellordering on N, T n has a smallest element, say n 1. Define f(n) = n 1. Notice that this definition insures that f(n) > n, for every n S. 29

30 This defines the function f : S S. Next, we define a function g : N S inductively as follows: Let g(0) be the smallest element of S, say g(0) = n 0. Assuming that g(k) has been defined, let g(k + 1) = f(g(k)). This gives an inductive definition for g. Notice that, for every natural number n, g(n + 1) = f(g(n)) > g(n). Exercise 23. Prove by induction that if m < n, then g(m) < g(n). (Hint: Since m < n, we have n = m + k, for some k 1. Proceed by induction on k If k = 1, the previous paragraph gives the desired inequality, hence proving the base case. Now do the inductive step.) The exercise proves that g : N S is an injection. If S were finite, we would have a bijection h : S N p, for some natural number p. But then the composite h g would be an injection N N p. Exercise 22 shows that this is a contradiction. The proof of Proposition 6 shows slightly more than what the proposition asserts. Not only does it show that every unbounded subset S of N is infinite, it also shows that #(N) #(S). Since S N, we also have #(S) #(N). Therefore, now applying the Schröder-Bernstein Theorem, we may conclude that any unbounded subset of N has the same cardinality as #(N). Since every subset of N is either bounded or unbounded, we see that the subsets of N have either finite cardinality or the same cardinality as that of N (which is infinite). We can state this in a slightly stronger way as follows: Let S be any set such that there is an injective map f : S N. Then, of course, f defines a bijection between S and the subset R(f) of N. So, the above statement can be rephrased as: Let S be any set such that #(S) #(N). Then, either #(S) is finite or #(S) = #(N). In other words #(N) is the smallest infinite cardinality. It is often denoted ℵ 0. The smallest infinite cardinality ℵ 0. 30

31 This leaves open one obvious question: Are there infinite cardinalities ℵ 0? The answer to this is yes, and a proof was given via the extra-credit exercise in the homework assignment for Week 6: specifically, for any set A it was shown that #(A) #(P(A)) (a result of G. Cantor). It follows that ℵ 0 #(P(N)). If we apply the exercise (Cantor s Theorem) with X = P(N), we get a strictly larger cardinality, #(P(P(N))), and so on. This shows that there are infinitely many infinite cardinalities. Of course, most of the mathematics we do involves more familiar sets, such as the set of real numbers R, the set of points in the plane, etc. The above exercise does not clarify what the cardinalities of these sets are. For example, is it possible that ℵ 0 = #(R)? Another important result of Cantor shows that the answer is no. We shall go through his proof later in the course. 31