Set theory as a foundation for mathematics

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1 Set theory as a foundation for mathematics Waffle Mathcamp 2011 In school we are taught about numbers, but we never learn what numbers really are. We learn rules of arithmetic, but we never learn why these rules are true. In this class we will see how we can construct numbers and rigorously prove the rules of arithmetic using only the notion of a set. What is a set? Informally, a set is a collection of objects. For example, there is a set {Waffle, Mathcamp, MCSP} which consists of me, Mathcamp, and the string MCSP. We could also have a set {0, 5, 2π, 734} consisting of the numbers 0, 5, 2π and 734. However, since we wish to define numbers in terms of sets, we cannot yet allow ourselves to talk about sets of numbers. Indeed, in order to build everything purely out of sets, we will require that all our sets be sets of sets. That is, every element of our sets is itself a set. This may sound circular if we can only consider sets of sets, how can we ever get started constructing any sets? However, one set that we can be sure exists is the empty set = {}, the set with no elements. Since has no elements, all of its elements are sets, so it is a set of sets. Once we have, we can then consider { }, {{ }}, {, { }}, and so on. In general, we will usually define sets using an expression like S = {x : P (x)}. This means that S is the set of all sets x such that the property P (x) is true. For example, assuming we already know what the integers Z are, we could define a set A = {x : x Z and x > 2} = {x Z : x > 2} of all integers greater than 2. We write x S to mean that x is an element of the set S. For example, for A as above, 3 A. Let X and Y be sets. Then we can construct the following sets: The union X Y = {x : x X or x Y }. The intersection X Y = {x : x X and x Y }. The difference X \ Y = {x : x X and x Y }. We say X is a subset of Y, or X Y, if every element of X is also an element of Y. I would like to emphasize that while we will be giving specific constructions of number systems as sets in these notes, there is nothing special about the exact constructions we use. There is more than one way (in fact, infinitely many ways) to exhibit sets that can serve as number systems, and what is important is not which one we choose but the fact that one does exist. For example, the natural numbers should not be defined as the set of Definition 1.6, but as any set equipped with an arithmetic structure which is isomorphic to that given in Section 2. Finally, it should be noted that for most of these notes we will be using what is called naive set theory, in which we are allowed to define any set we want to using an expression of the form {x : P (x)}. This approach to set theory is called naive because it doesn t actually work! At the end of the notes, we see why it doesn t work and why we must use more restrictive rules when constructing sets. Don t worry, though; all of the constructions made in these notes are in fact valid under these more restrictive rules (called the Zermelo-Fraenkel axioms for set theory). 1

2 1 The Natural Numbers The first, most basic mathematical object we will define using sets is the natural numbers (i.e., the nonnegative integers). We would like to define the set of natural numbers as N = {0, 1, 2, 3,...}, but there are a few difficulties with this. First, we need to define what 0, 1, 2, 3, and so on are as sets. Then we have to figure out what... means more precisely. Let s start at the very beginning: what is 0? We could take any set and call it 0, but perhaps the most natural choice is the empty set, the only set that has 0 elements. Thus we make a definition: Definition =. OK, we ve defined 0 as a set; how about 1? Just as we defined 0 to be the empty set, a set with 0 elements, perhaps we could define 1 to be a set with 1 element. Again, the choice is arbitrary, but we will choose: Definition = {0} = { }. That is, 1 is the set whose only element is 0. We can now similarly define all the natural numbers: Definition = {0, 1}, 3 = {0, 1, 2},..., n = {0, 1, 2,..., n 1},... That is, we one-by-one define each natural number as the set of all the smaller natural numbers. This allows us to talk about any natural number individually if you hand me a number, I can (laboriously) write it down as a set built up from the empty set. For example, 4 = {0, 1, 2, 3} = {, {0}, {0, 1}, {0, 1, 2}} = {, { }, {, { }}, {, { }, {, { }}}}. However, it is still not clear how we can talk about all natural numbers at once. Indeed, Definition 1.3 is plagued by the same... that we worried about before. However, we can begin to make this more precise by seeing that there s an explicit pattern we can write down describing how to get from n to n + 1. In fact, since n = {0, 1,..., n 1} and n + 1 = {0, 1,..., n 1, n}, we can see that n + 1 = n {n}. We thus make the following definition: Definition 1.4. Let x be a set. Then the successor x + 1 of x is defined as x + 1 = x {x}. In this definition, we really only care about the case that x is a natural number, but we haven t yet defined what a natural number is! We can now state more precisely what we mean by N = {0, 1, 2, 3,...} : 1. 0 N 2. If k N, then k + 1 N as well. 3. All elements of N can be obtained using (1) and (2). This is still not completely satisfactory, since we have not written down an expression for N; we ve just listed some properties it should have. However, with a little more cleverness we can turn these properties into a definition. Definition 1.5. A set S is inductive if 0 S and whenever k S, k + 1 S as well. Definition 1.6. The set of natural numbers is N = {n : n S for every inductive set S}. 2

3 That is, an inductive set is a set that satisfies conditions (1) and (2) above. Since N only contains things that must be contained in any inductive set, it automatically satisfies (3). Let s check that N also satisfies (1) and (2). Proposition 1.7. The natural numbers are inductive. Proof. First, 0 N since by definition, 0 is in every inductive set. Now suppose k N and let S be an inductive set. Then k S, so k + 1 S since S is inductive. Since this holds for every inductive set S, it follows that k + 1 N. Thus N is inductive. You may have noticed that everything we ve been doing looks suspiciously like induction. If you hadn t noticed, hopefully the name inductive at least heightened your suspicions. In fact, our definition of the natural numbers is perfectly designed to make induction work. Theorem 1.8 (Proof by Induction). Let P (n) be a property that a natural number n can have. Suppose that: 1. P (0), and 2. For all k N, P (k) implies P (k + 1). Then P (n) holds for all natural numbers n. Proof. Let S = {k N : P (k)}. Then the hypotheses say exactly that S is an inductive set. By definition of N, then, S contains all of N. That is, P (n) holds for all n. This is the first major fact about numbers that we usually take for granted that we have been able to turn into a rigorously proved theorem using sets. We will soon use induction to prove several more. Looking back at the definition of N and the proof of Theorem 1.8, the two are really equivalent. Given this, there is a very important idea to keep in mind when attempting to solve problems about N: Principle 1.9. Because the natural numbers are defined to be such that induction works, essentially the only way to prove statements about them is by induction. For example, we can prove several things that ought be be true but aren t clear from our abstract definition of N. For instance, we intuitively know that N should only contain the sets 0 =, 1 = {0}, 2 = {0, 1}, 3 = {0, 1, 2}, and so on. But how do we know that other sets which we do not expect to see in the and so on are not in N? We prove it by induction! Here s an example of how this works. Proposition The set {1} is not a natural number. Proof. We prove by induction that for all n N, n {1}. First, clearly 0 = is not {1}. Now suppose k {1}; we wish to show that k + 1 = k {k} is not {1}. Since k k + 1 and the only element of {1} is 1, if k + 1 = {1} we must have k = 1. But then k + 1 = = {0, 1} = {1}. Thus it is impossible for k + 1 to be {1}. Hence {1} N. Here s a definition we ll use in the exercises and in the next section. Definition Let m and n be natural numbers. Then we write m < n if m n. Intuitively, this makes sense, because n = {0, 1, 2, 3,..., n 1} is exactly the set of smaller natural numbers. 3

4 1.1 Exercises Remark Many of the exercises provided in these notes have many parts, and may occasionally seem long and tedious. The purpose of these exercises is to give you practice working with the ideas on your own and understand them better, and also help convince you that you could rigorously prove any basic fact about numbers if you had to. If you ve already done a few exercises and it s getting easy but tedious, you should not feel like you have to do the rest. Just convince yourself that you could solve them if you had to, and beware that we will often use results from the exercises in later sections. On the other hand, a few of the exercises are more difficult, and if there s an exercise that you re finding really hard, come talk to me at TAU and I ll gladly help you through it! Exercise 1.1. Prove that the set {1, {1}} is not a natural number. (Hint: Imitate the proof of Proposition 1.10.) Exercise 1.2. Prove that if n N then either n = 0 or there is some m N such that n = m + 1. (Hint: Use induction.) Exercise 1.3. Show that m < n + 1 iff m n, where means either < or = (recall that m < n + 1 means m n + 1 by Definition 1.11). Exercise 1.4. Prove that the relation < is transitive: if k < m and m < n, then k < n. induction on n and Exercise 1.3). (Hint: Use The following is a slightly different form of induction that we will use occasionally. Exercise 1.5 (Strong Induction). Let Q(n) be a property that a natural number n can have. Suppose that whenever k N and Q(m) holds for all m < k, then Q(k) holds as well. Then prove that Q(n) holds for all natural numbers n. (Hint: Let P (k) be the statement that Q(m) holds for all m < k, and prove P (k) by ordinary induction on k.) Exercise 1.6. The relation < is antireflexive: n n for all natural numbers n. (Hint: Use strong induction.) 4

5 2 Counting We start with an important property of the relation < introduced at the end of last time. Theorem 2.1. The relation < is a (strict) total order on N. That is, it satisfies the following properties: 1. Transitivity: if k < m and m < n, then k < n. 2. Antireflexivity: n n for all natural numbers n. 3. Totality: for any m, n N, either m < n, m = n, or m > n. Proof. The first two properties are just Exercises 1.4 and 1.6. For totality, we will use induction on both m and n. Say that m and n are comparable if m < n, m = n, or m > n; we want to show any two natural numbers are comparable. First fix m. Let s prove by induction on n that if m < n, then m + 1 n. If n = 0, there is no m such that m < n, so this is vacuous. Now suppose that m < n implies m + 1 n, and suppose m < n + 1; we want to show m + 1 n + 1. By Exercise 1.3, this means that m n. If m = n, then m + 1 = n + 1 so m + 1 n + 1. If m < n, then by the induction hypothesis, m + 1 n < n + 1. Hence we have shown by induction that m < n implies m + 1 n. Also, note that if m n, then m + 1 > m n. This together with the previous result implies that if m and n are comparable, then m + 1 and n are comparable. Now we prove the totality by induction on m. First, let m = 0. We prove by induction on n that 0 n for all n. For n = 0, we have 0 = n. For the induction step, if 0 n, then 0 n < n + 1. Thus 0 is comparable to every n. Furthermore, we remarked above that if m is comparable to n, so is m + 1. Thus by induction on m, every m and n are comparable. OK, now we have the natural numbers and we know when one number is bigger than another. Let s see what else numbers are good for. The primary purpose of the natural numbers is to count things. What does it mean to count? Well, when we count that a set S has (say) 4 elements, we are really matching up its elements with the numbers 1, 2, 3, and 4. That is, we are establishing a one-to-one correspondence, or bijection between S and the set {1, 2, 3, 4}. For us, it will be more convenient to start counting a 0, so instead we get a bijection with the set {0, 1, 2, 3}, which happens to be our set 4. What is a bijection more precisely? Well we could say we have a function f from S to 4 which sends an element of S to the number we re matching it with. This function has the property that for every n 4, there is exactly one s S such that f(s) = n. But what is a function? Traditionally, a function f : X Y between two sets is defined to be a set of ordered pairs (x, y) with x X and y Y satisfying certain properties. However, to make sense of this purely in terms of sets, we first have to define what an ordered pair is. There are many possible definitions; here is one. Definition 2.2. Let x and y be sets. Then the ordered pair (x, y) is the set {{x}, {x, y}}. The property that we really care about for ordered pairs is the following. Proposition 2.3. Let (x, y) and (a, b) be ordered pairs. Then (x, y) = (a, b) iff x = a and y = b. Proof. If (x, y) = (a, b) then x = a because x is uniquely defined by the property that x s for all s (x, y) = {{x}, {x, y}} and a is uniquely defined by the same property for (a, b). A similar argument shows that y = b (though we should consider separately the cases x = y and x y). Conversely, if x = a and y = b, then clearly (x, y) = (a, b). We can now define functions and bijections as usual. 5

6 Definition 2.4. Let X and Y be sets. Then a function from X to Y is a set f of ordered pairs (x, y) with x X and y Y such that for each x X, there is a unique y Y such that (x, y) f. We normally write f(x) = y instead of (x, y) f. We say f is an injection if for each y Y there is at most one x X such that f(x) = y, and f is a surjection if for each y Y there is at least one x X such that f(x) = y. We say f is a bijection if it is both an injection and surjection, i.e. for every y Y there is a unique x X such that f(x) = y. When there is a bijection between two sets, we can say they have the same size. If X is in bijection with Y and Y is in bijection with Z, X is in bijection with Z by composing the two bijections: if f : X Y and g : Y Z are bijections, then h(x) = g(f(x)) is a bijection from X to Z. Recall that we constructed the natural numbers so that for each n, we have n = {0, 1, 2, 3,..., n 1}. That is, n itself has n elements, informally speaking at least, so we can say a general set S has n elements if it is in bijection with n. Also, the number of elements in any finite set should be a natural number. This motivates the following definition. Definition 2.5. A set S is finite if it is in bijection with some n N. Here s another fact we may have thought we knew which we can now prove: the number of elements of a finite set is unique. In other words, counting works, and no matter what order you count a set in you get the same answer. Lemma 2.6. Let k be a natural number, n = k + 1, m < n, and let T = n \ {m} be the set of all elements of m except m. Then T is in bijection with k. Proof. Note that k = n \ {k}, as n = k + 1 = k {k}. Define f : k T as follows. If l < m, let f(l) = l. If l m, let f(l) = l + 1. It is not hard to check that this is a bijection. 1 Theorem 2.7 (Counting works). Let S be a finite set. Then there is only one natural number n that is in bijection with S. Proof. Suppose S is in bijection with both k and n and k n. Then by composing the bijections we obtain a bijection f : k n. By Theorem 2.1(d), we may assume k < n. We now prove by induction on n that there is no bijection from k to n for k < n. For n = 0, this statement is vacuous. Now suppose n = m + 1 and the statement is true for m. Suppose k < n and there is a bijection from n to k. Since n = m + 1 is nonempty, k is nonempty, so k 0. By Exercise 1.2, there is some l such that k = l + 1. Now restrict the bijection from n to k to the subset m n. We then get a bijection from m to the set T = k \ {f(m)}. By Lemma 2.6, T is also in bijection with l. But then m is in bijection with l < k m (by Exercise 1.3, k < n = m + 1 implies k m), contradicting the induction hypothesis. Hence there cannot be a bijection between k and n for k < n, so S cannot be in bijection with both k and n. Definition 2.8. Let S be a finite set. Then the cardinality S of S is the unique natural number that is in bijection with S. 2.1 Exercises Exercise 2.1 (Induction starting at n). Let P (k) be a property of natural numbers and n N. Suppose that P (n + 1) is true and for all n N, if P (k) is true then P (k + 1) is true. Show that P (k) is true for all k > n. (Hint: Prove by induction that either P (k) or k n. Exercise 2.2. Show that m < n iff m + 1 < n + 1. (Hint: Use Exercise 1.3.) 1 To see that this is a bijection, it is very important to know that k k so that n k. This follows from Exercise 2.1(c). 6

7 Exercise 2.3. Verify Lemma 2.6. That is, given m < n = k + 1, define f : k n \ {m} by f(l) = l if l < m and f(l) = l + 1 if m l < n. Show that for any j n \ {m}, there is a unique l < k such that f(l) = j. Exercise 2.4. Let n N. Show that there is no injective function f : n + 1 n. That is, if f : n + 1 n, then there are i, j n such that i j but f(i) = f(j). (Hint: Use induction on n and Lemma 2.6.) Exercise 2.5 (The Pigeonhole Principle). Suppose n < k are natural numbers. Show that there is no injective function f : k n. That is, if f : k n, then there are i, j n such that i j but f(i) = f(j). Exercise 2.6. When talking about cardinalities, it is customary to say that S T if there is an injection from S to T. Show that this agrees with our order on natural numbers: m n iff there is an injection from m to n. 7

8 3 Addition, Multiplication, and Subtraction We now know how to count a set using the natural numbers. Now let s do arithmetic. First, what is addition? Well, if you have two sets, you can add their sizes by just mushing them together to form one big set. Here s a more formal way of putting that. Definition 3.1. Let S and T be sets. Then the disjoint union S T is the set {(0, s) : s S} {(1, t) : t T }. If n, m N, then their sum is n + m = n m. Wait a minute! We ve already defined n + 1 for natural numbers, and now we ve given a different definition. Do these two definitions agree? They do, since we can define a bijection from n {n} to n 1 by sending m to (0, m) for m < n and sending n to (1, 0). Some basic properties of addition are easy to prove using this definition. Proposition 3.2. Addition of natural numbers is commutative and associative: n + m = m + n and n + (m + k) = (n + m) + k. Also, for any n N, n + 0 = n. Proof. For commutativity, we define a bijection from n m to m n via (i, k) (1 i, k) (for i = 0, 1). Associativity is similar and left as an exercise. For the last property, note that the map m (0, m) is a bijection from n to n 0. The following property, called cancellation, is a bit harder, but not too bad. Lemma 3.3. If n N is nonzero, then there is a unique m N such that m + 1 = n. Proof. By Exercise 1.2, there exists such an m, so it suffices to show that it is unique. Suppose m + 1 = k + 1 = n. Then by Lemma 2.6, both m and k are in bijection with the set n \ {0}. By Theorem 2.7, this implies m = k. Proposition 3.4. If l + n = k + n, then l = k. Proof. We use induction on n. For n = 0, this follows from Proposition 3.2. Now suppose it is true for m and n = m + 1. Then if l + n = k + n, then l + m + 1 = k + m + 1, or (l + 1) + m = (k + 1) + m. By the induction hypothesis, l + 1 = k + 1, and by Lemma 3.3, l = k. The cancellation property allows us to define subtraction. If k + m = n, we write n m = k, and this is well-defined (assuming any such k exists) by cancellation. In the exercises you will show that n m exists iff n m. Now let s look at multiplication. Again, we look to what multiplication means in terms of sets. When we multiply m by n, we take a copy of n for each element of m. We could write this as having ordered pairs, the first coordinate of which tells us which copy we re in and the second of which tells us which element of n to take. This motivates the following definition. Definition 3.5. Let S and T be sets. Then their (Cartesian) product is the set S T of all ordered pairs (s, t) where s S and t T. If n, m N, then their product is nm = n m. Again, lots of basic properties are quite easy from this definition. Proposition 3.6. Multiplication of natural numbers is commutative and associative: n(mk) = (nm)k. Also, for any n N, n 0 = 0 and n 1 = n. nm = mn and Proof. For commutativity, define a bijection from n m to m n by (k, l) (l, k). Associativity is similar and left as an exercise. As a set, n 0 = n =. A bijection from n to n 1 is given by k (k, 0). Proposition 3.7. Multiplication distributes over addition: m(n + k) = mn + mk. Proof. Define a bijection from m (n k) to (m n) (m k) by (j, (i, l)) (i, (j, l)). 8

9 In the exercises you will prove some more basic properties of arithmetic. Now that we ve learned how to work with natural numbers as sets, let s try to understand more complicated numbers. Let s start with the integers. There are lots of ways we could go about adding negative numbers to the natural numbers to get all integers. For example, we could simply arbitrarily construct a set n for each positive natural number n (say, we could define n = (N, n)). This would work, but it would make defining arithmetic operations complicated because we would have to constantly split everything into cases depending on whether our numbers were positive or negative. The method we will use to construct the integers will be more natural in that it does not treat positive and negative numbers separately and in that the technique generalizes to many other settings (for example, we will use the same method later to construct the rational numbers). The idea is not to simply construct new sets n, but to construct sets m n for all natural numbers m and n. To construct a set that we could call m n is simple enough we can just use the ordered pair (m, n). However, a single integer has many representations as a difference m n. For example, 0 2, 1 3, 2 4, and so on should all represent the same integer. Which one should we actually pick to represent that integer? Instead of arbitrarily picking one of them, we will represent the integer 2 as the set of all ways of writing it as a difference. This is best captured using the notion of equivalence classes. Definition 3.8. A relation on a set S is a set R S S. We usually write x R y instead of (x, y) R. A relation is an equivalence relation if it satisfies the following properties: 1. Reflexivity: x x 2. Symmetry: If x y then y x 3. Transitivity: If x y and y z then x z If x S, then the equivalence class of x is [x] = {y S : y x}. We write S/ for the set of equivalence classes of elements of S. We assume the reader is either familiar with or can verify for themselves the following fact. If not, you can ask me about it at TAU. Proposition 3.9. Let S be a set with an equivalence relation. Then every x S is contained in exactly one equivalence class, and [x] = [y] iff x y. In our case, the set S should be the set N N of ordered pairs of natural numbers, and we should have (m, n) (k, l) if m n should equal k l. Unfortunately, if m < n, we haven t defined what m n is yet. What we really want is a condition that is equivalent to m n = k l but only uses operations we know are well-defined. One thing we can do is add n + l to both sides of the equation to obtain the condition m + l = k + n. Lemma The relation (m, n) (k, l) if m + l = k + n is an equivalence relation on N N. Proof. For reflexivity, we just use the tautology m + n = m + n. For symmetry, we use the fact that m + l = k + n implies k + n = m + l. For transitivity, suppose (m, n) (k, l) and (k, l) (i.j). Then adding the equalities m + l = k + n and k + j = l + i together gives m + l + k + j = k + n + l + i. Cancelling k + l gives m + j = n + i, or (m, n) (i, j). We can thus make the following definition. Definition The integers are the quotient Z = (N N)/, for defined as in Lemma We write [m, n] for the equivalence class of (m, n) N N. 9

10 3.1 Exercises Exercise 3.1. Prove that = 4 by explicitly giving a bijection from 2 2 to 4. Exercise 3.2. Construct bijections to prove the associativity of addition and multiplication (i.e., (m + n) + k = m + (n + k) and (mn)k = m(nk)). Exercise 3.3. This exercise explores how arithmetic interacts with the ordering on the natural numbers. (a): Show that for any m, l < k iff l + m < k + m. (Hint: Use Exercise 2.2 and induction.) (b): Show that n m exists (i.e., there exists k such that m + k = n) iff m n. (Hint: For the forward direction, apply (b) with l = 0. For the reverse direction, prove m n and let k = n \ m.) (c): Show that if m < n and l k, then m + l < n + k. (Hint: Write n + k = n + l + (k l).) (d): Show that if m > 0, then l < k iff lm < km. (Hint: Use distributivity, induction, and (d).) Exercise 3.4. Show that multipication (except by 0!) has cancellation: if k 0 and nk = mk, then n = m. It follows that division by natural numbers other than 0, when it can be done at all, is well-defined. (Hint: Use Exercise 3.3(e).) Exercise 3.5 is optional, but you should at least think about it if you haven t seen equivalence relations before. Exercise 3.5. (a): Prove Proposition 3.9. (b): A partition of a set X is a collection P of nonempty subsets of X such that if A, B P, then A B =, and such that the union P of all the sets in P is all of X. Define x y if x, y A for some A P. Show that is an equivalence relation and P = X/. The following exercise is optional (but interesting!). Exercise 3.6. If S and T are sets, let Fun(S, T ) be the set of all functions from S to T. For n, m N, define n m = Fun(m, n). Explain why this definition should be the same as the familiar notion of exponentiation and see what you can prove about it (e.g., properties like n m n k = n m+k and (n m ) k = n mk ). What is 0 0? 10

11 4 Integers and Rational Numbers But wait! The integer 0 should be the equivalence class of pairs (m, n) with m n = 0, i.e. the set {(0, 0), (1, 1), (2, 2),...}. This is very different than the set 0 = which we defined as the natural number 0. Similarly, for any n N the equivalence class [n, 0] should represent n as an integer, but it is very different from n as a set. Is this a problem? The answer is no, for a reason we ve been periodically mentioning, which is that the specific way we chose to define the natural numbers is not the definition of the natural numbers but just a definition. The reason we care about the natural numbers is the properties they have when we do arithmetic and count with them, not the specific sets we use to represent them. The only reason that we chose to define them the way we did is that it made it easier to prove those properties. However, any other set that looks like the natural numbers with elements called 0, 1, 2, and so on, and operations called addition and multiplication and an order relation that behave in exactly the same way is just ast good. This idea of looking the same is captured more precisely by the notion of an isomorphism, which is a bijection which preserves all of the relevant structure (addition, multiplication, order). 2 We have not yet defined arithmetic and order on the integers, but we can start with the following result. Proposition 4.1. The function F (n) = [n, 0] is a bijection from N to {[n, 0] : n N} Z. Proof. This map is clearly surjective. For injectivity, suppose F (n) = F (m). Then [n, 0] = [m, 0] so (n, 0) (m, 0). But this just means that n + 0 = m + 0, so n = m. From now on, we will not normally distinguish notationally between the natural number n N and the integer F (n), and will pretend that N actually is a subset of Z. In Exercise 4.1, you will show that the definitions of arithmetic and order for integers agree with the definitions we already had for natural numbers. Let s now try to define addition of integers. Since (m n) + (k l) should equal (m + k) (n + l), we make the following definition. Definition 4.2. Let [m, n] and [k, l] be integers. Then their sum is [m, n] + [k, l] = [m + k, n + l]. But wait! The integer [m, n] is equal to the integer [m + 1, n + 1], so [m, n] + [k, l] should equal [m + 1, n + 1] + [k, l]. This is not obvious from the definition we have given. The property that addition should not depend on what representative (m, n) of the equivalence class [m, n] we choose is called addition being well-defined. Proposition 4.3. Addition of integers is well-defined: if [m, n] = [m, n ] and [k, l] = [k, l ], then [m + k, n + l] = [m + k, n + l ]. Proof. This is just algebra. We have m + n = n + m and k + l = k + l, and adding these two equations gives m + n + k + l = n + m + k + l, or Hence [m + k, n + l] = [m + k, n + l ]. (m + k) + (n + l ) = (n + l) + (m + k ). Note that addition of integers agrees with our earlier definition of addition of natural numbers: if we have integers [m, 0] and [n, 0] which should be the same as the natural numbers m and n, then [m, 0] + [n, 0] = [m + n, 0 + 0] = [m + n, 0] is the natural number m + n. Here are some basic properties of addition. Proposition 4.4. Addition of integers is commutative and associative. For any [m, n] Z, [m, n] + 0 = [m, n]. 2 In other words, an isomorphism a bijection f where m + n = k iff f(m) + f(n) = f(k), mn = k iff f(m)f(n) = f(k), and m < n iff f(m) < f(n) basically, everything works the same on either side of the bijection. 11

12 Proof. For commutativity, note that [m + k, n + l] = [k + m, l + n] by commutativity of addition of natural numbers. The other two similarly follow from the same properties for natural numbers (using that 0 = [0, 0]. Unlike the natural numbers, with the integers we can also always define subtraction. Definition 4.5. Let [m, n] Z. Then its negative is [m, n] = [n, m] (why does this definition make sense, and why is it well-defined?). We write [m, n] [k, l] = [m, n] + ( [k, l]). We intended for [m, n] to represent m n ; is it true that [m, n] = m n for this definition of subtraction? It is, because m n = [m, 0] + [0, n] = [m, n]. Note also that ( 1)[m, n] = [m, n], since [0, 1][m, n] = [m0 + n1, m1 + n0] = [n, m]. Proposition 4.6. For any integer [m, n], [m, n] [m, n] = 0. Proof. We have [m, n] [m, n] = [m, n] + [n, m] = [m + n, m + n] = 0. Remark 4.7. Propositions 4.4 and 4.6 together say that Z is an abelian group under addition, an abstract algebraic structure with addition and subtraction. This together with Proposition 4.11 says that Z is a commutative ring, an abstract algebraic structure with addition, subtraction, and multiplication. We can now show that Z indeed agrees with our intuitive idea of the integers. Proposition 4.8. Every integer is a natural number xor n for a unique nonzero natural number n. Proof. Let [k, l] be an integer. If k l, then k l N exists by Exercise 3.3(b), and it is easy to see that [k, l] = [k l, 0] is a natural number. If k < l, then [k, l] = [0, l k] = [l k, 0] and l > k so l k is a nonzero natural number. This n = l k is unique since if n = m, [0, n] = [0, m] so m = n. The proof of exclusivity of the two possibilities is similar and left to the reader. From now on, we will sometimes simply write n for an arbitrary integer. The results we have developed show that we can manipulate integers according to the ordinary rules of addition and subtraction. Now let s define multiplication. Since (m n)(k l) should equal (mk + nl) (ml + nk), we make the following definition. Definition 4.9. Let [m, n] and [k, l] be integers. Then their product is [m, n][k, l] = [mk + nl, ml + nk]. Proposition Multiplication of integers is well-defined: if [m, n] = [m, n ] and [k, l] = [k, l ], then [mk + nl, ml + nk] = [m k + n l, m l + n k ]. Proof. This is just algebra, but it s quite long and tedious. Basically, you make a bunch of formal manipulations starting from the equations m + n = n + m and k + l = k + l and eventually arrive at mk + nl + m l + n k = ml + nk + m k + n l. You are invited to work out the details if you enjoy messy algebra. Proposition Multiplication of integers is commutative and associative and distributes over addition. For any [m, n] Z, [m, n] 0 = 0 and [m, n] 1 = [m, n]. Proof. For commutativity, we have [m, n][k, l] = [mk + nl, ml + nk] = [km + ln, lm + kn] = [k, l][m, n]. The other assertions are similar but more complicated algebra and are left as exercises. The following facts will be important to us next section. Proposition Multiplication by integers other than 0 has cancellation: if mn = kn and n 0, then m = k. 12

13 Proof. If m, n, and k are all natural numbers, this is just Exercise 3.4. Suppose n N and m, k N. Then since mn = kn, m( n) = ( 1)mn = ( 1)kn = k( n), and we are done by the case where all the numbers are natural numbers. Suppose n, k N and m N. Then ( m)n N and mn = kn N, which implies mn = ( m)n = 0. Since n 0 and ( m) 0 = 0, Exercise 3.4 implies m = 0, and similarly we can show that k = 0. There are 5 other cases to consider, but they are all similar to the ones done above. There are a few other things to check to be sure that everything works as we expect, but their proofs are just as simple as all the proofs so far have been. Some of them are in the exercises, but others we leave without proof and trust that you can come up with a proof on the spot if we need to use them. To close this section, we define the order on the integers, whose properties are fleshed out in the exercises. Definition Let m, n Z. Then we say m n if n m N. 4.1 Exercises Exercise 4.1. Show that the map F of Proposition 4.1 is an isomorphism for all the structure we ve defined. That is, show that our definitions of addition, subtraction, multiplication, and order for integers all agree with the definitions in Section 2 for natural numbers. (The remark following Proposition 4.3 showed this for addition.) Exercise 4.2. Complete the proof of Proposition 4.4. Exercise 4.3. In this exercise we prove some basic properties of the ordering on the integers. (a): Show that n 0 iff n N. (b): Show that < is a strict total order, where < means and not = and strict total order means it satisfies the conditions of Theorem 2.1. (Hint: Use Proposition 4.8.) (c): Show that every part of Exercise 3.3 except (c) holds for integers, and that (c) fails. (d): Show that m n iff m n. You can skip the following exercise if we didn t get to talking about multiplication in class. Exercise 4.4. Finish the proof of Proposition

14 5 The Rational Numbers We will now construct the rational numbers from the integers in much the same way as we constructed the integers from the natural numbers. Just as integers were defined as differences of natural numbers, rational numbers will be quotients of integers. Just as we had m n = k l if m + l = k + n, we will have m/n = k/l if ml = kn. We must also not allow 0 as a denominator. Lemma 5.1. The relation (m, n) (k, l) if ml = kn is an equivalence relation on Z (Z \ {0}). Proof. The proof is exactly the same as that of Lemma 3.10 except we turn addition into multiplication everywhere. For example, for transitivity, suppose (m, n) (k, l) and (k, l) (i.j). Then multiplying the equalities ml = kn and kj = li together gives mlkj = knli. Cancelling kl (which is possible since k, l Z \ {0}) gives mj = ni, or (m, n) (i, j). Definition 5.2. The rational numbers are the quotient Q = (Z (Z \ {0})/, for defined as in Lemma 5.1. We write m/n for the equivalence class of (m, n) Z (Z \ {0}). Just as we could think of the natural numbers as sitting inside the integers, we can think of the integers as sitting inside the rational numbers. Proposition 5.3. The function G(n) = n/1 is a bijection from Z to {n/1 : n Z} Q. Proof. Again, the proof is exactly the same as Proposition 4.1, substituting multiplication for addition and 1 for 0. For example, for injectivity, suppose G(n) = G(m). Then (n, 1) (m, 1), so n 1 = m 1, so n = m. As before, we will not distinguish between n Z and G(n) Q. We will leave it as an exercise to see that this G preserves arithmetic and order as we define it below. Definition 5.4. Let m/n and k/l be rational. Then their sum is m/n + k/l = (ml + kn)/(nl). As for the integers, we must check that this definition is well-defined. Proposition 5.5. Addition of rational numbers is well-defined: if m/n = m /n and k/l = k /l, then (ml + kn)/(nl) = (m l + k n )/(n l ). Proof. We have (ml + kn)(n l ) = (mn ll + kl nn ) = (m nll + k lnn ) = (m l + k n )(nl). It is now straightforward to show the following. Proposition 5.6. Addition of rational numbers is commutative and associative. For any m/n Q, m/n + 0 = m/n and m/n + ( m)/n = 0. Proof. Exercise. Remark 5.7. Proposition 5.6 says that Q is an abelian group under addition. The treatment of multiplication is similar, and equally straightforward. Definition 5.8. Let m/n and k/l be rational. Then their product is (m/n)(k/l) = (mk)/(nl). 14

15 Proposition 5.9. Multiplication and reciprocals of rational numbers is well defined. Multiplication is commutative and associative and distributes over addition. For any m/n Q, (m/n) 0 = 0, (m/n) 1 = m/n. If m/n 0, we additionally have (m/n)(n/m) = 1. Proof. Exercise. Remark Proposition 5.9 together with Proposition 5.6 says that Q is a field, an abstract algebraic structure with addition, subtraction, multiplication, and division. Finally, we define the ordering on the rational numbers. Definition Let m/n Q. Then we write m/n > 0 if mn > 0 (Why does this definition make sense? Why is it well-defined?). For x, y Q, we write x > y if x y > 0. Proposition The relation < is a (strict) total order on Q. If x, y, z Q and x < y then x+z < y +z. If z > 0, we also have xz < yz, and if z < 0 we also have xz > yz. Proof. Exercise. Remark Proposition 5.12 together with the previous results says that Q is an ordered field, a field with a total order compatible with the field structure. 5.1 Exercises Except for Exercise 5.5, these exercises are all just checking basic properties of the rational numbers that we skipped in class. There are a lot of things to check but they are all pretty much straightforward, and it may get tedious and boring. If it is getting tedious and boring, don t feel like you have to do it all; what s important is that you understand how these proofs work, not that you do every single one of them. Exercise 5.1. Show that the map G of Proposition 5.3 is an isomorphism for all the structure we ve defined. That is, show that our definitions of addition, multiplication, and order for rationals all agree with the definitions in Section 4 for integers. Exercise 5.2. Prove Proposition 5.6. Exercise 5.3. Prove Proposition 5.9. Exercise 5.4. Prove Proposition Exercise 5.5. Prove that Q is an archimedean ordered field: for any x Q, there is an n N such that n > x. 15

16 6 The Real Numbers While the intuition behind constructing the integers and the rational numbers was fairly straightforward, the real numbers are another question entirely. It is not even necessarily clear what a real number is intuitively. In school, you ve probably seen real numbers presented as infinite decimals. Indeed, this is perfectly valid, and we could make it into a rigorous definition. However, this definition would be unrelated to our previous definition of the rationals and would unnaturally favor 10 as the base. Instead, we will look for a more natural definition. Passing from the natural numbers to the integers involved making addition better-behaved (so that additive inverses exist and subtraction is possible). Passing from the integers to the rationals involved making multiplication better-behaved (so that multiplicative inverses exist and division is possible). Passing from the rationals to the reals will involve making the order relation better-behaved, though in a rather subtle way. A way to geometrically think of the irrational numbers is that they are holes in the rational number line. That is, they are places on the line whose location we can specify but where there is not any rational number. For example, we can define 2 as being right in between all the positive rationals whose square is less than 2 and all the positive rationals whose square is greater than 2. We could define π as being between 3, 3.1, 3.14, 3.141,... and 4, 3.2, 3.15, 3.142,.... In general, we can define any real number by saying what rational numbers are less than it and what rational numbers are greater than it. Alternatively, we could just give the rational numbers that are less than it, since the rational numbers greater than it are then all the rest. This motivates the following definition. Definition 6.1. A (left) cut in the rational numbers (or in any totally ordered set) is a subset L Q such that whenever x L and y < x, y L as well. We should think of a cut as the set of all rational numbers less than some real number. However, this definition isn t yet quite right a cut does not exactly correspond to a real number. The first problem is simple: according to Definition 6.1, both and all of Q are cuts. However, these geometrically would correpond to the points and, which we do not want to call real numbers. Thus we must exclude these two possibilities. Definition 6.2. A cut L Q is proper if L and L Q. The second problem with the notion of a cut is more subtle: more than one cut might correspond to the same number. For example, let L = {x Q : x < 2} and M = {x Q : x 2}. Then L and M both ought to represent the number 2. We could construct a similar example with 2 replaced by any rational number. To put it another way, we have to decide whether we want our cuts to be all rationals strictly less than a real number or a rationals less than or equal to a real number. We will (arbitrarily) choose strictly less than, so we will choose L rather than M to represent 2. How do we detect if a cut is of type L instead of type M? We could do it in an ad hoc way, but instead we will introduce the following definition which is very important to studying the real numbers. Definition 6.3. Let S Q (or any totally ordered set). An upper bound for S is an x Q such that x s for all s S. A least upper bound, or supremum of S is an upper bound that is less than every other upper bound. The supremum of S, if one exists, is unique. Indeed, if x and y are both suprema, then they are both upper bounds so x y and y x so x = y. We write sup S for the supremum of S if it exists. In the case of our cuts L and M above, it is easy to see that sup L = sup M = 2. The difference between L and M can be stated as the fact that M contains its supremum but L does not. On the other hand, cuts corresponding to irrational numbers do not have suprema, because what should be the supremum is the missing irrational number itself. Definition 6.4. A cut L Q is open if sup L, if it exists, is not contained in L. 16

17 Note that under this definition, a cut without a supremum is automatically open. It turns out that these two corrections to our definition of cuts are enough to get the real numbers. Definition 6.5. The real numbers R are the set of all proper open cuts in Q. Once again, we must see how our old Q embeds in this new number system. Proposition 6.6. The map H(x) = {y Q : y < x} is a bijection from Q to {L R : sup L exists}. Proof. Since sup H(x) = x, H clearly maps into {L R : sup L exists}. It is also clear that H is injective. Finally, if L R and sup L = x, then L H(x) since x is an upper bound for L and L is open. Furthermore, no y < x is an upper bound, so for all y < x there is a z L with y < z. Since L is a cut, this implies y L, for any y < x. That is, H(x) L, so H(x) = L. Hence H is surjective. As before, we should also check that this map H is compatible with arithmetic and order that we will define. As before, we leave this as an exercise. As before, we will not distinguish between x Q and H(x) R. Since the real numbers are defined in terms of the order relation on Q, it is easy to define the order relation on R. Indeed, it is intuitively not hard to see that one cut is to the left of another cut iff it is a subset. Definition 6.7. Let L, M R. Then we write L < M if L M. Proposition 6.8. The relation < is a (strict) total order on R. Proof. The only part that is not routine is totality: if L M, then L < M or M < L. Suppose L M, i.e. L M. Then there is some x L that is not in M. Since M is a cut, this also implies y M for all y > x. Thus M {y : y < x}. But since L is a cut and x L, every y < x is also in L. Thus M L, so M < L. Since the real numbers were defined in attempt to fill in the holes in the rational numbers, we would like to know whether there are any holes left in the real numbers. The following concept captures the idea of not having holes. Definition 6.9. A totally ordered set X is complete if whenever S X is nonempty and has an upper bound, then it has a supremum. It can be shown, in fact, that we get an equivalent notion if we restrict to the case that S is a cut. As we saw, the cuts which had suprema were those that corresponded to rational numbers, so completeness says that every cut already corresponds to some point in your set there are no holes to fill. Theorem The real numbers are complete. Proof. Let S R be nonempty and have an upper bound, and let L = S, the union of all the elements of S. Then L is nonempty since S is nonempty. If K R is an upper bound for S and x Q \ K, then x > y for all y K. Since every M S is contained in K, so is L. Thus x > y for all y L, so L is not all of R. Hence L is proper. Finally, if x = sup L, then x is an upper bound for L and hence for all M S, and it follows that x M for all M S so x L, so L is open. Hence L R. We now claim that L = sup S in R. It is clear that L M for all M S, so it is an upper bound. Furthermore, if L M for all M S, then L S = L by definition of the union, so L is the least upper bound. We now define arithmetic on R. Definition Let L and M be real numbers. Then their sum is L + M = {x + y : x L and y M}. The negative of a real number L is L = { x : x L and x sup L}. The absolute value of L is L if L 0 and L if L < 0 17

18 In this case we have no equivalence relation to worry about for well-definedness, but we must check that L + M and L are indeed proper open cuts. This takes some work, but is too tedious to be worth going through here. Proposition Addition of real numbers is commutative and associative. For any L R, L + 0 = L and L + ( L) = 0. Proof. We prove that L+0 = L and leave the rest as an exercise. Note that here 0 = H(0) = {x Q : x < 0}. For any y L and x 0, x + y < y so x + y L, hence L + 0 L. Conversely, if x L, x sup L, so there is some z L such that z > x. Then y = x z 0, so x = z + y L + 0. Hence L L + 0, sol = L + 0. Remark Proposition 6.12 says that R is an abelian group under addition. Multiplication is rather more painful to define, and is most easily defined by splitting into cases depending on the signs of the factors Definition Let L 0 and M 0 be real. Then their product is LM = {xy : x L, y M, and x, y 0} {x : x < 0}. If L < 0 and M 0 or L 0 and M < 0, we define LM = L M, and if L < 0 and M < 0 we define LM = L M. The reciprocal of a real number L > 0 is L 1 = {x : xy < 1 for all y M such that y 0}. If L < 0, we define L 1 = L 1. Again, there is a fair amount of work to show that these products and reciprocals are themselves real numbers, and we omit this. Proposition Multiplication of real numbers is commutative and associative and distributes over addition. For any L R, L 0 = 0 and L 1 = L. If L 0, we additionally have LL 1 = 1. Proof. Exercise, though be warned that this is a fair deal of work. Remark Proposition 6.15 together with Proposition 6.12 says that R is a field. Finally, we record the fact that the ordering on R is compatible with the arithmetic. Proposition If K, L, M R and K < L then K + M < L + M. If M > 0, we also have KM < LM, and if M < 0 we also have KM > LM. Proof. Exercise. Remark Proposition 6.17 together with the previous results says that R is an ordered field. In fact, it can be shown that R is the only complete ordered field, up to isomorphism. This fact is very useful in showing that different methods of defining the real numbers are equivalent. 6.1 Exercises Exercise 6.1. Show that the map H of Proposition 6.6 is an isomorphism for all the structure we ve defined. That is, show that our definitions of addition, multiplication, and order for real numbers all agree with the definitions in Section 5 for rationals. The first three exercises have you fill in proofs that we skipped in class. I recommend that you do a few of these until you get the hang of how they go, but don t worry about doing all of them. Exercise 6.2. Complete the proof of Proposition Exercise 6.3. Prove Proposition Exercise 6.4. Prove Proposition Exercise 6.5. Prove that R is archimedean: for any L R, there is an n N such that n > L. Exercise 6.6. Define lower bounds and greatest lower bounds, or infima, in the same was as we defined upper bounds and suprema. Show that every nonempty subset of R with a lower bound has an infimum. 18

19 7 Problems with set theory Throughout these notes, we have worked with the assumption that a simple, intuitive notion of sets can and should be used as the basis to develop all of mathematics. There are various reasons why this assumption is not entirely correct, and in this section we will see the main problems with it. The first problem with the set theory we have been using thus far is that it is not self-consistent! Let us be more precise. It turns out that everything we have done has used only the rules of logic plus the following two axioms about sets, the first of which is harmless but the second of which turns out to be too strong. 1. A set is determined by its elements 2. For any property P (x) of sets, there is a set {x : P (x)} of all sets having property P. Let s see how we can get a contradiction, called Russell s paradox, out of this. We use Axiom (2) with the property x x. Thus let S = {x : x x}. Now we ask, is S S? If S S, then by definition of S, S S. But if S S, then similarly S S! This is a contradiction! Here s another similar paradox, called Cantor s paradox. As you may have seen before, the real numbers are uncountable: they are not in bijection with the natural numbers. This is usually proved by the diagonal argument using the representation of real numbers as decimal (or binary) expansions. This diagonal argument can be extended to produce a set which is strictly larger than any given set. Definition 7.1. Let X be a set. Then the power set of X is P(X) = {S : S X}. Theorem 7.2 (Cantor). For any X, X is not in bijection with P(X). Proof. Suppose f : X P(X) is a bijection. Define S = {x X : x f(x)}. Then S P(X), so S = f(y) for some y X. But then y f(y) = S iff y S, a contradiction. Hence no such bijection f can exist. Now let s apply Cantor s theorem to the set of all sets. For example, we could define this set as V = {x : x = x}, since x = x is true for any x. By Cantor s theorem, V is not in bijection with P (V ). But V contains all sets, so it is clear that P (V ) = V. This is a contradiction. 3 It is thus clear that set theory as we have been doing it, also known as naive set theory, is inconsistent. If we want to be able to consistently use set theory, we must somehow modify Axiom (2). The standard solution that set theorists have accepted is to reject the idea that any property can define a set and conclude that Russell s Paradox and Cantor s Paradox are instead proofs that no set S = {x : x x} or V = {x : x = x} can exist! After all, assuming they did exist, we reached a contradiction, so therefore they must not exist. By rejecting Axiom (2), however, we must come up with new axioms that tell us how we can construct sets. The main idea behind these axioms, the standard formulation of which is known as the Zermelo- Fraenkel Axioms (with Choice), or ZFC, is that instead of being about to create a set out of thin air, we must construct it out of set we already have. For example, if we know a set X exists, we are allowed to conclude that the power set P (X) exists. Or, in close analogy with Axiom (2), given a set X and any property P, the set {x X : P (x)} exists. If we knew the set V of all sets existed we could apply this with X = V to conclude Axiom (2), but V does not exist. While this proliferation of new axioms works, it certainly loses much of the aesthetic appeal of naive set theory. Instead of just using basic pure logic about how sets work to construct all of mathematics, we must now use some hodgepodge of arbitrary-seeming rules about what sets we can make. This sentiment is perhaps worsened by the fact that one of these axioms is explicitly the fact that N itself exists we cannot, in fact, construct any infinite set without just directly assuming that one exists. Furthermore, a famous result of Gödel states that there is no perfect set of axioms. More precisely, for any consistent collection of axioms in which we can express arithmetic which we can write down, there is 3 You may have noticed that the proof of Cantor s theorem looks a lot like the argument in Russell s paradox. In fact, if you unwind the argument of Cantor s paradox to include the proof of Cantor s theorem in the case X = V, you will see that it is really the same as Russell s paradox. 19

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