2.4 The Completeness Property of R

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1 2.4. THE COMPLETENESS PROPERTY OF R The Completeness Property of R In this section, we start studying what makes the set of real numbers so special, why the set of real numbers is fundamentally different from the set of rational numbers. The completeness property is also known as the least upper bound property. We will use both terms in this section. In this section we will use interval notation though we haven t defined intervals yet. A precise definition of intervals will come at the end of the section, after the least upper bound property. Until then, the reader can use the intuitive definition of an interval given in previous mathematics classes such as calculus Bounded Sets Maximum and Minimum of a Set Definition 119 (Extrema) Let S be a subset of R 1. An element x 0 of S is said to be a maximum of S if x 0 x for every other x in S. In this case, we say that x 0 is the largest element of S and we write x 0 = max S. 2. An element x 1 of S is said to be a minimum of S if x 1 x for every other x in S. In this case, we say that x 1 is the smallest element of S and we write x 1 = min S. 3. An extremum is either a maximum or a minimum. Remark 120 An extremum for a set S is always an element of S. Remark 121 To prove an element M is a maximum for a set S, we have to prove two things: 1. M S. 2. No other element of S is larger than M. Remark 122 It is similar for a minimum. Example 123 The smallest element (or minimum) of [0, 1] is 0. Its largest element (or maximum) is 1. More generally, if a and b are two real numbers such that a b then min [a, b] = a and max [a, b] = b. Example 124 The minimum of [0, 1) is 0. It does not have a maximum. To be a maximum, a number α would have to be in [0, 1). Thus, we would have α < 1. But then α < α + 1 < 1, so α + 1 is also an element of [0, 1) which is 2 2 larger than α. This contradicts the fact that α = max ([0, 1)).

2 48 CHAPTER 2. THE STRUCTURE OF R Example 125 Following the examples above, it is easy to see that if a and b are two real numbers such that a b then [a, b) does not have a maximum and min [a, b) = a, (a, b] does not have a minimum and max (a, b] = b, (a, b) has neither a minimum nor a maximum. Example 126 The largest element of [0, 1] {3} is 3. Example 127 Consider the set S = { q Q : q > 0 and q 2 < 2 }. This set has no maximum. To show this, we show that if p S, one can find q S such that p < q. Let p S. In particular, p > 0 and p 2 < 2. Define q = p + 2 p2 p + 2 = 2p + 2 p + 2 First, it should be clear that q Q and q > 0 (since p 2 < 2). Also, it is clear that q > p. We only need to establish that q 2 < 2 so that q S. q 2 2 = 4p2 + 8p + 4 (p + 2) 2 2 = 4p2 + 8p + 4 2p 2 8p 8 (p + 2) 2 = 2p2 4 (p + 2) 2 = 2 ( p 2 2 ) (p + 2) 2 < 0 Since p 2 < 2. Thus q 2 < 2. Therefore, given any p S, we have found q S such that p < q. Thus S has no largest element. Upper and Lower Bounds, Bounded Sets Definition 128 (Bounded) Let S be a subset of R. 1. S is said to be bounded above if there exists a number M in R such that x M for every x in S. M is called an upper bound for S. 2. S is said to be bounded below if there exists a number m in R such that x m for every x in S. m is called a lower bound for S. 3. S is said to be bounded if it is bounded above and below. 4. S is said to be unbounded if it lacks either an upper bound or a lower bound.

3 2.4. THE COMPLETENESS PROPERTY OF R 49 Remark 129 Not every subset of R has an upper bound. For example, (4, ) does not have an upper bound, it is unbounded. However, if a set S has an upper bound M, then every number larger than M is also an upper bound. That is, if S has an upper bound, then it has infinitely many. A similar result holds for lower bounds. Consider (0, 1). 1 is an upper bound. In fact, any number M 1 is an upper bound. Example and every number larger than 3 is an upper bound of [0, 3). On the other hand, 2.99 is not an upper bound. Example is also an upper bound of [0, 3]. So, an upper bound of a set can be in the set (more on this later). Since a set usually has an infinite number of upper bounds, a possible question is: given a set S, what is the set of upper bounds of S. We explore this question in the next two examples. Example 132 What is the set of upper bounds of [0, 1]? A number M is an upper bound of [0, 1] if and only if M 1 therefore, the set of upper bounds of [0, 1] is [1, ). Example 133 What is the set of upper bounds of [0, 1)? We see that if M 1, then M is an upper bound of [0, 1). Could [0, 1) have upper bounds smaller than 1? The answer is no. Obviously, to be an upper bound of [0, 1), a number would have to be greater than 0. So, we only need to see what happens between 0 and 1. But if α < 1 is an upper bound of [0, 1) then α < α + 1 < 1 which contradicts the fact that α is an upper bound. So, [0, 1) 2 cannot have an upper bound less than 1. It follows that the set of upper bounds of [0, 1) is [1, ). Remark 134 The reader should remember the argument we used above. It is a result which was established in theorem 103. Remark 135 [0, 1] and [0, 1) have the same set of upper bounds. Remark 136 More generally, if a and b are two real numbers such that a b, then [a, b], (a, b), (a, b] and [a, b) have the same set of upper bounds which is [b, ) and the same set of lower bounds which is (, a]. Remark 137 An upper bound M of a set S may or may not be in S. Example 138 The set of lower bounds of {2} [3, 4] is (, 2]. Example 139 The empty set,, presents an interesting case. Every real number is both an upper bound and a lower bound of. To see this, it is better to look at why a number may fail to be an upper bound. M will fail to be an upper bound of a set S if there exists an element of S larger than M. If x is any real

4 50 CHAPTER 2. THE STRUCTURE OF R number, no member of can be larger than x. Thus x is an upper bound of. Since this argument can be carried out for arbitrary x, it follows that any real number x is an upper bound of. A similar argument can be used for lower bounds. Example 140 The set N is clearly bounded below by 1. Intuitively, we know that it is not bounded above. Proving it is more diffi cult. We can see that it is not bounded above by an integer. If n were an upper bound of N, then n + 1 is also an integer larger than n. This contradicts the fact that n is an upper bound for N. However, there could exist a real number which is an upper bound for N. While this is not true, we cannot prove it at this stage. Example 141 Consider the set S = {0, 12, 23, 34 },... = {1 1n } : n N. This set is bounded below by any real number m 0. It is bounded above by any real number M 1. Lemma 142 Suppose that S is a non-empty subset of R, α is a lower bound of S and β is an upper bound of S. Then, we must have α β. Proof. Since S, we can choose x S. Because α is a lower bound of S, we have α x. Because β is an upper bound of S, we have x β. Using the transitivity property of, we obtain α β. Supremum and Infimum of a Set Definition 143 (Supremum and infimum) Let S be a subset of R 1. If S is bounded above, then an upper bound of S is said to be a supremum denoted sup S or a least upper bound denoted lub S if it is less than any other upper bound of S. If this number exists, we will denote it by sup S. 2. If S is bounded below, then a lower bound of S is said to be an infimum denoted inf S or a greatest lower bound denoted glb S if it is greater than any other lower bound of S. If this number exists, we will denote it by inf S. Remark 144 It should be noted that sup S and inf S are not necessarily elements of S. Remark 145 Obviously, if S is not bounded above then S does not have a supremum. Similarly, if S is not bounded below, it does not have an infimum. What about the converse? Another way of defining the supremum is as follows. A number u R is a supremum of a subset S of R if it satisfies the two conditions 1. s u s S 2. If v is any number such that s v s S then u v

5 2.4. THE COMPLETENESS PROPERTY OF R 51 The first condition says that u is an upper bound. The second says that u is less than any other upper bound that is u is the least upper bound. Remark 146 We see that to prove a real number β is an upper bound of a set S, we must prove: 1. β is an upper bound of S. 2. Any other upper bound of S is larger than β. This condition can be proven directly, that is we assume that γ is another upper bound of S and show we must have β γ. It can also be done by contradiction. We assume that γ is another upper bound such that γ < β and derive a contradiction. Example 147 Consider the set S = (0, 2). A number M 2 is an upper bound. So, the set of upper bounds is [2, ). The smallest element of this set is 2. Therefore, the least upper bound of S is 2 which is not in S. We would write sup (0, 2) = 2. You will also note that S does not have a largest element or a maximum. Example 148 Consider the set [0, 2]. A number M 2 is an upper bound. So, the set of upper bounds is [2, ). The smallest element of this set is 2. Therefore, the least upper bound is 2 which is in S. We would write sup [0, 2] = 2. You will also note that S has a maximum, 2. In this example, the maximum and the supremum are equal. This is in fact true for every set which has a maximum as we will see later. Example 149 More generally, if a and b are two real numbers such that a b, then sup (a, b) = sup [a, b] = b and inf (a, b) = inf [a, b] = a. Also, the following quantities do not exist: sup (a, ), sup [a, ), inf (, b) and inf (, b]. Example 150 N has a greatest lower bound, it is 1. So, inf N = 1. Since N is not bounded above, it does not have any upper bound thus it does not have a least upper bound. Another way of characterizing the supremum of a set is given below. It is a way we will use throughout this text. Make sure you understand it and remember it. Theorem 151 Let S be a non-empty subset of R. An upper bound M 0 of S satisfies M 0 = sup S, if and only if for each y < M 0, there exists an x in S for which y < x M 0 Proof. We need to prove both directions. 1. Let us assume that M 0 = sup S. We need to prove that for each y < M 0, there exists an x in S for which y < x M 0. We do a proof by contradiction. Let y < M 0 be given and assume that there is no element x of S such that y < x. Then, for every x in S, x y. Thus, y is an upper bound of S which is smaller than M 0 which contradicts the fact that M 0 is the supremum.

6 52 CHAPTER 2. THE STRUCTURE OF R 2. Let M 0 be an upper bound of S with the property that for each y < M 0, there exists an x in S for which y < x M 0. We need to show that M 0 = sup S. Since M 0 is already an upper bound, it is enough to show it is the smallest. If γ were an upper bound strictly smaller than M 0, then by assumption, there would exists an x in S for which γ < x M 0. But then γ would not be an upper bound of S, which contradicts our assumption. Thus, there cannot be an upper bound of S smaller than M 0. It follows that M 0 = sup S. Remark 152 This theorem says that one can get as close as one wants to the supremum of a set and still be in the set. This is obvious if the supremum is in the set. The theorem says it is also true if the supremum is outside of the set. Another way of understanding this is that the theorem implies that there is nothing between a set and its supremum because nothing can fit there. In other words, if the supremum of a set is not in the set, then it is the closest it can be to the set. Nothing else can fit in between. Remark 153 If we represent the set of real numbers by the real line and consider that the subset S in the theorem is a portion of the real line, then the theorem says that no element of S can be to the right of M 0 however, there is at least one element of S to the right of every element to the left of M 0. Example 154 Consider the set S = {0, 12, 23, 34 },... = {1 1n } : n N. Clearly inf A = 0 (see the next proposition). Intuitively, we think that sup A = 1 because 1 is an upper bound. If γ is any real number less than 1 (γ < 1) then one can find a natural number n 0 such that γ < 1 1 n 0 1. It would be a natural number satisfying n 0 > 1 1 γ. Thus 1 1 n 0 S. By theorem 151 this means that 1 = sup S. Example 155 Consider the set S = { q Q : q > 0 and q 2 < 2 }. Prove that if sup S exists then it cannot be a rational number. We do a proof by contradiction. Let α = sup S and assume that α Q.. Then, we know that α 2 2. It follows that either α 2 < 2 or α 2 > 2. If α 2 > 2, then 2 < α. By theorem 151, there exists s S such that 2 < s α. But then,we would have s 2 > 2 so that s / S which is a contradiction. So this case cannot occur. The only possibility left is that α 2 < 2. Since S has no largest element, there exists q S such that α < q thus α is not an upper bound of S hence cannot be its supremum. Since all the possible cases cannot happen, our assumption that α Q cannot be true. We will see later that S has indeed a supremum, it is 2. Proposition 156 Let S be a subset of R. 1. If S has a smallest element, then min S = inf S.

7 2.4. THE COMPLETENESS PROPERTY OF R If S has a largest element, then max S = sup S. Proof. We prove part 1 and leave part 2 as an exercise. Let m = min S. By definition, m s for any s S. Thus m is also a lower bound of S. If γ is another lower bound of S, then γ m since m S. Thus m is the greatest lower bound of S or m = inf S. Proposition 157 If the supremum exists, it is unique. A similar result holds for the infimum. Proof. See exercises. In the last example we did, we saw that the set S, which is a subset of Q, could not have a supremum in Q. This brings the questions "when do we know if a set has a supremum, and in which set is the supremum?". There is a similar question for infimum. We answer these questions in the next subsection. We will see that the answer is at the heart of the difference between R and Q The Axiom of Completeness We are now ready to state the axiom of completeness. This axiom is also known as the supremum or least upper bound property. There are several forms of this axiom; they are obviously equivalent. We state two versions of this axiom, one for supremum, one for infimum. Axiom 158 (Supremum Property) Every non-empty subset of R that is bounded above has a supremum in R. The second version is stated as a theorem because it can be proven using the first one. Theorem 159 (Infimum property) Every non-empty subset of R that is bounded below has an infimum in R. Proof. We do a direct proof. We will prove the infimum exists by finding it. Let S be a non-empty subset of R which is bounded below. Define L to be the set of lower bounds of S. Since S is bounded below, L. Furthermore, L is bounded above by elements of S. By the supremum property, L has a supremum. Call it α that is α = sup L. We will show that α = inf S. To prove that α = inf S, we first prove that α is a lower bound of S. We then prove that no lower bound greater than α can exist, making α the greatest lower bound of S. First, we prove that α is a lower bound of S. For this, we need to show that every element of S is larger than α. Let s S. Then s is an upper bound of L. Since α = sup L, that is α is the least upper bound of L, it follows that α s. We have proven that if s is an arbitrary element of S, then we had s α. It follows that α is a lower bound of S.

8 54 CHAPTER 2. THE STRUCTURE OF R Next, we show that α is the greatest of the lower bounds of S. This is straightforward. If γ is another lower bound of S, then γ is an element of L and therefore γ α since α is the least upper bound of L hence an upper bound of L. Therefore α is the greatest lower bound (or the infimum) of S. Remark 160 In the first part of the proof, where we proved that α is a lower bound of S, it would have been wrong to say α is a lower bound because α = sup L and L is the set of lower bounds of L. It is wrong because the supremum or the infimum of a set do not necessarily below to the set. Thus α is not necessarily a lower bound of S. It turns out that it is. But we know this after the proof we gave. Remark 161 The axiom and the theorem say that R is complete. We ll give a full definition of completeness in the next section. Remark 162 Recall that one diff erence between supremum and largest element of a set is that the latter is in the set while the former need not be. If we replace the word supremum by maximum or largest element in axiom 158, the result no longer holds. Consider (0, 5). This is clearly a non empty subset of R which is bounded above. It does indeed have a supremum, but no maximum. Definition 163 If S is a non-empty subset of R, we set: 1. sup S = if S is not bounded above. 2. inf S = if S is not bounded below. Remark 164 The case of is, once again, an interesting one. We have already established that every real number was both an upper bound and a lower bound of. Thus, from the definition above, it follows that sup = inf = We illustrate with an example how to work with suprema and infima. Example 165 Let S be a non-empty bounded subset of R. If a > 0, show that sup (as) = a sup S where as = {as : s S}. Let α = sup S. We need to show that sup (as) = aα. For this, we show that aα is an upper bound of S and that it is the smallest of the upper bounds of S. 1. aα is an upper bound of S. We need to show that aα as for any s S. Let s S. Since α = sup S, it follows that α is an upper bound of S. Thus, we have α s Since a > 0, it follows that thus aα is an upper bound of as. aα as

9 2.4. THE COMPLETENESS PROPERTY OF R aα is the least upper bound. We show that if γ is any other upper bound of as then γ aα. Clearly if γ = the result is true. Suppose that γ is finite. Then, as γ for any s S. Thus s γ a for any s S. This makes γ a follows that an upper bound of S. Since α = sup S, it α γ a and therefore aα γ Intervals We can now define precisely what an interval is. Definition 166 (interval) A subset S of R is an interval if whenever x, y are in S with x < y then every real number t satisfying x < t < y is also in S. The next theorem records the familiar possible forms of an interval. Theorem 167 An interval has one of the following nine forms: 1. (a, b) = {x R : a < x < b}, a bounded, open interval. 2. [a, b) = {x R : a x < b}, a bounded, half-open interval. 3. (a, b] = {x R : a < x b}, a bounded, half-open interval. 4. [a, b] = {x R : a < x < b}, a bounded, closed interval. 5. (, b) = {x R : x < b}, an unbounded, open interval. 6. (, b] = {x R : x b}, an unbounded, closed interval. 7. (a, ) = {x R : a < x}, an unbounded, open interval. 8. [a, ) = {x R : a x}, an unbounded, closed interval. 9. (, ) = R, an unbounded, both open and closed interval. Remark 168 If a < b, when we write [a, b] we mean an interval of real numbers. If we want an interval on rational numbers, we either use set notation that is {x Q : a x b}. We can also use [a, b] Q. Similarly for integers or natural numbers, we use [a, b] Z or [a, b] N.

10 56 CHAPTER 2. THE STRUCTURE OF R Exercises 1. In lemma 142, prove that if we add the condition that S has more than one element, then α < β. 2. Show that if S is bounded above and below, then there exists a number N > 0 for which N x N if x S. 3. Show that 1 (a + b + a b ) = max {a, b} 2 4. Suppose that A is a non-empty bounded set of real numbers that has no largest member and that a A. Explain why the sets A and A \ {a} have exactly the same upper bounds. 5. Give an example of a set A that has a largest member a such that the sets A and A \ {a} have exactly the same upper bounds. 6. Give an example of a set A that has a largest member a such that the sets A and A \ {a} do not have exactly the same upper bounds. 7. Answer each part below. (a) Given that S is a non empty subset of a given interval [a, b], explain why, for every member x of the set S, we have x a + b a. (b) Given that a set S of numbers is bounded and that T = { x : x S}, prove that the set T must also be bounded. 8. Is it possible for a set of numbers to have a supremum even though it has no largest member? 9. Show that if a subset S has a maximum, then the maximum is also the supremum. Similarly, show that if S has a minimum, then the minimum is also the infimum. 10. Show that if the supremum of a subset S exists, then it is unique. Prove the same result for the infimum. 11. Let S denote the set in brackets in each case below. Find sup S and inf S. (a) { x 2 3x < 4 }. (b) {3x + 5 < 4x 7}. 12. Given that A is a set of real numbers and that sup A A, explain why sup A = max A.

11 2.4. THE COMPLETENESS PROPERTY OF R Given that A is a set of real numbers and that inf A A, explain why inf A = min A. 14. Given that α is an upper bound of a set A and that α A, explain why α = sup A. 15. Explain why the empty set does not have a supremum. 16. Explain why the set [1, ) does not have a supremum. 17. Given that α = sup A and that x < α, what conclusions can you draw about the number x? 18. Given that α = inf A and that x > α, what conclusions can you draw about the number x? 19. State and prove the infimum version of theorem If A and B are sets of real numbers, then the sets A + B and A B are defined by A + B = {a + b : a A and b B} and A B = {a b : a A and b B} A = { a : a A} A.B = {ab : a A and b B} (a) Find A + B and A B in each case below. i. A = [0, 1] and B = [ 1, 0]. ii. A = [0, 1] and B = {1, 2, 3}. iii. A = (0, 1) and B = {1, 2, 3}. (b) Prove that if two sets A and B are bounded, then so are A + B and A B. (c) Prove that sup ( B) = inf (B) and inf ( B) = sup (B). (d) Prove that if A and B are non-empty and bounded above then sup (A + B) = sup A + sup B and sup (A B) = sup A inf B. (e) Show by example that in general sup (A.B) (sup A) (sup B). 21. Given that two sets A and B are bounded above and below. Answer the following questions: (a) Explain why their union A B is bounded above and below. (b) Prove that sup (A B) = max {sup A, sup B}. (c) Prove that inf (A B) = min {inf A, inf B}. (d) Prove that sup (A B) min {sup A, sup B}.

12 58 CHAPTER 2. THE STRUCTURE OF R 22. Suppose that A is a non-empty bounded set of real numbers that has no largest member and that a A. Prove that sup A = sup (A \ {a}). 23. Given that A and B are sets of numbers, that A is non-empty, that B is bounded above, and that A B, explain why sup A and sup B exist and why sup A sup B. 24. Given that A and B are non-empty subsets of R with A B and B bounded, show that inf B inf A sup A sup B. 25. Given that A is a non-empty bounded set of numbers, explain why inf A sup A. 26. Let S be a subset of R and let a R. Define a + S = {a + s : s S}. Assume that S is non-empty and bounded. Show that sup (a + S) = a + sup S and inf (a + S) = a + inf S. 27. Show that every non-empty finite subset of R contains both a maximum and a minimum element (hint: use induction). 28. Does (a, b) Z have a largest element, a smallest element? If yes, what are they and why? 29. Let S be a non-empty bounded subset of R. Let α = sup S and β = inf S. Let ɛ > 0 be given. (a) Explain why α and β exist. (b) Prove that there exists s 0 S such that α ɛ < s 0. (c) Prove that there exists s 1 S such that s 1 < β + ɛ.

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