The Real Numbers. Chapter The Completeness Property of R

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1 Chapter 1 The Real Numbers 1.1 The Completeness Property of R Example (Bartle (a) Page 39). A nonempty set S 1 with a finite number of elements will have a least element, say u, and a largest element, say w. Then, u = infs 1 and w = sups 1. Both the infinimum and supremum, that is u and v, are members of S 1. Example (Bartle (b) Page 39). Given S 2 = {x : 0 x 1}. Since 0 x 1 for all x S 2, we have a lower bound 0, and an upper bound 1. If v < 1 then there exists some s S 2 such that v < s, for example s = 1. Thus, v is not an upper bound of S 2, and since v < 1 is arbitrary, we conclude that sups 2 = 1. If w > 0 then there exists some s S 2 such that s < w, for example s = w/2 < w. Thus, w is not a lower bound of S 2, and since w > 0 is arbitrary, we conclude that infs 2 = 0. Note that the infimum and supremum are members of S 2. Example (Bartle (c) Page 39). Given S 3 = {x : 0 < x < 1}. Since 0 < x < 1 for all x S 3, we have a lower bound 0, and an upper bound 1. If v < 1 there exists some s S 3 such that v < s. Thus, v is not an upper bound of S 3, and since v < 1 is arbitrary, we conclude that sups 3 = 1. If w > 0 then there exists some s S 3 such that s < w, for example s = w/2 < w. Thus, w is not a lower bound of S 3, and since w > 0 is arbitrary, we conclude that infs 3 = 0. Note that the infimum and supremum are not members of S 3. Problem 1 (Bartle Exercise 1 for Section 2.3). Let S 1 := {x R : x 0}. Show in detail that the set has lower bounds, but no upper bounds. Show that infs 1 = 0. Since x 0 for all x S 1, u = 0 is a lower bound of S 1, while S 1 has no upper bounds. If v > 0 then there exists some s S 1 such that 0 < s < v, for example s = v/2 < v. Thus, v is not a lower bound of S 1, and since v is arbitrary infs 1 = 0. Also, infs 1 S 1. Since S 1 has no upper bounds sups 1 does not exist. Problem 2 (Bartle Exercise 2 for Section 2.3). Let S 2 := {x R : x > 0}. Does S 2 have lower bounds? Does S 2 have upper bounds? Does infs 2 exist? Does sups 2 exist? Prove your statements. Since S 2 and x > 0, the set S 2 is bounded below by 0. Then, from the Completeness Property S 2 also has an infimum. For every ε > 0, there exists some s ε S 2 such that 0 < s ε < 0 + ε; for example 1

2 s ε = ε/2 < ε. Thus, 0+ε is not a lower bound, and since ε is arbitrary, infs 2 = 0. Since x > 0, the set S 2 has no upper bounds, and by definition sups 2 does not exist. Problem 3 (Bartle Exercise 3 for Section 2.3). Let S 3 := {1/n : n N}. Show that sups 3 = 1 and infs 3 0. Since 0 < 1/n 1 for all n N, the set S 3 has a lower bound 0 and an upper bound 1, and by the Completeness Property has both an infimum and a supremum. For every ε > 0 there exists some s ε S 3 such that 1 ε < s ε 1; for example s ε = 1. Thus, 1 ε is not an upper bound of S 3, and since ε is arbitrary, sups 3 = 1. Also, for every ε > 0 there exists some s ε S 3 such that 0 s ε 0+ε, say s ε = 0 or s ε = ε/2. Thus, 0+ε is not a lower bound of S 3, and since ε > 0 is arbitrary, infs 3 = 0 1. Problem 4 (Bartle Exercise 4 for Section 2.3). Let S 4 := {1 ( 1) n /n : nn}. Find infs 4 and sups 4. Since 1/2 1 ( 1) n /n 2, the set S 4 has a lower bound 1/2 and an upper bound 2, and by the Completeness Property has an infimum and supremum. For every ε > 0, there exists some s ε S 4 such that 1/2 s ε < 1/2+ε; for example s ε = 1/2+ε/2. Thus, 1/2+ε is not a lower bound, and since ε > 0 is arbitrary infs 4 = 1/2. For every ε > 0 there exists s ε S 4 such that 2 ε < s ε 2, say s ε = 2. Thus, 2 ε is not an upper bound of S 4, and since ε > 0 is arbitrary, sups 4 = 2. Problem 5 (Bartle Exercise 5(a) for Section 2.3). Let A := {x R : 2x+5 > 0}. Find the infimum and supremum, if they exist. Since 2x+5 > 0 we have x > 5/2, and so A has a lower bound 5/2, but no upper bounds. Thus, by the Completeness Property, A also has an infimum, but the supremum does not exist. For every ε > 0 there exists a ε A such that 5/2 a ε < 5/2+ε, say a ε = 5/2+ε/2. Thus, 5/2+ε is not a lower bound of A, and since ε is arbitrary, infa= 5/2. Problem 6 (Bartle Exercise 5(b) for Section 2.3). Todo... Problem 7 (Bartle Exercise 5(c) for Section 2.3). Todo... Problem 8 (Bartle Exercise 5(d) for Section 2.3). Todo... Problem 9 (Bartle Exercise 6 for Section 2.3). Let S be a nonempty subset of R that is bounded below. Prove that infs = sup{ s : s S}. Let S be a nonempty subset of R that is bounded below. Then, by the Completeness Property of R an infimum of S exists, say w := infs. Thus w s for each s S. But this implies that s w for all s S, so that if we define S := { s : s S}, then w is an upper bound for S. We must show that u is the supremum of S. Let v < w for all v S. Then w < v, and since w = infs there must exist some t S such that w < t < v. But this implies that there exists some t S such that v < t < w. Thus v is not an upper bound of S, and since v is arbitrary, w = sup S. Hence, infs = sup{ s : s S}. Problem 10 (Bartle Exercise 7 for Section 2.3). If a set S R contains one of its upper bounds, show that this upper bound is the supremum of S. Let u S be an upper bound of S. If v is any upper bound of S, then s v for all s S. But u S, so u v, and since u is an upper bound then it must be the least upper bound; that is, u := sups. Problem 11 (Bartle Exercise 8 for Section 2.3). Let S R be nonempty. Show that u R is an upper bound of S if and only if the conditions t R and t > u imply that t / S. 1 The question asks to show that inf S 3 0. The infimum infs 3 = 0 is for the case of n. If n = 2, then infs 3 = 1/2. Perhaps ask lecturer to just elaborate a little more on this problem at consultation. 2

3 Let S R be nonempty. The proof proceeds as follows: Assume u R is an upper bound of S. Then s u for all s S. Thus, if t > u for all t R, then t > s for all s S. Hence t / S. Conversely, assume that t > u t / S, and suppose, to the contrary, that u is not an upper bound of S. Then, since S is nonempty there must exist some s S such that s > u. But s > u s / S, which is a contradiction. Hence, u is an upper bound of S. Problem 12 (Bartle Exercise 9 for Section 2.3). Let S R be nonempty. Show that if u = sups, then for every n N the number u 1/n is not an upper bound of S, but the number u+1/n is an upper bound of S 2. Let S R be a nonempty set, and let u := sups. For all n N we have 1/n > 0, so that u 1/n < u for all n N. Thus, there must exist some s S such that u 1/n < s < u and so u 1/n is not an upper bound of S. However, u < u+1/n for all n N, and since u := sups, we have that s u < u+1/n for all s S. Hence, u+1/n is an upper bound of S for all n N. Problem 13 (Bartle Exercise 10 for Section 2.3). Show that if A and B are bounded subsets of R, then A B is a bounded set. Show that sup(a B) = sup{supa,supb}. Let A R and B R. Let u := supa and v := supb. Also, let w := sup{u,v}. For all a A and all b B we have a u w and b v w respectively. Thus w is an upper bound for A B. We need to show that w is the supremum (least upper bound) of A B. If z is any upper bound of A B, then z is also an upper bound of A and B; that is, u z and v z. Hence w z, and since z is any upper bound of A B, w must be the least upper bound, and we conclude that sup(a B) = sup{supa,supb}. Problem 14 (Bartle Exercise 11 for Section 2.3). Let S be a bounded set in R and let S 0 be a nonempty subset of S. Show that infs infs 0 sups. Since the set S is bounded, the Completeness Property of R ensures that S has an infimum and a supremum. If s 0 S 0, then s 0 S, so that s 0 sups for all s 0 S 0. This implies that sups 0 sups. Similarly, infs s 0 for all s 0 S 0, so that infs infs 0. Hence, infs infs 0 sups 0 sups. Problem 15 (Bartle Exercise 12 for Section 2.3). Let S R and suppose that s := sups belongs to S. If u / S, show that sup(s {u}) = sup{s,u}. Since u / S and s S we have that u s. Thus we consider only the two cases that u < s and s < u. Now, if u < s then s = sup{s,u} since s is an upper bound which is contained in the set {s,u}. Also, for all s S we have that s u < s so that s is an upper bound of the set S {u}, and since s is contained in this set, s = sup(s {u}). Similarly, if s < u, then u = sup{s,u} and consequently u = sup(s {u}). Problem 16 (Bartle Exercise 13 for Section 2.3). Todo... Problem 17 (Bartle Exercise 14 for Section 2.3). Todo Applications of the Supremum Property Example (Bartle (a) Page 40). Let S be a nonempty subset of R that is bounded above, and let a be any number in R. Define the set a+s := {a+s : s S}. Show that sup(a+s) = a+sups. Solution: Let u := sups. Then for all s S we have a+s a+u, and so a+u is an upper bound of 2 The converse is also true. See Exercise in Bartle 3

4 the set a+s. Consequently, sup(a+s) a+u. Now, if v is any upper bound of a+s, then a+s v for all s S. Thus s v a for all s S, and so v a is an upper bound for S. But since u := sups, we have that u v a, and so a+u v. Since v is any upper bound of a+s we have a+u sup(a+s). Now we have that sup(a+s) a+u and a+u sup(a+s), which means that sup(a+s) = a+sups. Example (Bartle (b) Page 41). Suppose A and B are nonempty subsets of R that satisfy the property a b for all a A,b B Prove that supa supb Solution: For any given b B we have that a b for all a A. Thus b is an upper bound of A, and so supa b, where supa is the least upper bound of A. Since supa b holds for all b B, we have that supa is a lower bound of the set B. Hence, we conclude that supa supb. Example (Bartle Page 41). Suppose that f and g are real-valued functions with common domain D R. We assume that f and g are bounded. 1. Show that if f(x) g(x) for all x D, then sup(d) supg(d). Solution: Firstly, f(x) g(x) supg(d). This implies that supg(d) is an upper bound for f(d). Hence, supf(d) supg(d). Example (Bartle (b) Page 41). Todo... Example (Bartle (c) Page 41). Todo... Problem 18 (Bartle Exercise 1 for Section 2.4). Show that sup{1 1/n : n N} = 1. Let S = {1 1/n : n N}. Since 1 1/n < 1 for all n N we have an upper bound of 1 for S. Thus the Completeness Property ensures that a supremum of S exists. For every ε > 0, the Archimedean Property implies that there exists some n ε N such that 1/ε < n ε. Thus 1/n ε < ε so that 1 ε < 1 1/n ε < 1. Since 1 1/n ε S we know that 1 ε is not an upper bound of S, and since ε > 0 is arbitrary, sups = 1. Problem 19 (Bartle Exercise 2 for Section 2.4). If S := {1/n 1/m : n,m N}, find infs and sups. Since 1 1/n 1/m 1, the set S has a lower bound 1 and an upper bound 1. By the Completenes Property the set S has an infimum and a supremum. Infimum: We have that 1/n 1/m 1/n 1 > 1. Then 1 is indeed a lower bound. For every ε > 0 there exists an n N such that 0 < 1/n < ε. Thus 1 < 1/n 1 < ε 1 = 1+ε. But 1/n 1 S. Hence, 1+ε is not a lower bound of S, and since ε > 0 is arbitrary, infs = 1. Supremum: The set S has the property S = S. Thus, we can invoke the earlier result infs = sup{ s : s S}. Since infs = 1 we have 1 = sup{ s : s S} = sup( S) = sups. This implies that sups = 1. Problem 20 (Bartle Exercise 3 for Section 2.4). Let S R be nonempty. Prove that if a number u in R has the properties: (i) for every n N the number u 1/n is not an upper bound of S, and (ii) for every number n N the number u+1/n is an upper bound of S, then u = sups. 3 3 This is the converse of Exercise in Bartle 4

5 Todo... Problem 21 (Bartle Exercise 4 (a) for Section 2.4). Let S be a nonempty bounded set in R. Let a > 0, and let as := {as : s S}. Prove that inf(as) = ainfs, sup(as) = asups Supremum: Let u := sups. Then s u for all s S, so that as au; that is, au is an upper bound for the set as. Consequently, sup(as) au. If v is any upper bound of as, then as v for all s S, and so s v/a for all s S; that is, v/a is an upper bound for S. Thus, u v/a, so that au v, and since v is any upper bound of as, we have au sup(as). Combining the inequalities sup(as) au and au sup(as), we have sup(as) = asups Infimum: Todo... Problem 22 (Bartle Exercise 4 (b) for Section 2.4). Let S be a nonempty bounded set in R. Let b < 0, and let bs := {bs : s S}. Prove that inf(bs) = bsups, sup(bs) = binfs Since b < 0, we consider b > 0 and apply the results of the previous exercise (Exercise (a) Bartle). Define S := { s : s S}. Then inf(bs) = inf( b S). But inf( b S) = binf S, so that inf(bs) = binf S. Also from an earlier result (Exercise Bartle) inf S = sups. Thus, inf(bs) = b( sups) and we have our final result inf(bs) = bsups. Todo sup(bs) = binfs... Problem 23 (Bartle Exercise 5 for Section 2.4). Let S be a set of nonnegative real numbers that is bounded above and let T := {x 2 : x S}. Prove that if u = sups, then u 2 = supt. Give an example that shows that the conclusion may be false if the restriction against negative numbers is removed. Todo... Problem 24 (Bartle Exercise 6 for Section 2.4). Let X be a nonempty set and let f : X R have bounded range in R. If a R, show that Example 2.4.1(a) (in Bartle) implies that Show that we also have sup{a+f(x) : x X} = a+sup{f(x) : x X} inf{a+f(x) : x X} = a+inf{f(x) : x X} Supremum: Let u := supx. Then f(x) u for all x X, so that a+f(x) a+u for all x X; that is, a+u is an upper bound of a+f(x) for all x X. Thus sup{a+f(x) : x X} a+u. Todo: Finish... Problem 25 (Bartle Exercise 7 for Section 2.4). Let A and B be bounded nonempty subsets of R, and let A+B := {a+b : a A,b B}. Prove that sup(a+b) = supa+supb,inf(a+b) = infa+infb Supremum: Let u := supa and v := supb. The a u for all a A and b v for all b B. Thus, a + b u + v for all a A,b B so that u + v is an upper bound for the set A + B. Hence, from the Completeness Property, a supremum for A+B exists, and sup(a+b) u+v = supa+supb. It needs to be shown that u+v is the supremum (least upper bound) of A+B. We can do so by showing that for every ε > 0 we can find a s ε A+B such that (u+v) ε < s ε u+v. If we choose a ε A such that u ε/2 < a ε u and b ε B such that v ε/2 < b ε v then we have (u ε/2)+(v ε/2) < a ε +b ε u+v, or simplified (u+v) ε < a ε +b ε u+v 5

6 But a ε +b ε A+B, and since ε > 0 is arbitrary we know that (u+v) ε is not an upper bound of A+B, so that u+v is the supremum and sup(a+b) = supa+supb Problem 26 (Bartle Exercise 8 for Section 2.4). Let X be a nonempty set, and let f and g be defined on X and have bounded ranges in R. Show that and that sup{f(x)+g(x) : x X} sup{f(x) : x X}+sup{g(X) : x X} inf{f(x) : x X}+inf{g(X) : x X} inf{f(x)+g(x) : x X} Give examples to show that each of these inequalities can be either equalities or strict inequalitles. Supremum: Let u := supf(x) and v := supg(x). Then f(x) u and g(x) v for all x X. Thus f(x)+g(x) u+v for all x X, so that u+v is an upper bound of the set {f(x)+g(x) : x X}. This implies that {f(x)+g(x) : g X} sup{f(x)+g(x) : x X} u+v. Hence, sup{f(x)+g(x) : x X} sup{f(x) : x X}+sup{g(X) : x X} Infimum: Let u := inff(x) and v := infg(x). Then u f(x) and v g(x) for all x X, and consequently u +v f(x) +g(x) for all x X. Then u+v is a lower bound of the set {f(x) + g(x) : x X}, which implies that u+v inf{f(x)+g(x) : x X} {f(x)+g(x) : g X}. Hence, inf{f(x) : x X}+inf{g(X) : x X}inf{f(x)+g(x) : x X} Problem 27 (Bartle Exercise 9 for Section 2.4). Todo... Problem 28 (Bartle Exercise 10 for Section 2.4). Todo... Problem 29 (Bartle Exercise 11 for Section 2.4). Todo... Problem 30 (Bartle Exercise 12 for Section 2.4). Todo... Problem 31 (Bartle Exercise 13 for Section 2.4). Given any x R, show that there exists a unique Z such that n 1 x < n. In the trivial case where x = 0 then n = 1 is the unique element of Z. If x > 0, then by an earlier result (Corollary in Bartle), there exists an n x N such that n x 1 < x < n x. If x < 0, then we can apply the same argument to x > 0. Thus, we know that given an arbitrary x R, there exists some n x Z such that n x n < n x. We must show that such an n x Z is unique. Suppose, without loss of generality, that m x < n x also satisfies m x 1 x < m x. Also, we know that m x n x 1 since n x,m x Z. We now have the inequality m x 1 x < m x n x 1 x < n x, which is a contradiction as it implies x < x. Hence, the assumption that there also exists m x < n x satisfying m x 1 x < m x is false, and so we have uniqueness. Problem 32 (Bartle Exercise 14 for Section 2.4). If y > 0, show that there exists n N such that 1/2 n < y. For each n N we have that n < 2 n. Thus we also have that 1/2 n < 1/n. But from an earlier result (Corollary Bartle), we know that given y > 0 there exists some n y N such that 0 < 1/n y < y. Hence, for each y > 0 we have that 0 < 1/2 n < 1/n < y; that is, 1/2 n < y. Problem 33 (Bartle Exercise 15 for Section 2.4). Todo... Problem 34 (Bartle Exercise 16 for Section 2.4). Todo... Problem 35 (Bartle Exercise 17 for Section 2.4). Todo... Problem 36 (Bartle Exercise 18 for Section 2.4). Todo... Problem 37 (Bartle Exercise 19 for Section 2.4). Since x < y and u > 0, we know that x/u < y/u. Then, by the Density Theorem, there exists some r Q such that x/u < r < y/u. But this implies that x < ru < y. Hence the set {ru : r Q} is dense in R. 6

7 1.3 Intervals Problem 38 (Bartle Exercise 1 for Section 2.5). If I := [a,b] and I := [a,b ] are closed intervals in R, show that I I if and only if a a and b b. Let I I. Then a,b I so that a a,b b, or a a and b b. Conversely, suppose a lea and b b. Then for any x I we have a x b, so that a x b. Thus x I for all x I, and hence, I I. Problem 39 (Bartle Exercise 2 for Section 2.5). If S R is nonempty, show that S is bounded if and only if there exists a closed bounded interval I such that S I. Let S be bounded by a lower bound a and an upper bound b. Then for any s S we have a s b. Hence s [a,b], and so S I where I := [a,b]. Conversely, let S I where I is the closed bound interval I := [a,b]. But then a s b for all s S. Hence, S is bounded. Problem 40 (BartleExercise3forSection2.5). If S R is a nonempty bounded set, and I S := [infs,sups], show that S I s. Moreover, if J is any closed bounded interval containing S, show that I S J. The set S is nonempty and bounded, so the Completeness Property ensures that the infimum and supremum of S exist. Since infs s sups for all s S, we know that s I S = [infs,sups] for all s S. Hence, S I S. Since J is any closed bounded interval, let J := [a,b]. If S J then a s b for all s S. Thus a is a lower bound of S and b is an upper bound of S, and by the Completeness Property, the infimum and supremum of S exist. Thus a infs and sups b. But I S = [infs,sups], so by Exercise Bartle, I S J. Problem 41 (Bartle Exercise 4 for Section 2.5). Leave. Problem 42 (Bartle Exercise 5 for Section 2.5). Leave. Problem 43 (Bartle Exercise 6 for Section 2.5). Todo... Problem 44 (Bartle Exercise 7 for Section 2.5). Let I n := [0,1/n] for n N. Prove that n=1i n = {0}. Since 0 [0,1/n] for all n N, we know that {0} n=1 I n. Thus we know that n=1 I n is nonempty and contains {0}. Let x n=1 (x is arbitrary). But this is only true if x [0,1/n], so that 0 x 1/n for all n N. We must show that x = 0. From the Archimedean Property implies that for every ε > 0 there exists some n ε N such that 0 < 1/n ε < ε. Thus we now have that 0 x 1/n ε < ε. But since ε > 0 is arbitrary, we can make it as small as we like, and hence, x = 0. Problem 45 (Bartle Exercise 8 for Section 2.5). Todo... Problem 46 (Bartle Exercise 9 for Section 2.5). Todo... Problem 47 (Bartle Exercise 10 for Section 2.5). Todo... Problem 48 (Bartle Exercise 11 for Section 2.5). Todo... Problem 49 (Bartle Exercise 12 for Section 2.5). Todo... Problem 50 (Bartle Exercise 13 for Section 2.5). Todo... Problem 51 (Bartle Exercise 14 for Section 2.5). Todo... Problem 52 (Bartle Exercise 15 for Section 2.5). Todo... Problem 53 (Bartle Exercise 16 for Section 2.5). Todo... Problem 54 (Bartle Exercise 17 for Section 2.5). Todo... 7