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CONTENTS
- Chapter 1 Preliminaries
- Chapter 2 The Real Numbers
- Chapter 3 Sequences
- Chapter 4 Limits.
- Chapter 5 Continuous Functions
- Chapter 6 Differentiation.
- Chapter 7 The Riemann Integral
- Chapter 8 Sequences of Functions
- Chapter 9 Infinite Series
- Chapter 10 The Generalized Riemann Integral
- Chapter 11 A Glimpse into Topology
- Selected Graphs
CHAPTER 1
PRELIMINARIES
We suggest that this chapter be treated as review and covered quickly, without detailed classroom discussion. For one reason, many of these ideas will be already familiar to the students — at least informally. Further, we believe that, in practice, those notions of importance are best learned in the arena of real analysis, where their use and significance are more apparent. Dwelling on the formal aspect of sets and functions does not contribute very greatly to the students’ understanding of real analysis. If the students have already studied abstract algebra, number theory or com- binatorics, they should be familiar with the use of mathematical induction. If not, then some time should be spent on mathematical induction. The third section deals with finite, infinite and countable sets. These notions are important and should be briefly introduced. However, we believe that it is not necessary to go into the proofs of these results at this time.
Section 1.
Students are usually familiar with the notations and operations of set algebra, so that a brief review is quite adequate. One item that should be mentioned is that two sets A and B are often proved to be equal by showing that: (i) if x ∈ A, then x ∈ B, and (ii) if x ∈ B, then x ∈ A. This type of element-wise argument is very common in real analysis, since manipulations with set identities is often not suitable when the sets are complicated. Students are often not familiar with the notions of functions that are injective (= one-one) or surjective (= onto).
Sample Assignment: Exercises 1, 3, 9, 14, 15, 20.
Partial Solutions:
- (a) B ∩ C = { 5 , 11 , 17 , 23 ,.. .} = { 6 k − 1 : k ∈ N}, A ∩ (B ∩ C) = { 5 , 11 , 17 } (b) (A ∩ B) \ C = { 2 , 8 , 14 , 20 } (c) (A ∩ C) \ B = { 3 , 7 , 9 , 13 , 15 , 19 }
- The sets are equal to (a) A, (b) A ∩ B, (c) the empty set.
- If A ⊆ B, then x ∈ A implies x ∈ B, whence x ∈ A ∩ B, so that A ⊆ A ∩ B ⊆ A. Thus, if A ⊆ B, then A = A ∩ B. Conversely, if A = A ∩ B, then x ∈ A implies x ∈ A ∩ B, whence x ∈ B. Thus if A = A ∩ B, then A ⊆ B.
- If x is in A \ (B ∩ C), then x is in A but x /∈ B ∩ C, so that x ∈ A and x is either not in B or not in C. Therefore either x ∈ A \ B or x ∈ A \ C, which implies that x ∈ (A \ B) ∪ (A \ C). Thus A \ (B ∩ C) ⊆ (A \ B) ∪ (A \ C).
1
Chapter 1 — Preliminaries 3
that f −^1 (G) ∩ f −^1 (H) ⊆ f −^1 (G ∩ H). The opposite inclusion is shown in Example 1.1.8(b). The proof for unions is similar.
- If f (a) = f (b), then a/
a^2 + 1 = b/
b^2 + 1, from which it follows that a^2 = b^2. Since a and b must have the same sign, we get a = b, and hence f is injective. If − 1 < y < 1, then x := y/
1 − y^2 satisfies f (x) = y (why?), so that f takes R onto the set {y : − 1 < y < 1 }. If x > 0, then x =
x^2 <
x^2 + 1, so it follows that f (x) ∈ {y : 0 < y < 1 }.
- One bijection is the familiar linear function that maps a to 0 and b to 1, namely, f (x) := (x − a)/(b − a). Show that this function works.
- (a) Let f (x) = 2x, g(x) = 3x. (b) Let f (x) = x^2 , g(x) = x, h(x) = 1. (Many examples are possible.)
- (a) If x ∈ f −^1 (f (E)), then f (x) ∈ f (E), so that there exists x 1 ∈ E such that f (x 1 ) = f (x). If f is injective, then x 1 = x, whence x ∈ E. Therefore, f −^1 (f (E)) ⊆ E. Since E ⊆ f −^1 (f (E)) holds for any f , we have set equality when f is injective. See Example 1.1.8(a) for an example. (b) If y ∈ H and f is surjective, then there exists x ∈ A such that f (x) = y. Then x ∈ f −^1 (H) so that y ∈ f (f −^1 (H)). Therefore H ⊆ f (f −^1 (H)). Since f (f −^1 (H)) ⊆ H for any f , we have set equality when f is surjective. See Example 1.1.8(a) for an example.
- (a) Since y = f (x) if and only if x = f −^1 (y), it follows that f −^1 (f (x)) = x and f (f −^1 (y)) = y. (b) Since f is injective, then f −^1 is injective on R(f ). And since f is surjec- tive, then f −^1 is defined on R(f ) = B.
- If g(f (x 1 )) = g(f (x 2 )), then f (x 1 ) = f (x 2 ), so that x 1 = x 2 , which implies that g ◦ f is injective. If w ∈ C, there exists y ∈ B such that g(y) = w, and there exists x ∈ A such that f (x) = y. Then g(f (x)) = w, so that g ◦ f is surjective. Thus g ◦ f is a bijection.
- (a) If f (x 1 ) = f (x 2 ), then g(f (x 1 )) = g(f (x 2 )), which implies x 1 = x 2 , since g ◦ f is injective. Thus f is injective. (b) Given w ∈ C, since g ◦ f is surjective, there exists x ∈ A such that g(f (x)) = w. If y := f (x), then y ∈ B and g(y) = w. Thus g is surjective.
- We have x ∈ f −^1 (g−^1 (H)) ⇐⇒ f (x) ∈ g−^1 (H) ⇐⇒ g(f (x)) ∈ H ⇐⇒ x ∈ (g ◦ f )−^1 (H).
- If g(f (x)) = x for all x ∈ D(f ), then g ◦ f is injective, and Exercise 22(a) implies that f is injective on D(f ). If f (g(y)) = y for all y ∈ D(g), then Exercise 22(b) implies that f maps D(f ) onto D(g). Thus f is a bijection of D(f ) onto D(g), and g = f −^1.
Section 1.
The method of proof known as Mathematical Induction is used frequently in real analysis, but in many situations the details follow a routine patterns and are
4 Bartle and Sherbert
left to the reader by means of a phrase such as: “The proof is by Mathematical Induction”. Since may students have only a hazy idea of what is involved, it may be a good idea to spend some time explaining and illustrating what constitutes a proof by induction. Pains should be taken to emphasize that the induction hypothesis does not entail “assuming what is to be proved”. The inductive step concerns the validity of going from the assertion for k ∈ N to that for k + 1. The truth of falsity of the individual assertion is not an issue here.
Sample Assignment: Exercises 1, 2, 6, 11, 13, 14, 20. Partial Solutions:
- The assertion is true for n = 1 because 1/(1 · 2) = 1/(1 + 1). If it is true for n = k, then it follows for k + 1 because k/(k + 1) + 1/[(k + 1)(k + 2)] = (k + 1)/(k + 2).
- The statement is true for n = 1 because [ 12 · 1 · 2]^2 = 1 = 1^3. For the inductive step, use the fact that [ (^1) 2 k(k^ + 1)
] 2
[ 1
2 (k^ + 1)(k^ + 2)
] 2
- It is true for n = 1 since 3 = 4 − 1. If the equality holds for n = k, then add 8(k + 1) − 5 = 8k + 3 to both sides and show that (4k^2 − k) + (8k + 3) = 4(k + 1)^2 − (k + 1) to deduce equality for the case n = k + 1.
- It is true for n = 1 since 1 = (4 − 1)/3. If it is true for n = k, then add (2k + 1)^2 to both sides and use some algebra to show that
1 3 (4k
(^3) − k) + (2k + 1) (^2) = 1 3 [4k
(^3) + 12k (^2) + 11k + 3] = 1 3 [4(k^ + 1)
(^3) − (k + 1)],
which establishes the case n = k + 1.
- Equality holds for n = 1 since 1^2 = (−1)^2 (1 · 2)/2. The proof is completed by showing (−1)k+1[k(k + 1)]/2 + (−1)k+2(k + 1)^2 = (−1)k+2[(k + 1)(k + 2)]/2.
- If n = 1, then 1^3 + 5 · 1 = 6 is divisible by 6. If k^3 + 5k is divisible by 6, then (k + 1)^3 + 5(k + 1) = (k^3 + 5k) + 3k(k + 1) + 6 is also, because k(k + 1) is always even (why?) so that 3k(k + 1) is divisible by 6, and hence the sum is divisible by 6.
- If 5^2 k^ − 1 is divisible by 8, then it follows that 52(k+1)^ − 1 = (5^2 k^ − 1) + 24 · 52 k is also divisible by 8.
- 5k+1^ − 4(k + 1) − 1 = 5 · 5 k^ − 4 k − 5 = (5k^ − 4 k − 1) + 4(5k^ − 1). Now show that 5 k^ − 1 is always divisible by 4.
- If k^3 + (k + 1)^3 + (k + 2)^3 is divisible by 9, then (k + 1)^3 + (k+2)^3 + (k + 3)^3 = k^3 + (k + 1)^3 + (k + 2)^3 + 9(k^2 + 3k + 3) is also divisible by 9.
- The sum is equal to n/(2n + 1).
6 Bartle and Sherbert
- Part (b) Let f be a bijection of Nm onto A and let C = {f (k)} for some k ∈ Nm. Define g on Nm− 1 by g(i) := f (i) for i = 1,... , k − 1, and g(i) := f (i + 1) for i = k,... , m − 1. Then g is a bijection of Nm− 1 onto A\C. (Why?) Part (c) First note that the union of two finite sets is a finite set. Now note that if C/B were finite, then C = B ∪ (C \ B) would also be finite.
- (a) The element 1 can be mapped into any of the three elements of T , and 2 can then be mapped into any of the two remaining elements of T , after which the element 3 can be mapped into only one element of T. Hence there are 6 = 3 · 2 · 1 different injections of S into T. (b) Suppose a maps into 1. If b also maps into 1, then c must map into 2; if b maps into 2, then c can map into either 1 or 2. Thus there are 3 surjections that map a into 1, and there are 3 other surjections that map a into 2.
- f (n) := 2n + 13, n ∈ N.
- f (1) := 0, f (2n) := n, f (2n + 1) := −n for n ∈ N.
- The bijection of Example 1.3.7(a) is one example. Another is the shift defined by f (n) := n + 1 that maps N onto N \ { 1 }.
- If T 1 is denumerable, take T 2 = N. If f is a bijection of T 1 onto T 2 , and if g is a bijection of T 2 onto N, then (by Exercise 1.1.21) g ◦ f is a bijection of T 1 onto N, so that T 1 is denumerable.
- Let An := {n} for n ∈ N, so
An = N.
- If S ∩T = ∅ and f : N → S, g: N → T are bijections onto S and T , respectively, let h(n) := f ((n + 1)/2) if n is odd and h(n) := g(n/2) if n is even. It is readily seen that h is a bijection of N onto S ∪ T ; hence S ∪ T is denumerable. What if S ∩ T = ∅?
- (a) m + n − 1 = 9 and m = 6 imply n = 4. Then h(6, 4) = 12 · 8 · 9 + 6 = 42. (b) h(m, 3) = 12 (m + 1)(m + 2) + m = 19, so that m^2 + 5m − 36 = 0. Thus m = 4.
- (a) P({ 1 , 2 }) = {∅, { 1 }, { 2 }, { 1 , 2 }} has 2^2 = 4 elements. (b) P({ 1 , 2 , 3 }) has 2^3 = 8 elements. (c) P({ 1 , 2 , 3 , 4 }) has 2^4 = 16 elements.
- Let Sn+1 := {x 1 ,... , xn, xn+1} = Sn ∪ {xn+1} have n + 1 elements. Then a subset of Sn+1 either (i) contains xn+1, or (ii) does not contain xn+1. The induction hypothesis implies that there are 2n^ subsets of type (i), since each such subset is the union of {xn+1} and a subset of Sn. There are also 2n subsets of type (ii). Thus there is a total of 2n^ + 2n^ = 2 · 2 n^ = 2n^ + 1^ subsets of Sn+1.
- For each m ∈ N, the collection of all subsets of Nm is finite. (See Exercise 12.) Every finite subset of N is a subset of Nm for a sufficiently large m. Therefore Theorem 1.3.12 implies that F(N) =
m=1 P(Nm) is countable.
CHAPTER 2
THE REAL NUMBERS
Students will be familiar with much of the factual content of the first few sections, but the process of deducing these facts from a basic list of axioms will be new to most of them. The ability to construct proofs usually improves gradually during the course, and there are much more significant topics forthcoming. A few selected theorems should be proved in detail, since some experience in writing formal proofs is important to students at this stage. However, one should not spend too much time on this material. Sections 2.3 and 2.4 on the Completeness Property form the heart of this chapter. These sections should be covered thoroughly. Also the Nested Intervals Property in Section 2.5 should be treated carefully.
Section 2.
One goal of Section 2.1 is to acquaint students with the idea of deducing conse- quences from a list of basic axioms. Students who have not encountered this type of formal reasoning may be somewhat uncomfortable at first, since they often regard these results as “obvious”. Since there is much more to come, a sampling of results will suffice at this stage, making it clear that it is only a sampling. The classic proof of the irrationality of
2 should certainly be included in the discussion, and students should be asked to modify this argument for
3 , etc.
Sample Assignment: Exercises 1(a,b), 2(a,b), 3(a,b), 6, 13, 16(a,b), 20, 23.
Partial Solutions:
- (a) Apply appropriate algebraic properties to get b = 0 + b = (−a + a) + b = −a + (a + b) = −a + 0 = −a. (b) Apply (a) to (−a) + a = 0 with b = a to conclude that a = −(−a). (c) Apply (a) to the equation a + (−1)a = a(1 + (−1)) = a · 0 = 0 to conclude that (−1)a = −a. (d) Apply (c) with a = −1 to get (−1)(−1) = −(−1). Then apply (b) with a = 1 to get (−1)(−1) = 1.
- (a) −(a + b) = (−1)(a + b) = (−1)a + (−1)b = (−a) + (−b). (b) (−a) · (−b) = ((−1)a) · ((−1)b) = (−1)(−1)(ab) = ab. (c) Note that (−a)(−(1/a)) = a(1/a) = 1. (d) −(a/b) = (−1)(a(1/b)) = ((−1)a)(1/b) = (−a)/b.
- (a) Add −5 to both sides of 2x + 5 = 8 and use (A2),(A4),(A3) to get 2x = 3. Then multiply both sides by 1/2 to get x = 3/2. (b) Write x^2 − 2 x = x(x − 2) = 0 and apply Theorem 2.1.3(b). Alternatively, note that x = 0 satisfies the equation, and if x = 0, then multiplication by 1 /x gives x = 2.
7
Chapter 2 — The Real Numbers 9
- If a = 0, then 2.1.8(a) implies that a^2 > 0; since b^2 ≥ 0, it follows that a^2 + b^2 > 0.
- If 0 ≤ a < b, then 2.1.7(c) implies ab < b^2. If a = 0, then 0 = a^2 = ab < b^2. If a > 0, then a^2 < ab by 2.1.7(c). Thus a^2 ≤ ab < b^2. If a = 0, b = 1, then 0 = a^2 = ab < b = 1.
- (a) If 0 < a < b, then 2.1.7(c) implies that 0 < a^2 < ab < b^2. Then by Example 2.1.13(a), we infer that a =
a^2 <
ab <
b^2 = b. (b) If 0 < a < b, then ab > 0 so that 1 /ab > 0, and thus 1 /a − 1 /b = (1/ab)(b − a) > 0.
- (a) To solve (x − 4)(x + 1) > 0, look at two cases. Case 1: x − 4 > 0 and x + 1 > 0, which gives x > 4. Case 2: x − 4 < 0 and x + 1 < 0, which gives x < −1. Thus we have {x : x > 4 or x < − 1 }. (b) 1 < x^2 < 4 has the solution set {x : 1 < x < 2 or − 2 < x < − 1 }. (c) The inequality is 1/x − x = (1 − x)(1 + x)/x < 0. If x > 0, this is equiva- lent to (1 − x)(1 + x) < 0, which is satisfied if x > 1. If x < 0, then we solve (1 − x)(1 + x) > 0, and get − 1 < x < 0. Thus we get {x : − 1 < x < 0 or x > 1 } (d) the solution set is {x : x < 0 or x > 1 }.
- If a > 0, we can take ε 0 := a > 0 and obtain 0 < ε 0 ≤ a, a contradiction.
- If b < a and if ε 0 := (a − b)/2, then ε 0 > 0 and a = b + 2ε 0 > b + ε 0.
- The inequality is equivalent to 0 ≤ a^2 − 2 ab + b^2 = (a − b)^2.
- (a) If 0 < c < 1, then 2.1.7(c) implies that 0 < c^2 < c, whence 0 < c^2 < c < 1. (b) Since c > 0, then 2.1.7(c) implies that c < c^2 , whence 1 < c < c^2.
- (a) Let S := {n ∈ N : 0 < n < 1 }. If S is not empty, the Well-Ordering Property of N implies there is a least element m in S. However, 0 < m < 1 implies that 0 < m^2 < m, and since m^2 is also in S, this is a contradiction to the fact that m is the least element of S. (b) If n = 2p = 2q − 1 for some p, q in N, then 2(q − p) = 1, so that 0 < q − p < 1. This contradicts (a).
- (a) Let x := c − 1 > 0 and apply Bernoulli’s Inequality 2.1.13(c) to get cn^ = (1 + x)n^ ≥ 1 + nx ≥ 1 + x = c for all n ∈ N, and cn^ > 1 + x = c for n > 1. (b) Let b := 1/c and use part (a).
- If 0 < a < b and ak^ < bk, then 2.1.7(c) implies that ak^ + 1^ < abk^ < bk^ + 1^ so Induction applies. If am^ < bm^ for some m ∈ N, the hypothesis that 0 < b ≤ a leads to a contradiction.
- (a) If m > n, then k := m − n ∈ N, so Exercise 22(a) implies that ck^ ≥ c > 1. But since ck^ = cm^ −^ n, this implies that cm^ > cn. Conversely, the hypothesis that cm^ > cn^ and m ≤ n lead to a contradiction. (b) Let b := 1/c and use part (a).
10 Bartle and Sherbert
- Let b := c^1 /mn. We claim that b > 1; for if b ≤ 1, then Exercise 22(b) implies that 1 < c = bmn^ ≤ b ≤ 1, a contradiction. Therefore Exercise 24(a) implies that c^1 /n^ = bm^ > bn^ = c^1 /m^ if and only if m > n.
- Fix m ∈ N and use Mathematical Induction to prove that am^ +^ n^ = aman^ and (am)n^ = amn^ for all n ∈ N. Then, for a given n ∈ N, prove that the equalities are valid for all m ∈ N.
Section 2.
The notion of absolute value of a real number is defined in terms of the basic order properties of R. We have put it in a separate section to give it emphasis. Many students need extra work to become comfortable with manipulations involving absolute values, especially when inequalities are involved. We have also used this section to give students an early introduction to the notion of the ε-neighborhood of a point. As a preview of the role of ε-neighborhoods, we have recast Theorem 2.1.9 in terms of ε-neighborhhoods in Theorem 2.2.8.
Sample Assignment: Exercises 1, 4, 5, 6(a,b), 8(a,b), 9, 12(a,b), 15. Partial Solutions:
- (a) If a ≥ 0, then |a| = a =
a^2 ; if a < 0, then |a| = −a =
a^2. (b) It suffices to show that | 1 /b| = 1/|b| for b = 0 (why?). If b > 0, then 1 /b > 0 (why?), so that | 1 /b| = 1/b = 1/|b|. If b < 0, then 1/b < 0, so that | 1 /b| = −(1/b) = 1/(−b) = 1/|b|.
- First show that ab ≥ 0 if an only if |ab| = ab. Then show that (|a| + |b|)^2 = (a + b)^2 if and only if |ab| = ab.
- If x ≤ y ≤ z, then |x − y| + |y − z| = (y − x) + (z − y) = z − x = |z − x|. To establish the converse, show that y < x and y > z are impossible. For example, if y < x ≤ z, it follows from what we have shown and the given relationship that |x − y| = 0, so that y = x, a contradiction.
- |x − a| < ε ⇐⇒ −ε < x − a < ε ⇐⇒ a − ε < x < a + ε.
- If a < x < b and −b < −y < −a, it follows that a − b < x − y < b − a. Since a − b = −(b − a), the argument in 2.2.2(c) gives the conclusion |x − y| < b − a. The distance between x and y is less than or equal to b − a.
- (a) | 4 x − 5 | ≤ 13 ⇐⇒ − 13 ≤ 4 x − 5 ≤ 13 ⇐⇒ − 8 ≤ 4 x ≤ 18 ⇐⇒ − 2 ≤ x ≤ 9 /2. (b) |x^2 − 1 | ≤ 3 ⇐⇒ − 3 ≤ x^2 − 1 ≤ 3 ⇐⇒ − 2 ≤ x^2 ≤ 4 ⇐⇒ 0 ≤ x^2 ≤ 4 ⇐⇒ − 2 ≤ x ≤ 2.
- Case 1: x ≥ 2 ⇒ (x + 1) + (x − 2) = 2x − 1 = 7, so x = 4. Case 2: − 1 < x < 2 ⇒ (x + 1) + (2 − x) = 3 = 7, so no solution. Case 3: x ≤ − 1 ⇒ (−x − 1) + (2 − x) = − 2 x + 1 = 7, so x = −3. Combining these cases, we get x = 4 or x = −3.
12 Bartle and Sherbert
- If a ≤ b ≤ c, then mid{a, b, c} = b = min{b, c, c} = min{max{a, b}, max{b, c}, max{c, a}}. The other cases are similar.
Section 2.
This section completes the description of the real number system by introducing the fundamental completeness property in the form of the Supremum Property. This property is vital to real analysis and students should attain a working under- standing of it. Effort expended in this section and the one following will be richly rewarded later.
Sample Assignment: Exercises 1, 2, 5, 6, 9, 10, 12, 14.
Partial Solutions:
- Any negative number or 0 is a lower bound. For any x ≥ 0, the larger number x + 1 is in S 1 , so that x is not an upper bound of S 1. Since 0 ≤ x for all x ∈ S 1 , then u = 0 is a lower bound of S 1. If v > 0, then v is not a lower bound of S 1 because v/ 2 ∈ S 1 and v/ 2 < v. Therefore inf S 1 = 0.
- S 2 has lower bounds, so that inf S 2 exists. The argument used for S 1 also shows that inf S 2 = 0, but that inf S 2 does not belong to S 2. S 2 does not have upper bounds, so that sup S 2 does not exists.
- Since 1/n ≤ 1 for all n ∈ N, then 1 is an upper bound for S 3. But 1 is a member of S 3 , so that 1 = sup S 3. (See Exercise 7 below.)
- sup S 4 = 2 and inf S 4 = 1/2. (Note that both are members of S 4 .)
- It is interesting to compare algebraic and geometric approaches to these problems. (a) inf A = − 5 /2, sup A does not exist, (b) sup B = 2, inf B = −1, (c) sup C = 1, inf B does not exist, (d) sup D = 1 +
6, inf D = 1 −
- If S is bounded below, then S′^ := {−s : s ∈ S} is bounded above, so that u := sup S′^ exists. If v ≤ s for all s ∈ S, then −v ≥ −s for all s ∈ S, so that −v ≥ u, and hence v ≤ −u. Thus inf S = −u.
- Let u ∈ S be an upper bound of S. If v is another upper bound of S, then u ≤ v. Hence u = sup S.
- If t > u and t ∈ S, then u is not an upper bound of S.
- Let u := sup S. Since u is an upper bound of S, so is u + 1/n for all n ∈ N. Since u is the supremum of S and u − 1 /n < u, then there exists s 0 ∈ S with u − 1 /n < s 0 , whence u − 1 /n is not an upper bound of S.
- Let u := sup A, v := sup B and w := sup{u, v}. Then w is an upper bound of A ∪ B, because if x ∈ A, then x ≤ u ≤ w, and if x ∈ B, then x ≤ v ≤ w. If z is
Chapter 2 — The Real Numbers 13
any upper bound of A ∪ B, then z is an upper bound of A and of B, so that u ≤ z and v ≤ z. Hence w ≤ z. Therefore, w = sup(A ∪ B).
- Since sup S is an upper bound of S, it is an upper bound of S 0 , and hence sup S 0 ≤ sup S.
- Consider two cases. If u ≥ s∗, then u = sup(S ∪ {u}). If u < s∗, then there exists s ∈ S such that u < s ≤ s∗, so that s∗^ = sup(S ∪ {u}).
- If S 1 := {x 1 }, show that x 1 = sup S 1. If Sk := {x 1 ,... , xk} is such that sup Sk ∈ Sk, then preceding exercise implies that sup{x 1 ,... , xk, xk + 1} is the larger of sup Sk and xk + 1 and so is in Sk + 1.
- If w = inf S and ε > 0, then w + ε is not a lower bound so that there exists t in S such that t < w + ε. If w is a lower bound of S that satisfies the stated condition, and if z > w, let ε = z − w > 0. Then there is t in S such that t < w + ε = z, so that z is not a lower bound of S. Thus, w = inf S.
Section 2.
This section exhibits how the supremum is used in practice, and contains some important properties of R that will often be used later. The Archimedean Proper- ties 2.4.3–2.4.6 and the Density Properties 2.4.8 and 2.4.9 are the most significant. The exercises also contain some results that will be used later.
Sample Assignment: Exercises 1, 2, 4(b), 5, 7, 10, 12, 13, 14.
Partial Solutions:
- Since 1 − 1 /n < 1 for all n ∈ N, the number 1 is an upper bound. To show that 1 is the supremum, it must be shown that for each ε > 0 there exists n ∈ N such that 1 − 1 /n > 1 − ε, which is equivalent to 1/n < ε. Apply the Archimedean Property 2.4.3 or 2.4.5.
- inf S = −1 and sup S = 1. To see the latter note that 1/n − 1 /m ≤ 1 for all m, n ∈ N. On the other hand if ε > 0 there exists m ∈ N such that 1/m < ε, so that 1/ 1 − 1 /m > 1 − ε.
- Suppose that u ∈ R is not the supremum of S. Then either (i) u is not an upper bound of S (so that there exists s 1 ∈ S with u < s 1 , whence we take n ∈ N with 1/n < s 1 − u to show that u + 1/n is not an upper bound of S), or (ii) there exists an upper bound u 1 of S with u 1 < u (in which case we take 1 /n < u − u 1 to show that u − 1 /n is not an upper bound of S).
- (a) Let u := sup S and a > 0. Then x ≤ u for all x ∈ S, whence ax ≤ au for all x ∈ S, whence it follows that au is an upper bound of aS. If v is another upper bound of aS, then ax ≤ v for all x ∈ S, whence x ≤ v/a for all x ∈ S, showing that v/a is an upper bound for S so that u ≤ v/a, from which we conclude that au ≤ v. Therefore au = sup(aS). The statement about the infimum is proved similarly.
Chapter 2 — The Real Numbers 15
y + 1/n ∈ S 3. If y^2 > 3 and 1 /m < (y^2 − 3)/ 2 y show that y − 1 /m ∈ S 3. Therefore y^2 = 3.
- Case 1: If a > 1, let Sa := {s ∈ R : 0 ≤ s, s^2 < a}. Show that Sa is nonempty and bounded above by a and let z := sup Sa. Now show that z^2 = a. Case 2: If 0 < a < 1, there exists k ∈ N such that k^2 a > 1 (why?). If z^2 = k^2 a, then (z/k)^2 = a.
- Consider T := {t ∈ R : 0 ≤ t, t^3 < 2 }. If t > 2, then t^3 > 2 so that t /∈ T. Hence y := sup T exists. If y^3 < 2, choose 1/n < (2 − y^3 )/(3y^2 + 3y + 1) and show that (y + 1/n)^3 < 2, a contradiction, and so on.
- If x < 0 < y, then we can take r = 0. If x < y < 0, we apply 2.4.8 to obtain a rational number between −y and −x.
- There exists r ∈ Q such that x/u < r < y/u.
Section 2.
Another important consequence of the Supremum Property of R is the Nested Intervals Property 2.5.2. It is an interesting fact that if we assume the validity of both the Archimedean Property 2.4.3 and the Nested Intervals Property, then we can prove the Supremum Property. Hence these two properties could be taken as the completeness axiom for R. However, establishing this logical equivalence would consume valuable time and not significantly advance the study of real anal- ysis, so we will not do so. (There are other properties that could be taken as the completeness axiom.) The discussion of binary and decimal representations is included to give the student a concrete illustration of the rather abstract ideas developed to this point. However, this material is not vital for what follows and can be omitted or treated lightly. We have kept this discussion informal to avoid getting buried in technical details that are not central to the course.
Sample Assignment: Exercises 3, 4, 5, 6, 7, 8, 10, 11. Partial Solutions:
- Note that [a, b] ⊆ [a′, b′] if and only if a′^ ≤ a ≤ b ≤ b′.
- S has an upper bound b and a lower bound a if and only if S is contained in the interval [a, b].
- Since inf S is a lower bound for S and sup S is an upper bound for S, then S ⊆ IS. Moreover, if S ⊆ [a, b], then a is a lower bound for S and b is an upper bound for S, so that [a, b] ⊇ IS.
- Because z is neither a lower bound or an upper bound of S.
- If z ∈ R, then z is not a lower bound of S so there exists xz ∈ S such that xz ≤ z. Also z is not an upper bound of S so there exists yz ∈ S such that z ≤ yz. Since z belongs to [xz , yz ], it follows from the property (1) that z ∈ S.
