Introduction to Real Analysis 4th Edition Bartle Solutions Manual

CONTENTS Chapter 1 Preliminaries .....................................................1 Chapter 2 The Real Numbers ............................................... 7 Chapter 3 Sequences .......................................................17 Chapter 4 Limits ...........................................................28 Chapter 5 Continuous Functions ........................................... 33 Chapter 6 Differentiation ...................................................43 Chapter 7 The Riemann Integral ...........................................51 Chapter 8 Sequences of Functions ..........................................61 Chapter 9 Infinite Series ................................................... 68 Chapter 10 The Generalized Riemann Integral ............................. 77 Chapter 11 A Glimpse into Topology .......................................88 Selected Graphs .............................................................95 This sample only, Download all chapters at: 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 combinatorics, 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 1.1 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: 1. (a) B ∩ C ={5,11,17,23,...}={6k −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} 2. The sets are equal to (a) A, (b) A∩B, (c) the empty set. 3. 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. 4. 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 ∈ AB or x ∈ AC, which implies that x ∈ (AB) ∪ (AC). Thus A(B ∩ C) ⊆ (AB) ∪ (AC). 1 Conversely, if x is in (AB) ∪ (AC), then x ∈ AB or x ∈ AC. Thus x ∈ A and either x /∈ B or x /∈ C, which implies that x ∈ A but x /∈ B ∩ C, so that x ∈ A(B ∩ C). Thus (AB) ∪ (AC) ⊆ A(B ∩ C). Since the sets A(B∩C) and (AB)∪(AC) contain the same elements, they are equal. 5. (a) If x ∈ A∩(B ∪C), then x∈A and x∈B ∪C. Hence we either have (i) x ∈ A and x ∈ B, or we have (ii) x ∈ A and x ∈ C. Therefore, either x ∈ A ∩ B or x ∈ A ∩ C, so that x ∈ (A ∩ B) ∪ (A ∩ C). This shows that A ∩ (B ∪ C) is a subset of (A∩B)∪(A∩C). Conversely, let y be an element of (A∩B)∪(A∩C). Then either (j) y ∈ A∩B, or (jj) y ∈A∩C. It follows that y ∈A and either y ∈B or y ∈C. Therefore, y ∈A and y ∈B ∪C, so that y ∈A∩(B ∪C). Hence (A∩B)∪ (A∩C) is a subset of A∩(B ∪C). In view of Definition 1.1.1, we conclude that the sets A∩(B ∪C) and (A∩B)∪(A∩C) are equal. (b) Similar to (a). 6. The set D is the union of {x : x∈A and x /∈ B} and {x : x /∈ A and x∈B}. 7. Here An ={n+1,2(n+1),...}. (a) A1 ={2,4,6,8,...},A2 ={3,6,9,12,...},A1 ∩A2 = {6,12,18,24,...} = {6k : k ∈ N}=A5. , because if n>1, then n∈An−1; moreover 1 ∈/ An. Also , because n /∈An for any n∈N. 8. (a) The graph consists of four horizontal line segments. (b) The graph consists of three vertical line segments. 9. No. For example, both (0, 1) and (0,−1) belong to . 1 − − − 12. If 0 is removed from E and F, then their intersection is empty, but the intersection of the images under f is {y : 0<y ≤1}. ) is empty, and f(E F) = 14. If y ∈f(E ∩ F), then there exists x∈E ∩ F such that y =f(x). Since x∈E implies y ∈f(E), and x∈F implies y ∈f(F), we have y ∈f(E)∩f(F). This proves f(E ∩F) ⊆ f(E)∩f(F). 15. If x∈f−1(G) ∩ f−1(H), then x∈f−1(G) and x∈f−1(H), so that f(x)∈G and f(x)∈H. Then f(x)∈G ∩ H, and hence x∈f−1(G ∩ H)