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Solution Manual for Mathematical Proofs: A Transition to Advanced Mathematics, 4th Edition

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    Solutions Manual Mathematical Proofs A Transition to Advanced Mathematics Fourth Edition Gary Chartrand Western Michigan University Albert D. Polimeni State University of New York at Fredonia Ping Zhang Western Michigan University

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    iii Table of Contents 0. Communicating Mathematics 0.1 Learning Mathematics 0.2 What Others Have Said About Writing 0.3 Mathematical Writing 0.4 Using Symbols 0.5 Writing Mathematical Expressions 0.6 Common Words and Phrases in Mathematics 0.7 Some Closing Comments About Writing 1. Sets 1.1 Describing a Set 1.2 Subsets 1.3 Set Operations 1.4 Indexed Collections of Sets 1.5 Partitions of Sets 1.6 Cartesian Products of Sets Exercises for Chapter 1 2. Logic 2.1 Statements 2.2 Negations 2.3 Disjunctions and Conjunctions 2.4 Implications 2.5 More on Implications 2.6 Biconditionals 2.7 Tautologies and Contradictions 2.8 Logical Equivalence 2.9 Some Fundamental Properties of Logical Equivalence 2.10 Quantified Statements 2.11 Characterizations Exercises for Chapter 2 3. Direct Proof and Proof by Contrapositive 3.1 Trivial and Vacuous Proofs 3.2 Direct Proofs 3.3 Proof by Contrapositive 3.4 Proof by Cases 3.5 Proof Evaluations Exercises for Chapter 3 4. More on Direct Proof and Proof by Contrapositive 4.1 Proofs Involving Divisibility of Integers 4.2 Proofs Involving Congruence of Integers 4.3 Proofs Involving Real Numbers 4.4 Proofs Involving Sets 4.5 Fundamental Properties of Set Operations 4.6 Proofs Involving Cartesian Products of Sets Exercises for Chapter 4 5. Existence and Proof by Contradiction 5.1 Counterexamples 5.2 Proof by Contradiction

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    iv 5.3 A Review of Three Proof Techniques 5.4 Existence Proofs 5.5 Disproving Existence Statements Exercises for Chapter 5 6. Mathematical Induction 6.1 The Principle of Mathematical Induction 6.2 A More General Principle of Mathematical Induction 6.3 The Strong Principle of Mathematical Induction 6.4 Proof by Minimum Counterexample Exercises for Chapter 6 7. Reviewing Proof Techniques 7.1 Reviewing Direct Proof and Proof by Contrapositive 7.2 Reviewing Proof by Contradiction and Existence Proofs 7.3 Reviewing Induction Proofs 7.4 Reviewing Evaluations of Proposed Proofs Exercises for Chapter 7 8. Prove or Disprove 8.1 Conjectures in Mathematics 8.2 Revisiting Quantified Statements 8.3 Testing Statements Exercises for Chapter 8 9. Equivalence Relations 9.1 Relations 9.2 Properties of Relations 9.3 Equivalence Relations 9.4 Properties of Equivalence Classes 9.5 Congruence Modulo n 9.6 The Integers Modulo n Exercises for Chapter 9 10. Functions 10.1 The Definition of Function 10.2 One-to-one and Onto Functions 10.3 Bijective Functions 10.4 Composition of Functions 10.5 Inverse Functions Exercises for Chapter 10 11. Cardinalities of Sets 11.1 Numerically Equivalent Sets 11.2 Denumerable Sets 11.3 Uncountable Sets 11.4 Comparing Cardinalities of Sets 11.5 The Schr¨ oder-Bernstein Theorem Exercises for Chapter 11 12. Proofs in Number Theory 12.1 Divisibility Properties of Integers 12.2 The Division Algorithm 12.3 Greatest Common Divisors

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    v 12.4 The Euclidean Algorithm 12.5 Relatively Prime Integers 12.6 The Fundamental Theorem of Arithmetic 12.7 Concepts Involving Sums of Divisors Exercises for Chapter 12 13. Proofs in Combinatorics 13.1 The Multiplication and Addition Principles 13.2 The Principle of Inclusion-Exclusion 13.3 The Pigeonhole Principle 13.4 Permutations and Combinations 13.5 The Pascal Triangle 13.6 The Binomial Theorem 13.7 Permutations and Combinations with Repetition Exercises for Chapter 13 14. Proofs in Calculus 14.1 Limits of Sequences 14.2 Infinite Series 14.3 Limits of Functions 14.4 Fundamental Properties of Limits of Functions 14.5 Continuity 14.6 Differentiability Exercises for Chapter 14 15. Proofs in Group Theory 15.1 Binary Operations 15.2 Groups 15.3 Permutation Groups 15.4 Fundamental Properties of Groups 15.5 Subgroups 15.6 Isomorphic Groups Exercises for Chapter 15

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    Exercises for Chapter 1 Exercises for Section 1.1: Describing a Set 1.1 Only (d) and (e) are sets. 1.2 (a) A = { 1 , 2 , 3 } = { x ∈ S : x > 0 } . (b) B = { 0 , 1 , 2 , 3 } = { x ∈ S : x ≥ 0 } . (c) C = {− 2 , − 1 } = { x ∈ S : x < 0 } . (d) D = { x ∈ S : | x | ≥ 2 } . 1.3 (a) | A | = 5. (b) | B | = 11. (c) | C | = 51. (d) | D | = 2. (e) | E | = 1. (f) | F | = 2. 1.4 (a) A = { n ∈ Z : − 4 < n ≤ 4 } = {− 3 , − 2 , . . . , 4 } . (b) B = { n ∈ Z : n 2 < 5 } = {− 2 , − 1 , 0 , 1 , 2 } . (c) C = { n ∈ N : n 3 < 100 } = { 1 , 2 , 3 , 4 } . (d) D = { x ∈ R : x 2 − x = 0 } = { 0 , 1 } . (e) E = { x ∈ R : x 2 + 1 = 0 } = {} = ∅ . 1.5 (a) A = {− 1 , − 2 , − 3 , . . . } = { x ∈ Z : x ≤ − 1 } . (b) B = {− 3 , − 2 , . . . , 3 } = { x ∈ Z : − 3 ≤ x ≤ 3 } = { x ∈ Z : | x | ≤ 3 } . (c) C = {− 2 , − 1 , 1 , 2 } = { x ∈ Z : − 2 ≤ x ≤ 2 , x  = 0 } = { x ∈ Z : 0 < | x | ≤ 2 } . 1.6 (a) A = { 2 x + 1 : x ∈ Z } = {· · · , − 5 , − 3 , − 1 , 1 , 3 , 5 , · · ·} . (b) B = { 4 n : n ∈ Z } = {· · · , − 8 , − 4 , 0 , 4 , 8 , · · ·} . (c) C = { 3 q + 1 : q ∈ Z } = {· · · , − 5 , − 2 , 1 , 4 , 7 , · · ·} . 1.7 (a) A = {· · · , − 4 , − 1 , 2 , 5 , 8 , · · ·} = { 3 x + 2 : x ∈ Z } . (b) B = {· · · , − 10 , − 5 , 0 , 5 , 10 , · · ·} = { 5 x : x ∈ Z } . (c) C = { 1 , 8 , 27 , 64 , 125 , · · ·} = { x 3 : x ∈ N } . 1.8 (a) A = { n ∈ Z : 2 ≤ | n | < 4 } = {− 3 , − 2 , 2 , 3 } . (b) 5 / 2, 7 / 2, 4. (c) C = { x ∈ R : x 2 − (2 + √ 2) x + 2 √ 2 = 0 } = { x ∈ R : ( x − 2)( x − √ 2) = 0 } = { 2 , √ 2 } . (d) D = { x ∈ Q : x 2 − (2 + √ 2) x + 2 √ 2 = 0 } = { 2 } . (e) | A | = 4, | C | = 2, | D | = 1. 1.9 A = { 2 , 3 , 5 , 7 , 8 , 10 , 13 } . B = { x ∈ A : x = y + z, where y, z ∈ A } = { 5 , 7 , 8 , 10 , 13 } . C = { r ∈ B : r + s ∈ B for some s ∈ B } = { 5 , 8 } . 1

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    2 Exercises for Section 1.2: Subsets 1.10 (a) A = { 1 , 2 } , B = { 1 , 2 } , C = { 1 , 2 , 3 } . (b) A = { 1 } , B = {{ 1 } , 2 } . C = {{{ 1 } , 2 } , 1 } . (c) A = { 1 } , B = {{ 1 } , 2 } , C = { 1 , 2 } . 1.11 Let r = min( c − a, b − c ) and let I = ( c − r, c + r ). Then I is centered at c and I ⊆ ( a, b ). 1.12 A = B = D = E = {− 1 , 0 , 1 } and C = { 0 , 1 } . 1.13 See Figure 1. 6 2 7 U B 8 5 4 3 1 A         . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. ... ............... .. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. .. ............... ... .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 1: Answer for Exercise 1.13 1.14 (a) P ( A ) = {∅ , { 1 } , { 2 } , { 1 , 2 }} ; |P ( A ) | = 4. (b) P ( A ) = {∅ , {∅} , { 1 } , {{ a }} , {∅ , 1 } , {∅ , { a }} , { 1 , { a }} , {∅ , 1 , { a }}} ; |P ( A ) | = 8. 1.15 P ( A ) = {∅ , { 0 } , {{ 0 }} , A } . 1.16 P ( { 1 } ) = {∅ , { 1 }} , P ( P ( { 1 } )) = {∅ , {∅} , {{ 1 }} , {∅ , { 1 }}} ; |P ( P ( { 1 } )) | = 4. 1.17 P ( A ) = {∅ , { 0 } , {∅} , {{∅}} , { 0 , ∅} , { 0 , {∅}} , {∅ , {∅}} , A } ; |P ( A ) | = 8. 1.18 P ( { 0 } ) = {∅ , { 0 }} . A = { x : x = 0 or x ∈ P ( { 0 } ) } = { 0 , ∅ , { 0 }} . P ( A ) = {∅ , { 0 } , {∅} , {{ 0 }} , { 0 , ∅} , { 0 , { 0 }} , {∅ , { 0 }} , A } . 1.19 (a) S = {∅ , { 1 }} . (b) S = { 1 } . (c) S = {∅ , { 1 } , { 2 } , { 3 } , { 4 , 5 }} . (d) S = { 1 , 2 , 3 , 4 , 5 } . 1.20 (a) False. For example, for A = { 1 , { 1 }} , both 1 ∈ A and { 1 } ∈ A . (b) Because P ( B ) is the set of all subsets of the set B and A ⊂ P ( B ) with | A | = 2, it follows that A is a proper subset of P ( B ) consisting of exactly two elements of P ( B ). Thus P ( B ) contains at least one element that is not in A . Suppose that | B | = n . Then |P ( B ) | = 2 n . Since 2 n > 2, it follows that n ≥ 2 and |P ( B ) | = 2 n ≥ 4. Because P ( B ) ⊂ C , it is impossible that | C | = 4. Suppose that A = {{ 1 } , { 2 }} , B = { 1 , 2 } and C = P ( B ) ∪ { 3 } . Then A ⊂ P ( B ) ⊂ C , where | A | = 2 and | C | = 5.

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    3 (c) No. For A = ∅ and B = { 1 } , |P ( A ) | = 1 and |P ( B ) | = 2. (d) Yes. There are only three distinct subsets of { 1 , 2 , 3 } with two elements. 1.21 B = { 1 , 4 , 5 } . Exercises for Section 1.3: Set Operations 1.22 (a) A ∪ B = { 1 , 3 , 5 , 9 , 13 , 15 } . (b) A ∩ B = { 9 } . (c) A − B = { 1 , 5 , 13 } . (d) B − A = { 3 , 15 } . (e) A = { 3 , 7 , 11 , 15 } . (f) A ∩ B = { 1 , 5 , 13 } . 1.23 Let A = { 1 , 2 , . . . , 6 } and B = { 4 , 5 , . . . , 9 } . Then A − B = { 1 , 2 , 3 } , B − A = { 7 , 8 , 9 } and A ∩ B = { 4 , 5 , 6 } . Thus | A − B | = | A ∩ B | = | B − A | = 3. See Figure 2. B 9 7 8 6 5 4 3 2 1 A          . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . .. ... ............................................... ... . .. . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . .. .. ... ......................................... ...... .. .. . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 2: Answer for Exercise 1.23 1.24 Let A = { 1 , 2 } , B = { 1 , 3 } and C = { 2 , 3 } . Then B  = C but B − A = C − A = { 3 } . 1.25 (a) A = { 1 } , B = {{ 1 }} , C = { 1 , 2 } . (b) A = {{ 1 } , 1 } , B = { 1 } , C = { 1 , 2 } . (c) A = { 1 } , B = {{ 1 }} , C = {{ 1 } , 2 } . 1.26 (a) and (b) are the same, as are (c) and (d). 1.27 Let U = { 1 , 2 , . . . , 8 } be a universal set, A = { 1 , 2 , 3 , 4 } and B = { 3 , 4 , 5 , 6 } . Then A − B = { 1 , 2 } , B − A = { 5 , 6 } , A ∩ B = { 3 , 4 } and A ∪ B = { 7 , 8 } . See Figure 3. 2 A B 7 8 6 3 4 1 U 5     . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ... ... ........................... .... .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . .. ... .......................... ..... .. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .     Figure 3: Answer for Exercise 1.27

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    4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ................... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ................... A C B A C B ( C − B ) ∪ A C ∩ ( A − B ) (a) (b) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. ........... .. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. ............ . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. ........... ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . .............. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . ........... ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. ........... .. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 4: Answers for Exercise 1.28 1.28 See Figures 4(a) and 4(b). 1.29 (a) The sets ∅ and {∅} are elements of A . (b) | A | = 3. (c) All of ∅ , {∅} and {∅ , {∅}} are subsets of A . (d) ∅ ∩ A = ∅ . (e) {∅} ∩ A = {∅} . (f) {∅ , {∅}} ∩ A = {∅ , {∅}} . (g) ∅ ∪ A = A . (h) {∅} ∪ A = A . (i) {∅ , {∅}} ∪ A = A . 1.30 (a) A = { x ∈ R : | x − 1 | ≤ 2 } = { x ∈ R : − 2 ≤ x − 1 ≤ 2 } = { x ∈ R : − 1 ≤ x ≤ 3 } = [ − 1 , 3] B = { x ∈ R : | x | ≥ 1 } = { x ∈ R : x ≥ 1 or x ≤ − 1 } = ( −∞ , − 1] ∪ [1 , ∞ ) C = { x ∈ R : | x + 2 | ≤ 3 } = { x ∈ R : − 3 ≤ x + 2 ≤ 3 } = { x ∈ R : − 5 ≤ x ≤ 1 } = [ − 5 , 1] (b) A ∪ B = ( −∞ , ∞ ) = R , A ∩ B = {− 1 } ∪ [1 , 3], B ∩ C = [ − 5 , − 1] ∪ { 1 } , B − C = ( −∞ , − 5) ∪ (1 , ∞ ). 1.31 A = { 1 , 2 } , B = { 2 } , C = { 1 , 2 , 3 } , D = { 2 , 3 } . 1.32 A = { 1 , 2 , 3 } , B = { 1 , 2 , 4 } , C = { 1 , 3 , 4 } , D = { 2 , 3 , 4 } . 1.33 A = { 1 } , B = { 2 } . 1.34 A = { 1 , 2 } , B = { 2 , 3 } . 1.35 Let U = { 1 , 2 , . . . , 8 } , A = { 1 , 2 , 3 , 5 } , B = { 1 , 2 , 4 , 6 } and C = { 1 , 3 , 4 , 7 } . See Figure 5. Exercises for Section 1.4: Indexed Collections of Sets 1.36 ⋃ α ∈ A S α = S 1 ∪ S 3 ∪ S 4 = [0 , 3] ∪ [2 , 5] ∪ [3 , 6] = [0 , 6] . ⋂ α ∈ A S α = S 1 ∩ S 3 ∩ S 4 = { 3 } .

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    5 U B 4 3 1 2 7 6 8 5 C A         . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. .. .................... .... .. .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. .. .. ..................... ... .. . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . .. .. ..................... ... .. . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Figure 5: Answer for Exercise 1.35 1.37 ⋃ X ∈ S X = A ∪ B ∪ C = { 0 , 1 , 2 , . . . , 5 } and ⋂ X ∈ S X = A ∩ B ∩ C = { 2 } . 1.38 (a) ⋃ α ∈ S A α = A 1 ∪ A 2 ∪ A 4 = { 1 } ∪ { 4 } ∪ { 16 } = { 1 , 4 , 16 } . ⋂ α ∈ S A α = A 1 ∩ A 2 ∩ A 4 = ∅ . (b) ⋃ α ∈ S B α = B 1 ∪ B 2 ∪ B 4 = [0 , 2] ∪ [1 , 3] ∪ [3 , 5] = [0 , 5]. ⋂ α ∈ S B α = B 1 ∩ B 2 ∩ B 4 = ∅ . (c) ⋃ α ∈ S C α = C 1 ∪ C 2 ∪ C 4 = (1 , ∞ ) ∪ (2 , ∞ ) ∪ (4 , ∞ ) = (1 , ∞ ) . ⋂ α ∈ S C α = C 1 ∩ C 2 ∩ C 4 = (4 , ∞ ). 1.39 Since | A | = 26 and | A α | = 3 for each α ∈ A , we need to have at least nine sets of cardinality 3 for their union to be A ; that is, in order for ⋃ α ∈ S A α = A , we must have | S | ≥ 9. However, if we let S = { a, d, g, j, m, p, s, v, y } , then ⋃ α ∈ S A α = A . Hence the smallest cardinality of a set S with ⋃ α ∈ S A α = A is 9. 1.40 (a) ⋃ 5 i =1 A 2 i = A 2 ∪ A 4 ∪ A 6 ∪ A 8 ∪ A 10 = { 1 , 3 } ∪ { 3 , 5 } ∪ { 5 , 7 } ∪ { 7 , 9 } ∪ { 9 , 11 } = { 1 , 3 , 5 , . . . , 11 } . (b) ⋃ 5 i =1 ( A i ∩ A i +1 ) = ⋃ 5 i =1 ( { i − 1 , i + 1 } ∩ { i, i + 2 } ) = ⋃ 5 i =1 ∅ = ∅ . (c) ⋃ 5 i =1 ( A 2 i − 1 ∩ A 2 i +1 ) = ⋃ 5 i =1 ( { 2 i − 2 , 2 i } ∩ { 2 i, 2 i + 2 } ) = ⋃ 5 i =1 { 2 i } = { 2 , 4 , 6 , 8 , 10 } . 1.41 (a) { A n } n ∈ N , where A n = { x ∈ R : 0 ≤ x ≤ 1 /n } = [0 , 1 /n ]. (b) { A n } n ∈ N , where A n = { a ∈ Z : | a | ≤ n } = {− n, − ( n − 1) , . . . , ( n − 1) , n } . 1.42 (a) A n = [ 1 , 2 + 1 n ) , ⋃ n ∈ N A n = [1 , 3) and ⋂ n ∈ N A n = [1 , 2]. (b) A n = ( − 2 n − 1 n , 2 n ) , ⋃ n ∈ N A n = ( − 2 , ∞ ) and ⋂ n ∈ N A n = ( − 1 , 2). 1.43 ⋃ r ∈ R + A r = ⋃ r ∈ R + ( − r, r ) = R ; ⋂ r ∈ R + A r = ⋂ r ∈ R + ( − r, r ) = { 0 } . 1.44 For I = { 2 , 8 } , | ⋃ i ∈ I A i | = 8. Observe that there is no set I such that | ⋃ i ∈ I A i | = 10, for in this case, we must have either two 5-element subsets of A or two 3-element subsets of A and a 4-element subset of A . In each case, not every two subsets are disjoint. Furthermore, there is no set I such that | ⋃ i ∈ I A i | = 9, for in this case, one must either have a 5-element subset of A and a 4-element subset of A (which are not disjoint) or three 3-element subsets of A . No 3-element subset of A contains 1 and only one such subset contains 2. Thus 4 , 5 ∈ I but there is no third element for I .

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    6 1.45 ⋃ n ∈ N A n = ⋃ n ∈ N ( − 1 n , 2 − 1 n ) = ( − 1 , 2); ⋂ n ∈ N A n = ⋂ n ∈ N ( − 1 n , 2 − 1 n ) = [0 , 1]. 1.46 (a) ∞ ⋃ n =1 ( − 1 n , 1 n ) = ( − 1 , 1); ∞ ⋂ n =1 ( − 1 n , 1 n ) = { 0 } (b) ∞ ⋃ n =1 [ n − 1 n , n + 1 n ] = [0 , 2]; ∞ ⋂ n =1 [ n − 1 n , n + 1 n ] = { 1 } 1.47 (a) ∞ ⋃ n =1 { sin 2 nπ 2 + cos 2 nπ 2 } = ∞ ⋂ n =1 { sin 2 nπ 2 + cos 2 nπ 2 } = { 1 } (b) ∞ ⋃ n =1 { sin nπ 2 + cos nπ 2 } = {− 1 , 1 } ; ∞ ⋂ n =1 { sin nπ 2 + cos nπ 2 } = ∅ Exercises for Section 1.5: Partitions of Sets 1.48 (a) S 1 is a partition of A . (b) S 2 is not a partition of A because g belongs to no element of S 2 . (c) S 3 is a partition of A . (d) S 4 is not a partition of A because ∅ ∈ S 4 . (e) S 5 is not a partition of A because b belongs to two elements of S 5 . 1.49 (a) S 1 is not a partition of A since 4 belongs to no element of S 1 . (b) S 2 is a partition of A . (c) S 3 is not a partition of A because 2 belongs to two elements of S 3 . (d) S 4 is not a partition of A since S 4 is not a set of subsets of A . 1.50 S = {{ 1 , 2 , 3 } , { 4 , 5 } , { 6 }} ; | S | = 3. 1.51 A = { 1 , 2 , 3 , 4 } . S 1 = {{ 1 } , { 2 } , { 3 , 4 }} and S 2 = {{ 1 , 2 } , { 3 } , { 4 }} . 1.52 Let S = { A 1 , A 2 , A 3 } , where A 1 = { x ∈ N : x > 5 } , A 2 = { x ∈ N : x < 5 } and A 3 = { 5 } . 1.53 Let S = { A 1 , A 2 , A 3 } , where A 1 = { x ∈ Q : x > 1 } , A 2 = { x ∈ Q : x < 1 } and A 3 = { 1 } . 1.54 A = { 1 , 2 , 3 , 4 } , S 1 = {{ 1 } , { 2 } , { 3 , 4 }} and S 2 = {{{ 1 } , { 2 }} , {{ 3 , 4 }}} . 1.55 Let S = { A 1 , A 2 , A 3 , A 4 } , where A 1 = { x ∈ Z : x is odd and x is positive } , A 2 = { x ∈ Z : x is odd and x is negative } , A 3 = { x ∈ Z : x is even and x is nonnegative } , A 4 = { x ∈ Z : x is even and x is negative } .

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    7 1.56 Let S = {{ 1 } , { 2 } , { 3 , 4 , 5 , 6 } , { 7 , 8 , 9 , 10 } , { 11 , 12 }} and T = {{ 1 } , { 2 } , { 3 , 4 , 5 , 6 } , { 7 , 8 , 9 , 10 }} . 1.57 |P 1 | = 2, |P 2 | = 3, |P 3 | = 5, |P 4 | = 8, |P 5 | = 13, |P 6 | = 21. 1.58 (a) Suppose that a collection S of subsets of A satisfies Definition 1. Then every subset is nonempty. Every element of A belongs to a subset in S . If some element a ∈ A belonged to more than one subset, then the subsets in S would not be pairwise disjoint. So the collec- tion satisfies Definition 2. (b) Suppose that a collection S of subsets of A satisfies Definition 2. Then every subset is nonempty and (1) in Definition 3 is satisfied. If two subsets A 1 and A 2 in S were neither equal nor disjoint, then A 1  = A 2 and there is an element a ∈ A such that a ∈ A 1 ∩ A 2 , which would not satisfy Definition 2. So condition (2) in Definition 3 is satisfied. Since every element of A belongs to a (unique) subset in S , condition (3) in Definition 3 is satisfied. Thus Definition 3 itself is satisfied. (c) Suppose that a collection S of subsets of A satisfies Definition 3. By condition (1) in Defi- nition 3, every subset is nonempty. By condition (2), the subsets are pairwise disjoint. By condition (3), every element of A belongs to a subset in S . So Definition 1 is satisfied. Exercises for Section 1.6: Cartesian Products of Sets 1.59 A × B = { ( x, x ) , ( x, y ) , ( y, x ) , ( y, y ) , ( z, x ) , ( z, y ) } . 1.60 A × A = { (1, 1), (1, { 1 } ), (1, {{ 1 }} ), ( { 1 } , 1), ( { 1 } , { 1 } ), ( { 1 } , {{ 1 }} ), ( {{ 1 }} , 1), ( {{ 1 }} , { 1 } ), ( {{ 1 }} , {{ 1 }} ) } . 1.61 P ( A ) = {∅ , { a } , { b } , A } , A × P ( A ) = { ( a, ∅ ) , ( a, { a } ) , ( a, { b } ) , ( a, A ) , ( b, ∅ ) , ( b, { a } ) , ( b, { b } ) , ( b, A ) } . 1.62 P ( A ) = {∅ , {∅} , {{∅}} , A } , A × P ( A ) = { ( ∅ , ∅ ) , ( ∅ , {∅} ) , ( ∅ , {{∅}} ) , ( ∅ , A ) , ( {∅} , ∅ ) , ( {∅} , {∅} ) , ( {∅} , {{∅}} ) , ( {∅} , A ) } . 1.63 P ( A ) = {∅ , { 1 } , { 2 } , A } , P ( B ) = {∅ , B } , A × B = { (1 , ∅ ) , (2 , ∅ ) } , P ( A ) × P ( B ) = { ( ∅ , ∅ ) , ( ∅ , B ) , ( { 1 } , ∅ ) , ( { 1 } , B ) , ( { 2 } , ∅ ) , ( { 2 } , B ) , ( A, ∅ ) , ( A, B ) } . 1.64 { ( x, y ) : x 2 + y 2 = 4 } , which is a circle centered at (0, 0) with radius 2. 1.65 S = { (3 , 0) , (2 , 1) , (2 , − 1) , (1 , 2) , (1 , − 2) , (0 , 3) , (0 , − 3) , ( − 3 , 0) , ( − 2 , 1) , ( − 2 , − 1) , ( − 1 , 2) , ( − 1 , − 2) } . See Figure 6. 1.66 A × B = { (1 , 1) , (2 , 1) } , P ( A × B ) = {∅ , { (1 , 1) } , { (2 , 1) } , A × B } . 1.67 A = { x ∈ R : | x − 1 | ≤ 2 } = { x ∈ R : − 1 ≤ x ≤ 3 } = [ − 1 , 3], B = { y ∈ R : | y − 4 | ≤ 2 } = { y ∈ R : 2 ≤ y ≤ 6 } = [2 , 6], A × B = [ − 1 , 3] × [2 , 6], which is the set of all points on and within the square bounded by x = − 1, x = 3, y = 2 and y = 6.

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    8 (0 , − 3) (0 , 3) (3 , 0) (2 , − 1) ( − 1 , 2) (1 , 2) ( − 2 , 1) (2 , 1) ( − 3 , 0) ( − 2 , − 1) ( − 1 , − 2) (1 , − 2) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .              ........................................... ................ ............................... ............... ................................ ............... ............................... ................ ............................... ................ ............... ................ ............... ................ ................ ............... ............................................... ............... ................................ ............... ................ ............... ................ ................ ............... ..... ................... . . . . . . . . . . . . . . . . . . . Figure 6: Answer for Exercise 1.65 1.68 A = { a ∈ R : | a | ≤ 1 } = { a ∈ R : − 1 ≤ a ≤ 1 } = [ − 1 , 1], B = { b ∈ R : | b | = 1 } = {− 1 , 1 } , A × B is the set of all points ( x, y ) on the lines y = 1 or y = − 1 with x ∈ [ − 1 , 1], while B × A is the set of all points ( x, y ) on the lines x = 1 or x = − 1 with y ∈ [ − 1 , 1]. Therefore, ( A × B ) ∪ ( B × A ) is the set of all points lying on (but not within) the square bounded by x = 1, x = − 1, y = 1 and y = − 1. 1.69 (a)–(b) ( A × B ) ∩ ( B × A ) = ( A ∩ B ) × ( B ∩ A ) = { (2 , 2) , (2 , 3) , (3 , 2) , (3 , 3) } . 1.70 For A = { 1 , 2 } , B = { 1 , 2 , 3 } , C = { 1 , 2 , 3 , 4 } and D = { 2 , 3 } , it follows that (( A × B ) ∪ ( C × D )) − ( D × D ) = R . 1.71 Since ⋃ 3 i =1 ( A i × A i ) = { (1 , 1) , (1 , 2) , (2 , 1) , (2 , 2) , (2 , 3) , (3 , 2) , (3 , 3) , (3 , 4) , (4 , 3) , (4 , 4) } , it follows that ∣ ∣ ∪ 3 i =1 ( A i × A i ) ∣ ∣ = 10. 1.72 The set { A × A, A × B, B × A, B × B } is a partition of S × S . Chapter 1 Supplemental Exercises 1.73 (a) A = { 4 k + 3 : k ∈ Z } = { . . . , − 5 , − 1 , 3 , 7 , 11 , . . . } . (b) B = { 5 k − 1 : k ∈ Z } = { . . . , − 6 , − 1 , 4 , 9 , 14 , . . . } . 1.74 (a) A = { x ∈ S : | x | ≥ 1 } = { x ∈ S : x  = 0 } . (b) B = { x ∈ S : x ≤ 0 } . (c) C = { x ∈ S : − 5 ≤ x ≤ 7 } = { x ∈ S : | x − 1 | ≤ 6 } . (d) D = { x ∈ S : x  = 5 } .

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    9 1.75 (a) { 0 , 2 , − 2 } (b) { } (c) { 3 , 4 , 5 } (d) { 1 , 2 , 3 } (e) {− 2 , 2 } (f) { } (g) {− 3 , − 2 , − 1 , 1 , 2 , 3 } . 1.76 (a) | A | = 6 (b) | B | = 0 (c) | C | = 3 (d) | D | = 0 (e) | E | = 10 (f) | F | = 20. 1.77 A × B = { ( − 1 , x ) , ( − 1 , y ) , (0 , x ) , (0 , y ) , (1 , x ) , (1 , y ) } . 1.78 (a) ( A ∪ B ) − ( B ∩ C ) = { 1 , 2 , 3 } − { 3 } = { 1 , 2 } . (b) A = { 3 } . (c) B ∪ C = { 1 , 2 , 3 } = ∅ . (d) A × B = { (1 , 2) , (1 , 3) , (2 , 2) , (2 , 3) } . 1.79 Let S = {{ 1 } , { 2 } , { 3 , 4 } , A } and let B = { 3 , 4 } . 1.80 P ( A ) = {∅ , { 1 }} , P ( C ) = {∅ , { 1 } , { 2 } , C } . Let B = {∅ , { 1 } , { 2 }} . 1.81 Let A = {∅} and B = P ( A ) = {∅ , {∅}} . 1.82 Only B = C = ∅ and D = E . 1.83 U = { 1 , 2 , 3 , 5 , 7 , 8 , 9 } , A = { 1 , 2 , 5 , 7 } and B = { 5 , 7 , 8 } . 1.84 (a) A r is the set of all points in the plane lying on the circle x 2 + y 2 = r 2 . ⋃ r ∈ I A r = R × R (the plane) and ⋂ r ∈ I A r = ∅ . (b) B r is the set of all points lying on and inside the circle x 2 + y 2 = r 2 . ⋃ r ∈ I B r = R × R and ⋂ r ∈ I B r = { (0 , 0) } . (c) C r is the set of all points lying outside the circle x 2 + y 2 = r 2 . ⋃ r ∈ I C r = R × R − { (0 , 0) } and ⋂ r ∈ I C r = ∅ . 1.85 Let A 1 = { 1 , 2 , 3 , 4 } , A 2 = { 3 , 5 , 6 } , A 3 = { 1 , 3 } , A 4 = { 1 , 2 , 4 , 5 , 6 } . Then | A 1 ∩ A 2 | = | A 2 ∩ A 3 | = | A 3 ∩ A 4 | = 1, | A 1 ∩ A 3 | = | A 2 ∩ A 4 | = 2 and | A 1 ∩ A 4 | = 3. 1.86 (a) (i) Give an example of five sets A i (1 ≤ i ≤ 5) such that | A i ∩ A j | = | i − j | for every two integers i and j with 1 ≤ i < j ≤ 5. (ii) Determine the minimum positive integer k such that there exist four sets A i (1 ≤ i ≤ 4) satisfying the conditions of Exercise 1.79 and | A 1 ∪ A 2 ∪ A 3 ∪ A 4 | = k .

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    10 (b) (i) A 1 = { 1 , 2 , 3 , 4 , 7 , 8 , 9 , 10 } A 2 = { 3 , 5 , 6 , 11 , 12 , 13 } A 3 = { 1 , 3 , 14 , 15 } A 4 = { 1 , 2 , 4 , 5 , 6 , 16 } A 5 = { 7 , 8 , 9 , 10 , 11 , 12 , 13 , 14 , 15 , 16 } . (ii) The minimum positive integer k is 5. The example below shows that k ≤ 5. Let A 1 = { 1 , 2 , 3 , 4 } , A 2 = { 1 , 5 } , A 3 = { 1 , 4 } , A 4 = { 1 , 2 , 3 , 5 } . If k = 4, then since | A 1 ∩ A 4 | = 3, A 1 and A 4 have exactly three elements in common, say 1, 2, 3. So each of A 1 and A 4 is either { 1 , 2 , 3 } or { 1 , 2 , 3 , 4 } . They cannot both be { 1 , 2 , 3 , 4 } . Also, they cannot both be { 1 , 2 , 3 } because A 3 would have to contain two of 1, 2, 3 and so | A 3 ∩ A 4 | ≥ 2, which is not true. So we can assume that A 1 = { 1 , 2 , 3 , 4 } and A 4 = { 1 , 2 , 3 } . However, A 2 must contain two of 1, 2, 3 and so | A 1 ∩ A 2 | ≥ 2, which is impossible. 1.87 (a) | S | = | T | = 10. (b) | S | = | T | = 5. (c) | S | = | T | = 6. 1.88 Let A = { 1 , 2 , 3 , 4 } , A 1 = { 1 , 2 } , A 2 = { 1 , 3 } , A 3 = { 3 , 4 } . These examples show that k ≤ 4. Since | A 1 − A 3 | = | A 3 − A 1 | = 2, it follows that A 1 contains two elements not in A 3 , while A 3 contains two elements not in A 1 . Thus | A | ≥ 4 and so k = 4 is the smallest positive integer with this property. 1.89 (a) S = { ( − 3 , 4) , (0 , 5) , (3 , 4) , (4 , 3) } . (b) C = { a ∈ B : ( a, b ) ∈ S } = { 3 , 4 } D = { b ∈ A : ( a, b ) ∈ S } = { 3 , 4 } C × D = { (3 , 3) , (3 , 4) , (4 , 3) , (4 , 3) } . 1.90 A = { 1 , 2 , 3 } , B = {{ 1 , 2 } , { 1 , 3 } , { 2 , 3 }} , C = {{ 1 } , { 2 } , { 3 }} , D = P ( C ) = {∅ , {{ 1 }} , {{ 2 }} , {{ 3 }} , {{ 1 } , { 2 }} , {{ 1 } , { 3 }} , {{ 2 } , { 3 }} , C } . 1.91 S = { x ∈ R : x 2 + 2 x − 1 = 0 } = {− 1 + √ 2 , − 1 − √ 2 } . A − 1+ √ 2 = {− 1 + √ 2 , √ 2 } , A − 1 − √ 2 = {− 1 − √ 2 − √ 2 } . (a) A s = A − 1 − √ 2 and A t = A − 1+ √ 2 . A s × A t = { ( − 1 − √ 2 , − 1 + √ 2) , ( − 1 − √ 2 , √ 2) , ( − √ 2 , 1 + √ 2) , ( − √ 2 , √ 2) } . (b) C = { ab : ( a, b ) ∈ B } = {− 1 , − √ 2 − 2 , √ 2 − 2 , − 2 } . The sum of the elements in C is − 7. 1.92 (a) For | A | = 2, the largest possible value of | A ∩ P ( A ) | is 2. The set A = {∅ , {∅}} has this property. (b) For | A | = 3, the largest possible value of | A ∩ P ( A ) | is 3. The set A = {∅ , {∅} , {{∅}}} has this property. (c) For | A | = 4, the largest possible value of | A ∩ P ( A ) | is 4. The set A = {∅ , {∅} , {{∅}} , {{{∅}}}} has this property.

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    11 1.93 ∞ ⋃ n =1 S n = ∞ ⋂ n =1 S n = [ − √ 2 , √ 2 ] . First, observe that ∞ ⋂ n =1 S n = S 1 = { x + y : x, y ∈ R , x 2 + y 2 = 1 } . Let f ( x ) = x + y , where y = ± √ 1 − x 2 , say y = √ 1 − x 2 . Then f ( x ) = x + √ 1 − x 2 . Since f ′ ( x ) = 1 − x √ 1 − x 2 , it follows that f ′ ( x ) = 0 when x = 1 √ 2 and so the maximum value of x + y is √ 2. Since f ( − 1 √ 2 ) = 0 and f is continuous on [ − √ 2 , √ 2 ] , it follows that f takes on all values of [ 0 , √ 2 ] . If x + y = r ∈ [ 0 , √ 2 ] , it follows that ( − x ) + ( − y ) = − r ∈ [ − √ 2 , 0 ] . Hence, S 1 = [ − √ 2 , √ 2 ] . If a 2 + b 2 = r where 0 < r < 1, then a + b ∈ ( − √ 2 , √ 2 ) and so ∞ ⋃ n =1 S n = [ − √ 2 , √ 2 ] as well. 1.94 (a) No. For example, the elements 1, 2 and 5 belong to more than one subset of S . (b) Yes. (c) Yes. 1.95 In order for | A × ( B ∪ C ) | = | A × B | + | A × C | , the sets B and C must be disjoint.

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