enderton set theory solutions

Solutions to Elements of Set Theory [email protected] January 14, 2011

Contents1 Introduction 2 Axioms and Operations 3 Relations and Functions 3.1 Ordered Pairs . . . . . . . . 3.2 Relations . . . . . . . . . . 3.3 n-ary Relations . . . . . . . 3.4 Functions . . . . . . . . . . 3.5 Innite Cartesian Products 3.6 Ordering Relations . . . . . 4 Natural Numbers 4.1 Inductive Sets . . . 4.2 Peanos Postulates 4.3 Recursion on . . 4.4 Arithmetic . . . . . 4.5 Ordering on . . . 4.6 Review Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 3 11 11 12 13 14 20 25 30 30 30 31 33 36 39

5 Construction of the Real Numbers 42 5.1 Integers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42 5.2 Rational Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

1

Introduction

Exercise 1.1. Which of the following become true when is inserted in place of the blank? Which become true when is inserted? (a) {}

{, {}}.

by Herbert Enderton. I make no claim to correctness, as Im just learning as I write these.

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INTRODUCTION (b) {} (c) {{}} (d) {{}} (e) {{}} {, {{}}}. {, {}}. {, {{}}}. {, {, {}}}.

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Solution. Choices (a) and (d) become true when is inserted in place of the blank. Choices (b) and (c) become true when is inserted in place of the blank. Choice (e) is not true in either case. Exercise 1.2. Show that not two of the three sets , {}, and {{}} are equal to each other. Solution. Note that {} equal. , and {{}} . Also, {{}} {}. Hence no two are

Exercise 1.3. Show that if B C, then PB PC. Solution. Suppose A PB. Then A B, and thus A C, as containment is transitive. Hence A PC. Exercise 1.4. Assume that x and y are members of a set B. Show that {{x}, {x, y}} PPB. Proof. Since x and y are members of B, it follows that {x} B and {x, y} B. So {x} and {x, y} PB, and thus {{x}, {x, y}} PB, so {{x}, {x, y}} PPB. Exercise 1.5. Dene the rank of a set c to be the least such that c V . Compute the rank of {{}}. Compute the rank of {, {}, {, {}}}. Solution. Observe that V+1 = PV . Taking V0 = A = , it follows that V1 = PV0 = {, {}}. Hence the rank of {{}} is 1. Futhermore, V2 = PV1 = {, {}, {{}}, {, {}}}. Thus {, {}, {, {}}} has rank 2. Exercise 1.6. We have stated that V+1 = A PV . Prove this at least for < 3. Solution. By denition, for = 0, V1 = V0 PV0 = A PV0 . For = 1, The last equality follows from Exercise 1.3, as V0 V1 , and thus PV0 PV1 = PV1 . The case for = 2 follows similarly. V2 = V1 PV1 = A PV0 PV1 = A PV1 .

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AXIOMS AND OPERATIONS

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Exercise 1.7. List all the members of V3 . List all the members of V4 . Solution. Without listing them, V3 has 16 members, and V4 has 32 members.

2

Axioms and Operations

Exercise 2.1. Assume that A is the set of integers divisible by 4. Similarly assume that B and C are the sets of integers divisible by 9 and 10, respectively. What is in A B C? Solution. The set A B C consists precisely of the numbers divisible by the least common multiple of 4, 9 and 10. That is, it is the set of all multiples of 180. Exercise 2.2. Give an example of sets A and B for which Solution. Take A = {{a}, {b, c}} and B = {{a}, {b}, {c}}. Exercise 2.3. Show that every member of a set A is a subset of A. Solution. If A = , then the statement is vacuously true. So suppose A is nonempty, and take some b A. Again, if b is empty, then the statement holds. So suppose b is nonempty. Then for any t b, t A by the Union Axiom. In other words, t(t b t This is precisely the denition that b Exercise 2.4. Show that if A B, then A. A B. A). A= B but A = B.

Solution. Recall that x A (b A) x b. Take x A. So by denition, there exists b A such that x b. Since A B, b B as well. Then by denition, x B. Exercise 2.5. Assume that every member of A is a subset of B. Show that A B. Solution. Take a A . So by the Union Axiom, there exists b A such that a b. By assumption, b B, and thus by denition of subset, a B. Exercise 2.6. (a) Show that for any set A, PA = A. (b) Show that A P A. Under what conditions does equality hold?

Solution. For (a), take x PA. So there exists some b PA such that x b. Since b PA, b A, and thus x A. Also, since A A, by denition of the power set, A PA. Then by Exercise 2.3, it follows that A PA. For (b), take x A. Again by Exercise 2.3, x A, so by denition of the power set, x P A. Equality holds when A has form {}. If A = {}, then A = , and so P A = {}. If A = {a, } for some a A, then A = a, and so P A = Pa, which gives all subsets of a, which may not equal A. Note that if A = {a, b} for a, b A and a, b nonempty, then A = a b, so P A contains a b, and thus equality does not hold in this case.

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AXIOMS AND OPERATIONS (a) Show that for any sets A and B, PA PB = P(A B). (b) Show that PA PB P(A B). Under what conditions does equality hold?

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Exercise 2.7.

Solution. Take x PA PB. Then x A and x B, so in either case, x A B. Hence x P(A B). Take y P(A B). Thus y A B, and so y A and y B, so y PA PB. For (b), take x PA PB. Then x A or x B, so in either case, x A B. Hence x P(A B). Furthermore, equality holds if either A or B is a subset of the other. Without loss of generality, suppose B A. If x A B, then x A since A B = A. The same argument holds if B A. However, suppose B A. Consider then the case where B = {a, b} and A = {a, c}. Then let x = {b, c}. Then x P(A B), but x PA PB. Exercise 2.8. Show that there is no set to which every singleton (that is, every set of the form {x}) belongs. Solution. Suppose X is the set of all singletons. Then leads to Russells paradox. X is the set of all sets, which

Exercise 2.9. Give an example of sets a and B for which a B but Pa PB. Solution. Let a = {} and B = {{}}. Clearly a B. Then Pa = {, {}} and PB = {, {{}}}, so Pa PB. Exercise 2.10. Show that if a B, then Pa PP B. Solution. First, recall from Exercise 2.6 that for any set A, A P A. So in this case, B P B. Hence if a B, then a P B, that is, a B. Then observe that Pa P B, by Exercise 1.3. From Pa P B it follows that Pa PP B. Exercise 2.11. Show that for any sets A and B, A = (A B) (A B) and A (B A) = A B

Solution. A (AB)(AB) is obvious. Now take x (AB)(AB). If x AB, then x A. If x A B, again x A. Since B A B, we have A (B A) A B by the monotonicity properties. Take x A B. If x A, then clearly x A (B A). If x B, then by the rst part of the exercise, then either x A B, whence x A, or x B A. Exercise 2.12. Verify the following identity: C (A B) = (C A) (C B).

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Solution. Take x C (A B). Then x C, but x A B. Note that it is not the case that x A and x B. If x A but x B, then x C B. If x A but x B, then x C A. Finally, if x A and x B, then x C (A B) = (C A) (C B). Now take x (C A) (C B). If x C A, then x A, and thus x A B. Then x C (A B). A similar argument handles the case where x C B. Exercise 2.13. Show that if A B, then C B C A. Solution. Recall that if A B, then B A. Take x C B. Since x B, then x B, and so x A. It follows that x C A. Exercise 2.14. Show by exmaple that for some sets A, B, and C, the set A (B C) is dierent from (A B) C. Solution. Let A = {a, d, e, f }, B = {b, d, e, g}, and C = {c, e, f, g}. Then B C = {b, d}, so A (B C) = {a, e, f }. But A B = {a, f }, so (A B) C = {a}. Exercise 2.15. Dene the symmetric dierence A + B of sets A and B to be the set (A B) (B A). (a) Show that A (B + C) = (A B) + (A C). (b) Show that A + (B + C) = (A + B) + C. Solution. First observe that in general, A (B C) = (A B) (A C). Hence A (B + C) = A [(B C) (C B)] = [A (B C)] [A (C B)]

= (A B) + (A C)

= [(A B) (A C)] [(A C) (A B)]

Part (b) can best be shown with a membership table. First observe that A + (B + C) = A + [(B C) (C B)] and (A + B) + C = [(A B) (B A)] + C

= [A [(B C) (C B)]] [[(B C) (C B)] A]

= [[(A B) (B A)] C] [C [(A B) (B A)]]

From this it is easy to verify the following membership table.

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AXIOMS AND OPERATIONS A B C A+(B+C) (A+B)+C

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Exercise 2.16. Simplify: [(A B C) (A B)] [(A (B C)) A]. Solution. Since (A B) (A B C), it follows that (A B C) (A B) = (A B). For the same reason, (A (B C)) A = A. So the expression simplies to (A B) A, which further simplies to B A. Exercise 2.17. Show that the following four conditions are equivalent. (a) A B, (b) A B = , (c) A B = B, (d) A B = A. Solution. If A B, then there is no element in A that is not also in B, hence A B = . Also, if A B = , then if there is any x A, then x B as well. So (a) and (b) are equivalent. Also (a) is clearly equivalent to both (c) and (d). Exercise 2.18. Assume that A and B are subsets of S. List all of the dierent sets that can be made from these three by use of the binary operations , , and . Solution. A few that spring to mind are A, B, S, A B, A B, A B, B A, S A, S B, S (A B), S (A B), S (A B), S (B A), S S = . Exercise 2.19. Is P(A B) always equal to PA PB? Is it ever equal to PA PB? Solution. Recall that is in the power set of any set. So P(A B), as well as PB. It follows that PA PB, and thus the two will never be equal. Exercise 2.20. Let A, B, and C be sets such that A B = A C and A B = A C. Show that B = C.

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Solution. Take x B. There are two subcases. If x A, then x A B, and thus x A C, so x C. If x A, then x A B regardless, so x A C, from which it follows that x C. In either case, x C, so B C. A symmetric argument shows that C B, and so B = C. Exercise 2.21. Show that (A B) = A B. Solution. Take x (A B). Hence y (A B) such that x y. Now y A or y B, and so x A or x B. In either case, x A B. Take x A B. If x A, then y A such that x y, Since y A B, x (A B) as well. A similar argument handles the case where x B. Exercise 2.22. Show that if A and B are nonempty sets, then (A B) = A B. Solution. Take x (A B). So for all y A B, x y. In particular, for all a A, x a, so x A, and for all b B, x b, so x B. Now take x A B. So x A and x B. Then for any c A B, c A or c B. It follows that x c. Hence x (A B). Exercise 2.23. Show that if B is nonempty, then A B= {A X | X B}. Solution. First observe that for any X B, B X. Hence A B A X for all X B, and thus A B {A X | X B}. For the reverse containment, if x PX for all X A , then x X for all X, and thus x A . Hence x P A . Now take any x {A X | X B}. So x A X for all X B. If x A, the containment holds. Otherwise, if x A, we must have x X for all X B, that is, x B, and the equality follows. Exercise 2.24. (b) Show that {PX | X A } P Under what conditions does equality hold? Solution. Take x P A . So x A . But observe that for any X A , A X, and so x X. That is, x PX, so x {PX | X A }. For (b), take x {PX | X A }. So x PX for some X A . It follows that x X, but X A , so x A , so x P A . Now take x P A . So x A . In order for x {PX | X A }, one needs x X. If A X for some X A , then equality will hold. Exercise 2.25. Is A B always the same as what conditions does equality hold? {A X | X B}? If not, then under A. (a) Show that if A is nonempty, then P A = {PX | X A }.

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Solution. Yes, they are always the same. Take x A B. If x A, then x A X for all X B, so x {A X | X B}. If x B, then x X for some X B, and thus x A X for that particular X. If x {A X | X B}, then x A X for some X B. If x A we are done. If x A, then we must have x X for some X B, and thus x B. Exercise 2.26. Consider the following sets: A = {3, 4}, B = {4, 3} , C = {4, 3} {}, D = {x | x2 7x + 12 = 0}, E = {, 3, 4}, F = {4, 4, 3}, G = {4, , , 3}. For each pair of sets specify whether or not the sets are equal. Solution. First, simplify the sets. We have A = {3, 4}, B = {3, 4}, C = {3, 4, }, D = {3, 4}, E = {, 3, 4}, F = {3, 4}, G = {4, , 3}. From this, it is clear which sets are equal and which are not. Exercise 2.27. Give an example of sets A and B for which A B is nonempty and A B= (A B). A = , so

Solution. Take A = {, {}} and B = {{}}. Then A B = {{}}. But A B = . However, (A B) = {{}} = {}. Exercise 2.28. Simplify: {{3, 4}, {{3}, {4}}, {3, {4}}, {{3}, 4}}. Solution.

{{3, 4}, {{3}, {4}}, {3, {4}}, {{3}, 4}} = {3, 4, {3}, {4}, 3, {4}, {3}, 4} = {3, 4, {3}, {4}} Exercise 2.29. Simplify: (a) (b) {PPP, PP, P, }. {PPP{}, PP{}, P{}}.

Solution. For (a), since is included in the family, any member of the intersection must a member of , which gives {PPP, PP, P, } = . For (b), observe that P{} is a subset of both PPP{} and PP{}, and hence {PPP{}, PP{}, P{}} = P{}. Exercise 2.30. Let A be the set {{}, {{}}}. Evaluate the following: (a) PA, (b) A, A,

(c) P

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AXIOMS AND OPERATIONS (d) PA.

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Solution. Indeed, (a) PA = {, {{}}, {{{}}}, {{}, {{}}}}. (b) A = {, {}}. A = {, {}, {{}}, {, {}}}. (c) P

(d) Recall that taking a union of the power set of some set essentially returns said set. Hence PA = A.

Exercise 2.31. Let B be the set {{1, 2}, {2, 3}, {1, 3}, {}}. Evaluate the following sets. (a) (b) (c) (d) B, B, B, B.

Solution. Indeed, (a) (b) (c) (d) B = {1, 2, 3, }, B = , B= B= {1, 2, 3, } = {1, 2, 3, }, = .

Exercise 2.32. Let S be the set {{a}, {a, b}}. Evaluate and simplify: (a) (b) (c) S, S, S( S S). S = S {a} = a. For (c), rst note S) = (a b) (a b a) = ( S S) when a = b

Solution. First, S = {a, b} = a b, and S = a b and S = a. Thus S( (a b) (b a) = b.

Exercise 2.33. With S as in the preceding exercise, evaluate and when a = b.

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AXIOMS AND OPERATIONS S = {a, b} and S) = S = {a}, so ({b}) = b. S = , so ( S

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Solution. Suppose a = b. Then ( S

({a, b} {a}) = S

If a = b, then S = {{a}} = {{b}}. In either case, = .

S) =

Exercise 2.34. Show that {, {}} PPPS for every set S. Solution. Let S be any set. Since S, we have PS. It follows that {} PS, and since clearly PS, we have , {} PPS. Hence {, {}} PPS so {, {}} PPPS. Exercise 2.35. Assume that PA = PB. Prove that A = B. Solution. Take x A. Then {x} A, so {x} PA = PB. Thus {x} B or x B. So A B and a symmetric argument shows B A, so A = B. Note this show that any set has a unique power set. Exercise 2.36. Verify that for all sets the following are correct. (a) A (A B) = A B. (b) A (A B) = A B. Solution. Take x A (A B). Hence x A but x B, for if x B, then x A B, a contradiction. For the other containment, note A B B, and so A (A B) A B. For (b), take x A (A B), so x A but x A B, which implies x A or x B. Since x A, we must have x B, so x A B. For the reverse containment, if x A B, then x A, and since x B, clearly x A B. So x A (A B. Exercise 2.37. Show that for all sets the following equations hold. (a) (A B) C = (A C) (B C). (b) A (B C) = (A B) (A C). (c) (A B) C = A (B C). Solution. (a) Take x (A B) C. If x A, then x A C. If x B, then x B C. Conversely, if x A C, then x (A B) C, and similarly for the case if x B C. (b) Take x A (B C), if x B C, then either x B or x C. In the rst case, x A B, in the second, x A C. Conversely, if x A B, then x B, so x B C, so x A (B C). If x A C, then x B C. (c) If x (A B) C, then x A, but x B and x C, so x B C. Conversely, if x A (B C), then x B and x C, so x A B and x C.

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RELATIONS AND FUNCTIONS

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Exercise 2.38. Prove that for all sets the following are valid. (a) A C B C A B C. (b) C A C B C A B. Solution. Assuming A C B C, then for any x A B, x C if x A or if x B. Conversely, since both A and B are subsets of A B, we have A C and B C. Now assuming C A C B, for any x C one has x A and x B, so x A B. Conversely, A B A and A B B, and so C A C B.

33.1

Relations and FunctionsOrdered Pairs

Exercise 3.1. Suppose that we attempted to generalize the Kuratowski denitions of orrdered pairs to ordered triples by dening x, y, z

= {{x}, {x, y}, {x, y, z}}.

Show that this denition is unsuccessful by giving examples of objects u, v, w, x, y, z with x, y, z = u, v, w but with either y = v or z = w (or both). Solution. Take x = 1, y = 2, z = 1 and u = 1, v = 2, w = 2. Then x, y, z and u, v, w but z = w. Exercise 3.2. (a) Show that A (B C) = (A B) (A C).

= {{1}, {1, 2}, {1, 2, 1}} = {{1}, {1, 2}} = {{1}, {1, 2}, {1, 2, 2}} = {{1}, {1, 2}}

(b) Show that if A B = A C and A = , then B = C. Solution. Take x, y A (B C). So y B C. If y B, then x, y A B and if y C, then x, y A C. Conversely, if x, y A B, then y B C, so x, y A (B C), and similary for if y C. For (b), take x B, so for some a A, a, x A B. Hence a, x = a , c for some a A and c C. By Theorem 3A, x = c, so x C, and thus B C. A symmetric argument shows C B. Exercise 3.3. Show that A B= {A X | X B}. Solution. Take x, y A B. So y B, and thus there exists some X B such that y X. Hence x, y A X, and thus x, y {A X | X B}. Conversely, if x, y A X for some X B, then y B, and so x, y A B. Exercise 3.4. Show that there is no set to which every ordered pair belongs.

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Solution. Suppose that such a set exists. So in particular, for every set X, the ordered pair X, X = {{X}, {X, X}} = {{X}} is in this set. It follows by a subset axiom that this set contains the set of all singleton sets, which, by Exercise 2.8, has been shown to not exist. Hence we reach a contradiction, so there is no set to which every ordered pair belongs. Exercise 3.5. (a) Assume that A and B are given sets, and show that there exists a set C such that for any y, y C y = {x} B for some x in A.

In other words, show that {{x} B | x A} is a set. (b) With A, B, and C as above, show that A B = C. Solution. First observer that for any x A, {x} B A B. It follows that {x} B P(A B). Hence {{x} B | x A} P(A B). Now since A B is indeed a set, P(A B) is also a set, and thus by a subset axiom, we can construct {w P(A B) | w = {x} B for some x A}. More formally, the following is an axiom, ABDCw(w C w D a(a A w = {x} B)), and we can instantiate with A = A, B = B, and D = P(A B). Furthermore, A B = C. For take y A B. So y = a, b for some a A and b B. In particular, y {a} B, so y {{x} B | x A}. Conversely, take y {{x} B | x A}. So y {x} B for some x A. But {x} B A B, and thus y A B.

3.2

Relations

Exercise 3.6. Show that a set A is a relation i A domA ranA. Solution. Suppose that A is a relation. So by denition, A is a set of ordered pairs. If A is the empty relation, the containment holds trivially. Otherwise, take x, y A. Clearly, there exists y such that x, y A so x dom A, and similarly, there exists x such that x, y A, so y ran A. So A dom A ran A. Conversely, suppose A dom A ran A. So A is a set of ordered pairs, and by denition, a relation. Exercise 3.7. Show that if R is a relation, then d R = R.

Solution. Take xd R. Recall that d R = dom R ran R, and by Lemma 3D, dom R R and ran R R, so by Exercise 2.38, d R R. Conversely, take x R. Hence there exists some y R such that x y and there exists some z R such that y z. Now z = a, b for a dom R and b ran R. If

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y = {a}, then x = a, and so x dom R, and hence x d R. If y {a, b}, then either x = a, whence x dom R, and hence x d R, or x = b, whence x ran R, and hence x d R. Exercise 3.8. Show that for any set A : dom ran A = A = {dom R | R A }, {ran R | R A }.

Solution. Take x dom A. So y x, y A . Since x, y A , for some R A , x, y R, so x dom R, and thus x {dom R | R A }. For some R A , take x dom R, so y x, y R. Hence { x, y } R, but R A , and so { x, y } A , so x, y A , so x dom A . The argument for the range is similar. Take x ran A . So t t, x A. So for some R A , t, x R, so x ran R. Now for any R A , take x R. Hence t t, x R, so by the same reasoning as above, { t, x } R A , and hence t, x A , so x ran A. Exercise 3.9. Discuss the result of replacing the union operation by the intersection operation in the preceding problem. Solution. Suppose x A . So y x, y A , that is, x, y R for all R A . Hence x {dom R | R A }. However, suppose x {dom R | R A }. All one may deduce from this is that for each R A , there exists some yR such that x, yR R, but this does not imply that there is some y such that x, y R for all R A . Indeed, consider A = {{ 2, 3 }, { 2, 5 }}. So 2 dom R for all R A , but there is no single element such that 2, y R for all R. So all we may conclude is that dom A {dom R | R A }.

By similar reasoning, we see that ran A {ran R | R A }.

The reverse containment does not holds, for consider the set A = {{ 2, 3 }, { 5, 3 }}. Then 3ran R for all R A , but there is no such t such that t, 3 ran A.

3.3

n-ary Relations

Exercise 3.10. Show that an ordered 4-tuple is also an ordered m-tuple for every positive integer m less than 4. Solution. First observe that by denition, x, y, z = x, y, z = x, y , z and so

x, y , z = {{ x, y }, { x, y , z}}.

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So just as x, y is just a set, so is x, y, z . Hence by denition x1 , x2 , x3 , x4 = x1 , x2 , x3 , x4 ,

so x1 , x2 , x3 , x4 is an ordered pair, with rst coordinate x1 , x2 , x3 and second coordinate x4 . Note that it is indeed ordered, since x1 , x2 , x3 , x4 = y1 , y2 , y3 , y4 x1 , x2 , x3 , x4 = y1 , y2 , y3 , y4 x 1 = y1 , x 2 = y2 , x3 = y3 , x4 = y 4

x1 , x2 , x3 = y1 , y2 , y3 and x4 = y4 Similarly, by denition, x1 , x2 , x3 , x4 = = x1 , x2 , x3 , x4 x1 , x2 , x3 , x4 ,

where we have applied the fact that x, y, z = x, y , z , with x = x1 , x2 , y = x3 , and z = x4 . It follows that x1 , x2 , x3 , x4 is also an ordered triple. Finally, by denition, x1 , x2 , x3 , x4 = x1 , x2 , x3 , x4 , so it is also a 1-tuple.

3.4

Functions

Exercise 3.11. Prove the following version (for functions) of the extensionality principle: Assume that F and G are functions, dom F = dom G, and F (x) = G(x) for all x in the common domain. Then F = G. Solution. Take x, F (x) F . So x dom F = dom G, and thus x, G(x) G. But F (x) = G(x), and so x, F (x) = x, G(x) , so x, F (x) G and thus F G. A symmetric argument shows G F , and so F = G. Exercise 3.12. Assume that f and g are functions and show that f g dom f dom g (x dom f )f (x) = g(x). Solution. Suppose f g. Now take x dom f . Then x, f (x) f , so x, g(x) g. Thus x dom g, so dom f dom g. Moreover, x, g(x) g so since g is a function, one must have that x, f (x) = x, g(x) , and thus f (x) = g(x). Conversely, take x, f (x) f . Then x dom f , so by assumption, f (x) = g(x). It follows that x, f (x) = x, g(x) , so since x dom g, x, f (x) g, and thus f g. Exercise 3.13. Assume that f and g are functions with f g and dom g dom f . Show that f = g. Solution. By Exercise 3.12, it suces to show that (x dom g)g(x) = f (x) from which the containment g f follows. Take x dom g. Then x dom f so x, f (x) f . Since f g, x, f (x) g. Furthermore, since x dom g, x, g(x) g. Since g is a function, that fact that x, f (x) , x, g(x) g implies that g(x) = f (x), as desired.

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Exercise 3.14. Assume that f and g are functions. (a) Show that f g is a function. (b) Show that f g is a function i f (x) = g(x) for every x in (dom f ) (dom g). Solution. Assume that x, y f g and x, y f g. In particular, x, y f g and x, y f , so since f is a function, y = y , and hence f g is a function. For (b), rst suppose f g is a function. Take any x (dom f ) (dom g). Thus x, f (x) f and x, g(x) g. So x, f (x) , x, g(x) f g, and since f g is a function, f (x) = g(x). Conversely, suppose that x, y and x, y are in f g. There are several cases to consider. If x, y f and x, y f , then y = y since f is a function. Similarly, if both pairs are in g, then y = y since g is a function. Finally, if x, y is in f and the x, y is in g, then x (dom f ) (dom g), and so by hypothesis, y = f (x) = g(x) = y . The same holds if the pairs happen to be in the other set. In any case, y = y , and so f g is a function. Exercise 3.15. Let A be a set of functions such that for any f and g in A , either f g or g f . Show that A is a function. Solution. Take x, y , x, y A . Now for some f A , x, y f and for some g A , x, y g. Without loss of generality, suppose f g, and so x, y g. It follows that y = g(x) = y , and hence A is a function. Exercise 3.16. Show that there is no set to which every function belongs. Solution. Suppose such a set exists, so let F denote the set of all functions. Then by a subset axiom,Cf (f C f F (xy)((x dom f y dom f x = y) (z)(z dom f z, z f ))).

Hence such a subset C consists of functions f : {x} {x} : x x. That is, functions of the form f = { x, x } = {{x}, {x, x}} = {{{x}}}. Note then that f = {x}, and in particular, C is the set of all singletons, which has been shown to not exist. Hence no such set F exists. Exercise 3.17. Show that the composition of two single-rooted sets is again single-rooted. Conclude that the composition of two one-to-one functions is again one-to-one. Solution. Let F and G be two single-rooted sets. Suppose that x(F G)y and x (F G)y. So there exists t and t such that xGt tF y and x Gt t F y. Since F is single-rooted, t = t , and then since G is single-rooted, from xGt and x Gt we see x = x . Hence F G is single-rooted. Since injective functions are just special cases of single-rooted relations, we conclude that the composition of injections is again injective.

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Exercise 3.18. Let R be the set { 0, 1 , 0, 2 , 0, 3 , 1, 2 , 1, 3 , 2, 3 }. Evaluate the following: R R, R {1}, R1 Solution. By inspection, we have R R = { 0, 2 , 0, 3 , 1, 3 }. Also, R {1} = { 1, 2 , 1, 3 }, R1 = { 1, 0 }, R {1} = {2, 3}, and R1 {1} = {0}. Exercise 3.19. Let Evaluate each of the following: A(), A , A {} , A {, {}} , A1 , A A, A {}, A {, {}}, A. Solution. First, note that A is indeed a function. Hence, A() = {, {}}. However, there are no ordered pairs u, v such that u , thus, A = . Since A is a function, A {} = A(). Also, A {, {}} = {A(), A({})} = {{, {}}, }. Flipping the pairs, A1 = { {, {}}, , , {} }. Taking as the intermediary, A A = { {}, {, {}} }. By similar reasons to the computation of A , A = . However, A {} = { , {, {}} }. and A {, {}} = A. A = { , {, {}} , {}, }. , A {1}, R {1} , and R1 {1} .

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Finally, rst note that A = , {, {}} {}, = {{}, {, {, {}}}} {{{}}, {{}, }} = {{}, {, {, {}}}, {{}}, {{}, }} and thus A = {} {, {, {}}} {{}}] {{}, } = {, , {, {}}, {}, {}, } = {, {}, {, {}}}

Exercise 3.20. Show that F

A = F (A ran F ).

Solution. By denition, F A = { u, v | uF v u A}. Hence such u, v F , so F A F . Also, u A and thus v ran F , so u, v A ran F , so F A A ran F , and altogether, F F (A ran F ). Now take u, v F (A ran F ). So u, v A ran F , so u A. Also, u, v F , so uF v. Together, these imply u, v F A, and the equality follows. Exercise 3.21. Show that (R S) T = R (S T ) for any sets R, S, and T . Solution. Take x, y (R S) T . Hence there exists t such that x(R S)t tT y. This implies there exists s such that xRs sSt tT y. So xRs s(S T )y, and thus x, y R (S T ). The reverse containment follows similarly. Exercise 3.22. Show that the following are correct for any sets. (a) A B F A F B . (b) (F G) A = F G A . (c) Q (A B) = (Q A) (Q B). Solution. Indeed, (a) Take v F A . So u, v F for some u A. By assumption, A B, so u B as well, and thus v F B .

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(b) Take v (F G) A . So u(F G)v and u A for some u. Hence there exists some t such that uGt tF v. So t G A , and thus v F G A .

Conversely, take y F G A . So for some x G A , xF y and for some z A, zGx. Then z(F G)y, and so y (F G) A .

(c) Take v Q (A B). So for some u A B, uQv. If u A, v Q A, and if u B, then v Q B. In either case, Q (A B) (Q A) (Q B). Now take y (Q A) (Q B). If y Q A, then there is some x A such that xQy, and hence y Q (A B). The case where y B follows similarly.

Exercise 3.23. Let IA be the identity function on the set A. Show that for any sets B and C, B IA = B A and IA C = A C. Solution. First take x, y B IA . So there exists t such that xIA t tBy. But since IA is the identity function, t = x. So xBy, and so x, y B A. Conversely, take x, y B A. So xBy and x A. In particular, xIA x xBy, so x(B IA )y, that is, x, y B IA . For the other equality, rst take x IA C . So for some c C, c, x IA . Note that c dom IA = A, so c A. And since IA is the identity on A, we must have c = x, so x C, and x ran IA = A, so x A C. Conversely, take x A C. Since x A, x, x IA , but since x C, it follows that x IA C . Exercise 3.24. Show that for a function F , F 1 A = {x dom F | F (x) A}. Solution. Take x F 1 A . So for some y A, y, x F 1 and so x, y F . Thus y = F (x) A, and x dom F , as desired. Conversely, take v {x dom F | F (x) A}. So x, F (u) F , and thus F (u), u F 1 . Since F (u) A, it follows that u F 1 A . Exercise 3.25. (a) Assume that G is a one-to-one function. Show that G G1 is Iran G , the identity function on ran G. (b) Show that the result of part (a) holds for any function G, not necessarily one-to-one. Solution. Suppose that x(G G1 )y. Hence there exists t such that xG1 t tGy. So tGx tGy. Now x ran G, and since G is a function, x = y. Hence G G1 is Iran G . Note that nowhere is injectivity used, and thus (a) holds for any function G. Exercise 3.26. Prove the second halves of parts (a) and (b) of Theorem 3K. That is, prove that F A = {F A | A A } and F A = {F A | A A }.

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Solution. Take v F A . So there exists u A such that uF v. Since u A , there exists an A A such that u A. So v F A for this particular A, so u {F A | A A }. Conversely, take v {F A | A A }. So for some A A , v F A , and thus there is some u A such that uF v. Since u A and A A , u A , and thus v F A . For (b), take v F A . So uF v for some u A . That is, for every A A , u A. It follows that v F A for each A A , and so the containment holds. To see that equality holds when F is single-rooted, take y {F A | A A }. So for each A A , y F A , and thus there exists uA A such that uA , y F . Since F is single-rooted, these uA are all equal, so denoting uA = u, one has u A , and hence y F A . Exercise 3.27. Show that dom (F G) = G1 dom F for any sets F and G. Solution. Take x dom (F G). So there exists y such that x(F G)y. Moreover, there exists t, such that xGttF y, and thus tG1 x. Since t dom F , we have x G1 dom F . Now take x G1 dom F . So for some t dom F , tG1 x, and so xGt. Since t dom F , there exists some y such that tF y, and thus x(F G)y, so x dom (F G). Exercise 3.28. Assume that f is a one-to-one function from A into B, and that G is the function with dom G = PA dened by the equation G(X) = f X . Show that G maps PA one-to-one into PB. Solution. First note that G indeed maps PA into PB. For X PA, G(X) = f X = {v | uf v, u X}, so v B, and thus f X B, so f X PB. Now suppose G(X) = G(Y ) for X, Y PA, and thus f X = f Y . Take x X. Since f is dened on all of A, and X A, we have x, f (x) f , so f (x) f X . Then f (x) f Y , and thus for some y Y , y, f (x) f , but since f is injective, y = x. So x Y , and thus X Y . A parallel argument shows Y X, and so X = Y , and thus G : PA PB is injective. Exercise 3.29. Assume that f : A B and dene a function G : B PA by G(b) = {x A | f (x) = b}. Show that if f maps A onto B, then G is one-to-one. Does the converse hold? Solution. Suppose that G(b) = G(b ). Note that G(b) is nonempty, since f is surjective. Now take x G(b), so f (x) = b. Since x G(b ), f (x) = b also, so b = b . Essentially, and b B has a unique set of preimages. However, the converse does not hold. Take A = {1}, and B = {1, 2}, and suppose G(1) = {1} and G(2) = . So G is injective, however, from the denition of G, we have that f (1) = 1, but 2 has no preimage under f , so f does not map A onto B. Exercise 3.30. Assume that F : PA PA and that F has the monotonicity property: X Y A F (X) F (Y ).

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Dene B= {X A | F (X) X} and C= {X A | X F (X)}.

(a) Show that F (B) = B and F (C) = C. (b) Show that if F (X) = X, then B X C. Solution. Observe, by Theorem 3K, for X A, F (B) = F F (X)X

Hence F (B) B, and thus by the monotonicity property, F (F (B)) F (B), which implies F (B) B, and thus F (B) = B. Again by Theorem 3K, So C F (C), and thus by montonicity, F (C) F (F (C)), from which it follows that F (C) C, so F (C) = C. For (b), if F (X) = X, then F (X) X, and so B X, as X is one of the sets in the intersection. Also, X F (X), so X C, as X is one of the sets in the union. Thus B X C. F (C) = F XF (X)

X

F (X)X

F (X)

X = B.F (X)X

X =

XF (X)

F (X)

X = C.

XF (X)

3.5

Innite Cartesian Products

Exercise 3.31. Show that from the rst form of the axiom of choice we can prove the second form, and conversely. Solution. Assume the rst form. Let I be any set and let H be any function such that dom H = I, and H(i) = for all i I. Dene a relation R I iI H(i) by i, x R x H(i). By assumption, there exists a function G R with dom G = dom R = I, as for each i I, i dom R since H(i) is nonempty. So for all i, G(i) G, i, G(i) R, and thus by the denition of R, G(i) H(i). It follows that G iI H(i), so iI H(i) = . Thus the second form follows from the rst. Conversely, let R be any relation, and denote dom R = I. Dene a function H : I P(ran R) : i H(i) := {x ran R | iRx}. In particular, H is a function with domain I, and H(i) = for all i I. So by the second form, iI H(i) = , so take G iI H(i). Hence dom G = I, and for all i I, G(i) H(i). Also, for any i, G(i) G, G(i) H(i) ran R, and so i, G(i) R, so G R. Hence the two statements of the Axiom of Choice are equivalent.11

Thanks to Arturo Magidin for his hints on this exercise.

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Exercise 3.32. (a) Show that R is symmetric i R1 R. (b) Show that R is transitive i R R R. Solution. Suppose R is symmetric. Take x, y R1 . So xR1 y, and thus yRx, but since R is symmetric, xRy, so x, y R. Conversely, suppose R1 R. Suppose xRy, then yR1 x, and so y, x R1 R, so yRx, and R is symmetric. Now suppose R is transitive. Take x, z R R. Thus there exists y such that xRy yRz, and since R is transitive, xRz, so x, z R. Conversely, suppose R R R, and suppose xRy yRz. Thus x, z R R, so x, z R, and thus xRz, so R is transitive. Exercise 3.33. Show that R is a symmetric and transitive relation i R = R1 R. Solution. First suppose that R is symmetric and transitive. Take x, y R. So xRy, and thus yRx. By transitivity, we have yRy. Thus yR1 y and from xRy y R 1y we have x(R1 R)y, so R R1 R. Now take x, z R1 R. Hence there exists y such that xRy yR1 z. But from yR1 z implies zRy, which implies yRz, as R is symmetric, and thus from xRy yRz we have xRz, and thus R = R1 R. Conversely, assume R = R1 R. Suppose xRy. So x(R1 R)y, and thus there exists z such that xRz zR1 y. Hence xRz zR1 y xRz yRz zR1 x yRz y(R1 R)x yRx and so R is symmetric. Now suppose xRy yRz. Hence x(R1 R)y y(R1 R)z. Hence there exist s and t such that xRs sR1 y and yRt tR1 z. In particular, since R is symmetric, we have sRy yR1 t, and so s(R1 R)t, so sRt. Then observe, sRt tR1 s zRt tR1 s zRs sR1 z1

xRs sR1 z x(R so R is transitive. xRz R)z

since xRs

Exercise 3.34. Assume that A is a nonemptyset, every member of which is a transitive relation. (a) Is the set (b) Is A a transitive relation?

A a transitive relation?

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Solution. Yes, A is a transitive relation. Suppose x, y , y, z A . It follows that for every A A , x, y , y, z A as well, and since A is transitive, x, z A as well. Hence x, z A , so A is transitive. This is not so for A . Suppose A = {A1 , A2 }, where A1 = { a, b , b, c , a, c } and A2 = { b, d , d, e , b, e }. So both A1 and A2 are transitive, but A = { a, b , b, c , a, c , b, d , d, e , b, e }, which is not transitive, since a, d A.

Exercise 3.35. Show that for any R and x, we have [x]R = R {x} . Solution. Take y [x]R . So x, y R and by denition, y R {x} . Conversely, take y R {x} . Then for some t {x}, t, y R. Clearly t = x, so x, y R, and thus y [x]R . Exercise 3.36. Assume that f : A B and that R is an equivalence relation on B. Dene Q to be the set { x, y A A | f (x), f (y) R}. Show that Q is an equivalence relation on A. Solution. For any x A, f (x) B, so f (x), f (x) R. Hence x, x Q, so Q is reexive on A. Now suppose x, y Q. Then f (x), f (y) R, so f (y), f (x) R since R is symmetric, so y, x Q, and hence Q is symmetric. Finally, suppose x, y , y, z Q. So f (x), f (y) , f (y), f (z) R, and thus f (x), f (z) R since R is transitive, so x, z Q and Q is transitive. Thus Q is an equivalence relatoin on A. Exercise 3.37. Assume that is a partition of a set A. Dene the relation R as follows: xR y (B )(x B y B). Show that R is an equivalence relation on A. Solution. For any x A, there exists some B such that x B, but the denition of a partition. Hence xR x, so R is reexive. Now suppose that xR y. So there exists some B such that x B y B, which obviously implies y B x B, so yR x, so R is symmetric. Now suppose xR y and yR z. Hence there exist B, C such that x B y B and y C z C. But note that y B C, and hence we must have B = C, since no two distinct sets in a partition intersect. Thus x C z C, so xR z, and R is transitive.

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Exercise 3.38. Theorem 3P shows that A/R is a partition of A whenever R is an equivalence relation on A. Show that if we start with the equivalence relation R of the preceding exercise, then the partition A/R is just . Solution. Take any [x] A/R . Note for any y [x], there exists a B such that x, y B, by the denition of R . Fix this B. For any z [x], we have that yR z, since R is an equivalence relation. By the reasoning the previous exercise, z B as well. So any two elements of [x] are in this xed B, and thus [x] B. Moreover, for any b B, clearly bR x, and thus b [x]. Hence [x] = B. So it follows that any equivalence class in A/R equals some set in the partition . On the other hand, take any C . Now C is nonempty, so take m C. By denition, this C is a subset of [m], since any element in C are R -related to m. However, by the same reasoning in the above paragraph, this m C, and thus [m] C. This shows that any equivalence class in A/R is equal to some C , and any C is equal to some equivalence class in A/R . So and A/R are families with the same sets as members, and thus the same. Exercise 3.39. Assume that we start with an equivalence relation R on A and dene to be the partition A/R. Show that R , as dened in Exercise 3.37 is just R. Solution. First take x, y R. Then since is a partition consisting of equivalences classes of R, x, y B for some unique B . Then by the denition of R , so x, y R , and thus R R . Now take x, y R , and so there exists some B such that x B and y B. Now B is an equivalence class of R, so B = [x]R = [y]R , and thus by Lemma 3N, xRy, so x, y R. Hence R = R . Exercise 3.40. Dene an equivalence relation R on the set P of positive integers by mRn m and n have the same number of prime factors. Is there a function f : P/R P/R such that f ([n]R ) = [3n]R for each n? Solution. Dene a function F : P P : n 3n. Then recall from Theorem 3Q that such a function f exists i F is compatible with R. However, consider the positive integers 2 and 3. Note 2R3 since both 2 and 3 have exactly one prime factor. However, F (2) = 6 = 2 3 and F (3) = 9 = 32 , so F (2) has 2 prime factors, but F (3) still only has 1. Thus F (2), F (3) R, so F is not compatible with R, so no such function f exists. Exercise 3.41. Let R be the set of real numbers and dene the relation Q on R R by u, v Q x, y i u + y = x + v. (a) Show that Q is an equivalence relation on R R. (b) Is there a function G : (R R)/Q (R R)/Q satisfying the equation G([ x, y ]Q ) = [ x + 2y, y + 2x ]Q ?

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Solution. Q is reexive on R R due to the commutativity of + on R. Suppose u, v Q x, y , so u + y = x + v. Then clearly x + v = u + y, so x, y Q u, v , so Q is symmetric. Now assume u, v Q x, y x, y Q w, z . Thus u+y =x+vx+z =w+y from which we see that u + z = x + v y + z = (x + z y) + v = w + v which implies u, v Q w, z . Hence Q is transitive. Dene a function F : R R R R : x, y x + 2y, y + 2x . Then if u, v Q x, y , we have u + y = x + v. This implies u + y + 2(x + v) = x + v + 2(u + y), which in turn implies (u + 2v) + (y + 2x) = (x + 2y) + (v + 2u). That is, F ( u, v )QF ( x, y ). So F is compatible with Q, and thus there exists such a function G by Theorem 3Q. Exercise 3.42. State precisely the analogous results mentioned in Theorem 3Q. Solution. The analogous results are: Assume that R is an equivalence relation on A and that F : A A A. If F is compatible with R, then there exists a unique F : A/R A/R A/R such that F ([x], [y]) = [F (x, y)] for all x, y A. If F is not compatible with R, then no such F exists. Of course, we must extend the denition of compatibility. Note that we would like to have the following commutative diagram: AA F

A/R A/R A/R F

A

From this, we see that if x, y and u, v have the same image under A A (A/R) (A/R), that is, [x], [y] = [u], [v] , then we we also like [F x, y ] = [F u, v ]. This suggests that we dene that F is compatible with R if for any x, y, u, v R, xRy uRv = F x, y RF u, v .

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Assume that F is compatible with R, and we will prove that such a F exists. We , so dene F as require that [x], [y] , [F (x, y)] F F = { [x], [y] , [F (x, y)] | x, y A}. To see that F to indeed be a function, consider the pairs [u], [v] , [F u, v ] in F . Then we have the implications [x], [y] = [u], [v] = [x] = [u] [y] = [v] = F x, y RF u, v = [F x, y ] = [F u, v ] So F is a function. It is clear that dom F = A/R A/R and ran F A/R. Also F [x], [y] = [F x, y ], since F [x], [y] = [F x, y ] F . Also, F is unique, for suppose for some G : A/R A/R A/R the same condition holds. Then for any x, y A, G [x], [y] = [F [x], [y] ] = F [x], [y] so F = G. Now suppose that F is not compatible. By incompatibility, there are some pairs x, y , u, v A A such that xRu yRv but it is not case that F x, y RF u, v . Hence [x] = [u], [y] = [v], but [F x, y ] = [F u, v ]. But we must have F [x], [y] = [F x, y ] and F [u], [v] = [F u, v ] = xRu yRv [x], [y] , [F x, y ] and

which is impossible since the left sides are equal, since F is assumed to be a function, but the right sides are not equal.

3.6

Ordering Relations

Exercise 3.43. Assume that R is a linear ordering on a set A. Show that R1 is also a linear ordering on A. Solution. Suppose that xR1 y yR1 z. Then yRx zRy, and since R is transitive, zRx and so xR1 z. Furthermore, for any x, y, exactly one of xRy, x = y, or yRx holds, and thus exactly one of yR1 x, x = y, or xR1 y holds. Exercise 3.44. Assume that < is a linear ordering on a set A. Assume that f : A A and that f has the property that whenever x < y, then f (x) < f (y). Show that f is one-to-one and that whenever f (x) < f (y), then x < y. Solution. Suppose that f (x) = f (y). So f (x) < f (y), and thus x < y. Similarly, f (y) < f (x), and thus y < x. Since < satisies the trichotomy property, we must have that x = y, so f is injective. Now suppose that f (x) < f (y), but x < y. If x = y, then f (x) = f (y), a contradiction. If y < x, then f (y) < f (x), again a contradiction to the trichotomy property. Hence x < y is the only possibility.

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Exercise 3.45. Assume that

enderton set theory solutions

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