Solutions to Elements of Set Theory ∗ [email protected] January 14, 2011
Contents 1 Introduction
1
2 Axioms and Ope...

Author:
Herbert B. Enderton

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

Contents 1 Introduction

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2 Axioms and Operations

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3 Relations and Functions 3.1 Ordered Pairs . . . . . . . . 3.2 Relations . . . . . . . . . . 3.3 n-ary Relations . . . . . . . 3.4 Functions . . . . . . . . . . 3.5 Infinite Cartesian Products 3.6 Ordering Relations . . . . .

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11 11 12 13 14 20 25

4 Natural Numbers 4.1 Inductive Sets . . . 4.2 Peano’s Postulates 4.3 Recursion on ω . . 4.4 Arithmetic . . . . . 4.5 Ordering on ω . . . 4.6 Review Exercise .

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30 30 30 31 33 36 39

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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 I’m just learning as I write these.

1

1

INTRODUCTION (b) {∅}

2

{∅, {{∅}}}.

(c) {{∅}}

{∅, {∅}}.

(d) {{∅}}

{∅, {{∅}}}.

(e) {{∅}}

{∅, {∅, {∅}}}.

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 {∅} ! ∅, and {{∅}} ! ∅. Also, {{∅}} ! {∅}. Hence no two are equal. 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. Define the rank of a set c to be the least α such that c ⊆ Vα . Compute the rank of {{∅}}. Compute the rank of {∅, {∅}, {∅, {∅}}}. Solution. Observe that Vα+1 = PVα . 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 Vα+1 = A ∪ PVα . Prove this at least for α < 3. Solution. By definition, for α = 0, V1 = V0 ∪ PV0 = A ∪ PV0 . For α = 1, V2 = V1 ∪ PV1 = A ∪ PV0 ∪ PV1 = A ∪ PV1 .

The last equality follows from Exercise 1.3, as V0 ⊆ V1 , and thus PV0 ∪ PV1 = PV1 . The case for α = 2 follows similarly.

2

AXIOMS AND OPERATIONS

3

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 A = B but A '= B. 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 ∈ A). ! This is precisely the definition that b ⊆ A. ! ! Exercise 2.4. Show that if A ⊆ B, then A ⊆ B. ! ! Solution. Recall that x ∈ A ⇔ (∃b ∈ A) x ∈ b. Take x ∈ A. So by definition, there!exists b ∈ A such that x ∈ b. Since A ⊆ B, b ∈ B as well. Then by definition, 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 definition 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 definition 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 definition ! 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.

2

AXIOMS AND OPERATIONS

Exercise 2.7.

4

(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? 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 X is the set of all sets, which leads to Russell’s paradox. 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 ! that is, a ⊆ B. Then observe ! 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 ⊆ (A∪B)∪(A−B) is obvious. Now take x ∈ (A∩B)∪(A−B). If x ∈ A∩B, 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 first 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).

2

AXIOMS AND OPERATIONS

5

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 different 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. Define the symmetric difference 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 ∩ C) − (A ∩ B)]

= (A ∩ B) + (A ∩ C)

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

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

and (A + B) + C = [(A − B) ∪ (B − A)] + C

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

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

2

AXIOMS AND OPERATIONS A ! ! ! !

B ! ! ! !

6 C ! ! ! !

A+(B+C) !

(A+B)+C !

! ! !

! ! !

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 simplifies to (A ∪ B) − A, which further simplifies 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 different 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.

2

AXIOMS AND OPERATIONS

7

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 Since y ∈ A ∪ B, ! x ∈ A ∪ B. If x ∈ A, then ∃y ∈ A such that x ∈ y, ! 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. (a) Show that if A is nonempty, then P A = {PX | X ∈ A }. (b) Show that

"

{PX | X ∈ A } ⊆ P

"

A.

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 {A ∪ X | X ∈ B}? If not, then under what conditions does equality hold?

2

AXIOMS AND OPERATIONS

8

! 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).

Solution. # # Take A = {∅, {∅}} # and B =#{{∅}}. Then A ∩ B = {{∅}}. But A ∩ B = ∅. However, (A ∩ B) = {{∅}} = {∅}. Exercise 2.28. Simplify: "

Solution. "

#

A = ∅, so

{{3, 4}, {{3}, {4}}, {3, {4}}, {{3}, 4}}.

{{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) {PPP∅, PP∅, P∅, ∅}. # (b) {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, ! (c) P A,

2

AXIOMS AND OPERATIONS (d)

!

9

PA.

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

(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, # (b) B, #! (c) B, !# (d) B.

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

Exercise 2.32. Let S be the set {{a}, {a, b}}. Evaluate and simplify: !! (a) S, ## (b) S, #! !! !# (c) S∪( S− S). !! ! ## # Solution. First, ! # S = {a, b} = # a! ∪ b, and! ! S ! = # {a} = a. For (c), first note # ! S = a ∩ b and S = a. Thus S∪( S− S) = (a ∩ b) ∪ (a ∪ b − a) = (a ∩ b) ∪ (b − a) = b. !! # Exercise 2.33. With S as in the preceding exercise, evaluate ( S − S) when a '= b and when a = b.

2

AXIOMS AND OPERATIONS

10

! # Solution. Suppose a '= b. Then S = {a, b} and S = {a}, so "" $ " " ( S− S) = ({a, b} − {a}) = ({b}) = b.

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

!

S−

#

S = ∅, so

!! # ( S − 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 first 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.

3

RELATIONS AND FUNCTIONS

11

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.

3 3.1

Relations and Functions Ordered Pairs

Exercise 3.1. Suppose that we attempted to generalize the Kuratowski definitions of orrdered pairs to ordered triples by defining /x, y, z0∗ = {{x}, {x, y}, {x, y, z}}. Show that this definition is unsuccessful by giving examples of objects u, v, w, x, y, z with /x, y, z0∗ = /u, v, w0∗ 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, z0∗ = {{1}, {1, 2}, {1, 2, 1}} = {{1}, {1, 2}} and /u, v, w0∗ = {{1}, {1, 2}, {1, 2, 2}} = {{1}, {1, 2}} but z '= w. Exercise 3.2.

(a) Show that A × (B ∪ C) = (A × B) ∪ (A × C).

(b) Show that if A × B = A × C and A '= ∅, then B = C. Solution. Take /x, y0 ∈ A × (B ∪ C). So y ∈ B ∪ C. If y ∈ B, then /x, y0 ∈ A × B and if y ∈ C, then /x, y0 ∈ A × C. Conversely, if /x, y0 ∈ A × B, then y ∈ B ∪ C, so /x, y0 ∈ A × (B ∪ C), and similary for if y ∈ C. For (b), take x ∈ B, so for some a ∈ A, /a, x0 ∈ A × B. Hence /a, x0 = /a" , c0 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, y0 ∈ A × B. So y ∈ B, and thus ! there exists some X ∈ B such that y ∈ X. Hence /x, y0 ∈ A × X, and thus /x, y0 ∈ {A × X | X ! ∈ B}. Conversely, if ! /x, y0 ∈ A × X for some X ∈ B, then y ∈ B, and so /x, y0 ∈ A × B. Exercise 3.4. Show that there is no set to which every ordered pair belongs.

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12

Solution. Suppose that such a set exists. So in particular, for every set X, the ordered pair /X, X0 = {{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, ∀A∀B∀D∃C∀w(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, b0 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 iff A ⊆ domA × ranA. Solution. Suppose that A is a relation. So by definition, A is a set of ordered pairs. If A is the empty relation, the containment holds trivially. Otherwise, take /x, y0 ∈ A. Clearly, there exists y such that /x, y0 ∈ A so x ∈ dom A, and similarly, there exists x such that /x, y0 ∈ 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 definition, a relation. !! Exercise 3.7. Show that if R is a relation, then fld R = R.

Solution. Take xfld!R.! Recall that fld R = dom R ∪ ran ! R,!and by Lemma 3D, dom R ⊆ !! R and ran R ⊆ R, so by Exercise 2.38, fld 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, b0 for a ∈ dom R and b ∈ ran R. If

3

RELATIONS AND FUNCTIONS

13

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

" " ran A = {ran R | R ∈ A }. ! ! ! Solution. Take x ∈ dom A. So ∃y/x, y0!∈ A . Since /x, y0 ∈ A , for some R ∈ A , /x, y0 ∈ R, so x ∈ dom R, and thus x ∈ {dom R | R ∈ ! A }. For some R ∈ A ,!take x ∈ dom R,!so ∃y/x, y0 ∈ R. ! Hence {/x, y0} ⊆ R, but R ⊆ A , and so {/x, y0} ⊆ A , so /x, y0 ∈ A , so x ∈ dom A . ! ! The argument for the range is similar. Take x ∈ ran A . So ∃t/t, x0 ∈ A. So for some R ∈ A , /t, x0 ∈ R, so x ∈ ran R. Now for any R ∈ A !, take x ∈ R. Hence ∃t/t, x0 ∈ R, so by the same reasoning as above, {/t, x0} ⊆ R ⊆ A , and hence ! ! /t, x0 ∈ 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, y0 ∈ A# , that is, /x, y0 ∈ 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 0 ∈ R, but this does not imply that there is some y such that /x, y0 ∈ R for all R ∈ A . Indeed, consider A = {{/2, 30}, {/2, 50}}. So 2 ∈ dom R for all R ∈ A , but there is no single element such that /2, y0 ∈ 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, # 30}, {/5, 30}}. Then 3ran R for all R ∈ A , but there is no such t such that /t, 30 ∈ 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 definition, /x, y, z0 = //x, y0, z0 and so /x, y, z0 = //x, y0, z0 = {{/x, y0}, {/x, y0, z}}.

3

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14

So just as /x, y0 is just a set, so is /x, y, z0. Hence by definition /x1 , x2 , x3 , x4 0 = //x1 , x2 , x3 0, x4 0, so /x1 , x2 , x3 , x4 0 is an ordered pair, with first coordinate /x1 , x2 , x3 0 and second coordinate x4 . Note that it is indeed ordered, since //x1 , x2 , x3 0, x4 0 = //y1 , y2 , y3 0, y4 0 ⇔ /x1 , x2 , x3 , x4 0 = /y1 , y2 , y3 , y4 0

⇔ x 1 = y1 , x 2 = y2 , x3 = y3 , x4 = y 4

⇔ /x1 , x2 , x3 0 = /y1 , y2 , y3 0 and x4 = y4 Similarly, by definition, /x1 , x2 , x3 , x4 0 = ///x1 , x2 0, x3 0, x4 0 = //x1 , x2 0, x3 , x4 0,

where we have applied the fact that /x, y, z0 = //x, y0, z0, with x = /x1 , x2 0, y = x3 , and z = x4 . It follows that /x1 , x2 , x3 , x4 0 is also an ordered triple. Finally, by definition, //x1 , x2 , x3 , x4 00 = /x1 , x2 , x3 , x4 0, 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)0 ∈ F . So x ∈ dom F = dom G, and thus /x, G(x)0 ∈ G. But F (x) = G(x), and so /x, F (x)0 = /x, G(x)0, so /x, F (x)0 ∈ 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)0 ∈ f , so /x, g(x)0 ∈ g. Thus x ∈ dom g, so dom f ⊆ dom g. Moreover, /x, g(x)0 ∈ g so since g is a function, one must have that /x, f (x)0 = /x, g(x)0, and thus f (x) = g(x). Conversely, take /x, f (x)0 ∈ f . Then x ∈ dom f , so by assumption, f (x) = g(x). It follows that /x, f (x)0 = /x, g(x)0, so since x ∈ dom g, /x, f (x)0 ∈ 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 suffices 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)0 ∈ f . Since f ⊆ g, /x, f (x)0 ∈ g. Furthermore, since x ∈ dom g, /x, g(x)0 ∈ g. Since g is a function, that fact that /x, f (x)0, /x, g(x)0 ∈ g implies that g(x) = f (x), as desired.

3

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15

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 iff f (x) = g(x) for every x in (dom f ) ∩ (dom g). Solution. Assume that /x, y0 ∈ f ∩ g and /x, y " 0 ∈ f ∩ g. In particular, /x, y0 ∈ f ∩ g and /x, y " 0 ∈ f , so since f is a function, y = y " , and hence f ∩ g is a function. For (b), first suppose f ∪ g is a function. Take any x ∈ (dom f ) ∩ (dom g). Thus /x, f (x)0 ∈ f and /x, g(x)0 ∈ g. So /x, f (x)0, /x, g(x)0 ∈ f ∪ g, and since f ∪ g is a function, f (x) = g(x). Conversely, suppose that /x, y0 and /x, y " 0 are in f ∪ g. There are several cases to consider. If /x, y0 ∈ f and /x, y " 0 ∈ 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, y0 is in f and the /x, y " 0 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, y0, /x, y " 0 ∈ A . Now for some f ∈ A , /x, y0 ∈ f and for some g ∈ A , /x, y " 0 ∈ g. Without loss of generality, suppose f ⊆ g, and so /x, y0 ∈ 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, ∃C∀f (f ∈ C ⇔ f ∈ F ∧ (∀x∀y)((x ∈ dom f ∧ y ∈ dom f ⇒ x = y) ∧ (∀z)(z ∈ dom f ⇒ /z, z0 ∈ f ))).

Hence such a subset C consists of functions f : {x} → {x} : x 4→ x. That is, functions of the form f = {/x, x0} = {{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.

3

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16

Exercise 3.18. Let R be the set {/0, 10, /0, 20, /0, 30, /1, 20, /1, 30, /2, 30}. Evaluate the following: R ◦ R, R " {1}, R−1 " {1}, R!{1}", and R−1 !{1}". Solution. By inspection, we have

R ◦ R = {/0, 20, /0, 30, /1, 30}. Also, R " {1} = {/1, 20, /1, 30}, R−1 "= {/1, 00}, R!{1}" = {2, 3}, and R−1 !{1}" = {0}.

Exercise 3.19. Let

A = {/∅, {∅, {∅}}0, /{∅}, ∅0}.

−1 Evaluate each of !the ! following: A(∅), A!∅", A!{∅}", A!{∅, {∅}}", A , A ◦ A, A " ∅, A " {∅}, A " {∅, {∅}}, A.

Solution. First, note that A is indeed a function. Hence, A(∅) = {∅, {∅}}.

However, there are no ordered pairs /u, v0 such that u ∈ ∅, thus, A!∅" = ∅. Since A is a function, A!{∅}" = A(∅). Also, A!{∅, {∅}}" = {A(∅), A({∅})} = {{∅, {∅}}, ∅}. Flipping the pairs, A−1 = {/{∅, {∅}}, ∅0, /∅, {∅}0}. Taking ∅ as the intermediary, A ◦ A = {/{∅}, {∅, {∅}}0}. By similar reasons to the computation of A!∅", A " ∅ = ∅. However, and

A " {∅} = {/∅, {∅, {∅}}0}. A " {∅, {∅}} = A.

3

RELATIONS AND FUNCTIONS

Finally, first note that "

17

A = /∅, {∅, {∅}}0 ∪ /{∅}, ∅0 = {{∅}, {∅, {∅, {∅}}}} ∪ {{{∅}}, {{∅}, ∅}} = {{∅}, {∅, {∅, {∅}}}, {{∅}}, {{∅}, ∅}}

and thus ""

A = {∅} ∪ {∅, {∅, {∅}}} ∪ {{∅}}] ∪ {{∅}, ∅} = {∅, ∅, {∅, {∅}}, {∅}, {∅}, ∅} = {∅, {∅}, {∅, {∅}}}

Exercise 3.20. Show that F " A = F ∩ (A × ran F ). Solution. By definition, F " A = {/u, v0 | uF v ∧ u ∈ A}. Hence such /u, v0 ∈ F , so F " A ⊆ F . Also, u ∈ A and thus v ∈ ran F , so /u, v0 ∈ A × ran F , so F " A ⊆ A × ran F , and altogether, F "⊆ F ∩ (A × ran F ). Now take /u, v0 ∈ F ∩ (A × ran F ). So /u, v0 ∈ A × ran F , so u ∈ A. Also, /u, v0 ∈ F , so uF v. Together, these imply /u, v0 ∈ 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, y0 ∈ (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, y0 ∈ 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, v0 ∈ F for some u ∈ A. By assumption, A ⊆ B, so u ∈ B as well, and thus v ∈ F !B".

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18

(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, y0 ∈ 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, y0 ∈ B " A. Conversely, take /x, y0 ∈ B " A. So xBy and x ∈ A. In particular, xIA x ∧ xBy, so x(B ◦ IA )y, that is, /x, y0 ∈ B ◦ IA . For the other equality, first take x ∈ IA !C". So for some c ∈ C, /c, x0 ∈ 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, x0 ∈ 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, x0 ∈ F −1 and so /x, y0 ∈ 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)0 ∈ F , and thus /F (u), u0 ∈ 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 ◦ G−1 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 ◦ G−1 )y. Hence there exists t such that xG−1 t ∧ tGy. So tGx ∧ tGy. Now x ∈ ran G, and since G is a function, x = y. Hence G ◦ G−1 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 ,# y0 ∈ 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) = G−1 !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 xGt∧tF y, and thus tG−1 x. Since t ∈ dom F , we have x ∈ G−1 !dom F ". Now take x ∈ G−1 !dom F ". So for some t ∈ dom F , tG−1 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 defined 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 defined on all of A, and X ⊆ A, we have /x, f (x)0 ∈ f , so f (x) ∈ f !X". Then f (x) ∈ f !Y ", and thus for some y ∈ Y , /y, f (x)0 ∈ 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 define 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 definition 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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20

Define B=

$

{X ⊆ A | F (X) ⊆ X}

and

C=

(a) Show that F (B) = B and F (C) = C.

"

{X ⊆ A | X ⊆ F (X)}.

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

F (X)⊆X

$

X = B.

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, " " " F (C) = F X = F (X) ⊇ X = C. X⊆F (X)

X⊆F (X)

X⊆F (X)

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.

3.5

Infinite Cartesian Products

Exercise 3.31. Show that from the first form of the axiom of choice we can prove the second form, and conversely. Solution. Assume the first form. Let I be any set and let H be ! any function such that dom H = I, and H(i) '= ∅ for all i ∈ I. Define a relation R ⊆ I × i∈I H(i) by /i, x0 ∈ 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)0)∈ G, /i, G(i)0)∈ R, and thus by the definition of R, G(i) ∈ H(i). It follows that G ∈ i∈I H(i), so i∈I H(i) '= ∅. Thus the second form follows from the first. Conversely, let R be any relation, and denote dom R = I. Define a function H : I → P(ran R) : i 4→ 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, i∈I H(i) '= ∅, so take G ∈ i∈I H(i). Hence dom G = I, and for all i ∈ I, G(i) ∈ H(i). Also, for any /i, G(i)0 ∈ G, G(i) ∈ H(i) ⊆ ran R, and so /i, G(i)0 ∈ R, so G ⊆ R. Hence the two statements of the Axiom of Choice are equivalent.1 1

Thanks to Arturo Magidin for his hints on this exercise.

3

RELATIONS AND FUNCTIONS

21

Exercise 3.32. (a) Show that R is symmetric iff R−1 ⊆ R. (b) Show that R is transitive iff R ◦ R ⊆ R. Solution. Suppose R is symmetric. Take /x, y0 ∈ R−1 . So xR−1 y, and thus yRx, but since R is symmetric, xRy, so /x, y0 ∈ R. Conversely, suppose R−1 ⊆ R. Suppose xRy, then yR−1 x, and so /y, x0 ∈ R−1 ⊆ R, so yRx, and R is symmetric. Now suppose R is transitive. Take /x, z0 ∈ R ◦ R. Thus there exists y such that xRy ∧ yRz, and since R is transitive, xRz, so /x, z0 ∈ R. Conversely, suppose R ◦ R ⊆ R, and suppose xRy ∧ yRz. Thus /x, z0 ∈ R ◦ R, so /x, z0 ∈ R, and thus xRz, so R is transitive. Exercise 3.33. Show that R is a symmetric and transitive relation iff R = R−1 ◦ R. Solution. First suppose that R is symmetric and transitive. Take /x, y0 ∈ R. So xRy, and thus yRx. By transitivity, we have yRy. Thus yR−1 y and from xRy ∧ y R −1y we have x(R−1 ◦ R)y, so R ⊆ R−1 ◦ R. Now take /x, z0 ∈ R−1 ◦ R. Hence there exists y such that xRy ∧ yR−1 z. But from yR−1 z implies zRy, which implies yRz, as R is symmetric, and thus from xRy ∧ yRz we have xRz, and thus R = R−1 ◦ R. Conversely, assume R = R−1 ◦ R. Suppose xRy. So x(R−1 ◦ R)y, and thus there exists z such that xRz ∧ zR−1 y. Hence xRz ∧ zR−1 y ⇒ xRz ∧ yRz ⇒ zR−1 x ∧ yRz ⇒ y(R−1 ◦ R)x ⇒ yRx and so R is symmetric. Now suppose xRy ∧ yRz. Hence x(R−1 ◦ R)y ∧ y(R−1 ◦ R)z. Hence there exist s and t such that xRs ∧ sR−1 y and yRt ∧ tR−1 z. In particular, since R is symmetric, we have sRy ∧ yR−1 t, and so s(R−1 ◦ R)t, so sRt. Then observe, sRt ⇒ tR−1 s ⇒ zRt ∧ tR−1 s

⇒ zRs ⇒ sR−1 z

⇒ xRs ∧ sR−1 z ⇒ x(R

−1

⇒ xRz

since xRs

◦ R)z

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

3

RELATIONS AND FUNCTIONS

22

# # Solution. Yes, A is a transitive relation. Suppose /x, y0, /y, z0 ∈ A . It follows that for every A ∈ A # , /x, y0,#/y, z0 ∈ A as well, and since A is transitive, /x, z0 ∈ A as well. Hence /x, z0 ∈ A , so ! A is transitive. This is not so for A . Suppose A = {A1 , A2 }, where A1 = {/a, b0, /b, c0, /a, c0}

and A2 = {/b, d0, /d, e0, /b, e0}. So both A1 and A2 are transitive, but " A = {/a, b0, /b, c0, /a, c0, /b, d0, /d, e0, /b, e0}, which is not transitive, since /a, d0 '∈

!

A.

Exercise 3.35. Show that for any R and x, we have [x]R = R!{x}". Solution. Take y ∈ [x]R . So /x, y0 ∈ R and by definition, y ∈ R!{x}". Conversely, take y ∈ R!{x}". Then for some t ∈ {x}, /t, y0 ∈ R. Clearly t = x, so /x, y0 ∈ R, and thus y ∈ [x]R . Exercise 3.36. Assume that f : A → B and that R is an equivalence relation on B. Define Q to be the set {/x, y0 ∈ A × A | /f (x), f (y)0 ∈ R}. Show that Q is an equivalence relation on A. Solution. For any x ∈ A, f (x) ∈ B, so /f (x), f (x)0 ∈ R. Hence /x, x0 ∈ Q, so Q is reflexive on A. Now suppose /x, y0 ∈ Q. Then /f (x), f (y)0 ∈ R, so /f (y), f (x)0 ∈ R since R is symmetric, so /y, x0 ∈ Q, and hence Q is symmetric. Finally, suppose /x, y0, /y, z0 ∈ Q. So /f (x), f (y)0, /f (y), f (z)0 ∈ R, and thus /f (x), f (z)0 ∈ R since R is transitive, so /x, z0 ∈ Q and Q is transitive. Thus Q is an equivalence relatoin on A. Exercise 3.37. Assume that Π is a partition of a set A. Define the relation RΠ as follows: xRΠ y ⇔ (∃B ∈ Π)(x ∈ B ∧ y ∈ B). Show that RΠ is an equivalence relation on A. Solution. For any x ∈ A, there exists some B ∈ Π such that x ∈ B, but the definition of a partition. Hence xRΠ x, so RΠ is reflexive. Now suppose that xRΠ y. So there exists some B ∈ Π such that x ∈ B ∧ y ∈ B, which obviously implies y ∈ B ∧ x ∈ B, so yRΠ x, so RΠ is symmetric. Now suppose xRΠ y and yRΠ z. Hence there exist B, C ∈ Π 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 xRΠ z, and RΠ is transitive.

3

RELATIONS AND FUNCTIONS

23

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 RΠ of the preceding exercise, then the partition A/RΠ is just Π. Solution. Take any [x] ∈ A/RΠ . Note for any y ∈ [x], there exists a B ∈ Π such that x, y ∈ B, by the definition of RΠ . Fix this B. For any z ∈ [x], we have that yRΠ z, since RΠ is an equivalence relation. By the reasoning the previous exercise, z ∈ B as well. So any two elements of [x] are in this fixed B, and thus [x] ⊆ B. Moreover, for any b ∈ B, clearly bRΠ x, and thus b ∈ [x]. Hence [x] = B. So it follows that any equivalence class in A/RΠ equals some set in the partition Π. On the other hand, take any C ∈ Π. Now C is nonempty, so take m ∈ C. By definition, this C is a subset of [m], since any element in C are RΠ -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/RΠ is equal to some C ∈ Π, and any C ∈ Π is equal to some equivalence class in A/RΠ . So Π and A/RΠ 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 define Π to be the partition A/R. Show that RΠ , as defined in Exercise 3.37 is just R. Solution. First take /x, y0 ∈ R. Then since Π is a partition consisting of equivalences classes of R, x, y ∈ B for some unique B ∈ Π. Then by the definition of RΠ , so /x, y0 ∈ RΠ , and thus R ⊆ RΠ . Now take /x, y0 ∈ RΠ , and so there exists some B ∈ Π 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, y0 ∈ R. Hence R = RΠ . Exercise 3.40. Define 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. Define a function F : P → P : n 4→ 3n. Then recall from Theorem 3Q that such a function f exists iff 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)0 '∈ 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 define the relation Q on R × R by /u, v0Q/x, y0 iff 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, y0]Q ) = [/x + 2y, y + 2x0]Q ?

3

RELATIONS AND FUNCTIONS

24

Solution. Q is reflexive on R × R due to the commutativity of + on R. Suppose /u, v0Q/x, y0, so u + y = x + v. Then clearly x + v = u + y, so /x, y0Q/u, v0, so Q is symmetric. Now assume /u, v0Q/x, y0 ∧ /x, y0Q/w, z0. Thus u+y =x+v∧x+z =w+y from which we see that u + z = x + v − y + z = (x + z − y) + v = w + v which implies /u, v0Q/w, z0. Hence Q is transitive. Define a function F : R × R → R × R : /x, y0 4→ /x + 2y, y + 2x0. Then if /u, v0Q/x, y0, 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, v0)QF (/x, y0). 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 Fˆ : A/R × A/R → A/R such that Fˆ ([x], [y]) = [F (x, y)]

for all x, y ∈ A.

If F is not compatible with R, then no such Fˆ exists. Of course, we must extend the definition of compatibility. Note that we would like to have the following commutative diagram: A×A +

F

−−−−→

A +

A/R × A/R −−−−→ A/R Fˆ

From this, we see that if /x, y0 and /u, v0 have the same image under A × A → (A/R) × (A/R), that is, /[x], [y]0 = /[u], [v]0, then we we also like [F /x, y0] = [F /u, v0]. This suggests that we define that F is compatible with R if for any x, y, u, v ∈ R, xRy ∧ uRv =⇒ F /x, y0RF /u, v0.

3

RELATIONS AND FUNCTIONS

25

Assume that F is compatible with R, and we will prove that such a Fˆ exists. We require that //[x], [y]0, [F (x, y)]0 ∈ Fˆ , so define Fˆ as Fˆ = {//[x], [y]0, [F (x, y)]0 | x, y ∈ A}. To see that Fˆ to indeed be a function, consider the pairs //[x], [y]0, [F /x, y0]0 and //[u], [v]0, [F /u, v0]0 in Fˆ . Then we have the implications /[x], [y]0 = /[u], [v]0 =⇒ [x] = [u] ∧ [y] = [v] =⇒ xRu ∧ yRv

=⇒ F /x, y0RF /u, v0

=⇒ [F /x, y0] = [F /u, v0]

So Fˆ is a function. It is clear that dom Fˆ = A/R × A/R and ran Fˆ ⊆ A/R. Also Fˆ /[x], [y]0 = [F /x, y0], since Fˆ /[x], [y]0 = [F /x, y0] ∈ Fˆ . Also, Fˆ 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]0 = [F /[x], [y]0] = Fˆ /[x], [y]0 so F = G. Now suppose that F is not compatible. By incompatibility, there are some pairs /x, y0, /u, v0 ∈ A × A such that xRu ∧ yRv but it is not case that F /x, y0RF /u, v0. Hence [x] = [u], [y] = [v], but [F /x, y0] '= [F /u, v0]. But we must have Fˆ /[x], [y]0 = [F /x, y0]

and Fˆ /[u], [v]0 = [F /u, v0]

which is impossible since the left sides are equal, since Fˆ 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 R−1 is also a linear ordering on A. Solution. Suppose that xR−1 y ∧ yR1− z. Then yRx ∧ zRy, and since R is transitive, zRx and so xR−1 z. Furthermore, for any x, y, exactly one of xRy, x = y, or yRx holds, and thus exactly one of yR−1 x, x = y, or xR−1 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 < satisifies 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.

3

RELATIONS AND FUNCTIONS

26

Exercise 3.45. Assume that

Contents 1 Introduction

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2 Axioms and Operations

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3 Relations and Functions 3.1 Ordered Pairs . . . . . . . . 3.2 Relations . . . . . . . . . . 3.3 n-ary Relations . . . . . . . 3.4 Functions . . . . . . . . . . 3.5 Infinite Cartesian Products 3.6 Ordering Relations . . . . .

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4 Natural Numbers 4.1 Inductive Sets . . . 4.2 Peano’s Postulates 4.3 Recursion on ω . . 4.4 Arithmetic . . . . . 4.5 Ordering on ω . . . 4.6 Review Exercise .

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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 I’m just learning as I write these.

1

1

INTRODUCTION (b) {∅}

2

{∅, {{∅}}}.

(c) {{∅}}

{∅, {∅}}.

(d) {{∅}}

{∅, {{∅}}}.

(e) {{∅}}

{∅, {∅, {∅}}}.

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 {∅} ! ∅, and {{∅}} ! ∅. Also, {{∅}} ! {∅}. Hence no two are equal. 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. Define the rank of a set c to be the least α such that c ⊆ Vα . Compute the rank of {{∅}}. Compute the rank of {∅, {∅}, {∅, {∅}}}. Solution. Observe that Vα+1 = PVα . 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 Vα+1 = A ∪ PVα . Prove this at least for α < 3. Solution. By definition, for α = 0, V1 = V0 ∪ PV0 = A ∪ PV0 . For α = 1, V2 = V1 ∪ PV1 = A ∪ PV0 ∪ PV1 = A ∪ PV1 .

The last equality follows from Exercise 1.3, as V0 ⊆ V1 , and thus PV0 ∪ PV1 = PV1 . The case for α = 2 follows similarly.

2

AXIOMS AND OPERATIONS

3

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 A = B but A '= B. 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 ∈ A). ! This is precisely the definition that b ⊆ A. ! ! Exercise 2.4. Show that if A ⊆ B, then A ⊆ B. ! ! Solution. Recall that x ∈ A ⇔ (∃b ∈ A) x ∈ b. Take x ∈ A. So by definition, there!exists b ∈ A such that x ∈ b. Since A ⊆ B, b ∈ B as well. Then by definition, 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 definition 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 definition 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 definition ! 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.

2

AXIOMS AND OPERATIONS

Exercise 2.7.

4

(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? 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 X is the set of all sets, which leads to Russell’s paradox. 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 ! that is, a ⊆ B. Then observe ! 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 ⊆ (A∪B)∪(A−B) is obvious. Now take x ∈ (A∩B)∪(A−B). If x ∈ A∩B, 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 first 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).

2

AXIOMS AND OPERATIONS

5

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 different 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. Define the symmetric difference 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 ∩ C) − (A ∩ B)]

= (A ∩ B) + (A ∩ C)

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

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

and (A + B) + C = [(A − B) ∪ (B − A)] + C

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

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

2

AXIOMS AND OPERATIONS A ! ! ! !

B ! ! ! !

6 C ! ! ! !

A+(B+C) !

(A+B)+C !

! ! !

! ! !

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 simplifies to (A ∪ B) − A, which further simplifies 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 different 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.

2

AXIOMS AND OPERATIONS

7

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 Since y ∈ A ∪ B, ! x ∈ A ∪ B. If x ∈ A, then ∃y ∈ A such that x ∈ y, ! 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. (a) Show that if A is nonempty, then P A = {PX | X ∈ A }. (b) Show that

"

{PX | X ∈ A } ⊆ P

"

A.

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 {A ∪ X | X ∈ B}? If not, then under what conditions does equality hold?

2

AXIOMS AND OPERATIONS

8

! 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).

Solution. # # Take A = {∅, {∅}} # and B =#{{∅}}. Then A ∩ B = {{∅}}. But A ∩ B = ∅. However, (A ∩ B) = {{∅}} = {∅}. Exercise 2.28. Simplify: "

Solution. "

#

A = ∅, so

{{3, 4}, {{3}, {4}}, {3, {4}}, {{3}, 4}}.

{{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) {PPP∅, PP∅, P∅, ∅}. # (b) {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, ! (c) P A,

2

AXIOMS AND OPERATIONS (d)

!

9

PA.

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

(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, # (b) B, #! (c) B, !# (d) B.

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

Exercise 2.32. Let S be the set {{a}, {a, b}}. Evaluate and simplify: !! (a) S, ## (b) S, #! !! !# (c) S∪( S− S). !! ! ## # Solution. First, ! # S = {a, b} = # a! ∪ b, and! ! S ! = # {a} = a. For (c), first note # ! S = a ∩ b and S = a. Thus S∪( S− S) = (a ∩ b) ∪ (a ∪ b − a) = (a ∩ b) ∪ (b − a) = b. !! # Exercise 2.33. With S as in the preceding exercise, evaluate ( S − S) when a '= b and when a = b.

2

AXIOMS AND OPERATIONS

10

! # Solution. Suppose a '= b. Then S = {a, b} and S = {a}, so "" $ " " ( S− S) = ({a, b} − {a}) = ({b}) = b.

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

!

S−

#

S = ∅, so

!! # ( S − 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 first 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.

3

RELATIONS AND FUNCTIONS

11

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.

3 3.1

Relations and Functions Ordered Pairs

Exercise 3.1. Suppose that we attempted to generalize the Kuratowski definitions of orrdered pairs to ordered triples by defining /x, y, z0∗ = {{x}, {x, y}, {x, y, z}}. Show that this definition is unsuccessful by giving examples of objects u, v, w, x, y, z with /x, y, z0∗ = /u, v, w0∗ 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, z0∗ = {{1}, {1, 2}, {1, 2, 1}} = {{1}, {1, 2}} and /u, v, w0∗ = {{1}, {1, 2}, {1, 2, 2}} = {{1}, {1, 2}} but z '= w. Exercise 3.2.

(a) Show that A × (B ∪ C) = (A × B) ∪ (A × C).

(b) Show that if A × B = A × C and A '= ∅, then B = C. Solution. Take /x, y0 ∈ A × (B ∪ C). So y ∈ B ∪ C. If y ∈ B, then /x, y0 ∈ A × B and if y ∈ C, then /x, y0 ∈ A × C. Conversely, if /x, y0 ∈ A × B, then y ∈ B ∪ C, so /x, y0 ∈ A × (B ∪ C), and similary for if y ∈ C. For (b), take x ∈ B, so for some a ∈ A, /a, x0 ∈ A × B. Hence /a, x0 = /a" , c0 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, y0 ∈ A × B. So y ∈ B, and thus ! there exists some X ∈ B such that y ∈ X. Hence /x, y0 ∈ A × X, and thus /x, y0 ∈ {A × X | X ! ∈ B}. Conversely, if ! /x, y0 ∈ A × X for some X ∈ B, then y ∈ B, and so /x, y0 ∈ A × B. Exercise 3.4. Show that there is no set to which every ordered pair belongs.

3

RELATIONS AND FUNCTIONS

12

Solution. Suppose that such a set exists. So in particular, for every set X, the ordered pair /X, X0 = {{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, ∀A∀B∀D∃C∀w(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, b0 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 iff A ⊆ domA × ranA. Solution. Suppose that A is a relation. So by definition, A is a set of ordered pairs. If A is the empty relation, the containment holds trivially. Otherwise, take /x, y0 ∈ A. Clearly, there exists y such that /x, y0 ∈ A so x ∈ dom A, and similarly, there exists x such that /x, y0 ∈ 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 definition, a relation. !! Exercise 3.7. Show that if R is a relation, then fld R = R.

Solution. Take xfld!R.! Recall that fld R = dom R ∪ ran ! R,!and by Lemma 3D, dom R ⊆ !! R and ran R ⊆ R, so by Exercise 2.38, fld 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, b0 for a ∈ dom R and b ∈ ran R. If

3

RELATIONS AND FUNCTIONS

13

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

" " ran A = {ran R | R ∈ A }. ! ! ! Solution. Take x ∈ dom A. So ∃y/x, y0!∈ A . Since /x, y0 ∈ A , for some R ∈ A , /x, y0 ∈ R, so x ∈ dom R, and thus x ∈ {dom R | R ∈ ! A }. For some R ∈ A ,!take x ∈ dom R,!so ∃y/x, y0 ∈ R. ! Hence {/x, y0} ⊆ R, but R ⊆ A , and so {/x, y0} ⊆ A , so /x, y0 ∈ A , so x ∈ dom A . ! ! The argument for the range is similar. Take x ∈ ran A . So ∃t/t, x0 ∈ A. So for some R ∈ A , /t, x0 ∈ R, so x ∈ ran R. Now for any R ∈ A !, take x ∈ R. Hence ∃t/t, x0 ∈ R, so by the same reasoning as above, {/t, x0} ⊆ R ⊆ A , and hence ! ! /t, x0 ∈ 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, y0 ∈ A# , that is, /x, y0 ∈ 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 0 ∈ R, but this does not imply that there is some y such that /x, y0 ∈ R for all R ∈ A . Indeed, consider A = {{/2, 30}, {/2, 50}}. So 2 ∈ dom R for all R ∈ A , but there is no single element such that /2, y0 ∈ 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, # 30}, {/5, 30}}. Then 3ran R for all R ∈ A , but there is no such t such that /t, 30 ∈ 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 definition, /x, y, z0 = //x, y0, z0 and so /x, y, z0 = //x, y0, z0 = {{/x, y0}, {/x, y0, z}}.

3

RELATIONS AND FUNCTIONS

14

So just as /x, y0 is just a set, so is /x, y, z0. Hence by definition /x1 , x2 , x3 , x4 0 = //x1 , x2 , x3 0, x4 0, so /x1 , x2 , x3 , x4 0 is an ordered pair, with first coordinate /x1 , x2 , x3 0 and second coordinate x4 . Note that it is indeed ordered, since //x1 , x2 , x3 0, x4 0 = //y1 , y2 , y3 0, y4 0 ⇔ /x1 , x2 , x3 , x4 0 = /y1 , y2 , y3 , y4 0

⇔ x 1 = y1 , x 2 = y2 , x3 = y3 , x4 = y 4

⇔ /x1 , x2 , x3 0 = /y1 , y2 , y3 0 and x4 = y4 Similarly, by definition, /x1 , x2 , x3 , x4 0 = ///x1 , x2 0, x3 0, x4 0 = //x1 , x2 0, x3 , x4 0,

where we have applied the fact that /x, y, z0 = //x, y0, z0, with x = /x1 , x2 0, y = x3 , and z = x4 . It follows that /x1 , x2 , x3 , x4 0 is also an ordered triple. Finally, by definition, //x1 , x2 , x3 , x4 00 = /x1 , x2 , x3 , x4 0, 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)0 ∈ F . So x ∈ dom F = dom G, and thus /x, G(x)0 ∈ G. But F (x) = G(x), and so /x, F (x)0 = /x, G(x)0, so /x, F (x)0 ∈ 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)0 ∈ f , so /x, g(x)0 ∈ g. Thus x ∈ dom g, so dom f ⊆ dom g. Moreover, /x, g(x)0 ∈ g so since g is a function, one must have that /x, f (x)0 = /x, g(x)0, and thus f (x) = g(x). Conversely, take /x, f (x)0 ∈ f . Then x ∈ dom f , so by assumption, f (x) = g(x). It follows that /x, f (x)0 = /x, g(x)0, so since x ∈ dom g, /x, f (x)0 ∈ 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 suffices 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)0 ∈ f . Since f ⊆ g, /x, f (x)0 ∈ g. Furthermore, since x ∈ dom g, /x, g(x)0 ∈ g. Since g is a function, that fact that /x, f (x)0, /x, g(x)0 ∈ g implies that g(x) = f (x), as desired.

3

RELATIONS AND FUNCTIONS

15

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 iff f (x) = g(x) for every x in (dom f ) ∩ (dom g). Solution. Assume that /x, y0 ∈ f ∩ g and /x, y " 0 ∈ f ∩ g. In particular, /x, y0 ∈ f ∩ g and /x, y " 0 ∈ f , so since f is a function, y = y " , and hence f ∩ g is a function. For (b), first suppose f ∪ g is a function. Take any x ∈ (dom f ) ∩ (dom g). Thus /x, f (x)0 ∈ f and /x, g(x)0 ∈ g. So /x, f (x)0, /x, g(x)0 ∈ f ∪ g, and since f ∪ g is a function, f (x) = g(x). Conversely, suppose that /x, y0 and /x, y " 0 are in f ∪ g. There are several cases to consider. If /x, y0 ∈ f and /x, y " 0 ∈ 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, y0 is in f and the /x, y " 0 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, y0, /x, y " 0 ∈ A . Now for some f ∈ A , /x, y0 ∈ f and for some g ∈ A , /x, y " 0 ∈ g. Without loss of generality, suppose f ⊆ g, and so /x, y0 ∈ 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, ∃C∀f (f ∈ C ⇔ f ∈ F ∧ (∀x∀y)((x ∈ dom f ∧ y ∈ dom f ⇒ x = y) ∧ (∀z)(z ∈ dom f ⇒ /z, z0 ∈ f ))).

Hence such a subset C consists of functions f : {x} → {x} : x 4→ x. That is, functions of the form f = {/x, x0} = {{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.

3

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16

Exercise 3.18. Let R be the set {/0, 10, /0, 20, /0, 30, /1, 20, /1, 30, /2, 30}. Evaluate the following: R ◦ R, R " {1}, R−1 " {1}, R!{1}", and R−1 !{1}". Solution. By inspection, we have

R ◦ R = {/0, 20, /0, 30, /1, 30}. Also, R " {1} = {/1, 20, /1, 30}, R−1 "= {/1, 00}, R!{1}" = {2, 3}, and R−1 !{1}" = {0}.

Exercise 3.19. Let

A = {/∅, {∅, {∅}}0, /{∅}, ∅0}.

−1 Evaluate each of !the ! following: A(∅), A!∅", A!{∅}", A!{∅, {∅}}", A , A ◦ A, A " ∅, A " {∅}, A " {∅, {∅}}, A.

Solution. First, note that A is indeed a function. Hence, A(∅) = {∅, {∅}}.

However, there are no ordered pairs /u, v0 such that u ∈ ∅, thus, A!∅" = ∅. Since A is a function, A!{∅}" = A(∅). Also, A!{∅, {∅}}" = {A(∅), A({∅})} = {{∅, {∅}}, ∅}. Flipping the pairs, A−1 = {/{∅, {∅}}, ∅0, /∅, {∅}0}. Taking ∅ as the intermediary, A ◦ A = {/{∅}, {∅, {∅}}0}. By similar reasons to the computation of A!∅", A " ∅ = ∅. However, and

A " {∅} = {/∅, {∅, {∅}}0}. A " {∅, {∅}} = A.

3

RELATIONS AND FUNCTIONS

Finally, first note that "

17

A = /∅, {∅, {∅}}0 ∪ /{∅}, ∅0 = {{∅}, {∅, {∅, {∅}}}} ∪ {{{∅}}, {{∅}, ∅}} = {{∅}, {∅, {∅, {∅}}}, {{∅}}, {{∅}, ∅}}

and thus ""

A = {∅} ∪ {∅, {∅, {∅}}} ∪ {{∅}}] ∪ {{∅}, ∅} = {∅, ∅, {∅, {∅}}, {∅}, {∅}, ∅} = {∅, {∅}, {∅, {∅}}}

Exercise 3.20. Show that F " A = F ∩ (A × ran F ). Solution. By definition, F " A = {/u, v0 | uF v ∧ u ∈ A}. Hence such /u, v0 ∈ F , so F " A ⊆ F . Also, u ∈ A and thus v ∈ ran F , so /u, v0 ∈ A × ran F , so F " A ⊆ A × ran F , and altogether, F "⊆ F ∩ (A × ran F ). Now take /u, v0 ∈ F ∩ (A × ran F ). So /u, v0 ∈ A × ran F , so u ∈ A. Also, /u, v0 ∈ F , so uF v. Together, these imply /u, v0 ∈ 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, y0 ∈ (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, y0 ∈ 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, v0 ∈ F for some u ∈ A. By assumption, A ⊆ B, so u ∈ B as well, and thus v ∈ F !B".

3

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18

(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, y0 ∈ 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, y0 ∈ B " A. Conversely, take /x, y0 ∈ B " A. So xBy and x ∈ A. In particular, xIA x ∧ xBy, so x(B ◦ IA )y, that is, /x, y0 ∈ B ◦ IA . For the other equality, first take x ∈ IA !C". So for some c ∈ C, /c, x0 ∈ 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, x0 ∈ 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, x0 ∈ F −1 and so /x, y0 ∈ 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)0 ∈ F , and thus /F (u), u0 ∈ 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 ◦ G−1 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 ◦ G−1 )y. Hence there exists t such that xG−1 t ∧ tGy. So tGx ∧ tGy. Now x ∈ ran G, and since G is a function, x = y. Hence G ◦ G−1 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 }.

3

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19

! ! ! 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 ,# y0 ∈ 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) = G−1 !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 xGt∧tF y, and thus tG−1 x. Since t ∈ dom F , we have x ∈ G−1 !dom F ". Now take x ∈ G−1 !dom F ". So for some t ∈ dom F , tG−1 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 defined 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 defined on all of A, and X ⊆ A, we have /x, f (x)0 ∈ f , so f (x) ∈ f !X". Then f (x) ∈ f !Y ", and thus for some y ∈ Y , /y, f (x)0 ∈ 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 define 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 definition 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 ).

3

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20

Define B=

$

{X ⊆ A | F (X) ⊆ X}

and

C=

(a) Show that F (B) = B and F (C) = C.

"

{X ⊆ A | X ⊆ F (X)}.

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

F (X)⊆X

$

X = B.

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, " " " F (C) = F X = F (X) ⊇ X = C. X⊆F (X)

X⊆F (X)

X⊆F (X)

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.

3.5

Infinite Cartesian Products

Exercise 3.31. Show that from the first form of the axiom of choice we can prove the second form, and conversely. Solution. Assume the first form. Let I be any set and let H be ! any function such that dom H = I, and H(i) '= ∅ for all i ∈ I. Define a relation R ⊆ I × i∈I H(i) by /i, x0 ∈ 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)0)∈ G, /i, G(i)0)∈ R, and thus by the definition of R, G(i) ∈ H(i). It follows that G ∈ i∈I H(i), so i∈I H(i) '= ∅. Thus the second form follows from the first. Conversely, let R be any relation, and denote dom R = I. Define a function H : I → P(ran R) : i 4→ 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, i∈I H(i) '= ∅, so take G ∈ i∈I H(i). Hence dom G = I, and for all i ∈ I, G(i) ∈ H(i). Also, for any /i, G(i)0 ∈ G, G(i) ∈ H(i) ⊆ ran R, and so /i, G(i)0 ∈ R, so G ⊆ R. Hence the two statements of the Axiom of Choice are equivalent.1 1

Thanks to Arturo Magidin for his hints on this exercise.

3

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21

Exercise 3.32. (a) Show that R is symmetric iff R−1 ⊆ R. (b) Show that R is transitive iff R ◦ R ⊆ R. Solution. Suppose R is symmetric. Take /x, y0 ∈ R−1 . So xR−1 y, and thus yRx, but since R is symmetric, xRy, so /x, y0 ∈ R. Conversely, suppose R−1 ⊆ R. Suppose xRy, then yR−1 x, and so /y, x0 ∈ R−1 ⊆ R, so yRx, and R is symmetric. Now suppose R is transitive. Take /x, z0 ∈ R ◦ R. Thus there exists y such that xRy ∧ yRz, and since R is transitive, xRz, so /x, z0 ∈ R. Conversely, suppose R ◦ R ⊆ R, and suppose xRy ∧ yRz. Thus /x, z0 ∈ R ◦ R, so /x, z0 ∈ R, and thus xRz, so R is transitive. Exercise 3.33. Show that R is a symmetric and transitive relation iff R = R−1 ◦ R. Solution. First suppose that R is symmetric and transitive. Take /x, y0 ∈ R. So xRy, and thus yRx. By transitivity, we have yRy. Thus yR−1 y and from xRy ∧ y R −1y we have x(R−1 ◦ R)y, so R ⊆ R−1 ◦ R. Now take /x, z0 ∈ R−1 ◦ R. Hence there exists y such that xRy ∧ yR−1 z. But from yR−1 z implies zRy, which implies yRz, as R is symmetric, and thus from xRy ∧ yRz we have xRz, and thus R = R−1 ◦ R. Conversely, assume R = R−1 ◦ R. Suppose xRy. So x(R−1 ◦ R)y, and thus there exists z such that xRz ∧ zR−1 y. Hence xRz ∧ zR−1 y ⇒ xRz ∧ yRz ⇒ zR−1 x ∧ yRz ⇒ y(R−1 ◦ R)x ⇒ yRx and so R is symmetric. Now suppose xRy ∧ yRz. Hence x(R−1 ◦ R)y ∧ y(R−1 ◦ R)z. Hence there exist s and t such that xRs ∧ sR−1 y and yRt ∧ tR−1 z. In particular, since R is symmetric, we have sRy ∧ yR−1 t, and so s(R−1 ◦ R)t, so sRt. Then observe, sRt ⇒ tR−1 s ⇒ zRt ∧ tR−1 s

⇒ zRs ⇒ sR−1 z

⇒ xRs ∧ sR−1 z ⇒ x(R

−1

⇒ xRz

since xRs

◦ R)z

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

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RELATIONS AND FUNCTIONS

22

# # Solution. Yes, A is a transitive relation. Suppose /x, y0, /y, z0 ∈ A . It follows that for every A ∈ A # , /x, y0,#/y, z0 ∈ A as well, and since A is transitive, /x, z0 ∈ A as well. Hence /x, z0 ∈ A , so ! A is transitive. This is not so for A . Suppose A = {A1 , A2 }, where A1 = {/a, b0, /b, c0, /a, c0}

and A2 = {/b, d0, /d, e0, /b, e0}. So both A1 and A2 are transitive, but " A = {/a, b0, /b, c0, /a, c0, /b, d0, /d, e0, /b, e0}, which is not transitive, since /a, d0 '∈

!

A.

Exercise 3.35. Show that for any R and x, we have [x]R = R!{x}". Solution. Take y ∈ [x]R . So /x, y0 ∈ R and by definition, y ∈ R!{x}". Conversely, take y ∈ R!{x}". Then for some t ∈ {x}, /t, y0 ∈ R. Clearly t = x, so /x, y0 ∈ R, and thus y ∈ [x]R . Exercise 3.36. Assume that f : A → B and that R is an equivalence relation on B. Define Q to be the set {/x, y0 ∈ A × A | /f (x), f (y)0 ∈ R}. Show that Q is an equivalence relation on A. Solution. For any x ∈ A, f (x) ∈ B, so /f (x), f (x)0 ∈ R. Hence /x, x0 ∈ Q, so Q is reflexive on A. Now suppose /x, y0 ∈ Q. Then /f (x), f (y)0 ∈ R, so /f (y), f (x)0 ∈ R since R is symmetric, so /y, x0 ∈ Q, and hence Q is symmetric. Finally, suppose /x, y0, /y, z0 ∈ Q. So /f (x), f (y)0, /f (y), f (z)0 ∈ R, and thus /f (x), f (z)0 ∈ R since R is transitive, so /x, z0 ∈ Q and Q is transitive. Thus Q is an equivalence relatoin on A. Exercise 3.37. Assume that Π is a partition of a set A. Define the relation RΠ as follows: xRΠ y ⇔ (∃B ∈ Π)(x ∈ B ∧ y ∈ B). Show that RΠ is an equivalence relation on A. Solution. For any x ∈ A, there exists some B ∈ Π such that x ∈ B, but the definition of a partition. Hence xRΠ x, so RΠ is reflexive. Now suppose that xRΠ y. So there exists some B ∈ Π such that x ∈ B ∧ y ∈ B, which obviously implies y ∈ B ∧ x ∈ B, so yRΠ x, so RΠ is symmetric. Now suppose xRΠ y and yRΠ z. Hence there exist B, C ∈ Π 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 xRΠ z, and RΠ is transitive.

3

RELATIONS AND FUNCTIONS

23

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 RΠ of the preceding exercise, then the partition A/RΠ is just Π. Solution. Take any [x] ∈ A/RΠ . Note for any y ∈ [x], there exists a B ∈ Π such that x, y ∈ B, by the definition of RΠ . Fix this B. For any z ∈ [x], we have that yRΠ z, since RΠ is an equivalence relation. By the reasoning the previous exercise, z ∈ B as well. So any two elements of [x] are in this fixed B, and thus [x] ⊆ B. Moreover, for any b ∈ B, clearly bRΠ x, and thus b ∈ [x]. Hence [x] = B. So it follows that any equivalence class in A/RΠ equals some set in the partition Π. On the other hand, take any C ∈ Π. Now C is nonempty, so take m ∈ C. By definition, this C is a subset of [m], since any element in C are RΠ -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/RΠ is equal to some C ∈ Π, and any C ∈ Π is equal to some equivalence class in A/RΠ . So Π and A/RΠ 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 define Π to be the partition A/R. Show that RΠ , as defined in Exercise 3.37 is just R. Solution. First take /x, y0 ∈ R. Then since Π is a partition consisting of equivalences classes of R, x, y ∈ B for some unique B ∈ Π. Then by the definition of RΠ , so /x, y0 ∈ RΠ , and thus R ⊆ RΠ . Now take /x, y0 ∈ RΠ , and so there exists some B ∈ Π 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, y0 ∈ R. Hence R = RΠ . Exercise 3.40. Define 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. Define a function F : P → P : n 4→ 3n. Then recall from Theorem 3Q that such a function f exists iff 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)0 '∈ 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 define the relation Q on R × R by /u, v0Q/x, y0 iff 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, y0]Q ) = [/x + 2y, y + 2x0]Q ?

3

RELATIONS AND FUNCTIONS

24

Solution. Q is reflexive on R × R due to the commutativity of + on R. Suppose /u, v0Q/x, y0, so u + y = x + v. Then clearly x + v = u + y, so /x, y0Q/u, v0, so Q is symmetric. Now assume /u, v0Q/x, y0 ∧ /x, y0Q/w, z0. Thus u+y =x+v∧x+z =w+y from which we see that u + z = x + v − y + z = (x + z − y) + v = w + v which implies /u, v0Q/w, z0. Hence Q is transitive. Define a function F : R × R → R × R : /x, y0 4→ /x + 2y, y + 2x0. Then if /u, v0Q/x, y0, 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, v0)QF (/x, y0). 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 Fˆ : A/R × A/R → A/R such that Fˆ ([x], [y]) = [F (x, y)]

for all x, y ∈ A.

If F is not compatible with R, then no such Fˆ exists. Of course, we must extend the definition of compatibility. Note that we would like to have the following commutative diagram: A×A +

F

−−−−→

A +

A/R × A/R −−−−→ A/R Fˆ

From this, we see that if /x, y0 and /u, v0 have the same image under A × A → (A/R) × (A/R), that is, /[x], [y]0 = /[u], [v]0, then we we also like [F /x, y0] = [F /u, v0]. This suggests that we define that F is compatible with R if for any x, y, u, v ∈ R, xRy ∧ uRv =⇒ F /x, y0RF /u, v0.

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Assume that F is compatible with R, and we will prove that such a Fˆ exists. We require that //[x], [y]0, [F (x, y)]0 ∈ Fˆ , so define Fˆ as Fˆ = {//[x], [y]0, [F (x, y)]0 | x, y ∈ A}. To see that Fˆ to indeed be a function, consider the pairs //[x], [y]0, [F /x, y0]0 and //[u], [v]0, [F /u, v0]0 in Fˆ . Then we have the implications /[x], [y]0 = /[u], [v]0 =⇒ [x] = [u] ∧ [y] = [v] =⇒ xRu ∧ yRv

=⇒ F /x, y0RF /u, v0

=⇒ [F /x, y0] = [F /u, v0]

So Fˆ is a function. It is clear that dom Fˆ = A/R × A/R and ran Fˆ ⊆ A/R. Also Fˆ /[x], [y]0 = [F /x, y0], since Fˆ /[x], [y]0 = [F /x, y0] ∈ Fˆ . Also, Fˆ 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]0 = [F /[x], [y]0] = Fˆ /[x], [y]0 so F = G. Now suppose that F is not compatible. By incompatibility, there are some pairs /x, y0, /u, v0 ∈ A × A such that xRu ∧ yRv but it is not case that F /x, y0RF /u, v0. Hence [x] = [u], [y] = [v], but [F /x, y0] '= [F /u, v0]. But we must have Fˆ /[x], [y]0 = [F /x, y0]

and Fˆ /[u], [v]0 = [F /u, v0]

which is impossible since the left sides are equal, since Fˆ 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 R−1 is also a linear ordering on A. Solution. Suppose that xR−1 y ∧ yR1− z. Then yRx ∧ zRy, and since R is transitive, zRx and so xR−1 z. Furthermore, for any x, y, exactly one of xRy, x = y, or yRx holds, and thus exactly one of yR−1 x, x = y, or xR−1 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 < satisifies 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

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