Solutions to Elements of Set Theory ∗
yunone.se@gmail.com
January 14, 2011
Contents
1 Introduction
1
2 Axioms and Operations
3
3 Relations and Functions
11
3.1 Ordered Pairs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
3.2 Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
3.3 n-ary Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13
3.4 Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
3.5 Infinite Cartesian Products . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
3.6 Ordering Relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
4 Natural Numbers
30
4.1 Inductive Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
4.2 Peano’s Postulates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
4.3 Recursion on ω . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
4.4 Arithmetic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
4.5 Ordering on ω . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36
4.6 Review Exercise . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
5 Construction of the Real Numbers
42
5.1 Integers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
5.2 Rational Numbers . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
1
Introduction
Exercise 1.1. Which of the following become true when ∈ is inserted in place of the
blank? Which become true when ⊆ is inserted?
(a) {∅}
∗
{∅, {∅}}.
by Herbert Enderton. I make no claim to correctness, as I’m just learning as I write these.
1
1
INTRODUCTION
2
(b) {∅}
{∅, {{∅}}}.
(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
Relations and Functions
3.1
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
RELATIONS AND FUNCTIONS
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
RELATIONS AND FUNCTIONS
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
RELATIONS AND FUNCTIONS
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
RELATIONS AND FUNCTIONS
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 F̂ : A/R × A/R → A/R such that
F̂ ([x], [y]) = [F (x, y)]
for all x, y ∈ A.
If F is not compatible with R, then no such F̂ 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
F̂
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 F̂ exists. We
require that //[x], [y]0, [F (x, y)]0 ∈ F̂ , so define F̂ as
F̂ = {//[x], [y]0, [F (x, y)]0 | x, y ∈ A}.
To see that F̂ to indeed be a function, consider the pairs //[x], [y]0, [F /x, y0]0 and
//[u], [v]0, [F /u, v0]0 in F̂ . 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 F̂ is a function. It is clear that dom F̂ = A/R × A/R and ran F̂ ⊆ A/R. Also
F̂ /[x], [y]0 = [F /x, y0], since F̂ /[x], [y]0 = [F /x, y0] ∈ F̂ . Also, F̂ 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] = F̂ /[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
F̂ /[x], [y]0 = [F /x, y0]
and F̂ /[u], [v]0 = [F /u, v0]
which is impossible since the left sides are equal, since F̂ 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 <A and <B are linear orderings on A and B, respectively.
Define the binary relation <L on the Cartesian product A × B by:
/a1 , b1 0 <L /a2 , b2 0
iff either
a1 <A a2 ∨ (a1 = a2 ∧ b1 <B b2 ).
Show that <L is a linear ordering on A × B.
Solution. Suppose /a1 , b1 0 <L /a2 , b2 0∧/a2 , b2 0 <L /a3 , b3 0. So a1 <A a2 ∨(a1 = a2 ∧b1 <B
b2 ) and a2 <A a3 ∨ (a2 = a3 ∧ b2 <B b3 ). First suppose that a1 <A a2 and a2 <A a3 , so
by transitivity, a1 <A a3 , and thus /a1 , b1 0 <L /a3 , b3 0. Now suppose that a1 <A a2 and
a2 = a3 ∧ b2 <B b3 , and hence a1 <A a3 , so again /a1 , b1 0 <L /a3 , b3 0.
Now suppose a1 = a2 ∧ b1 <B b2 and a2 <A a3 . Thus a1 <A a3 , so /a1 , b1 0 <L /a3 , b3 0.
Finally, if a1 = a2 ∧ b1 <B b2 and a2 = a3 ∧ b2 <B b3 , then a1 = a3 , and b1 <B b3 , so
/a1 , b1 0 <L /a3 , b3 0, and thus <L is transitive.
To show the trichotomy property, assume that /a1 , b1 0 '<L /a2 , b2 0 and /a1 , b1 0 '<L
/a2 , b2 0. In particular, a1 '<A a2 and a1 '= a2 , so since <A is a linear ordering, one
must have a2 <A a1 , and so /a2 , b2 0 <L /a1 , b1 0. Thus we see at least one of the three
possibility holds. To see that at most one possibility holds, suppose /a1 , b1 0 <L /a2 , b2 0.
Now clearly /a1 , b1 0 =
' /a2 , b2 0, since we must have that either a1 <A a2 or b1 <B b2 . Also,
/a2 , b2 0 <
' L /a1 , b1 0 since if a1 <A a2 , then a2 '<A a1 and a2 '= a1 . And if a1 = a2 ∧b1 <B b2 ,
then a2 '<A a1 and b2 '<B b1 . A symmetric argument show that if /a2 , b2 0 <L /a1 , b1 0 holds,
then neither of the other two possibilities do.
Also, if /a1 , b1 0 = /a2 , b2 0, then a1 '<A a2 and b1 '<B b2 , and a2 '<A a1 and b2 '<B b1 , so
neither of the other two possibilities hold. Hence <L is a linear ordering.
Exercise 3.46. Evaluate the following sets:
##
(a)
/x, y0.
###
(b)
{/x, y0}−1 .
Solution.
$$
$$
/x, y0 =
{{x}, {x, y}}
$
= {x} = x
Denote R = {/x, y0}, so R−1 = {/x, y0}−1 = {/y, x0}. Thus,
$$$
$$$
{/x, y0}−1 =
{/y, x0}
$$
=
/y, x0
=y
Exercise 3.47.
by part (a)
3
RELATIONS AND FUNCTIONS
27
(a) Find all of the functions from {0, 1, 2} into {3, 4}.
(b) Find all of the functions from {0, 1, 2} into {3, 4, 5}.
Solution. Without listing them, there are 23 functions from {0, 1, 2} into {3, 4} and 3·2·1
functions from {0, 1, 2} into {3, 4, 5}.
Exercise 3.48. Let T be the set {∅, {∅}}.
(a) Find all of the ordered pairs, in any, in PT .
(b) Evaluate and simplify: (PT )−1 ◦ (PT " {∅})
Solution. Firstly,
PT = {∅, {∅}, {{∅}}, {∅, {∅}}}.
Hence there are no ordered pairs in PT . As such, there are no v such that ∅PT v, and
thus PT " {∅} = ∅. It follows immediately that (PT )−1 ◦ (PT " {∅}) = ∅ as well.
Exercise 3.49. Find as many equivalence relations as you can on the set {0, 1, 2}.
Solution. Recall that the number of equivalence relations on a set is the same as the
number of distinct partitions of that set. We can partition {0, 1, 2} in 5 ways, namely:
• {0}, {1}, {2}
• {0}, {1, 2}
• {1, }, {0, 2}
• {2}, {0, 1}
• {0, 1, 2}
These correspond to the following equivalence relations, with each set in the partition
corresponding to an equivalence class. We have
• R1 = {/0, 00, /1, 10, /2, 20}
• R2 = {/0, 00, /1, 10, /2, 20, /1, 20, /2, 10}
• R3 = {/1, 10, /0, 00, /2, 20, /0, 20, /2, 00}
• R4 = {/2, 20, /0, 00, /1, 10, /0, 10, /1, 00}
• R5 = {/0, 00, /1, 10, /2, 20, /0, 10, /1, 00, /0, 20, /2, 00, /1, 20, /2, 10}
Exercise 3.50.
(a) Find a linear ordering on {0, 1, 2, 3} that contains the ordered pairs /0, 30 and /2, 10.
(b) Now find a different one meeting the same conditions.
3
RELATIONS AND FUNCTIONS
28
Solution. In an abuse of notation, consider the linear ordering 0 < 2 < 1 < 3. Indeed,
0 < 3 and 2 < 1, as desired. Another possibility is the linear ordering 2 < 0 < 1 < 3.
Exercise 3.51. Find as many linear orderings as possible on the set {0, 1, 2} that contain
the pair /2, 10.
Solution. By the trichotomy property, we see that any two distinct elements must be
related in some way. Hence we have three possible linear orderings with 2 < 0, in particular
they are 1 < 2 < 0, 2 < 1 < 0, and 2 < 0 < 1.
Exercise 3.52. Suppose that A × B = C × D. Under what conditions can we conclude
that A = C and B = D?
Solution. We must have that A, B, C, D are all nonempty. If this is the case, for any
/a, b0 ∈ A × B, /a, b0 ∈ C × D, so a ∈ C and b ∈ D and thus A ⊆ C and B ⊆ D. A
parallel argument shows that C ⊆ A and D ⊆ B. However, if some of the sets are empty,
say B = D = ∅, then, for example, Q × B = R × D = ∅, but clearly Q '= R.
Exercise 3.53. Show that for any set R and S we have (R∪S)−1 = R−1 ∪S −1 , (R∩S)−1 =
R−1 ∩ S −1 , and (R − S)−1 = R−1 − S −1 .
Solution. Suppose /r, s0 ∈ (R ∪ S)−1 , so /s, r0 ∈ R ∪ S. If /s, r0 ∈ R, then /r, s0 ∈ R−1 ,
and if by the same reasoning, if /s, r0 ∈ S, then /r, s0 ∈ S −1 . Hence (R∪S)−1 ⊆ R−1 ∪S −1 .
Conversely, if /r, s0 ∈ R−1 ∪ S −1 , then if /r, s0 ∈ R−1 , then /s, r0 ∈ R, so /s, r0 ∈ R ∪ S,
so /r, s0 ∈ (R ∪ S)−1 , thus implying equality. A completely parallel argument shows the
equality (R ∩ S)−1 = R−1 ∩ S −1 .
Take /r, s0 ∈ (R − S)−1 , so /s, r0 ∈ R − S, and thus /s, r0 ∈ R and /s, r0 '∈ S. Hence
/r, s0 ∈ R−1 but /r, s0 '∈ S −1 , so /r, s0 ∈ R−1 − S −1 . The reverse containment follows,
since each of these prior steps is reversable.
Exercise 3.54. Prove that the following equations hold for any sets.
(a) A × (B ∩ C) = (A × B) ∩ (A × C).
(b) A × (B ∪ C) = (A × B) ∪ (A × C).
(c) A × (B − C) = (A × B) − (A × C).
Solution. The proofs of (a) and (b) are straightforward. For (c) take /x, y0 ∈ A×(B −C).
Then x ∈ A and y ∈ B −C, and thus y ∈ B but y '∈ C, so /x, y0 ∈ A×B but /x, y0 '∈ A×C.
The converse follows similarly.
Exercise 3.55. Answer “yes” or “no.” Where the answer is negative, supply a counterexample.
(a) Is it always true that (A × A) ∪ (B × C) = (A ∪ B) × (A ∪ C)?
(b) Is it always true that (A × A) ∩ (B × C) = (A ∩ B) × (A ∩ C)?
3
RELATIONS AND FUNCTIONS
29
Solution. The answer to (a) is no. Take A = {1}, B = {2}, C = {3}, and consider the
pair /1, 30 ∈ (A ∪ B) × (A ∪ C). However, clearly /1, 30 '∈ A × A and /1, 30 '∈ B × C since
3 '∈ A and 1 '∈ B, respectively. It is not hard to see that the answer to (b) is yes.
Exercise 3.56. Answer “yes” or “no.” Where the answer is negative, supply a counterexample.
(a) Is dom (R ∪ S) always the same as dom R ∪ dom S?
(b) Is dom (R ∩ S) always the same as dom R ∩ dom S?
Solution. The answer to (a) is yes. Take x ∈ dom R ∪ S, then there is some y such
that /x, y0 ∈ R ∪ S. If /x, y0 ∈ R, then x ∈ dom R, and similarly for S. Conversely, if
x ∈ dom R ∪ dom S, then if x ∈ dom R, there exists a y such that /x, y0 ∈ R ⊆ R ∪ S,
and similarly /x, y0 ∈ S ⊆ R ∪ S if x ∈ dom S.
For (b), the answer is no. Consider R = {/1, 20}, S = {/1, 30}, and thus R∩S = ∅. Now
1 ∈ dom R ∩ dom S, but 1 '∈ dom R ∩ S, since R ∩ S = ∅, and hence dom R ∩ S = ∅.
Exercise 3.57. Answer “yes” or “no.” Where the answer is negative, supply a counterexample.
(a) Is R ◦ (S ∪ T ) always the same as (R ◦ S) ∪ (R ◦ T )?
(b) Is R ◦ (S ∩ T ) always the same as (R ◦ S) ∩ (R ◦ T )?
Solution. The answer to (a) is yes. For (b), let R = {/3, 20, /4, 20}, S = {/1, 30} and
T = {/1, 40}. So S ∩ T = ∅, and thus R ◦ (S ∩ T ) = ∅ as well. However by inspection,
/1, 20 ∈ (R ◦ S) ∩ (R ◦ T ), and thus the two sets are not equal.
Exercise 3.58. Give an example to show that F !F −1 !S"" is not always the same as S.
Solution. Take S = {1} and F = {/2, 30}, so F −1 = {/3, 20}. Now F −1 !S" = ∅, as 1 is
not in the domain of F . It follows then that F !F −1 !S"" = ∅ as well.
Exercise 3.59. Show that for any sets Q " (A∩B) = (Q " A)∩(Q " B) and Q " (A−B) =
(Q " A) − (Q " B).
Solution. Take /u, v0 ∈ Q " (A∩B), so there exists u ∈ A∩B such that uQv. Since u ∈ A
and u ∈ B, /u, v0 ∈ Q " A and /u, v0 ∈ Q " B. Conversely, if /u, v0 ∈ (Q " A) ∩ (Q " B),
then uQv and u ∈ A and u ∈ B, so u ∈ A ∩ B, so /u, v0 ∈ Q " (A ∩ B).
If /u, v0 ∈ Q " (A − B). So uQv and u ∈ A but u '∈ B. So /u, v0 ∈ Q " A,
but /u, v0 '∈ Q " B so /u, v0 ∈ (Q " A) − (Q " B). Conversely, take /u, v0 ∈ Q "
(A − B) = (Q " A) − (Q " B), so uQv but u ∈ A and u '∈ B. So u ∈ A − B, and thus
/u, v0 ∈ Q " (A − B).
Exercise 3.60. Prove that for any sets (R ◦ S) " A = R ◦ (S " A).
4
NATURAL NUMBERS
30
Solution. Take /u, v0 ∈ (R ◦ S) " A. Then u ∈ A and u(R ◦ S)v, so there exists some t
such that uSt ∧ tRv. Now /u, t0 ∈ S " A and /t, v0 ∈ R. Hence /u, v0 ∈ R ◦ (S " A).
Conversely, take /u, v0 ∈ R ◦ (S " A). Hence there exists t such that u(S " A)t and
tRv. Now /u, t0 ∈ (S " A) implies u ∈ A and uSt, and so u(R ◦ S)v, which implies
/u, v0 ∈ (R ◦ S) " A.
4
Natural Numbers
4.1
Inductive Sets
Exercise 4.1. Show that 1 '= 3, i.e., that ∅+ '= ∅+++ .
Solution. By defintion of the successor, we have
∅+ = {∅} and
∅+++ = {∅, {∅}, {∅, {∅}}}.
Now take, for instance, {∅} ∈ ∅+++ . However, {∅} '∈ ∅+ , and thus by Extensionality,
∅+ '= ∅+++ .
4.2
Peano’s Postulates
Exercise 4.2. Show that if a is a transitive set, then a+ is also a transitive set.
Solution. It suffices to show that for all x ∈ a+ , x ⊆ a. Recall a+ = a∪{a}. Take x ∈ a+ .
If x ∈ a, then x ⊆ a ⊆ a+ , so x ⊆ a+ . If x ∈ {a}, then x = a, and again x ⊆ a+ . Hence
a+ is transitive.
Exercise 4.3.
(a) Show that if a is a transitive set, then Pa is also a transitive set.
(b) Show that if Pa is a transitive set, then a is also a transitive set.
Solution. Recall that if a is transitive, then a ⊆ Pa. So for any x ∈ Pa, we have x ⊆ a,
and hence x ⊆ Pa. Thus Pa is a transitive
set.
!
!
For (b), if Pa is transitive, then Pa ⊆ Pa. Since Pa = a, we have a ⊆ Pa. So
for any x ∈ a, x ∈ Pa, and thus x ⊆ a, so a is a transitive set.
!
Exercise 4.4. Show that if a is a transitive set, then a is also a transitive set.
!
!
Solution. Take
x
∈
a.
Since
a
is
transitive,
a ⊆ a, and thus
!
! x ∈ a. Now take y ∈ x.
Recall
a = {b | (∃c ∈ a)(b ∈ c)}. It follows that y ∈ a, as y ∈ x ∈ a. Hence
! that !
x ⊆ a, and a is a transitive set.
Exercise 4.5. Assume that every member of A is a transitive set.
!
(a) Show that A is a transitive set.
#
(b) Show that A is a transitive set (assuming that A is nonempty.)
4
NATURAL NUMBERS
31
!
Solution. Take x ∈ A , so x ∈ A for some A ∈ A . Now !
take y ∈ x, then y ∈
!x ∈ A
and
so
y
∈
A
since
A
is
a
transitive
set.
It
follows
that
y
∈
A
,
and
thus
x
⊆
A , so
!
A is a transitive#set.
Now take x ∈ A . So for all A ∈ A , x ∈ #
A. Take any #
y ∈ x, and thus since each A
is a transitive set, y ∈ A for all A ∈ A , so y ∈ A . Hence A is a transitive set.
!
Exercise 4.6. Prove the converse to Theorem 4E: If (a+ ) = a, then a is a transitive
set.
!
Solution. Suppose that (a+ ) = a, we then have the following equalities:
"
"
"
"
"
a = (a+ ) = (a ∪ {a}) =
a ∪ {a} =
a ∪ a.
It follows that
4.3
!
a ⊆ a, which is equivalent to saying a is a transitive set.
Recursion on ω
Exercise 4.7. Complete part 4 of the proof of the recursion theorem on ω.
Solution. If h1 and h2 satisfy the conclusion of the theorem, then
h1 (0) = h2 (0) = a
so 0 ∈ S. Now suppose that k ∈ S, so that h1 (k) = h2 (k). We then have
h1 (k + ) = F (h1 (k)) = F (h2 (k)) = h2 (k + ).
So k + ∈ S, and hence S is inductive. So S = ω, so h1 = h2 .
Exercise 4.8. Let f be a one-to-one function from A into A, and assume that c ∈ A −
ran f . Define h : ω → A by recursion:
h(0) = c,
h(n+ ) = f (h(n)).
Show that h is one-to-one.
Solution. Let
T = {n ∈ ω | (∀m, n)m '= n =⇒ h(m) '= h(n)}.
Suppose m '= 0. Then by Theorem 4C, m = p+ for some p. Note then that h(m) =
h(p+ ) = f (h(p)) '= c, since c '∈ ran f . So h(0) = c '= h(m), and thus 0 ∈ T . Now suppose
k ∈ T , and that h(k + ) = h(m). By the same reasoning above, m '= 0, so m = p+ for some
p. Then,
f (h(k)) = h(k + ) = h(m) = h(p+ ) = f (h(p)).
Since f is injective, h(k) = h(p), and thus k = p, since k ∈ T . So k + = p+ = m, and thus
k + ∈ T . So T is inductive, and thus T = ω, so h is one-to-one.
4
NATURAL NUMBERS
32
Exercise 4.9. Let f be a function from B into B, and assume that A ⊆ B. We have two
possible methods for constructing the “closure” C of A under f . First define C ∗ to be the
intersection of the closed supersets of A:
$
C ∗ = {X | A ⊆ X ⊆ B ∧ f !X" ⊆ X}.
Alternatively, we could apply the recursion theorem to obtain the function h for which
h(0) = A
Define C∗ to be
!
h(n+ ) = h(n) ∪ f !h(n)".
ran h; in other words
C∗ =
"
h(i).
i∈ω
Show that C ∗ = C∗ .
Solution. Note that for all i ∈ ω, f !h(i)" ⊆ h(i+ ). Thus,
"
"
"
f !h(i)" ⊆
h(i+ ) ⊆
h(i) = C∗ .
f !C∗ " =
i∈ω
i∈ω
i∈ω
It follows that C ∗ ⊆ C∗ , since clearly A ⊆ C∗ ⊆ B, and so C∗ is one of the sets being
intersected to form C ∗ .
Now let
T = {n ∈ ω | h(n) ⊆ C ∗ }.
Note 0 ∈ T , since h(0) = A ⊆ C ∗ , as each set X being intersected contains A. Now
assume k ∈ T . Since h(k + ) = h(k) ∪ f !h(k)", and h(k) ⊆ C ∗ by hypothesis, it suffices
to show that f !h(k)" ⊆ C ∗ . Since h(k) ⊆ C ∗ , h(k) ⊆ X for all X being intersected, and
thus f !h(k)" ⊆ f !X" ⊆ X. Hence f !h(k)" ⊆ C ∗ for similar reasoning as above. Hence
h(k + ) ⊆ C ∗ and k + ∈ T . So T is inductive, and thus T = ω. Since h(n) ⊆ C ∗ for all n,
C∗ ⊆ C ∗ , and equality follows.
Exercise 4.10. In Exercise 4.9, assume that B is the set of real numbers, f (x) = x2 , and
A is the closed interval [ 12 , 1]. What is the set called C ∗ and C∗ ?
Solution. Note that any set X in the intersection forming C ∗ must contain 1/2. Since X
is also closed under f , 1/4 ∈ X, and thust 1/16 ∈ X, and so on. Essentially, if 1/2n ∈ X,
then 1/2n+1 ∈ X. Since this sequence converges to 0, any set in the intersection must
contain the interval [0, 1]. Since [0, 1] itself is closed under f , and is contained in every set
in the intersection, we have C ∗ = C∗ = [0, 1].
Exercise 4.11. In Exercise 4.9, assume that B is the set of real numbers, f (x) = x − 1,
and A = {0}. What is the set called C ∗ and C∗ ?
Solution. Since 0 ∈ X for each X in the intersection, −1 ∈ X as well, and thus −2 ∈ X
an so on. So each set must contain the set {n ∈ Z | n ≤ 0}. Since this set is closed under
f , we have that it is equal to C ∗ .
4
NATURAL NUMBERS
33
Exercise 4.12. Formulate an analogue to Exercise 4.9 for a function f : B × B → B.
Solution. An analogue for the closure of A under f would be the set
C ∗ = {X | A ⊆ X ⊆ B ∧ f !X × X" ⊆ X}.
4.4
Arithmetic
Exercise 4.13. Let m and n be natural numbers such that m · n = 0. Show that either
m = 0 or n = 0.
Solution. By the contrapositive, assume m '= 0 and n '= 0. Since n '= 0, n = p+ for some
p. Then
m · n = Mm (n) = Mm (p+ ) = Mm (p) + m '= 0.
The last inequality holds, for if it were equality, then 0 = Mm (p+ ) = Mm (p)+ = σ(Mm (p)),
and thus 0 ∈ ran σ, contrary to the first of the Peano axioms.
Exercise 4.14. Call a natural number even if it has the form 2 · m for some m. Call it
odd if it has the form (2 · p) + 1 for some p. Show that each natural number is either even
or odd, but never both.
Solution. Recall from Theorem 4K that + and · are commutative. First, let us make
two observations.
• For all n ∈ ω, n+ = n + 1. To see this, let
T = {n ∈ ω | n+ = n + 1}.
Note that 0 ∈ T , since 0+ = 1 = 0+1, with this last equality following from Theorem
4K(2). Suppose k ∈ T , so k + = k + 1. Then k ++ = (k + 1)+ = k + + 1, again by
Theorem 4K(2). Hence T = ω.
• For all n ∈ ω, n · 1 = n. To see this, let
S = {n ∈ ω | n · 1 = n}.
Note 0 ∈ S by (M1). Suppose k ∈ S, so k · 1 = k. Then, by (M2) and the first
observation,
k+ · 1 = 1 · k+ = 1 · k + 1 = k · 1 + 1 = k + 1 = k+ .
Hence S = ω.
4
NATURAL NUMBERS
34
Now let
A = {n ∈ ω | ∃m(n = 2 · m) ∨ ∃p(n = 2 · p + 1)}.
This is the set of natural numbers that are either even or odd. Note that 0 ∈ A, since
0 = 2 · 0. So assume k ∈ A. If k is even, then k = 2 · m, so k + = (2 · p)+ = 2 · p + 1 by the
first observation. So k + is odd. If k is odd, then k = 2 · p + 1, so by the second observation
and the distributive law,
k + = (2 · p + 1)+ = 2 · p + 1+ = 2 · p + 2 = 2 · p + 2 · 1 = 2 · (p + 1).
Hence k + is even, so k + ∈ A. Hence A = ω, and thus all natural numbers are either even
or odd.
Now let
B = {n ∈ ω |¬(∃m(n = 2 · m) ∧ ∃p(n = 2 · p + 1))},
the set of all natural numbers that are not both even and odd. We have seen that 0 is
even. 0 is not odd, for if 0 = 2 · p + 1, then 0 = (2 · p)+ = σ(2 · p), but then 0 ∈ ran σ,
contrary to the first Peano axiom. Hence 0 ∈ B. Suppose k ∈ B. Suppose k is odd but
not even, so k = 2 · p + 1 for some p. As seen before, k + is thus even. However, k + is not
odd, for if k + = 2 · m + 1 for some m, then since the successor function is injective, we
have
k + = 2 · m + 1 = (2 · m)+ =⇒ k = 2 · m
contrary to the fact that k is not even. Suppose k is even, but not odd. So suppose
k = 2m '= 2n + 1. Then since σ is injective,
k + = (2·m)+ '= (2·n+1)+ ⇒ k + = 2·m+1 '= 2·n+1+ = 2·n+2 = 2·n+2·1 = 2·(n+1).
So k + is odd, but not even. So by induction, B = ω.
Exercise 4.15. Prove the associative law of addition
m + (n + p) = (m + n) + p.
Solution. Let
A = {p ∈ ω | m + (n + p) = (m + n) + p}.
Note that m + (n + 0) = m + n = (m + n) + 0, so 0 ∈ A. So assume k ∈ A. Then,
m + (n + k + ) = m + (n + k)+ = (m + (n + k))+ = ((m + n) + k)+ = (m + n) + k + .
So k + ∈ A, and thus A = ω.
Exercise 4.16. Prove the commutative law for multiplication
m · n = n · m.
Solution. First, we prove two preliminary facts.
4
NATURAL NUMBERS
35
• Let
A = {n ∈ ω | 0 · n = 0}.
Note that 0 · 0 = 0 by (M1), so 0 ∈ A. Assume that k ∈ A, so 0 · k = 0. Thus
0 · k + = 0 · k + 0 = 0 + 0 = 0,
so k ∈ A, and thus A = ω.
• Now let
B = {n ∈ ω | m+ · n = m · n + n}.
Note by (M1), m+ · 0 = 0 = m0̇ = m · 0 + 0, so 0 ∈ B. Assume k ∈ B, so
m+ · k = m · k + k. Then
m+ · k + = m+ · k + m+
=m·k+k+m
by M2
+
+
= m · k + (k + m )
= m · k + (k + + m)
=m·k+m+k
+
=m·k +k
+
since k ∈ B
by the second result
commutativity of addition
+
Now let
C = {m ∈ ω | m · n = n · m}.
By the first result, we have 0 · n = 0 = n · 0, so 0 ∈ C. Suppose k ∈ C, so k · n = n · k.
Then by (M2),
k+ · n = k · n + n = n · k + n = n + k+ .
Hence k + ∈ C, so C = ω.
Exercise 4.17. Prove that mn+p = mn · mp .
Solution. Let
A = {p ∈ ω | mn+p = nn · mp }.
Using the results from the previous exercises, observe that
mn+0 = mn = mn · 1 = mn · m0
so 0 ∈ A. Now assume that k ∈ A, so mn+k = mn · mk . Then by (A2) and (E2),
+
+
+
mn+k = m(n+k) = mn+k · m = mn · mk · m = mn · mk .
Hence k + ∈ A, so A = ω.
4
NATURAL NUMBERS
4.5
36
Ordering on ω
Exercise 4.18. Simplify: ∈−1
ω !{7, 8}".
Solution. Observe by definition that
−1
∈−1
ω !{7, 8}" = {v | ∃u ∈ {7, 8}, u ∈ω v} = {v | ∃u ∈ {7, 8}, v ∈ω u}.
Hence ∈−1
ω !{7, 8}" = {0, 1, 2, 3, 4, 5, 6, 7}.
Exercise 4.19. Prove that if m is a natural number and d is a nonzero number, then
there exist numbers q and r such that m = (d · q) + r and r is less than d.
Solution. Let
B = {m ∈ ω | ∃q, r((m = (d · q) + r) ∧ r ∈ d)}.
Since 0 = d · 0 + 0, and since 0 ∈ d, and so 0 ∈ B. Now assume that k ∈ B, so there exist q
and r such that k = (d · q) + r for some r ∈ d. It follows that k + = ((d · q) + r)+ = d · q + r+ .
By Lemma 4L, since r ∈ d, r+ ∈ d+ , and thus r+ ∈ d. Now if r+ ∈ d, the conditions are
satisfied, so k + ∈ B. If r+ = d, then
k + = (d · q) + d = (d · q) + d · 1 = d · (q + 1) = d · (q + 1) + 0,
and thus k + ∈ B, so B is inductive, and thus B = ω.
Exercise 4.20. Let A be a nonempty subset of ω such that
!
A = A. Show that A = ω.
Solution.
! First, observe that 0 ∈ A, since A is nonempty,
! for any such p ∈ A, 0 ∈ p, and
thus 0 ∈ A = A. Now suppose that k ∈ A. Then k ∈ A, and so k ∈ a for some a ∈ A.
If a = k + , we have k + ∈ A. Note that we can not have a ∈ k +!
, otherwise, a ∈ k, which
+
+
contradicts the trichotomy law. Hence k ∈ a, and so k ∈ A = A. In either case,
k + ∈ A, so A is inductive. Hence A = ω.
Exercise 4.21. Show that no natural number is a subset of any of its elements.
Solution. Let n be any natural number. If n = 0, the conclusion holds vacuously, so
suppose n '= 0. Let m ∈ n be any member of n. If n ⊆ m, then by Corollary 4M,
n ∈ m, which contradicts trichotomy. Hence no natural number is a subset of any of its
elements.
Exercise 4.22. Show that for any natural numbers m and p we have m ∈ m + p+ .
Solution. Since p+ '= 0, we have 0 ∈ p+ . Then by Theorem 4N, 0 + m ∈ p+ + m. Then
by commutativity of addition and (A1), we have m ∈ m + p+ .
Exercise 4.23. Assume that m and n are natural numbers with m less than n. Show that
there is some p in ω for which m + p+ = n.
4
NATURAL NUMBERS
37
Solution. First note that n '= 0, since m ∈ n. If m = 0, then we can take 0+p+ = p+ = n,
and we know such a p exists since n is nonzero. Now suppose k ∈ n, and that for some
p, k + p+ = n. Since k ∈ n, k + ∈ n. If k + = n, then the conclusion holds trivially, since
k + '∈ n. If k + ∈ n, then observe that
k + p+ = (k + p)+ = k + + p = n.
Since k + '= n, we have p '= 0, and thus p = q + for some q. Hence k + + q + = n, and thus
conclusion holds for all m ∈ n. Essentially, since we know 0 + n = n, we can then find a p
such that 1 + p+ = n, and from this we can find a q such that 2 + q + = n, and so on for
all m ∈ n. This must eventually terminate, as n is finite.
Exercise 4.24. Assume that m + n = p + q. Show that
m ∈ p ⇔ q ∈ n.
Solution. Suppose m ∈ p. By the previous exercise, there exists some r such that
m + r+ = p. Hence m + n = m + r+ + q. Due to the cancellation and commutativity
of addition, we then have n = q + r+ . It follows immediately from the previous exercises
that q ∈ n. A completely parallel argument shows that if q ∈ n, then m ∈ p.
Exercise 4.25. Assume that n ∈ m and q ∈ p. Show that
(m · q) + (n · p) ∈ (m · p) + (n · q).
Solution. Since q ∈ p, from Exercise 23 we have that p = q + s+ for some s. Then note
n ∈ m =⇒ n · s+ ∈ m · s+
=⇒ m · q + n · q + n · s+ ∈ m · q + n · q + m · s+
=⇒ m · q + n · (q + s+ ) ∈ m · (q + s+ ) + n · q
=⇒ (m · q) + (n · p) ∈ (m · p) + (n · q).
Exercise 4.26. Assume that n ∈ ω and f : n+ → ω. Show that ran f has a largest
element.
Solution. Let
A = {n ∈ ω | f : n+ → ω has a largest element.}.
If n = 0, then n+ = 1, so ran f = {f (0)}, and thus trivially possesses a largest element.
So suppose k ∈ A, and thus for f : k + → ω, max(ran f ) exists. Recall k ++ = k + ∪ {k + },
and consider f : k + ∪ {k + } → ω. Not f !k + " possesses a largest element, call it M . The
only other element in ran f is f (k + ). If f (k + ) = M , then max ran f = M = f (k ) . If
f (k + ) ∈ M , then max(ran f ) = M . If M ∈ f (k + ), then by transitivity, max(ran f ) =
f (k + ). In any case, ran f possesses the largest element, so k + ∈ A. So for any n ∈ ω and
f : n+ → ω, ran f has a largest element.
4
NATURAL NUMBERS
38
Alternatively, recall that any natural number is just the set of smaller natural numbers.
As such, n+ is a set with n+ = n + 1 elements. Since f is a function, ran f can have at
most n + 1 elements. Thus ran f is a finite set, and let us say that it has j + elements. If
ran f has 1 element, then trivially it has a largest element. Otherwise, take two distinct
elements m and n in ran f . By the trichotomy law, {m, n} has a largest element. So
suppose that any subset A of j or fewer elements of ran f has a largest element, say
max A. Denoting k as the element in the set ran f \ A, we see that the largest element of
ran f is the larger of max A and k, which is sure to exist by trichotomy.
Exercise 4.27. Assume that A is a set, G is a function, and f1 and f2 map ω into A.
Further assume that for each n in ω both f1 " n and f2 " n belong to dom G and
f1 (n) = G(f1 " n) ∧ f2 (n) = G(f2 " n).
Show that f1 = f2 .
Solution. It suffices to show that f1 and f2 agree on all elements in ω. Let
B = {m ∈ ω | f1 (m) = f2 (m)}.
Note that f1 " 0 = {/u, v0 | f (u) = v ∧ u ∈ 0} = ∅, since there are no elements in 0.
Similarly, f2 " 0 = ∅, and so
f1 (0) = G(f1 " 0) = G(∅) = G(f2 " 0) = f2 (0)
and thus 0 ∈ B. So assume that k ∈ B, and hence f1 (k) = f2 (k). Then observe that
f1 (k + ) = G(f1 " k + ) = G(f1 " k) ∪ G(f1 " {k}) = G(f1 " k) ∪ G(/k, f1 (k)0)
and
f2 (k + ) = G(f2 " k + ) = G(f2 " k) ∪ G(f2 " {k}) = G(f2 " k) ∪ G(/k, f2 (k)0).
By the induction hypothesis, G(f1 " k) = G(f2 " k), and since f1 (k) = f2 (k), then
G(/k, f1 (k)0) = G(/k, f2 (k)0). It follows that f1 (k + ) = f2 (k + ), and so k + ∈ B. So B = ω,
and f1 = f2 .
Exercise 4.28. Rewrite the proof of Theorem 4G using, in place of induction, the well
ordering of ω.
Solution. Let
T = {n ∈ ω | n '⊆ ω}.
Suppose m := min T exists, so m '⊆ ω. Now m '= 0, since 0 ⊆ ω. Thus m = p+ for some
p. Then m = p ∪ {p}. Since p ∈ m, we must have p ⊆ ω. Clearly {p} ⊆ ω. It follows
that m ⊆ ω, a contradiction. Hence T has no least element, and thus by the well-ordering
principle, T = ∅. It follows that for all n ∈ ω, n ⊆ ω, and thus ω is transitive.
4
NATURAL NUMBERS
4.6
39
Review Exercise
Exercise 4.29. Write an expression for the set 4 using only symbols ∅, {, }, and commas.
Solution. Recall 4 = {0, 1, 2, 3}. Hence
4 = {∅, {∅}, {∅, {∅}}, {∅, {∅}, {∅, {∅}}}}.
Exercise 4.30. What is
!
4? What is
#
4?
Solution. By Theorem 4E, ∪(a+ ) = a, and thus ∪4 = 3 = {∅, {∅}, {∅, {∅}}}. Also,
∩4 = 0 ∩ 1 ∩ 2 ∩ 3 = 0, since any set intersected with ∅ is again ∅.
!!
Exercise 4.31. What is
7?
Solution. Again by Theorem 4E, we have
""
"
7=
6 = 5.
Exercise 4.32.
(a) Let A = {1}. Calculate A+ and
!
(b) What is ({2}+ )?
!
(A+ ).
Solution. First,
A+ = {1} ∪ {{1}} = {1, {1}} = {{∅}, {{∅}}}.
Hence
"
(A+ ) = {∅} ∪ {{∅}} = {∅, {∅}} = 2.
Also, {2}+ = {2} ∪ {{2}} = {2, {2}}. So
"
({2}+ ) = 2 ∪ {2} = {∅, {∅}} ∪ {{∅, {∅}}} = {∅, {∅}, {∅, {∅}}} = 3.
Exercise 4.33. Which of the following
sets are transitive? (For each set S that is not
!
transitive, specify a member of S not belonging to S.)
(a) {0, 1, {1}}.
(b) {1}.
(c) /0, 10.
Solution.
4
NATURAL NUMBERS
40
(a) Note
∪{0, 1, {1}} = 0 ∪ 1 ∪ {1} = 1 ∪ {1} = {∅} ∪ {{∅}} = {∅, {∅}} = 2 = {0, 1}.
Since {0, 1} ⊆ {0, 1, {1}}, this set is transitive.
!
(b) {1} = 1 = {∅}. Note that {∅} '⊆ {1} = {{∅}}, since ∅ '∈ {{∅}}.
!
!
!
(c) Note /0, 10 = {{0}, {0, 1}} = {2} = 2. Since 2 = {0, 1} '⊆ {{0}, {2}}, since
∅ ∈ 2, but ∅ '∈ {{0}, {2}}.
Exercise 4.34. Find suitable a, b, etc. making each of the following sets transitive.
(a) {{{∅}}, a, b}.
(b) {{{{∅}}}, c, d, e}.
Solution. For (a), let a = ∅ and b = {∅}. Then
"
{{{∅}}, ∅, {∅}} = {∅, {∅}}.
Since {∅, {∅}} ⊆ {{{∅}}, ∅, {∅}}, the set is transitive.
For (b), let c = ∅, d = {∅}, and e = {{∅}}. Then
"
{{{{∅}}}, ∅, {∅}, {{∅}}} = {∅, {∅}, {{∅}}}.
Since {∅, {∅}, {{∅}}} ⊆ {{{{∅}}}, ∅, {∅}, {{∅}}}, the set is transitive.
Exercise 4.35. Let S be the set /1, 00.
(a) Find a transitive set T1 for which S ⊆ T1 .
(b) Find a transitive set T2 for which S ∈ T2 .
Solution. Let
and
T1 = {{{∅}}, {∅, {∅}}, ∅, {∅}}
T2 = {{{{∅}}, {∅, {∅}}}, {∅, {∅}}, {{∅}}.
It is straightforward to verify that both these sets satisfy the conditions of the problem.
Exercise 4.36. There is a function h : ω → ω for which h(0) = 3 and h(n+ ) = 2 · h(n).
What is h(4)?
Solution. One has the following set of implications:
h(0) = 3 =⇒ h(1) = 2 · h(0) = 6
=⇒ h(2) = 2 · h(1) = 12
=⇒ h(3) = 2 · h(2) = 24
=⇒ h(4) = 2 · h(3) = 48
4
NATURAL NUMBERS
41
Exercise 4.37. We will say that a set S has n elements (where n ∈ ω) iff there is a oneto-one function from n onto S. Assume that A has m elements and B has n elements.
(a) Show that A and B are disjoint, then A ∪ B has m + n elements.
(b) Show that A × B has m · n elements.
Solution. Let f1 : m → A and f2 : n → B be two possible bijections. Define a new
bijection g : m + n → A ∪ B as such:
,
f1 (a) if a ∈ m
g(a) =
f2 (b) otherwise, where a = m + b.
This equation is a bijection, and since A and B are disjoint, there will be a unique image
for each a ∈ m + n.
Define also a bijection h : m · n → A × B as such:
g(a) = /f1 (i), f2 (j)0
if m · i ∈ a ∈ m · i+ and a = (m · i) + j.
Essentially for all intervals of m, we holds and element of A fixed, and lets the second
coordinate range through all n possibilities for B, before moving to the next interval.
Exercise 4.38. Assume that h is the function from ω into ω for which h(0) = 1 and
h(n+ ) = h(n) + 3. Give an explicit expression for h(n).
Solution. I claim that h(n) = 3 · n + 1 for all n ∈ ω. Note that h(0) = 3 · 0 + 1. Assume
that h(k) = 3 · k + 1. Then h(k + ) = h(k) + 3 = 3 · k + 1 + 3 = 3 · (k + 1) + 1 = 3 · k + + 1.
So by induction, h(n) = 3 · n + 1 for all n ∈ ω.
Exercise 4.39. Assume that h is the function from ω into ω for which h(0) = 1 and
h(n+ ) = h(n) + (2 · n) + 1. Give an explicit (not recursive expression for h(n).
Solution. After calculating the first few values, I claim h(n) = n · n + 1. Note h(0) =
0 · 0 + 1 = 1. So assume h(k) = k · k + 1. Then
h(k + ) = h(k) + 2 · k + 1
=k·k+1+2·k+1
=k·k+2·k+2
= k · k + k + (k + 1) + 1
= k · (k + 1) + (k + 1) + 1
= (k + 1) · (k + 1) + 1
= k+ · k+ + 1
Hence h(n) = n · n + 1 for all n ∈ ω.
Exercise 4.40. Assume that h is the function from ω into ω defined by h(n) = 5 · n + 2.
Express h(n+ ) in terms of h(n) as simply as possible.
5
CONSTRUCTION OF THE REAL NUMBERS
42
Solution. Note that
h(n+ ) = 5 · n+ + 2 = 5 · n + 5 + 2 = 5 · n + 2 + 5 = h(n) + 5.
5
Construction of the Real Numbers
5.1
Integers
Exercise 5.1. Is there a function F : Z → Z satisfying the equation
F ([/m, n0]) = [/m + n, n0]?
Solution. Let
F̂ : ω × ω → ω × ω : /m, n0 4→ /m + n, n0.
By Theorem 3Q, it is enough to show that G is not compatible with ∼ to show that such a
function F does not exist. Take for example /5, 20 and /3, 00. Then clearly /5, 30 ∼ /3, 00.
Note that Ĝ(/5, 20) = /5+2, 20 = /7, 20 and F̂ (/3, 00) = /3, 00. Hence F̂ (/5, 20) '∼ F̂ (/3, 00),
so F̂ is not compatible with ∼, and thus no such F exists.
Exercise 5.2. Is there a function G : Z → Z satisfying the equation
G([/m, n0]) = [/m, m0]?
Solution. Let
Ĝ : ω × ω → ω × ω : /m, n0 4→ /m, m0.
Suppose /m, n0 ∼ /m" , n" 0. Note that G(/m, n0) ∼ G(/m" , n" 0), since /m, n0 ∼ /m" , n" 0 as
clearly m + m" = m" + m. Thus Ĝ is compatible with ∼, and so such a function G exists.
In fact, it is simply the function that maps all integers to 0.
Exercise 5.3. Is there a function H : Z → Z satisfying the equation
H([/m, n0]) = [/n, m0]?
Solution. Let
Ĥ : ω × ω → ω × ω : /m, n0 4→ /n, m0.
Note that if /m, n0 ∼ /m" , n" 0 then m + n" = m" + n, and so n + m" = n" + m, and hence
/n, m0 ∼ /n" , m" 0, or Ĝ(/m, n0) ∼ Ĝ(/m" , n" 0). Since Ĥ is compatible with ∼, and thus by
Theorem 3Q, such a function H exists.
Exercise 5.4. Prove that +Z is associative.
5
CONSTRUCTION OF THE REAL NUMBERS
43
Solution. Let a = [/m, n0], b = [/p, q0], and c = [/r, s0]. Then
(a +Z b) +Z c = ([/m, n0] +Z [/p, q0]) +Z [/r, s0]
= ([/m + p, n + q0]) +Z [/r, s0]
= [/(m + p) + r, (n + q) + s0]
= [/m + (p + r), n + (q + s)0]
by the associativity of natural numbers
= [/m, n0] +Z [/p + r, q + s0]
= [/m, n0] +Z ([/p, q0] +Z [/r, s0])
= a +Z (b +Z c)
Exercise 5.5. Give a formula for subtraction of integers:
[/m, n0] − [/p, q0] =?
Solution. Informally, we would like to have that
(m − n) − (p − q) = (m + q) − (n + p)
which suggests that we should define subtraction of integers by the formula
[/m, n0] − [/p, q0] = [/m + q, n + p0].
Note this is well defined. Suppose /m, n0 ∼ /m" , n" 0 and /p, q0 ∼ /p" , q " 0. So
m + n " = m" + n
and
p + q " = p" + q.
Adding the first equation to the reverse of the second yields
m + q + n " + p " = m" + q " + n + p
which implies /m + q, n + p0 ∼ /m" + q " , n" + p" 0. This justifies the definition of −Z .
Exercise 5.6. Show that a ·Z 0Z = 0Z for every integer a.
Solution. Let a = [/m, n0]. Then
a ·Z 0Z = [/m, n0] ·Z [/0, 00] = [/m · 0 + n · 0, m · 0 + n · 00] = [/0, 00] = 0Z ,
where we use the fact that 0 · k = 0 for k ∈ ω, and 0 + 0 = 0.
Exercise 5.7. Show that
a ·Z (−b) = (−a) ·Z b = −(a ·Z b)
for all integers a and b.
5
CONSTRUCTION OF THE REAL NUMBERS
44
Solution. Let a = [/m, n0] and b = [/p, q0]. Then
a ·Z (−b) = [/m, n0] ·Z [/q, p0] = [/mq + np, mp + nq0]
and
(−a) ·Z b = [/n, m0] ·Z [/p, q0] = [/np + mq, nq + mp0].
Hence by the commutativity of + and · on ω, we have a ·Z (−b) = (−a) ·Z b. Also, observe
a ·Z b + (−a) ·Z b = [/mp + nq + np + mq, mq + np + nq + mp0] = [/0, 00] = 0Z .
Since inverses are unique, we must have that (−a) ·Z b = −(a ·Z b) and so
a ·Z (−b) = (−a) ·Z b = −(a ·Z b).
Exercise 5.8. Prove parts (a), (b), and (c) of Theorem 5ZL.
Solution. For m, n ∈ ω,
(a) E(m + n) = [/m + n, 00] = [/m, 00] +Z [/n, 00] = E(m) + E(n).
(b) Since 0 · k = 0 for k ∈ ω,
E(mn) = [/mn, 00] = [/m, 00] ·Z [/n, 00] = E(m) ·Z E(n).
(c) Note
m ∈ n ⇐⇒ m + 0 ∈ n + 0 ⇐⇒ [/m, 00] <Z [/n, 00] ⇐⇒ E(m) <Z E(n).
Exercise 5.9. Show that
[/m, n0] = E(m) − E(n)
for all natural numbers m and n.
Solution. Indeed,
[/m, n0] = [/m, 00] +Z [/0, n0] = [/m, 00] − [/n, 00] = E(m) − E(n).
5
CONSTRUCTION OF THE REAL NUMBERS
5.2
45
Rational Numbers
Exercise 5.10. Show that r ·Q 0Q = 0Q for every rational number r.
Solution. Let r = [/a, b0], with b '= 0Z . Then
r ·Q 0Q = [/a, b0] ·Q [/0, 10] = [/a · 0, b · 10] = [/0, b0] = [/0, 10] = 0Q .
Exercise 5.11. Give a direct proof that if r ·Q s = 0Q , then either r = 0Q or s = 0Q .
Solution. Let r = [/a, b0] and s = [/c, d0], and suppose r ·Q s = 0Q . Then
r ·Q s = 0Q ⇐⇒ [/ac, bd0] = [/0, 10]
⇐⇒ ac ·Z 1 = bd ·Z 0
⇐⇒ ac = 0Z
since Z is an integral domain.
⇐⇒ a = 0Z ∨ b = 0Z
If a = 0Z , then r = 0Q , and likewise for s if b = 0Z .
Exercise 5.12. Show that
r < Q 0Q
iff
0Q <Q −r.
Solution. Let r = [/a, b0] with b '= 0Z . Suppose r <Q 0Q , so
[/a, b0] <Q [/0, 10] ⇐⇒ a · 1 < 0 · b = 0
⇐⇒ a < 0
⇐⇒ 0 < −a
⇐⇒ 0 · b < (−a) · 1
⇐⇒ 0Q <Q −r
since −r = [/−a, b0].
Exercise 5.13. Give a new proof of the cancellation law for +Z using Theorem 5ZD
instead of Theorem 5ZJ.
Solution. Theorem 5ZD tells us that additive inverses for integers exist. So suppose
a +Z c = b +Z c. The additive inverse of c, call it −c, exists, and adding it to both sides
yields the implications
a +Z c +Z (−c) = b +Z c +Z (−c) =⇒ a +Z 0Z = b +Z 0Z =⇒ a = b.
Exercise 5.14. Show that the ordering of rations is dense, i.e., between any two rationals
there is a third one:
p <Q s ⇒ (∃r)(p <Q r <Q s).
5
CONSTRUCTION OF THE REAL NUMBERS
46
Solution. Let p = [/a, b0] and s = [/c, d0], with b, d > 0Z . Note that this implies ad <Z cb,
as well as abd <Z bbc and add <Z bcd since b and d are positive. Now let
r = (p +Q s) ÷ [/2, 10] = [/ad + cb, bd0] ·Q [/1, 20] = [/ad + cb, 2bd0].
Using the inequalities mentioned above, observe
2abd = abd + abd <Z abd + bbc = b · (ad + cb)
from which it follows that p <Q r. Also,
(ad + cb) · d = add + bcd < bcd + bcd = 2bcd
from which it follows that r <Q s, and so p <Q r <Q s, showing that the ordering of the
rational numbers is dense.
!
Exercise 5.15. In Theorem 5RB, show that A is closed downward and has no largest
element.
!
Solution. Suppose q ∈ A, so for some x !
∈ A, q ∈!A. Take any r < q, then since x is
downward closed, r ∈ !
x. It follows that r ∈ A, so A is closed downward.
Now take any p ∈ A. So p ∈ x for some x!∈ A. Since x has no largest element,
there
!
is some q ∈ x such that p < q. !Of course, q ∈ A as!well. Hence for any p ∈ A, we can
always find some element q ∈ A larger than p, so A has no largest element.
Exercise 5.16. In Lemma 5RC, show that x +R y has no largest element.
Solution. Take any q + r ∈ x +R y. So q ∈ x and r ∈ y. Since neither x nor y has a
largest element, there exist q " ∈ x and r" ∈ y such that q < q " and r < r" . Since addition
preserves order in the rationals, q + r < q " + r" ∈ x +R y. Hence x +R y has no largest
element.
Exercise 5.17. Assume that a is a positive integer. Show that for any integer b there is
some k in ω with
b < a · E(k).
Solution. In the argument that follows, < shall be used to denote <Z . Firstly, if b ≤ 0,
then clearly b < a, so take k = 1. Then b < a = a · 1Z = a · E(1).
Now suppose that b > 0. Let b = [/m, n0]. Since b = [/m, n0] > [/0, 00], we have
0 + n = n ∈ m = 0 + m. Then b = [/m, n0] < [/m + 1, 00] = E(m + 1), since m + 0 =
m ∈ (m + 1) + n, as m ∈ m + 1 and 0 ∈ n. Since 0 < a, we have by Theorem 5ZJ that
b · a < E(m + 1) · a. Now since 1 ≤ a, we also have b = b · 1 ≤ b · a. It follows then that
b ≤ b · a < E(m + 1) · a.
By transitivity and commutativity of multiplication, we then have that b < a · E(m + 1),
so take k = m + 1.
5
CONSTRUCTION OF THE REAL NUMBERS
47
Exercise 5.18. Assume that p is a positive rational number. Show that for any rational
number r there is some k in ω with
r < p · E(E(k)).
(Here k is in ω, E(k) is the corresponding integer, and E(E(k)) is the corresponding
rational.)
Solution. First, if r ≤ 0, then r < p, and so r < p = p · 1Q = p · E(E(1)) since
E(E(1)) = E([/1, 00]) = E(1Z ) = [/1, 10] = 1Q .
Now suppose 0 < r. Denote r = [/m, n0]. Note that [/0, 10] < [/m, n0] implies 0 · n < m · 1,
so m is positive. Furthermore, denote p = [/s, t0], with s > 0 since p is positive. Consider
E(E((m + 1) · t)) = [/(m + 1) · t, 10]. I claim r < p · E(E((m + 1) · t)). Observe
p · E(E((m + 1) · t)) = [/s, t0] · [/(m + 1) · t, 10]
= [/s · (m + 1) · t, t0]
= [/s · (m + 1), 10]
by canceling the t.
Now m < m + 1, and since n and s are both positive integers, 1 ≤ n · s. Then by Theorem
5QJ we have
m = m · 1 ≤ m · (n · s) < (m + 1) · n · s.
from which it follows that [/m, n0] < [/s · (m + 1), 10], so r < p · E(E((m + 1) · t)). Hence
take k = (m + 1) · t.
Exercise 5.19. Assume that p is a positive rational number. Show that for any real
number x there is some rational q in x such that
p+q ∈
/ x.
Solution. We will assume that the following properties hold for the rationals: if p > 0
and q < 0, then pq < 0. Also, (−p)q = p(−q) = −(pq), both of which is easily verifiable
by appling the definitions of orderings on Q, Z, and ω.
Now, a modification of the result of the previous exercise. I claim that if p is a positive
rational, and r is any rational, then there exists an integer n ∈ Z such that p · E(n) < r. If
r ≥ 0Q , then we make take any n < 0Z , and hence E(n) < 0Q , so p · E(n) < r. If r < 0Q ,
then observe that
p · E(n) < r ⇐⇒ −r < −(p · E(n)) = p · (−E(n))
as we may add the additive inverse of each to both sides, and the equality follows from
one of our originally stated properties. Now −r > 0Q , so it follows that −E(n) > 0Q as
well. So −E(n) = E(E(k)) for some k ∈ ω, so we know −r < p · (−E(n)) holds by the
previous exercise. Hence p · E(n) < r for some n ∈ Z.
Now since x '= Q, we know there is some r ∈
/ x. By the previous exercise, there is some
k ∈ ω such that r < p · E(E(k)). Hence p · E(E(k)) ∈
/ x. Similarly, since x '= ∅, there is
5
CONSTRUCTION OF THE REAL NUMBERS
48
some rational s ∈ x, and we know there is some n ∈ Z such that p · E(n) < s, and thus
p · E(n) ∈ x. Now consider the set
A = {m ∈ Z | p · E(m) ∈
/ x}.
Note that this set is nonempty, but bounded below by n, since for any m such that
p · E(m) ∈
/ x, we must have m > n, since x is downward closed. It follows that there
is a least such m such that p · E(m) ∈
/ x, and thus p · E(m − 1) ∈ x. We may take
q = p · E(m − 1), and thus
p + q = p + p · E(m − 1) = p · (1Q + E(m − 1)) = p · (E(1) + E(m − 1)) = p · E(m),
and thus p + q ∈
/ x. 2
Exercise 5.20. Show that for any real number x, we have 0R ≤R |x|.
Solution. To prove this statement is equivalent to proving
{r ∈ Q | r < 0} ⊆ {p ∈ Q | (∃s > p) − s ∈
/ x}.
Take r ∈ 0R , so r < 0. If r ∈ x, we are done, so suppose r ∈
/ x. Since r < 0, it follows that
for all y ∈ x, y < 0, for if y ≥ 0, then we would have that r ∈ x. Hence x contains only
negative rational numbers. Let s = 0. Then s > r, −s = 0 (since 0 + 0 = 0), and hence
−s ∈
/ x. So r ∈ −x, thus proving the containment.
Exercise 5.21. Show that if x <R y, then there is a rational number r with
x <R E(r) <R y.
Solution. First, we will prove the archimedean property of R. That is, if x, y ∈ R and
x > 0R , then there exists a positive n ∈ Z such that y <R x · E(E(n)).
Let
A = {x · E(E(n)) | n ∈ Z ∧ n > 0Z }.
Suppose that the above claim is not true, so it is not the case that there exists an n such
that x · E(E(n)) >R y. That is, x · E(E(n)) ≤R y for all n, so y is an upper bound of
A. By Theorem 5RB, A has a supremum, so let α = sup A. Since x > 0R , it follows that
α − x < α, and thus α − x is not an upper bound of A. So
α − x < x · E(E(m))
for some m ∈ Z.
So
α < x · E(E(m)) + x · E(E(1)) = x · (E(E(m)) + E(E(1))) = x · E(E(m + 1)),
as the composition of homomorphisms is again a homomorphism. Hence we reach a
contradiction, since α is an upper bound of A, so the archimedean property holds.
2
Thanks to Andres Caicedo for his suggestions and sketch proof.
5
CONSTRUCTION OF THE REAL NUMBERS
49
Now suppose x <R y, so 0 <R y − x. By the archemedean property, there exists an
n ∈ Z such that
E(E(n)) · (y − x) >R 1.
Furthermore, there exist m1 , m2 ∈ Z such that m1 >R x · E(E(n)) and E(E(m2 )) >R
x · (−E(E(n))). Hence
−E(E(m2 )) <R E(E(n)) · x <R E(E(m1 )).
Now x · E(E(n)) is a real number, and so taking p = 1 in Exercise 5.18, we see there is an
integer k such that E(E(k)) ∈ x but E(E(k + 1)) ∈
/ x. That is,
E(E(m − 1)) <R x · E(E(n)) <R E(E(m))
where we write k = m + 1. So
x · E(E(n)) <R E(E(m)) ≤R x · E(E(n)) + 1 <R y · E(E(n))
and thus
x · E(E(n)) <R E(E(m)) <R y · E(E(n)).
Now since n is nonzero, both E(n) and E(E(n)) are also nonzero, since it is the composition of injections. Then by Theorem 5RI(d), there is a nonzero real y, such that x · y = 1R .
Denote such a y as E(E(n))−1 . Note that E(E(n)−1 ) · E(E(n)) = E(E(n)−1 · E(n)) =
E(1Q ) = 1R , since E is a homomorphism, and then since E is injective, we have that
E(E(n))−1 = E(E(n)−1 ). So altogether,
E(E(n)−1 ) · E(E(n)) = E(E(n)−1 · E(n)),
with E(n)−1 · E(n) ∈ Q. So multiplying through yields
x <R (E(n)−1 · E(n)) · E(m) <R y
Taking this to be the desired r, we have
x <R E(E(n)−1 · E(n) · E(m)) <R y.
Exercise 5.22. Assume that x ∈ R. How do we know that |x| ∈ R?
Solution. Recall |x| = x ∪ −x. Since x '= ∅ and −x '= ∅, clearly |x| '= ∅. Suppose now
that x <R 0R . Then for any q ∈ x, 0 > q, and −0 = 0 ∈
/ x, since 0 ∈
/ 0R , and consequently
not in x. Hence q ∈ −x, and thus x ⊆ −x. Hence |x| = x ∪ −x = −x '= Q.
Suppose instead that x ≥R 0. Suppose r ≥ 0, then if s > r, then s > 0, and so −s < 0.
Thus s ∈ x, so r ∈
/ −x. So if r ∈ −x, then necessarily r < 0, and so r ∈ x. Hence −x ⊆ x,
and so |x| = x ∪ −x = x '= Q. Hence ∅ =
' |x| =
' Q.
Now suppose q ∈ |x|. Suppose r < q. If q ∈ x, then r ∈ x since x is downward closed,
so r ∈ |x|. Similarly, if q ∈ −x, then r ∈ −x ,since −x is downward closed, so r ∈ |x|.
Now take any p ∈ |x|. If p ∈ x, there exists some p" ∈ x such that p" > p, since x has
no greatest element. If p ∈ −x, there exists some p"" ∈ −x such that p"" > p, since −x has
no greatest element. In either case, p" or p"" is in |x|, and so |x| has no greatest element.
Altogether, |x| ∈ R.