Real Analysis with Economic Applications - Chapter B ppt

Chapter B
Countabilit y
This chapter is about Cantor’s coun tability theory whic h i s a standard prerequisite
for eleme ntary real analysis. Our tre atment is incom p lete in that we cover only th ose
results that are im m ediately r elevant for the p resent course. In particular, we have
little t o sa y here a bout ca rdinalit y theory a nd ordinal numbers.
1
We shall, however,
cover tw o relativ ely ad vanced topics here, namely, the theory of o rder isomorphisms
and the Schröder-Bernstein Theorem on the “equivalence” of infinite sets. If you are
familiar with countable and uncountable sets, you might wan t to skip Section 1 and
jump directly to the d is cu s sion of these two topics. The former o n e is put to good
use in the last section of the chapter which pro v ides an introduction to ordinal utilit y
theory, a topic that we s hall revisit a few m o re times later. The Schröder-Bernstein
Theorem is, in turn, proved via the Tars ki F ixed Point Theorem, the first of the many
fixe d point th eorems that are discussed in this book. Th is theorem should certainly
be inclu d ed in the tool kit of an economic th eo rist, for it has recen tly found a number
of impor tant application s in game the ory.
1 Countable and Uncountable Sets
In this section we revisit set theory, and begin to sketch a syste matic meth od of
thinking about the “size” o f any given set. The issue is n ot problematic in th e c ase of
finite sets, for w e can simply count the number of members of a giv en finite set, and
use this number as a measure o f its size. Thus, quite simply, one finite set is “more
crow ded” than another, if it contains more elemen ts than the other. But how can
one extend t his method t o the case o f infinite sets? O r, how c an we decide whether
or not a given infinite set is “more crowded” than another suc h set? C learly, things
get icy with infinite sets. O ne may even t hink at firstthatanytwoinfin ite se ts are
equally crowded, or ev en that the question is meaningless. There is, how ever, an
in tu itive way to approach the problem of ranking the sizes of infin ite sets, an d it is
this approa ch that th e current section is about. As a first pass, we will talk about
the sense in which an infin ite set can be thought o f as “small.” We will then a pp ly

our discussion to compare the sizes of the sets of in tegers, rational numbers, and real
n u mbers. Various other app lica t ion s will be given as we proceed.
2
1
For a rigorous introduction to these topics, and to set theory in general, you should consult on
outlets like Halmos (1960), Kaplan sky (1977), Enderton (1977), and/or Devlin (1993). My favorite
is, by far, Enderton (1977).
2
We ow e our abilit y to compare the sizes of infinite sets to the great German mathematician Georg
Cantor (1845-1818). While the notion of infinity was debated in philosophy for over two thousand
years, it was Cantor who pr ovided a precise manner in which infinity can be understood, studied and
63
Let us begin b y defining the two fundamen tal concepts that will play a fundamen-
tal role in what follows.
Dhilqlwlrq. AsetX is ca lled coun tably infinite if there exists a bijection f that
maps X onto the set N o f natural numbers.Xis called coun table if it is either finite
or countably infinite.Xis called uncoun table if it is not countable.
So, quite intuitively, we can “cou nt” the members of a countable set just like
w e could coun t the natural n umbers. An infinite set like X = {x
1
,x
2
, } is thus
countable, for w e can put the members of X in to one-to-one correspondence with the
natural numbers:
x
1
←→ 1 x
2
←→ 2 ··· x

m
←→ m ···
Conv ersely, if X is countably infinite, then it can be enumerated as X = {x
1
,x
2
, }.
To see this for mally, let f be a bijection from X on to N. Then f is invertible (Propo-
sition A.2), and the inverse fun ct ion f
−1
is a bijection f r om N onto X. But this means
that we must have X = {f
−1
(1),f
−1
(2), }. Th us, if we let x
i
= f
−1
(i),i=1, 2, ,
we may write X = {x
1
,x
2
, } nice and easy.
Remark 1. Let X be a counta ble subset of R. Can we meaningfully talk about
the sum of the elements of X? Well, it de pends! If X is finite, ther e is o f course
no problem ; the s um of t he elements of X, denoted

x∈X

x is w e ll-defined.
3
When
X is coun tably infinite, w e need to be a bit m ore careful. The natural inclination
is to take an enumeration {x
1
,x
2
, } of X,anddefine

x∈X
x as

∞
x
i
, provided
that

∞
x
i
∈ R. Unfortunately, this need not w ell-define

x∈X
x —whatifweuseda
different enumeration of X? You s ee, the issue is non e other th an th e rearran gem ent of
infinite series that w e discussed in Section A.3.5. The right definition is:

x∈X

x :=

∞
x
i
where {x
1
,x
2
, } is any en u m e ration of X, provided that

∞
x
i
∈ R and

∞
x
i
is invariant under rearrangements. (If any of the latter two c o nd ition s fails,
we say that

x∈X
x is undefined.) In particular, thanks to Proposition A.10,

x∈X
x
is well-defined when X ⊆ R
+
. 

even m anipula ted. It would not be an exaggeration to say that Cantor’s ideas were deca des ahead
of his contemporaries. His strength did not stem from his capability to do hard proofs, but from
his ability to think “outside the box.” Most mathematicians of his cohort, including som e eminent
figures (such as Klein, Kronecker and Poincaré), found Cantor’s theory of infinite sets nonsensical,
which has lat e r found to provide a sound foundation for muc h of mathematics a t large. After Cantor,
so says James (2002), p. 214, “mathematics was never to be the same again.” There are not a whole
lot of people in history about whom o ne can say something like this.
There a re m any references that detail Cantor’s life and work. My favorite is Dauben (1980) from
which one learns some mathematics as well.
3
Recall that we handle the exceptional case X = ∅ by convention:

x∈∅
x =0.
64
Letuslookatafewexamplesofcountablesets. N is countable, a s it is evident
from the definition of countability. It is also easy to see that any subset of N is
countable. For instance, the set of all even natural numbers is countably infinite,
because we can put this set in one-to-one correspondence with N:
2 ←→ 14←→ 2 ··· 2m ←→ m ···
Similarly, the set of all prime nu mbers is countable.
4
So, in a sense, there are as
man y ev en (or prime) numbers as there are natural numbers.
5
Since the set of all
even numbers is a proper subset of N, thi s may appear counter-intuitive at first.
Nonetheless, this sort of a thing is really in the natu re of the notion of “infinit y,”
and it simply tells us that one has to dev e lop a new kind of “intuition” to deal with
infinite sets. The following result is a step to wards this directio n.

Proposition 1. Every subset of a co untable set is countable.
Pro of. Let X be a coun table set, and tak e any su bset S of X. If S is finite, there
isnothingtoprove,sosay|S| = ∞. Then X m ust be countably infinite, and thus we
can enumerate it as X = {x
1
,x
2
, }. No w define the self-map f on N inductively as
follows:
f(1) = min{i ∈ N : x
i
∈ S}
f(2) = min{i ∈ N : i>f(1) and x
i
∈ S}
f(3) = min{i ∈ N : i>f(2) and x
i
∈ S}

(So x
f(1)
is the first term of the sequence (x
1
,x
2
, ) that belongs to S, x
f (2)
is the
second, and so o n.) Now c onsider g ∈ S
N

defined by g(i):=x
f(i)
. Since g is a bijection
(isn’t it?), we are done. 
Exercise 1 . Show that if B is a countable set, and if there exists an injection from
aset
A into B, then A must be countable.
4
This set is, in fact, countably infinite, fo r according to a celebrated theorem of Euclid, there ar e
infinitely many prime numbers. Euclid’s proof of this theorem is almost as famous as the theorem
itself: Suppose there are finitely many primes, say, x
1
, , x
m
, and show that 1+

m
x
i
is prime.
(One needs t o use in the a rgument the fact that every integer k>1 is either prime o r a product of
primes, but this ca n easily be proved by using the Principle of Mathematical Induction.)
5
This observation is popularized by means of the following anectode. One night countably infi-
nitely man y passengers came to Hilbert’s Hotel, each looking for a room. Now Hilbert’s Hotel did
contain c ountably infinitely man y rooms, but that night all of the rooms w ere occupied. This was
no problem for Hilbert. He asked ev erybody staying in the hotel to come down to the lobb y, and
reallocated e ach of them using only the even-numbered rooms. This wa y all of the newco mers could
be accommodated in the odd-numbered rooms. (Here Hilbert refers to David Hilbert, one of the
most prominent figures in the history o f mathematics. We will meet him later in the course.)

65
Exercise 2.
H
Show that every infinite set has a countably infinite subset.
Proposition 1 shows how one can obtain new coun t ab le sets by “shrinking” a
countable set. It is also possible to “expand” a coun table set in o rder to obtain
another countable set. For instance, Z
+
is coun table, for f : i → i +1 defines a
bijection from N ∪ {0} on to N.
6
Similarly, th e set Z of all integers is countab le.
Indeed, the map f : N → Z\{0}, defined by f(i):=
i+1
2
if i is odd, and f (i):=−
i
2
if
i is even, is a bijection. M ore generally, we have the follo wing useful resu lt.
Proposition 2. A countable union o f countable sets is coun ta ble.
Pro of. Let X
i
be a countable set for each i ∈ N. Amoment’sreflection will co nvince
y ou that we would be done if we could sho w that X := X
1
∪X
2
∪··· is a co untable set.
(But make su re you are really convinced.) It is without loss of gen erality t o assume

that X
i
∩ X
j
= ∅ for ve r y distinct i and j. (Right?) Since each X
i
is coun table, we
ma y enumerate it as X
i
= {x
i
1
,x
i
2
, } (but note t hat X
i
may be finite). Now d efine
the mapping f : X → N
2
by f(x
i
k
):=(i, k). (Is f well-defined? Why?). Clearly, f is
injectiv e, a nd h ence, by Exercise 1, we m ay conclude that X is coun table, pro vided
that N
2
is countable. The proof is thus complete in view of the next exercise .
7


Exercise 3 . Counting the members of A × B as indicated in Figure 1, show that if
A and B are countable sets, t hen so is A × B.
Exercise 4.(A nother Proof for Proposition 2 ) Using the notation of the proof of
Prop osition 2, define
g ∈ N
X
by g(x
k
i
):=2
i
3
k
. Show that g is injective and invoke
Exercise 1 t o conclude that
X is countable.
∗∗∗∗FIGURE B.1 ABOUT HERE ∗∗∗∗
An importa nt implication of these results con c erns the c ou ntability of the set Q of
all rational n u mbers. Indeed, we have Q =

{X
n
: n ∈ N} where X
n
= {
m
n
: m ∈ Z}
for a ll n ∈ N. Bu t, o bviously, there is an injection from each X
n

in to Z, so since Z
is countable, so is each X
n
(Exercise 1 ). Therefore, by Pr oposition 2, we find that
Q must be countably infinite. (By Proposition 1, Q ∩ I is a lso countably i nfinite for
an y interval I in R.)
Corollary 1. (Cantor) Q is countable.
6
What would Hilbert do if a new customer came to h is hotel w hen all rooms were occupied?
7
Hidden in the proof is the Axiom of Choice. Quiz. Where is it? (Hint. This is subtle. Let A
i
be the set of all enumerations of X
i
(that is, the set of all bijections from N (or a finite set) on to
X
i
),i=1, 2, True, I know that each A
i
is nonempty, because each X
i
is countable. But in the
proof I work with an element of X
∞
A
i
, don’t I?)
66
So, there are as many rational num bers as there are natural num bers! This is
truly a “deep” observation. It tells us that Q can be enumerate d as {q

1
,q
2
, }, while
w e do not at all know ho w to “construct” such an enumeration. Suffices it to note
that Cantor himself was fascina ted with this theorem about which he is often quoted
as saying “I see it, but I don’t believe it.”
8
The coun ta bility of Q has profound implications. For instance, it implies that any
(nondegenera te) interval partition of R must be countab le, an observation which is
certainly worth keeping in min d. T h is is because in any giv en nondegenerate in terval,
there exists at least one rational number (in fact, there are infinitely many of them).
Th us, if there exi sted uncoun tably many nono verlapping intervals in R, we could
deduce that Q contains an uncoun table subset, which co n tradicts Proposition 1. In
fact, we can say som ething a little stronger in this regard.
Proposition 3. Let I be a set of nondegenerate in tervals in R suc h that |I ∩ J| ≤ 1
for any I, J ∈ I.ThenI is countable.
Pro of. Given the nondegeneracy hy pothesis, for each I ∈ I, there exist real num-
bers a
I
<b
I
with (a
I
,b
I
) ⊆ I, s o b y Proposition A.6, we can find a q
I
∈ Q with
q

I
∈ I. Define f : I → Q by f(I):=q
I
. Since any two mem bers of I overlap at most
at one point, f is injective. (Yes?) T hus the claim follows from the countability of Q
and Exercis e 1. 
So much for counta ble sets, what of uncountable ones? As you might guess, there
are plenty of them. Foremost, our belov ed R is uncoun table. Indeed, one cannot
enumerate R as {x
1
,x
2
, }, or what is the same thing, one cannot exhaust all real
numbers by counting them lik e one coun ts the natural numbers. Put differently, R is
“more crowded” than any coun table set.
Proposition 4. (Can tor) R is unco un table.
There are various w ays of proving this result. The proof we presen t here has
the a dvantage of making transparen t the reliance of t he result o n t he Completeness
Axiom. It is ba sed on the f ollow ing useful fact.
Can tor’s Nested In terval Lemma. Let I
m
:= [a
m
,b
m
] be a closed interval for each
m ∈ N.IfI
1
⊇ I
2

⊇ ···,then

∞
I
i
= ∅. If, in addition, b
m
− a
m
→ 0, then

∞
I
i
is
asingleton.
8
To be more precise, however, I should note that Cantor made this remark about the possibility
of constructing a bijection between a given line and a plane. While the spirit of this is simila r to
the argument that yields the countability of Q, it is considerably deeper. See D auben (1980) for a
detailed historical account of the matter.
67
Pro of. I
1
⊇ I
2
⊇ ··· im plies that (a
m
) is a bounded and increasing sequence
while (b

m
) is a bounded and decreasing sequence. By Proposition A.8, both of t hese
sequences conv erge (and this we o we to the Com p leteness Axiom), so lim a
m
= a and
lim b
m
= b for some real n u mbers a and b. It is easy to see that a
m
≤ a ≤ b ≤ b
m
for
all m, and that [a, b]=

∞
I
i
. (Check!) The second claim follows from the fact that
b − a = lim b
m
− lim a
m
= lim(b
m
− a
m
). 
We can now prov e the uncountability of the reals by mea n s of a method that w e
will later use on a few oth er occasions as well. (Note. This method is sometimes
referred to as “butterfly h unting.” You’ll s ee wh y in a s econd.)

ProofofProposition4. To deriv e a c ontradiction, suppose th at [0, 1] = {a
1
,a
2
, }
=: I
0
. We wish to find a real n umber (a “butterfly”) w hich is not e qual to a
i
for any
i ∈ N.
Divid e I
0
in to an y three closed nondegenerate intervals (say, [0, 1/3], [1/3, 2/3]
and [2/3, 1]). Clearly, a
1
does not belong to at l east one of these thre e in tervals, call
an y on e such interval I
1
. (There ha s to be a “butterfly” in I
1
.) Now play the same
game with I
1
, tha t is, divide I
1
in to three closed nondegenera te intervals, and observe
that a
2
does not belong to at least one of these subin tervals, say I

2
. (There has to be
a “butterfly” in I
2
.) Continuing this w ay, w e obtain a nested sequence (I
m
) of closed
in tervals, so Can tor’s Nested Interval Lemm a y ields

∞
I
i
= ∅. (Aha! We caught
our “ bu tterfly” in

∞
I
i
.) But, by construction, a
i
/∈ I
i
for e ach i, so we must have

∞
I
i
= ∅, a contradiction. 
Wh ile the the ory of countability is full of surprisin g results, it is extremely useful
and its basics are not v ery difficult to m aster. H owever, to develop a good intuition

about the theory, one needs to play around with a good num ber of coun table and
uncountable sets. The follow ing exercises provide an opportunity to do precisely this.
Exercise 5.
H
(a)ShowthatN
m
is countable for any m ∈ N.
(b) Prov e or dispro ve:
N
∞
is countable.
Exercise 6.Let
A and B be any sets such that A is countable, B is uncountable and
A ⊆ B. Can B\A = {x ∈ B : x/∈ A} be countable? Is the set of all irrational
numbers countable?
Exercise 7.
H
Let A and B be any sets such that A is countable and B is uncountable.
Show that
A ∪ B is uncountable. In fact, show that there is a bijection from B onto
A ∪ B.
Exercise 8.
H
For any self-map f on R,wesaythatx ∈ R is a point of disc onti-
nuity of
f if f is not continuous at x. Show that if f is monotonic, then it can have
at most countably many points of discontinuit y. Give an example of a real function
on
R which has uncountably many points of discontinuity.
68

∗
Exercise 9.
H
Let I be an open interv al and f ∈ R
I
.LetD
f
stand for the set of all
x ∈ I such that f is differentiable at x.Provethatiff is concave, then I\D
f
is
countable.
Exercise 10.
H
For any sets A and B, we write A 
card
B, if there exists an injection
from
B into A but there does not exist an injection from A into B.
(a)ShowthatA 
card
B need not hold even if B is a proper subset of A.
(b) Show that R 
card
Q.
(c)Showthat
2
A

card

A for any nonempty set A.
(d)Showthat2
N
is uncountable.
(e)(Cantor’s Par adox )Let
X = {x : x is a s et}. Use part (c) and the fact that
2
X
⊆ X to establish that X cannot be considered as a “set.”
2LosetsandQ
The set of rational numbers is not only countable, but it is also ordered in a natur al
wa y (Proposition A.4). These two properties of Q combine nicely in various applica-
tions of r eal a na lysis. It may th us be a g ood idea t o see what kinds o f s ets we can
in general not only count like Q but also order like Q. In fac t, we will see in Sect ion
4 that this issue is closely related t o a fundamen tal problem in decision theory. In
this section, there fore, w e will make its statement precise, and then outline Cantor’s
solutio n for it.
Let u s first s e e in what w ay we can relate a giv e n linearly ordered coun tab le set
to the set of rational n umbers.
Proposition 5. (Can to r) Let X be a countable set and  alinearorderonX. There
exists a function f : X → Q such that
f(a) ≥ f(b) if and only if a  b
for any a, b ∈ X.
Pro of. The claim is trivial when X is finite,soweassumethatX is countably
infinite. Owing to their countability, we may enumerate X and Q as
X = {x
1
,x
2
, } an d Q = {q

1
,q
2
, }.
We co nstruct the function f ∈ Q
X
as follows. First let f (x
1
):=q
1
. If x
1
 x
2
(x
2
 x
1
), then set f(x
2
) as the first elemen t (with respect to the subscript s) in
{q
2
, } such that q
1
≥ f(x
2
) (f(x
2
) ≥ q

1
, respectively). Proceeding inductively, for
an y m =2, 3, , we set f(x
m
) as the first elemen t of {q
1
, }\{f(x
1
), , f(x
m−1
)}
which has the same or der relation (with respect to ≥)tof(x
1
), ,f(x
m−1
) as x
m
has
to x
1
, , x
m−1
(w it h res pect to ). (Why is f well-defined?) I t follows readily from
this construction that, for any a, b ∈ X, we have f(a) ≥ f(b) iff a  b. 
69
So is it then true that the order-theoret ic properties of any coun tab le loset is
identical to that of the rational numbers? No! While, according to Proposition 5,
we can embed the loset (N, ≥) in (Q, ≥) in an order-preserving manner, it is not true
that (N, ≥) and (Q, ≥) ar e identical with respect to all order-theoretic properties. For
instance, wh ile N has a ≥-minim um, Q does not have a ≥-minim um . The p roblem

is that we cannot em bed (Q, ≥) back in (N, ≥). Formally speaking, these two losets
are not order-isomorphic.
Dhilqlwlrq. Let (X, 
X
) and (Y,
Y
) be two posets. A map f ∈ Y
X
is s a id to be
order-preserving (or isotonic), provided that
f(x) 
Y
f(y) if and only if x 
X
y
for any x, y ∈ X. If f ∈ Y
X
is an order-pr eser vin g injection, then it called an order-
embedd in g from X into Y. If such an f exists, then we say th at (X, 
X
) can be
order-embedded i n (Y,
Y
). Finally, if f is an order-preservin g bije ction , then it is
called an order-isomorphism.Ifsuchanf exists , (X, 
X
) and (Y,
Y
) are said to
be order-isomorphic.

If t wo posets are order-isomorphic, th ey are i n distinguisha b le from each other
insofar as their order-theoretic properties are concerned; on e can simp ly be thought
of as the relabeling of the other in this regard. In con cert w ith this view , “being order-
isomorphic” acts as an equivalence relation on any given class of posets. (Proof?) In
this sense, the order structures of any two finite losets of the same size are identical,
but those of two infinite losets ma y be different from each other. Consider, for
instance, (Z
+
, ≥) an d (Z
−
, ≥); th e ≥-m in imum of the former is t h e ≥-maximum
of the latter. Similarly, the posets ({1, 3, 7, }, ≥) and ({2, 4, 5, }, ≥) are order-
theoretically iden tical, but (N, ≥) and (Q, ≥) are n ot. In fact, Proposition 5 sa ys
that every c ou ntable loset can be order-embedded in Q, butofcourse,noteveryloset
is order-isomorphic t o Q. Here are some more exam p les.
Exercise 11.LetA :=

1
2
,
1
3
,
1
4
,

and B :=

3

2
,
5
3
,
7
4
,

. Prove:
(a)
(A, ≥) and (N, ≥) are not order-isomorphic, but (A, ≥) and (A ∪ {1}, ≥) are.
(b)
(A, ≥) and (A ∪ B, ≥) are not order-isomorphic.
Exercise 12.Let
(X, 
X
) and (Y, 
Y
) be two order-isomorphic posets. Show that
X has a 
X
-maximum iff Y has a 
Y
-maximu m. (The same also applies to the
minimum elements, of course.)
Exercise 13.
H
Prove that any poset (S, ) is order-isomorphic to (X,⊇) for some
nonempty set

X.
It is now time to iden tify those countable losets whose order structures are indis-
tinguishab le from that of Q. The following property turns o u t to be crucial for th is
purpose.
70
Dhilqlwlrq. Let (X,) be a preordered set and S ⊆ X. If, for any x, y ∈ X such
that x  y, there exists an element s of S suc h that x  s  y, we say that S is
-dense (or, order-dense) in X. If S = X here, we simply say that X is -dense
(or, order-dense ).
Proposition 6 . (Can tor) Let (X, 
X
) and (Y,
Y
) be two countab le losets with
neither maximum nor minimum elements (with respect to t h eir respective linear
orders).Ifboth(X, 
X
) and (Y, 
Y
) are order-dense, then they are order-isomorphic.
Exercise 14.
H
Pro ve Proposition 6.
A special case of Proposition 6 solves the m otivating p rob lem of this section by
c haracterizing the losets th at are order-isomorp hic to Q. This findin g is important
enough to deserve separate mention, so we state it as a co rollary below. In Section
4, we will see how important this observation is for utility theory.
Corollary 2. Any countable and order-den s e loset with neith e r maximum nor mini-
mumelementsisorder-isomorphictoQ (and, therefore, to Q ∩ (0, 1)).
3 SomeMoreAdvancedSetTheory

3.1 The Cardinality Ordering
We now turn to the problem of com paring the “size” of tw o infinite sets. We hav e
already had a head start on this front in Section 1. That section taugh t us that
N and Q are “ equ ally crow d ed” (Corollary 1), wh ereas R is “more crowded” than
N (Proposition 4). This is because we can put all members of N in a one-to -one
cor r espondence w it h all members of Q while mapping ea ch natural number to a real
n umber can never exhaust the en tire R. If we wish to generalize this reasoning to
compare t he “size” of a ny tw o sets, then we arrive at the f ollowing notion.
Dhilqlwlrq. Let A an d B be an y t wo sets. We say that A is cardinally larger
than B, denoted A :
card
B, if t here e xists an injection f from B into A. If, on t he
other han d, we can find a bijection from A onto B, then we say t ha t A an d B are
cardin ally equiva lent, and deno te this by A ∼
card
B.
So, a set is countably infinite iff it is cardinally equivalent to N. Moreover, we have
learnedinSection1thatN ∼
card
N
2
∼
card
Q while [0, 1] :
card
N but not N :
card
[0, 1].
Similarly, 2
S

:
card
S but not convers ely, for any nonem pty set S (Exercise 10). Fo r
another e xample, w e n ote that 2
N
∼
card
{0, 1}
∞
, that is, the class of all subsets of the
natural numbers is cardinally equ ivalen t to the set of all 0-1 seq uences. Indeed, the
71
map f :2
N
→ {0, 1}
∞
define d by f(S):=(x
S
m
), where x
S
m
=1if m ∈ S and x
S
m
=0
otherwise , is a bijection. He re are some other examples.
Exercise 15.
H
Prove the following facts:

(a)
R ∼
card
R\N;(b) [0, 1] ∼
card
2
N
;(c) R ∼
card
R
2
;(d) R ∼
card
R
∞
.
Exercise 16.
H
Prove: A set S is infinite iff there is a set T ⊂ S with T ∼
card
S.
It is readily verified that A :
card
B and B :
card
C imp ly A :
card
C for a n y sets
A, B an d C, that is, :
card

is a preorder on an y given class of sets. But there is still a
potential problem with taking the relation :
card
as a basis for comparing the “sizes”
of two sets. This is because at the momen t we do not kno w what to mak e of the
case A :
card
B and B :
card
A. Intuitive ly, we would like t o sa y in this cas e that A
and B are equally cro w d ed, or formally, w e w ould like to declar e that A is cardinally
equ ivalent t o B. But, given our defin itions, th is is not at all an ob v ious conclusion.
What is more, if it didn’t hold, w e would be in serious trouble. Assume for a momen t
that we can find a set S suc h that S :
card
N and N :
card
S but not S ∼
card
N.
According t o our interpretation , we would like to s ay in this case that th e “size” of
S isneitherlargernorsmallerthanN, but S :
card
N and not S ∼
card
N en tail that S
is an uncountable set! Obviously, this would be a problematic situation which wou ld
forbid thinking uncoun table sets as being much more “crowded” than countable ones.
But, of cour s e , such a problem never arise s.
The Sc h röder-Bern stein Theorem .

9
For any two sets A and B such that A :
card
B and B :
card
A, we have A ∼
card
B.
Intherestofthissectionwewillprovideaproofofthisimportanttheorem.Our
proof is base d on the following resu lt wh ich is of interest in and of itself. It is the
first of man y fixed point theorems that you will encounter in this text.
Tarski’s Fixed Point T he o re m. Let (X, ) be a poset such that
(i) if S ∈ 2
X
\{∅} has an -upper bound in X, then sup

(S) ∈ X;
(ii) X has a -maximum and a -min imum.
If f is a self-map on X such that x  y implie s f(x)  f(y) for an y x, y ∈ X, then
it has a fixed point, tha t is, f(x)=x for som e x ∈ X.
10
9
This result is sometimes referred to as Bernstein’s Theorem, or the Cantor-Bernstein Theorem.
This is because Cantor has actually conjectured the result publicly in 1897, but he was unable to
prove it. Cantor’s conjecture was proved that year by (then 19 years old) Felix Bernstein. This
proof was never published; it w as popularized instead by an 18 98 book of Emile Borel. In the same
year Friedric h Schröder published an independent proof of the fact (but his argument con tained a
(relatively minor) e rror which he corrected in 1911).
10
A special case of this result is the famous Knaster-Tarski Fixed Point Theorem: Every order-

preserving self-map on a complete lattice has a fixed point. (See Exercise A .16 to recall the definition
of a c omplete lattice.)
72
Pro of. Let f be a self-map on X such that f(x)  f(y) when ever x  y.Define
S := {z ∈ X : f (z)  z}, and observ e that S is nonemp ty and bound ed from abov e
by (ii). (Why nonempty?) By (i), on the other hand, x := sup

S exists. Observe
that, for any y ∈ S, we have f(y)  y an d x  y. The latter expression implies that
f(x)  f(y), so combining t his with the for m e r , we find f(x)  y for any y ∈ S.
Thus f(x) is an -upper bound for S in X, which implies f (x)  x. In turn, this
yields f(f(x))  f(x) which m eans that f(x) ∈ S. But then, since x =sup

S, we
have x  f(x). Since  is antisymmetric, we mu st then have f(x)=x. 
Th is theorem is sur p r isin gly general. It guarantees the existence of a fixed point
by using the order structure at ha nd and nothing mor e. For instance, it implies, as
an immed iate cor ollary, the following beautiful result (whic h is, of c ou r s e, n ot valid
for decr easing function s ; see F igure 2).
∗∗∗∗FIGU R E B.2 ABOU T HE RE ∗∗∗∗
Corollary 3. Ev ery increasing self-map on [0, 1] has a fixed poin t.
Or how about t he following? If n ∈ N, ϕ
i
:[0, 1]
n
→ [0, 1] is an increasing func -
tion, i =1, , n, and Φ is a s el f-m ap on [0, 1]
n
defined by Φ(t):=(ϕ
1

(t), , ϕ
n
(t)),
then Φ has a fixe d point. Noticing the similarit y of these results w ith those obtained
from the B ro uwer Fixed Point T h eorem ( wh ich you have probab ly seen so mew here
before — if not, do n’t worry, we’ll talk about it quite a bit in Section D.8) should give
you an idea about how useful Tarski’s Fixed Point Theorem is. Inde e d , w h en one
lacks the continuity of the inv olv ed individuals’ objective functions, but has rather
the monotonicity of them, this theorem does the work of the Brou wer Fixed Point
Theorem to ensure the existe nce of equilibrium in a variety of economic models.
11
Exercise 17.Let(X, ) and f satisfy the hypotheses of Tarski’s Fixed Point Theo-
rem. Let
Fix(f) stand for the set of all fixed points of f on X. Prove or disprove:
(Fix(f), ) is a poset with a -maximum and -minimum element.
Let us no w turn b ack to our p rima ry goal of pr oving the S chröder-Bernstein
Theorem. It turn s out that a clever applic ation of Tarski’s Fixed Point Theorem
delivers th is result in a silver platter.
12
We will first derive the following intermediate
step which makes the l ogic of t he proof transparen t.
11
If you don’t know what I am talking about here, that’s just fine. The importance of fixed point
theory will become crystal clear as you learn m ore economics. This text will help too; pretty much
every c hapter henceforth co ntains a bit of fixed point theory and its applications.
12
I l earned t his method of proof from Gleason (1998), whic h is also use d by Carothers (2000).
There are many alternative proofs. See, for instance, Halmos (1960), Cox (1968) or Kaplansky
(1979).
73

Banac h’s Deco mpositio n Theo rem. Let X an d Y be a n y tw o nonempt y sets and
tak e any f ∈ Y
X
and g ∈ X
Y
. Then X can be w ritten as the union of tw o disjoint
sets X
1
and X
2,
and Y canbewrittenastheunionoftwodisjointsetsY
1
and Y
2
such that f(X
1
)=Y
1
and g(Y
2
)=X
2
.
Pro of. Clearly, (2
X
, ⊇) is a poset that has a ⊇-maximum and a ⊇-minimum.
Moreov er, for any (nonempt y) class A ∈ 2
X
, we have sup
⊇

A =

A. No w define the
self-map Φ on 2
X
by
Φ(S):=X\g(Y \f(S)),
and observ e that B ⊇ A implies Φ(B) ⊇ Φ(A) for any A, B ⊆ X. We may thus
apply Ta rsk i’s Fixed Point Theorem to c on clude that there e x ists a set S ⊆ X such
that Φ(S)=S.Theng(Y \f(S)) = X\S, and hence defining X
1
:= S, X
2
:= X\S,
Y
1
:= f(S), and Y
2
:= Y \f(S) completes the proof.
13

Now the stage is set for settling the score, for the Schröder-Bernstein Theorem is
an almost imm ediate corollary of Banach’s Decomposition Th eorem .
ProofoftheSchröder-BernsteinTheorem. Let A and B be two nonempty sets,
and tak e any two injections f : A → B and g : B → A. By Banach’s Decomposition
Theorem, there exist a doubleton p artition {A
1
,A
2
} of A and a doubleton partition

{B
1
,B
2
} of B such that F := f|
A
1
is a bijection from A
1
onto B
1
, and G := g|
B
2
is a
bijection fr o m B
2
onto A
2
. But then h : A → B, defined by
h(x):=

F (x), if x ∈ A
1
G
−1
(x), if x ∈ A
2
,
is a bijection, and w e are done. 

This pr oof is a beautiful illustration of the power of fixed poin t theorems. Like
most proofs that are based on a fixed point argumen t , however, it is no t con structive.
That is, it does n ot tell us ho w to map A onto B in a o ne -to-one manner, it ju s t
says tha t such a th in g is possible to do. If this worries you at all, the n you w ill be
pleased to hear that there a re other wa ys of proving the Schröder-Bernstein Theorem,
and s om e of these actua lly “con struct” a tangible b ijection between the involved sets.
Two such proofs are reported in Halmos (1960) and Kaplans ky (1977).
13
Whoa! Talk about a rabbit-out-of-the-hat proof! What is going on here? Let me explain.
Suppose the assertion is true, so there exists such a decomposition of X and Y. Then X is such a set
that when you subtract the image f(X
1
) from Y, you end up with a set the image of which under
g is X\X
1
, that is, g(Y \f(X
1
)) = X\X
1
. But the converse is also true, that is, if we can get our
hands on an X
1
⊆ X with g(Y \f(X
1
)) = X\X
1
, then we are done. That is, the assert ion is true
iff we can find a (possibly empty) subset S of X with g (Y \f(S)) = X\S, that is, a fixed point of
φ! And how do you find suc h a fixed point? Well, with this little structure at hand, Alfred Tarski
should be the first person we should ask help from.

74
3.2 The We ll-Ord ering Principle
We now turn to the concept of w ell-ordering. Since this notion will pla y a limited
role in this book, ho wever, our exposition will be brief.
Dhilqlwlrq. Let (X,) be a poset. The relation  is said to be a well-ordering
if ev ery nonempty subset of X has a -minimum. In this case X is said to be
w e ll-ordered by ,and(X, ) is called a well-ord e red s et,orshortlyawoset.
Well-ordered sets are in general very useful because t hey possess an inductive
structure. In particular, for any such set we ha ve the following result, which is much
in the same s pirit with t he Principle of Mathematical Induction.
Exercise 18.(T he Principle of Transfinite Induction)Let(X, ) be a woset. Prove:
If
S is a nonempty subset of X such th at, for any x ∈ X, {y ∈ X : x  y} ⊆ S
implies x ∈ S, then S = X.
It is obvious that the set of all natural numbers is well-or dered by the usual order
≥. On the other hand, (2
{1,2}
, ⊇) isnotawosetsincetheset{{1}, {2}} does not have
a ⊇-m in imum. Sim ilar ly, ([0, 1], ≥) is no t a woset. Ho wev er, a fundamental theorem
(or axiom, if you will) of set theory says that, on any given set, we can a lways define
a partial or der that well-orders th at set. So, for instance, we can well-order 2
{1,2}
, say,
b y using the partial o rder  on 2
{1,2}
that satisfies {1, 2}  {2}  {1} ∅.(Notice
that {{1}, {2}} does have a -minimum, which is {1}.) To giv e another example,
cons ider t he se t {1, 2}×N. We can well-order this set by using the partial order 
on {1, 2}×N with
··· (2, 2)  (2, 1)  ··· (1, 2)  (1, 1).

Much less trivially, thanks to the Axiom of Choice, i t is possible to m ake [0, 1] awoset
as w ell. (Think about it! Ho w can this be?) In fact, w e hav e the following sensational
result.
The Well-Order ing Princip le . (Zer melo) Every set can be well-ordered.
This is one o f the most surprising consequences of the Axiom of C hoice.
14
Our
interest in the Well-Ordering Principle here stems from the fact that this principle
allows us to answer the follow ing question: Can we alw ays rank the cardin a lities of
an y two sets? The answer is affirmative, and w h en equipped with the We ll-Ord er ing
Principle, not difficult to prove.
14
The Well-Ordering principle is in fact equivalent to the Axiom of Choice (given the standard
axioms of set theory), ev en t hough the former see ms impossible, and the latter self-evident.
75
Exercise 19. Sho w that, for any tw o sets A and B, we have either A :
card
B or
B :
card
A. Conclude that :
card
acts as a complete preorder on any class of sets.
Exercise 20. Prove that the diagonal relation
D
X
on any nonempty set X (Example
A.1.[1]) can be written as the in tersection of two linear orders on
X.
4 Application: Ordinal U tilit y Th eory

In th is section we outline the ba sic framework of ordinal utility theory, and sh ow that
the notion of countability plays an importan t role in this theory. We will later revisit
this topic i n Chapters C and D to present a number of classic results th at lie in the
foundation s of economic theory.
4.1 Pr efe r en ce Relatio ns
Throughout this section the set X is viewed a s a nonempt y set o f outcomes (or
prizes, o r alternatives). As noted in Ex ample A.3, by a pref erence relation  on
X, we mean a preorder on X. As you already know well, the preference relation of an
individual con tains all the information that concerns he r tastes about t he o utcomes
in X.Ifx  y holds, w e unde rstand that t he individual with preference relation 
views the alternative x at least as good as the alter na tive y. Induced from  are the
strict preference relation  on X, and the indifference relation ∼ on X, defined
by x  y iff x  y bu t not y  x, and x ∼ y iff x  y and y  x, respectiv ely. Both
of these relations are readily checked t o be transitive (Exercise A.7). Furthe rmore, it
is easy to v e rify that if either x  y and y  z, or x  y and y  z, th en x  z.
Wh ile r eflexivity is trivial, the transitivity requ irem e nt for a p ref ere n ce re lation is
an immediate reflection of th e hypothesis of rationality that is almost alwa y s impose d
on economic agen ts . A good illustra tion of this is through what is called the “money
pump ” argum ent. Suppose that the strict preference relation of a certain individual
on X is given b y a non-transitiv e binary relation R. Th en we ma y find a triplet
(x, y, z ) ∈ X
3
suc h that x is strictly better than y, an d y is weakly better than z,
while the agen t either cannot compare x and z, or finds z at least as good as x. Now
suppose that t he individual owns x to begin with. Then it is quite conceivable that
she m a y exchange x with z, barring an y status quo bias. But she likes y better than
z, so it shouldn’t be too difficult to c onvince her t o t rad e z with y. What is more, she
would certainly be willing to pa y at least a small amount of money to exc h an g e y with
x. So, after all this t rade, t he agen t ends up where she started, but she is now a b it
poorer. In fact, repeating this argumen t, we may extract arbitrarily large amoun ts

of money from her; something a “rational” individual should not allow. Thus, so
the argum e nt co n clud es, there is good reason to view transitivity (of at least that of
one’s strict preference relation) as a basic tenet of rationalit y. We subscribe to this
76
positio n in what follows.
15
Another property commonly imposed on a preference relation is that of complete-
ness (i.e. the requiremen t that e ither x  y or y  x holds for each x, y ∈ X). This
property is , h owever, less appealing tha n transitivit y on intuitive grounds in that it
appears muc h less link ed to the notion of rationalit y. Why can’t a rational individual
be indecisive on occasion? Can’t a rational person ev er say “I don’t k now”? (You
might wan t to read Aum a n n (1962) and Mandler (1999) for illuminating discussions
about the conc e ptual distinction between the notion of rationa lity an d the Complete-
ness Axio m , and Eliaz and O k (2005) to see how one may behaviorally distinguish
between indifference and indecisiveness in general.) A t the v ery least, there are in-
stances in which a decision maker is actually composed of sev eral agents eac h w ith
a possibly distinct objective function. For instance, in coalitional bargaining games,
it may be quite natural to specify the preferences of eac h coalition by means of a
vector of utility functions (on e for each m ember of t h e c oalition ), and this requir es
one to view the preference relati o n of each coalitio n as an i ncomp let e preference re-
lation. The s a me reason in g applies to s ocial c hoice problems — the most comm on ly
used social w elfare or dering in economics, the Pareto dominance (Example A.3), is
an incomplete preorder (unless the society consists of a single individual). At any
rate, at least some of the main results of individual choice theory can be obtained
without imposing the completeness property. Thus, in th is book, we ch oose not to
impose this require ment on a preferenc e r elation at the onset. T h e only price to pay
for this is the n e ed to q u alify a preference relation with the adjectiv e “complete” in
the statements of s ome of the subsequent results.
Before proceeding further, let u s define the following bit of terminolog y that will
be used repeatedly in what follo ws.

Dhilqlwlrq. Let  be a preference relation on X. For any x ∈ X, the weak and
strict upper -con tour sets of x are t he sets de fined as
U

(x):={y ∈ X : y  x} and U

(x):={y ∈ X : y  x},
15
Having said this, however, be warned that trans itivity of ∼ is a somewhat problematic postulate.
The problem is that an individual may be indifferent between two alternatives just because she fails
to perceive a difference between them. The now-classical argument goes as follows: “I’m indifferent
between a cup of coffee with no sugar, and the same cup of coffee with one grain of sugar added.
I’m also indifferent bet ween the latter and the same cup of coffee with another grain of sugar added;
I simply cannot tell the difference. And this, ad infinitum. But then transitivit y would require that
I be indifferent between a cup of coffee with no sugar and the same cup with an enormous amount
of sugar added in, and t hat I am not.” Moreov er, there are other appealing reasons for studying
non transitive preferences, especially in sequential (intertemporal) choice co ntexts (cf. Fishburn
(1991), and Ok and Masatlioglu (2005)). Nev e rtheless, it is almost universal practice to work with
transitive preferences, and this is not the place to depart from this tradition. In what follows,
therefore, I will work only with suc h preferences, and will, furthermore, ignore the issues related
to the d escription of choice problems (the so-called framing effects ), and/or the computational
limitations of decision makers (the bounded rationality theory).
77
respectively. The weak and strict lower -contour sets of x are defined analo-
gously:
L

(x):={y ∈ X : x  y} and L

(x):={y ∈ X : x  y}.

4.2 Utility Rep resentation of Com plete Prefere nce Re lations
While the p referenc e relation of a n agent contains all the information that c on cerns
her tastes, this is really not the most convenient way of s ummariz ing this info rmation.
Maximizing a b in ary relation (while a well-defined matter) is a much less friendlier
exercis e than maximizing a (utility) function. Thus, it would be quite useful if we
knew how and when one can find a real function that attac hes to an a lternativ e x
a (strictly) hi gher va lue than a n alternative y iff x is rank e d (strictly) above y by a
given preference relation. As you w ill surely recall, such a function is called the utility
function of the individual w ho possesses this p reference relation. A fundamen tal
question in the theory of individual choice is therefore the following: What sort of
preferen ce relations can be described b y means of a utility function ?
We begin by f ormally defining what it m eans to “describe a preference relation
b y means of a utility function.”
Dhilqlwlrq. Let X be a nonempty set, and  a preference relation on X. For any
∅ = S ⊆ X,wesaythatu ∈ R
S
represents  on S, if u is an order-preserving
function , that is, if
x  y if and only if u(x) ≥ u(y)
for an y x, y ∈ S. If u represents  on X,wesimplysaythatu represents .If
such a function exists, then  is said to b e represen table,andu is called a utility
function for .
Thus, if u represents , and u(x) >u(y), we understand that x is strictly preferred
to y by an agent with the preference relation . (Notice that if u represents , then
 is complete , an d u(x) >u(y) iff x  y, and u(x)=u(y) iff x ∼ y, for any
x, y ∈ X.) It is com m onplace to say in this case that the a gent deriv es m ore u tility
from obtain in g altern ative x than y; hence the term utility function. However, one
should be care fu l in ad op ting this interpretation , for a utility fun c tion that r e p res e nts
a preference relation  is not unique. Th e refore, a utility function cannot be thou g ht
of as measuring the “util” con tent of an alternative. It is rather ordinal in the sen se

that if u repr esents , then so does f ◦ u for any strictly increasing self-m ap f on
R. More forma lly, we sa y that an ordinal utility function is unique up to strictly
increasing transformations.
Proposition 7. Let X beanynonemptysetandu ∈ R
X
represent the preference
relation  on X. Then, v ∈ R
X
represents  if, and only if, the re exis ts a strictly
increasing function f ∈ R
u(X)
such that v = f ◦ u.
78
Exercise 21. Prove Proposition 7.
Exercise 22.
H
Let  be a complete preference relation on a nonempty set X, and let
∅ = B ⊆ A ⊆ X. If u ∈ [0, 1]
A
represents  on A and v ∈ [0, 1]
B
represents  on
B, then there exists an extension of v that represents  on A. True or false?
We now proceed to the analysis of preference relations t hat a ctually admit a
representation by a utility function. It is instr uctive to begin with the trivial case in
whic h X is a finite set. A m om ent’s reflection will be enough to convince yourself
that any com p lete p reference relation  on X is represen table in t his case. Indeed,
if |X| < ∞, thenallwehavetodoistofind the s et of least preferred elements of X
(whic h exist by finiteness), say S, and assign the utility value 1 to any member of S.
Next we choose the least preferred elemen ts in X\S, and assign t he utility value 2 to

any suc h element. Con tin uing this way, w e even tually exhaust X (since X is finite)
and hence obtain a representa tion of  as w e sought . (Hidden in the argumen t is the
Principle of Mathematical Induction, right?)
In fact, Proposition 5 bring s u s ex ceeding ly close to co n cluding that any complete
preference relation  on X is representable whenev er X is a countable set. The only
reason w hy we cannot conclude this immediately is because this proposition is pro ved
for linear ord ers w h ile a p ref eren ce relation need not be antisymmetric (that is, its
indifference classes need not be singletons). However, this is not a serious difficult y;
allwehavetodoistomakesureweassignthesameutilityvaluetoallmembersthat
belongtothesameindifference class. (We ha ve used the same trick when proving
Corollary A .1, rem ember?)
Proposition 8. Let X be a nonempt y countable set, and  a complete preference
relation on X.Then i s represen table.
Pro of. Recall that ∼ is an equivalence relation, so the quotient set X/
∼
is well-
defined. Define the linear order 
∗
on X/
∼
by [x]
∼

∗
[y]
∼
iff x  y. (Why is 
∗
well-defin ed?) By Proposition 5, th ere exists a fu n ction f : X/
∼

→ Q that represen ts

∗
.Butthenu ∈ R
X
, defined by u(x):=f([x]
∼
), represents  . 
This proposition also p aves wa y towards the f ollowing i nteresting result that ap-
plies to a rich class of p reference relations.
Proposition 9. Let X be a nonempty set, a nd  a complete preference relation on
X. If X c o ntains a countable -dense subset, then  can be represented by a utilit y
function u ∈ [0, 1]
X
.
Pro of. If  = ∅, then it is enough to tak e u as any constant f un ction , so we may
assume  = ∅ to concen trate on the non trivial case. Assume that there is a countable
-dense set Y in X. By Proposition 8, there exists a w ∈ R
Y
suc h that w(x) ≥ w(y)
79
iff x  y for any x, y ∈ Y. Clearly, the function v ∈ [0, 1]
Y
define d by v(x):=
w(x)
1+|w(x)|
,
also represents  on Y. (W hy?)
Now take any x ∈ X with L


(x) = ∅, and define
α
x
:= sup{v(t):t ∈ L

(x) ∩ Y }.
By -denseness of Y and boundedness of v, α
x
is well-defined for any x ∈ X. (Wh y ?)
Define next the function u ∈ [0, 1]
X
by
u(x):=
⎧
⎨
⎩
1, if U

(x)=∅
0, if L

(x)=∅
α
x
, otherwise
.
 = ∅ implies that u is we ll-d e fined. (Wh y ?) The rest of the proof is to check that u
actually represen ts  on X.
16
We leave verifying this as an exercise, but just to get

you going, let’s show that x  y imp lie s u(x) >u(y) for any x, y ∈ X with L

(y) = ∅.
Since Y is -dense in X, th ere m u st exist z
1
,z
2
∈ Y suc h t hat x  z
1
 z
2
 y.
Since z
1
∈ L

(x) ∩ Y and v represen t s  on Y, we have α
x
≥ v(z
1
) >v(z
2
). On the
other hand, since v(z
2
) >v(t) for all t ∈ L

(y) ∩ Y (why?), we also hav e v(z
2
) ≥ α

y
.
Combining these observations yields u(x) >u(y). 
Exercise 23. Complete the proof of Proposition 9.
The next exercise provides a generalization of Proposition 9 by offering an alternative
denseness condition that is actually necessary and sufficient for the representability
of a linear order.
∗
Exercise 24.
H
Let X be a nonempt y set, and  a linear order on X. Then,  is
representable iff
X contains a countable set Y such that, for each x, y ∈ X\Y with
x  y, there exists a z ∈ Y such that x  z  y.
The characterization result giv en in Exercise 24 can be used to iden tify certain
preference relations t ha t a re not rep res entable by a u tility function. Here is a standard
example o f such a relation.
E{dpsoh 1. (The Lexicographic Preference Relation ) Consider the linear order 
lex
on R
2
defined as x 
lex
y iff either x
1
>y
1
or x
1
= y

1
and x
2
≥ y
2
. (This relation is
called the lexicographic order.) We write x 
lex
y whenev er x 
lex
y and x = y.
For instance, (1, 3) 
lex
(0, 4) and (1, 3) 
lex
(1, 2). As you probably r ecall from
a microecono m ic s class you’ve taken befor e, 
lex
is not representable b y a utility
function . We now provide two proofs of this fact.
16
Alternative representations may be obtained b y replacing the role of α
x
in this proof w ith
λα
x
+(1− λ)inf{v(y):y ∈ U

(x) ∩ Y } for a ny λ ∈ [0, 1].
80

First pro of. Suppose that there exists a set Y ⊂ R
2
suc h that, for all x, y ∈ R
2
\Y
with x 
lex
y, we have x 
lex
z 
lex
y for some z ∈ Y. We shall show t hat Y must
then be uncoun table, whic h is enough to conclude that 
lex
is not rep res e ntable in
view of Ex ercise 24. Tak e any real a, an d p ick any z
a
∈ Y such that (a, 1) 
lex
z
a
and z
a

lex
(a, 0). Clearly, we mu st have z
a
1
= a and z
a

2
∈ (0, 1). But then z
a
= z
b
for
an y a = b, while {z
a
: a ∈ R} ⊆ Y. It follo ws tha t Y is not coun tab le (Proposition 1).
The following is a more direct proof of the fact that 
lex
is not represe ntable.
Second proof. Let u : R
2
→ R represent 
lex
. Then, for any a ∈ R,wehave
u(a, a +1)>u(a, a) so that I(a):=(u(a, a),u(a, a +1))is a nondegenerate interval
in R.Moreover,I(a) ∩ I(b)=∅ for any a = b, for we have
u(b, b) >u(a, a +1) whenever b>a,
and
u(b, b +1)<u(a, a) whenever b<a.
Therefore, the map a → I(a) is an in je c tion from R in to {I(a):a ∈ R}. But
since {I(a):a ∈ R} is coun tab le (Proposition 3), this entails that R is countable
(Proposition 1), a contradiction. 
Th e class of all preference relations that do not possess a utility repres entation is
recently characterized by Be ard on et al. (2002). As the next exam p le illu strates, this
class includes non-lexicographic preferences as well.
17
Exercise 25. (Dubra-Echenique) Let X be a nonempt y set, and let P(X) denote the

class of all partitions of
X. Assume that  is a complete preference relation on P(X)
such that A ⊂ B implies B  A.
18
Pro ve:
(a)If
X =[0, 1], then  is not representable by a utility function.
(b)If
X is uncountable, then  is not representable by a utility function.
4.3 Utility Representation of Incomplete Preference Rela-
tions
As noted earlier, there is a conc e p tua l adva ntage in taking a (possibly in c omplete)
preorder as the p rimitive of analysis in the theory of ration al choice. Yet a n incomplete
preorde r cann ot be represented by a utilit y function — if i t did, it would not be
incomp lete . Thus it seems like t h e analytical scope of adopting such a point of
view i s, per force , rat her limited. Fortunate ly, however, it i s still poss ible to provide a
utility r ep re sentation for an incomple te preference relation, provided tha t w e suitably
generalize the notion of a “utilit y fun ction.”
17
For another example in an in teresting economic context, see Basu and Mitra (2003).
18
Sziplrajn’s Theorem assures us t hat there exists su ch a preference relation. (Why?)
81
To begin w ith , let us note that while it is ob viou s ly not possib le to find a u ∈ R
X
for an incomplete preorde r  such that x  y iff u(x) ≥ u(y) for any x, y ∈ X,there
ma y nev ertheless exist a real f unction u on X such that
x  y imp lie s u(x) >u(y) and x ∼ y implies u(x)=u(y)
for an y x, y ∈ X. Among o thers, th is approac h wa s e x plored first by Richter (1966)
and Peleg (1970). We shall t h us refer to suc h a real function u as a Ric h ter-P eleg

utility fun ct ion for . Recall that Sziplrajn’s Theorem guaran tees that  can be
extended to a complete preference relation (C orollary A.1). If  has a Richter-P eleg
utility function u then, and on ly then,  can be extended to a c omplete preference
relation that is represen ted by u in the ordinar y sen se.
The follo wing result shows that Proposition 9 can be extended to the case of
incomp lete preord e r s if one is willing to accept this particular notion of utility repre-
sentat io n . (You sho uld contras t the associated pr oofs.)
Lem ma 1. (Ric hter) Let X be a n onempt y set, and  a preference r elation on X.
If X contains a countable -dense subset, then there exists a Richter-Peleg utilit y
function for .
Pro of. (Peleg) We w ill prove the c laim assum in g that  is a p artial order; the
extension to the case of preorders is carried out as in the proof of Proposition 8.
Obviously, if  = ∅, then there is nothing to prove. So l et  = ∅, and assume
that X contains a countable -denseset—let’scallthissetY. Clearly, there must
then exist a, b ∈ Y such that b  a. Thus {(a, b)

: a, b ∈ Y and b  a} is a
countably infinite set, where (a, b)

:= {x ∈ X : b  x  a}. We enumerate this
set as {(a
1
,b
1
)

, (a
2
,b
2

)

, }. Ea ch (a
i
,b
i
)

∩ Y is p artially ordered by  so that,
by the Hausdorff Maximal Principle, it contains a ⊇-maximal loset, say (Z
i
, ). By
-densene ss of Y , Z
i
has neither a -maxim um nor a -minimum. Moreove r, by its
maximalit y, it i s -d en s e in itself. By Cor ollary 2, theref ore, there exists a bijection
f
i
: Z
i
→ (0, 1) ∩ Q such that x  y iff f
i
(x) ≥ f
i
(y) for an y x, y ∈ Z
i
. Now define the
map ϕ
i
∈ [0, 1]

X
by
ϕ
i
(x):=

sup{f
i
(t):x  t ∈ Z
i
}, if L

(x) ∩ Z
i
= ∅
0, otherwise.
Clearly, we ha ve ϕ
i
(x)=0for all x ∈ L

(a
i
), and ϕ
i
(x)=1for all x ∈ U

(b
i
). Using
this observation an d th e definition of f

i
, one can show that, for an y x, y ∈ X with
x  y we have ϕ
i
(x) ≥ ϕ
i
(y). (Verify!) To complete t he proof, t hen, define
u(x):=
∞

i=1
ϕ
i
(x)
2
i
for all x ∈ X.
(Since the r ange of each ϕ
i
is contained in [0, 1] and

∞
1
2
i
=1(Examp le 8. [2]), u
is well-defined.) N otice that, fo r any x, y ∈ X with x  y, there exists a j ∈ N with
82
x  b
j

 a
j
 y so that ϕ
j
(x)=1> 0=ϕ
j
(y). Since x  y implies ϕ
i
(x) ≥ ϕ
i
(y) for
every i, therefore, w e find that x  y im p lie s u(x) >u(y), and the proof is complete.

Unfort un ate ly, the Richter-Peleg formulation of utility representation h as a s er iou s
shortcoming i n t h at it may result in a substan tial information loss. Indeed, o ne cannot
recover the original preference relation  from a Richter- Peleg utility function u;the
inform ation contained in u is strictly less than that contained in .Allwecandeduce
from the statement u(x) >u(y) is that it is not th e case that y is strictly better th an
x for t he subject individual. We cannot tell if this agent actually lik es x better than
y, or that she is unable to rank x and y. (That is, we are una ble to capture the
“indecisiveness” of a decision m ak er by u sing a Rich ter-P eleg utility f unction.) The
problem is, o f course, due to the fact that the ran ge of a real function is completely
ordered while its d omain is not. One way o f overcoming this problem is by using a
poset- valu ed utility fu n ct io n, or better, by u sin g a set of real-value d utility functions
in the follo wing w ay.
Dhilqlwlrq. Let X be a nonempt y s et, and  a preference relation on X. We say
that the set U ⊆ R
X
represen ts ,if
x  y if and only if u(x) ≥ u(y) for all u ∈ U,

for any x, y ∈ X.
Here are some immediate examples.
E{dpsoh 2. [1] Let n ∈ {2, 3, }. Obviously, we cannot represen t the partial order
≥ on R
n
b y a single utility functio n, bu t we can r epresent it by a (finite) set of utility
function s . Indeed, defining u
i
(x):=x
i
for each x ∈ R
n
,i=1, , n, we find
x ≥ y iff u
i
(x) ≥ u
i
(y) i =1, , n.
for any x, y ∈ R
n
.
[2] Let Z := {1, , m}, m ≥ 2 and let L
Z
stand for the set of all probabilit y dis-
tributions (lotteries) on Z, that is, L
Z
:= {(p
1
, ,p
m

) ∈ R
m
+
:

m
p
i
=1}. Then the
first-order stochastic dom inance ordering on L
Z
, denoted by :
FSD
, is defined
as follo ws:
p :
FSD
q iff
k

i=1
p
i
≤
k

i=1
q
i
k =1, , m − 1.

(H ere we interpret m as th e best p rize, m − 1 as the s econ d best prize, and so on.)
The p a r tial order :
FSD
on L
Z
is represen ted by the set {u
1
, , u
m−1
} of real functions
on L
Z
, where u
k
(p):=−

k
p
i
,k=1, , m − 1.
83
[3] For an y nonempty set X, the diagonal relation D
X
can be represen ted b y a
set of two utility function s . Th is follows from Exer c ise 20 and Proposition 8. (Note.
If X is countable, we do not have t o use the Axiom of Choice t o pro ve this fact.) 
Exercise 26.LetX := {x ∈ R
2
: x
2

1
+ x
2
2
=1and x
1
=0} and define the p artial
order
 on X as
x  y iff x
1
y
1
> 0 and x
2
≥ y
2
.
Show that  can be represented by a set of two continuous real functions on X.
Exercise 27. Prove: If there exists a countable set of bounded utility functions that
represent a preference relation, then there is a Richter-Peleg utility function for that
preference relation.
Exercise 28. Pro ve: If there exists a countable set of utilit y functions that represents a
complete preference relation, then t his r elation is representable in the ordinary sense.
Exercise 29.
H
Let  be a reflexive partial order on R such that x  y iff x>y+1
for any x, y ∈ R. Is there a U ⊆ R
R
that represents ?

Exercise 30.Define the partial order
 on N × R by (m, x)  (n, y) iff m = n and
x ≥ y. Is there a U ⊆ R
N×R
that represen ts ?
Suppose  i s represented by {u, v} ⊆ R
X
. Oneinterpretationwecangiveto
this situation is that the ind iv idual w ith th e preferenc e relatio n  is a person who
deems two dimensions relevant for comparing the alternatives in X.(Thinkofa
potential g raduate studen t who com pares the graduate programs that she is admitted
in according to the amou nt of financial aid t hey provide a nd the reputation of their
programs.) Her preferences over the first dimension are represen ted by the utilit y
function u, and the second by v. She then judges the value of an alternativ e x on
the basis of its performance on both dimensions, that is, by the 2-vector (u(x),v(x)),
and prefers this alternative to y ∈ X iff x perform s better than y in both of the
dime nsions, that is, (u(x),v(x)) ≥ (u(y),v(y)). T he utility representation notio n w e
advance above is th us closely related to decision-mak ing with m ultip le objective s.
19
19
Of course, this is an “as if” interpretation. The primitive of the model is , so when U represents
, wemayonlythink“asif”eachmemberofU measures (completely) how the agent feels about a
particular dimension of the alternatives.
Let me elaborate on this a bit. Suppose the agent indeed attributes two dimensions to the
alternatives, and ranks the first one wit h respect to u and the second with respect to v, but she
can compare some of the alternatives even in the absence of dominance in both alterna tives. More
concretely, suppose  is given as: x  y iff U
α
(x) ≥ U
α

(y) for all α ∈ [1/3, 2/3], where U
α
:=
αu +(1− α)v for all real α. Then,  is represented by U := {U
α
:1/3 ≤ α ≤ 2/3}, and while we
may interpret “a s i f” the agent views every member of U measuring the value of a dimension of a n
alternative (so it is “as if” there are uncountably many dimensions for her), w e in fact know here
that each member of U corresponds instead to a potential aggregation of the values o f the actual
dimensions relevant to the p roblem.
84
At any rate , this fo rmulation generalizes the usual utility rep re se ntation notion
w e studied abo ve. Moreov er, it does not cause an y information loss; the potential
incompleteness of  is fully reflectedinthesetU that rep resents . Finally, it m akes
working with inc omplete prefer enc e relatio ns analytically less difficult, for it is often
easier to manipulate vector-valued f unctions than preorders.
So, what sort of preference relations can be represen ted b y means of a set of
utility functions? It turns out that the answer is not ver y difficu lt, especially if one
is prepared to adopt our usual order-denseness requiremen t. In what follows, given
an y preference relation  on a n on em pty set X, we write x  y when x and y are
-incomparable, that is, when neither x  y nor y  x holds. Observe that this
defines  as an irreflexive binar y relation on X.
Proposition 10.
20
Let X be a nonempt y set, and  a preference relation on X. If
X contains a countable -dense subset, then there exists a nonempty set U ⊆ R
X
whic h represents .
Pro of. O nce again w e will pro ve the claim assuming that  is a partial order;
the extension to the case of preord ers is straightforwa rd. Assume that X contains a

countable -dense subset, and let U be the co llec tion of all u ∈ R
X
suc h that x  y
implies u(x) >u(y) for an y x, y ∈ X. By Lemma 1, U is nonempt y. We wish to
show tha t U actually rep resents . Ev idently, this means that, for any x, y ∈ X with
x  y, ther e exist at least two functions u and v in U such that u(x) >u(y) an d
v(y) >v(x). (Why?)
Fix any x
∗
,y
∗
∈ X with x
∗
 y
∗
, and pick an arbitrary w
o
∈ U. De fine w ∈ [0, 1]
X
by w(z):=
w
o
(z)
1+|w
o
(z)|
, and note that w is also a R icht er- Peleg utility function f or .
(We have also used the same trick when pro v in g Proposition 9, rem e mber?) Now let
Y := {z ∈ X : z  x
∗

or z  y
∗
},
and define u, v ∈ R
X
as
u(x):=
⎧
⎪
⎪
⎨
⎪
⎪
⎩
w(z)+4, if z ∈ Y
3, if z = x
∗
2, if z = y
∗
w(z), otherwise
and v(x):=
⎧
⎪
⎪
⎨
⎪
⎪
⎩
w(z)+4, if z ∈ Y
2, if z = x

∗
3, if z = y
∗
w(z), otherw ise
.
Weleaveitforyoutoverifythatbothu and v are Ric hter-Peleg utility functions for
.Thusu, v ∈ U and we ha ve u(x
∗
) >u(y
∗
) while v(y
∗
) >v(x
∗
). 
Exercise 31. Complete the proof of Proposition 10.
20
A special case of this result was obtained in Ok (2002) where you can also find some results that
guarantee the finiteness of the representing set of utility functions.
85
Exercise 32.
H
Define the partial order  on R
+
as: x  y iff x ∈ Q
+
and y ∈ R
+
\Q.
Show that there is no countable -dense subset of R

+
, but there exists an U ⊆ R
R
+
with |U| =2that represents .
All this is nice, but it is clear that we need str onger utility representation results
for applic ation s. For instanc e, at present w e h ave no w ay of ev e n speaking about
representing a p reference relation by a continuous utilit y function (or a set of such
functions). In fact, a lot can be said about this issue, but this requires us first to go
throu gh a number of top ics in re al analy sis . We w ill com e b ack to this problem in
Chapters C and D when we are better prepared to tackle it.
86