Real Analysis with Economic Applications - Chapter C pps

Chapter C
M etric Spaces
This chapter provides a self-contained review of the basic theory of m etric spaces.
Chances are good that y ou are familiar with the rudiments of this theory, so our
exposition starts a b it faster than us ual. But don’t wo rry, we slow do wn wh en we
get to the “real stuff,” that being the analy sis of the properties of connec ted n e ss ,
separab ility, compactness and completeness for metric spaces.
Con n ecte dne s s is a ge ometric propert y th a t w ill be of limited u se in this course.
Consequently, its discussion here is quite b rief; all we do is to iden tify the connected
subsets of R, and p repare fo r t he Intermediate Value T heorem that will be given in
the next chapt er. Our treatme nt o f separability is also r e latively short, even though
this c o n cep t will be important for us later on. Because s e p arab ility u s ua lly makes
an appearance only in relatively advan c ed contexts, we will study th is propert y in
greater detail later.
Utility theory that w e sketc he d out in Section B.4 can be tak e n to the next le vel
withthehelpofevenanelementaryinvestigationofconnectedandseparablemetric
spaces. As a brief application, therefore, we formulate here the “metric” v ersions of
some of the utility represen tation theorems that w ere p rov e d in that section. The
story will be brought to its conclusion in Chapter D.
The bulk of this chapter i s de voted to t he analysis of met ric spaces t hat a re e ither
compact or c om p lete. A good understanding of these two properties is essen tial
for real analysis and optimization theory, so w e spend quite a bit of time studying
them. In particular, we consider several examples, give t wo proofs of the Heine-
Borel Theorem for good measure, and discuss wh y closed and bounded spaces need
not be compact in general. Totally bounded sets, the sequential characterization of
com pactnes s, and the rela t i o nship between compactne ss and com pleteness, are also
studied with care.
Most of the results established in this chapter are relatively preliminary observa-
tions whose main purpose is to create good grounds to deriv e a n umber of “ deeper”
facts in later chapters. But there is one major exception: the Banach Fixed Poin t
Theorem . While elementary, and has an amazingly simple proof, this result is of sub-

stantial in tere st, and has nu merous applica tion s. We th u s exp lore it here at len gth .
In p articular, w e consider so me of th e variants of this celebr ated theorem, an d sho w
how it can be used to prov e the “existence” of a solution to certain t ypes of func-
tional equ ation s. As a major application, we prove here both th e local a nd global
versions of the fun da me ntal existe nce theorem of Emile Picard for differential e qua-
tions. Two m ajor generalizations of the Ban ach Fixed P o int Theorem , along w ith
further applications, will be considered in subsequen t chapters.
1
1
Among the excellent introductory references for the analysis of metric spaces are Sutherland
87
1BasicNotions
Recall that w e think of a real function f on R as con t inuous at a given point a ∈ R
iff the image of a poin t (under f)whichisclosetoa is itse lf close to f(a). So, for
instance, the indicator function 1
{
1
2
}
on R is not continuous a t
1
2
, because points that
are arbitrarily close t o
1
2
are not mapped by th is fu nc tion to points tha t are arbitrarily
closetoitsvalueat
1
2

. On the other hand, this function is continuous at every other
point in its domain .
It is crucial to un derstand a t the outset that this “geom etr ic” wa y of thinking
about continuity depends intrinsically on the “distance” bet ween two points on the
real line. While there is an obvious measur e of distance in R, this observation is
importan t precisely because it paves way towards thinking about the contin u ity of
function s defined on more complicated sets on whic h the meaning of the te rm “close”
is not transparen t. As a prerequisite for a suitably general analysis of continuous
functions, therefore, we need to elaborate on the notion of d istance between two
elements of an arbitrary set. This is precisely what we intend to d o in this section.
1.1 M etric Spaces: Definitions and Examples
We begin with the for mal definition of a metric space.
Dhilqlwlrq. Let X beanynonemptyset.Afunctiond : X × X → R
+
that satisfies
the following properties is called a distance function (or a metric)onX:Forany
x, y, z ∈ X,
(i) d(x, y)=0if an d only if x = y,
(ii) (Symmetry) d(x, y)=d(y, x),
(iii) (Triangle Inequality ) d(x, y) ≤ d(x, z)+d(z, y).
If d is a d istance func tion on X, we say that (X, d) is a metric space,andreferto
the elements of X as points in (X, d). If d satisfie s (ii) and (iii), and d(x, x)=0for
an y x ∈ X, then w e say that d is a semim etric on X, and (X, d) is a semimetric
space.
(1975), Rudin (1976), Kaplansky (1977), and Haaser and Sulliva n (1991). O f the more recent
expositions, my personal favorite is Carothers (2000). The firstpartofthatbeautifullywritten
book not only pro vides a much broader perspective of metric spaces than I am able to do here,
but it a lso cov ers additional topics (s uch as compactificationandcompletionofmetricspaces,and
category-type theorems), and sheds light to the historica l dev elopmen t of the material. For a more
advanced (but still very readable) account, I should refer you to Royden (1994), which is a classic

text on real analysis.
88
Rec all that we th in k of the distance bet wee n t wo points x and y on the real line
as |x − y| . Thus the map (x, y) → |x −y| serves as a function that tells us how muc h
apart are any two elements of R from each other. Among others, this function satisfies
properties (i)-(iii) of the definition above (Exam p le A.7). By wa y of abstraction , the
notion of distance function is built only on these three properties. It is remarkable
that these properties are strong enough to introduce to an arbitrary nonem pty set a
geometry ric h enough to build a satisfactory theory of continuous functions.
2
Notation. When th e (semi)metric und er c onsider ation is ap pa rent from the context,
it is customa ry to dispense with the notation (X, d), and refer to X as a metric space.
We also adhere to th is convention here (and spare the notation d for a generic metric
on X). But when we feel that there is a danger of confusion, or we endo w X with a
particular metric d, t h en we shall reve rt bac k to the more d escriptive notation (X, d).
Let us look at some standard examples of metric spaces.
E{dpsoh 1. [1] Let X be any nonem pty set. A trivial w ay of making X ametric
spaceistousethemetricd : X × X → R
+
whic h is defined by
d(x, y):=

1,x= y
0,x= y
.
It is easy to chec k that (X, d) is indeed a metric space. Here d is called t h e discrete
metric on X, and (X, d) is called a discrete space.
[2] Let X := {x ∈ R
2
: x

2
1
+ x
2
2
=1}, and define d ∈ R
X×X
by letting d(x, y) be
thelengthoftheshorterarcinX that join x and y. It is easy to see that this defines
d as a metric on X, and thus (X, d) is a m etric spac e.
[3] Given any n ∈ N, there are various ways of metrizing R
n
. Indeed, (R
n
,d
p
) is
a m etric space for eac h 1 ≤ p ≤∞, where d
p
: R
n
× R
n
→ R
+
is defined by
d
p
(x, y):=


n

i=1
|x
i
− y
i
|
p

1
p
for 1 ≤ p<∞,
and
d
p
(x, y):=max{|x
i
− y
i
| : i =1, , n} for p = ∞.
It is easy to see that each d
p
sat i sfies the first two axioms of being a distan c e func tion .
The verification of the triangle inequalit y in the case of p ∈ [1, ∞) is, on the other
2
The c oncept of metric space was first introduced in the 1906 dissertation of Maurice Fréchet
(1878-1973). (We owe the term “metric space” to Felix Hausdorff, however.) Considered as one
of the major founders of modern real (and functional) analysis, Fréchet is also the mathematician
who first introduced the abstract formulation of compactness and completeness properties. (See

Dieudonné (1981) and Taylor (1982)).
89
hand, not a trivial matter. It rath e r follows from the followin g celebr ate d result of
Herman n Minkowski:
Min kowsk i’s Ineq u ality 1. For any n ∈ R, a
i
,b
i
∈ R,i=1, , n, and any 1 ≤ p<
∞,

n

i=1
|a
i
+ b
i
|
p

1
p
≤

n

i=1
|a
i

|
p

1
p
+

n

i=1
|b
i
|
p

1
p
.
To be able to move faster, we postpone the p r oof o f this importan t inequ ality to t he
end of this subsection . You are invited at this point, how ever, to sho w that (R
n
,d
p
)
is not a m etric space for p<1.
It may be instructive to examine the geometry of the unit “circle”
C
p
:= {x ∈ R
2

: d
p
(0,x)=1}
for various choice s of p. (Here 0 stands for the 2-vector (0, 0).)ThisisdoneinFigure
1, which suggests that the sets C
p
in some sense “converges” to the set C
∞
. Indeed,
for eve r y x, y ∈ R
2
, we have d
m
(x, y) → d
∞
(x, y). (Proof?)
∗∗∗∗FIGURE C.1 ABOUT HERE ∗∗∗∗
The space (R
n
,d
2
) is ca lled the n-dimensional Euclidean space in analysis.
When we refer to R
n
in th e sequel without specifying a particular metric, you sh ou ld
understand that w e view th is set as metrized by the metric d
2
. That is to say, the
notation R
n

is spared fo r the n-dimensional Euclidean space in what follows. If we
wish to endow R
n
with a metric d ifferent than d
2
, w e will be explicit about it.
Notation. Throughout this text we denote the metric space (R
n
,d
p
) as R
n,p
for a ny
1 ≤ p ≤∞. Ho wever, alm ost always, w e use the notatio n R
n
inst ead o f R
n,2
.
Before leaving this ex a mple, let’s see h ow w e can metrize an extended Euclide an
space, say R. For this purpose, w e define the function f : R → [−1, 1] by f(−∞):=
−1,f(∞):=1, and f(x):=
x
1+|x|
for all x ∈ R. The standa rd metric d
∗
on R is the n
defined by
d
∗
(x, y):=|f(x) − f(y)| .

The im portant thing to obser ve he re is that this makes R ametricspacewhichis
essentially identical to [−1, 1]. This is because f is a biject ion from R onto [−1, 1]
that leaves the distance between any two points in tact: d
∗
(x, y)=d
1
(f(x),f(y)) for
an y x, y ∈ R. So, w e should expect that an y metric property that is true in [−1, 1] is
also true in R.
3
3
This point may be somewhat v ague rig ht now. That’s okay, i t will become clearer bit by bit as
we move on.
90
[4] For any 1 ≤ p<∞, we define

p
:=

(x
m
) ∈ R
∞
:
∞

i=1
|x
i
|

p
< ∞

.
This set is metrized b y means of the metric d
p
: 
p
× 
p
→ R
+
with
d
p
((x
m
) , (y
m
)) :=

∞

i=1
|x
i
− y
i
|
p


1
p
.
(When we speak of 
p
as a metric space, we always have this metric in m in d !) O f
course,wehavetocheckthatd
p
is well-defined as a real-valued function, and that
it satisfies the triang le inequ ality. But no w orries, these facts follow r e ad ily from th e
following generalization of Mink owsk i’s Inequality 1:
Min kowsk i’s Ine q ua lity 2. For any (x
m
) , (y
m
) ∈ R
∞
and 1 ≤ p<∞,

∞

i=1
|x
i
+ y
i
|
p


1
p
≤

∞

i=1
|x
i
|
p

1
p
+

∞

i=1
|y
i
|
p

1
p
. (1)
We will pro ve this inequality at the end of this subsection. You should assume its
validity for now, and v e r ify that d
p

is a m etric on 
p
for any 1 ≤ p<∞.
By 
∞
, we mean the s et of all bounded real sequences, th at is,

∞
:= {(x
m
) ∈ R
∞
:sup{|x
m
| : m ∈ N} < ∞} .
It is implicitly understood that this set i s e ndowed with the metric d
∞
: 
∞
×
∞
→ R
+
with
d
∞
((x
m
) , (y
m

)) := sup{|x
m
− y
m
| : m ∈ N}.
That d
∞
is indeed a metric will be verified below. This m etric is called the sup-
metric on the set of all bounded r eal sequences.
Before we leave th is example let us stress that a ny 
p
space is smaller than the set
of all r eal sequences R
∞
since the mem bers of such a space are real sequences that are
either bounde d or that satisfy some form of a summability condition (that ensures
that d
p
is real-valued). Indeed, no d
p
defines a distance function on the en tire R
∞
.
(Why?) But this does not m ean that we cannot metrize th e set of all real sequences
in a us e fu l wa y. We c an, and we will, later in th is c h a p ter.
[5] Let T beanynonemptyset. By B(T ) we mean th e set o f all bounded real
function s defined on T, that is,
B(T ):=

f ∈ R

T
:sup{|f(x)| : x ∈ T } < ∞

.
We will always think of this space as metrized by the sup-metric d
∞
: B(T) ×
B(T ) → R
+
whic h is defined by
d
∞
(f,g):=sup{|f(x) −g(x)| : x ∈ T }.
91
It is easy to see that d
∞
is real-valued. Indeed, for any f, g ∈ B(T),
d
∞
(f,g) ≤ sup{|f(x)| : x ∈ T } +sup{|g(x)| : x ∈ T} < ∞.
It is also readily checked that d
∞
satisfies the first two req uirem ents o f being a distance
function. As for the triangle inequalit y, all we need is to invoke the cor responding
property of the absolute value function (Example A.7). After all, if f,g,h ∈ B(T ),
then
|f(x) − g(x)| ≤ |f(x) − h(x)| + |h(x) −g(x)|
≤ sup{|f(y) −h(y)| : y ∈ T} +sup{|h(y) −g(y)| : y ∈ T }
= d
∞

(f,h)+d
∞
(h, g)
for any x ∈ T, so
d
∞
(f,g)=sup{|f(x) − g(x)| : x ∈ T} ≤ d
∞
(f,h)+d
∞
(h, g) .
Given that a sequence and/or an n-vector can alwa ys be thought of as special
functions (Se c tion A.1 .6), it is plain that B({1, , n}) co inc i d es w ith R
n,∞
(for any
n ∈ N) while B(N) coincides w ith 
∞
. (Right?) Th e ref ore , the inequality we just
established pro ves in one strok e that both R
n,∞
and 
∞
are metric spaces. 
Remark 1. Distance functions need not bounded. However, giv en any metric space
(X, d), we can always find a bounded metric d

on X that orders th e distances between
points in the space ordinally the same way as the original metric. (That is, such a
metric d


satisfies: d(x, y) ≥ d(z, w) iff d

(x, y) ≥ d

(z,w) for all x, y, z, w ∈ X.)
Indeed, d

:=
d
1+d
is such a d istance function. (Note. We have 0 ≤ d

≤ 1.) As we
proceed further, it will becom e clear that there is a good sense in which (X, d) and
(X,
d
1+d
) c an be though t of as “equivale nt” in terms of certain characteristics (and
not so in terms of others).
4

If X i s a metric space (with metric d )and∅ = Y ⊂ X, we can view Y as a
metric space in its own righ t by using the distance function induced b y d on Y. More
precisely, we mak e Y a metric space by means of the distance function d|
Y ×Y
. We
then say that (Y,d|
Y ×Y
),orsimplyY, is a metric subspace of X. Fo r instance, we
think of a ny in terval, say [0, 1], as a metric subspace of R; this means simply that

the distance between any two elements x and y of [0, 1] is calculate d by viewing x
and y as points in R: d
1
(x, y)=|x − y| . Of course, we can also think of [0, 1] as a
4
This is definitely a good point to keep in mind. When there is a “natural” unbounded metric
on the space that you are working with, but for some reason you need a bounded metric, you can
alw ays modify the original metric to get a new bounded and ordinally equivalent metric on your
space. (More on this in Section 1.5 below.) Sometimes, and we will encounter such an instance
later, this little trick does wonders.
92
metric subspace of R
2
. Formally, we would d o this by “identifying” [0, 1] with th e set
[0, 1] ×{0} (or wit h {0}×[0, 1], or with [0, 1] ×{47}, etc.), and consider [0, 1] ×{0}
as a metric subspace of R
2
. This would render the distance between x and y equal
to, again, |x − y| (for, d
2
((x, 0), (y, 0)) = |x −y|).
5
Con v ention. Throughout this book, when we consider a nonempty subset S of
a Euclidean space R
n
as a metric spac e without e xp lic itly men t ion ing a particula r
metric, you should understand that we view S a m etric subspace of R
n
.
E{dpsoh 2. [1] For any positiv e integ er n, we may think of R

n
as a m etric subspace
of R
n+1
by identify ing it with th e subset R
n
×{0} of R
n+1
. By induction, therefore,
R
n
can be though t of as a m etric subspace of R
m
for any m, n ∈ N with m>n.
[2] Let −∞ <a<b<∞, an d consider t he metric space B[a, b] introduced in
Example 1.[5]. R ecall that ev e ry contin uous function on [a, b] is bounded (Exerc ise
A.53), and h ence C[a, b] ⊆ B[a, b]. Consequently, w e can consider C[a, b] as a metric
subspace of B[a, b]. Indeed, throughout this text, whenever w e talk about C[a, b] as
a metric space, we think of the distance bet ween any f and g in C[a, b] as d
∞
(f,g),
unless otherwise is explic itly mention ed.
[3] Let −∞ <a<b<∞, and recall that we denote the set of all con tin uously
differentiable functions on [a, b] by C
1
[a, b] (Sectio n A .4.2). The m e tric tha t is used
for this space is usually not the sup-metric. That is, we do not define C
1
[a, b] as a
metric subspace of B[a, b]. (There are a good reasons for this, but w e’ll get to them

later.) Instea d, C
1
[a, b] is commonly metrized by means of the distance function
D
∞
: C
1
[a, b] × C
1
[a, b] → R
+
define d by
D
∞
(f,g):=d
∞
(f,g)+d
∞
(f

,g

).
It is th is metric that we have in mind w hen talking about C
1
[a, b] as a metric space.

Exercise 1.
H
If (X, d) and (X, ρ) aremetricspaces,is(X, max{d, ρ}) necessarily a

metric space? Ho w about
(X, min{d, ρ})?
Exercise 2. For any metric space X, show that |d(x, y) −d(y, z)| ≤ d(x, z) for all
x, y, z ∈ X.
Exercise 3. For an y semimetric space X,define the binary relation ≈ on X by
x ≈ y iff d(x, y)=0. Now define [x]:={y ∈ X : x ≈ y} for all x ∈ X, and let
X := {[x]:x ∈ X}. Finally, define D : X
2
→ R
+
by D([x], [y]) = d(x, y).
(a) Show that ≈ is an equivalence relation on X.
(b)Provethat(X ,D) is a metric space.
5
Quiz. Is the metric space g iven in Example 1.[2] a metric subspace of R
2
?
93
Exercise 4. Show that (C
1
[0, 1],D
∞
) is a metric space.
Exercise 5. Let
(X, d) be a metric space and f : R
+
→ R a concave and strictly
increasing function with
f(0) = 0. Show that (X, f ◦ d) is a metric space.
The final orde r of busin es s in this sub s ection is to pr ove Minkowski’s Inequalities

whichwehaveinvokedabovetoverifythatR
n,p
and 
p
aremetricspacesforany
1 ≤ p<∞. Since the first one is a special case of the second (yes?), all we need is to
establish Minkowsk i’s Inequa lity 2.
Proof of Minkow ski’s Inequality 2. Take any (x
m
) , (y
m
) ∈ R
∞
and fixany1 ≤
p<∞. If either

∞
|x
i
|
p
= ∞ or

∞
|y
i
|
p
= ∞, then (1) becomes trivial, so we
assume that


∞
|x
i
|
p
< ∞ and

∞
|y
i
|
p
< ∞. (1) is also trivially true if either (x
m
)
or (y
m
) equals (0, 0, ), so we focus on the case where both α := (

∞
|x
i
|
p
)
1
p
and
β := (


∞
|y
i
|
p
)
1
p
are positiv e real nu mbers.
Define the real sequences (ˆx
m
) or (ˆy
m
) by ˆx
m
:=
1
α
|x
m
| an d ˆy
m
:=
1
β
|y
m
| . (Notice
that


∞
|ˆx
i
|
p
=1=

∞
|ˆy
i
|
p
.) Using the triangle inequality for the absolute value
function (Examp le A.7), and the fact that t → t
p
is an inc re a s ing map on R
+
, we
find
|x
i
+ y
i
|
p
≤ (|x
i
| + |y
i

|)
p
=(α |ˆx
i
| + β |ˆy
i
|)
p
=(α + β)
p

α
α+β
|ˆx
i
| +
β
α+β
|ˆy
i
|

p
for eac h i =1, 2 , But since t → t
p
is a convex m a p on R
+
, we have

α

α+β
|ˆx
i
| +
β
α+β
|ˆy
i
|

p
≤
α
α+β
|ˆx
i
|
p
+
β
α+β
|ˆy
i
|
p
,i=1, 2, ,
and hence
|x
i
+ y

i
|
p
≤ (α + β)
p

α
α+β
|ˆx
i
|
p
+
β
α+β
|ˆy
i
|
p

,i=1, 2,
Summing over i, th en,
∞

i=1
|x
i
+ y
i
|

p
≤ (α + β)
p

α
α+β
∞

i=1
|ˆx
i
|
p
+
β
α+β
∞

i=1
|ˆy
i
|
p

=(α + β)
p

α
α+β
+

β
α+β

1
p
.
Thus

∞
|x
i
+ y
i
|
p
≤ (α + β)
p
which is equ ivalen t to (1). 
We conclude by noting that (1) holds as an equalit y (for an y given 1 <p<∞)
iff either x =(0, 0, ) or y = λx for som e λ ≥ 0. The proof is left as an exercise.
94
1.2 Open and Closed Sets
We no w review a n umber of fundamental concepts regarding metric spaces.
Dhilqlwlrq. Let X be a metric (or a sem imetric) space. Fo r an y x ∈ X and ε > 0,
we define the ε-neigh borhood of x in X as the set
N
ε,X
(x):={y ∈ X : d(x, y) < ε}.
In turn, a neighborhood of x in X is any subset of X that con tains at least one
ε-neighborhood of x in X.

The first thing that you should note about the ε-neighborhoo d of a point x in
a ( s e m i)metric sp ace is that such a set is n ever empty, for it contains x. Secondly,
mak e sure y ou understand that this notion is based on four primitive s. Ob v iou sly,
the ε-neighborhood o f x in a metric s pace X depends on ε and x. But it also depends
on the set X and the distance function d used to metrize this set. For instance, the
1-ne ighborhood of 0 :=(0, 0) in R
2
is {(x
1
,x
2
) ∈ R
2
: x
2
1
+ x
2
2
< 1}, whereas the
1-ne ighborhood of 0 in R ×{0} (vie wed as a m e tric subspace of R
2
)is{(x
1
, 0) ∈ R
2
:
−1 <x
1
< 1}. Similarly, the 1-neighborhood of 0 in R

2
is d istin c t from that in R
2,p
for p =2.
The notion of ε-neighborhoods pla ys a m ajor role in real analysis mainly through
the following definition.
Dhilqlwlrq. A subset S of X is said to be open in X (or an open subset of X)
if, for each x ∈ S, th ere exists an ε > 0 suc h that N
ε,X
(x) ⊆ S. AsubsetS of X is
said to be closed in X (or a closed subset of X)ifX\S is open in X.
Because an ε-neighborhood of a point is inheren tly connected to the underlying
metric space, so does the notions of open and c losed sets. P lease keep in mind that
c han ging the metric on a giv e n set, or concentrating on a metric subspace of the
original metric space, would in general yield differentclassesofopen(andhence
closed) sets.
Dhilqlwlrq. Let X be a metric space and S ⊆ X. The largest open set in X that is
contained in S (th at is, the ⊇-m ax imum of the class of all open subsets of X contained
in S) is ca lled the inte rior of S (r e lative to X), and is denoted by int
X
(S). On the
other hand, the closure of S (relative to X), denoted by cl
X
(S), is defined as the
sma llest closed set in X that contains S (that is, the ⊇-minim um of the class of all
closed subsets of X that con t ain S). The boundary of S (relative to X), denoted
by bd
X
(S), is de fined as
bd

X
(S):=cl
X
(S)\int
X
(S).
95
Let X be a metric s pace, and Y a metric subspace of X. Fo r any subset S of Y, we
ma y think o f the interior of S as lying i n X or in Y. (And y e s , these may well be quite
different!) It is for th is reason t ha t we use the notation int
X
(S), instead of int(S), to
mean the interior of S relative to the m etric spa ce X. However, if there is only one
metric space under consideration, or the context lea ves no room for confusion, w e may,
and will, simply write int(S) to denote the interior of S relative to the appropriate
space. (The sam e comments a p p l y to the closure and boun d ary operators as well.)
E{dpsoh 3. [1] In a ny metric space X, the sets X and ∅ are both open and closed.
(The sets which a re both open and closed are sometim es called clopen in analysis.)
On the o ther hand, for any x ∈ X and ε > 0, the set N
ε,X
(x) is open, and the set
{x} is closed.
To prov e that N
ε,X
(x) is open, t ake any y ∈ N
ε,X
(x) and d efine δ := ε−d(x, y) > 0.
We have N
δ,X
(y) ⊆ N

ε,X
(x) because, by the triangle inequalit y,
d(x, z) ≤ d(x, y)+d(y, z) <d(x, y)+ε − d(x, y)=ε
for an y z ∈ N
δ,X
(y). (See Figure 2 to see the intuition of the argument.)
To prove that {x} is closed, we need to show that X\{x} is open. If X = {x},
there is nothing to p rove. (Yes?) On the o ther ha nd , if there exists a y ∈ X\{x }, we
then have N
ε,X
(y) ⊆ X\{x}, where ε := d(x, y). It follows that X\{x} is open, and
{x} is closed.
∗∗∗∗FIGURE C.2 ABOUT HERE ∗∗∗∗
[2] Any subset S of a nonem p ty set X is open with respect to the discrete metric.
For, if x ∈ S ⊆ X, then w e ha ve N
1
2
,X
(x)={x} ⊆ S, where the discrete m etric is
used in computing N
1
2
,X
(x). Th us: A ny subset of a discrete space is clopen.
[3] It is possible for a set in a metric sp ac e to be neither open nor closed . In
R, for instance, (0, 1) is open, [0, 1] is clos ed, and [0, 1) is neither open n or closed.
But observe that the structure of the mother metric space is crucial for the v alidity
of these statements. For instance, the set [0, 1) is open when considered as a set in
themetricspaceR
+

. (Indeed, re lative to this metric subspa ce of R, 0 belongs to the
in terior of [0, 1), and the boundary of [0, 1) equals {1}.) More ge n erally, th e follo wing
fact is tru e :
Exercise 6 . Given any metric space X, let Y beametricsubspaceofX,andtake
any
S ⊆ Y. Show that S is open in Y iff S = O ∩ Y forsomeopensubsetO of X,
anditisclosedinY iff S = C ∩Y forsomeclosedsubsetC of X.
Warnin g. Giv en an y m etric space X, let Y beametricsubspaceofX, and take any
U ⊆ Y. An immediate application of Exe rcis e 6 shows that
U is open in X on ly if U is open in Y.
96
(Yes?) But the converse is false! For instance, (0, 1) ×{0} is an open subset of
[0, 1] ×{0} but it is not an open subset of [0, 1]
2
. Yet,ifthemetricsubspaceunder
consideration is an open subset of the mother metric space, all goes well. Put precisely,
provided that Y is open in X,
U is open in X iff U is open in Y.
(Proof. If U is open in Y, then by Exercise 6, there is an open subset O of X such
that U = O ∩ Y, so if Y is o pen in X as well, U must be open in X (because the
in tersection of tw o open sets is open).)
Sim ilar remarks apply to closed sets as well, of c ou r se. If C ⊆ Y, th en
C is closed in X on ly if C is closed in Y,
and, provided that Y is closed in X,
C is close d in X iff C is closed in Y.
(Proofs?)
[4] How do w e know that int
X
(S) is well-defined for any subset S of a metric
space X? (P erhaps the class {O ∈ 2

X
: O is open in X and O ⊆ S} does not have
a ⊇- maximum, t h a t is, there is no largest open subset of X that is contained in S!)
Thereasonisthattheunionofanycollectionofopensetsisopeninametricspace.
(Yes?) Th us in t
X
(S) is well-defined, since, th ank s to this p ropert y, we ha ve
int
X
(S)=

{O ∈ O
X
: O ⊆ S},
where O
X
is the class of all open subsets of X.
6,7
By contrast, the intersection of a ny
collection of closed subsets of X is closed (why?), a nd hence cl
X
(S) is w e ll-d e fined
for an y S ∈ 2
X
:
cl
X
(S)=

{C ∈ C

X
: S ⊆ C},
where C
X
is the c lass of all closed subsets of X.
Warnin g. While the intersection of a finite collectionofopensetsisopen(why?),
an arbitrary intersection of open sets need not be open in general. For instance,
(−1, 1) ∩ (−
1
2
,
1
2
) ∩ ···= {0} isnotanopensubsetofR. Similarly, the union of a
finite collection of closed sets is closed, but an arbitrary union of closed sets need not
be closed.
[5] It is obvious that a set S in a metric space X is closed iff cl
X
(S)=S. (Is it?)
Similarly, S is open iff int
X
(S)=S. Also observ e that, for an y subset S of X, we
6
This presumes that {O ∈ O
X
: O ⊆ S} = ∅; how do I know that this is the case?
7
A useful Corollary. x ∈int
X
(S) iff there exists an ε > 0 such that N

ε,X
(x) ⊆ S.
97
have x ∈ bd
X
(S) iff S ∩N
ε,X
(x) and (X\S) ∩N
ε,X
(x) are nonempty for any ε > 0.
(Proofs?)
8
[6] A se t is open in a given metric s pace (X, d) iff it is open in (X,
d
1+d
).(Recall
Remark 1.) So, in terms of their o pen set structures, these t wo m etric spaces are
identical (even though the “ distance” between an y t wo poin t s in X would be assessed
differently by d and
d
1+d
).
[7] In contrast to metric spaces, semimetric spaces may h av e very few open (th us
close d) sets. For instan ce, (X, d
o
) is a semim e tric space if d
o
(x, y):=0for all x, y ∈ X.
This space is not a metric space unless |X| =1, and the only open and/or closed sets
in it are ∅ and X. One may thus view such a space as a polar opposite of a discrete

space — it is called an ind i screte space. 
Exercise 7.
H
Can you find a metric on N such that ∅ = S ⊆ N is open iff N\S is
finite?
Exercise 8.
Pro ve all the assertions made in Examples 3.[4] and 3.[5].
Exercise 9.
H
Show that, for an y subset S of a metric space X, t he in terior of bd
X
(S)
equals S iff S = ∅. Find an example of a metric space X that contains a nonempty
set
S with int
X
(bd
X
(S)) ⊇ S.
Exercise 10.
H
Given a metric space X, let Y be a metric subspace of X, and S ⊆ X.
Show that
in t
X
(S) ∩ Y ⊆ int
Y
(S ∩ Y ) and cl
X
(S) ∩ Y ⊇ cl

Y
(S ∩ Y ),
and giv e examples to show that the conv erse containmen ts do not hold in general.
Also pro ve that
int
X
(S) ∩ Y = in t
Y
(S ∩ Y ),
provided that Y is open in X. Similarly, cl
X
(S) ∩ Y = cl
Y
(S ∩ Y ) holds if Y is
closed in
X.
Exercise 11. Let S be a closed subset of a metric space X, and x ∈ X\S. Show that
thereexistsanopensubset
O of X such that S ⊆ O and x ∈ X\O.
1.3 Conv ergent Sequences
The notion of closedness (and hence openness) of a set in a metric space can be
c haracterized by means of the sequences th at live i n that space. Since th is char acter-
ization often sim plifies things consid era bly, it will be a good idea to provide it here
before proceeding further into the theory of metric spaces. Let us firs t recall what it
means for a sequen ce to converge in a metric space.
8
Quiz. What is the boundary of the unit “circle” C
∞
definedinExample1.[3]?
98

Dhilqlwlrq. Let X be a metric (or a semimetric) space, x ∈ X, and (x
m
) ∈ X
∞
.
9
We say that (x
m
) con verges to x if, for ea ch ε > 0, there exists a real number M
(that may depend on ε) such that d(x
m
,x) < ε for all m ≥ M. (Note. Thisisthe
sam e thing as saying d(x
m
,x) → 0.) In this case , we sa y that (x
m
) conv e rges in X,
or that it is convergent in X, we refer to x as the lim it of (x
m
), and write either
x
m
→ x or lim x
m
= x.
A sequence (x
m
) in a metric space X can conve rge to at most one limit. Indeed, if
both x
m

→ x and x
m
→ y held true, then, by symmetry a n d the t rian gle inequality,
we would have
d(x, y) ≤ d(x, x
m
)+d(x
m
,y) for all m =1, 2,
while
d(x, x
m
)+d(x
m
,y) → 0,
which implies x = y. (Why?)
10
Th e situation is different in the case of semimetric
spaces.
Exercise 12.
H
Show that a conv ergen t sequence may have m ore than one limit in a
semimetric space.
To recap, a sequence (x
m
) in a metric space X converges to a point x in this space,
if, for an y ε > 0, all but finitely man y term s of the sequence (x
m
) belong to N
ε,X

(x).
One way of th in kin g about this intuitively is view ing the sequen ce (x
m
) as “staying
in N
ε,X
(x) eventu ally” n o ma tter h ow small ε is. Equ ivalen tly, we hav e x
m
→ x iff,
for every open neighborhood O of x in X, there exists a positive in te ge r M such that
x
m
∈ O for all m ≥ M. So, for instance, the sequence (1,
1
2
,
1
3
, ) ∈ R
∞
+
con verges to
zero, because, for any open n eighborhood O of 0 in R
+
,thereexistsanM ∈ N such
that
1
m
∈ O for all m ≥ M. (Yes?)
E{dpsoh 4. [1] A s equence (x

m
) is con vergent in a discrete space iff it is e ventually
constant (th at is, there ex ists an M ∈ N such tha t x
M
= x
M+1
= ···.)
[2] A c onstant sequence in any m e tric space is conv ergent.
[3] Take any n ∈ N, and let (x
m
)=((x
m
1
, , x
m
n
)) be a sequence in R
n
. It is easy
to sh ow that x
m
→ (x
1
, , x
n
) iff (x
m
i
) converges to x
i

for each i =1, , n. (Prove!)
9
In this book, as a notational convention, I denote a generic sequence in a given (abstract) metric
space X by (x
m
), (y
m
) etc (This convention becomes particular ly useful when, for instance, the
terms of (x
m
) are themselves sequences.) The generic real (or extended real) sequences are de not ed
as (x
m
), (y
m
), etc., and generic sequences of real functions are denoted a s (f
m
), (g
m
), etc
10
I could write the argument more compactly as d(x, y) ≤ d(x, x
m
)+d(x
m
,y) → 0. I will use this
sort of a s horthand expression quite frequently in what follows.
99
[4] Consider the follo wing real sequences: For each m ∈ N,
x

m
:=

0, ,0,
1
m
, 0,

,y
m
:= (0, , 0, 1, 0, ),z
m
:=

1
m
, ,
1
m
, 0,

,
where the only nonzero term of the sequences x
m
and y
m
is the mth on e, and all
but the first m terms of z
m
are zero. Since d

p
((x
m
), (0, 0, )) =
1
m
for each m, we
have (x
m
) → (0, 0, ) in 
p
for any 1 ≤ p ≤∞. (Don’t forget that here (x
m
) is
a sequence of real sequences.) In contrast, it is easily checked that the sequence
(y
1
,y
2
, ) isnotconvergentinany
p
space ( 1 ≤ p ≤∞). On the other hand,
we have d
∞
(z
m
, (0, 0, )) =
1
m
→ 0, so (z

m
) converges to (0, 0, ) in 
∞
.Yet
d
1
(z
m
, (0, 0, )) = 1 for ea ch m, so (z
m
) does not converge to (0, 0, ) in 
1
. Is
(z
m
) convergen t in a ny 
p
, 1 <p<∞?
[5] Let f
m
∈ B[0, 1] be defined b y f
m
(t):=t
m
,m=1, 2, Sincef
m
(t) → 0 for
all t ∈ [0, 1) and f
m
(1) = 1 for all m, it may at first s eem plausible that f

m
→ f where
f(t)=0for all 0 ≤ t<1 and f(1) = 1. But this is false, because d
∞
(f
m
,f)=1for
all m ∈ N. T his example shows, again, ho w detrimental the c hoice of the m etric may
be in studying the convergence of sequences. (We will come back to this example in
due course.) 
Exercise 13. For any given metric space (X, d), show that, for any (x
m
) ∈ X
∞
and
x ∈ X, we have x
m
→ x in (X, d) iff x
m
→ x in (X,
d
1+d
).
1.4 Sequ ential Chara cterization of Closed Sets
Here is th e sequ ential ch a rac te rization of closed sets we p romised above.
Proposition 1. AsetS in a metric space X is c lose d if, a n d only if, every seque nce
in S that con v erges in X con verges to a poin t in S.
Proof. Let S be a closed subset of X, and take any (x
m
) ∈ S

∞
with x
m
→ x fo r
some x ∈ X. If x ∈ X\S, then we can find an ε > 0 with N
ε,X
(x) ⊆ X\S, because
X\S is open. But since d(x
m
,x) → 0, there m us t exist a large enough M ∈ N
such that x
M
∈ N
ε,X
(x), contradicting that all terms of the sequence (x
m
) lies in S.
Con versely, suppose that S is not closed i n X.ThenX\S is n o t open, so we can find
an x ∈ X\S such that every ε-neighborhood around x intersects S. Thus, for an y
m =1, 2, , there is an x
m
∈ N
1
m
,X
(x) ∩ S. But then (x
m
) ∈ S
∞
and lim x

m
= x,
and yet x/∈ S. Thus if S was not closed, there would ex ist at least one se quence in S
that converges to a poin t outside of S. 
To understand what this result sa ys (or better, what it does not say), consider
themetricspace(0, 1) and ask yo u r se lf if (0, 1) is a closed subset of th is space. A
common mistake is to answer this q u estion in the n egative, and use Proposition 1
100
to suggest a p roof. The fallaciou s arg ument goes as follows: “B y Proposition 1, the
interval (0, 1) cannot be closed, because the sequ ence (1,
1
2
,
1
3
, ) in (0, 1) converges
to 0, a point which is outside of (0 , 1).” The problem with this argument is that it
works with a non-convergent seq uence in (0, 1). Indeed, the sequence (1,
1
2
,
1
3
, ) does
not con v erge anywhere in the space (0 , 1). A f ter all, the only poss ible limit for th is
seque nce is 0, but 0 does not liv e in the mother space. (Note. (1,
1
2
,
1

3
, ) would
be convergent, for instance, if our mother space w as [0, 1).) In fact, any con v ergen t
seque nce in (0, 1) must converge in (0, 1) (because of the funny s truc tur e of this metric
space), a n d therefore we m u st conclude that (0, 1) is closed as a subset of itself, which
is, of course, a triviality (Example 3.[1]). Th is observa tion poin ts once again to the
fact that the metric properties of sets (such as the convergence of sequences) depend
crucially on the structure of the mother metric space under consideration.
Exercise 14 .
H
Prove that, for an y subset S of a metric space X, the following are
equivalent:
(a)
x ∈ cl
X
(S);
(b) Every open neighborhood of x in X in tersects S;
(c) There exists a sequence in S which conv erges to x.
Exercise 15. Let
(x
m
) be a sequence in a metric space X. We say that x ∈ X is a
cluster point of
(x
m
) if, for each ε > 0,N
ε,X
(x) contains infinitely many terms of
(x
m

).
(a) Show that any con vergent sequence has exactly one cluster poin t. (Note.The
con verse is not true; consider, for instance,
(1, 0, 2, 0, 3, 0, ) ∈ R
∞
.)
(b) For any
k ∈ Z
+
, give an example of a sequence (in some metric space) with
exactly
k many cluster points.
(c) Show that
x is a cluster point of (x
m
) iff there is a subsequence of (x
m
) that
con verges to
x.
1.5 Eq uivalence of Me tr ics
When would endowing a given nonempty set X with tw o different metrics d and
D yield metric spaces that could reasonably be considered as “equiva lent”? Wh ile
this que r y is rather vague at present, w e can still com e up with a fe w benchmark
responses to it. For instance, it makes perfect s ens e to vie w two me tric sp ac es o f
the form (X, d) and (X, 2d) as “identical.” The se co nd s p ac e s imp ly equals a version
of the form er spa ce in whic h the distances are measured in a different scale. Fine,
how about (X, d) and (X,
√
d)?

11
This com pa r ison se ems a bit more subtle . While
d and
√
d are ordinally identical, one is not a sim p le rescaling of the other, so the
metric spaces they induce may in principle look rather differen t fr om certain points
of view. A t the v ery least, it seems that the connection bet ween (X, d) and (X, 2d)
11
If d is a metric on X, then so is
√
d. Why?
101
is tighter than that bet ween (X, d) and (X,
√
d), even though we would not expect
the properties of the latter two spaces to be vastly differen t from each other.
Let us be more precise now.
Dhilqlwlrq. Let d and D be two metrics on a nonempty set X, and denote the classes
of all open subsets of X with respect to d and D as O(d) and O(D), respectively. We
say tha t d and D (and/o r (X, d) and (X, D ))areequivalen t if O (d)=O(D),and
that they are strongly equivalent if
αd ≤ D ≤ βd
for some real numbers α, β ≥ 0.
As we procee d further in the course, it will become clear t hat the class o f all open
subsets of a given metric space determines a good deal of the properties of this space
(at least insofar as the basic questions that concern real analysis). Consequently,
if two metrics on a giv en nonempt y set generate precisely the same class of open
sets, then the resulting metric spaces are bound to look “identical” from a va riety
of viewpoints. For ins tance, the cla ss es of all closed subsets of two equivalent m e tric
spaces are the same. Moreo ver, if a sequence in one metric space conv erges, then it

also does so in a ny equivalent metric space. (W hy?)
The reasons for considering two strongly equivalent spaces as “identical” are even
more compelling. Notice first that any two strongly equivalent m etrics are equivalent.
(Why?)Toshowthattheconverseisfalse,weshallusethefollowingconcept.
Dhilqlwlrq. A subset S of a metric space X is called bounded (in X) if there
exists an ε > 0 suc h that S ⊆ N
ε,X
(x) for som e x ∈ S. If S is not bounded, then it is
said to be unbounded.
It is clear that if (X, d) and (X, D) are strongly equ ivalent metric spaces, t hen a
subset S of X is boun ded in (X,d) iff it is bounded in (X, D). (Yes?) By contrast,
boundedness is not a propert y that is in varian t under e quivalence of metrics. Indeed,
if (X, d) is an unbounded metric space, then (X,
d
1+d
) is a bounded metric space,
whereas d and
d
1+d
are equivalent. (Recall Ex am ple 3.[6]). Thus , d and
d
1+d
are
equivalent metrics on X that are not strongly equivalent.
In fact, s tron g equivalence of metrics i s substantially more deman d ing tha n their
equivalence. We used the boundedness propert y here only for illustrative purposes.
Tw o equivalen t metric spaces (X, d) and (X, D) need not be strongly equivalent,
even if X is rendered bounded by both of these metrics. Fo r instance, ([0, 1],d
1
)

and ([0, 1],
√
d
1
) areequivalent(bounded)metricspaces.
12
Yet, d
1
and
√
d
1
are not
12
More generally, the following is true for any metric sp ace (X, d): If f : R
+
→ R is strictly
increasing, continuous and subadditive, then (X, d) and (X, f ◦ d) are e quivalent metric spaces.
Proof? (Note. Here subadditivity of f means that f (a + b) ≤ f (a)+f(b) for all a, b ≥ 0.)
102
strongly equivalent metrics on [0, 1]. (Proof. There is no α > 0 such that α

d
1
(x, y) ≤
d(x, y) for a ll 0 ≤ x, y ≤ 1.)
To give another exam p le, let n ∈ N, and take an y 1 ≤ p ≤∞. Clearly, for any
metric d
p
on R

n
we have
d
∞
(x, y) ≤ d
p
(x, y) ≤ nd
∞
(x, y) for all x, y ∈ R
n
.
Thus, R
n,p
is strongly equiva lent to R
n,∞
for any 1 ≤ p<∞. Since “being strongly
equivalent to” is an equivale n ce relation on the class of all metrics on R
n
, we conclude:
d
p
and d
q
are strongly equ iva lent m etrics on R
n
for any 1 ≤ p, q ≤∞.
13
2 C o nnectedness and Separab ility
The notion of metric space alone is too general to be useful in applications. Indeed,
some met r ic spaces can be quite ill-behav e d (e.g. d i sc re te space s), so we need to

“find” those sp aces that possess certain regularit y properties. We consider two suc h
properties in this section. Th e first of these, connectedness, gives one a glimpse o f
ho w one would study the geometry of an a rbitr ary metric sp ace. Th e second one,
separab ility, iden ti fies those metric spaces that have relatively “few” open sets. While
they are important in other contexts, these properties pla y a limited role in this book.
We thus proceed at somewhat of a quick pace here.
2.1 Connected Metric Spaces
Intuitively speaking, a connected subset of a metric space is one that cannot be
partitioned in to two (or more) separate pieces, it is rather in one whole piece. In R,
for instance, w e lik e to think (0, 1) as connected and [0, 1] ∪ [2, 3) as disconnected.
The d efin ition below formalizes this sim ple geometric in tu ition.
Dhilqlwlrq. We say that a metric space X is connected if there d o not exist tw o
nonempt y and disjoint open subsets O and U of X such that O ∪ U = X. In turn,
asubsetS of X is said to be connected in X (or a connected subset of X)if
S is a connected metric subspace of X. (So, S ⊆ X is connected in X iff it cannot be
written as a disjoin t union of tw o nonempty sets that are open in S.)
The follow in g simple re sult pro v ide s an interesting chara cte riz ation of conn ecte d
metric spaces.
13
Due to this fact, it simply does not matter which d
p
metric is used to metrize R
n
, for the
purposes of this text. I do not, of course, claim that all properties of interest are shared by R
n,p
and
R
n,q
for any p, q ≥ 1. If we were interested in the shape of the unit circles, for example, then there

is no way we would view R
n,1
and R
n
as identical (Figure 1).
103
Proposition 2. Let X be a m etric space. T hen, X is connected if, and only if, the
only clopen sub sets of X are ∅ an d X.
Proof. If S/∈ {∅,X} is a clopen sub set of X,thenX cannot be connected since
X = S ∪ (X\S). Conversely, assume that X is not connected. In this case we can
find nonempty and disjoin t open subsets O and U of X such that O ∪ U = X. But
then U = X\O so that O must be both open and closed. Since O/∈ {∅,X}, th is
proves th e claim. 
So, for instance, a discrete space is not connected unless it contains only one
element, because any subset of the discrete space is clopen. Similarly, Q is not
connected in R, for Q =(−∞,
√
2) ∪ (
√
2, ∞). In fact, the o nly connected subsets o f
R are the intervals, as we show next.
E{dpsoh 5. We claim that any interval I is connected in R. To derive a con tra-
diction, suppose we could wr ite I = O ∪ U for nonempty and disjoin t open subsets
O and U of I. Pick an y a ∈ O and b ∈ U and let a<bwithout loss of generality.
Define c := sup{t ∈ O : t<b}, and note that a ≤ c ≤ b and hence c ∈ I (because
I is an in terval). If c ∈ O, th en c = b (since O ∩ U = ∅), so c<b.But, since O is
open, there exists an ε > 0 such that b>c+ ε ∈ O, which c ontradicts the choice of
c. (W hy?) If, on the other hand, c/∈ O, then a<c∈ U (since I = O ∪ U). T hen,
given that U is open, (c − ε,c) ⊆ U, and hence (c − ε,c) ∩ O = ∅, which means that
sup{t ∈ O : t<b} ≤ c − ε, con tradiction.

Conv ersely, let I be a nonempty connected set in R, but assume that I is not an
in te rval. The latter hy pothesis imp lies that there exist tw o points a and b in I suc h
that (a, b)\I = ∅. (Wh y ?) Pick any c ∈ (a, b)\I, and define O := I ∩ (−∞,c) and
U := I ∩(c, ∞). Observ e that O and U are nonempty and disjoint open subsets of I
(Exercise 6) while O ∪ U = I, con tradicting the connectedness of I. Conclusion: A
nonempty s ubset of R is connected iff it is an interval. 
We won’t elaborate on the importance of the notion of connectedness just as yet.
This is best seen when one considers the properties of continuou s functions defined
on connected metric spaces, and th us we relegate further discussion of connectedness
to the next chapter (Section D.2).
Exercise 16. Show that the closure of any connected subset of a metric space is
connected.
Exercise 17. Show that if
S is a finite (nonempty) class of connected subsets of a
metric space such that

S = ∅, then

S m ust be connected.
Exercise 18. For any given
n ∈ {2, 3, }, prove that a convex subset of R
n
is
necessarily connected, but not conversely.
104
∗
Exercise 19.
H
Show that a metric space which has countably many points is con-
nected i ff it contains only one point.

2.2 Separable Metric Spaces
Dhilqlwlrq. Let X be a m etric space and Y ⊆ X. If cl
X
(Y )=X, then Y is s aid to
be dense in X (or a dense subset of X). In turn, X is said to be sepa rable if it
contains a countable dense set.
Intuitively speaking, one m ay thin k o f a separable metric space as a space which is
“not very large.” Af ter all, in such a spa ce, there is a coun tab le set which is “almost”
equal to the en tire space.
Than ks to Exercise 14, it is readily observed that a set Y ⊆ X is dense in a metric
space X if, and only if, a ny point in the grand spa ce X can be approached by means
of a sequence that is contained entirely in Y. So, X is a sep ara ble metric space i f, and
only if, it contains a countable set Y such that x ∈ X iff there exists a (y
m
) ∈ Y
∞
with y
m
→ x. This characterization of separability often p roves useful in proofs. For
instance, it allows us to use Lemma A.1 to conclude that Q is dense in R. (By the
way , bd
R
(Q)=R, righ t?)Thus,R is a separable me tric space . This fact allows us
to find other examples of separable spaces.
E{dpsoh 6. [1 ] R
n
is sepa rable, n =1, 2, T his is because any x ∈ R
n
can be
approached by a sequence in R

n
all c omponents of whic h are rational; that is, Q
n
is
dense in R
n
. (Q
n
is countable, right?) In fact, R
n,p
is separable for any 1 ≤ p ≤∞.
(Why?)
[2] A d iscrete space is separable iff it is countable.
∗
[3] Anymetricsubspaceofaseparablemetricspaceisseparable.
14
To prove this,
let X be any metric space, and Y a coun table dense subset of X. Take any metric
subspace Z of X, whic h we wish to pro v e to be separable. Define
Y
m
:= {y ∈ Y : N
1
m
,X
(y) ∩ Z = ∅}.
(Note. Each Y
m
is nonempty, th a nks to the densenes s of Y ). N ow p ick an arbitrary
z

m
(y) ∈ N
1
m
,X
(y) ∩ Z for each m ∈ N and y ∈ Y
m
, and d efine
W := {z
m
(y):y ∈ Y
m
and m ∈ N}.
Clearly, W is a coun tab le subset of Z. (Yes?) Now take a ny z ∈ Z. By densen e ss of
Y, fo r each m ∈ N we can find a y
m
∈ Y with d(z,y
m
) <
1
m
.Soz ∈ N
1
m
,X
(y
m
) ∩ Z,
and hence y
m

∈ Y
m
. Therefore,
d(z,z
m
(y
m
)) ≤ d(z,y
m
)+d(y
m
,z
m
(y
m
)) <
1
m
+
1
m
=
2
m
.
14
The proof is a bit subtle – it ma y be a good idea to skip it in t he first reading.
105
It follow s that any element of Z is in fact a limit of some sequ en c e in W, that is, W
is dense in Z.

[4] C[a, b] is separable for any −∞ <a≤ b<∞. (R ecall that C[a, b] is endo wed
with the sup -m etr ic (Ex ample 2.[2]).) An elementary proof of this claim could be
given by u sing the p iece w ise linear fun ction s with kin k s occurring at rational poin ts
with rationa l values. Since the constru ctio n is a bit te d ious we omit th e det ails he re.
The separab ility of C[a, b] can also be seen as a c orollary of the follow ing famous
result of Karl Weierstrass (whic h he proved in 188 5).
The Weierstrass Approx imation T heore m. The set of all polynomials defined
on [a, b] is dense in C[a, b]. That is,
cl
C[a,b]
(P[a, b]) = C[a, b].
We shall later ob tain this t he ore m as a special case of a sub st antially more general
result (in Section D.6.4). What is importan t for us at present is th e obs ervation that
the separa bility of C[a, b] follo ws read ily from this theorem an d the fact that the s e t
of all polynomials on [a, b] with ration al coefficients is dense in P[a, b] .
15
You should
not have much difficult y in supplying the d etails of the argument.
[5] 
p
is separa ble for any 1 ≤ p<∞. It may be tempting to suggest 
p
∩Q
∞
as a
candidate for a coun table dense subset of 
p
. But this wouldn’t w ork, because, while
this set is indeed dense in 
p

, it is not countable (s ince Q
∞
is uncoun table). Instead
w e consider the set of those sequences in 
p
∩ Q
∞
all but finitely man y components
of w h ich are zero.
16
It is easy to check that this set, call it Y, is countable. (Proof?)
To verify that Y is dense, tak e any sequence (x
m
) ∈ 
p
and fixanyε > 0. Since

∞
|x
i
|
p
< ∞, theremustexistanM ∈ N such that

∞
i=M+1
|x
i
|
p

<
ε
p
2
.
17
Moreover,
since Q is dense in R, we can find a rational r
i
suc h that |r
i
− x
i
|
p
<
ε
p
2M
for eac h
i =1, , M. But then
d
p
((x
m
) , (r
1
, , r
M
, 0, 0, )) =


M

i=1
|r
i
−x
i
|
p
+
∞

i=M+1
|x
i
|
p

1
p
<

ε
p
2
+
ε
p
2


1
p
= ε.
15
Observe that the cardinality of this set is equal to that of Q ∪Q
2
∪···(and not Q
∞
). Since a
countable union of countable sets i s count able (Proposition B.2), the set of all polynomials on [a, b]
with rational coefficien ts is thus countable.
16
Idea of proof. Since (the pth po wer of) a sequence in 
p
is summable, the tail of any such
sequence must be “very close” to the zero sequence. (Warning. Thiswouldnotbethecasein
∞
.)
But since Q is dense in R, we can approximate the earlier (finitely many) terms of such a sequence
by using the rational num bers. Thus, it seems, all we need are those sequences of rational numbers
whose all but finitely many terms are zero.
17
Why do we have lim
M→∞

∞
i=M+1
|x
i

|
p
=0?If you solved Exercise A.45, you know the answer.
Otherwise, note that since (

M
|x
i
|
p
) converges to some n umber as M →∞,saya, we can find,
for any δ > 0, a large enough M ∈ N such that

M
|x
i
|
p
≥ a −δ, that is,

∞
i=M+1
|x
i
|
p
≤ δ.
106
Thus a ny elem ent of 
p

can be approac hed by a sequence (of sequences) in Y. 
One major rea son for wh y separable metric spaces are useful is t hat in such spaces
all open sets can be described in terms of a coun table set of open s ets. For instanc e, all
open subsets of R
n
can be expressed by using the members of the countable collection
{N
ε,R
n
(x):x ∈ Q
n
and ε ∈ Q
++
}. Indeed, by using the denseness of Q in R,itcan
easily be verified that
U =


N
ε,R
n
(x) ∈ 2
U
: x ∈ Q
n
and ε ∈ Q
++

for an y open subset U of R
n

. T h us, there is sense in whic h all open subsets of R
n
are genera ted by a given countable collection of open sets in R
n
. The following result
generalizes th is observation to the case of an arbitra ry separable metric space.
Proposition 3. Let X be metric space. If X is sepa rab le, then there e xists a
countable class O of open subs ets of X suc h that
U =

{O ∈ O : O ⊆ U}
for an y open subset U of X.
Proof. Assume that X is separable, pic k an y coun table dense subset Y of X, and
define
O := {N
ε,X
(y):y ∈ Y and ε ∈ Q
++
}.
O is coun table since it has the same cardinality with Y × Q
++
. Now pick an y open
subset U of X, and take any x ∈ U. We claim that x ∈ O for some O ∈ O with
O ⊆ U. Since U is open in X,thereexistsanε ∈ Q
++
such that N
ε,X
(x) ⊆ U.
But since cl
X

(Y )=X, there must exist a y ∈ Y with d(x, y) <
ε
2
. But then x ∈
N
ε
2
,X
(y) ⊆ N
ε,X
(x) ⊆ U. (Yes?) T his proves that U ⊆

{O ∈ O : O ⊆ U}. The
converse containment is ob vious. 
Keep in min d that many inte restin g properties of a m e tric space are defined
through the notion of an open set. For instance, if we kno w all the open subsets
of a metric space, then we know all the closed and/or connected sets in this space.
(Why?) Sim ilarly, if we know all the open subsets of a metric spa ce , then we know
whic h sequences in th is space a re con vergen t and whic h ones are n ot. (W hy?) Conse-
quen tly, a lot can be learned about the gen eral structu re of a metric space by studying
the class of all open subsets of this space. But the significance of this observation
w ould be limited, if, in a sense, w e had “too man y” open sets lying around. In the
case of a separable metric space this is not the case, for such a space has only a
countable n umber of open sets “that matter” in the sense that all other open subsets
of this space can be described using only these (countably many) open sets. This is
107
the g ist of Proposition 3, th e importance of which will become clearer when you see
it action in the following subsection .
18
No w you may a sk: Given this m otivation, wh y don’t we forget a bout separabilit y,

and rather deal with spaces that satisfy the conclusion of Proposition 3 directly?
Good question! The answ er is giv en in the next exercise.
Exercise 20. Prove the converse of Proposition 3.
Exercise 21. (a)Let
X be a metric s pace such that there exists an ε > 0 and an
uncountable set
S ⊆ X such that d(x, y) > ε for all distinct x, y ∈ S. Show that X
cannot be separable.
(b) Show that

∞
is not a separable metric space.
Exercise 22.
H
Show that R is cardinally larger than any separable metric space. Also
show that
C[0, 1] is cardinally equivalent to R.
19
2.3 Applica tions to Utility Theory
To giv e a quick application, we now go bac k to the decision theoretic setting described
in Section B .4, and se e ho w one may be a ble to improv e the utility repre se ntation
resultswehaveobtainedthereinthecasewherethealternativespaceX has a metric
structur e. We need the following terminology which b u ild s on the definition in tro-
duced at the end of Section B.4.1.
Dhilqlwlrq. Let X be a metric space and  a preference relation on X.Wesay
that  is upper semicontin u ous if L

(x) is an open subset of X for each x ∈ X,
and that  is lowe r semicontinuou s if U


(x) is an open subset of X for eac h x. In
turn,  is called con tinuous if it is both upper and lower semicontinuous.
Intuitively speakin g, if  is an upper semicon tin uous preference relation on a
metric space X, and if x  y, then an alternativ e z wh ic h is “v e ry close” to y should
also be deemed stri ctly worse than x. Put differently, if the sequence (y
m
) ∈ X
∞
converges to y, then there exists an M ∈ N such that x  y
m
for each m ≥ M.
(Notice how the “metric” in question, which is a p ur ely ma th ematical term, and the
preferen ce relation, which is an psychological concept, are link e d tight by the notion
of s e m ic ontinuity.) Lower semicontin uity is inter p r e ted sim ilarly.
If  is a complete preference relation, then  is upper semicontin uous iff U

(x) is
closed for all x ∈ X, and it is lower semicontinuous iff L

(x) is closed for all x ∈ X.
(Why?) So, given any x, y ∈ X, if (y
m
) is a seque nce i n X with y
m
→ y and y
m
 x
18
By the way, this proposition also sho ws t hat if d and D are equivalen t metrics on a nonempty
set X, then (X, d) is a separable iff (X, D) is separable. (The same also goes for the connectedness

property, of course.)
19
This exercise presumes familiarity with th e cardinality theory we reviewed i n Section B.3.1.
108
for each m =1, 2, , then we have y  x, provided that  is a complete and upper
semicontinuous preference relation on X.
20
As we shall elaborate fur t h er in the nex t chapter, fin ding a utility repre sentation
for a contin uous and complete preference relation is a relativ ely easy matter, provided
that the metric space X (of alternatives) under consideration is w ell-beha ved. To
illustrate this poin t, le t’s consider the case where X is both connected and separable.
Let Y be a countable dense subset of X, and pick an y x, y ∈ X with x  y.(Ifthere
is no such x and y in X, then an y constant function would represent .) Since X is
connected, and L

(x) and U

(y) are open in X,wemusthaveL

(x) ∩ U

(y) = ∅.
But then L

(x)∩U

(y) is a nonempt y open subset of X, and hence, since a dense set
intersects ev ery nonempty open set (why?), we must hav e Y ∩L

(x)∩U


(y) = ∅. This
proves that Y is -de nse in X (Section B.2). Applyin g Proposition B.9, therefore, w e
find: E v ery continuou s complete preference relation on a connected separable metric
space can be represented by a u tility fun c tion.
This is very nice already, but we c an do much better. By means o f a different
argument, we can show that we don’t in fact need the connected h y pothesis in this
stateme nt, an d continuity can be relaxed to upper (or lower) semic ontinuity in it. This
argument, due to Trout Rader, illustrates a powerful techn iqu e whic h is frequen tly
used in utility representation exercises . (We ha ve in fact already used this techniqu e
when proving Lemm a B.1.)
Ra der’s Utilit y Represe ntatio n Theorem 1. Let X be a separable metr ic space,
and  a complete prefere nce relation on X. If  is upper semicon tinuous, then it
can be represented b y a utilit y function u ∈ [0, 1]
X
.
Proof. Since X is a s eparab le metr ic space, there mu st exist a countable collection
O of open subsets of X such that U =

{O ∈ O : O ⊆ U} for an y open set U in
X (Proposition 3). Since O is coun table, we may enumerate it as O = {O
1
,O
2
, }.
No w let
M(x):={i ∈ N : O
i
⊆ L


(x)}
and define u ∈ [0, 1]
X
as
u(x):=

i∈M( x)
1
2
i
.
Notice that
x  y iff L

(x) ⊇ L

(y) only if M(x) ⊇ M(y) iff u(x) ≥ u(y)
20
But if  is not c omplete, then this conclusion need not be true; upper semicontinuity does not,
in general, imply that U

(x) is a closed subset of X for each x ∈ X. Besides, as you are asked to
demonstrate in Exercise 25, it is often s imply impossible to demand the closedness of all weak upper
and lower -contour sets fro m an incomplete and continuous preference relation .
109
for any x, y ∈ X.
21
Therefore, w e will be done if we can show that M(x) ⊇ M(y)
implies L


(x) ⊇ L

(y) for any x, y ∈ X. To this end, take any x and y in X such
that L

(x) ⊇ L

(y) is false, that is, there is a z ∈ L

(y)\L

(x). Then since L

(y)
is an open subset of X by hypothesis, there is an O ∈ O that contain s z and that
is contained in L

(y). (Why ?) Since z/∈ L

(x),Ois not contained in L

(x),which
proves th at M(x) ⊇ M(y) is false. Therefore M(x) ⊇ M(y) implies L

(x) ⊇ L

(y),
and we are done.
22


Exercise 23. Is ≥ an upper semicontinuous preference relation on R
2
+
? How about

lex
of Example B.1?
Exercise 24. Let
X := R, and show that the utility function constructed in the proof
of Rader’s Utility Representation Theorem 1 need not be contin uous.
∗
Exercise 25.
H
(Sc hmeidler) Let X be a connected metric space and  a preference
relation on
X with  = ∅. Prove that if  is continuous, and U

(x) and L

(x) are
closed in
X for each x ∈ X, then  must be complete.
We will turn to issues related to the problem of represen ting a complete preference
relation by means of a continuous utility function in S ec tion D.5.2.
3Compactness
We now c om e to one of the most f un damen ta l co n cepts of real analysis, an d one tha t
plays an important role in optimization theory: compactness. Our immediate task is
to outline a basic analysis of those metric spac es that possess this property. Plenty
of applications will be given later.
3.1 Basic Definitions and the Heine-Borel Theorem

Dhilqlwlrq. Let X be a metric space and S ⊆ X. AclassO of subsets of X is said
to cov er S if S ⊆

O. If all members of O are open in X, then we sa y th at O is an
open co ver of S.
Here co mes the definition of the compactness property. If it seems a bit unnatural
to y ou , that’s oka y. Part of our task later on will be to explain why in fact this is
such a fundamental property.
21
Recall my con vention about summ ing over the empty set:

i∈∅
(whatever)=0. So, if M(x)=∅,
we have u(x)=0here.
22
Note the slick use of separability and upper semicontinuity together in the argumen t. A good
idea to better understand what is going on here is to check that u would not be well-definedifI
used the lexicographic order (on R
2
) in its construction.
110
Dhilqlwlrq. A metric space X is s aid to be compact if every open cover of X has
a finite subset that a lso co vers X. AsubsetS of X is said to be compact in X (or a
compact subset of X) if every open cover of S has a finite subset that also co vers
S.
Warnin g. Compactness of a subset S of X means, by definition, the follo wing: If
O is a class of open sets in X such that S ⊆

O, then there is a finite U ⊆ O
with S ⊆


U. ButwhatifweregardS a s a metric subspace of X? In that case S
is a metric s pace in its o wn right, and h ence its comp actness means the following:
If O is a c lass of open sets in S such that S ⊆

O, then there is a finite U ⊆ O
with S ⊆

U. Th us, “S is a compa ct subset of X” and “S is a com p act me tric
subspace of X” are distinct statements. Fortunately, th is is only academic , for these
two statements are in fa ct equ iva lent. Tha t is, S is compact iff every open cover of
S (with sets open in S)hasafinite subset that covers S.
23
Th us, for any subset S
of X, the phr ase “S is compact” is unam biguous.
(An immediate im p lica tion : Compactness is a property that is invariant und e r
equiva lence of metrics.)
(Another immediate implication: If S is a compact subset of a metric spa ce Y,
and Y is a m etric subspace of X, then S is c ompact i n X.)
As a first pass, let us examine a sp ace whic h is no t compact, n amely, the open
interval (0, 1). Consider the collection O := {(
1
i
, 1) : i =1, 2, } and ob s erve that
(0, 1) = (
1
2
, 1) ∪ (
1
3

, 1) ∪ ···,thatis,O is an open co ver of (0, 1). Does O ha v e a
finite subset that covers (0, 1)?No,becausetheinf of any finite subset of O is
bounded away from 0, so no such subset can possibly cov er (0, 1) en tirely. Therefore,
w e conclude that (0, 1) is not a compact subset of R (or that (0, 1) is not a compact
metric space).
24
For a positiv e example, note that a finite subset of any metric space is necessarily
comp ac t. Much less trivially, [0, 1] is a compact subset of R. Let us prov e this
claim b y means of a bisection argument that parallels the one w e gave in Section
B.1 to estab lis h t he uncountability of R. (This is the me th od of “butt erflyhunting,”
remem ber?) Suppose there exists an open cover O of [0, 1] no finite subset of w hic h
covers [0, 1].
25
Then, either [0,
1
2
] or [
1
2
, 1] is not covered by an y finite subset of
O. (Wh y ?) Pic k any one of these intervals with this pr opert y, call it [a
1
,b
1
]. (The
“butterfly” m ust be in [a
1
,b
1
].)Then,either[a

1
,
1
2
(b
1
− a
1
)] or [
1
2
(b
1
− a
1
),b
1
] is not
covered by any finite subset of O. Pick any one of these intervals with this pr o perty,
call it [a
2
,b
2
]. (The “butterfly” m ust be in [a
2
,b
2
].) Con tin uing this wa y inductively,
w e obtain two sequences (a
m

) and (b
m
) in [0, 1] such that
23
There is som ething to be proved her e. Please recall Exercise 6, and supply a proof.
24
Quiz. How about R?
25
Again, it doesn’t matter whether the e lements of O areopeninthemothermetricspaceR or
in [0, 1].
111