Chap01 Metric and Normed Spaces
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C h a p t e r 1M e t r i c a n d N o r m e d S p a c e s
We are all familiar with the geometrical properties of ordinary, threedimensionalEuclidean space. A persistent th em e in m athe m atics is the gro uping of various kindsof objects into abstract spaces. This grouping enables us to extend our intuit ionof the relationship between points in Euclidean space to the relationship betweenmore general kinds of objects , leading to a clearer and deeper understanding ofthose objects .
Th e simplest sett ing for the stu dy of man y problem s in analysis is th at of a metricspac e. A metric space is a set of po ints with a suitab le notion of th e distan ce betwe enpo ints . We can use the metric , or distance function, to define the fund amen talconcepts of analysis, such as convergence, continuity, and compactness.
A metric space need not have any kind of algebraic structure defined on it. Inman y ap plications, however, the m etric space is a l inear space with a m etric d erivedfrom a norm th at gives th e " length " of a vecto r. Such spaces are called norm edlinear spaces. For example, ndimensional Euclidean space is a normed linear space(after the choice of an arbitrary point as the origin). A central topic of this bookis the study of infinitedimensional normed linear spaces, including function spacesin which a single point rep resen ts a function. As we will see, th e geo me trical intuition derived from finitedimensional Euclidean space remains essential, althoughcompletely new features arise in the case of infinitedimensional spaces.
In this chapter, we define and study metric spaces and normed linear spaces.Along the way, we review a number of definitions and results from real analysis.
1 .1 M e tr i c s a n d n o r m sLet X be an arbi t rary nonempty se t .Def in i t ion 1 .1 A metric, or distance function, on X is a function
d.XxX^R,l

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2 Metric and Normed Spaces
Fig . 1 .1 A family tre e used in th e defini t ion of the ul t ram etr ic in Ex am ple 1 .3 .
with the following properties:(a) d(x, y) > 0 for all x, y X, and d(x, y) = 0 if and only if x = y;(b ) d(x,y) = d(y,x), for all x,y X;(c) d(x,y) < d(x,z) + d(z,y), for all x,y,z 6 X.
A metric space (X, d) is a set X equipped with a metric d.When the me t r ic d is understood from the context, we denote a metric space
simply by the set X. In words, the definit ion states that:(a) distances are nonnegative, and the only point at zero distance from x is x
itself;(b) the distance is a symmetric function;(c) distances satisfy the triangle inequality.
For points in the Euclidean plane, the tr iangle inequality states that the length ofone side of a triangle is less than the sum of the lengths of the other two sides.E x a m p l e 1 . 2 The set of real numbers R with the distance function d{x,y) =\x y\ is a metric space. The set of complex numbers C with the distance functiond(z,w) = \z w\ is also a metric space.E x a m p l e 1 . 3 Let X be a set of people of the same generation with a commonancestor, for example, all the grandchildren of a grandmother (see Figure 1.1). Wedefine the distance d(x,y) between any two individuals x and y as the number ofgenerations one has to go back along female lines to find the first common ancestor.

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Metrics and norms 3
For example, the distance between two sisters is one. It is easy to check that d is ametric. In fact, d satisfies a s tronger condition than the tr iangle inequality, namely
d(x,y) < max{d(x,z),d(z,y)} for all x,y,z X. (1.1)A metr ic d which satisfies (1.1) is called an ultrametric. Ultrametrics have beenused in taxonomy to characterize the genetic proximity of species.E x a m p l e 1 . 4 Let X be the set of nletter words in a A;character alphabet A ={ 0 1 , 0 2 , . . . ,afc}, meaning tha t X {(x\,X2, ,xn)  Xi A}. We define thedis tance d(x, y) between two words x = (x\,..., xn) and y = ( j / i , . . . , yn) to be thenumber of places in which the words have different letters. That is,
d(x,y) = # { i \x i ^ 1 / i } .T he n (X, d) is a metric space.E x a m p l e 1.5 Suppose (X, d) is any m etric space and Y is a subset of X. We definethe distance between points of Y by restr icting the metric d to Y. The resul t ingmetric space (Y,d\y), or (Y,d) for short, is called a metric subspace of (X, d), orsimply a subspace when it is clear th at we are talking a bo ut m etric spaces. Forexam ple, (E,  ) is a metr ic su bspa ce of (C,   ) , and the space of rational numbers(Q ,  ) is a m etr ic su bsp ace of (K,   ) .Example 1 .6 I f X and Y are sets , then the Cartesian product I x F i s th e set ofordered pairs (x, y) with x X and y Y. If dx and dy are metrics on X and Y ,respectively, then we may define a metric dxxY on the product space by
dxxY ((xi,yi),(x2,y2)) = dx{x 1,x 2) +dY (2/1,2/2)for all x\, #2 E l an d 2/1,2/2 Y.
We recall the definition of a linear, or vector, space. We consider only real orcomplex linear spaces.Def in i t ion 1 .7 A linear space X over the scalar field K (or C) is a set of points, orvectors, on which are defined o peration s of vector addition and scalar m ultiplicationwith the following properties:
(a) the set X is a commutative group with respect to the operation + of vectoraddition, meaning that for all x,y,z X, we have x + y = y + x andx + (y + z) = (x + y) + z, there is a zero vector 0 such that x + 0 = xfor all a; e X, and for each x X there is a unique vector x such tha tx+{x) = 0;
(b) for all x, y X and A,\i E (or C), we have lx = x, (A + fj,)x = Xx + fix,X(fix) = (\(j,)x, and A(x + y) = Aa; + Xy.

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4 Metric and Normed Spaces
We assume th at t h e reader is familiar with th e elemen tary theor y of l inear spaces.Some references are given in Section 1.9.
A norm on a linear space is a function that gives a notion of the "length" of avector.De f i n i t i on 1 . 8 A norm on a linear space X is a fun ction   : X > E w ith th efollowing properties:
(a ) 11 a; 11 > 0, for all x X (nonnegative);(b) Aa; = Aa;, for all x X and A 1 (or C) (ho mog eneous);(c) a; + 2/ <  z   + HJ/II, for all x,y G X ( tr iangle inequality) ;(d ) 11a;11 = 0 implies that x = 0 (strictly positive).
A normed linear space (X, \\  ) is a linear space X equipped with a norm    .A normed linear space is a metric space with the metric
d{x,y)\\xy\\. (1.2)All the concepts we define for metric spaces therefore apply, in particular, to normedlinear spaces. The metric associated with a norm in this way has the special properties of trans lation invariance, mean ing th at for all z 6 X, d(x + z,y + z) = d(x, y),and hom ogeneity, m eanin g th at for all A E (or C), d(Xx,Xy) = \X\d(x,y).
T he closed unit ball B of a normed linear space X is the setB = {xeX :  z  < 1 } .
A subse t C of a linear space is convex iftx + (l t)y e C (1.3)
for all x,y e C and all real numbers 0 < t < 1, meaning that the l ine segmentjoining any two poin ts in the set l ies in the set . Th e tr iangle inequality implies th atthe unit ball is convex, and its shape gives a good picture of the norm's geometry.E x a m p l e 1 . 9 The set of real numbers R with the absolute value norm  x   = xis a oned imensio nal real no rme d linear space. M ore generally, W 1, where n =1 ,2 ,3 , . . . , is an ndimension al l inear space. We define the Euclidean norm of apoin t x = (xi,x2,. , xn) W 1 by
11*11 = ^xj+xl + ^'+xl,and call W 1 equipped with the Euclidean norm ndimensional Euclidean space. W ecan also define other norms on M. For example, the su m or 1norm is given by
IMIi = x i + a;2 +  + a ;n .

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Fig. 1.2 Th e un it balls in Inorm (B m a x ) .1 for the Euclidean norm (JB2), the sum norm (Bi), and the maximum
T he maximum norm is given byIMImax = max{ xi  , F2I , . . . , \x\}.
We also call the maximum norm the oonorm, and denote it by a;oo The unitballs in R2 for each of these norms are shown in Figure 1.2. We will equip W 1 withthe Euclidean norm, unless stated otherwise.E x a m p l e 1 . 1 0 A linear subspace of a linear space, or simply a subspace when itis clear we are talking about linear spaces, is a subset that is itself a linear space.A subse t M of a linear space X is a subs pa ce if an d on ly if Arc + ny e M for allA, \x R (or C) and all x, y M . A subspace of a normed linear space is a normedlinear space with norm given by the restr iction of th e norm on I to M .
We will see later on that all norms on a finitedimensional linear space lead toexactly the same notion of convergence, so often it is not important which normwe use. Different norms on an infinitedimensional linear space, such as a functionspace, may lead to completely different notions of convergence, so the specificationof a norm is crucial in this case.
We will always regard a normed linear space as a metric space with the metricdefined in equ ation (1.2), unless we explicitly sta te otherw ise. Nev ertheless, thisequation is not the only way to define a metric on a normed linear space.E xam p l e 1 . 11 I f (X ,   ) is a normed linear space, then
d(x,y) = IF 2/11l + \\xy\ (1.4)

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Fig . 1.3 A sequenc e (x) converging to x.defines a nonhomogeneous, translation invariant metric on X. In this metric , thedistance between two points is always less than one.
1 .2 C o n v e r g e n c eWe first consider the convergence of sequences of real numbers. A sequence of realnum bers is a ma p from th e na tu ra l numbers N = { 1 ,2 ,3 , . . . } to R. Th at is , wi theach n N, we associate a real number xn E. We deno te a sequence by (x n),or ( i n ) ^ . ! when we want to indica te the range of the index n. The index n is a"dummy" index, and we may also write the sequence as (xk) or (zfc)Si
Another common notation for a sequence is {x n}. This notation is a l i t t leambiguous because a sequence is not the same thing as a set . For example,( 0 , 1 , 0 , 1 , 0 , . . . ) a n d ( 1 , 0 , 0 , 0 , 0 , . . . )
are different sequences, but the set of terms is {0,1} in each case.A subsequence of a sequence (x n) is a sequence of the form (x nk), where foreach k N we have n* N, and nk < rik+i for all k. Tha t i s , k H* n^ is a
str ictly increasing function from the set of natural numbers to itself. For example,( l / f c 2 ) ^ = 1 is a subsequence of (l/n)^L1.The most important concept concerning sequences is convergence.De f in i t io n 1 . 1 2 A sequence (a;n) of real numbers converges to x M. if for everye > 0 there is an N 6 N such tha t \xn x\ < t for all n > N. The point x is calledth e limit of (x n).
In this definition, the integer N depends on e, since smaller e 's usually requirelarger N's, and we could write N(e) to make the depen dence explicit . Com monways to write the convergence of (x n) to x are
xn  x as n  oo, lim xn x.n>ooA sequence that does not converge is said to diverge. If a sequence diverges
because i ts terms eventually become larger than any number, i t is often convenient

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Convergence 7
to regard the sequence as converging to oo. That is, we say xn > oo if for everyM e M. there is an N 6 N such that xn > M for all n > N. Similarly, we sayxn > oo if for every M EM. there is an TV N such tha t xn < M for all n> N.E x a m p l e 1 . 1 3 Here are a few examples of the limits of convergent sequences:
lim = 0, l im ns in I ) = 1, lim [ 1 + 1 = e.nvoo n nvoo \ n I noo "TThe sequence (logn) diverges because logn > oo as n > oo. The sequence ((l) n )diverges becau se its term s oscillate between 1 and 1, and it does no t converge t oeither oo or oo.
A sequence is said to be Cauchy if its terms eventually get arbitrarily closetogether .De f i n i t i on 1 . 14 A sequence (x n) is a Cauchy sequence if for every e > 0 there isan N N such tha t \x m xn\ < e for all m,n> N.
Suppose tha t (x n) converges to x. Given e > 0, there is an integer N such tha t\x n x\ < e/2 when n> N. If m, n > N, then use of the triangle inequality impliest h a t
so (x n) is Cauchy. Thus, every convergent sequence is a Cauchy sequence. For thereal numbers, the converse is also true, and every Cauchy sequence is convergent.The convergence of Cauchy sequences is a fundamental defining property of thereal numbers, called completeness. We will discuss completeness for general metricspaces in greater detail below.E x a m p l e 1 . 1 5 The sequence (x n) with xn = logn is not a Cauchy sequence, sincel o g n oo. Nevertheless, we have
\x n+i  xn\ = log (1 +  j > 0as n > oo. This example shows that it is not sufficient for successive terms in asequence to get arbitrari ly close together to ensure that the sequence is Cauchy.
We can use the definition of the convergence of a sequence to define the sum ofan infinite series as the limit of its sequence of partia l sum s. Let (x n) be a sequencein E. The sequence of partial sums (sn) of the series ]T xn is defined bysn = ^2xk (15)
fc=i

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8 Metric and Normed SpacesIf (s n ) converges to a limit s, then we say that the series ^ xn converges to s, andwri te
oo
n = lIf the sequence of partial sums does not converge, or converges to infinity, then wesay that the series diverges. The series ^2 xn is said to be absolutely convergentif the series of absolute values ^ \xn\ converges. Ab solute convergen ce impliesconvergen ce, bu t no t conversely. A useful pro pe rty of an abso lutely co nvergent seriesof real (or complex) numbers is that any series obtained from it by a permutationof its terms converges to the same sum as the original series.
The definitions of convergent and Cauchy sequences generalize to metric spacesin an obvious way. A sequence {x n) in a metric space (X, d) is a m a p n H i w hichassociates a point xn X with each natural number n G N.De f i n i t i on 1 . 16 A sequence (x n) in X converges to x X if for every e > 0 thereis an iV E N such that d(x n,x) < e for all n > N. The sequence is Cauchy if forevery e > 0 there is an N N such tha t d(x m,xn) < e for all m,n> N.
Figure 1.3 shows a convergent sequence in the Eu clidean plan e. Pr op erty (a)of the metric in Definition 1.1 implies that if a sequence converges, then its limitis unique. That is, if xn > x and xn 4 y, then x = y. The fact that convergentsequences are Cauchy is an immediate consequence of the triangle inequality, as before. The property that every Cauchy sequence converges singles out a particularlyuseful class of metric spaces, called complete metric spaces.De f i n i t i on 1 . 17 A metric space (X, d) is complete if every Cauchy sequence in Xconverges to a limit in X. A subset Y of X is complete if the metric subspace(Y, d\y) is comp lete. A norm ed linear space th at is com plete with respect to th emetric (1.2) is called a Banach space.E x a m p l e 1 . 1 8 The space of rational numbers Q is not complete, s ince a sequenceof rational numbers which converges in R to an irrational number (such as v2 or7r) is a Cauchy sequence in Q, but does not have a limit in Q.E x a m p l e 1.19 The finitedimensional linear space K" is a Banach space with respect to the sum , maxim um , and Euclidean norm s defined in Ex am ple 1.9. (SeeExercise 1.6.)
Series do not make sense in a general metric space, because we cannot add pointstogether. We can, however, consider series in a normed linear space X. Just as forreal numbers, if (x n) is a sequence in X, then the series 2~2=i xn converges tos X if the sequence (s n) of partial sums, defined in (1.5), converges to s.

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10 Metric and Normed Spaces
If A does not have an upper bound, we define sup A oo, and if A does nothave a lower bound, we define inf A = oo. The convention that every number isboth an upper and a lower bound of the empty set 0 is sometimes convenient, soth at su p0 = oo and inf 0 = oo.
The supremum of a set A may, or may not, belong to A itself. If it does, thensup A is called the maximum of A, and is also den oted by m ax A Similarly, if theinnmum belongs to A, then inf A is called the minimum of A, and is also denotedby min A. The illustration in Figure 1.4 shows an example.
T hus , provided we allow the values oo, every set of real numbers has a supremum and an infimum, but i t does not necessarily have a maximum or a minimum.
Ne xt, we define th e limin f and lim su p of a real sequen ce. Fi rst, we considermonotone sequences. A sequence (x n) is said to be monotone increasing if xn xn+ i, for every n. A monotonesequence is a sequence th at is mo noton e increasing or m ono tone decreasing. Amonotone increasing sequence converges to its supremum (which could be oo), and amo no ton e decreasing sequence converges to its infimum (which could be oo). T hus ,provided that we allow for convergence to oo, all monotone sequences converge.
Now suppose tha t (x n) is an arbitrary sequence of real numbers. We construct anew sequence (y n) by taking the supremum of successively truncated "tails" of theoriginal sequence, yn = sup {xk \ k > n). The sequence (y n) is monotone decreasingbeca use the suprem um is tak en over smaller sets for larger n 's . Th erefore, th esequence (y n) has a limit, which we call the limsup of the sequence (x n ) , and denoteby lim sup z n . Similarly, taking the infimum of the successively truncated "tails" of(xn), we get a monotone increasing sequence. We call the limit of that sequence, theliminf of (x n), and denote i t by l iminf xn. Thus, we have the following definition.De f i n i t i on 1 . 23 Let (x n) be a sequence of real numbers. Then
l im s upa ;n = lim [sup {xk \ k > n}],n Kx> n>ool iminf xn lim [inf {xk \ k > n}].n^oo nfooAn other comm on n otatio n for the l im sup an d l im inf is
l im s upa ; n = l ima;n , liminf xn = lim x.We make the natural convention that if
sup {xk  k > n} oo, or inf {xk \ k >n} oo,for every n , the n lim sup xn = oo, or lim inf xn oo, respectively. In con trastto the l imit , the l iminf and l imsup of a sequence of real numbers always exist ,provided th at we allow the values oo . Th e l im sup of a sequence whose terms arebounded from above is finite or oo, and the lim inf of a sequence whose terms arebounded from below is finite or oo.

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Continuity 11It follows from the definition that
l iminf xn < l im s upa ; n .n>oo n+ooMoreover, a sequence (xn) converges if and only if
lim inf xn = lim su p xn,n  x x ) n  n x iand, in that case, the limit is the common value of l iminf xn and lim sup xn.E x a m p l e 1.24 If xn = (  1 ) " , t h e n
liminf xn = 1 , l im s upa ;n = 1.n>oo n>ooT he lim inf and lim sup have different values and the sequence does not have a limit.E x a m p l e 1.25 If {x i Q G E  n G N, a G A} is a set of real numbers indexed bythe na tura l numbers N and an arbi t rary set A, then
sup lim inf xn a \ < lim inf SUpXnaeASee Exercise 1.10 for the proof, and the analogous inequality with inf and lim sup .
Suppose tha t A is a nonem pty subset of a general metric space X. The diameterof A is
d i a m A = sxvp{d(x,y) \ x,y G A}.T h e set A is bounded if its diameter is finite. It follows that A is bounde d if andonly if there is an M 6 1 and an x0 X such tha t d(xo, x) < M for all x A. Thedistance d(x, A) of a point x G X from the set A is defined by
d{x,A) =M{d(x,y) \ y e A}.The s ta tement d(x, A) = 0 does not imply tha t x . A.
We say t ha t a function / : X ) Y is bounded if its range f(X) is bounded. Forexample , a realvalued function / : X > R is bounded if the re is a f inite number Msuch tha t  / (a ; )  < M for all x G X. We say t h a t / : X > E is bounded from aboveif there is an M 6 E such tha t f{x) < M for all x X, and bounded from below ifthe re is an M E such tha t f(x) > M for all x G X.
1.4 C o n t i n u i t yA real function / : E E is continuous at a point xo G E if for every e > 0 the reis a S > 0 such tha t \x xQ\ < S implies \f(x) f(x0)\ < e. Thus , cont inui ty of /at XQ is the proper ty tha t the value of / at a point close to XQ is close to the value

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of / at Xo The definition of continuity for functions between metric spaces is anobvious generalization of the definition for real functions. Let (X, dx) and (Y, dy)be two metric spaces.De f in i t io n 1 . 2 6 A function / : X > Y is continuous a t x0 6 X if for every e > 0there is a S > 0 such tha t dx(x,x0) < S implies dy (f(x),f(x0)) < e. The function/ is continuous on X if it is continuous at every point in X.
If / is not continuous at x, then we say th at / is discontinuous a t x. There arecontinuous functions on any metric space. For example, every constant function iscont inuous .E x a m p l e 1 . 2 7 Let a X, and define / : X  R by f(x) = d(x,a). Then / iscontinuous on X.
We can also define continuity in term s of lim its. If / : X > Y, we say tha tf(x) 2/0 a s x > xo, or
lim f(x) = 2/0,X>X Q
if for every e > 0 there is a S > 0 such that 0 < dx (x ,xo) < S implies thatdy (f(x),yo) < t. M ore gene rally, if / : D C X > Y has domain D, and XQ is alimit of points in D, then we say f(x) > yo as x t xo in D if for every e > 0 thereis a 6 > 0 such that 0 < dx {X,XQ) < S and x D implies that dy (f(x),yo) < e. Afunction / : X > Y is continuous at XQ X if
lim f(x) = f(x 0),X^XQmeaning that the l imit of f(x) as x > xo exists and is equ al to th e value of / at xo.Ex a m p le 1 . 2 8 I f / : (0, a) > Y for some a > 0, and f(x)  L as x > 0, then wewrite
lim f(x) = L.x*0+Sim ilarly, if / : (  a , 0) > Y, and /(a;) > L as x > 0, then we write
lim /(ar) = L.x>oIf / : X y and i? is a sub set of X , the n we say th a t / is continuous on E if
it is continuous at every point x 6 E. This property is , in general , not equivalentto the continuity of the restr iction f\s of / on E.E x a m p l e 1 . 2 9 Let / : K >ffibe the characterist ic function of the ration als , whichis defined by
f 1 i f x e Q ,/ W ~ \ 0 i f x ^ Q .

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Continuity 13
Th e function / is discontinuous at every point of R, bu t /  Q : Q > R is the cons tantfunction f\q{x) = 1, so /  Q is continuous on Q.
A subtle , but important, s trengthening of continuity is uniform continuity.De f i n i t i on 1 . 30 A function / : X > Y is uniformly continuous on X if for everye > 0 there is a 8 > 0 such tha t dx(x,y) < 8 implies dy(f'(x), f'(y)) < e for allx,y G X.
The crucial difference between Definition 1.30 and Definition 1.26 is that thevalue of 8 does not d epend on the point a; I , so th at f(y) gets closer to f(x) ata uniform rate as y gets closer to x.
In the following, we will denote all metrics by d when it is clear from the contextwhich metric is meant.E x a m p l e 1 . 3 1 The function r : (0,1) > R defined by r(x) = 1/x is continuous on(0,1) but not uniformly continuous. The function s : R > R defined by s(x) = x2is continuous on R but not uniformly continuous. If [a , b] is any bounded interval,then s[a,6] is uniformly continuous on [a , b].E x a m p l e 1 . 3 2 A function / : R" > R m is affine if
/ (tx + (1  t)y) = tf(x) + (1  t)f(y) for all x, y G R" and t G [0,1].Ev ery affine function is uniform ly con tinu ou s. A n affine function / can be wri tte nin the form f(x) = Ax + b, where A is a constant mxn matr ix and & is a constantmv ecto r. Affine functions are more general tha n linear function s, for which 6 = 0.
There is a useful equivalent way to characterize continuous functions on metricspaces in terms of sequences.De f i n i t i on 1 . 33 A function / : X > Y is sequentially continuous at x G X if forevery sequence (x n) that converges to x in X, the sequence (f(xn)) converges tof(x) in Y.P r o p o s i t i o n 1 . 3 4 Let X, Y be m etric spaces. A function / : X > Y is continuousa t x if and only if it is sequentially continuous at x.Proof. Fir st, we show th at if / is con tinuo us, the n it is sequen tially con tinuo us.Suppose tha t / is continuous at x, and xn > x. Let e > 0 be given. By th econtinuity of /, we can choose 0 so tha t d(x, xn) < 6 implies d(f(x), f{xn)) < e.By the convergence of (x n), we can choose N so tha t n > N implies d(x, xn) < 8.Therefore, n> N implies d(f(x),f(x n)) < e, and f(xn) > f(x).
To prove the con verse, we show tha t if / is disco ntinu ous, th en it is no t seq uentially con tinuo us. If / is discontinu ous at x, then there is an e > 0 such that forevery n G N th ere exists xn G X with d(x,xn) < \/n and d(f(x), f{xn)) > e. T h esequence (x n) converges to x bu t ( / (# ) ) does not converge to f{x).

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f(x)
xF ig . 1.5 Th e func t ion / i s upp er semico n t inuous , bu t no t con t inu ous , a t the po in t x.
There are two kinds of "halfcontinuous" realvalued functions, denned as follows.Def in i t ion 1 .35 A function / : X > M. is upper semicontinuous on X if for allx X and every sequence xn > x, we have
l i m s u p / ( x n ) < f(x).ntooA function / is lower semicontinuous on X if for all x G X and every sequencexn > x, we have
liminf f(xn) > f(x).n o oT he definition is illustr ated in Figu re 1.5. A function / : X > R is continuous
if and only if it is upper and lower semicontinuous.
1 .5 O p e n a n d c lo s e d s e t sOpen sets provide another way to formulate the concepts of convergence and continuity. In this section, we define open sets in a metric space. We will discuss opensets in the more general context of topological spaces in Chapter 4.
Let (X, d) be a metric space. The open ball, Br(a), with radius r > 0 and centera X is the setBr(a) = {x X  d(x, a) < r}.
T h e closed ball, B r{a), is the setB r(a)  {x X  d(x, a) < r}.

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Open and closed sets 15Definition 1.36 A subset G of a metric space X is open if for every x G thereis an r > 0 such that B r{x ) is contained in G. A subset F of X is closed if itscomplement Fc = X \ F is open.
For example, an open ball is an open set, and a closed ball is a closed set. Thefollowing properties of open and closed sets are easy to prove from the definition.P r o p o s i t i o n 1 . 3 7 Let X be a metric space.
(a) Th e empty set 0 and th e whole set X are open and closed.(b) A finite intersectio n of open sets is open .(c) An arbitrary union of open sets is open.(d) A finite unio n of closed sets is closed.(e) An arbitrary intersection of closed sets is closed.
E x a m p l e 1 . 3 8 The interval J n = (1/n, 1) is open in E for every n N, but theintersectionOOfV = [o,i)
7 1 = 1
is not open. Thus, an infinite intersection of open sets need not be open.E x a m p l e 1 . 3 9 Let {q n \ n S N} be an enumerat ion of the ra t ional numbers Q,and e > 0. We define the open interval In in E by
T he n G =  J "J=1 In is an open set which contains Q. T he su m of th e length s ofthe intervals In is 2e, which can be made as small as we wish. Nevertheless, everyinterval in E contains infinitely many rational numbers, and therefore infinitely
many intervals 7.A subset of E has Lebesgue measure zero if for every e > 0 there is a countable
collection of open intervals whose union contains the subset such that the sum ofthe lengths of the intervals is less than e. Thus, the previous example shows thatthe set of rational numbers Q, or any other countable subset of E, has measure zero.A property which holds everywhere except on a set of measure zero is said to holdalmost everywhere, abbrevia ted a.e. For example, the function XQ : E > K thatis one on the rational numbers and zero on the irrational numbers is zero almosteverywhere.
Every open set in E is a countable union of disjoint open intervals . T he stru ctu reof open sets in E n for n > 2 may be much more complicated.

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E x a m p l e 1 . 4 0 We define a closed set F\ in R by removing the "middle third"(1/3,2/3) of the interva l [0,1]. That is ,
F1 = [0,1/3] U [2 /3 ,1 ] .We define F2 by removing the middle thirds of the intervals in Fi, so tha t
F2 = [0,1/9] U [2/9,1/3 ] U [2/3 ,7/9] U [ 8 / 9 ,1 ] .Continuing this removal of middle thirds, we obtain a nested sequence of closed sets(F n). The intersection F = f^Li Pn is a closed set called the Cantor set. A numberx [0,1] belongs to the Cantor set if and only if it has a base three expansion thatcontains no l ' s . Th e endpoin ts of the closed intervals in the Fn's do not have auniqu e expansion. For exam ple, we can write 1/3 6 F in base three as 0.1000. . .and as 0.0222 Th e Can tor set is an unco untab le set of Lebesgue mea sure zerowhich contains no open intervals, and is a simple example of a fractal. Heuristically,any part of the set for example, the left part contained in the interval [0,1/3] is a scaled version of th e whole set. T he n am e fractal refers to th e fact th at ,with a suitable definition of the Hausdorff dimension of a set, the Cantor set has afractional dimen sion of log 2 / log 3 PS 0.631. The Hausdorff dimension of the Cantorset lies between that of a point, which has dimension 0, and an interval, which hasdimension 1.
Closed sets in a metric space can be given an alternative, sequential characterization as sets that contain their l imit points .P r o p o s i t i o n 1 . 4 1 A subset F of a metric space is closed if and only if everyconvergent sequence of elements in F converges to a limit in F. That is, if xn  xand xn E F for all n, then x F.E x a m p l e 1 . 4 2 A subset of a complete metric space is complete if and only if it isclosed.
T h e closure A of a set A C X is the smallest closed set containing A. Fromproperty (e) of Proposition 1.37, the closure A is the intersection of all closed setstha t conta in A. In a metric space, the closure of a set A can also be obtained byadding to A all limits of convergent sequences of elements of A. That is ,
A = {x X  there exist an A such tha t an > x}. (1.6)The closure of the set of rational numbers Q in the space of real numbers R is
the whole space R. Sets with this property are said to be dense.De f i n i t i on 1 . 43 A subset A of a metric space X is dense in X if A = X.

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The completion of a metric space 17It follows from (1.6) that A is a dense subset of the m etric space X if and only if
for every x G X there is a sequence (a) in A such tha t an  x. Thus , every pointin X can be approximated arbitrari ly closely by points in the dense set A. We willencou nter man y dense sets later on. The orem 2.9, the W eierstrass app roxim ationtheorem, gives one example.De f i n i t i on 1 . 44 A metric space is separable if it has a countable dense subset.
For example, K with i ts usual me tric is separable because Q is a countable densesubset. On the other hand, R with the discrete metric d(x, y) = 1 when x y isnot separable .De f i n i t i on 1 . 45 Let a; be a point in a metric space X. A set U C X is a neighborhood of x if there is an open set G C U with x G G.
Equivalently, a set U is a neighborhood of x if U contains a ball BT (x ) centereda t x for some r > 0. Definition 1.16 for the convergence of a sequence can thereforebe reph rased in th e following way. A sequence (x n) converges to x if for everyneighborhood U of x there is an N G N such t ha t xn G U for all n> N.The following proposit ion characterizes continuous functions as functions that"pull back" open sets to open sets .P r o p o s i t i o n 1 . 4 6 Let X, Y be metric spaces and / : X > Y. T he function / iscontinuous on X if and only if f~ 1(G) is open in X for every open set G in Y.Proof. Suppose th at / is continuous and G C Y is op en . If a G / _ 1 ( G ) , t h e nthere is a b G G with b = f(a). Since G is open, there is an e > 0 with B ((b) C G.Since / is con tinuo us, there is a 6 > 0 such tha t d(x,a) < 6 implies d(f(x),b) < e.It follows that B s(a ) C / _ 1 ( G ) > s o / _ 1 ( G ) i s P e n Conversely, suppose that / is discontinuous at some point a in X. Then the reis an e > 0 such that for every S > 0, there is an x G X with d(x,a) < 5 andd(f(x)>f(a)) > e It follows that, although a belongs to the inverse image of theopen set B t(f(a)) under / , the inverse image does not contain Bg(a) for any 6 > 0,so it is no t ope n. E x a m p l e 1 . 4 7 If s : E> E is th e function s(x) = x2, then s _ 1 ( (  4 , 4 ) ) = (2,2)is open , as required by continuity. On th e other hand , s(( 2,2)) = [0,4) is notopen. Thus, continuous functions need not map open sets to open sets .
1 .6 T h e c o m p l e t i o n o f a m e t r i c s p ac eWorking with incomplete metric spaces is very inconvenient. For example, supposewe wish to solve an equation for which we cannot write an explicit expression forthe solution. We may instead construct a sequence (x n) of approximate solutions,

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for example, by use of an i terative method or some kind of numerical scheme. Ifthe approximate solutions get closer and closer together with increasing n, meaningthat they form a Cauchy sequence in a metric space, then we would like to concludethat the approximate solutions have a l imit , and then try to show that the l imit isa solution. We cannot do this unless the metric space in which the approximationslie is complete.
In this section we explain how to extend an incomplete metric space X to a larger, complete metric space, called the completion of X. We construct the completionof X as a set of equivalence classes of Cauchy sequences in X which "ought" to converge to the same point. For a brief review of equivalence relations and equivalenceclasses, see Exercise 1.22. A point x G X is naturally identified with the class ofCauchy sequences in X that converge to x, while classes of Cauchy sequences thatdo not converge in X correspond to new points in the completion. In effect, weconstruct the completion by filling the "holes" in X that are detected by i ts Cauchysequences.E x a m p l e 1 . 4 8 The completion of th e set of ra t ional numbers Q is the set of realnumbers K. A real number x is identified with the equivalence class of ra t ionalCauchy sequences that converge to x. When we wri te a rea l number in decimalnotation, we give a Cauchy sequence of rational numbers that converges to it.
In order to give a formal definition of the completion, we require the notion ofan isometry between two metric spaces (X , dx) and (Y, dy)De f i n i t i on 1 . 49 Am ap i: X > Y which satisfies
dY(i(xi),i(x2)) = dx(x!,x2) (1.7)for all xi,X2 E. X is called an isometry or an isometric embedding of X into Y.An isometry which is onto is called a metric space isomorphism, or an isomorphismwhen it is clear from the context that we are dealing with metric spaces. Two metricspaces X and Y are isomorphic if there is an isomorphism i: X > Y.
Equation (1.7) implies that an isometry i is onetoone and cont inuous . Wethink of i as "identifying" a point x G X with i ts image i{x) G Y, so tha t i(X) is a"copy" of X embedded in Y. Two isomorphic metric spaces are indistinguishableas metric spaces, although they may differ in other ways.E x a m p l e 1 . 5 0 The m ap i : C ) M2 denned by t(x + iy) = (x,y) is a metr icspace isomorphism between the comp lex numb ers (C,  ) and the Euclidean plane(R 2 ,    ). In fact, since i is linear, the spaces C and R2 are isomorphic as realnormed linear spaces.
We can now define the completion of a metric space. The example of the realand ra t ional numbers is helpful to keep in mind while reading this definit ion.

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The completion of a metric space 19
De f i n i t i on 1 . 51 A metric space {X,d) is called the completion of (X,d) if thefollowing conditions are satisfied:
(a) there is an isometr ic embedding i : X X;(b) the image space i{X) is dense in X;(c) the space (X,d) is complete.
The main theorem about the completion of metr ic spaces is the fol lowing.T h e o r e m 1 . 5 2 Every metr ic space has a comp letion. Th e completion is uniqueup to isomorphism.Proof. First , we prove that the completion is unique up to isomorphism, if i texists. Suppose th at {X\,dx) and (X2,d2) are two completions of (X,d), withcorresponding isometr ic embeddings ii : X > Xi a nd i2 : X > X2. We will use i\to extend i2 from X to the completion X\ and ob ta in an i somorph i sm?: X\ > X^To define Ton x 6 Xi, we pick a sequ ence (x n) in X such that (ii(xn)) convergesto x in Xi. Such a sequence exists because ii(X) is dense in X\. The sequence{ii{xn)) is Cauchy because it converges. Since %i and 2 axe isometries, it followst h a t (x n) an d (i2(xn)) are also Cauchy. The space X2 is complete, hence (i2(xn))converges in X2. We define
T(a?) = lim i2(xn) (18)n>ooIf (x 'n) is another sequence in X such that ( i i (a4) ) converges to x i n Xi , t hen
d2 ( 2 K ) , 2 ( iB ) ) = dCi^arn ) = di (ti(x'n),ii(xn)) > 0as n > oo. T h us , (i2(x 'n)) and (2(a;7i)) converge to the same limit, and J(x) iswelldefined.
If x, y belong to X2, an dJ(x) = lim i2(x n), T(y) = lim i2(y n),n >oo nfoot hen
d2 (T(i),T (j/)) = lim d2(i2(x n), i2(y)) = lim d(x n,y n) = di{x,y).n>oo nooTherefore?is an i sometry of X\ into X2. By using constant sequences in X, we seeth at To i i (x ) = i2(x ) for all a; e X, so that Tidentifies the image of X in X j underii with the image of X in X 2 under i2.To show th at T is onto, we observe th at Xx contains the limit of all Cauchysequences in ix (X ), so th e isom orphic space T(X i) con tains the limit of all Cauchysequences in 12(X). Therefore i2(X) C T(X i) . By assum pt ion, i2(X ) is dense inX2, so i2{X) X2, and T(Xi) = X.2 This shows that any two completions areisomorphic.

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Second, we prove the completion exists. To do this, we construct a completionfrom Cauchy sequences in X. We define a relation ~ between Cauchy sequencesx = (xn) and y (y n) in X by
x ~ y if an d only if lim d(x n, yn) 0.n>ooTwo convergent Cauchy sequences x, y satisfy x ~ y if and only if they have thesame limit. It is straightforward to check that ~ is an equivalence relation on theset C of Cauchy sequences in X. Let X be the set of equivalence classes of ~ in C.We call an element (a;n) x of an equivalence class x X, a representative of x.W e define d : l x l  > t b y
d(x,y)= lim d(x n,y n), (1.9)n>oowhere (a;) and (y n) are any two representatives of x and y, respectively. The limitin (1.9) exists because {d{xn,y n))^=l is a Cauchy sequence of real nu m be rs. Forthis definition to make sense, it is essential that the limit is independent of whichrepresentatives of x and y are chosen. Suppose that (x n), (x'n) represent x an d (y n),(y 'n) represent y. Then, by the tr iangle inequality, we have
d(x n,yn) < d(xn,x 'n)+d(x'n,y 'n) + d(y'n,y n),d(x n,yn) > d(x'n,y 'n)d(xn,x 'n)~d(y'n,y n).
Taking the limit as n > oo of these inequalit ies , and using the assumption that(x n) ~ (x 'n) and (y n) ~ (y'n), we find that
lim d(x n,y n)= lim d(x'n,y 'n).n >o o n>ooT h u s , the limit in (1.9) is independent of the representatives, and d is welldefined.It is s traightforward to check that d is a metric on X.
To show that the metric space {X,d) is a completion of (X, d), we define anembedding i : X > X as the map that takes a point x 6 X to the equivalenceclass of Cauchy sequences that contains the constant sequence (x n) with xn = x forall n. This map is an isometric embedding, since if (x n) and (y n) are the constantsequences with xn = x and yn = y, we have
d(i(x),i(y)) = lim d(x n,y n) = d(x,y).n>ooThe image i(X) consists of the equivalence classes in X which have a constant
representative Cauchy sequence. To show the density of i{X) in X, le t (x n) be arepresenta t ive of an a rb i t rary poin t i E l . We define a sequence (y n) of constantsequences by yn = (yn.k)^! where yn^ = xn for all n,k N. From the definitionof (y n) and the fact that (x n) is a Cauchy sequence, we have
lim d(y n,x) = lim lim d(x n,Xk) = 0.n>oo n>oo &>oo

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The completion of a metric space 21
T h u s , i(X) is dense in X.Finally, we prove that (X,d) is com plete. We will use C an tor 's "diagonal"
argument, which is useful in many other contexts as well. Let (x n) be a Cauchysequence in X. In order to prove that a Cauchy sequence is convergent, it is enoughto prove that it has a convergent subsequence, because the whole sequence convergesto the limit of any subsequence. Picking a subsequence, if necessary, we can assumet h a t (x n) satisfies
d(x m,x n) < for al l m , n > N. (110)For each term xn, we choose a representat ive Cauchy sequence in X, denoted by(xn,k)kLi Any subsequence of a representat ive Cauchy sequence of xn is also arepresentat ive of xn. We can therefore choose the representat ive so that
d(x n,k,xi) <  for all k,l>n. (111)nWe claim that the "diagonal" sequence (x ktk)kxL l 1S a Cauchy sequence, and thatthe equivalence class x to which it belongs is the limit of (x n) in X. The fact thatwe can obtain the limit of a Cauchy sequence of sequences by taking a diagonalsequence is the key point in proving the existence of the completion.To prove that the diagonal sequence is Cauchy, we observe that for any i N,
d(x k,k,xi,i) k,x kti) + d(xkti,xi>i) +d(x lti,xij). (1.12)The definition of d and (1.10) imply that for all k,l > N,
d(x k,x ) = lim d(x ki,x,i) < . (113)Tak ing th e lim sup of (1.12) as i > oo, an d u sing (1.11) and (1.13) in th e res ult, wefind that for all k, l> N,
3d(x k:k ,x u) < .Therefore (x k,k) is Cauchy.By a similar argument, we find that for all k,n> N,
3d(x tk,x k< k) < limswp{d(xn:k,x n!i) +d(xnii ,x k,i) + d(xkt i,x kik)} < .i  + o o lyTherefore, for n > N, we have
~ 3d(x n,x ) = lim d(x n,k,x k,k) < .Hence, the Cauchy sequence (x n) converges to x as n > oo, and X is complete. D

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It is slightly annoying that the completion X is constructed as a space of equivalence classes of sequences in X, rather than as a more direct extension of X. Forexample, if X is a space of functions, then there is no guarantee that its completioncan be identified with a space of functions that is obtained by adding more functionsto the original space.E x a m p l e 1 . 5 3 Let C ([0,1]) be the set of continu ous functions / : [0,1] > E. Wedefine the L2novm of / by
/ r1II/II = [jo \m \2 dx
The associated metric d(f,g) = \\f g\\ is a very useful one, analogous to theEuclidean metric on E n , but the space C([0,1]) is not complete with respect toit . Th e completion is denoted by L2([0,1]), and it can nearly be identified withth e space of Lebesgue meas urab le, squa reinteg rable functions. M ore precisely, apoint in L 2([0,1]) can be identified with an equivalence class of squareintegrablefunctions, in which two functions that differ on a set of Lebesgue measure zero areequivalen t. According to th e RieszFisher theore m , if (/ ) is a Cauchy sequence withrespect to the L 2 norm, then there is a subsequence (fnk) that converges pointwisea.e. to a squareintegrable function, and this fact provides one way to identify anelement of th e comp letion with an equivalence class of functions. M any of th eusual operations on functions can be defined on equivalence classes, independentlyof which representative function is chosen, but the pointwise value of an element/ 6 L2([0,1]) can no t be defined unam biguo usly.
In a similar way, the space L2 (E) of equivalence classes of Lebesgue measurable,square integrable functions on E is the completion of the space C C(E) of continuousfunctions on E with compact support (see Definition 2.6) with respect to the L2norm
II/II =(Jjf(x) \ 2dx^ .We will see later on that these L2 spaces are fundamental examples of infinite
dimensional Hilbert spaces. We discuss measure theory in greater detail in Chapter 12. We will use facts from th at chap ter as needed thro ug ho ut t he bo ok, includingFubini 's theorem for the exchange in the order of integration, and the dominatedconvergence theorem for passage to the limit under an integral sign.
1.7 C o m p a c t n e s sCompactness is one the most important concepts in analysis . A simple and usefulway to define compact sets in a metric space is by means of sequences.

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Compactness 23D e f i n i t i o n 1.54 A subset K of a metric space X is sequentially compact if everysequence in K has a convergent subsequence whose limit belongs to K.
W e can take K = X in this definition, so t h a t X is sequential ly compact if everysequence in X has a convergent subsequence. A subset K of (X, d) is sequentiallycompac t if and only if the metric subspace (K, d\j 1 for all m ^ n. The closed, bounded interval [0,1] is a sequential lycompact subset of R, as we prove below. The halfopen interval (0,1] is not asequential ly compact subset of M, because the sequence (1/n) converges to 0, andtherefore has no subsequence with l imit in (0 ,1]. The limit does, however, belongto [0,1].
The full importance of compact sets will become clear only in the set t ing ofinfinitedimensional normed spaces. It is nevertheless interest ing to s tar t wi th thef ini tedimensional case. Compact subsets of Rn have a simple, explicit characterizat ion.T h e o r e m 1.56 ( H e i n e  B o r e l ) Asubset of Rn is sequential ly compact if and onlyif it is closed and bounded .
The fact that closed, bounded subsets of W1 are sequential ly compact is a consequence of the following theorem, called the BolzanoWeierstrass theorem, eventhough Bolzano had little to do with its proof. We leave it to the reader to use thistheorem to complete the proof of the HeineBorel theorem.T h e o r e m 1.57 ( B o l z a n o  W e i e r s t r a s s ) Every bounded sequence in Rn has aconvergent subsequence.Proof. We will construct a Cauchy subsequence from an arbi t rary bounded sequence. Since Rn is complete , the subsequence converges.
Let (a;*) be a bounded sequence in R". There is an M > 0 such tha t Xk e[M,M]n for all k. The set [M,M]n is an ndimensional cube of side 2M. Wedenote this cube by Co We par t i t ion C0 into 2 cubes of side M. We denote by C\one of the smaller cubes that contains inf ini tely many terms of the sequence (a;*),meaning tha t x^ C\ for infinitely many k N. Such a cube exists because thereis a finite number of cubes and an infinite number of t e rms in the sequence. Letfci b e the smallest index such that x^ C\. We pick the first term of thesubsequence.
To choose the second term, we form a new sequence (yk) by deleting from (a;*)the term x^ and all terms which do not belong to C\. We r epea t the proceduredescribed in the previous parag raph , but with (xk) replaced by (yk), and Co replaced

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Fi g. 1.6 A set w it h a finite enet for it.by C\. This procedure gives a subcube C j where Cj is a cube of sideM / 2  7  1 . The refore ( x ^ ) is a Cauch y sequence, and hence it converg es. D
The following criterion for the sequential compactness of a metric space is ofteneasier to verify than the definition. Let A be a subset of a metric space X. We saythat a collection {G a \ a e A} of subsets of X is a cover of A if its union containsA, meaning tha t Ac{jG a.aAThe number of sets in the cover is not required to be countable. If every Ga in thecover is open, then we say that {G a} is an open cover of A.Let e > 0. A subse t {x a \ a A} of X is called an enet of the subset A ifthe family of open balls {B c{x a) \ a A} is an open cover of A. If the set {x a} isfinite, then we say that {x a} is a finite enet of A (see Figure 1.6).D ef in i t i on 1.58 A subset of a metr ic space is totally bounded if it has a finite enetfor every e > 0.
That is , a subset A of a metric space X is totally bounded if for every e > 0there is a finite set of points {11,12,   j ^ n } i n X such tha t A c UiLi Bc( x')

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T h e o r e m 1.59 A subset of a metric space is sequentially compact if and only if itis complete and tota l ly bounded.Proof. The proof that a complete, totally bounded set K is sequentially compactis the same as the proof of the B olzanoWeierstrass theorem 1.57. Suppose tha t (a;n)is a sequence in K. Then, since K is tota l ly bounded, there is a sequence of balls{Bk) such tha t Bk has radius 1/2* and every intersection Au 0^=1 &i containsinfinitely many terms of the sequence. We can therefore choose a subsequence(x nk) such tha t xnk Ak for every k. This subsequence is Cauchy, and, since K iscomplete, it converges.
To prove the converse, we show that a sequentially compact space is complete,and tha t a space which is not tota l ly bounded is not sequentially compact.If (xn) is a Cauchy sequence in a sequentially compact space K, then it has aconvergent subsequence. The whole Cauchy sequence converges to the limit of anyconvergent subsequence. Hence K is complete.
Now suppose tha t K is not tota l ly bounded. Then there is an e > 0 such tha tK has no finite enet. For every finite subset {xi,... ,xn} of K, the re is a pointxn+i K such tha t xn+\ UILi Bc{xi). Consequently, we can find an infinitesequence (xn) in K such tha t d(x m,xn) > e for all m / n. This sequence does notcontain a Cauchy subsequence, and hence has no convergent subseq uence . Th ereforeK is not sequentially compact.
Anothe r way to define compactness is in t e rms of open sets . We say t ha ta cover {Ga} of A has a finite subcover if the re is a finite subcollection of sets{G ai,...,G aJ such tha t A C U"=i G ;D e f i n i t i o n 1.60 A subset K of a metric space X is compact if every open cover ofK has a finite subcover.E x a m p l e 1.61 The space of real numbers R is not compact, s ince the open cover{(n  l , n + 1)  n Z} of M. has no finite subcover. The halfopen interval (0,1]is not compact, s ince the ope n c ove r { ( l / 2n , 2 /n )  n 6 N} has no finite subcover.If this open cover is extended to an open cover of [0,1], then the extension mustcontain an open neighborhood of 0. This open neighborhood, together with a finitenum be r of sets from the cover of (0 ,1], is a finite subcover of [0,1].
For metric spaces, compactness and sequential compactness are equivalent.T h e o r e m 1.62 Asubset of a me tric space is compact if and only if it is sequentiallycompact .Proof. Firs t , we prove that sequential compactness implies compactness. We willshow tha t an arbitrary open cover of a sequentially compact set has a countablesubcover, and t h a t a countable cover has a finite subcover.

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L e m m a 1 . 6 3 A sequentially compact metric space is separable.Proof. By Theorem 1.59, there is a f inite ( l /n)net An of a sequentially compactspace K for every n G N. Let A = U ^L i ^n Th en A is countable, because it is acountable union of finite sets, and A is dense in K by con struction . D
Suppose tha t {G a \ a G A} is an arbitrary open cover of a sequentially compactspace K. From Lemma 1.63, the space K has a countable dense subset A. Let Bbe the collection of open balls with rational radius and center in A, and let C bethe subcollection of balls in B that are contained in at least one of the open setsG a. The collection B is countable because i t is a countable union of countable sets .Hence, the subcollection C is also countable.
For every x G K, there is a set G a in the open cover of K with x G Ga. SinceG a is open , there is an e > 0 such tha t B ((x) C G a. Since A is dense in K, there isa point y A such tha t d(x,y) < e /3 . Then x Be/3(y), and B2e/3(y ) C G a. (Itma y help to draw a picture!) Th us , if q is a rational number with e/3 < q < 2 e / 3 ,then x G B9(2/) and 5 g (y) C G a. It follows that Bq(y ) G C, so any point a; in ifbelongs to a ball in C. Hence C is an open cover of K. For every B G C, we pick ana s G A such that B C G aB . T he n { G a B  B C} is a coun tabl e subcove r of if,because \JBC GaB conta ins  J B C B, which contains K.We will show by contradiction that a countable open cover has a finite subcover.Suppose tha t {G n  n G N} is a countab le open cover of a sequen tially com pactspace K that does not have a finite subcover. Then the finite union J n = 1 G n doesnot contain K for any N. We can therefore construct a sequence (xk) in if asfollows. We pick a point x\ G K. Since {G n} covers K, there is an iVi such thatxi GJVJ We pick x2 K such tha t x2 U ^ i i ^ m a n ^ choose iV2 such tha tx2 G/v2. Then we pick X3 G K such tha t x3 fi U n = i ^ " ' a n ( ^ s o o n  Since
Xk e G;vfc, and xfc g ( J G ,7 1 = 1
the open set Gjvfc is not equal to Gn for any n < JVjti. Thus, the sequence (N k)is strictly increasing, and Nk 00 as A; 4 00. It follows th a t, for any n , t he reis an integer Kn such tha t Xk $L G when k > Kn. If x G G n , then all points ofthe sequence eventually leave the open neighborhood G n of a;, so no subsequenceof (xk) can converge to x. Since the collection {G n } covers K, the sequence (x n )has no subsequence that converges to a point of K. This contradicts the sequentialcompactness of K, and proves that sequential compactness implies compactness.
To prove the converse, we show that if a space is not sequentially compact,the n it is not com pact. Suppo se th at K has a sequence (x n) with no convergentsubsequence. Such a sequence must contain an infinite number of dist inct points ,so we can assume without loss of generality that xm ^ xn for m ^ n.Let x G K. If the open ball B t{x ) contains a point in the sequence th at is dist inctfrom x for every e > 0, then x is the limit of a subsequence, which contradicts the

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Compactness 27assumption that the sequence has no convergent subsequence in K. Hence, there isan ex > 0 such that the open ball B tx (x) contains either no points in the sequence,if x itself does not belong to the sequence, or one point, if x belongs to the sequence.The collection of open balls {B ei (x)  x K} is an open cover of K. Everyfinite subcollection of n open balls contains at most n terms of the sequence. Sincethe terms of the sequence are distinct, no finite subcollection covers K. T hus , Khas an open cover with no finite subcover, and K is not com pact.
In future, we will abb reviate "sequentially com pact" to "com pact" w hen referring to metric spaces. The following terminology is often convenient.Def in i t ion 1 .64 A subset A of a metric space X is precompact if its closure in Xis compact.
Th e term "relatively com pact" is frequently used instead of "precom pact." Th isdefinit ion means that A is precompact if every sequence in A has a convergentsubsequence. Th e l imit of the subsequence can be any poin t in X, and is notrequired to belong to A. Since compact sets are closed, a set is compact if and onlyif it is closed and precompact. A subset of a complete metric space is precompactif and only if it is totally bounded.E x a m p l e 1 . 6 5 A subset of W 1 is precompact if and only if it is bounded.
Continuous functions on compact sets have several nice properties. Prom Proposition 1.34, continuous functions preserve the convergence of sequences. It followsimmediately from Definit ion 1.54 that continuous functions preserve compactness.T h e o r e m 1 . 6 6 Let / : K > Y be continuous on K, where K is a compact metricspace and Y is any metric space. Then f(K) is compact.
Since compact sets are bounded, continuous functions on a compact set arebou nde d. Moreover, continuous functions on comp act sets are uniformly con tinuou s.T h e o r e m 1 . 6 7 Let / : K > Y be a continuous function on a compact set K.Th en / is uniformly con tinuous.Proof. Supp ose tha t / is not uniformly continuou s. T he n ther e is an e > 0 suchthat for all 6 > 0, there are x,y X with d(x,y) < S and d(f(x),f(y)) > e. TakingS = 1/ra for n N, we find that there are sequences (x n) an d (y) in X such tha t
d(x n,yn) e . ( 1 . 1 4 )Since K is compact there are convergent subsequences of (x n) and (y n) which,for simplicity, we again denote by (x n ) and (y n). From (1.14), the subsequencesconverge to the same limit , but the sequences (f(xn)) an d (f(yn)) either diverge orconverge to different lim its. Th is con trad icts th e con tinuity of / .

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28 Metric and Normed Spaces
1 .8 M a x i m a a n d m i n i m aM aximum and minimu m problems are of central im porta nce in application s. Forexample, in many physical systems, the equilibrium state is one which minimizesenergy or maximizes entropy, and in optimization problems, the desirable state ofa system is one which minimizes an app rop riate cost function. Th e ma them aticalformulation of these problems is the maximization or minimization of a realvaluedfunction / on a state space X. Each point of the state space, which is often a metricspace, represents a possible state of the system. The existence of a maximizing, orminimizing, point of / in X may not be at all clear; indeed, such a point mayno t exis t. T he following the or em gives sufficient c on ditio ns for th e existen ce ofmaximizing or minimizing points namely, th at the function / is continuou s andthe state space X is compact. Although these conditions are fundamental , they aretoo strong to be useful in many applications. We will return to these issues lateron .T h e o r e m 1 . 6 8 Let K be a comp act me tric space and / : K > E a continuous,realvalued function. Th en / is bo un ded on K and a t ta ins i ts maximum and minimum. That is , there are points x,y K such tha t
fix) = inf f(z), f(y) = sup f(z). (1.15)z K zeK
Proof. From Theorem 1.66, the image f(K) is a compact subset of E, and therefore / is bounded by the HeineBorel theorem in Theorem 1.56.
It is enoug h to prove th at / att ain s its infimum , beca use th e applica tion of thisresult to / implies th at / a t tai ns i ts suprem um . Since / is bou nde d, i t is bou ndedfrom below, and the infimum m of / on K is finite. By the definition of the infimum,for each n N there is an xn K such tha t
m < f(xn) ooThe sequence (xn)^=1 need not converge, but since K is compact the sequencehas a convergent subsequence, which we denote by (xnk)k^=1 . We denote the l imitof the subsequence by x. Th en, since / is con tinuo us, we have from (1.16) t h a t
f{x) = lim f(xn) =m.f cooThere fore, / a t t a ins i ts in fimum m a t i .
The strategy of this proof is typical of many compactness arguments. We construct a sequence of approximate solutions of our problem, in this case a minimizing

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Maxima and minima 29
sequence (x n) that satisfies (1.16). We use compactness to extract a convergent subsequence, and show that the limit of the convergent subsequence is a solution of ourprob lem, in this case a poin t wh ere / att ain s its infimum . Th e following ex amp lesillustrate Theorem 1.68 and some possible behaviors of minimizing sequences.E x a m p le 1 .69 The funct ion f(x) = x*/4  x2/2 is continuous and bounded on[2,2] . I t a t ta ins i ts minimum at x = 1 . An example of a minimizing sequence(x n) is given by xn = (  1 )" . In fac t, f(xn) = inf f(x) for all n. This minimizingsequence does not converge because i ts terms jump back and forth between x = 1and x 1. The subsequences (x 2k+i) and (x 2k) converge, to x =  1 a n d x = 1,respectively.
As this example shows, the compactness argument does not imply that a pointwhere / a t ta ins i ts minim um is unique. Th ere are man y possible minimizing sequences, and there may be subsequences of a given minimizing sequence that converge to different lim its. If, however, th e function / at tai ns its min imu m at aunique point, then it follows from Exercise 1.27 that every minimizing sequencemust converge to that point.E x a m p le 1 .70 The func tion f(x) = e~ x is continuous and bounded from belowon th e non com pac t set E. T he infimum of / on E is zero, bu t / does no t a tta inits infimum . An exam ple of a minim izing sequence (x n) is given by T heterms of the minimizing sequence "escape" to infinity, and it has no convergentsubsequence.E x a m p l e 1.71 Th e discontinuous function / on the com pact set [0,1] defined by
, , . f lo g x if 0 < x < 1,f{x) = { 0 if z = 0,is not bounded from below. A sequence (x n) is a minimizing sequence if xn 0 asn oo. In that case, f(xn) > co a s n > oo, but / is discontinuous at the limitpoint x = 0.
Some of the conclusions of Theorem 1.68 still hold for semicontinuous functions.An almost identical proof shows the following result.T h e o r e m 1 .7 2 L et K be a com pact me tric space. If / : K > E is upper semicontinuo us, then / is bou nded from above and attain s its sup remu m. If / : K > E islower semicontinuous, then / is bounded from below and attains its infimum.E x a m p le 1 .73 We define f,g : [0,1] > E by
., . f x if 0 < a; < 1, , . f x if 0 < x < 1,f{x) = { 1 i f x = 0 , " 9M = \  l i f* = 0.~
The function / is upper semicontinuous, and does not attain its infimum, while gis lower semicontinuous and attains i ts minimum at x = 0.

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30 Metric and Normed Spaces
1 . 9 Re f e r e n c e sFor introd ucti on s to basic real analy sis, see M arsde n an d Hoffman [37] or R ud in[47]. Simmons [50] gives a clear and accessible discussion of metric, normed, andtopolo gical spaces. For linear algeb ra, see Halm os [19] an d Lax [30]. Tw o oth erbooks with a similar purpose to this one are Naylor and Sell [40] and Stakgold [52].
1 . 10 E xe r c i s e sE xe r c i s e 1 . 1 A set A is countably infinite if the re is a oneto one, on to m ap from Ato N. A set is countable if it is finite or countably infinite, otherwise it is uncountable.
(a) Prove that the set Q of rational numbers is countably infinite .(b) Prove that the set E of real numbers is uncountable.
E xe r c i s e 1 . 2 Give an e6 proof thatoo5>
n = 0when \x \ < 1.Exerc i se 1 .3 I f x, y, z are points in a metric space (X, d), show tha t
d(x,y) > \d{x,z) d{y,z)\.E xe r c i s e 1 . 4 Suppose tha t (X, dx) and (Y, dy) are metric spaces. Prove that theCar tes ian product Z = X x Y is a metric space with metric d defined by
d(zi,z2) =dx(xi,x2) +dY(yi,y2),where zi = (x i . j / i ) and z2 = {x 2,y2).E xe r c i s e 1 . 5 Suppose tha t (X ,    ) is a norm ed linear spac e. Pro ve th at (1.2)and (1.4) define metrics on X.E xe r c i s e 1 . 6 Starting from the fact that R equipped with i ts usual distance function is complete, prove that Rn equipped with the sum, maximum, or Euclideannorm is comple te .E xe r c i s e 1 . 7 Show that the series f* (ir
2 ^ nn = l
is not absolutely convergent. Show that by permuting the terms of this series onecan obtain series with different limits.
11^7'

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Exercises 31
E x e r c i s e 1.8 Let (x n ) be a sequence of r ea l numb ers . A poin t c l U {00 } iscalled a cluster point of (x n) if there is a convergent subsequence of (x n) with l imitc. Let C denote the set of cluster points of (x n). Prove tha t C is closed and
lim s up xn = max C and lim inf xn m in C.E x e r c i s e 1.9 Let (x n) be a bounded sequence of real numbers.
(a) Pro ve th at for every e > 0 and every N N there are n i , n 2 > N, such tha tlim s up xn < xn, + e, xn2 e < lim inf xn.nxx, n^
(b) Pro ve th at for every e > 0 th ere is an N N such tha txm < lim sup xn + e, xm > lim inf xn en>oo n*oo
for all m>N.(c) Prove th at (x n) converges if and only if
lim inf xn = lim sup xn.nvoo n^ooE xe r c i s e 1 . 10 Consider a family {x n,a} of real numbers indexed by n N anda A. Prove the following relations:
lim sup (inf i n , a ) < inf I lim sup xn,a 1 ;sup (lim inf xn a) < lim inf I su p x n ] 0 . 1 .
E xe r c i s e 1 . 11 If (x n) is a sequence of real numbers such thatlim xn x,
n >ooand an oo
E xe r c i s e 1 . 12 Let (X, dx), (Y,dy), and (Z,dz) be metric spaces and let / : X Y, and g : Y > Z be continuous functions. Show that the composition
h = gof:X+Z,defined by h(x) = g(f(x)), is also continuous.E xe r c i s e 1 . 13 A function / : E > K is said to b e differentiable at x if the followinglimit exists and is finite:
m = lim f ( ^h )  f ( x )ft>o h

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32 Metric and Normed Spaces
(a) Pr ov e th a t if / is differentiable at x, the n / is continuo us at x.(b) Show th at the function
, , . f x2 sin (l/x2) iix^O,f{x) = \ 0 if* = 0.is differentiable at x = 0 but the derivative is not continuous at x = 0.
(c) Prove that if / is differentiable at x and has a local maximum or minimuma t x, then f'(x) = 0.
Exerc i se 1 .14 I f / : [a , b] > K is continuous on [a,b] and differentiable in (a, b),then prove that there is a a < < b such tha t
f (b) f (a) = f'(0(ba).This result is called the mean value theorem. Deduce that if f'(x) = 0 for alla < x < b th en / is a con stan t function.E xe r c i s e 1 . 15 Prove that every compact subset of a metric space is closed andbounded. Prove that a closed subset of a compact space is compact.E xe r c i s e 1 . 16 Suppose tha t F and G are closed and open subsets of M", respectively, such that F C G. Show th at th ere is a con tinuou s function / : W 1 > E sucht h a t :
(a) 0 < f(x) < 1;(b ) f(x) = 1 for x G F;(c) f(x) = 0 for x G G c.
H I N T . Consider the functiond{x,G c)f ( x ) = d(x, G c) + d(x, F) '
This result is called Urysohn's lemma.E xe r c i s e 1 . 17 Let (X, d) be a complete metric space, and Y c X. Prove tha t(Y, d) is complete if and only if Y is a closed subset of X.E xe r c i s e 1 . 18 Let (X,d) be a metric space, and let (x n) be a sequence in X.Prove that if (x n) has a Cauchy subsequence, then, for any decreasing sequence ofposit ive tk > 0, there is a subsequence (x nk) of (x n) such tha t
d(x nk ,xni) < tk for all k < I.E xe r c i s e 1 . 19 Following the construction of the Cantor set C by the removal ofmiddle thirds, we define a function F on the com plem ent of th e C an tor set [0,1] \ Cas follows. F irs t, we define F(x) = 1/2 for 1/3 < x < 2 /3 . T he n F(x) = 1/4 for1/9 < x < 2/9 and F(x)  3/4 for 7/9 < x < 8/9 , and so on. Prove tha t F extends

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Exercises 33
to a unique continuous function F : [0,1] > E. Pro ve th at F is differentiable atevery i E l \ C and F'(x) = 0. This function is called the Cantor function. I tsgraph is sometimes called the devil's staircase.E xe r c i s e 1 . 20 L e t X be a normed linear space. A series ^x n in X is absolutelyconvergent if Y^ \\xn\\ converges to a finite value in E. Pro ve th at X is a Banachspace if and only if every absolutely convergent series converges.E x er c i se 1 .21 Suppose tha t X is a Banach space, and (x mn) is a doubly indexedsequence in X such tha t
oo ooY l ^2\ \Xmn\\ < OO.m=ln= l
Prove tha to o / o o \ o o / o o \
2s I /L i Xm n ) ~ Z*i I A^i Xm n J '7 n =l \ n = l / n=l \m=l /E xe r c i s e 1 . 22 L e t S be a set. A relation ~ betw een poin ts of S is called anequivalence relation if, for all a,b,c S, we have:
(a ) a ~ a;(b) a ~ & imp lies 6 ~ a;(c) a ~ b an d 6 ~ c implies a ~ c.
Define the equivalence class C a associated with a 5 byC a = {6 S  a ~ b} .
Prove that two equivalence classes are either disjoint or equal, so ~ partitions S intoa union of disjoint equivalence classes. Show that the relation ~ between Cauchysequences defined in the proof of Theorem 1.52 is an equivalence relation.E xe rc i se 1 .23 Suppose tha t / : X > E is lower semicontinuous and M is a realnum ber . Define / M : X > E by
fM(x) =min(f(x),M).Prove tha t / M is lower semicontinuou s.E xe rc i se 1 .24 Le t / : X > E b e a realvalu ed fun ction on a set X. T h e epigraphepi / of / is th e subse t of X x E consisting of points that l ie above the graph of / :
epi / = {(x, t) X x E  t > f(x)}.Prove that a function is lower semicontinuous if and only if its epigraph is a closedset.

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E x er c i se 1.25 A funct ion / : Rn > R is coercive iflim f(x) = oo. (1.17)
Explicitly, this condition means that for any M > 0 there is an R > 0 such tha ta; > R implies f(x) > M. Pro ve tha t if / : E n > R is lower semicontinuous andcoercive, the n / is bo un ded from below and a tta in s its infimum.E xe r c i s e 1 . 26 L e t p : R2 > R be a polynomial function of two real variables.Suppose tha t p(x,y) > 0 for all x,y 6 R. Does every such function attain itsinfimum? Prove or disprove.E x er c i se 1 .27 Suppose tha t (x n) is a sequence in a compact metric space withthe property that every convergent subsequence has the same limit x. Prove tha txn  x as n  oo.