Introductory notes in analysis

Stephen Semmes

Rice University

Contents

I Basic notions and results 4

1 Metric spaces 4

1.1 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

2 Least upper bounds 5

2.1 Additional properties . . . . . . . . . . . . . . . . . . . . . . . . . 5

3 Open sets 6

3.1 Other metrics on Rn . . . . . . . . . . . . . . . . . . . . . . . . . 63.2 Norms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

4 Closed sets 8

4.1 Complements of subsets . . . . . . . . . . . . . . . . . . . . . . . 84.2 Unions and intersections . . . . . . . . . . . . . . . . . . . . . . . 94.3 Closure and dense sets . . . . . . . . . . . . . . . . . . . . . . . . 9

5 Compactness 10

5.1 The limit point property . . . . . . . . . . . . . . . . . . . . . . . 105.2 Unions and closed subsets . . . . . . . . . . . . . . . . . . . . . . 11

6 Relatively open sets 11

6.1 Compact sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

7 Totally bounded sets 12

8 Closed intervals 13

8.1 Compactness properties . . . . . . . . . . . . . . . . . . . . . . . 138.2 Cells in Rn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 148.3 Another construction . . . . . . . . . . . . . . . . . . . . . . . . . 14

9 Countable sets 15

9.1 Unions and products . . . . . . . . . . . . . . . . . . . . . . . . . 159.2 Uncountable sets . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

1

10 Separable metric spaces 16

10.1 Lindelof’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 17

11 Connected sets 18

11.1 The real line . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

II Sequences, series, and functions 18

12 Sequences 19

13 Complex numbers 19

13.1 Sequences in C . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

14 Subsequences and sequential compactness 20

15 Cauchy sequences 21

16 Infinite series 22

16.1 The Cauchy condensation test . . . . . . . . . . . . . . . . . . . . 22

17 Alternating series 23

18 Power series 24

19 Extended real numbers 25

20 Upper and lower limits 25

20.1 Sequences of sets . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

21 Root and ratio tests 27

22 Continuous mappings 27

22.1 Another characterization . . . . . . . . . . . . . . . . . . . . . . . 28

23 Continuity and compactness 29

23.1 One-to-one mappings . . . . . . . . . . . . . . . . . . . . . . . . . 29

24 Continuity and connectedness 30

25 Uniform continuity 30

25.1 Compact spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

26 Uniform convergence 31

26.1 Complex-valued functions . . . . . . . . . . . . . . . . . . . . . . 32

27 The supremum metric 33

27.1 The supremum norm . . . . . . . . . . . . . . . . . . . . . . . . . 33

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28 The contraction mapping theorem 34

29 Limits of functions 34

29.1 Limits and sequences . . . . . . . . . . . . . . . . . . . . . . . . . 35

30 One-sided limits 36

30.1 Monotone functions . . . . . . . . . . . . . . . . . . . . . . . . . 36

III Some additional topics 37

31 Cauchy products 37

32 The exponential function 38

33 Diameters of bounded sets 39

34 Compactness and completeness 40

34.1 Cauchy subsequences . . . . . . . . . . . . . . . . . . . . . . . . . 41

35 The Baire category theorem 41

36 Diameters of compact sets 42

37 Another fixed-point theorem 42

38 An extension theorem 43

39 An embedding theorem 44

40 Completions 45

41 Integral metrics 46

42 Double sums 48

43 Rearrangements 49

44 Sums and limits 50

44.1 Monotone convergence . . . . . . . . . . . . . . . . . . . . . . . . 5244.2 Fatou’s lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

Appendix

A Three homework assignments 54

A.1 Maximizing distances . . . . . . . . . . . . . . . . . . . . . . . . . 54A.2 Minimizing distances . . . . . . . . . . . . . . . . . . . . . . . . . 55A.3 Positive lower bounds . . . . . . . . . . . . . . . . . . . . . . . . 55

3

References 55

Part I

Basic notions and results

1 Metric spaces

A metric space is a (nonempty) set M equipped with a distance function ormetric d(x, y) defined for x, y ∈ M such that d(x, y) is a nonnegative realnumber for every x, y ∈ M , d(x, y) = 0 if and only if x = y,

d(y, x) = d(x, y)

for every x, y ∈ M , and

d(x, z) ≤ d(x, y) + d(y, z)

for every x, y, z ∈ M . This last condition is known as the triangle inequality.These notes will be largely concerned with metric spaces, loosely following

the first four chapters and selections from other parts of Rudin’s famous text[20]. Of course, the discussion here will be rather informal, and much moreinformation can be found in Rudin’s book. We shall frequently use slightlydifferent definitions, notations, and conventions, and skip over some topics,while reorganizing the presentation of other topics. We may also elaboratea bit more on some points, mention different proofs or examples, and so on.There are numerous other excellent texts on essentially the same subject, a fewof which may be found in the references at the end.

1.1 Examples

As usual, the real line is denoted R. If x is a real number, then the absolute

value of x is denoted |x| and defined to be equal to x when x ≥ 0 and to −xwhen x ≤ 0. One can check that

|x y| = |x| |y|

and|x + y| ≤ |x| + |y|

for all x, y ∈ R. It follows that |x − y| defines a metric on R, known as thestandard metric on the real line. For each positive integer n, Rn consists of then-tuples x = (x1, . . . , xn) of real numbers, i.e., xi ∈ R for i = 1, . . . , n. TheEuclidean distance between x = (x1, . . . , xn), y = (y1, . . . , yn) ∈ Rn is

( n∑

i=1

(xi − yi)2)1/2

,

4

which reduces to |x − y| when n = 1. One can show that this defines a metricon Rn. If (M,d(x, y)) is any metric space and E ⊆ M , then the restriction ofd(x, y) to x, y ∈ E defines a metric on E, which is to say that E becomes ametric space too.

2 Least upper bounds

Let A be a set of real numbers, i.e., A ⊆ R. We say that a real number b isan upper bound for A if a ≤ b for every a ∈ A. We say that a real number c isthe least upper bound or supremum of A if (i) c is an upper bound of A and (ii)c ≤ b for every b ∈ R which is an upper bound for A. It follows immediatelyfrom the definition that the supremum is unique when it exists. For if c, c′ ∈ R

both satisfy the conditions for the supremum of A, then c′ ≤ c because c is anupper bound for A and c′ is less than or equal to any upper bound of A, andsimilarly c ≤ c′ because c′ is an upper bound for A and c is less than or equal toany upper bound of A. The supremum of a set A ⊆ R is denoted supA whenit exists.

If A is the empty set ∅, then every b ∈ R is an upper bound of A, and thesupremum of A does not exist. Some sets of real numbers have no upper boundin R, like A = R. The completeness property of the real numbers states thatevery nonempty A ⊆ R with an upper bound has a supremum, and one canview the real numbers as a completion of the rationals.

2.1 Additional properties

A real number β is said to be a lower bound for a set A ⊆ R if β ≤ a for everya ∈ A. Similarly, γ ∈ R is the greatest lower bound or infimum of A if (i) γ is alower bound for A and (ii) β ≤ γ for every β ∈ R which is a lower bound for A.It is easy to see that the infimum of A is unique when it exists, in which eventit is denoted inf A. The completeness property of the real numbers implies thatevery nonempty set A ⊆ R with a lower bound has an infimum. For if B is theset of real numbers which are lower bounds for A, then every element of A isan upper bound for B. The hypotheses that A be nonempty and have a lowerbound imply that B is nonempty and has an upper bound, and one can checkthat inf A = supB. Alternatively, if −A is the set of real numbers of the form−a, a ∈ A, then the negatives of the lower bounds for A are the same as theupper bounds for −A. One can use this to show that the infimum of A is equalto the negative of the supremum of −A.

Let us mention some other properties of real numbers. First, if x, y ∈ R

and x < y, then there is a rational number r such that x < r < y. Second,for any two positive real numbers u, v, there is is a positive integer n suchthat u ≤ n v. These two statements are basically equivalent to each otherand can be derived from completeness or considered as consequences of theway in which the real numbers are obtained from the rationals by completion.Third, every positive real number has a positive square root. For if y > 0 and

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Ay = {x ∈ R : x > 0 and x2 < y}, then one can show that Ay 6= ∅, Ay has anupper bound in R, and that (supA)2 = y.

3 Open sets

Let (M,d(x, y)) be a metric space. For each x ∈ M and positive real number r,the open ball in M with center x and radius r is denoted B(x, r) and defined by

B(x, r) = {y ∈ M : d(x, y) < r}.

A set U ⊆ M is said to be an open set if for every x ∈ U there is an r > 0such that B(x, r) ⊆ U . For every z ∈ M and t > 0 the open ball B(z, t) in Mis an open set. To see this, let x be any element of B(z, t), which thus satisfiesd(x, z) < t. Put r = t − d(x, z), and let us check that B(x, r) ⊆ B(z, t). Ify ∈ B(x, r), then the triangle inequality implies that

d(y, z) ≤ d(y, x) + d(x, z) < r + d(x, z) = t,

which is to say that d(y, z) < t and y ∈ B(z, t), as desired.Similarly, for z ∈ M and t ≥ 0, consider

V (z, t) = {u ∈ M : d(u, z) > t}.

Let x be an element of V (z, t), and put r = d(x, z) − t > 0. We would like toshow that B(x, r) ⊆ V (z, t), and hence that V (z, t) is an open set. If y ∈ M ,d(x, y) < r, and d(y, z) ≤ t, then the triangle inequality implies that

d(x, z) ≤ d(x, y) + d(y, z) < r + t = d(x, z),

a contradiction. Therefore d(y, z) > t for every y ∈ B(x, r), as desired.

3.1 Other metrics on Rn

Let d2(x, y) be the Euclidean metric on Rn, i.e.,

d2(x, y) =( n∑

i=1

(xi − yi)2)1/2

,

where x = (x1, . . . , xn), y = (y1, . . . , yn), as usual. One can check that

d∞(x, y) = max1≤i≤n

|xi − yi|

also defines a metric on Rn, and that

d∞(x, y) ≤ d2(x, y) ≤ √n d∞(x, y)

for every x, y ∈ Rn. The open balls in Rn with respect to the metric d∞(x, y)are cubes with sides parallel to the standard axes. It follows from the preceding

6

inequalities that d∞(x, y) determines the same class of open subsets of Rn asthe Euclidean metric.

Similarly,

d1(x, y) =

n∑

i=1

|xi − yi|

defines a metric on Rn. It is easy to see that

d∞(x, y) ≤ d1(x, y) ≤ nd∞(x, y)

for every x, y ∈ Rn. One can also show that d2(x, y) ≤ d1(x, y) ≤ √n d2(x, y).

At any rate, d1(x, y) determines the same class of open subsets of Rn as d∞(x, y)and the Euclidean metric.

3.2 Norms

Let V be a vector space over the real numbers. A norm on V is a function N(v)defined for v ∈ V such that N(v) is a nonnegative real number for every v ∈ V ,N(v) = 0 if and only if v = 0,

N(t v) = |t|N(v)

for every t ∈ R and v ∈ V , and

N(v + w) ≤ N(v) + N(w)

for every v, w ∈ V . If N(v) is a norm on V , then

d(v, w) = N(v − w)

defines a metric on V . For example,

‖x‖2 =( n∑

i=1

x2i

)1/2

defines a norm on Rn, the Euclidean norm, which corresponds to the Euclideanmetric on Rn. Similarly,

‖x‖1 =

n∑

i=1

|xi|

and‖x‖∞ = max

1≤i≤n|xi|

are norms on Rn, corresponding to the metrics d1(x, y), d∞(x, y), respectively.One can show that

‖x‖p =( n∑

i=1

|xi|p)1/p

is a norm on Rn for every real number p ≥ 1. Moreover, ‖x‖∞ is equal to thelimit of ‖x‖p as p → ∞.

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4 Closed sets

Let (M,d(x, y)) be a metric space. A point p ∈ M is said to be a limit point

of a set E ⊆ M if for every positive real number r there is a q ∈ E such thatd(p, q) < r and p 6= q.

Suppose that p ∈ M is a limit point of E ⊆ M , and let r > 0 be given. Letus check that there are infinitely many elements of E in B(p, r). Otherwise,there is a positive integer n such that there are exactly n elements x1, . . . , xn

of E in B(p, r) which are different from p. Let t be the minimum of d(xi, p),1 ≤ i ≤ n. Hence t is a positive real number and d(xi, p) ≥ t for i = 1, . . . , n. Ifq ∈ E and q 6= p, then either d(q, p) ≥ r > t, or q = xi for some i and d(q, p) ≥ t.In short, d(q, p) ≥ t when q ∈ E and q 6= p, which contradicts the hypothesisthat p be a limit point of E. Therefore, there are infinitely many elements of Ein B(p, r) for every r > 0 when p is a limit point of E. In particular, if E hasonly finitely many elements, then E has no limit points.

A set E ⊆ M is said to be closed if E contains all of its limit points in M ,i.e., if for every p ∈ M which is a limit point of E we have that p ∈ E. If E hasonly finitely many elements, then E is automatically closed, by the remarks ofthe previous paragraph. For every x ∈ M and t ≥ 0, the closed ball B(x, t) inM with center x and radius t, defined by

B(x, t) = {y ∈ M : d(x, y) ≤ t},

is a closed set. For if p is a limit point of B(x, t) and r > 0, then there is ay ∈ M such that d(x, y) ≤ t and d(p, y) < r, which implies that d(p, x) < t + rby the triangle inequality. Since this holds for every positive real number r, itfollows that d(p, x) ≤ t, as desired.

4.1 Complements of subsets

Let X be a set. If A, B are subsets of X, then A\B denotes the set of x ∈ Asuch that x 6∈ B. In particular, if E ⊆ X, then the complement of E in X isdefined to be X\E, and may be denoted Ec. Note that X\(X\E) = E whenE ⊆ X.

Now let (M,d(x, y)) be a metric space, and let us show that a set E ⊆ M isclosed if and only if its complement M\E in M is an open set. If E is a closedset in M and p ∈ M\E, then p is also not a limit point of E. It follows that thereis an r > 0 such that d(p, x) ≥ r for every x ∈ E. Equivalently, B(p, r) ⊆ M\E,which shows that M\E is an open set, since p is an arbitrary element of M\E.Next, suppose that M\E is an open set in M , and that p ∈ M is a limit pointof E. If p is not in E, then there is an r > 0 such that B(p, r) ⊆ M\E, becauseM\E is an open set. This contradicts the hypothesis that p be a limit pointof E. Hence p ∈ E, which implies that E is a closed set. Similarly, U ⊆ Mis an open set in M if and only if M\U is a closed set in M , by applying theprevious statement to E = M\U . As a special case, the empty set and M itselfare automatically open and closed subsets of M .

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4.2 Unions and intersections

Let A, X be sets, and suppose that for every α ∈ A, Eα is a subset of X, sothat {Eα}α∈A is a family of subsets of X indexed by A. The union

⋃α∈A Eα

and intersection⋂

α∈A Eα of the Eα’s can be defined in the usual way, and onecan check that

X\( ⋃

α∈A

Eα

)=

⋂

α∈A

(X\Eα) and X\( ⋂

α∈A

Eα

)=

⋃

α∈A

(X\Eα).

Suppose now that (M,d(x, y)) is a metric space. If {Eα}α∈A is a family ofclosed subsets of M , then their intersection

⋂α∈A Eα is also a closed set. For if

p ∈ M is a limit point of⋂

α∈A Eα, then is a limit point of each Eα, and hencean element of each Eα. Similarly, if {Uα}α∈A is a family of open subsets of M ,then their union

⋂α∈A Uα is an open set. For any element p of the union, there

is an α ∈ A such that p ∈ Uα, and hence an r > 0 such that B(p, r) is containedin Uα and therefore in the union.

If U1, . . . , Un are finitely many open subsets of M , then their intersection⋂ni=1 Ui is also an open set in M . To see this, let p be an element of the

intersection, and for each i = 1, . . . , n, let ri be a positive real number such thatB(p, ri) ⊆ Ui. If r = min(r1, . . . , rn), then B(p, r) ⊆ ⋂n

i=1 Ui. It follows thatthe union of finitely many closed subsets of M is a closed set in M too.

4.3 Closure and dense sets

Let (M,d(x, y)) be a metric space. If E ⊆ M , then E′ denotes the set of p ∈ Msuch that p is a limit point of E. The closure of E is denoted E and defined byE = E ∪ E′. By definition, E ⊆ M is a closed set if and only if E′ ⊆ E, whichis equivalent to E = E. For any E ⊆ M , one can check that E is a closed set inM . One can do this by showing that every limit point of E is also a limit pointof E and hence an element of E. One can also argue that M\E is an open set.Similarly, E′ is a closed set in M for any E ⊆ M .

One can also characterize E as the set of p ∈ M such that for every positivereal number r there is a q ∈ E with d(q, p) < r. Every p ∈ E automatically hasthis property, with q = p for each r > 0. If p ∈ M\E, then p has this propertyif and only if p is a limit point of E. One sometimes calls a p ∈ M with thisproperty an accumulation point of E in M .

A set E ⊆ M is said to be dense in M if E = M . Equivalently, E is densein M if for every x ∈ M and positive real number t there is a y ∈ E such thatd(x, y) < t. For example, the set Q of rational numbers is dense in the real linewith respect to the standard metric.

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5 Compactness

Let (M,d(x, y)) be a metric space. An open covering of a set K ⊆ M is a family{Uα}α∈A of open subsets of M , for some indexing set A, such that

K ⊆⋃

α∈A

Uα.

We say that K ⊆ M is compact if every open covering of K can be reduced toa finite subcovering, i.e., if for every open covering {Uα}α∈A of K in M thereare finitely many indices α1, . . . , αn ∈ A such that

K ⊆ Uα1∪ Uα2

∪ · · · ∪ Uαn.

If K has only finitely many elements, then K is compact.For example, if p ∈ M , then the family of open balls B(p, r), where r runs

through all positive real numbers, or all positive integers, is an open covering ofM , and hence of any K ⊆ M . If K is compact, then it follows that there arefinitely many r1, . . . , rn > 0 such that K ⊆ ⋃n

i=1 B(p, ri). Hence K ⊆ B(p,R),where R = max(r1, . . . , rn). In general, a set E ⊆ M is said to be bounded if Eis contained in a ball, and thus we conclude that compact sets are bounded.

Similarly, if p ∈ M , then the sets V (p, r) = {x ∈ M : d(x, p) > r} form anopen covering of M\{p} as r runs through all positive real numbers, or if werestrict our attention to r = 1/l for positive integers l. If K ⊆ M is compactand p 6∈ K, then we get an open covering of K, and compactness implies thatthere are finitely many r1, . . . , rn > 0 such that K ⊆ ⋃n

i=1 V (p, ri). ThereforeK ⊆ V (p, t) with t = min(r1, . . . , rn) > 0, and p cannot be a limit point of K.It follows that compact sets are closed.

5.1 The limit point property

Let (M,d(x, y)) be a metric space. A set E ⊆ M satisfies the limit point property

if for every set L ⊆ E with infinitely many elements there is a p ∈ E which isa limit point of L. If E has only finitely many elements, then E has the limitpoint property automatically. Let us check that a compact set K ⊆ M has thelimit point property. Suppose for the sake of a contradiction that L ⊆ K hasinfinitely many elements and that no p ∈ K is a limit point of L. This meansthat for every p ∈ K there is an r(p) > 0 such that B(p, r(p)) ∩ L is eitherempty or contains only p. The family of these open balls B(p, r(p)), p ∈ K,forms an open covering of K, since p is an element of B(p, r) for every r > 0.Compactness of K implies that there are finitely many elements p1, . . . , pn of Ksuch that K ⊆ ⋃n

i=1 B(pi, r(pi)). This implies that L has at most n elements,a contradiction.

As a partial converse, suppose that E ⊆ M has the limit point property,and that U1, U2, . . . is a sequence of open subsets of M such that E ⊆ ⋃∞

i=1 Ui.Suppose for the sake of a contradiction that for every l ≥ 1 there is an xl inE\⋃l

i=1 Ui. Let L be the set consisting of the xl’s, l ≥ 1. For every y ∈ E,

10

y ∈ Uk for some k, and thus y 6= xl when l ≥ k. Hence y = xl for only finitelymany l, and it follows that L has infinitely many elements. Because E satisfiesthe limit point property, there is a p ∈ E which is a limit point of L. We alsohave that p ∈ Ui for some i, which implies that infinitely many elements of Lare in Ui, since Ui is an open set. However, xl 6∈ Ui when l ≥ i, so that thereare fewer that i elements of L in Ui, a contradiction. Therefore E ⊆ ⋃l

i=1 Ui forsome l. One can use this partial converse to show that E is closed and bounded,in practically the same way as for compact sets.

5.2 Unions and closed subsets

Let (M,d(x, y)) be a metric space. Suppose that K1,K2 ⊆ M are compact, andlet us show that K1 ∪ K2 is compact. Let {Uα}α∈A be an arbitrary coveringof K1 ∪ K2 by open subsets of M . Hence {Uα}α∈A is an open covering ofK1 and K2 individually. Since K1 is compact, there are finitely many indicesα1, . . . , αl ∈ A such that K1 ⊆ ⋃l

i=1 Uαi. Similarly, since K2 is compact, there

are finitely many indices β1, . . . , βn ∈ A such that K2 ⊆ ⋃nj=1 Uβj

. Therefore

K1 ∪ K2 ⊆ ⋃li=1 Uαi

∪ ⋃nj=1 Uβj

, as desired.Suppose now that K ⊆ M is compact and that E ⊆ K is a closed set in M .

We would like to show that E is compact too. Let {Vβ}β∈B be an arbitraryopen covering of E in M . Because E ⊆ M is closed, M\E is an open set. Itfollows that the Vβ ’s, β ∈ B, together with M\E, form an open covering of Mand hence of K. Compactness of K implies that there are finitely many indicesβ1, . . . , βn ∈ B such that K ⊆ ⋃n

i=1 Vβi∪ (M\E), and thus E ⊆ ⋃n

i=1 Vβi.

There are analogous results for the limit point property. Specifically, ifE1, E2 ⊆ M have the limit point property and L ⊆ E1 ∪E2 has infinitely manyelements, then at least one of L ∩ E1, L ∩ E2 has infinitely many elements andhence a limit point in E1, E2 ⊆ E, as appropriate. If E ⊆ M has the limit pointproperty, A ⊆ E is a closed set in M , and L ⊆ A has infinitely many elements,then there is a limit point p of L in E, and p ∈ A because p is also a limit pointof A and A is closed.

6 Relatively open sets

Let (M,d(x, y)) be a metric space, and suppose that Y ⊆ M . We can considerY as a metric space too, using the restriction of the metric d(x, y) to x, y ∈ Y .Let us say that E ⊆ Y is relatively open in Y if E is an open set as a setcontained in Y . Explicitly, this means that for every p ∈ E there is an r > 0such that

{y ∈ Y : d(y, p) < r} ⊆ E.

If U is an open set in M and E = U∩Y , then it is easy to see that E is relativelyopen in Y . Conversely, suppose that E ⊆ Y is relatively open in Y . For everyp ∈ E, let r(p) be a positive real number such that E contains every y ∈ Ywith d(y, p) < r(p). Equivalently, B(p, r(p))∩Y ⊆ E, where B(x, t) is the usual

11

open ball in M with center x and radius t. Consider

U =⋃

p∈E

B(p, r(p)).

Clearly U is an open set in M , being the union of open subsets of M . Eachp ∈ E is an element of B(p, r(p)) and hence of U , and it is not difficult to checkthat E = U ∩ Y .

6.1 Compact sets

Let (M,d(x, y)) be a metric space, and suppose that Y ⊆ M . If K ⊆ Y ,then K is compact as a set contained in Y , considered as a metric space usingthe restriction of d(x, y) to x, y ∈ Y , if and only if K is compact as a setin M . For suppose that K is compact relative to Y , and let {Uα}α∈A be anarbitrary covering of K by open subsets of M . Each Uα∩Y is relatively open inY , and hence {Uα ∩ Y }α∈A is a covering of K by relatively open subsets of Y .Compactness relative to Y implies that there are finitely many indices α1, . . . , αl

in A such that K ⊆ ⋃li=1 Uαi

∩ Y . In particular, K ⊆ ⋃li=1 Uαi

. Conversely,suppose that K is compact as a set contained in M , and let {Vβ}β∈B be anarbitrary covering of K by relatively open subsets of Y . For every β ∈ B, letWβ be an open set in M such that Vβ = Wβ ∩ Y . Thus {Wβ}β∈B is an opencovering of K in M , and compactness of K in M implies that there are finitelymany indices β1, . . . , βn in B such that K ⊆ ⋃n

i=1 Wβi. Therefore K ⊆ ⋃n

i=1 Vβi,

since K ⊆ Y by hypothesis. Similarly, one can check that E ⊆ Y has the limitpoint property relative to Y if and only if E has the limit point property as aset contained in M .

7 Totally bounded sets

Let (M,d(x, y)) be a metric space. A set E ⊆ M is said to be totally bounded

if for every r > 0 there are finitely many elements x1, . . . , xl of E such that

E ⊆l⋃

i=1

B(xi, r).

Totally bounded sets are automatically bounded, since the union of finitely manybounded sets is bounded. If K ⊆ M is compact, then K is totally bounded,because for every r > 0 one can cover K by the open balls B(x, r), x ∈ K,and use compactness to cover K by B(x, r) with finitely many x ∈ K. Forany E ⊆ M and r > 0 one can look for a covering of E by balls with radiusr in the following way. Let x1 be any element of E, assuming that E 6= ∅. IfE ⊆ B(x1, r), then we stop. Otherwise, let x2 be an element of E which is notin B(x1, r). If E ⊆ B(x1, r) ∪ B(x2, r), then we stop. Otherwise, we continuethe process. The process may stop after finitely many steps and yield a coveringof E by finitely many open balls centered at elements of E and with radius r.

12

If not, then there is an infinite sequence x1, x2, . . . of elements of E such thatd(xi, xl) ≥ r when i < l. If L is the set of the xi’s, i ≥ 1, then L is an infiniteset contained in E which has no limit point in M . If E has the limit pointproperty, then this is impossible, and E is totally bounded.

8 Closed intervals

If a, b are real numbers with a ≤ b, then the closed interval [a, b] from a to bis the set of real numbers x such that a ≤ x ≤ b. The length of this interval isb− a. Closed intervals are clearly closed subsets of the real line with respect tothe standard metric.

Suppose that I1, I2, . . . is a sequence of closed intervals in the real line suchthat Ij+1 ⊆ Ij for every j ≥ 1. Let aj , bj be the endpoints of Ij , so that aj ≤ bj

in particular. The hypothesis that [aj+1, bj+1] ⊆ [aj , bj ] implies that aj ≤ aj+1

and bj+1 ≤ bj for every j ≥ 1. This implies in turn that

aj ≤ bl

for all positive integers j, l. For if j ≤ l, then aj ≤ al ≤ bl, and if j ≥ l, thenaj ≤ bj ≤ bl. Let A be the set of the aj ’s. Since aj ≤ b1 for every j ≥ 1, Ahas an upper bound, and hence a supremum. Each bl is an upper bound for A,and thus supA ≤ bl for every l ≥ 1. Of course the supremum of A is an upperbound for A, which is to say that aj ≤ supA for every j ≥ 1. It follows that

supA ∈ Ij

for every j ≥ 1.

8.1 Compactness properties

Let a, b be real numbers with a ≤ b, and consider the interval I = [a, b] in thereal line. Suppose that E ⊆ I has infinitely many elements. Put I1 = I, andchoose I2 to be [a, (a + b)/2] or [(a + b)/2, b], in such a way that E ∩ I2 alsohas infinitely many elements. In general, if a closed interval Ij = [aj , bj ] ⊆ Ihas been chosen such that E ∩ Ij has infinitely many elements, then let Ij+1

be one of the two closed intervals [aj , (aj + bj)/2], [(aj + bj)/2, bj ] such thatE ∩ Ij+1 has infinitely many elements. Hence Ij+1 ⊆ Ij and the length of Ij+1

is equal to one-half the length of Ij for every j. We know from a previousargument that there is a real number t such that t ∈ Ij for every j. Observethat |x − t| ≤ 2−j+1 (b − a) when x ∈ Ij , since the length of Ij is equal to2−j+1 times the length of I. It follows that for every r > 0, there are infinitelymany y ∈ E such that |y − t| < r, since y ∈ E ∩ Ij has this property when jis sufficiently large that 2−j+1 (b − a) < r. Therefore, t is a limit point of E,which shows that I has the limit point property.

Similarly, I is compact, as a set contained in the real line equipped withthe standard metric. For suppose that {Uα}α∈A is any covering of I by open

13

subsets of R. Suppose also for the sake of a contradiction that there is nofinite subcovering of I from this covering. If there were a finite subcovering foreach of the intervals [a, (a + b)/2], [(a + b)/2, b], then there would be a finitesubcovering for I. Hence there is no finite subcovering for at least one of thesetwo subintervals of I. As before, we put I1 = I, and choose closed subintervalsI2, I3, etc., so that Ij+1 ⊆ Ij , the length of Ij+1 is equal to one-half the lengthof Ij , and Ij cannot be covered by finitely many Uα’s for any j. There is a realnumber t in

⋂∞

j=1 Ij , and an α0 ∈ A such that t ∈ Uα0. For j sufficiently large,

Ij ⊆ Uα0, since Uα0

is an open set in R. This is a contradiction, since there isnot supposed to be a covering of Ij by finitely many Uα’s for any j.

8.2 Cells in Rn

Fix a positive integer n. If a1, . . . , an, b1, . . . , bn are real numbers such thatai ≤ bi for i = 1, . . . , n, then the corresponding cell is the set C ⊆ Rn consistingof x = (x1, . . . , xn) such that ai ≤ xi ≤ bi, 1 ≤ i ≤ n. One can show that cellssatisfy the limit point property and are compact as subsets of Rn equipped withthe Euclidean metric in much the same way as for closed intervals in the realline. There are two main changes in the argument, as follows. First, if C is acell in Rn associated to the real numbers a1, . . . , an, b1, . . . , bn as above, thenthe diameter of C, or maximal distance between two elements of C, is equal to

√√√√n∑

i=1

(bi − ai)2.

Just as an interval is the union of two subintervals of half the length, a cellC ⊆ Rn is the union of 2n cells of one-half the diameter of C. Second, ifC1, C2, . . . is a sequence of cells in Rn such that Cj+1 ⊆ Cj for every j ≥ 1,then there is a t ∈ Rn such that t ∈ ⋂∞

j=1 Cj . This is basically the same as theone-dimensional case repeated n times for the n components of t.

8.3 Another construction

Let (M,d(x, y)) be a metric space, let p be an element of M , and let E1, E2, . . .be a sequence of subsets of M . Suppose that En ⊆ B(p, 1/n) for each n, andlet E be the set consisting of p and the union of the En’s. One can check thatE is a closed set in M when the En’s are closed subsets of M . If En has thelimit point property for each n, then E does too. For if L ⊆ E has infinitelymany elements, then either p is a limit point of L, or L∩En has infinitely manyelements for some n and therefore L has a limit point in the same En.

Similarly, E is compact in M when the En’s are compact subsets of M . Tosee this, let {Uα}α∈A be an open covering of E in M . Since p ∈ E, there is anα0 ∈ A such that p ∈ Uα0

, and there is an r > 0 such that B(p, r) ⊆ Uα0because

Uα0is an open set in M . Thus En ⊆ Uα0

when n > 1/r. For each n there is alsoa set An ⊆ A with finitely many elements such that En ⊆ ⋃

α∈AnUα, by the

compactness of En. If A0 is the set consisting of α0 and the union of the An’s

14

such that n ≤ 1/r, then it follows that E ⊆ ⋃α∈A0

Uα. Also, A0 has only finitelymany elements, and so {Uα}α∈A0

is a finite subcovering of E from {Uα}α∈A,as desired. One can use these remarks to give examples of infinite-dimensionalcompact sets, by taking En to be n-dimensional for each n.

9 Countable sets

A set A has exactly n elements for some positive integer n if there is a listx1, . . . , xn of n elements of A such that every element of A is on the list exactlyonce. A set A is said to be countably infinite if there is an infinite sequencex1, x2, . . . of elements of A such that for every a ∈ A there is exactly one pos-itive integer i such that xi = a. By definition, the set Z+ of positive integersis countably infinite, with xi = i for each i. The set Z of all integers is count-ably infinite, because one can enumerate the integers with the sequence {xi}∞i=1

defined by xi = i/2 when i is even and xi = −(i − 1)/2 when i is odd.Suppose that A, B are sets and that {yj}∞j=1 is a sequence of elements of

B such that every a ∈ A is equal to yj for at least one j. One can list theelements of A by going through the yj ’s and keeping the ones that are in Awhen they occur the first time. The resulting list may be finite or infinite, andthe conclusion is that A is either finite or countably infinite. In particular, ifA ⊆ B and B is countably infinite, then A is finite or countably infinite.

9.1 Unions and products

Suppose that A1, A2, . . . is a sequence of finite sets and that A =⋃∞

i=1 Ai. It iseasy to make a list {xl}∞l=1 of the elements of A such that every a ∈ A is equalto xl for at least one l by first listing the elements of A1, then the elements ofA2, and so on. This list of the elements of A may have repetitions, and so Amay have finitely or countably-infinitely many elements.

If A, B are sets, then their Cartesian product A×B consists of the orderedpairs (a, b) such that a ∈ A and b ∈ B. For each positive integer n, let Cn bethe set of (i, j) ∈ Z+ ×Z+ such that i + j = n + 1. It is easy to see that Cn hasexactly n elements and that

⋃∞

n=1 Cn = Z+ × Z+. It follows that Z+ × Z+ iscountably infinite. Hence A × B is countably infinite whenever A, B are.

If A, B are finite or countably-infinite sets, then their union A ∪ B is too.One can first list the elements of A∪B by a doubly-infinite sequence {yl}∞l=−∞,and then convert that into an ordinary sequence using a listing of the elementsof Z. Similarly, if A1, A2, . . . is a sequence of finite or countably-infinite sets andA =

⋃∞

i=1 Ai, then A is finite or countably-infinite. For one can list the elementsof A with a doubly-indexed sequence {zp,q}∞p,q=1, and then convert that into anordinary sequence using a listing of the elements of Z+×Z+. As an application,the set Q of rational numbers is countably infinite, since Q =

⋃∞

n=1 Dn whereDn is the set of rational numbers of the form k/n, k ∈ Z.

15

9.2 Uncountable sets

Let B be the set of all binary sequences, which is to say the set of sequencesx = {xi}∞i=1 such that xi = 0 or 1 for every i ≥ 1. Suppose that

x(1) = {xi(1)}∞i=1, x(2) = {xi(2)}∞i=1, . . .

is a sequence of elements of B. If y = {yi}∞i=1 is the binary sequence defined byyi = 1−xi(i), then y 6= x(l) for every positive integer l. In general, a set is saidto be uncountable if it is neither finite nor countably infinite. The precedingargument shows that B is uncountable.

Every binary sequence corresponds to a real number in [0, 1] in the usualway, and every element of [0, 1] has at least one binary expansion. Sometimesa real number corresponds to two binary sequences, which are both eventuallyconstant. One can use the uncountability of B to show that [0, 1] is uncountabletoo.

10 Separable metric spaces

Let (M,d(x, y)) be a metric space. We say that M is separable if there is adense set E ⊆ M such that E has only finitely or countably many elements.For example, the real line equipped with the standard metric is separable, sincethe rationals are countable and dense in R. We say that E ⊆ M is ǫ-dense inM for some ǫ > 0 if for every x ∈ M there is a y ∈ E such that d(x, y) < ǫ.Thus E is dense in M if and only if E is ǫ-dense in M for every ǫ > 0. If forevery ǫ > 0 there is an ǫ-dense set Eǫ in M with only finitely or countably manyelements, then

⋃∞

n=1 E1/n is a dense set in M with only finitely or countablymany elements, and M is separable. Note that M is totally bounded if andonly if for every ǫ > 0 there is an ǫ-dense set Eǫ ⊆ M with only finitely manyelements, in which event M is separable in particular.

A collection B of open subsets of M is said to be a base for the topologyof M if for every point p ∈ M and radius r > 0 there is a V ∈ B such thatp ∈ V and V ⊆ B(p, r). Equivalently, B is a base for the topology of M if everyopen set W ⊆ M can be expressed as a union of elements of B. If E ⊆ M isdense, then the collection of open balls B(x, 1/n) with x ∈ E and n ∈ Z+ isa base for the topology of M . Consequently, if M is separable, then there isa base for the topology of M with only finitely or countably many elements.Conversely, suppose that B is a base for the topology of M with only finitelyor countably many elements. If V ∈ B and V 6= ∅, then let p(V ) be an elementof V . If E is the set of the points p(V ) chosen in this way, then E has onlyfinitely or countably many elements, and E is dense in M , which implies thatM is separable.

16

10.1 Lindelof ’s theorem

Let (M,d(x, y)) be a metric space, let B be a base for the topology of M , andlet {Uα}α∈A be any family of open subsets of M . Put

Bα = {V ∈ B : V ⊆ Uα}

for every α ∈ A, so that

Uα =⋃

V ∈Bα

V

for every α ∈ A, because B is a base for the topology of M . If we put

B′ =⋃

α∈A

Bα,

then we get that ⋃

V ∈B′

V =⋃

α∈A

( ⋃

V ∈Bα

V)

=⋃

α∈A

Uα.

If V ∈ B′, then V ∈ Bα for some α ∈ A, and we let α(V ) be an element of Awith this property. Thus V ⊆ Uα(V ) for every V ∈ B′, and we let A1 be the setof α(V ) ∈ A chosen in this way. It follows that

⋃

V ∈B′

V ⊆⋃

V ∈B′

Uα(V ) =⋃

α∈A1

Uα,

so that ⋃

α∈A

Uα =⋃

V ∈B′

V ⊆⋃

α∈A1

Uα.

More precisely, we have that

⋃

α∈A1

Uα =⋃

α∈A

Uα,

because the left side is automatically contained in the right side, since A1 ⊆ Aby construction. If B has only finitely or countably many elements, then B′ ⊆ Bhas only finitely or countably many elements, and it is easy to see that A1 hasonly finitely or countably many elements as well.

If M has the limit point property, then M is totally bounded, and henceseparable. This implies that there is a base for the topology of M with onlyfinitely or countably many elements, as before. It follows that every open cov-ering of M can be reduced to a finite or countable subcover, by the argumentin the previous paragraph. The limit point property also implies that everycountable open covering of M can be reduced to a finite subcovering, so thatM is compact. Similarly, if E ⊆ M has the limit point property, then one canshow that E is compact, by reducing to the case where E = M .

17

11 Connected sets

Let (M,d(x, y)) be a metric space. A pair of sets A,B ⊆ M are said to beseparated if

A ∩ B = A ∩ B = ∅.A set E is connected if there are no nonempty separated sets A,B ⊆ M suchthat A ∪ B = E.

Suppose that Y ⊆ M , which can also be considered as a metric space usingthe restriction of d(x, y) to x, y ∈ Y . If A ⊆ Y and p ∈ Y , then p is a limit pointof A relative to Y if and only if p is a limit point of A relative to M . Hencethe closure of A relative to Y is equal to the intersection of Y with the closureof A relative to M . It follows that A,B ⊆ Y are separated relative to Y if andonly if they are separated relative to M , and therefore that E ⊆ Y is connectedrelative to Y if and only if E is connected relative to M .

Let us say that E ⊆ M is ǫ-connected for some ǫ > 0 if for every p, q ∈ Ethere is a finite sequence x1, . . . , xn of elements of E such that x1 = p, xn = q,and d(xi, xi+1) < ǫ for 1 ≤ i < n. For any E ⊆ M , p ∈ E, and ǫ > 0,let Aǫ(p) be the set of q ∈ E which can by connected to p in this way, andput Bǫ(p) = E\Aǫ(p). It is easy to see that d(u, v) ≥ ǫ when u ∈ Aǫ(p) andv ∈ Bǫ(p), and hence that Aǫ(p), Bǫ(p) are separated as subsets of M . If E isconnected, then it follows that Aǫ(p) = E and that E is ǫ-connected for everyǫ > 0. The converse does not work in general, but it does hold for compact sets.

11.1 The real line

Let E be a set contained in the real line. If there are x, z ∈ E and y ∈ R\Esuch that x < y < z, then E is not connected. This is because the sets A, B ofw ∈ E such that w < y, y < w, respectively, are separated in R.

Conversely, suppose that E ⊆ R is not connected, and let A, B be nonemptyseparated subsets of R such that A∪B = E. Let x, z be elements of A, B, andlet us suppose without loss of generality that x < z, since we can interchangethe roles of A and B. Let y be the supremum of the set of t ∈ A such thatx ≤ t < z. Observe that y ∈ A, and hence y 6∈ B. In particular, y < z. Ify 6∈ A, then y 6∈ E, and x < y < z. If y ∈ A, then y 6∈ B, and there is a y′ ∈ R

such that y < y′ < z and y′ 6∈ E. In both cases there is a point between x andz which is not an element of E. It follows that intervals are connected subsetsof the real line, and in particular that the real line is connected.

18

Part II

Sequences, series, and functions

12 Sequences

Let (M,d(x, y)) be a metric space. A sequence {xj}∞j=1 of elements of M is saidto converge to x ∈ M if for every ǫ > 0 there is a positive integer L such that

d(xj , x) < ǫ for every j ≥ L.

In this event we call x the limit of the sequence {xj}∞j=1 and express this bylimj→∞ xj = x. The limit of a convergent sequence is unique, for if {xj}∞j=1

converges to x, x′ ∈ M , then

d(x, x′) ≤ d(x, xj) + d(xj , x′) < ǫ + ǫ = 2 ǫ

for every ǫ > 0 and sufficiently large j, depending on ǫ, which implies thatd(x, x′) = 0 and hence x = x′. A sequence in M is said to be bounded if the termsin the sequence are contained in a bounded set in M . Convergent sequencesare bounded, since all but finitely many terms in a convergent sequence arecontained in a ball of radius 1. If {xj}∞j=1 is a sequence of real numbers whichis bounded and monotone increasing, which is to say that xj ≤ xj+1 for everyj, then one can check that {xj}∞j=1 converges to the supremum of the set ofxj ’s. Similarly, a bounded monotonically decreasing sequence of real numbersconverges to the infimum of the set of terms in the sequence.

13 Complex numbers

The complex numbers are denoted C, and every z ∈ C can be expressed asx + y i, where x, y are real numbers and i2 = −1. In this case, x and y areknown as the real and imaginary parts of z, and the complex conjugate of z isdenoted z and defined to be x − y i. One can check that

z + w = z + w and z w = z w

for every z, w ∈ C. The modulus of a complex number z = x + y i, x, y ∈ R,is the nonnegative real number |z| =

√x2 + y2. Observe that |z|2 = z z, and

hence|z w| = |z| |w|

for every z, w ∈ C. If we think of a complex number z = x+y i as correspondingto (x, y) ∈ R2, then the modulus of z is the same as the ordinary Euclideannorm of the vector (x, y). It follows that

|z + w| ≤ |z| + |w|

19

for every z, w ∈ C, since the sum of two complex numbers corresponds to theusual sum of the corresponding vectors in R2. This implies that |z −w| definesa metric on C, which corresponds exactly to the standard Euclidean metric onR2.

13.1 Sequences in C

Suppose that {zj}∞j=1, {wj}∞j=1 are sequences of complex numbers which con-verge to z, w ∈ C, respectively. Let us check that {zj+wj}∞j=1 converges to z+w.Let ǫ > 0 be given, and let L1, L2 be positive integers such that |zj − z| < ǫ/2when j ≥ L1 and |wj − w| < ǫ/2 when j ≥ L2. If j ≥ max(L1, L2), then

|(zj + wj) − (z + w)| ≤ |zj − z| + |wj − w| <ǫ

2+

ǫ

2= ǫ,

as desired. Similarly, if {zj}∞j=1 is a sequence of complex numbers which con-verges to z ∈ C, and if a ∈ C, then {a zj}∞j=1 converges to a z. If {zj}∞j=1 is asequence of complex numbers which converges to 0 and {wj}∞j=1 is a bounded se-quence of complex numbers, then {zj wj}∞j=1 converges to 0 too. Consequently,if {zj}∞j=1, {wj}∞j=1 are sequences of complex numbers converging to z, w ∈ C,respectively, then {zj wj}∞j=1 converges to z w, since limj→∞ z wj = z w andlimj→∞(zj − z)wj = 0 by the previous statements. If {zj}∞j=1 is a sequence ofnonzero complex numbers which converges to z ∈ C with z 6= 0, then {1/zj}∞j=1

converges to 1/z. To see this, one can first check that {1/zj}∞j=1 is bounded,and then apply the previous statements to 1/zj − 1/z = −(zj − z)/(z zj).

14 Subsequences and sequential compactness

If {xj}∞j=1 is a sequence of elements of some set, and {jl}∞l=1 is a strictly increas-ing sequence of positive integers, then {xjl

}∞l=1 is a subsequence of {xj}∞j=1. Let(M,d(x, y)) be a metric space. If {xj}∞j=1 is a sequence of elements of M whichconverges to x ∈ M , then every subsequence {xjl

}∞l=1 of {xj}∞j=1 also convergesto x.

Observe that a point p ∈ M is an element of the closure E of a set E ⊆ M ifand only if there is a sequence {pn}∞n=1 of elements of E which converges to p.Similarly, p ∈ M is a limit point of E if and only if there is a sequence {pn}∞n=1 ofelements of E which converges to p and satisfies pn 6= p for every n. Let {xj}∞j=1

be a sequence of elements of M , and let Λ be the set of y ∈ M for which thereis a subsequence {xjl

}∞l=1 of {xj}∞j=1 which converges to y. This is somewhatanalogous to the closure of a set, or the set of limit points of a set. One cancheck that y ∈ Λ if and only if for every r > 0 there are infinitely many positiveintegers j such that d(xj , y) < r. For if y ∈ M has this property, then onecan choose jl for l ≥ 1 in such a way that d(xjl

, y) < 1/l and jl > j1, . . . , jl−1

when l ≥ 2. It follows easily from this characterization that Λ is automaticallya closed set in M .

A set K ⊆ M is said to be sequentially compact if for every sequence {xj}∞j=1

of elements of K there is a subsequence {xjl}∞l=1 which converges to an element

20

of K. If K ⊆ M is sequentially compact, then K has the limit point property.For if L ⊆ K has infinitely many elements, and if {xj}∞j=1 is any sequence ofdistinct elements of K, then the limit of any convergent subsequence of {xj}∞j=1

is a limit point of L. Conversely, if K ⊆ M has the limit point property, then Kis sequentially compact. To see this, let {xj}∞j=1 be any sequence of elements ofK. If there is an x ∈ K such that xj = x for infinitely many j, then {xj}∞j=1 hasa constant subsequence which converges trivially. Otherwise, the set L of thexj ’s has infinitely many elements, and hence a limit point in K. By the remarksof the previous paragraph, a limit point of L is also the limit of a subsequenceof {xj}∞j=1, as desired.

15 Cauchy sequences

Let (M,d(x, y)) be a metric space. A sequence {xj}∞j=1 of elements of M is saidto be a Cauchy sequence if for every ǫ > 0 there is a positive integer L such thatd(xj , xl) < ǫ for every j, l ≥ L. If {xj}∞j=1 converges to a point x ∈ M , then{xj}∞j=1 is a Cauchy sequence. For let ǫ > 0 be given, and let L be a positiveinteger such that d(xj , x) < ǫ/2 when j ≥ L. If j, l ≥ L, then

d(xj , xl) ≤ d(xj , x) + d(x, xl) <ǫ

2+

ǫ

2= ǫ,

as desired. If {xn}∞n=1 is a Cauchy sequence in M and {xnk}∞k=1 is a subsequence

of {xn}∞n=1 which converges to some x ∈ M , then {xn}∞n=1 converges to x. Forlet ǫ > 0 be given, and let L be a positive integer such that d(xj , xl) < ǫ/2 whenj, l ≥ L. If j ≥ L, then

d(xj , x) ≤ d(xj , xnk) + d(xnk

, x) <ǫ

2+

ǫ

2= ǫ

when k is sufficiently large that nk ≥ L and d(xnk, x) < ǫ/2. It follows that

d(xj , x) < ǫ for every j ≥ L, as desired. In particular, a Cauchy sequence ofelements of a compact set K ⊆ M converges to an element of K.

We say that M is complete if every Cauchy sequence in M converges to anelement of M . Cauchy sequences are always bounded, and hence the real lineis complete since bounded sets are contained in compact sets. Alternatively,suppose that {xj}∞j=1 is a Cauchy sequence, and let aj , bj be the infimum andsupremum of xl, l ≥ j, respectively. Thus aj ≤ aj+1 ≤ bj+1 ≤ bj for everyj, and hence if Ij = [aj , bj ], then Ij+1 ⊆ Ij for every j. Because {xj}∞j=1 is a

Cauchy sequence, the lengths of the Ij ’s tend to 0 as j → ∞. Therefore⋂∞

j=1 Ij

contains exactly one element, which is the limit of {xj}∞j=1. Similarly, the com-plex numbers and Rn equipped with the Euclidean metric are complete. Notethat sequences of complex numbers or elements of Rn are Cauchy sequencesor converge if and only if the corresponding sequences of components in R areCauchy sequences or converge.

21

16 Infinite series

Let {al}∞l=1 be a sequence of complex numbers. We say that the infinite series∑∞

l=1 al converges if the partial sums sn =∑n

l=1 al converge as a sequence ofcomplex numbers, in which event the sum

∑∞

l=1 al is defined to be limn→∞ sn.It will sometimes be convenient to have an a0 term in the series, which is notimportant for matters of convergence. The Cauchy criterion states that

∑∞

l=1 al

converges if and only if for every ǫ > 0 there is a positive integer L such that

∣∣∣∣n∑

j=l

aj

∣∣∣∣ < ǫ

when n ≥ l ≥ L. Indeed, this condition holds exactly when the correspondingsequence of partial sums {sn}∞n=1 is a Cauchy sequence. If

∑∞

l=1 al converges,then {al}∞l=1 converges as a sequence of complex numbers to 0, as one can see bytaking n = l in the Cauchy criterion. If

∑∞

l=1 al,∑∞

l=1 bl are convergent seriesof complex numbers and α, β ∈ C, then

∑∞

l=1(α al + β bl) converges, and

∞∑

l=1

(α al + β bl) = α

∞∑

l=1

al + β

∞∑

l=1

bl,

by the analogous results for sequences applied to the partial sums.A series

∑∞

l=1 al is said to converge absolutely if∑∞

l=1 |al| converges. Abso-lute convergence implies ordinary convergence, as a consequence of the Cauchycriterion. If {bl}∞l=1 is a sequence of nonnegative real numbers, then the partialsums for the series

∑∞

l=1 bl are monotone increasing and converge if and onlyif they are bounded. The comparison test states that if

∑∞

l=1 al is a series ofcomplex numbers,

∑∞

l=1 bl is a convergent series of nonnegative real numbers,and |al| ≤ bl for every l, then

∑∞

l=1 al converges absolutely.

16.1 The Cauchy condensation test

Let z be a complex number, and consider the series∑∞

l=0 zl. Here zl is inter-preted as being equal to 1 for every z ∈ C when l = 0. It is well known andeasy to check that

(1 − z)n∑

l=0

zl = 1 − zn+1

for each nonnegative integer n, so that

n∑

l=0

zl =1 − zn+1

1 − z

when z 6= 1. If |z| < 1, then limn→∞ zn = 0, and it follows that∑∞

l=0 zl

converges, with∞∑

l=0

zl =1

1 − z.

22

If |z| ≥ 1, then |zl| = |z|l ≥ 1 for every l, which implies that {zl}∞l=0 does notconverge to 0, and hence that

∑∞

l=0 zl diverges.Suppose now that {al}∞l=1 is a monotone decreasing sequence of nonnegative

real numbers. The Cauchy condensation test states that∑∞

l=1 al converges ifand only if

∑∞

k=0 2k a2k converges. More precisely, an infinite series of nonneg-ative real numbers converges if and only if the partial sums are bounded, andso the point of the Cauchy condensation test is that the partial sums of

∑∞

l=1 al

are bounded if and only if the partial sums of∑∞

k=0 2k a2k are bounded. In onedirection, we have that

2k+1−1∑

l=2k

al ≤ 2k a2k

for each k ≥ 0, because there are 2k terms on the left side, each of which isbounded by a2k , by monotonicity. This implies that

2n+1−1∑

l=1

al =

n∑

k=0

( 2k+1−1∑

l=2k

al

)≤

n∑

k=0

2k a2k

for each n ≥ 0, which shows that the partial sums of∑∞

l=1 al are bounded whenthe partial sums of

∑∞

k=0 2k a2k are bounded. In the other direction, we havethat

2k a2k ≤ 2

2k∑

l=2k−1+1

al

for each k ≥ 1, because there are 2k−1 terms on the right side, and a2k ≤ al

when l ≤ 2k by monotonicity again. Hence

n∑

k=0

2k a2k = a1 +

n∑

k=1

2k a2k ≤ a1 +2

n∑

k=1

( 2k∑

l=2k−1+1

al

)= a1 +2

2n∑

l=2

al ≤ 2

2n∑

l=1

al,

which shows that the partial sums of∑∞

k=0 2k a2k are bounded when the partialsums of

∑∞

l=1 al are bounded. As an application, for each positive real numberp,

∑∞

n=1 1/np converges if and only if∑∞

k=0 2(1−p) k converges, which happensexactly when p > 1.

17 Alternating series

Let {bj}∞j=1 be a monotone decreasing sequence of nonnegative real numberswhich converges to 0. The Leibniz alternating series test states that

∞∑

j=1

(−1)j bj

23

converges under these conditions. To see this, observe that

∣∣∣∣n∑

j=l

(−1)j bj

∣∣∣∣ ≤ bl

when n ≥ l, which implies that∑∞

j=1(−1)j bj satisfies the Cauchy criterion. For

example, it follows that∑∞

n=1(−1)n/np converges for every positive real numberp. More generally, suppose that {aj}∞j=1 is a sequence of complex numbers for

which the partial sums An =∑n

j=1 aj are bounded. If {bj}∞j=1 is as before, then∑∞

j=1 aj bj converges. For if A0 = 0, then

n∑

j=1

aj bj =

n∑

j=1

(Aj − Aj−1) bj =

n∑

j=1

Aj bj −n∑

j=1

Aj−1 bj

=

n∑

j=1

Aj bj −n−1∑

j=0

Aj bj+1

=

n∑

j=1

Aj (bj − bj+1) + An bn+1.

It suffices to show that∑∞

j=1 Aj (bj − bj+1) converges, since {An bn+1}∞n=1 con-

verges to 0 by hypothesis. For each n ≥ 1,∑n

j=1(bj − bj+1) = b1 − bn, which

implies that∑∞

j=1(bj −bj+1) converges. The convergence of∑∞

j=1 Aj (bj −bj+1)now follows from the comparison test, since the Aj ’s are bounded and the bj ’sare monotone decreasing. For example, if z ∈ C satisfies |z| = 1 and z 6= 1,then al = zl has bounded partial sums, and therefore

∑∞

l=1 zl/lp converges forevery p > 0.

18 Power series

Let {al}∞l=0 be a sequence of complex numbers, and consider the correspondingpower series

∑l=0 al z

l. Suppose that∑∞

l=0 al wl converges for some w ∈ C,

w 6= 0. This implies that liml→∞ al wl = 0, and in particular that {al w

l}∞l=0 isa bounded sequence of complex numbers. Let A be a nonnegative real numbersuch that |al w

l| ≤ A for every l ≥ 0. For every z ∈ C,

|al zl| ≤ A

( |z||w|

)l

,

and it follows from the comparison test that∑∞

l=0 al zl converges absolutely

when |z| < |w|. Depending on the coefficients al, it may be that∑∞

l=0 al zl only

converges in the trivial case where z = 0. At the other extreme, it may be that∑∞

l=0 al zl converges for every z ∈ C, in which event it converges absolutely

for every z ∈ Z by the previous remarks. Otherwise, the set of positive realnumbers r for which there is a w ∈ C such that |w| = r and

∑∞

l=0 al wl converges

24

is nonempty and bounded from above. The supremum R of this set is knownas the radius of convergence of the power series

∑∞

l=0 al zl, and we put R = 0

when the series converges only when z = 0 and R = +∞ when the seriesconverges for every z ∈ C. Thus

∑∞

l=0 al zl converges absolutely when |z| < R

and diverges when |z| > R, and moreover {al zl}∞l=0 is unbounded when |z| > R

by similar reasoning. The convergence of∑∞

l=0 al zl when |z| = R depends on

the situation.

19 Extended real numbers

The extended real numbers consist of the real numbers together with two ad-ditional elements, ±∞, such that −∞ < x < +∞ for every x ∈ R. One cancheck that every nonempty set of extended real numbers has an infimum andsupremum in the extended real line. Let us put x + ∞ = +∞ for x ∈ R,∞ + ∞ = +∞, and so on, but leave the sum of −∞ and +∞ undefined. Sim-ilarly, the product of a nonzero real number x and +∞ is equal to +∞ whenx > 0 and to −∞ when x < 0, etc. We can also set 1/ + ∞ = 1/ − ∞ = 0,and leave 1/0 undefined, although it makes sense to let 1/0 = +∞ in situationswhere the quantities are nonnegative.

If {xj}∞j=1 is a sequence of real numbers, then xj → +∞ as j → +∞ meansthat for every N ≥ 1 there is an L ≥ 1 such that xj ≥ N when j ≥ L. Similarly,xj → −∞ as j → +∞ means that for every N ≥ 1 there is an L ≥ 1 such thatxj ≤ −N when j ≥ L. The usual results about sums and products of limitsalso work for infinite limits when the sum or product of limits makes sense. If{xj}∞j=1 is any sequence of real numbers, then there is a subsequence {xjl

}∞l=1 of{xj}∞j=1 and an extended real number x such that xjl

→ x as l → ∞, where thisrefers to ordinary convergence when x ∈ R. If {xj}∞j=1 is a bounded sequence ofreal numbers, then there is a subsequence which converges to a real number bycompactness of closed intervals. Otherwise, there may not be an upper boundfor the xj ’s, or a lower bound, or either one. One can check that there is noupper bound for the xj ’s if and only if there is a subsequence which tends to+∞, and that there is no lower bound if and only if there is a subsequencewhich tends to −∞.

20 Upper and lower limits

Let {xj}∞j=1 be a sequence of real numbers, and let E be the set of extendedreal numbers x for which there is a subsequence {xjl

}∞l=1 of {xj}∞j=1 such thatxjl

→ x as l → ∞. We have seen that E 6= ∅. The upper limit of {xj}∞j=1 isdenoted lim supj→∞ xj and defined to be the supremum of E, and the lowerlimit of {xj}∞j=1 is denoted lim infj→∞ xj and defined to be the infimum of E.Hence

lim infj→∞

xj ≤ lim supj→∞

xj

25

automatically. If xj → x as j → ∞ for some extended real number x, thenevery subsequence also tends to x, x is the only element of E, and

x = lim infj→∞

xj = lim supj→∞

xj .

Put u = lim supj→∞ xj . If v is a real number such that v > u, then xj < vfor all but finitely many j. Otherwise there is a subsequence {xjl

}∞l=1 of {xj}∞j=1

such that xjl≥ v for every l, and a subsubsequence {xjln

}∞n=1 which tends to anextended real number z. This subsubsequence is also a subsequence of {xj}∞j=1,and we get a contradiction since z ≥ v > u, the supremum of the subsequentiallimits of {xj}∞j=1. If t < u, then xj > t for infinitely many j, because otherwiset is an upper bound for the subsequential limits of {xj}∞j=1 which is strictly lessthan the supremum. If u′, u′′ are extended real numbers which satisfy both ofthese properties and u′ < u′′, then we can get a contradiction by consideringa real number y such that u′ < y < u′′ and observing that either xj > y foronly finitely many j or for infinitely many j but not both. Thus u is uniquelycharacterized by these two properties. There is an analogous characterizationfor the lower limit.

20.1 Sequences of sets

Let X be a set, and let A1, A2, . . . be a sequence of subsets of X. The upperand lower limits of {An}∞n=1 are defined by

lim supn→∞

An =∞⋂

n=1

∞⋃

l=n

Al

and

lim infn→∞

An =∞⋃

n=1

∞⋂

l=n

Al.

By definition, lim supn→∞ An consists of the x ∈ X such that x ∈ An forinfinitely many n, while lim infn→∞ An consists of the x ∈ X such that x ∈ An

for all but finitely many n, which is the same as saying that x ∈ An for allsufficiently large n. In particular,

lim infn→∞

An ⊆ lim supn→∞

An.

One might say that {An}∞n=1 converges to A ⊆ X when

lim supn→∞

An = lim infn→∞

An = A.

If An ⊆ An+1 for each n, then {An}∞n=1 converges in this sense to A =⋃∞

n=1 An,and if An+1 ⊆ An for each n, then {An}∞n=1 converges in this sense to A =⋂∞

n=1 An. For any set E ⊆ X, let 1E(x) be the indicator function of E on X,which is equal to 1 when x ∈ E and to 0 when x ∈ X\E. It is easy to checkthat the upper and lower limits of 1An

(x) as a sequence of real numbers arethe same as the indicator functions associated to the upper and lower limits of{An}∞n=1 evaluated at x for every x ∈ X, respectively.

26

21 Root and ratio tests

Let {an}∞n=1 be a sequence of complex numbers, and consider

α = lim supn→∞

|an|1n .

The root test asserts that∑∞

n=1 an converges absolutely when α < 1 and di-verges when α > 1. For if α < 1, then we can let β be a real number such thatα < β < 1, and we get that |an|1/n < β for all but finitely many n. Hence|an| < βn for all but finitely many n, and

∑∞

n=1 an converges absolutely bycomparison with

∑∞

n=1 βn. If α > 1, then the an’s are unbounded.As an application, consider the power series

∑∞

n=1 an zn. If α is as above,then one can check that R = 1/α is the radius of convergence of this powerseries. When α = 0, this means that R = +∞, which is to say that the powerseries converges for every z ∈ C.

Suppose now that an 6= 0 for every n ≥ 1, and consider

γ = lim supn→∞

|an+1||an|

.

If γ < 1, then the ratio test asserts that∑∞

n=1 an converges absolutely. For ifγ < δ < 1, then there is a positive integer N such that |an+1|/|an| < δ whenn ≥ N . This implies that |an| ≤ δn−N |aN | when n ≥ N , and hence that∑∞

n=1 an converges absolutely since∑∞

n=1 δn converges. In the other direction,if

lim infn→∞

|an+1||an|

> 1,

then |an| → ∞ as n → ∞, and∑∞

n=1 an diverges.

22 Continuous mappings

Let (M,d(x, y)) and (N, ρ(u, v)) be metric spaces, and let f be a function definedon M with values in N . We say that f is continuous as a point x ∈ M if forevery positive real number ǫ there is a positive real number δ such that

ρ(f(y), f(x)) < ǫ

for every y ∈ M with d(y, x) < δ. Suppose that f : M → N is continuous atx ∈ M , and that {xj}∞j=1 is a sequence of elements of M that converges to x.Let ǫ > 0 be given, and let δ > 0 be as in the definition of continuity. Because{xj}∞j=1 converges to x, there is a positive integer L such that d(xj , x) < δ whenj ≥ L. It follows that ρ(f(xj), f(x)) < ǫ when j ≥ L, and hence that {f(xj)}∞j=1

converges to f(x) in N . Conversely, suppose that f is not continuous at x ∈ M .This means that there is an ǫ > 0 such that for every δ > 0 there is a y ∈ M withd(y, x) < δ and ρ(f(y), f(x)) ≥ ǫ. In particular, for each positive integer l thereis a yl ∈ M which satisfies d(yl, x) < 1/l and ρ(f(yl), f(x)) ≥ ǫ. Thus {yl}∞l=1

27

is a sequence of elements of M which converges to x for which {f(yl)}∞l=1 doesnot converge to f(x) in N . This shows that f is continuous at x if and only iffor every sequence {xj}∞j=1 of elements of M that converges to x, {f(xj)}∞j=1

converges to f(x) in N .We say that f is a continuous mapping from M to N if f is continuous

at every element of M . Equivalently, f : M → N is continuous if for everyconvergent sequence {xj}∞j=1 of elements of M , {f(xj)}∞j=1 converges in N , and

limj→∞

f(xj) = f(

limj→∞

xj

).

One can check that sums and products of continuous complex-valued functionsare continuous, using the analogous results for sequences and the characteriza-tion of continuity in terms of sequences, or using the same arguments as forsequences directly.

22.1 Another characterization

Suppose that X, Y are sets, and that f is a mapping from X to Y . For eachA ⊆ X, put f(A) = {y ∈ Y : y = f(a) for some a ∈ A}. Similarly, for E ⊆ Y ,put f−1(E) = {x ∈ X : f(x) ∈ E}. Note that X\f−1(E) = f−1(Y \E) forevery E ⊆ Y .

Let (M,d(x, y)), (N, ρ(u, v)) be metric spaces, and let f be a mapping fromM to N . If f : M → N is continuous and U is an open set in N , thenf−1(U) is an open set in M . To see this, let x be any element of f−1(U). Thusf(x) ∈ U , and there is an ǫ > 0 such that U contains every z ∈ N which satisfiesρ(z, f(x)) < ǫ. Let δ be a positive real number such that ρ(f(y), f(x)) < ǫ wheny ∈ M satisfies d(y, x) < δ, as in the definition of continuity. It follows thaty ∈ f−1(U) when y ∈ M and d(y, x) < δ, so that f−1(U) contains an openball in M centered at x, as desired. Conversely, if f−1(U) is an open set in Mwhenever U is an open set in N , then f : M → N is continuous. For let x ∈ Mand ǫ > 0 be given, and let U be the set of z ∈ N such that ρ(z, f(x)) < ǫ.Clearly x ∈ f−1(U) and U is an open set in N , and so there is a δ > 0 such thaty ∈ f−1(U) when y ∈ M and d(y, x) < δ, by hypothesis. This is the same assaying that ρ(f(y), f(x)) < ǫ when y ∈ M and d(y, x) < δ, as required by thedefinition of continuity. As a corollary, f : M → N is continuous if and only iff−1(E) is a closed set in M for every closed set E in N .

Let (M1, d1(x, y)), (M2, d2(u, v)), and (M3, d3(w, z)) be metric spaces, andsuppose that f1 : M1 → M2, f2 : M2 → M3 be continuous mappings. As usual,the composition f2 ◦ f1 is the mapping from M1 → M2 defined by

(f2 ◦ f1)(x) = f2(f1(x)), x ∈ M1.

One can show that f2 ◦ f1 is a continuous mapping from M1 to M3 using thedefinition of continuity in terms of ǫ’s and δ’s, or using the characterization ofcontinuity in terms of convergent sequences, or using the characterization ofcontinuity in terms of open sets.

28

23 Continuity and compactness

Let (M,d(x, y)) and (N, ρ(u, v)) be metric spaces, and suppose that f is acontinuous mapping from M to N . If K ⊆ M is compact, then f(K) is acompact set in N . For suppose that {Vα}α∈A is an open covering of f(K) inN . If Uα = f−1(Vα) for every α ∈ A, then {Uα}α∈A is an open covering of Kin M . Since K is compact in M , there are finitely many indices α1, . . . , αl ∈ Asuch that K ⊆ ⋃l

i=1 Uαi. This implies that f(K) ⊆ ⋃n

i=1 Vαi, as desired. As

a consequence, if f is a continuous real-valued function on M and K ⊆ M isnonempty and compact, then there are p, q ∈ K such that

f(p) ≤ f(x) ≤ f(p)

for every x ∈ K, which is to say that the maximum and minimum of f on Kare attained. More precisely, this follows from the fact that f(K) is closed andbounded, since it is compact. This statement is known as the extreme valuetheorem.

If f : M → N is continuous and K ⊆ M is sequentially compact, then wecan also argue directly that f(K) is sequentially compact in N . Let {wj}∞j=1 beany sequence of elements of f(K). For each j, let xj be an element of K suchthat f(xj) = wj . Because K is sequentially compact, there is a subsequence{xjl

}∞l=1 of {xj}∞j=1 which converges to a point x ∈ K. The continuity of fimplies that the corresponding subsequence {wjl

}∞l=1 of {wj}∞j=1 converges tof(x) in N .

23.1 One-to-one mappings

Let X, Y be sets, and let f be a function defined on X with values in Y . Wesay that f : X → Y is one-to-one if for every x, x′ ∈ X with x 6= x′ we havethat f(x) 6= f(x′). We say that f maps X onto Y if for every y ∈ Y there isan x ∈ X such that f(x) = y. If both conditions hold, then there is an inversemapping f−1 : Y → X such that f−1(y) = x when y = f(x).

Let (M,d(x, y)) and (N, ρ(u, v)) be metric spaces, and let f be a one-to-onecontinuous mapping from M onto N . If M is compact, then f−1 is continuousas a mapping from N onto M . For if E ⊆ M is closed, then E is compact in M ,and f(E) is compact in N . This implies that f(E) is a closed set in N , whichis to say that (f−1)−1(E) is a closed set in N for every closed set E ⊆ M , andhence f−1 is continuous. Alternatively, let {wj}∞j=1 be a sequence of elements

of N which converges to a point w ∈ N . Put xj = f−1(wj) and x = f−1(w),and suppose for the sake of a contradiction that {xj}∞j=1 does not converge tox in M . This means that there is an ǫ > 0 such that

d(xj , x) ≥ ǫ

for infinitely many j, and we may as well suppose that this inequality holdsfor all j by passing to a subsequence if necessary. Because M is sequentiallycompact, there is a subsequence {xjl

}∞l=1 of {xj}∞j=1 which converges to a point

29

y ∈ M , y 6= x. Thus wjl= f(xjl

) → f(y) 6= f(x) = w as l → ∞ by thecontinuity of f , a contradiction.

24 Continuity and connectedness

Let (M,d(x, y)) and (N, ρ(u, v)) be metric spaces, and suppose that f is acontinuous mapping from M into N . If E ⊆ M is connected, then f(E) isa connected set in N . For suppose to the contrary that there are nonemptyseparated sets A,B ⊆ N such that f(E) = A ∪ B. Using the continuity of f ,one can show that A1 = f−1(A)∩E, B1 = f−1(B)∩E are nonempty separatedsets in M such that E = A1 ∪B1, a contradiction. As an application, if a, b arereal numbers with a < b, f is a continuous real-valued function on the interval[a, b] ⊆ R, and c is a real number such that f(a) < c < f(b) or f(b) < c < f(a),then there is an x ∈ R such that a < x < b and f(x) = c. Otherwise, f([a, b])would not be a connected set in R, a contradiction. This fact is known asthe intermediate value theorem. For another application, let us say that a setE ⊆ M is pathwise connected if for every y, z ∈ E there are a, b ∈ R with a ≤ band a continuous mapping p : [a, b] → M such that p([a, b]) ⊆ E, p(a) = y,and p(b) = z. One can show that pathwise connected sets are connected, usingthe connectedness of p([a, b]) when p : [a, b] → M is continuous. If E ⊆ M ispathwise connected and f : M → N is continuous, then it is easy to see thatf(E) is pathwise connected in N . One can also show that connected open setsin Rn are pathwise connected, using the observation that open subsets of Rn

are locally pathwise connected.

25 Uniform continuity

Let (M,d(x, y)) and (N, ρ(u, v)) be metric spaces. A mapping f : M → N issaid to be uniformly continuous if for every ǫ > 0 there is a δ > 0 such thatρ(f(x), f(y)) < ǫ for every x, y ∈ M with d(x, y) < δ. Uniformly continuousmappings are continuous in particular. One can check that f : M → N isuniformly continuous if and only if for every pair of sequences {xj}∞j=1, {yj}∞j=1

of elements of M such that limj→∞ d(xj , yj) = 0,

limj→∞

ρ(f(xj), f(yj)) = 0.

It is easy to see that the composition of two uniformly continuous mappings isuniformly continuous, using the definition of uniform continuity in terms of ǫ’sand δ’s, or the characterization of uniform continuity in terms of sequences.

If f : M → N is uniformly continuous and {wl}∞l=1 is a Cauchy sequence ofelements of M , then {f(wl)}∞l=1 is a Cauchy sequence in N . If f is uniformlycontinuous and E ⊆ M is totally bounded, then f(E) is totally bounded inN . The sum of two uniformly continuous complex-valued functions is uniformlycontinuous, as is the product of such a function and a constant. The product

30

of two bounded uniformly continuous complex-valued functions is uniformlycontinuous.

25.1 Compact spaces

Let (M,d(x, y)) and (N, ρ(u, v)) be metric spaces, and let f be a continuousmapping from M to N . If M is compact, then f is uniformly continuous. Tosee this, let ǫ > 0 be given. For each x ∈ M , there is a δ(x) > 0 such that

ρ(f(y), f(x)) <ǫ

2

when y ∈ M and d(y, x) < δ(x), by continuity. If B(x) is the open ball inM with center x and radius δ(x)/2, then the open balls B(x), x ∈ M , coverM . By compactness, there are finitely many elements x1, . . . , xk of M suchthat M ⊆ ⋃k

i=1 B(xi). Put δ = min(δ(x1)/2, . . . , δ(xk)/2), and let x, y bearbitrary elements of M such that d(x, y) < δ. There is an i, 1 ≤ i ≤ k, suchthat x ∈ B(xi), and for which d(y, xi) < δ(xi)/2 + δ ≤ δ(xi), by the triangleinequality. It follows that

ρ(f(x), f(y)) ≤ ρ(f(x), f(xi)) + ρ(f(xi), f(y)) <ǫ

2+

ǫ

2= ǫ,

as desired.Alternatively, suppose that {xj}∞j=1, {yj}∞j=1 are sequences of elements of

M such that limj→∞ d(xj , yj) = 0, but ρ(f(xj), f(yj)) does not converge to 0.This means that there is an ǫ > 0 such that ρ(f(xj), f(yj)) ≥ ǫ for infinitelymany j. Without loss of generality, we may suppose that this holds for all j,since otherwise we can replace our sequences with the subsequences where itdoes hold. By compactness, there is a strictly increasing sequence {jl}∞l=1 ofpositive integers such that {xjl

}∞l=1 converges to a point x ∈ M , and we alsoget that {yjl

}∞l=1 converges to x too, since d(xjl, yjl

) → 0 as l → ∞. Continuityof f at x implies that {f(xjl

)}∞l=1 and {f(yjl)}∞l=1 both converge to f(x) in N ,

and hence that liml→∞ ρ(f(xjl), f(yjl

)) = 0, a contradiction.

26 Uniform convergence

Let E be a set, let (N, ρ(u, v)) be a metric space, and let fj , j ≥ 1, and f befunctions on E with values in N . If {fj(x)}∞j=1 converges to f(x) in N for everyx ∈ E, then we say that the sequence of functions {fj}∞j=1 converges pointwise

to f on E. We say that {fj}∞j=1 converges to f uniformly on E if for every ǫ > 0there is a positive integer L such that

ρ(fj(x), f(x)) < ǫ for every x ∈ E

when j ≥ L. It is easy to see that uniform convergence implies pointwiseconvergence.

31

Now suppose that (M,d(x, y)), (N, ρ(u, v)) are metric spaces, and let {fj}∞j=1

be a sequence of continuous mappings from M to N . If {fj}∞j=1 convergesuniformly to a mapping f : M → N , then f is continuous too. For let x ∈ Mand ǫ > 0 be given. Since {fj}∞j=1 converges uniformly to f , there is a positiveinteger L such that

ρ(fj(y), f(y)) <ǫ

3for every y ∈ M

when j ≥ L. Because fL is continuous at x, there is a δ > 0 such thatρ(fL(y), fL(x)) < ǫ/3 when y ∈ M and d(y, x) < δ. Therefore,

ρ(f(y), f(x)) ≤ ρ(f(y), fL(y)) + ρ(fL(y), fL(x)) + ρ(fL(x), f(x))

<ǫ

3+

ǫ

3+

ǫ

3= ǫ

when y ∈ M and d(y, x) < δ, as desired. The same argument shows that f isuniformly continuous if the fj ’s are. As another variant, if {xj}∞j=1 is a sequenceof elements of M that converges to x ∈ M , then one can show that {fj(xj)}∞j=1

converges to f(x) in N under these conditions.

26.1 Complex-valued functions

Let E be a set, and let us consider complex-valued functions on E. If a pair ofsequences {fj}∞j=1, {fj}∞j=1 of functions on E converge uniformly to functions f ,

f on E, respectively, then it is easy to see that {fj + fj}∞j=1 converges uniformly

to f + f on E. If {fj}∞j=1 converges uniformly to f on E and c is a complexnumber, then {c fj}∞j=1 converges uniformly to c f on E. If a pair of uniformlybounded sequences of functions on E converge uniformly, then the correspondingsequence of products converges uniformly to the product of limits.

Suppose that {aj}∞j=1 is a sequence of functions on E and {Aj}∞j=1 is asequence of nonnegative real numbers such that |aj(x)| ≤ Aj for every j ≥ 1and x ∈ E. If

∑∞

j=1 Aj converges, then∑∞

j=1 aj(x) converges absolutely forevery x ∈ E by the comparison test. Weierstrass made the nice observation thatthe partial sums of

∑∞

j=1 aj converge uniformly on E under these conditions.The main points are that

∣∣∣∣∞∑

j=1

aj(x) −n∑

j=1

aj(x)

∣∣∣∣ =

∣∣∣∣∞∑

j=n+1

aj(x)

∣∣∣∣ ≤∞∑

j=n+1

|aj(x)| ≤∞∑

j=n+1

Aj

for every x ∈ E and n ≥ 1, and that∑∞

j=n+1 Aj does not depend on x andtends to 0 as n → ∞, by hypothesis.

Let∑∞

l=0 al zl be a power series with complex coefficients. If r is a positive

real number and∑

l=0 |al| rl converges, then Weierstrass’ observation impliesthat

∑∞

l=0 al zl converges uniformly on the closed disk consisting of the z ∈ C

with |z| ≤ r. It follows that∑∞

l=0 al zl defines a continuous function on this

closed disk. If∑∞

l=0 al zl has radius of convergence R > 0, then one can check

32

that∑∞

l=0 al zl defines a continuous function on the open disk consisting of

z ∈ C such that |z| < R. More precisely, one can show that this function iscontinuous at a point z0 ∈ C such that |z0| < R, by applying the previousargument to a positive real number r such that |z0| < r < R.

27 The supremum metric

Let (M,d(x, y)), (N, ρ(u, v)) be metric spaces. A mapping f : M → N is saidto be bounded if f(M) is a bounded set in N . Let Cb(M,N) be the spaceof bounded continuous mappings from M into N . The supremum metric onCb(M,N) is defined by

θ(f1, f2) = sup{ρ(f1(x), f2(x)) : x ∈ M}

for f1, f2 ∈ Cb(M,N). It is easy to see that this is a metric on Cb(M,N), andthat a sequence {fj}∞j=1 of elements of Cb(M,N) converges to f ∈ Cb(M,N) inthe supremum metric if and only if {fj}∞j=1 converges to f uniformly.

Suppose that N is complete, and let us show that Cb(M,N) is completewith respect to the supremum metric. Let {fj}∞j=1 be a Cauchy sequence inCb(M,N), so that for every ǫ > 0 there is a positive integer L(ǫ) such that

θ(fj , fl) < ǫ

when j, l ≥ L(ǫ). In particular, {fj(x)}∞j=1 is a Cauchy sequence in N for everyx ∈ M , which converges to an element f(x) of N since N is complete. One cancheck that

ρ(fj(x), f(x)) ≤ ǫ for every x ∈ M

when j ≥ L(ǫ), which means that {fj}∞j=1 converges uniformly to f and that fis continuous, as desired.

27.1 The supremum norm

Let (M,d(x, y)) be a metric space, and let Cb(M) be the space of boundedcontinuous real-valued functions on M , i.e., Cb(M) = Cb(M,R). Thus Cb(M)is a vector space over the real numbers with respect to the usual operationsof pointwise addition of functions and multiplication of functions by constants,and moreover Cb(M) is a commutative algebra with respect to the operationof pointwise multiplication of functions. The supremum norm of a functionf ∈ Cb(M) is

‖f‖∗ = sup{|f(x)| : x ∈ M},and one can check that

‖f1 + f2‖∗ ≤ ‖f1‖∗ + ‖f2‖∗

and‖f1 f2‖∗ ≤ ‖f1‖∗ ‖f2‖∗

33

for every f1, f2 ∈ Cb(M). It is easy to see that ‖f1 − f2‖∗ is the same as thesupremum metric on Cb(M). Using the triangle inequality, one can check thatfp(x) = d(x, p) is a continuous function on M for every p ∈ M . These functionsare bounded when M is bounded, and otherwise min(fp(x), r) is a boundedcontinuous function on M for every r ≥ 0. This shows that there are alwaysa lot of bounded continuous real-valued functions on any metric space, and inparticular these functions separate elements of M , in the sense that for everyx, y ∈ M with x 6= y there is an f ∈ Cb(M) such that f(x) 6= f(y).

28 The contraction mapping theorem

Let (M,d(x, y)) be a metric space. Suppose that φ : M → M is a strictcontraction in the sense that there is a positive real number c < 1 such that

d(φ(x), φ(y)) ≤ c d(x, y)

for every x, y ∈ M . If x, x′ ∈ M are fixed by φ, which is to say that φ(x) = xand φ(x′) = x′, then d(x, x′) = d(φ(x), φ(x′)) ≤ c d(x, x′), which implies thatd(x, x′) = 0 since c < 1 and hence that x = x′. If M is complete, thenthe contraction mapping theorem states that there is an x ∈ M such thatφ(x) = x. To see this, let z be any point in M , and consider the sequence{zn}∞n=1 of elements of M defined recursively by z1 = z, zn+1 = φ(zn). Thusd(zn+2, zn+1) ≤ c d(zn+1, zn) for each n. By repeating this, we get

d(zn+1, zn) ≤ cn−1 d(z2, z1)

for every n ≥ 1, and hence

d(zn+l, zn) ≤l−1∑

i=0

d(zn+i+1, zn+i) ≤l−1∑

i=0

cn+i−1 d(z2, z1)

≤ cn−1( ∞∑

i=0

ci)

d(z2, z1) =cn−1

1 − cd(z2, z1)

for l, n ≥ 1. This implies that {zn}∞n=1 is a Cauchy sequence in M , whichconverges because M is complete. Moreover,

φ(

limn→∞

zn

)= lim

n→∞φ(zn) = lim

n→∞zn+1 = lim

n→∞zn

since φ is continuous.

29 Limits of functions

Let (M,d(x, y)) and (N, ρ(u, v)) be metric spaces. Suppose that E ⊆ M andthat p ∈ M is a limit point of E. Suppose also that f is a function defined on

34

E with values in N and that z ∈ N . We say that the limit of f(x) as x → pwith x ∈ E is equal to z if for every ǫ > 0 there is a δ > 0 such that

ρ(f(x), z) < ǫ

for every x ∈ E with x 6= p and d(x, p) < δ. Note that p may not be an elementof E, and the value of f at p is not involved in the limit even when p ∈ E. Thelimit is denoted

limx→p

x∈E

f(x)

when it exists. It may also be denoted more simply as

limx→p

f(x)

if E = M or the choice of E is clear from the context.Suppose that E = M . If p is a limit point of M , then

limx→p

f(x) = f(p)

if and only if f is continuous at p. If p is not a limit point of M , then there is aδ > 0 such that x ∈ M and d(x, p) < δ imply that x = p, and f is automaticallycontinuous at p.

29.1 Limits and sequences

Let (M,d(x, y)), (N, ρ(u, v)) be metric spaces, let E be a subset of M , and letp ∈ M be a limit point of E. Also let f be a mapping from E into N , and letz be an element of N . Under these conditions,

limx→p

x∈E

f(x) = z

if and only if {f(xj)}∞j=1 converges in N to z for every sequence {xj}∞j=1 of ele-ments of E that converges to p and satisfies xj 6= p for each j. This is analogousto the characterization of continuity in terms of convergence of sequences.

Suppose that N is the space of real or complex numbers, with the standardmetric. Let f1, f2 be functions on E for which the corresponding limits at pexist. In this case, the limits also exist for the sum f1 + f2 and product f1 f2,and satisfy

limx→p

x∈E

(f1(x) + f2(x)) = limx→p

x∈E

f1(x) + limx→p

x∈E

f2(x)

andlimx→p

x∈E

f1(x) f2(x) =(

limx→p

x∈E

f1(x))(

limx→p

x∈E

f2(x)).

This follows from the characterization of limits in terms of sequences and theanalogous statements for sequences, and it can also be shown directly in thesame way as for sequences.

35

30 One-sided limits

Let a, b be real numbers with a < b, and let f be a real-valued function on theopen interval (a, b). If a ≤ p < b, then the limit of f(x) as x → p with p < x < bmay be denoted

limx→p+

f(x) or f(p+)

when it exists. Similarly, if a < p ≤ b, then the limit of f(x) as x → p witha < x < p may be denoted

limx→p−

f(x) or f(p−)

when it exists. If a < p < b, then

limx→p

f(x)

exists if and only if the one-sided limits exist are equal.Suppose now that f : (a, b) → R is also monotone increasing. In this case,

the one-sided limits exist at each point in (a, b), with

f(p+) = inf{f(x) : p < x < b}

andf(p−) = sup{f(x) : a < x < p}.

In particular,f(p−) ≤ f(p) ≤ f(p+).

If f is bounded from below on (a, b), then f(a+) exists and is equal to theinfimum of f on (a, b), and if f is bounded from above on (a, b), then f(b−)exists and is equal to the supremum of f on (a, b). Note that

f(p+) ≤ f(t) ≤ f(q−)

when a ≤ p < t < q ≤ b.

30.1 Monotone functions

Let f be a monotone increasing real-valued function on an open interval (a, b)in the real line. Thus the one-sided limits f(p+), f(p−) exist at each pointp ∈ (a, b) and satisfy

f(p−) ≤ f(p) ≤ f(p+),

and f is continuous at p if and only if equality holds in both of these inequalities.Equivalently, f is not continuous at p if and only if

f(p−) < f(p+).

36

Let I(p) be the open interval (f(p−), f(p+)) for each p ∈ (a, b) at which f isdiscontinuous. If a < p < q < b, then f(p+) ≤ f(q−), and hence

I(p) ∩ I(q) = ∅

if f is also discontinuous at each of p and q. For each p ∈ (a, b) at which f isdiscontinuous, let r(p) be a rational number contained in I(p). It follows fromthe previous observation that

r(p) < r(q)

when a < p < q < b and f is discontinuous at both p and q. This implies that fcan be discontinuous at only finitely or countably many elements of (a, b), sincethe set of rational numbers is countably infinite.

Part III

Some additional topics

31 Cauchy products

If∑∞

j=0 aj ,∑∞

l=0 bl are two series of complex numbers, then their Cauchy prod-

uct is defined to be the series∑∞

n=0 cn, where

cn =

n∑

j=0

aj bn−j .(31.1)

For instance, if aj = bl = 0 for all but finitely many j, l ≥ 0, then cn = 0 for allbut finitely many n,

∞∑

n=0

cn =( ∞∑

j=0

aj

) ( ∞∑

l=0

bl

).(31.2)

Another reason for this definition is that the Cauchy product of the power series∑∞

j=0 aj zj ,∑∞

l=0 bl zl is the power series

∑∞

n=0 cn zn when the coefficients cn

are as in (31.1).Suppose for the moment that aj , bl are nonnegative real numbers for each

j, l ≥ 0, so that cn is also a nonnegative real number for each n ≥ 0. In thiscase, it is easy to see that

( [N/2]∑

j=0

aj

)( [N/2]∑

l=0

bl

)≤

N∑

n=0

cn ≤( N∑

j=0

aj

)( N∑

l=0

bl

)(31.3)

for every nonnegative integer N . Here [N/2] is the integer part of N/2, whichis the largest integer which is less than or equal to N/2. If

∑∞

j=0 aj ,∑∞

l=0 bl

37

converge, then it follows easily that∑∞

n=0 cn converges too, and that (31.2)holds.

If aj , bl are real or complex numbers such that∑∞

j=0 aj ,∑∞

l=0 bl converge

absolutely, then∑∞

n=0 cn also converges absolutely, and (31.2) holds again. Oneway to see this is to reduce to the previous case by expressing

∑∞

j=0 aj ,∑∞

l=0 bl

as linear combinations of convergent series of nonnegative real numbers. Alter-natively, for each N ≥ 0, it is easy to see that

N∑

n=0

|cn| ≤( N∑

j=0

|aj |) ( N∑

l=0

|bl|).(31.4)

This implies that∑∞

n=0 cn converges absolutely when∑∞

j=0 aj ,∑∞

l=0 bl converge

absolutely. To get (31.2), one can approximate∑∞

j=0 aj ,∑∞

l=0 bl by series withonly finitely many nonzero terms, and use inequalities like (31.4) to estimatethe remainders.

32 The exponential function

Put

E(z) =

∞∑

j=0

zj

j!(32.1)

for each complex number z. Here j! is “j factorial”, equal to the product of thepositive integers from 1 to j when j ≥ 1, and equal to 1 when j = 0. It is easyto see that this series converges absolutely for every z ∈ C, using the ratio test,for instance. It follows that the partial sums for E(z) converge uniformly onbounded subsets of C, and hence that E(z) is a continuous function on C.

If z, w ∈ C, then

E(z)E(w) =

∞∑

n=0

( n∑

j=0

zj

j!

wn−j

(n − j)!

),(32.2)

using Cauchy products, as in the previous section. Remember that

(z + w)n =n∑

j=0

n!

j! (n − j)!zj wn−j ,(32.3)

by the binomial theorem, so that

E(z)E(w) =

∞∑

n=0

(z + w)n

n!= E(z + w).(32.4)

In particular,E(z)E(−z) = E(0) = 1(32.5)

for each z ∈ C, which implies that E(z) 6= 0 and 1/E(z) = E(−z).

38

If x is a real number, then E(x) ∈ R, and E(x) ≥ 1 when x ≥ 0. If x ≤ 0,then 1/E(x) = E(−x) ≥ 1, and hence 0 < E(x) ≤ 1. It is easy to see from theseries expansion that E(x) is strictly increasing for x ≥ 0, and that E(x) → +∞as x → +∞. It follows that E(x) is also strictly increasing for x ≤ 0, and henceon the whole real line, and that E(x) → 0 as x → −∞.

If z is any real number again, then

E(z) =

∞∑

j=0

zj

j!= E(z).(32.6)

Thus|E(z)|2 = E(z)E(z) = E(z)E(z) = E(z + z) = E(2Re z),(32.7)

where Re z denotes the real part of z. Of course, E(2 rez) = E(Re z)2, by (32.4),so that

|E(z)| = E(Re z).(32.8)

33 Diameters of bounded sets

Let (M,d(x, y)) be a metric space. The diameter of a nonempty bounded setE ⊆ M is defined by

diam E = sup{d(p, q) : p, q ∈ E}.(33.1)

If A ⊆ B ⊆ M are nonempty and bounded, then

diam A ≤ diam B.(33.2)

If E ⊆ M is nonempty and bounded, then it is easy to see that the closure Eof E is also bounded, and one can check that

diam E = diam E.(33.3)

Suppose that {Ej}∞j=1 is a sequence of nonempty bounded subsets of M suchthat Ej+1 ⊆ Ej for each j and diamEj → 0 as j → ∞. If xj ∈ Ej for each j,then {xj}∞j=1 is a Cauchy sequence in M . If M is complete, then it follows that

{xj}∞j=1 converges to an element x of M . Note that x ∈ El for each l, sincexj ∈ El when j ≥ l. Conversely, suppose that {xj}∞j=1 is a Cauchy sequence inM , and let El be the set of xj with j ≥ l for each positive integer l. Thus El 6= ∅and El+1 ⊆ El for each l, and one can check that the El’s are bounded subsetsof M with diam El → 0 as l → ∞, because {xj}∞j=1 is a Cauchy sequence. If

x ∈ ⋂∞

l=1 El, then it is easy to see that {xj}∞j=1 converges to x in M .Suppose that φ : M → M is a strict contraction, in the sense that

d(φ(x), φ(y)) ≤ c d(x, y)(33.4)

39

for some nonnegative real number c < 1 and every x, y ∈ M . If E is a nonemptybounded subset of M , then it follows that φ(E) is also nonempty and bounded,and satisfies

diam φ(E) ≤ c diam E.(33.5)

Let φn be the n-fold composition of φ for each positive integer n, so that φ1 = φ,φ2 = φ ◦ φ, and φn+1 = φ ◦ φn = φn ◦ φ for every n ≥ 1. Put En = φn(M) foreach n ≥ 1, and observe that

En+1 = φn(φ(M)) ⊆ φn(M) = En(33.6)

for each n. If M is bounded, then

diam En ≤ cn diam M(33.7)

for each n, and diamEn → 0 as n → ∞ in particular. If M is also complete, thenthere is a point p ∈ ⋂∞

n=1 En, as in the previous paragraph. Note that φ(En) ⊆φ(En) = En+1 for each n, using the continuity of φ in the first step. Henceφ(p) ∈ En+1 for each n as well. This implies that φ(p) = p, since p, φ(p) ∈ En+1

for each n and diamEn+1 = diam En+1 → 0 as n → ∞. This gives another wayto look at the contraction mapping theorem when M is bounded.

34 Compactness and completeness

Let (M,d(x, y)) be a metric space. We have seen that compact subsets of M areclosed and totally bounded. Conversely, if M is complete and K ⊆ M is closedand totally bounded, then K is compact. One can show this in much the sameway as for compactness of closed intervals in the real line, or cells in Rn. Moreprecisely, let {Uα}α∈A be an open covering of K in M , and suppose for the sakeof a contradiction that K cannot be covered by finitely many Uα’s. Because Kis totally bounded, K can be expressed as the union of finitely many subsetswith diameter less than 1. At least one of these subsets of K cannot be coveredby finitely many Uα’s, since otherwise K could be covered by finitely many Uα’s.Thus we get a subset K1 of K with diam K1 < 1 such that K1 cannot be coveredby finitely many Uα’s. Repeating the process, we get a sequence K1,K2, . . . ofsubsets of K such that Kn+1 ⊆ Kn, diam Kn < 1/n, and Kn cannot be coveredby finitely many Uα’s, for each n. In particular, Kn 6= ∅ for each n, and it followsthat there is an x ∈ M such that x ∈ Kn for each n, because M is complete, asin the previous section. We also have that x ∈ K, because Kn ⊆ K for each nand K is closed. Hence x ∈ Uα0

for some α0 ∈ A, since K is covered by the Uα’sby hypothesis. As usual, there is an r > 0 such that B(x, r) ⊆ Uα0

, becauseUα0

is an open set in M , by hypothesis. This implies that

Kn ⊆ B(x, r) ⊆ Uα0(34.1)

when n > 1/r, since x ∈ Kn and diamKn < 1/n, by construction. Of course,(34.1) contradicts the fact that Kn cannot by covered by finitely many Uα’s,from which we conclude that K can be covered by finitely many Uα’s, as desired.

40

34.1 Cauchy subsequences

Alternatively, for any metric space M , a set E ⊆ M is totally bounded if andonly if every sequence of elements of E has a subsequence which is a Cauchysequence. For if E is not totally bounded, then we have seen that there is anǫ > 0 and a sequence {xl}∞l=1 of elements of E such that d(xl, xn) ≥ ǫ for everyl, n such that l 6= n. Every subsequence of this sequence has the same propertyand is therefore not Cauchy. Conversely, suppose that E is totally bounded,and that {xl}∞l=1 is a sequence of elements of E. For every ǫ > 0, E can beexpressed as the union of finitely many subsets of diameter less than ǫ, fromwhich it follows that there is a subsequence of {xl}∞l=1 whose terms are containedin a set of diameter less than ǫ. By repeating this argument, for every positiveinteger n there is a subsequence of {xl}∞l=1 whose terms are contained in a set ofdiameter less than 1/n, and which is a subsequence of the previous subsequencewhen n ≥ 2. The diagonal sequence whose nth term is the nth term of the nthsubsequence is a subsequence of {xl}∞l=1 which is a Cauchy sequence. If M iscomplete and E is totally bounded, then every sequence of elements of E hasa Cauchy subsequence which converges to an element of M , and hence to anelement of E when E is also closed. Thus E is sequentially compact when M iscomplete and E ⊆ M is closed and totally bounded.

35 The Baire category theorem

Let (M,d(x, y)) be a metric space. A set E ⊆ M is dense in M if and only iffor every nonempty open set V ⊆ M , E ∩ V 6= ∅. If E ⊆ M is dense in M andU ⊆ M is a dense open set in M , then one can use this to show that E ∩ U isdense in M . In particular, the intersection of two dense open subsets of M is adense open set in M .

The Baire category theorem states that if M is complete and U1, U2, . . . isa sequence of dense open subsets of M , then the intersection

⋂∞

n=1 Un is densein M . To see this, let p ∈ M and r > 0 be given, and let us show that there isa q ∈ ⋂∞

n=1 Un such that d(p, q) ≤ r. Since U1 ⊆ M is dense and open, thereis a q1 ∈ U1 and an r1 > 0 such that B(q1, r1) ⊆ B(p, r) ∩ U1 and r1 ≤ 1.Similarly, there is a q2 ∈ U2 and an r2 > 0 such that B(q2, r2) ⊆ B(q1, r1) ∩ U2

and r2 ≤ 1/2. By repeating the process, we get for each n ≥ 3 a point qn ∈ Un

and an rn > 0 such that B(qn, rn) ⊆ B(qn−1, rn−1) ∩ Un and rn ≤ 1/n. Byconstruction, {qn}∞n=1 is a Cauchy sequence in M , and hence converges to apoint q in M . Each qn is an element of B(p, r), and thus d(p, q) ≤ r too.Similarly, ql is an element of B(qn, rn) when l ≥ n, which implies that q is anelement of B(qn, rn) and therefore of Un for every n, as desired.

41

36 Diameters of compact sets

Let (M,d(x, y)) be a metric space, and let K be a nonempty compact subset ofM . Also let {pj}∞j=1, {qj}∞j=1 be sequences of elements of K such that

diam K < d(pj , qj) + 1/j(36.1)

for each j. By sequential compactness, there is a strictly increasing sequence{jl}∞l=1 of positive integers such that the subsequence {pjl

}∞l=1 of {pj}∞j=1 con-verges to an element p of K. Similarly, there is a strictly increasing sequenceof positive integers {ln}∞n=1 such that {qjln

}∞n=1 converges to q ∈ K. Since{pjln

}∞n=1, {qjln}∞n=1 both converge to p, q, respectively, it follows that the

diameter of K is equal to d(p, q).Suppose now that K1,K2, . . . is a sequence of nonempty compact subsets

of M such that Kn+1 ⊆ Kn for each n, and put K =⋂∞

n=1 Kn. If {xn}∞n=1

is a sequence of elements of M such that xn ∈ Kn for each n, then the limitof any convergent subsequence of {xj}∞j=1 is an element of K, and K 6= ∅ inparticular. Alternatively, if K is empty, then {M\Kn}n≥1 is an open coveringof K1 for which there is no finite subcovering, a contradiction. Similarly, ifU is an open set in M such that K ⊆ U , then Kn ⊆ U for some n, sinceotherwise the M\Kn’s together with U form an open covering of K1 with nofinite subcovering, a contradiction. Using this, one can show that

limn→∞

diam Kn = diam K(36.2)

under these conditions. Of course,

diam K ≤ diam Kn+1 ≤ diam Kn(36.3)

for each n, since K ⊆ Kn+1 ⊆ Kn, and so the point is to show that diamKn

is not too much larger than diamK when n is sufficiently large. To do this, letǫ > 0 be given, and consider

U(ǫ) =⋃

x∈K

B(x, ǫ/2).(36.4)

This is an open set in M that contains K and satisfies diam U(ǫ) ≤ diam K + ǫ.The previous argument implies that Kn ⊆ U(ǫ) for some n, which implies thatdiam Kn ≤ diam K + ǫ, as desired. One can also get the same conclusion byconsidering sequences of pairs of elements of the Kn’s, and then passing tosuitable subsequences to get corresponding pairs of elements of K.

37 Another fixed-point theorem

Let (M,d(x, y)) be a metric space, and suppose that φ : M → M is a contractionin the sense that

d(φ(x), φ(y)) < d(x, y)(37.1)

42

for every x, y ∈ M with x 6= y. If x, x′ ∈ M , φ(x) = x, φ(x′) = x′, and x 6= x′,then

d(x, x′) = d(φ(x), φ(x′)) < d(x, x′),(37.2)

a contradiction, which is to say that there can be at most one fixed point of φin M . If K ⊆ M is compact, then φ(K) is compact too, since φ is continuous.Hence the diameter of φ(K) is equal to the distance between two of its elements,and one can use this to show that

diam φ(K) < diam K(37.3)

when K has more than one element.If M is compact, then φ has a fixed point in M . A well-known trick for

showing this is to minimize

f(x) = d(φ(x), x).(37.4)

Specifically, there is a p ∈ M such that f(p) ≤ f(x) for every x ∈ M , sincef : M → R is continuous and M is compact. The contractivity property ofφ implies that f(φ(x)) < f(x) when φ(x) 6= x, and consequently φ(p) = p, asdesired.

As another argument, let φn be the n-fold composition of φ for each positiveinteger n, so that φ1 = φ and φn+1 = φ ◦ φn = φn ◦ φ for every n ≥ 1. Considerthe compact sets Kn = φn(M) and K =

⋂∞

n=1 Kn. Observe that

Kn+1 = φn(K1) ⊆ φn(M) = Kn(37.5)

for each n. It follows that K 6= ∅, since the Kn’s are nonempty and compact.Clearly φ(K) ⊆ K, and we would like to show that φ(K) = K. Let x be anyelement of K. For each n, x ∈ Kn+1, which means that

φ−1({x}) ∩ Kn 6= ∅.(37.6)

Note that φ−1({x}) is a closed set in M , because φ is continuous, so thatφ−1({x}) ∩ Kn is compact for each n. This implies that

∞⋂

n=1

(φ−1({x}) ∩ Kn) = φ−1(x) ∩ K 6= ∅,(37.7)

since φ−1({x})∩Kn+1 ⊆ φ−1({x})∩Kn for each n, and thus x ∈ φ(K). Henceφ(K) = K, which implies that K has exactly one element, since otherwisediam φ(K) < diam K, as before. If p is this element of K, then it follows thatφ(p) = p, as desired. An advantage of this argument is that for each r > 0, wealso get that Kn = φn(M) ⊆ B(p, r) for all sufficiently large n.

38 An extension theorem

Let (M,d(x, y)) and (N, ρ(u, v)) be metric spaces, and let E be a dense subsetof M . If f , g are continuous mappings from M into N such that f(x) = g(x)for every x ∈ E, then one can check that f(x) = g(x) for every x ∈ E.

43

Suppose now that f is a uniformly continuous mapping from E into N , andthat N is complete. Under these conditions, there is an extension of f to auniformly continuous mapping from M into N , which is unique by the remarkin the previous paragraph. To see this, let p be an element of M , and let {xj}∞j=1

be a sequence of elements of E that converges to p, which exists because E isdense in M . If p ∈ E, then {f(xj)}∞j=1 converges to f(p) in N , because f iscontinuous at p. Otherwise, {xj}∞j=1 may be considered as a Cauchy sequencein E. Because f : E → N is uniformly continuous, it follows that {f(xj)}∞j=1 isa Cauchy sequence in N . This implies that {f(xj)}∞j=1 converges to an elementof N , since N is supposed to be complete. If {yj}∞j=1 is another sequence ofelements of E that converges to p, then

limj→∞

d(xj , yj) = 0.

Using the uniform continuity of f : E → N again, one can check that

limj→∞

ρ(f(xj), f(yj)) = 0.

This implies that {f(xj)}∞j=1 and {f(yj)}∞j=1 converge to the same element ofN . Thus one can define the extension of f at p to be the limit of {f(xj)}∞j=1,where {xj}∞j=1 is any sequence of elements of E that converges to p in M . Notethat the uniform continuity of f : E → N may be formulated as saying that foreach ǫ > 0 there is a δ = δ(ǫ) > 0 such that

ρ(f(x), f(y)) ≤ ǫ

for every x, y ∈ E such that d(x, y) < δ(ǫ). When formulated this way, onecan check that the extension of f to a mapping from M into N just describedis also uniformly continuous, with the same choice of δ(ǫ) for each ǫ > 0, byapproximating elements of M by elements of E as before.

39 An embedding theorem

Let (M,d(x, y)) be a metric space, and let Cb(M) be the space of boundedcontinuous real-valued functions on M , equipped with the supremum metric.Suppose for the moment that M is bounded, and put

fp(x) = d(p, x)

for each p, x ∈ M . It is easy to see that fp(x) is continuous as a function ofx ∈ M for each p ∈ M , using the triangle inequality. Also, fp(x) is boundedas a function of x ∈ M for each p ∈ M , because M is bounded by hypothesis.Thus p 7→ fp may be considered as a mapping from M into Cb(M). Using thetriangle inequality again, one can check that

|fp(x) − fq(x)| ≤ d(p, q)

44

for every p, q, x ∈ M , and of course equality holds when x = p or q. This impliesthat p 7→ fp is an isometric embedding of M into Cb(M), in the sense that thedistance between fp and fq with respect to the supremum metric is equal tod(p, q) for every p, q ∈ M .

If M is not bounded, then we can modify the construction slightly to getthe same conclusion. Let a be a fixed element of M , and put

fp(x) = fp(x) − fa(x) = d(p, x) − d(a, x)

for every p, x ∈ M . Thus fp(x) is continuous as a function of x ∈ M for eachp ∈ M , since it is the difference of two continuous functions on M . It is easy tosee that fp(x) is bounded on M for each p ∈ M , even when M is unbounded,using the triangle inequality as in the previous paragraph. If p, q ∈ M , then

fp(x) − fq(x) = fp(x) − fq(x)

for each x ∈ M . This permits one to show that the distance between fp and fq

with respect to the supremum metric is equal to d(p, q) for every p, q ∈ M , for

the same reasons as before. Hence p 7→ fp is an isometric embedding of M intoCb(M), even when M is unbounded.

40 Completions

Let (M,d(x, y)) and (M1, d1(u, v)) be metric spaces. As usual, a mapping φ1

from M into M1 is said to be an isometric embedding if

d1(φ1(x), φ1(y)) = d(x, y)

for every x, y ∈ M , in which case it is obviously uniformly continuous. Bya completion of M we mean a complete metric space M1 and an isometricembedding φ1 : M → M1 such that φ1(M) is dense in M1. If φ1(M) is notdense in M1, then one can simply replace M1 with the closure of φ1(M) in M1,since a closed subset of a complete metric space is also complete as a metricspace, using the restriction of the metric from the larger space. In particular,one can get a completion of any metric space M using an isometric embeddingof M into Cb(M), as in the previous section.

If M is already complete, and φ1 is an isometric embedding of M onto a densesubset of a metric space M1, then φ1(M) = M1. For if z is any element of M1,then there is a sequence {φ1(xj)}∞j=1 of elements of φ1(M) that converges to z,because φ1(M) is dense in M1. In particular, {φ1(xj)}∞j=1 is a Cauchy sequencein M1, which implies that {xj}∞j=1 is a Cauchy sequence in M , because φ1 is anisometric embedding. Hence {xj}∞j=1 converges to an element x of M , becauseM is complete. This implies that {φ1(xj)}∞j=1 converges to φ1(x) in M1, so thatz = φ1(x) ∈ φ1(M), as desired.

If {xj}∞j=1, {yj}∞j=1 are Cauchy sequences in M , then one can check that{d(xj , yj)}∞j=1 is a Cauchy sequence in the real line, with respect to the standard

45

metric. This implies that {d(xj , yj)}∞j=1 converges in R, since the real line iscomplete. The limit of {d(xj , yj)}∞j=1 can be used as a measurement of distancebetween the two Cauchy sequences {xj}∞j=1, {yj}∞j=1. This definition of thedistance between two Cauchy sequences satisfies the requirements of a metric,except that the distance between distinct Cauchy sequences may be equal to 0.

Let us say that two Cauchy sequences {xj}∞j=1, {yj}∞j=1 in M are equivalentwhen

limj→∞

d(xj , yj) = 0.

This defines an equivalence relation on the set of all Cauchy sequences in M , andwe let M∗ be the corresponding set of equivalence classes of Cauchy sequencesin M . If {xj}∞j=1, {yj}∞j=1, {x′

j}∞j=1, and {y′j}∞j=1 are Cauchy sequences in M

such that {xj}∞j=1 is equivalent to {x′j}∞j=1 and {yj}∞j=1 is equivalent to {y′

j}∞j=1,then one can check that the distance between {xj}∞j=1 and {yj}∞j=1 mentionedearlier is equal to the distance between {x′

j}∞j=1 and {y′j}∞j=1. It follows that

this distance determines a metric on M∗.If x ∈ M , then the constant sequence {xj}∞j=1 with xj = x for each j is

a Cauchy sequence in M . This defines an embedding of M into M∗, which iseasily seen to preserve distances. It is also easy to see that the image of M inM∗ is dense with respect to the metric on M∗ discussed previously. In addition,one can show that M∗ is complete, and hence a completion of M .

Suppose that (M1, d1) and (M2, d2) are complete metric spaces, and thatφ1 : M → M1 and φ2 : M → M2 are isometric embeddings of M onto densesubsets of M1 and M2, respectively. Consider the mapping ψ from φ1(M) ontoφ2(M) defined by

ψ(φ1(x)) = φ2(x).

This is well-defined, because the isometric embeddings φ1, φ2 are automaticallyone-to-one. This is also an isometry with respect to the restrictions of d1, d2 toφ1(M), φ2(M), since

d2(φ2(x), φ2(y)) = d(x, y) = d1(φ1(x), φ1(y))

for every x, y ∈ M .In particular, ψ : φ1(M) → φ2(M) is uniformly continuous with respect to

the restrictions of d1, d2 to φ1(M), φ2(M). Because φ1(M) is dense in M1

and M2 is complete, there is a unique extension of ψ to a uniformly continuousmapping from M1 into M2. It is easy to see that this extension is also anisometry, and hence that it maps M1 onto M2, since M1 is complete and φ2(M)is dense in M2. This shows that the completion of a metric space is unique upto isometric equivalence in a natural way.

41 Integral metrics

Let f , g be continuous real-valued functions on the unit interval [0, 1], and put

d1(f, g) =

∫ 1

0

|f(x) − g(x)| dx,(41.1)

46

using the standard Riemann integral on the real line. It is easy to check thatthis defines a metric on the space C([0, 1]) of continuous real-valued functionson [0, 1]. Remember that the supremum metric on C([0, 1]) is defined by

d∞(f, g) = sup0≤x≤1

|f(x) − g(x)|,(41.2)

and that C([0, 1]) is complete with respect to this metric. Of course,

d1(f, g) ≤ d∞(f, g)(41.3)

for all continuous functions f , g on [0, 1], and one can give examples whered1(f, g) is much smaller than d∞(f, g). One can also give examples to showthat C([0, 1]) is not complete with respect to d1(f, g).

For instance, let fj(x) be the continuous function defined on [0, 1] by

fj(x) = 0 when 0 ≤ x ≤ 1/2 − 1/(2j),(41.4)

= 1 when 1/2 ≤ x ≤ 1,

and taking fj(x) to be linear between 1/2 − 1/(2j) and 1/2. If

f(x) = 0 when 0 ≤ x < 1/2,(41.5)

= 1 when 1/2 ≤ x ≤ 1,

then it is easy to see that

limj→∞

∫ 1

0

|fj(x) − f(x)| dx = 0.(41.6)

This is basically the same as saying that {fj}∞j=1 converges to f with respect tod1, even though f is not a continuous function on [0, 1]. In particular, {fj}∞j=1 isa Cauchy sequence with respect to d1, but it does not converge to a continuousfunction on [0, 1] with respect to d1. Note that {fj(x)}∞j=1 converges to f(x)pointwise on [0, 1], but not uniformly.

If a sequence {gj}∞j=1 of continuous real-valued functions on [0, 1] convergesuniformly to a continuous function g on [0, 1], then it is easy to see that

limj→∞

∫ 1

0

gj(x) dx =

∫ 1

0

g(x) dx.(41.7)

This is basically because of (41.3), and one can give examples to show that(41.7) may not hold if {gj(x)}∞j=1 only converges to g(x) pointwise on [0, 1]. If{gj(x)}∞j=1 converges to g(x) pointwise on [0, 1] and gj(x) is uniformly boundedfor 0 ≤ x ≤ 1 and j ≥ 1, then Lebesgue’s dominated convergence theoremimplies that (41.7) holds again. Lebesgue’s theory of integration also providesa description of the completion of C([0, 1]) with respect to d1. Another well-known result in this area states that monotone functions on the real line aredifferentiable “almost everywhere”.

47

42 Double sums

If aj,l is a nonnegative real number for every pair of positive integers j, l, thenwe would like to check that

∞∑

j=1

( ∞∑

l=1

aj,l

)=

∞∑

l=1

( ∞∑

j=1

aj,l

).(42.1)

More precisely, if∑∞

l=1 aj,l does not converge to a finite real number for somej, then we interpret the sum as being +∞, and we also interpret the left sideof (42.1) as being +∞. If

∑∞

l=1 aj,l converges for each j, but the sum of thesesums over j does not converge, then again we interpret the left side of (42.1)as being +∞. There are similar interpretations for the right side of (42.1), andpart of the statement is that one side of (42.1) is finite if the other side is finite.

In order to prove (42.1), it suffices to show that

∞∑

j=1

( ∞∑

l=1

aj,l

)≤

∞∑

l=1

( ∞∑

j=1

aj,l

)(42.2)

and∞∑

l=1

( ∞∑

j=1

aj,l

)≤

∞∑

j=1

( ∞∑

l=1

aj,l

).(42.3)

These two inequalities are essentially the same, but with the roles of j and linterchanged, and so it suffices to check the first one. To do this, it is enoughto verify that

N∑

j=1

( ∞∑

l=1

aj,l

)≤

∞∑

l=1

( ∞∑

j=1

aj,l

)(42.4)

for every positive integer N . This follows from

N∑

j=1

( ∞∑

l=1

aj,l

)=

∞∑

l=1

( N∑

j=1

aj,l

)≤

∞∑

l=1

( ∞∑

j=1

aj,l

),(42.5)

by the usual rules for dealing with finite sums of infinite series.If bj,l is a real or complex number for each j, l ∈ Z+, then (42.1) implies that

∞∑

j=1

( ∞∑

l=1

|bj,l|)

=

∞∑

l=1

( ∞∑

j=1

|bj,l|),(42.6)

with aj,l = |bj,l|. In particular, if one side of (42.6) is finite, then the other sideof (42.6) is also finite. In this case,

∑∞

l=1 bj,l converges absolutely for each j andsatisfies ∣∣∣∣

∞∑

l=1

bj,l

∣∣∣∣ ≤∞∑

l=1

|bj,l|,(42.7)

48

and hence∑∞

j=1

(∑∞

l=1 bj,l

)converges absolutely and satisfies

∣∣∣∣∞∑

j=1

( ∞∑

l=1

bj,l

)∣∣∣∣ ≤∞∑

j=1

∣∣∣∣∞∑

l=1

bj,l

∣∣∣∣ ≤∞∑

j=1

∞∑

l=1

|bj,l|.(42.8)

Similarly,∑∞

j=1 bj,l converges absolutely for each l and satisfies

∣∣∣∣∞∑

j=1

bj,l

∣∣∣∣ ≤∞∑

j=1

|bj,l|,(42.9)

so that∑∞

l=1

( ∑∞

j=1 bj,l

)converges absolutely and satisfies

∣∣∣∣∞∑

l=1

( ∞∑

l=1

bj,l

)∣∣∣∣ ≤∞∑

l=1

∣∣∣∣∞∑

j=1

bj,l

∣∣∣∣ ≤∞∑

l=1

∞∑

j=1

|bj,l|.(42.10)

We would like to check that

∞∑

j=1

( ∞∑

l=1

bj,l

)=

∞∑

l=1

( ∞∑

j=1

bj,l

),(42.11)

under these conditions. If bj,l ∈ R for each j, l ∈ Z+, then (42.11) can bederived from (42.1) by taking aj,l to be the positive and negative parts of bj,l.Similarly, the complex case can be obtained from the real case by passing tothe real and imaginary parts of bj,l. Alternatively, (42.11) clearly holds whenbj,l = 0 for all but finitely many (j, l) ∈ Z2

+. Otherwise, if (42.6) is finite, thenfor each ǫ > 0 there is a J(ǫ) ≥ 1 such that

∞∑

j=J(ǫ)+1

( ∞∑

l=1

|bj,l|)

<ǫ

2.(42.12)

One can then get an L(ǫ) ≥ 1 such that

J(ǫ)∑

j=1

( ∞∑

l=L(ǫ)+1

|bj,l|)

<ǫ

2.(42.13)

This permits one to reduce to the case where bj,l = 0 for all but finitely many(j, l) ∈ Z+, by an approximation argument.

43 Rearrangements

Let∑∞

j=1 aj be an infinite series of complex numbers, and let π be a permutationon the set Z+ of positive integers, which is to say a one-to-one mapping from Z+

onto itself. If bj = aπ(j) for each j, then∑∞

j=1 bj is said to be a rearrangement

49

of∑∞

j=1 aj . If∑∞

j=1 aj converges, then one might like to say that∑∞

j=1 bj alsoconverges and has the same sum, but this does not always work. Of course, thisdoes work when aj = 0 for all but finitely many j, and when π(j) = j for allbut finitely many j.

Another case where this works is when aj is a nonnegative real number foreach j. To see this, observe first that

n∑

j=1

bj ≤L(n)∑

l=1

al(43.1)

for n ≥ 1, where L(n) = max(π−1(1), . . . , π−1(n)). If∑∞

l=1 al converges, so thatthe partial sums of

∑∞

l=1 al are bounded, then it follows that the partial sumsof

∑∞

j=1 bj are bounded, and hence that∑∞

j=1 bj converges. This argument alsoshows that

∞∑

j=1

bj ≤∞∑

l=1

al,(43.2)

under these conditions. To get the opposite inequality, observe that

L∑

l=1

al ≤n(L)∑

j=1

bj(43.3)

for each L ≥ 1, where n(L) = max(π(1), . . . , π(L)).If aj are arbitrary real or complex numbers, and

∑∞

l=1 al converges abso-lutely, then

∑∞

j=1 bj also converges absolutely, and satisfies

∞∑

j=1

|bj | =

∞∑

l=1

|aj |,(43.4)

since one can apply the argument in the previous paragraph to |aj |. Of course,one would also like to say that

∑∞

j=1 bj =∑∞

l=1 al. If aj ∈ R for each j, thenthis can be derived from the argument in the previous paragraph, applied to thepositive and negative parts of aj . Otherwise, if the aj ’s are complex numbers,then one can reduce to the real case by considering the real and imaginary partsof aj . Alternatively, one can show that the two infinite sums are the same usingabsolute convergence to approximate them by finite sums that are the same.

If∑∞

j=1 aj is an infinite series of real numbers that converges but does notconverge absolutely, then it can be shown that there are rearrangements of∑∞

j=1 aj that do not converge. It can also be shown that there are rearrange-

ments of∑∞

j=1 that converge with sum equal to any given real number.

44 Sums and limits

Let Ak,l be a real or complex number for each pair of positive integers k, l,and suppose that

∑∞

l=1 Ak,l converges for each k. Suppose also that {Ak,l}∞k=1

50

converges to a real or complex number Al, as appropriate, for each l. One mightlike to say that

∑∞

l=1 Al converges under these conditions, and that

limk→∞

∞∑

l=1

Ak,l =

∞∑

l=1

Al.(44.1)

Of course, we do have that

limk→∞

L∑

l=1

Ak,l =

L∑

l=1

Al(44.2)

for each L ≥ 1, and the problem is basically to interchange the order of thelimits in k and L.

To do this, let us make the additional hypothesis that Bl is a nonnegativereal number for each l ≥ 1 such that

|Ak,l| ≤ Bl(44.3)

for each k, l ≥ 1, and that∑∞

l=1 Bl converges. This implies that

|Al| ≤ Bl(44.4)

for every l ≥ 1 too, and hence that∑∞

l=1 Al converges as well. Of course,

∣∣∣∣∞∑

l=1

Ak,l −∞∑

l=1

Al

∣∣∣∣ =

∣∣∣∣∞∑

l=1

(Ak,l − Al)

∣∣∣∣ ≤∞∑

l=1

|Ak,l − Al|(44.5)

for each k. Let ǫ > 0 be given, and let us check that

∞∑

l=1

|Ak,l − Al| < ǫ(44.6)

for all sufficiently large k. Because∑∞

l=1 Bl converges, there is an L(ǫ) ≥ 1 suchthat

∞∑

l=L(ǫ)+1

Bl =∞∑

l=1

Bl −L(ǫ)∑

l=1

Bl <ǫ

3.(44.7)

Thus

∞∑

l=L(ǫ)+1

|Ak,l − Al| ≤∞∑

l=L(ǫ)+1

(|Ak,l| + |Al|) ≤∞∑

l=L(ǫ)+1

2Bl <2 ǫ

3(44.8)

for each k. It remains to observe that

L(ǫ)∑

l=1

|Ak,l − Al| <ǫ

3(44.9)

51

for all sufficiently large k, which follows from the hypothesis that {Ak,l}∞k=1

converges to Al for each k. This is the analogue of Lebesgue’s dominated con-vergence theorem in the context of infinite series.

Alternatively, let E be the set of real numbers of the form 1/k for k ∈ Z+

together with 0. If l is a positive integer, then let fl be the real or complex-valued function on E defined by fl(1/k) = Ak,l for each k ≥ 1 and fl(0) = Al.Thus the hypothesis that {Ak,l}∞k=1 converges to Al for each l is equivalent tosaying that fl is continuous on E with respect to restriction of the standardmetric on the real line to E for each l. The hypothesis (44.3) then says that|fl| ≤ Bl on E for each l. The convergence of

∑∞

l=1 Bl implies that the partialsums of

∑∞

l=1 fl converge uniformly on E, by the observation of Weierstrass. Itfollows that

∑∞

l=1 fl defines a continuous function on E. Hence

limk→∞

∞∑

l=1

fl(1/k) =∞∑

l=1

fl(0),(44.10)

as before.Suppose now that aj,l is a real or complex number for each pair of positive

integers j, l, and put

Ak,l =

k∑

j=1

aj,l(44.11)

for each k, l ≥ 1. In this case, the previous question about limits of sumsbecomes one of interchanging order of summation, as in Section 42. In the proofof Theorem 8.3 on p175 of [20], Rudin essentially uses the argument describedin the previous paragraph to deal with interchanging the order of summation ofa double sum under suitable conditions of absolute convergence.

44.1 Monotone convergence

Let Ak,l be a nonnegative real number for each pair of positive integers k, l, andsuppose that Ak,l ≤ Ak+1,l for each k, l. This implies that for each l, Ak,l tendsto a nonnegative extended real number Al as k → ∞, which is the same as thesupremum of Ak,l over k as an extended real number. Under these conditions,

∞∑

l=1

Ak,l →∞∑

l=1

Al as k → ∞.(44.12)

This is the analogue of Lebesgue’s monotone convergence theorem for sums.More precisely, if

∑∞

l=1 Ak,l does not converge in the usual sense for some k,then we interpret its sum as being equal to +∞. In this case,

∑∞

l=1 Ak′,l = +∞when k′ ≥ k, by monotonicity,

∑∞

l=1 Al = +∞, and the convergence is trivial.Similarly, if Al0 = +∞ for some l0, then we interpret

∑∞

l=1 Al as being +∞.This means that Ak,l0 → +∞ as k → ∞, so that

∑∞

l=1 Ak,l → +∞ as k → ∞too. Otherwise, we may as well suppose that

∑∞

l=1 Ak,l is finite for every k, and

52

that Al is finite for every l. This still leaves the possibility that∑∞

l=1 Al = +∞,in which case the conclusion is that

∑∞

l=1 Ak,l → +∞ as k → ∞.Of course,

∞∑

l=1

Ak,l ≤∞∑

l=1

Al(44.13)

for each k, because of monotonicity. We also have that

L∑

l=1

Ak,l →L∑

l=1

Al as k → ∞(44.14)

for each L ≥ 1, and that

L∑

l=1

Al →∞∑

l=1

Al as L → ∞.(44.15)

It is easy to complete the proof using these remarks, since∑L

l=1 Ak,l ≤∑∞

l=1 Ak,l

for each k, L.If aj,l is a nonnegative real number for each pair of positive integers j, l, then

Ak,l =∑k

j=1 aj,l is monotone increasing in k. In this case, the preceding ques-tion about limits of sums corresponds to interchanging the order of summationin a double sum of nonnegative real numbers, as in Section 42.

44.2 Fatou’s lemma

If {aj}∞j=1, {bj}∞j=1 are sequences of real numbers, then it is easy to see that

supj≥1

(aj + bj) ≤(

supj≥1

aj

)+

(supj≥1

bj

).(44.16)

More precisely, if {aj}∞j=1 and {bj}∞j=1 have upper bounds in R, then their sumdoes too, and both sides of (44.16) are finite. Otherwise, the right side of (44.16)is interpreted as being +∞, and the inequality is trivial.

Similarly, one can show that

lim supj→∞

(aj + bj) ≤(

lim supj→∞

aj

)+

(lim sup

j→∞

bj

),(44.17)

as long as the right side is well-defined, in the sense that it is not of theform ∞ + (−∞) or (−∞) + ∞. One way to do this is to use the definitionof lim supj→∞(aj + bj) as the supremum of the set of subsequential limits of{aj + bj}∞j=1. By passing to suitable subsequences, it suffices to consider sub-sequences of {aj + bj}∞j=1 for which the corresponding subsequences of {aj}∞j=1

and {bj}∞j=1 also have limits. Alternatively, one can use the fact that if

lim supj→∞

aj < v and lim supj→∞

bj < w(44.18)

53

for some real numbers v, w, then aj < v and bj < w for all but finitely many j.This implies that aj + bj < v + w for all but finitely many j, and hence

lim supj→∞

(aj + bj) ≤ v + w.(44.19)

In the same way,(

lim infj→∞

aj

)+

(lim infj→∞

bj

)≤ lim inf

j→∞(aj + bj),(44.20)

as long as the left side is not of the form ∞ + (−∞) or (−∞) + ∞. This canalso be obtained by applying (44.17) to −aj , −bj . In particular, (44.20) holdswhen aj , bj ≥ 0 for each j, because

0 ≤ lim infj→∞

aj , lim infj→∞

bj ≤ +∞(44.21)

in this case.If aj,l is a nonnegative real number for each pair j, l of positive integers,

then∞∑

l=1

lim infj→∞

aj,l ≤ lim infj→∞

∞∑

l=1

aj,l.(44.22)

This is the analogue of Fatou’s lemma for sums. As usual, the sum of a divergentseries of nonnegative terms is interpreted as being +∞, as well as the sum onthe left when any of its terms is equal to +∞. To get (44.22), observe that

L∑

l=1

lim infj→∞

aj,l ≤ lim infj→∞

L∑

l=1

aj,l(44.23)

for every L ≥ 1, by (44.22). This implies that

L∑

l=1

lim infj→∞

aj,l ≤ lim infj→∞

∞∑

l=1

aj,l,(44.24)

because∑L

l=1 aj,l ≤∑∞

l=1 aj,l for each L, and (44.22) follows from this by takingthe supremum over L.

A Three homework assignments

These homework assignments should be given in the context of Part I, sincethey could be handled quite differently using results from Part II.

A.1 Maximizing distances

Let (M,d(x, y)) be a metric space, let K be a nonempty set contained in M ,and let p be an element of M . If K is compact, then there is a point q ∈ Ksuch that

d(p, x) ≤ d(p, q)

54

for every x ∈ K, which is to say that q maximizes the distance to p amongelements of K. For the homework assignment, one is asked to give two proofsof this statement, one based on the definition of compactness in terms of opencoverings and the other on the limit point property.

A.2 Minimizing distances

Let (M,d(x, y)) be a metric space, let K be a nonempty set contained in M ,and let p be an element of M . If K is compact, then there is a point z ∈ Ksuch that

d(p, z) ≤ d(p, y)

for every y ∈ K, which is to say that z minimizes the distance to p amongelements of K. For the homework assignment, one is asked to give two proofsof this statement, one based on the definition of compactness in terms of opencoverings and the other on the limit point property.

A.3 Positive lower bounds

Let (M,d(x, y)) be a metric space. Suppose that E, K are nonempty subsets ofM such that E is closed, K is compact, and E, K are disjoint, i.e.,

E ∩ K = ∅.

Show that there is a positive real number η such that

d(x, y) ≥ η

for every x ∈ E and y ∈ K.

References

[1] R. Beals, Analysis: An Introduction, Cambridge University Press, 2004.

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[3] A. Browder, Mathematical Analysis: An Introduction, Springer-Verlag,1996.

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[5] T. Gamelin and R. Greene, Introduction to Topology, 2nd edition, Dover,1999.

[6] R. Goldberg, Methods of Real Analysis, 2nd edition, Wiley, 1976.

[7] I. Kaplansky, Set Theory and Metric Spaces, 2nd edition, Chelsea, 1977.

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[8] T. Korner, A Companion to Analysis: A Second First and a First Second

Course in Analysis, American Mathematical Society, 2004.

[9] S. Krantz, The Elements of Advanced Mathematics, 2nd edition, Chapman& Hall / CRC, 2002.

[10] S. Krantz, Real Analysis and Foundations, 2nd edition, Chapman & Hall/ CRC, 2005.

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[12] S. Krantz, A Guide to Topology, Mathematical Association of America,2009.

[13] S. Krantz, Essentials of Topology with Applications, CRC Press, 2010.

[14] S. Lang, Undergraduate Analysis, 2nd edition, Springer-Verlag, 1997.

[15] B. Mendelson, Introduction to Topology, 3rd edition, Dover, 1990.

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the Calculus of Variations, American Mathematical Society, 2005.

[18] M. Reed, Fundamental Ideas of Analysis, Wiley, 1998.

[19] M. Rosenlicht, Introduction to Analysis Dover, 1986.

[20] W. Rudin, Principles of Mathematical Analysis, 3rd edition, McGraw-Hill,1976.

[21] R. Strichartz, The Way of Analysis, revised edition, Jones and Bartlett,2000.

[22] K. Stromberg, Introduction to Classical Real Analysis, Wadsworth, 1981.

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