Davidson K.R., Donsig A.P. Real Analysis with Real Applications

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8.6 Compactness and Subsets of C(K) 243

8.6.9. ARZELA–ASCOLI THEOREM.

Let K be a compact subset of R

. A subset F of C(K, R

) is compact if and only

if it is closed, bounded, and equicontinuous.

PROOF. The easy direction has been established: If F is compact, then it is closed,

bounded and equicontinuous.

So assume F has these three properties and let (f

) be a sequence in F. We

will construct a convergent subsequence. By Corollary 8.6.8, there is a sequence

) so that for each r > 0, there is an integer N so that {x

, . . . , x

} forms an

r-net for K.

We claim that there is a subsequence of (f

), call it (f

), so that

lim

k→∞

) = L

exists for all i ≥ 1.

To prove this, letΛ

denote the set of positive integers. Since

)

is a bounded

sequence, the Bolzano–Weierstrass Theorem (Theorem 2.6.4) provides a conver-

gent subsequence. That is, there is an inﬁnite subset Λ

⊂ Λ

so that

lim

n∈Λ

) = L

exists.

Next,

)

n∈Λ

is bounded sequence, so there is an inﬁnite subset Λ

⊂ Λ

that

lim

n∈Λ

) = L

exists.

Continuing in this way, we obtain a decreasing sequence Λ

⊃ Λ

⊃ ··· of

inﬁnite sets so that

lim

n∈Λ

) = L

converges for each i ≥ 1.

We now use a diagonalization method, similar to the proof that R is uncountable,

Theorem 2.8.7. Let n

be the kth entry of Λ

; and let Λ = {n

: k ∈ N}. Then for

each i, there are at most i − 1 entries of Λ that are not in Λ

. Thus,

lim

k→∞

) = lim

n∈Λ

) = L

for all i ≥ 1,

proving the claim.

For simplicity of notation, we use g

for f

in the remainder of the proof.

Now ﬁx ε > 0. By uniform equicontinuity, there is r > 0 so that

kf(x) − f(y)k <

for all f ∈ F and kx − yk < r.

Choose N so that {x

, . . . , x

} is an r-net for K. Since the g

converge at each of

these N points, there is some integer M so that

) − g

)| ≤

for all k, l ≥ M and 1 ≤ i ≤ N.

244 Limits of Functions

Let k, l ≥ M and pick x ∈ K. Since {x

, . . . , x

} is an r-net for K, there is

some i ≤ N such that kx −x

k < r. We need an ε/3-argument to ﬁnish the proof.

(x) − g

(x)k ≤ kg

(x) − g

)k + kg

) − g

)k + kg

) − g

(x)k

≤

= ε

Thus g

is uniformly Cauchy, and so converges uniformly by Theorem 8.2.2. The

limit g belongs to F because F is closed. Finally, as every sequence in F has a

convergent subsequence, it follows that F is compact. ¥

In particular, this theorem shows that if a sequence of functions (f

) in C[a, b]

forms a bounded equicontinuous subset, then (f

) hasa subsequence that converges

uniformly to some function in C[a, b].

8.6.10. EXAMPLE. Consider the subset K of C[0, 1] consisting of all functions

f ∈ C[0, 1] such that |f(0)| ≤ 5 and f has Lipschitz constant at most 47.

Notice that K is closed. For if f

∈ K converge uniformly to a function f,

then

|f(0)| = lim

n→∞

(0)| ≤ 5,

and

|f(x) − f(y)| = lim

n→∞

(x) − f

(y)| ≤ lim

n→∞

47|x − y| = 47|x − y|.

In particular,

|f(x)| ≤ |f(x) − f(0)|+ |f(0)| ≤ 47 + 5 = 52.

So K is bounded.

Finally, observe that K is equicontinuous. For if ε > 0, take r = ε/47. Then

if |x − y| < r,

|f(x) − f(y)| ≤ 47|x − y| < 47r = ε.

Therefore, this is a compact subset of C[0, 1].

Exercises for Section 8.6

A. Use f

(x) = x

on [0, 1] to show that B = {f ∈ C[0, 1] : kf k ≤ 1} is not compact.

B. Show that U = {f ∈ C[0, 1] : f (x) > 0 for all x ∈ [0, 1]} is open.

HINT: You need the Extreme Value Theorem.

C. What is the interior of V = {f ∈ C

(R) : f(x) > 0 for all x ∈ R}?

HINT: Compare with the previous exercise.

D. Prove that the family {sin(nx) : n ≥ 1} is not an equicontinuous subset of C[0, π].

E. (a) Show that

F =

F (x) =

f(t) dt : f ∈ C[0, 1], kfk

∞

≤ 1

is a bounded and equicontinuous subset of C[0, 1].

(b) Why is F not closed?

8.6 Compactness and Subsets of C(K) 245

such that f(0) = 0.

HINT: Construct F

in F so that F

−n

k) =

n+1

f(2

−n

k) for 0 ≤ k ≤ 2

F. (a) Let F be a subset of C[0, 1] that is closed, bounded, and equicontinuous. Prove

that there is a function g ∈ F such that

g(x) dx ≥

f(x) dx for all f ∈ F.

(b) Construct a closed bounded subset F of C[0, 1] for which the conclusion of the

previous problem is false.

G. Let K be a compact subset of R

. Show that a subset S of C(K) is compact if and

only if it is closed and totally bounded.

HINT: Show that totally bounded sets are bounded and equicontinuous.

H. Let F be an equicontinuous family of functions in C(K), where K is a compact subset

of R

. Suppose that for each x ∈ K,

sup{f(x) : f ∈ F} = M

is ﬁnite. Prove that F is bounded.

HINT: Use equicontinuity to bound F by M

+ 1 on a ball about x. Suppose that

)| tends to +∞. Extract a convergent subsequence of {x

I. Let F be a family of continuous functions deﬁned on R that is (i) equicontinuous and

satisﬁes (ii) sup{f(x) : f ∈ F} = M

< ∞ for every x. Show that every sequence

)

∞

n=1

has a subsequence that converges uniformly on [−k, k] for every k > 0.

HINT: Find a subsequence (f

1,n

)

∞

n=1

that converges uniformly on [−1, 1]. Then extract

a subsubsequence (f

2,n

)

∞

n=1

that converges uniformly on [−2, 2], and so on. Now use

a diagonal argument.

CHAPTER 9

Metric Spaces

This text focuses on subsets of a normed space, as this is the natural setting for

most of our applications. In this chapter, we introduce an apparently more general

framework, metric spaces, and some new ideas that are somewhat more advanced.

They play an occasional role in the advanced sections of the applications.

9.1. Deﬁnitions and Examples

In a normed vector space, the distance between elements is found using the

norm of the difference. However, a distance function can be deﬁned abstractly on

any set using the idea of a metric. Most of the arguments we used in the normed

context will also work for metric spaces, with only minimal changes. The crucial

difference is that in a metric space, we do not work in a vector space, so we cannot

use the addition or scalar multiplication.

9.1.1. DEFINITION. Let X be a set. A metric on a set X is a function ρ deﬁned

on X × X taking values in [0, ∞) with the following properties:

(1) (positive deﬁniteness) ρ(x, y) = 0 if and only if x = y,

(2) (symmetry) ρ(x, y) = ρ(y, x) for all x, y ∈ X,

(3) (triangle inequality) ρ(x, z) ≤ ρ(x, y) + ρ(y, z) for all x, y, z ∈ X.

A metric space is a set X with a metric ρ, denoted (X, ρ). If the metric is under-

stood, we use X alone.

9.1.2. EXAMPLES.

(1) If X is a subset of a normed space V , deﬁne ρ(x, y) = kx − yk. This is our

standard example.

(2) Put a metric on the surface of the sphere by setting ρ(x, y) to be the length of

the shortest path from x to y (known as a geodesic). This is the length of the shorter

arc of the great circle passing through x and y. More generally, we can deﬁne such

a metric on any smooth surface.

246

9.1 Deﬁnitions and Examples 247

(3) The discrete metric on a set X is given by

d(x, y) =

(

0 if x = y

1 if x 6= y.

(4) Deﬁne a metric on Z by ρ

(n, n) = 0 and ρ

(m, n) = 2

−d

, where d is the

largest power of 2 dividing m − n. It is trivial to verify properties (1) and (2). If

(l, m) = 2

−d

and ρ

(m, n) = 2

−e

, then 2

min{d,e}

divides l − n and so

ρ(l, n) ≤ 2

−min{d,e}

= max{ρ

(l, m), ρ

(m, n)}.

This metric is known as the 2-adic metric. Replacing 2 with another prime p, we

can deﬁne similarly the p-adic metric.

(5) If X is a closed subset of R

, let K(X) denote the collection of all nonempty

compact subsets of X. If A is a compact subset of X and x ∈ X, we deﬁne

dist(x, A) = inf

a∈A

kx − ak.

Then we deﬁne the Hausdorff metric on K(X) by

(A, B) = max{sup

a∈A

dist(a, B), sup

b∈B

dist(b, A)}

= max{sup

a∈A

inf

b∈B

ka − bk, sup

b∈B

inf

a∈A

ka − bk}.

Since A is closed, dist(x, A) = 0 if and only if x ∈ A. In particular, we see

that d

(A, B) = 0 if and only if A = B. So d

is positive deﬁnite and is evidently

symmetric. For the triangle inequality, let A, B, C be three compact subsets of X.

For each a ∈ A, the Extreme Value Theorem yields the existence of a closest point

b ∈ B, so ka − bk = dist(a, B). Then there is a closest point c ∈ C to b with

kb − ck = dist(b, C). Therefore,

dist(a, C) ≤ ka − ck ≤ ka − bk + kb − ck

= dist(a, B) + dist(b, C)

≤ d

(A, B) + d

(B, C).

Therefore, sup

a∈A

dist(a, C) ≤ d

(A, B)+d

(B, C). Reversing the roles of A and C

and combining the two inequalities, we obtain d

(A, C) ≤ d

(A, B)+d

(B, C).

Let A

:= {x ∈ R

: dist(x, A) ≤ ε}. Note that d

(A, B) ≤ ε if and only if

A ⊂ B

and B ⊂ A

This example will be examined in more detail in Section 11.7, when we study

fractal sets.

The notions of convergence and open set can be carried over to metric spaces

by replacing kx − yk with ρ(x, y). Once this is done, most of the other deﬁnitions

do not need to be changed at all.

248 Metric Spaces

9.1.3. DEFINITION. The ball B

(x) of radius r > 0 about a point x is deﬁned

as {y ∈ X : ρ(x, y) < r}. We write B

(x) if the metric is ambiguous. A subset U

is open if for every x ∈ U, there is an r > 0 so that B

(x) is contained in U .

A sequence (x

) is said to converge to x if lim

n→∞

ρ(x, x

) = 0. A set C is

closed if it contains all limit points of sequences of points in C.

If X is a subset of a normed space, then these deﬁnitions agree with our old

deﬁnitions; so the language is consistent. It is easy to see that a set is open precisely

when the complement is closed (adapt the proof of Theorem 4.3.8 for normed vec-

tor spaces).

9.1.4. DEFINITION. A sequence (x

)

∞

n=1

in a metric space (X, ρ) is a Cauchy

sequence if for every ε > 0, there is an integer N so that ρ(x

, x

) < ε for all

i, j ≥ N.

A metric space X is complete if every Cauchy sequence converges (in X).

9.1.5. EXAMPLES.

(1) As in Proposition 2.7.1, every convergent sequence is Cauchy.

(2) It is easy to show that a subset of a complete metric space is complete if and

only if it is closed. So one useful way to construct many complete metric spaces is

to take a closed subset of a complete normed space. The purpose of Exercise 9.1.L

is to show that every metric space arises in this manner, although it is not always a

natural context.

(3) If X has the discrete metric, the only way a sequence may converge to x is if

the sequence is eventually constant (i.e., x

= x for all n ≥ N ). This is because

the ball B

1/2

(x) = {x}. So every subset of X is both open and closed. Also X is

complete.

(4) Consider the 2-adic metric of Example 9.1.2(4) again. The balls have the form

−d

(n) = {m ∈ Z : m ≡ n (mod 2

)}. The sequence (2

)

∞

n=1

converges to 0

because ρ

, 0) = 2

−n

→ 0. Observe that 1 − (−2)

is an odd multiple of 3 for

all n ≥ 1. The sequence a

= (1 − (−2)

)/3 is Cauchy because if n > m ≥ N,

then a

− a

= (−2)

n−m

and therefore ρ

, a

) = 2

−m

≤ 2

−N

. This

sequence does not converge.

(5) When X is a closed subset of R

, the metric space (K(X), d

) is complete.

We will prove this in Theorem 11.7.2.

Continuous functions are deﬁned by analogy with the norm case.

9.1.6. DEFINITION. A function f from a metric space (X, ρ) into a metric

space (Y, d) is continuous if for every x

∈ X and ε > 0, there is a δ > 0 so that

d(f(x), f(x

)) < ε whenever ρ(x, x

) < δ.

9.1 Deﬁnitions and Examples 249

The proof of Theorem 5.3.1 goes through without change.

9.1.7. THEOREM. Let f map a metric space (X, ρ) into (Y, σ). The following

are equivalent:

(1) f is continuous on X;

(2) for every convergent sequence (x

)

∞

n=1

with lim

n→∞

= a in X, we have

lim

n→∞

f(x

) = f (a); and

(3) f

−1

(U) = {x ∈ X : f (x) ∈ U} is open in X for every open set U in Y .

As in the norm case, if the domain of a continuous function is not compact, the

function need not be bounded. However, we have the same solution: Consider the

normed vector space C

(X) of all bounded continuous functions f : X → R with

the sup norm kfk

∞

= sup

x∈X

|f(x)|.

9.1.8. THEOREM. The space C

(X) of all bounded continuous functions on a

metric space X with the sup norm kfk = sup{|f(x)| : x ∈ X} is complete.

PROOF. The proof of Theorem 8.2.1 works with S replaced by any metric space.

Likewise, Theorem 8.2.2 works with C

(X) because compactness is used only to

ensure that the sup norm is ﬁnite. Therefore, C

(X) is a complete normed vector

space. The details are left as an exercise. ¥

Exercises for Section 9.1

A. Show that ρ(x, y) = |e

− e

| is a metric on R.

B. Show that every subset of a discrete metric space is both open and closed.

C. Prove that U is open in (X, ρ) if and only if X \ U is closed.

D. Prove Theorem 9.1.7.

E. Given a metric space (X, ρ), deﬁne a new metric on X by σ(x, y) = min{ρ(x, y), 1}.

(a) Show that σ is a metric on X. Observe that X has ﬁnite diameter in the σ metric.

(b) Show that lim

n→∞

= x in (X, ρ) if and only if lim

n→∞

= x in (X, σ).

) is Cauchy in (X, ρ) if and only if it is Cauchy in (X, σ). Hence

completeness is the same for these two metrics.

F. Suppose that V is a normed vector space. If (X, ρ) is a metric space, observe that the

space C

(X, V ) of all bounded continuous functions from X to V is a vector space.

Show that kfk

∞

= sup

x∈X

kf(x)k is a norm on C

(X, V ).

G. Two metrics ρ and σ on a set X are topologically equivalent if for each x ∈ X and

r > 0, there is an s = s(r, x) > 0 so that B

(x) ⊂ B

(x) and B

(x) ⊂ B

(x).

(a) Prove that topologically equivalent metrics have the same open and closed sets.

(b) Prove that topologically equivalent metrics have the same convergent sequences.

250 Metric Spaces

H. Deﬁne a function on M

×M

by ρ(A, B) = rank(A −B). Prove that ρ is a metric

that is topologically equivalent to the discrete metric.

I. Put a metric ρ on all the words in a dictionary by deﬁning the distance between two

distinct words to be 2

−n

if the words agree for the ﬁrst n letters and are different at

the (n+1)st letter. Here we agree that a space is distinct from a letter. For example,

ρ(car, cart) = 2

−3

and ρ(car, call) = 2

−2

(a) Verify that this is a metric.

(b) Suppose that words w

, w

and w

are listed in alphabetical order. Show that

ρ(w

, w

) ≤ ρ(w

, w

and w

are listed in alphabetical order. Find a formula

for ρ(w

, w

) in terms of ρ(w

, w

) and ρ(w

, w

J. Recall the 2-adic metric of Examples 9.1.2(4) and 9.1.5(4). Extend it to Q by setting

(a/b, a/b) = 0 and, if a/b 6= c/d, then ρ

(a/b, c/d) = 2

−e

, where e is the unique

integer such that a/b − c/d = 2

(f/g) and both f and g are odd integers.

(a) Prove that ρ

is a metric on Q.

(b) Show that the sequence of integers a

= (1 − (−2)

)/3 converges in (Q, ρ

n! + 1

in this metric.

K. Complete the details of Theorem 9.1.8 as follows:

(a) Prove that Theorem 8.2.1 is valid when S is replaced by a metric space X.

(b) Prove that C

(X) is a complete normed vector space. HINT: Theorem 8.2.2

L. Suppose that (X, ρ) is a nonempty metric space. Let C

(X) be the normed space of all

bounded continuous functions on X with the sup norm kfk

∞

= sup{|f(x)| : x ∈ X}.

(a) Fix x

in X. For each x ∈ X, deﬁne f

(y) = ρ(x, y) − ρ(x

, y) for y ∈ X. Show

that f

is a bounded continuous function on X.

(b) Show kf

− f

∞

= ρ(x, y).

identiﬁes X with a

subset F of C

(X) that induces the same metric.

M. (a) Give an example of a decreasing sequence of closed balls in a complete metric

space with empty intersection. Compare with Exercise 7.2.J.

HINT: Use a metric on N topologically equivalent to the discrete metric so that

{n ≥ k} are closed balls.

(b) Show that a metric space (M, d) is complete if and only if every decreasing se-

quence of closed balls with radii going to zero has a nonempty intersection.

9.2. Compact Metric Spaces

As we have mentioned in the previous chapter, in general, a closed and bounded

set need not be compact. There are several useful properties that are equivalent to

compactness. In this section, we deﬁne these properties and prove they are equiva-

lent.

We have to change our language slightly. Our old notion of compactness will

be renamed sequential compactness. Although we introduce a new notion and

call it compactness, we will prove that the two notions are equivalent in normed

spaces and in metric spaces. There is a more general setting, topological spaces,

9.2 Compact Metric Spaces 251

where compactness and sequential compactness differ, but we never deal with it in

this book. As you will immediately recognize, the new notion also makes sense in

normed spaces and could have been introduced there.

9.2.1. DEFINITION. A collection of open sets {U

: α ∈ A} in X is called an

open cover of Y ⊂ X if Y ⊂

α∈A

. A subcover of {U

: α ∈ A} is just a

subcollection {U

: α ∈ B} for some B ⊂ A that is still a cover. In particular, it is

a ﬁnite subcover of Y if it is a ﬁnite collection of open subsets that covers Y .

A collection of closed sets {C

: α ∈ A} has the ﬁnite intersection property

if every ﬁnite subcollection has nonempty intersection.

For example, consider the (ﬁnite) cover indicated in Figure 9.1.

FIGURE 9.1. An open cover for X = [a, b].

9.2.2. DEFINITION. A metric space X is compact if every open cover of X

has a ﬁnite subcover.

A metric space X is sequentially compact if every sequence of points in X

has a convergent subsequence.

A metric space X is totally bounded if for every ε > 0, there are ﬁnitely many

points x

, . . . , x

∈ X so that {B

) : 1 ≤ i ≤ k} is an open cover.

9.2.3. BOREL–LEBESGUE THEOREM.

For a metric space X, the following are equivalent.

(1) X is compact.

(2) Every collection of closed subsets of X with the ﬁnite intersection property

has nonempty intersection.

(3) X is sequentially compact.

(4) X is complete and totally bounded.

PROOF. First assume (1) that X is compact, and let us prove (2). Suppose that

: α ∈ A} is a collection of closed sets such that

= ∅. Consider the

open sets U

= C

, the complements of C

. Then

= X.

Thus there is a ﬁnite subcover X = U

∪ ··· ∪ U

. Consequently,

∩ ··· ∩ C

∪ ··· ∪ U

= ∅.

So no collection of closed sets with empty intersection has the ﬁnite intersection

property.

Now assume (2) and let (x

) be a sequence in X. Deﬁne C

: i ≥ n}.

This is a decreasing sequence of closed sets. The collection {C

: n ≥ 1} has the

252 Metric Spaces

ﬁnite intersection property since the intersection of C

, . . . , C

contains the point

where n = max{n

, . . . , n

}. By hypothesis, there is a point x in

n≥1

Recursively deﬁne a sequence as follows. Let n

= 1. If n

k−1

is deﬁned, use the

fact that x ∈

: i > n

k−1

} to choose n

> n

k−1

with ρ(x, x

) < 1/k. By

construction, lim

k→∞

= x converges.

Assume sequential compactness (3), and consider (4). If (x

) is a Cauchy se-

quence, select a convergent subsequence, say lim

k→∞

= x. Given ε > 0, use the

Cauchy property to ﬁnd N so that i, j ≥ N implies that ρ(x

, x

) < ε/2. Then use

the convergence to ﬁnd n

> N so that ρ(x, x

) < ε/2. Then if i ≥ N,

ρ(x, x

) ≤ ρ(x, x

) + ρ(x

, x

) <

= ε.

So lim

i→∞

= x. Thus X is complete.

If X were not totally bounded, then there would be some ε > 0 so that no

ﬁnite collection of ε-balls can cover X. Recursively select points x

∈ X so that

/∈ B

) ∪ ··· ∪ B

k−1

) for all k ≥ 2. Consider the sequence (x

). We will

obtain a contradiction by showing that there is no convergent subsequence. Indeed,

if (x

) were a convergent subsequence, then it is Cauchy. So for some N large, all

i > N satisfy ρ(x

, x

) < ε, contrary to the fact that x

/∈ B

). Therefore, X

must be totally bounded.

Finally, we show that (4) implies (1). For each k ≥ 1, choose a ﬁnite set

, . . . x

so that B

1/k

) for 1 ≤ i ≤ n

covers X. Suppose that there is an

open cover U = {U

: α ∈ A} of X with no ﬁnite subcover. We will recursively

deﬁne a sequence of points y

= x

so that

j=1

1/j

) does not have a ﬁnite

subcover from U. At the stage k = 0, X does not have a ﬁnite subcover. Suppose

that we have this property for k − 1. If each

k−1

j=1

1/j

) ∩ B

1/k

) had a

ﬁnite subcover from U, then combining them for 1 ≤ i ≤ n

would yield a ﬁnite

subcover of

k−1

j=1

1/j

), contrary to fact. So we may pick y

= x

so that

j=1

1/j

) has no ﬁnite subcover.

Now we show that this leads to a contradiction. The sequence (y

) is Cauchy.

To see this, note that

k−1

j=1

1/j

) is nonempty because the empty set is covered.

Thus there is a point x ∈ B

1/j

) ∩ B

1/k

). Hence

ρ(y

, y

) ≤ ρ(y

, x) + ρ(x, y

) ≤

Given ε > 0, choose N so large that N > 2/ε. Then for j, k ≥ N, it follows that

ρ(y

, y

) ≤ 2/N < ε.

But (4) states that X is complete, and thus y = lim

k→∞

exists. Hence there

is some α so that U

contains y. Therefore, there is an ε > 0 so that B

(y) is

contained in U

. Choose any k > 2/ε for which ρ(y

, y) < ε/2. Then since

1/k < ε/2,

1/k

) ⊂ B

(y) ⊂ U

. This contradicts the fact that B

1/k

) does

not have a ﬁnite subcover. Thus it must be the case that X is compact. ¥

In general, being complete and bounded is not sufﬁcient to imply that a metric

space is compact. For example, by Exercise 9.1.E the real line has a bounded metric