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

Подождите немного. Документ загружается.

9.2 Compact Metric Spaces 253

topologically equivalent to the usual one. In this new metric it is bounded and

complete but is not compact.

Perhaps a more natural example is the closed unit ball B of C[0, 1] in the max

norm. This is complete because it is a closed subset of a complete space. However,

it is not totally bounded. To see this, let f

be the piecewise linear functions that

take the value 0 on [0, 1/(n + 1)] ∪[1/(n −1), 1] and 1 at 1/n. Since kf

∞

= 1,

theyall lie in B. But kf

−f

∞

= 1 if n 6= m. Thus no 1/2-ball can contain more

than one of them. Hence no ﬁnite family of 1/2-balls covers B. We already have a

complete description of the compact subsets of C[0, 1], thanks to the Arzela–Ascoli

Theorem (Theorem 8.6.9).

All of our basic theorems on continuous functions go through for continuous

functions on metric spaces. In particular, Theorem 5.4.3 can be established with

the same proof. We will give a proof based on open covers. In Exercise 9.2.H, this

will yield the Extreme Value Theorem.

9.2.4. THEOREM. Let C be a compact subset of a metric space (X, ρ). Sup-

pose that f is a continuous function of X into (Y, σ). Then the image set f(C) is

compact.

PROOF. Let U = {U

: α ∈ A} be an open cover of f(C) in Y . Since f is

continuous, V

:= f

−1

) are open sets in X. The collection V = {V

: α ∈ A}

is an open cover of C. Indeed, for each x ∈ C, f(x) ∈ f(C) and thus f(x) ∈ U

for some α. Hence x belongs to V

. By the compactness of C, select a ﬁnite

subcover V

, . . . , V

. Then

f(C) ⊂

[

i=1

f(V

) ⊂

[

i=1

So U

, . . . , U

is a ﬁnite subcover of f(C). Therefore, f(C) is compact. ¥

Exercises for Section 9.2

A. Let (X, ρ) be a metric space and let a subset (Y, ρ) be considered as a metric space

with the same metric.

(a) Show that every open set U in Y has the form V ∩ Y for some open set V in X.

HINT: This is easy for balls.

(b) Show that Y is compact if and only if every collection {V

: α ∈ A} of open sets

in X that covers Y has a ﬁnite subcover.

B. Show that if Y is a subset of a complete metric space X, then Y is compact if and only

if it is closed and totally bounded.

C. Show that a closed subset of a compact metric space is compact.

D. Show that every compact metric space is separable (i.e., it has a countable dense

subset).

E. (a) Prove that every open subset U of R

is the countable union of compact subsets.

HINT: Use the distance to U

and the norm to deﬁne the sets.

(b) Show that every open cover of an open subset of R

has a countable subcover.

254 Metric Spaces

F. Prove Cantor’s Intersection Theorem: A decreasing sequence of nonempty compact

subsets A

⊃ A

⊃ ··· of a metric space (X, ρ) has nonempty intersection.

G. Show that a continuous function from a compact metric space (X, ρ) into a metric

space (Y, d) is uniformly continuous. HINT: Fix ε > 0. For x ∈ X, choose δ

> 0

so that ρ(x, t) < 2δ

implies d(f(x), f (t)) < ε/2. Then {B

(x):x ∈ X} covers X.

H. If f is a continuous function from a compact metric space (X, ρ) into R, prove that

there is a point x

∈ X such that |f (x

)| = sup{|f(x)| : x ∈ X}.

HINT: Compare to the proof of the Extreme Value Theorem.

I. Let S

for n ≥ 1 be a ﬁnite union of disjoint closed balls in R

of radius at most 2

−n

such that S

n+1

⊂ S

and S

n+1

has at least two balls inside each ball of S

. Prove that

C =

n≥1

is a perfect, nowhere dense compact subset of R

HINT: Compare with Example 4.4.8.

J. If f is a continuous one-to-one function of a compact metric space X onto Y , show

that f

−1

is continuous. HINT: Theorem 9.2.4

K. Show that the previous exercise is false if X is not compact.

HINT: Map (0, 1] onto a circle with a tail (i.e., the ﬁgure 6).

L. We say (X, ρ) is a second countable metric space if there is a countable collection

U of open balls in X so that for every x ∈ X and r > 0, there is a ball U ∈ U with

x ∈ U ⊂ B

(x). Prove that (X, ρ) is second countable if and only if it is separable.

HINT: For ⇒, take the centres of balls in U. For ⇐, take all balls of radius 1/k, k ≥ 1,

about each point in a countable dense set.

9.3. Complete Metric Spaces

Now we turn to an important consequence of completeness. Some more termi-

nology is required.

9.3.1. DEFINITION. A subset A of a metric space (X, ρ) is nowhere dense

if int(

A) = ∅ (i.e., the closure of A has no interior). A subset B is said to be

ﬁrst category if it is the countable union of nowhere dense sets. A subset Y of a

complete metric space is a residual set if the complement Y

is ﬁrst category.

Nowhere dense sets are small in a certain sense, and thus sets of ﬁrst category

are also considered to be small. For example, the Cantor set is nowhere dense in R.

Our main result, which has surprisingly powerful consequences, is that complete

metric spaces are never ﬁrst category.

9.3.2. BAIRE CATEGORY THEOREM.

Let (X, ρ) be a complete metric space. Then the union of countably many nowhere

dense subsets of X has no interior, and in particular is a proper subset of X.

Equivalently, the intersection of countably many dense open subsets of X is dense

in X.

9.3 Complete Metric Spaces 255

PROOF. Consider a sequence (A

)

∞

n=1

of nowhere dense subsets of X. To show

that the complement of

n≥1

is dense X, take any ball

). We will con-

struct a point in this ball that is not in any

. It will then follow that this union has

no interior, whence the complement contains points arbitrarily close to each point

x ∈ X, and so is dense.

As A

has no interior, it does not contain B

). Pick x

in B

) \A

Since A

is closed, dist(x

, A

) > 0. So we may choose an 0 < r

< r

/2 so that

) is disjoint from

. Note that B

) ⊂

). Proceed recursively

choosing a point x

n+1

∈ B

) and an r

n+1

∈ (0, r

/2) so that

n+1

)

is disjoint from

n+1

. Clearly, B

n+1

) ⊂

The sequence (x

)

∞

n=1

is Cauchy. Indeed, given ε > 0, choose N so that

−N

< ε/2. Then r

< ε/2. However, all x

for n ≥ N lie in

). Thus

for n, m ≥ N ,

ρ(x

, x

) ≤ ρ(x

, x

) + ρ(x

, x

) <

= ε.

Because X is a complete space, there is a limit x

∞

= lim

n→∞

in X. The point x

∞

belongs to

n≥1

). Hence it is disjoint from every

for n ≥ 1.

If U

are dense open subsets of X, their complements A

are closed and

nowhere dense. By the previous paragraphs,

n≥1

has dense complement. This

complement is exactly

n≥1

. ¥

Here is one interesting and unexpected consequence. It says that most continu-

ous functions are nowhere differentiable. This is rather nonintuitive, as it was hard

work to construct even one explicit example of such a function in Example 8.4.9.

9.3.3. PROPOSITION. The set of continuous, nowhere differentiable functions

on an interval [a, b] is a residual set and in particular is dense in C[a, b].

PROOF. Say that a function f is Lipschitz at x

if there is a constant L so that

|f(x) − f(x

)| ≤ L|x − x

| for all x ∈ [a, b]. Our ﬁrst observation is that if f is

differentiable at x

, then it is also Lipschitz at x

. From the deﬁnition of derivative,

lim

x→x

f(x) − f(x

)

x − x

= f

). Choose δ > 0 so that for |x − x

| < δ,

|f(x) − f(x

|x − x

≤ |f

)| + 1.

Then for |x − x

| < δ, |f(x) − f(x

)| ≤ (|f

)| + 1)|x − x

|. If |x − x

| ≥ δ,

|f(x) − f(x

)| ≤ 2kf k

∞

≤

2kfk

∞

|x − x

So L = max

)| + 1,

2kfk

∞

is a Lipschitz constant at x

Let A

consist of all functions f ∈ C[a, b] such that f has Lipschitz constant

L ≤ n at some point x

∈ [a, b]. Let us show that A

is closed. Suppose that

256 Metric Spaces

)

∞

k=1

is a sequence of functions in A

converging uniformly to a function f.

Each f

has Lipschitz constant n at some point x

∈ [a, b]. Since [a, b] is compact,

there is a subsequence x

converging to a point x

∈ [a, b]. Then

|f(x) − f(x

)| ≤ |f (x) − f

(x)| + |f

(x) − f

+ |f

) − f

)| + |f

) − f(x

≤ kf − f

∞

+ n|x − x

| + n|x

− x

| + kf − f

∞

= 2kf − f

∞

+ n

|x − x

| + |x

− x

Take a limit as i → ∞ to obtain |f(x) − f(x

)| ≤ n|x − x

|. Thus A

is closed.

Next we show that A

has no interior and hence is nowhere dense. Fix a

function f ∈ A

and an ε > 0. Let us look for a function g in B

(f) that is not in

. By Theorem 5.5.9, f is uniformly continuous. Choose δ > 0 so that |x−y| < δ

implies that |f(x)−f(y)| < ε/4. Construct a piecewise linear continuous function

h which agrees with f on a sequence a = x

< x

< ··· < x

= b, where

− x

k−1

< δ for 1 ≤ k ≤ N . A simple estimate shows that kf − hk

∞

< ε/2.

The function h is Lipschitz with constant L, say. Choose M > 4π(L + n)/ε and

set g = h +

sinMx. Then

kf − gk

∞

≤ kf − hk

∞

+ kh − gk

∞

< ε.

To see that g is not in A

, take any point x

. We can alwayschoose a point x ∈ [a, b]

with |x − x

| < 2π/M so that sinMx = ±1 has sign opposite to the sign of

sinMx

. Therefore,

|g(x) − g(x

)| ≥

|sinMx − sinMx

| − |h(x) − h(x

≥

− L|x − x

| ≥

Mε

4π

− L

|x − x

| > n|x − x

So g does not have Lipschitz constant n at any point x

∈ [a, b].

Recall from Theorem 8.2.2 that C[a, b] is complete. By the Baire Category

Theorem, it is not the union of countably many nowhere dense sets. The ﬁrst cate-

gory set

n≥1

contains all functions that are differentiable at any single point.

Thus the complement, consisting entirely of nowhere differentiable functions, is a

residual set. Therefore, nowhere differentiable functions are dense in C[a, b]. ¥

Exercises for Section 9.3

A. Show that A is nowhere dense in X if and only if X \

A is dense in X.

B. Show that R

is not the union of countably many lines.

C. Suppose that X is a countable complete metric space. Show that X has isolated points

[i.e., points that are open (and closed)].

HINT: Write X as a union of single points and apply the Baire Category Theorem.

D. Show that a complete metric space containing more than one point that has no isolated

points is uncountable. HINT: Use the previous exercise.

9.4 Connectedness 257

E. A G

set is the intersection of a countable family of open sets.

(a) Show that if A ⊂ R is closed, then A is a G

set.

(b) Show that Q is not a G

subset of R. HINT: Show that R \Q is not ﬁrst category.

F. (a) If f is a real-valued function on a metric space X, show that the set of points at

which f is continuous is a G

set. HINT: Show that the set U

of points x for

which there are i ∈ N and δ > 0 so that |f(y) −

| <

for y ∈ B

(x) is open.

(b) Show that no function on [0, 1] can be continuous just on the rational points. Com-

pare this with Example 5.2.10.

G. Suppose that f

is a sequence of continuous real-valuedfunctions on a complete metric

space X which converge pointwise to a function f.

(a) Prove that there is a constant M and anopen set U ⊂ X such that sup

n≥1

(x)| ≤ M

for all x ∈ U. HINT: Let A

= {x ∈ X : sup

n≥1

(x)| ≤ k}.

(b) If f is continuous and ε > 0, show that there is an open set U and an integer N so

that |f(x) − f

(x)| < ε for all x ∈ U and n ≥ N.

HINT: Let B

= {x ∈ X : sup

n≥k

|f(x) − f

(x)| ≤ ε/2}.

H. (a) Show that the set of compact nowhere dense subsets of R

is dense in K(X), where

K(X) is equipped with the Hausdorff metric of Example 9.1.2(5).

HINT: If C is compact and ε > 0, use a ﬁnite ε-net. Finite sets are nowhere dense.

(b) Show that the set A

of those compact sets in K(X) that contain a ball of radius

1/n is a closed set with no interior.

I. Banach–Steinhaus Theorem. Suppose that X and Y are complete normed vector

spaces, and {T

: α ∈ A} is a family of continuous linear maps of X into Y such that

for each x ∈ X, K

= sup

α∈A

xk < ∞.

(a) Let A

= {x ∈ X : K

≤ n}. Show that A

is closed.

(b) Prove that there is some n

so that A

has interior, say containing B

has Lipschitz constant L.

HINT: If kxk < 1, then x

+ εx ∈ A

. Estimate kT

xk.

J. Show that [0, 1] is not the disjoint union of a countably inﬁnite family of nonempty

closed sets {A

: n ≥ 1}.

HINT: If U

= intA

, observe that X := [0, 1] \

n≥1

\ U

) is

complete. Find an integer n

and an open interval V so that ∅ 6= X ∩V ⊂ A

. Show

∩ V = ∅ for n 6= n

K. A function on [0, 1] that is not monotonic on any interval is called a nowhere mono-

tonic function. Show that these functions are a residual subset of C[0, 1].

HINT: Let A

= {±f : there is x ∈ [0, 1], (f(y)−f (x))(y−x) ≥ 0 for |y −x| ≤

9.4. Connectedness

The Intermediate Value Theorem shows that continuous real-valued functions

map intervals onto intervals (see Corollary 5.6.2). This should mean that inter-

vals have some special property not shared by other subsets of the line. This fact

may appear to be dependent on the order structure. However, this property can be

described in topological terms that allows a generalization to higher dimensions.

258 Metric Spaces

9.4.1. DEFINITION. A subset A of a metric space X is not connected if there

are disjoint open sets U and V such that A ⊂ U ∪ V and A ∩ U 6= ∅ 6= A ∩ V .

Otherwise, the set A is said to be connected.

A set A is totally disconnected if for every pair of distinct points x, y ∈ A,

there are two disjoint open sets U and V so that x ∈ U, y ∈ V and A ⊂ U ∪ V .

9.4.2. EXAMPLES.

(1) A = [−1, 0] ∪[3, 4] is not connected in R because U = (−2, 1) and V = (2, 5)

are disjoint open sets which each intersect A and jointly contain A.

(2) The set A = {x ∈ R : x 6= 0} is not connected because U = (−∞, 0)

and V = (0, +∞) provide the necessary separation. The sets U and V need not

be a positive distance apart. Even the small gap created by omitting the origin

disconnects the set.

(3) Q is totally disconnected. For if x < y ∈ Q, choose an irrational number z

with x < z < y. Then U = (−∞, z) 3 x and V = (z, ∞) 3 y are disjoint open

sets such that U ∪ V contains Q.

(4) It is more difﬁcult to construct a closed set that is totally disconnected. How-

ever, the Cantor set C is an example. Suppose that x < y ∈ C. Choose N so that

y − x > 3

−N

. By Example 4.4.8, C is obtained by repeatedly removing middle

thirds from each interval. At the N th stage, C is contained in the set S

that con-

sists of 2

intervals of length 3

−N

. Thus x and y belong to distinct intervals in S

Therefore, there is a point z in the complement of S

such that x < z < y. Then

U = (−∞, z) 3 x and V = (z, ∞) 3 y are disjoint open sets such that U ∪ V

contains S

and therefore contains C.

To obtain an example of a connected set, we must appeal to the Intermediate

Value Theorem.

9.4.3. PROPOSITION. The interval [a, b] is connected.

PROOF. Suppose to the contrary that there are disjoint open sets U and V such that

[a, b] ∩ U and [a, b] ∩ V are both nonempty, and [a, b] ⊂ U ∪ V . Deﬁne a function

f on [a, b] by

f(x) =

(

1 if x ∈ [a, b] ∩ U

−1 if x ∈ [a, b] ∩ V.

We claim that this function is continuous. Indeed, let x be any point in [a, b]

and let ε > 0. If x ∈ U, then since U is open, there is a positive number r > 0

such that B

(x) is contained in U . Thus |x − y| < r implies that

|f(y) − f(x)| = |1 − 1| = 0 < ε.

9.4 Connectedness 259

So f is continuous at x. Similarly, if x ∈ V , there is some ball B

(x) contained in

V ; and thus |x − y| < r implies that

|f(y) − f(x)| = | − 1 −(−1)| = 0 < ε.

This accounts for every point in [a, b], and therefore f is a continuous function.

Thus we have constructed a continuous function on [a, b] with range equal to

{−1, 1}. This contradicts the Intermediate Value Theorem. So our assumption that

[a, b] is not connected must be wrong. Therefore, the interval is connected. ¥

To obtain a useful criterion for connectedness, we need to know that it is a

property that is preserved by continuous functions. This is readily established using

the topological form of continuity based on open sets.

9.4.4. THEOREM. The continuous image of a connected set is connected.

PROOF. This theorem is equivalent to the contrapositive statement: If the image of

a continuous function is not connected, then the domain is not connected. To prove

this, suppose that f is a continuous function from a metric space (X, ρ) into (Y, σ)

such that the range f(X) is not connected. Then there are disjoint open sets U and

V in Y such that f(X) ∩U and f(S) ∩V are both nonempty, and f (X) ⊂ U ∪V .

Let S = f

−1

(U) and T = f

−1

(V ). By Theorem 9.1.7, the inverse image of an

open set is open; and thus both S and T are open. Since U and V are disjoint and

cover f(X), each x ∈ X has either f(x) ∈ U or f(x) ∈ V but not both. Therefore,

S and T are disjoint and cover X. Finally, since U ∩ f (X) is nonempty, it follows

that S ∩X is also nonempty; and likewise, T ∩X is not empty. This shows that X

is not connected, as required. ¥

9.4.5. EXAMPLE. In Section 5.6, we deﬁned a path in R

. Similarly, we deﬁne

a path in any metric space as a continuous image of a closed interval. A path is

connected because a closed interval is connected. In particular, if f is a continuous

function of [a, b] into R

, the graph G(f) = {(x, f(x)) : a ≤ x ≤ b} is connected.

9.4.6. DEFINITION. A subset S of a metric space X is said to be path con-

nected if for each pair of points x and y in S, there is a path in S connecting x to y,

meaning that there is a continuous function f from [0, 1] into S such that f(0) = x

and f(1) = y.

9.4.7. COROLLARY. Every path connected set is connected.

PROOF. Let S be a path connected set, and suppose to the contrary that it is not

connected. Then there are disjoint open sets U and V such that

S ⊂ U ∪ V and S ∩ U 6= ∅ 6= S ∩ V.

Pick points u ∈ S ∩ U and v ∈ S ∩ V . Since S is path connected, there is a path

P connecting u to v in S. It follows that P ⊂ U ∪ V . And P ∩ U contains u and

260 Metric Spaces

P ∩V contains v. Thus both are nonempty. Consequently, this shows that the path

P is not connected. This contradicts the previous example. Therefore, S must be

connected. ¥

9.4.8. EXAMPLES.

(1) R

is path connected and therefore connected.

(2) Recall that a subset of R

is convex if the straight line between any two points

in the set belongs to the set. Therefore, any convex subset of R

is path connected.

(3) The annulus {x ∈ R

: 1 ≤ kxk ≤ 2} is path connected and hence connected.

(4) There are connected sets that are not path connected. An example is the com-

pact set

S = Y ∪ G = {(0, t) : |t| ≤ 1} ∪ {(x, sin

) : 0 < x ≤ 1}.

To see that S is connected, suppose that U and V are two disjoint open sets

such that S is contained in U ∪ V . The point (0, 0) belongs to one of them, which

may be called U. Therefore, U intersects the line Y . Since Y is path connected, it

must be completely contained in U (for if V intersects Y , this would show that Y

was disconnected).

Since U is open, it contains a ball of positive radius r > 0 around the origin.

But (0, 0) is a limit point of G, and therefore U contains points of G. Since G is

a graph, it is path connected. So by the same argument, U contains G. Therefore,

V ∩ S is empty. This shows that S is connected.

If S were path connected, there would be a continuous function f of [0, 1] into

S with f (0) = (0, 0) and f (1) = (1, sin1). Let a = sup{x : f(x) ∈ Y }. Since f

is continuous, there is a δ > 0 such that kf (x) − f(a)k ≤ .5 if |x − a| ≤ δ. Set

b = a + δ. Let f (a) = (0, y) and f(b) = (u, sin

) =: u. A similar argument

shows that there is a point a < c < b such that f(c) = (t, sin

) =: t, where

t ≤

1 + 2πu

. Therefore, f([c, b]) is a connected subset of S containing both t and

u. The graph G

= {(x, sin

) : t ≤ x ≤ u} is connected, but removing any point

will disconnect it. So f([c, b]) contains G

. However, sin

takes both values ±1 on

[t, u], and so there is a point v = (v, sin

) on G

with kv−f (a)k > |sin

−y| ≥ 1.

This contradicts the estimate kf(x) − f(a)k ≤ .5 for all x ∈ [c, b]. Therefore, S is

not path connected.

Exercises for Section 9.4

A. Show that the only connected subsets of R are intervals (which may be ﬁnite or inﬁnite

and may or may not include the endpoints).

B. Show that if S is a connected subset of X, then so is

C. Let A ⊂ R

and B ⊂ R

be connected sets. Prove that A ×B is connected in R

m+n

HINT: Recall that A × B = {(a, b) : a ∈ A, b ∈ B}.

9.5 Metric Completion 261

D. Show that a connected open set U ⊂ R

is path connected.

HINT: Fix u

∈ U. The set V of points in U path connected to u

is open, as is the set

W of points in U not path connected to u

E. Find an example of a decreasing sequence A

⊃ A

··· of closed connected sets

in R

such that

k≥1

is not connected.

F. Let A

⊃ A

··· be a decreasing sequence of connected compact subsets of R

Show that

k≥1

is connected.

HINT: Show that if U ∪ V is an open set containing the intersection, then it contains

some A

for n sufﬁciently large.

G. Show that there is no continuous bijection of [−1, 1] onto the circle.

HINT: Such a function would map [−1, 0) ∪ (0, 1] onto a connected set.

H. Let (Y, ρ) be a subset of a metric space (X, ρ). Show that Y is connected in itself if

and only if it is connected in X.

HINT: If U and V are nonempty disjoint open subsets of Y with Y = U ∪ V , let

f(y) = max{dist(y, U), dist(y, V )}. Then U

f(y)/2

(y):y ∈ U} is open in X.

9.5. Metric Completion

Completeness is such a powerful property that we attempt to work in a com-

plete context whenever possible. The purpose of this section is to show that every

metric space may be naturally imbedded in a unique complete metric space known

as its completion. We do this by an efﬁcient, if sneaky, argument based on the

completeness of C

(X), the space of bounded real functions on X.

In principle, this method could be applied to the rational numbers to obtain

the real numbers in a base independent way. You should be concerned that our

argument uses the real numbers in its proof. Because of this circularity, we cannot

construct the real numbers this way. Nevertheless, we can show that the real line

is unique and hence the various different constructions we have mentioned all give

the same object. This puts the real numbers on a ﬁrmer theoretical footing.

In the next section, the completeness theorem will be applied to C[a, b] with

the L

norms to obtain the L

spaces. The completeness of these normed spaces

allows us to use the full power of analysis in studying them.

9.5.1. DEFINITION. Let (X, ρ) be a metric space. A completion of (X, ρ)

is a complete metric space (Y, d) together with a map T of X into Y such that

d(T x

, T x

) = ρ(x

, x

) for all x

, x

∈ X (an isometry) and T X is dense in Y .

When the metric space and its completion are concretely represented, often the

map T is an inclusion map, as in the next example. By an inclusion map, we mean

that T : X → Y , where X ⊂ Y and T is given by T (x) = x. Except for changing

the codomain, T is the identity map on X.

262 Metric Spaces

9.5.2. EXAMPLE. Let A be an arbitrary subset of R

with the usual Euclidean

distance. Then the closure A of A is a complete metric space (see Exercise 4.3.H).

Evidently, the distance functions coincide on these two sets, so the inclusion map

is trivially an isometry. As A is dense in A, this is a completion of A.

9.5.3. THEOREM. Every metric space (X, ρ) has a completion.

PROOF. Pick a point x

∈ X. For each x ∈ X, deﬁne f

(y) = ρ(x, y) − ρ(x

, y).

If ρ(y

, y

) < ε/2, then |ρ(x, y

) − ρ(x, y

)| ≤ ρ(y

, y

) < ε/2. So

) − f

)| ≤ |ρ(x, y

) − ρ(x, y

)| + |ρ(x

, y

) − ρ(x

, y

)| < ε.

Therefore, f

is (uniformly) continuous.

Observe that f

(y) −f

(y) = ρ(x

, y) −ρ(x

, y). By the triangle inequality,

ρ(x

, y) ≤ ρ(x

, x

)+ρ(x

, y), and thus f

(y) −f

(y) ≤ ρ(x

, x

). Interchang-

ing x

and x

yields kf

− f

∞

≤ ρ(x

, x

). However, substituting y = x

shows that this is an equality. In particular as f

= 0, we see that f

is bounded.

Thus T x = f

is an isometric imbedding of X into C

(X).

Let F = T X = {f

: x ∈ X}, and consider

F , the closure of F in C

(X).

This is a closed subset of a complete normed space, and therefore it is complete.

Evidently, F is a metric completion of X. ¥

It is important that this completion is unique in a natural sense. This will justify

our use of the terminology the completion and will allow us to consider X as a

subset of C without explicit use of the isometry T .

To prove uniqueness, we ﬁrst need a signiﬁcant result about continuous func-

tions on the metric completion. To simplify notation, we drop the map T and

consider X as a subset of its completion C and use the same notation for the two

metrics. A continuous function g on C is a continuous extension of a function f

on X if g|

= f .

9.5.4. EXTENSION THEOREM.

Let (X, ρ) be a metric space with metric completion (C, ρ). Let f be a uniformly

continuous function from (X, ρ) into a complete metric space (Y, σ). Then f has a

unique uniformly continuous extension g mapping C into Y .

PROOF. Let c be a point in C. Since X is dense in C, we may choose a sequence

) in X converging to c. We claim that (f(x

)) is a Cauchy sequence in Y .

Indeed, let ε > 0 be given. By the uniform continuity of f , there is a δ > 0 so

that ρ(x, x

) < δ implies σ(f(x), f (x

)) < ε. Since (x

) is Cauchy, there is an

integer N so that σ(x

, x

) < δ for all m, n ≥ N. Hence σ(f(x

), f(x

)) < ε

for all m, n ≥ N. Therefore, (f(x

)) is Cauchy and we may deﬁne g(c) to be

lim

n→∞

f(x

). Moreover, this is the only possible choice for a continuous extension

of f.

We must verify that g is well deﬁned, meaning that any other sequence con-

verging to c will yield the same value for g(c). Consider a second sequence (x

)