<< . .

. 41
( : 70)



. . >>


We therefore reformulate Question A as follows.
Question B: If (X, d) is a metric space, can we ¬nd a most economical com-
plete metric space (Z, ρ) such that Z ⊇ X and d(u, v) = ρ(u, v) for all u,
v ∈ X?
We can formulate an appropriate meaning for ˜most economical™ with the
aid of the notion of density.

De¬nition 14.1.3. If (X, d) is a metric space, we say that a subset E is
dense in X if, given any x ∈ X, we can ¬nd xn ∈ E with d(xn , x) ’ 0 as
n ’ ∞.

Exercise 14.1.4. Show that the following statements about a subset E of a
metric space (X, d) are equivalent.
(i) E is dense in X.
(ii) Given x ∈ X and > 0, we can ¬nd a y ∈ E such that d(x, y) < .
(iii) If F is a closed subset of X with F ⊇ E, then F = X.

The notion of density is widely used in analysis since, in some sense,
a dense subset of a metric space acts as a ˜skeleton™ for the space. For
the moment, let us see why it provides the appropriate de¬nition of ˜most
economical™ in the context of this chapter.

Lemma 14.1.5. Suppose (X, d), (Y, ρ) and (Y , ρ ) are metric spaces with
the following properties.
(i) (Y, ρ) and (Y , ρ ) are complete,
(ii) There exist maps θ : X ’ Y and θ : X ’ Y such that ρ(θu, θv) =
ρ (θ u, θ v) = d(u, v).
(iii) θX is dense in Y and θ X is dense in Y .
Then we can ¬nd a bijection φ : Y ’ Y such that φθ = θ and ρ (φw, φz) =
ρ(w, z) for all w, z ∈ Y .

Exercise 14.1.6. (i) Convince yourself that Lemma 14.1.5 can be roughly
restated in terms of Question A as follows. Suppose (X, d) is a metric space
and (Z, δ) and (Z , δ ) are complete metric spaces such that X ⊆ Z, X ⊆ Z ,
δ(u, v) = δ (u, v) = d(u, v) for all u, v ∈ X and X is dense in (Z, δ) and in
(Z , δ ). Then (Z, δ) and (Z , δ ) have the same metric structure and X sits
in both metric spaces in the same way. [Note, you must be able to describe
the purpose of the statement φθ = θ in this account.]
357
Please send corrections however trivial to twk@dpmms.cam.ac.uk

(ii) Convince yourself that the questions involved in Lemma 14.1.5 are
better treated in the language of Question A than in the language of Ques-
tion A.

Proof of Lemma 14.1.5. If y ∈ Y , then, since θX is dense in (Y, ρ), we can
¬nd xn ∈ X with ρ(θxn , y) ’ 0. Since θxn converges, it forms a Cauchy
sequence in (Y, ρ). But

ρ (θ xn , θ xm ) = d(xn , xm ) = ρ(θxn , θxm ),

so θ xn is a Cauchy sequence in (Y , ρ ) and so converges to a limit x, say.
˜
Suppose zn ∈ X with ρ(θzn , y) ’ 0. By the argument of the previous
paragraph, θ zn is a Cauchy sequence in (Y , ρ ) and so converges to a limit
z , say. We wish to show that x = z . To do this, observe that
˜ ˜˜

ρ (˜, z ) ¤ ρ (˜, θ xn ) + ρ (˜, θ zn ) + ρ (θ xn , θ zn )
x˜ x z
= ρ (˜, θ xn ) + ρ (˜, θ zn ) + ρ(θxn , θzn )
x z
¤ ρ (˜, θ xn ) + ρ (˜, θ zn ) + ρ(θxn , y) + ρ(θzn , y)
x z
’ 0 + 0 + 0 + 0 = 0,

as n ’ ∞. Thus ρ (˜, z ) = 0 and x = z . We have shown that we can de¬ne
x˜ ˜˜
φy unambiguously by φy = x. ˜
We have now de¬ned φ : Y ’ Y . To show that ρ (φw, φz) = ρ(w, z),
choose wn ∈ X and zn ∈ X such that ρ(θwn , w), ρ(θzn , z) ’ 0 as n ’ ∞.
Then ρ(θzn , θwn ) ’ ρ(z, w) as n ’ ∞. (See Exercise 14.1.7 if necessary.)
By de¬nition, ρ (θ zn , φz) ’ 0 and ρ (θ wn , φw) ’ 0, so ρ (θ zn , θ wn ) ’
ρ (φz, φw). Since

ρ (θ zn , θ wn ) = d(zn , wn ) = ρ(θzn , θwn ),

it follows that ρ (φw, φz) = ρ(w, z).
To show that φθ = θ , choose any x ∈ X. If we set xn = x for each n,
we see that ρ(θxn , θx) = 0 ’ 0 and ρ (θ xn , θ x) = 0 ’ 0 as n ’ ∞ so, by
de¬nition, φθx = θ x. Since x was arbitrary, it follows that φθ = θ .
Finally, we must show that φ is a bijection. There are several ways of
approaching this. We choose an indirect but natural approach. Observe that,
interchanging the roles of Y and Y , the work done so far also shows that
˜ ˜ ˜˜
there is a map φ : Y ’ Y such that φθ = θ and ρ(φw, φz) = ρ (w, z) for all
w, z ∈ Y . Since

˜ ˜ ˜
(φφ)θ = φ(φθ) = φθ = θ,
358 A COMPANION TO ANALYSIS

˜
we have φφ(y) = y for all y ∈ θX. Now θX is dense in Y , so, if y ∈ Y , we
can ¬nd yn ∈ θX such that ρ(yn , y) ’ 0. Thus
˜ ˜ ˜ ˜
ρ(φφ(y), y) ¤ ρ(φφ(y), φφ(yn )) + ρ(φφ(yn ), yn ) + ρ(yn , y)
˜ ˜
= ρ(φφ(y), φφ(yn )) + ρ(yn , y) = ρ (φ(y), φ(yn )) + ρ(yn , y)
= ρ(y, yn ) + ρ(yn , y) = 2ρ(yn , y) ’ 0
˜ ˜
as n ’ ∞. Thus φφ(y) = y for all y ∈ Y . Similarly, φφ(y ) = y for all
˜
y ∈ Y . Thus φ is the inverse of φ, and φ must be bijective.
Exercise 14.1.7. Let (X, d) be a metric space.
(i) Show that

d(x, y) ’ d(u, v) ¤ d(x, u) + d(y, v),

and deduce that

|d(x, y) ’ d(u, v)| ¤ d(x, u) + d(y, v)

for all x, y, u, v ∈ X.
(ii) If xn ’ x and yn ’ y, show that d(xn , yn ) ’ d(x, y) as n ’ ∞.
Exercise 14.1.8. The proof of Lemma 14.1.5 looks quite complicated but the
di¬culty lies in asking the right questions in the right order, rather than an-
swering them. Outline the questions answered in the proof of Lemma 14.1.5,
paying particular attention to the ¬rst two paragraphs.
We can now repose question B more precisely.
Question C: If (X, d) is a metric space, can we ¬nd a complete metric space
(Z, ρ) such that Z ⊇ X, X is dense in (Z, ρ) and d(u, v) = ρ(u, v) for all u,
v ∈ X?
Question C : If (X, d) is a metric space, can we ¬nd a complete metric
˜ ˜
space (Y, d) and a map θ : X ’ Y such that θX is dense in (Y, d) and
˜
d(θu, θv) = d(u, v) for all u, v ∈ X?
We shall leave these questions aside for the moment. Instead, we shall
give some examples of how behaviour on a dense subset may force behaviour
on the whole space. (Note, however that Exercise 5.7.1 and its preceeding
discussion shows that this need not always be the case.)
Lemma 14.1.9. Suppose (X, d) is a complete metric space with a dense
subset E. Suppose that E is a vector space (over F where F = R or F = C)
E such that d(x, y) = x ’E y E for all x, y ∈ E. Then X can
with norm
be given the structure of a normed vector space with norm such that E is
a vector subspace of X and d(x, y) = x ’ y for all x, y ∈ X.
359
Please send corrections however trivial to twk@dpmms.cam.ac.uk

Proof. Suppose1 that (E, +E , —E , E , F) is a normed vector space with ad-
dition denoted by +E and scalar multiplication denoted by —E .
Suppose x, y ∈ X. We can ¬nd xn , yn ∈ E such that d(xn , x), d(yn , y) ’
0. We know that xn and yn must be Cauchy sequences, and so
d(xn +E yn , xm +E ym ) = (xn +E yn ) ’E (xm +E ym ) E
= (xn ’E xm ) +E (yn ’E ym ) E ¤ xn ’E xm + y n ’ E ym
E E
= d(xn , xm ) + d(yn , ym ) ’ 0
as n, m ’ ∞. Thus xn +yn is a Cauchy sequence in X and must have a limit
z, say. Suppose now that xn , yn ∈ E are such that d(xn , x), d(yn , y) ’ 0.
As before, xn + yn must have a limit z , say. We want to show that z = z .
To do this, observe that, since
d(z, z ) ¤ d(z, xn +E yn ) + d(xn +E yn , xn +E yn ) + d(xn +E yn , z)
= d(z, xn +E yn ) + d(z , xn +E yn ) + (xn +E yn ) ’E (xn +E yn )
= d(z, xn +E yn ) + d(z , xn +E yn ) + (xn ’E xn ) +E (yn ’E yn ) E
¤ d(z, xn +E yn ) + d(z , xn +E yn ) + xn ’E xn E + yn ’E yn E
= d(z, xn +E yn ) + d(z , xn +E yn ) + d(xn , xn ) + d(yn , yn )
¤ d(z, xn +E yn ) + d(z , xn +E yn ) + d(xn , x) + d(xn , x) + d(yn , y) + d(yn , y)
’0+0+0+0+0+0=0
as n ’ ∞, it follows that z = z . We can, therefore, de¬ne x + y as the limit
of xn +E yn when xn , yn ∈ E and d(xn , x), d(yn , y) ’ 0. Observe that, if
x, y ∈ E, then, setting xn = x, yn = y, it follows that x +E y = x + y.
We leave it as an exercise for the reader to show that if » ∈ F and x ∈ X,
we can de¬ne »x unambiguously as the limit of » —E xn when xn ∈ E and
d(xn , x) ’ 0 and that, with this de¬nition, » —E x = »x whenever » ∈ F and
x ∈ E.
Now we must check that X with the newly de¬ned operations is indeed
a vector space. This is routine. For example, if x, y ∈ X and » ∈ F we can
¬nd xn , yn ∈ E such that xn ’ x and yn ’ y. Since E is vector space,
» —E (xn +E yn ) = » —E xn +E » —E yn
But, by our de¬nitions, xn +E yn ’ x+y, so »—E (xn +E yn ) ’ »(x+y). Again,
by de¬nition, »—E xn ’ »x and »—E yn ’ »y, so »—E xn +E »—E yn ’ »x+»y.
The uniqueness of limits now gives us
»(x + y) = »x + »y.
1
This proof with its plethora of subscript E™s may help explain why mathematicians
are prepared to put up with some ambiguity in order to achieve notational simplicity.
360 A COMPANION TO ANALYSIS

The remaining axioms are checked in the same way.
Finally, we must check that there is a norm on X such that d(x, y) =
x ’ y . First, observe that, if x, y ∈ X, we can ¬nd xn , yn ∈ E such that
xn ’ x and yn ’ y. We have

d(xn , yn ) = xn ’E yn = d(xn ’E yn , 0)
E


where 0 is the zero vector in E (and so in X). Since

xn ’E yn = xn +E (’1) — yn ’ x + (’1)y = x ’ y,

we have d(xn ’E yn , 0) ’ d(x ’ y, 0). But d(xn , yn ) ’ d(x, y) and the limit
is unique, so we have

d(x, y) = d(x ’ y, 0).

We now set a = d(a, 0) and check that is indeed a norm.

Exercise 14.1.10. Fill in the gaps in the proof of Lemma 14.1.9.
(i) Show that, if » ∈ F and x ∈ X, we can de¬ne »x unambiguously
as the limit of » —E xn when xn ∈ E and d(xn , x) ’ 0 and that, with this
de¬nition, » —E x = »x whenever » ∈ F and x ∈ E.
(ii) If you are of a careful disposition, check all the axioms for a vector
space hold for X with our operations. Otherwise, choose the two which seem
to you hardest and check those.
(iii) Check that de¬ned at the end of the proof is indeed a norm.

Lemma 14.1.11. Let (X, d) be a complete metric space with a dense subset
E. Suppose that E is a vector space (over F where F = R or F = C) with
inner product , E such that d(x, y) = x ’E y E for all x, y ∈ E, where
E is the norm induced by the inner product. Then X can be given the
structure of a vector space with inner product , such that E is a vector
subspace of X and x, y E = x, y for all x, y ∈ E.

Proof. By Lemma 14.1.9, X can be given the structure of a normed vector
space with norm such that E is a vector subspace of X and d(x, y) =
x ’ y for all x, y ∈ X. We need to show that it can be given an inner
product consistent with this norm.
First, observe that, if x, y ∈ X, we can ¬nd xn , yn ∈ E such that xn ’ x
and yn ’ y. Automatically, xn and yn form Cauchy sequences. Since any
Cauchy sequence is bounded, we can ¬nd a K such that xn , yn ¤ K for
361
Please send corrections however trivial to twk@dpmms.cam.ac.uk

all n. Thus, using the Cauchy-Schwarz inequality,

| xn , yn ’ xm , ym E| = | x n , yn ’ ym E + x n ’ xm , ym E |
E
¤ | x n , yn ’ ym E | + | x n ’ xm , ym E |
¤ x n y n ’ ym + x n ’ xm y m
¤ K y n ’ ym + K x n ’ xm ’ 0 + 0 = 0

and the xn , yn E form a Cauchy sequence in F converging to t.
Suppose now that xn , yn ∈ E are such that xn ’ x and yn ’ y. As
before, xn , yn E must have a limit t , say. We want to show that t = t .
Note that, also as before, we can ¬nd a K such that xn , yn ¤ K for
all n. Thus, using the Cauchy-Schwarz inequality again,

| xn , yn ’ xn , yn E| = | x n , yn ’ yn E + x n ’ xn , yn E |
E
¤ | x n , yn ’ yn E | + | x n ’ xn , yn E |
¤ x n y n ’ yn + x n ’ xn y n
¤ K yn ’ yn + K xn ’ xn ’ 0 + 0 = 0,

so t = t . We can thus de¬ne x, y unambiguously by choosing xn , yn ∈ E
such that xn ’ x and yn ’ y and taking x, y to be the limit of xn , yn E .
Setting xn = x, yn = y for all n we see that x, y = x, y E whenever
x, y ∈ E.
We observe that, if x ∈ X and we choose xn ∈ E such that xn ’ x, then
2
’ x, x
xn = xn , xn E


and
2
= d(xn , 0)2 ’ d(x, 0)2 = x 2
xn

so, by the uniqueness of limits, x 2 = x, x . The veri¬cation that , is
an inner product on X is left to the reader.
(An alternative proof which uses an important link between inner prod-
ucts and their derived norms is oulined in Exercises K.297 and K.298.)

Exercise 14.1.12. Complete the proof of Lemma 14.1.11 by showing that
, is an inner product on X.

By now the reader should have the formed the opinion that what seemed
to be a rather di¬cult proof technique when ¬rst used in the proof of Lemma 14.1.5
is, in fact, rather routine (though requiring continuous care). If she requires
further convincing, she can do Exercise K.299.
362 A COMPANION TO ANALYSIS

14.2 The solution
In this section we show how, starting from a metric space (X, d), we can
˜
construct a complete metric space (Y, d) and a map θ : X ’ Y such that θX
˜ ˜
is dense in (Y, d) and d(θu, θv) = d(u, v) for all u, v ∈ X. We give a direct
proof which can serve as a model in similar circumstances. (Exercise K.300
gives a quick and elegant proof which, however, only applies in this particular
case.)
We start by considering the space X of Cauchy sequences x = (xn )∞ . n=1
We write x ∼ y if x, y ∈ X and d(xn , yn ) ’ 0 as n ’ ∞. The reader should
verify the statements made in the next exercise.

Exercise 14.2.1. Suppose x, y, z ∈ X. Then
(i) x ∼ x.
(ii) If x ∼ y, then y ∼ x.
(ii) If x ∼ y and y ∼ z, then x ∼ z.

We write [x] = {y : y ∼ x} and call [x] the equivalence class of x. We
take Y to be the set of all such equivalence classes. The reader should verify
the statement made in the next exercise.

Exercise 14.2.2. Each x ∈ X belongs to exactly one equivalence class in Y .

We now want to de¬ne a metric on Y . The de¬nition follows a pattern
which is familiar from the previous section. Suppose a, b ∈ Y . If a = [x] and
b = [y], then

|d(xn , yn ) ’ d(xm , ym )| ¤ |d(xn , yn ) ’ d(xn , ym )| + |d(xn , ym ) ’ d(xm , ym )|
¤ d(yn , ym ) + d(xn , xm ) ’ 0 + 0 = 0

as n, m ’ ∞. Thus d(xn , yn ) is a Cauchy sequence in R and tends to a limit
t, say. Suppose now that a = [x ] and b = [y ]. Since

|d(xn , yn ) ’ d(xn , yn )| ¤ |d(xn , yn ) ’ d(xn , yn )| + |d(xn , yn ) ’ d(xn , yn )|
¤ d(yn , yn ) + d(xn , xn ) ’ 0 + 0 = 0,

˜
we have d(xn , yn ) ’ t as n ’ ∞. We can thus de¬ne d([x], [y]) unambigu-
ously by

˜
d([x], [y]) = lim d(xn , yn ).
n’∞

The reader should verify the statements made in the next exercise.
363
Please send corrections however trivial to twk@dpmms.cam.ac.uk

˜
Exercise 14.2.3. (i) (Y, d) is a metric space.
(ii) If we set θ(x) = [(x, x, x, . . . )], then θ is a map from X to Y such
˜
that d(θu, θv) = d(u, v).
˜
(iii) If a ∈ Y , then a = [x] for some x ∈ X. Show that d(θ(xn ), a) ’ 0
˜
as n ’ ∞, and so θ(X) is dense in (Y, d).
Lemma 14.2.4. If we adopt the hypotheses and notation introduced in this
˜
section, then (Y, d) is complete.
˜
Proof. Suppose [y(1)], [y(2)], . . . is a Cauchy sequence in (Y, d). Since θ(X)
˜ ˜
is dense in (Y, d), we can ¬nd xj ∈ X such that d(θ(xj ), [y(j)]) < j ’1 . We
observe that
˜
d(xj , xk ) = d(θ(xj ), θ(xk ))
˜ ˜ ˜
¤ d(θ(xj ), [y(j)]) + d(θ(xk ), [y(k)]) + d([y(j)], [y(k)])
˜
< j ’1 + k ’1 + d([y(j)], [y(k)]),

and so the xj form a Cauchy sequence in (X, d).
˜
We now wish to show that d([y(n)], [x]) ’ 0 as n ’ ∞. Let > 0 be
given. Since the sequence [y(n)] is Cauchy, we can ¬nd an M such that
˜
d([y(j)], [y(k)]) < /2 for all j, k ≥ M .

We now choose N ≥ M such that N ’1 < /6 and observe that the inequality
proved in the last paragraph gives

d(xj , xk ) < 5 /6 for all j, k ≥ N .

Thus
˜
d([x], θ(xk )) ¤ 5 /6 for all k ≥ N ,

and
˜ ˜ ˜
d([x], [y(k)]) ¤ d([x], θ(xk )) + d(θ(xk ), [y(k)]) < 5 /6 + k ’1 ¤ 5 /6 + N ’1 <

for all k ≥ N , so we are done.
Combining the results of this section with Lemma 14.1.5, we have the
following theorem.
Theorem 14.2.5. If (X, d) is a metric space, then there exists an essentially
˜ ˜ ˜
unique complete metric space (Y, d) such that X is dense in (Y, d) and d
restricted to X 2 is d.
364 A COMPANION TO ANALYSIS

˜
We call (Y, d) the completion of (X, d). Lemma 14.1.9 and Lemma 14.1.11
now give the following corollaries.

Lemma 14.2.6. (i) The completion of a normed vector space is a normed
vector space.
(ii) The completion of an inner product vector space is an inner product
vector space.

Exercise 14.2.7. Readers of a certain disposition (with which the author
sometimes sympathises and sometimes does not) will ¬nd the statements of
Theorem 14.2.5 and Lemma 14.2.6 unsatisfactorily lax. Redraft them in the
more rigorous style of Lemma 14.1.5 and Lemmas 14.1.9 and 14.1.11.

Exercise 14.2.8. If (X, d) is already complete, then the uniqueness of the
˜
completion means that (essentially) (X, d) = (Y, d). Go through the construc-
˜
tion of (Y, d) in this section and explain in simple terms why the construction
does indeed work as stated.


Why do we construct the reals? ™
14.3
Mathematicians know from experience that it is usually easier to ask ques-
tions than to answer them. However, in some very important cases, asking
the correct question may represent a greater intellectual breakthrough than
answering it.
From the time the study of Euclid™s Elements was reintroduced into Eu-
rope until about 1750, few of those who studied it can have doubted that it
described the geometry of the world, or, accepting that it did describe the
geometry of the world, asked why it did so. Amongst those who did was the

<< . .

. 41
( : 70)



. . >>