normed vector space as in Lemma 14.1.9, then we can ¬nd a map M : X 2 ’

X such that M (x, y) = ME (x, y) for all x, y ∈ E and M has properties (i)

to (iv) (with E replaced by X). Show that, if ME (x, y) = ME (y, x) for all

x, y ∈ E, then M (x, y) = M (y, x) for all x, y ∈ X. Show that, if there

exists an e ∈ E such that ME (x, e) = ME (e, x) = x for all x ∈ E, then

M (x, e) = M (e, x) = x for all x ∈ X.

Exercise K.300. [14.1, T] This neat proof that every metric space (E, d)

can be completed is due to Kuratowski14 .

(i) Choose e0 ∈ E. For each e ∈ E de¬ne fe (t) = d(e, t) ’ d(e0 , t) [t ∈ E].

Show that fe ∈ C(E), where C(E) is the space of bounded continuous function

g : E ’ R.

(ii) Give C(E) the usual uniform norm. Show that

fu ’ fv = d(u, v)

for all u, v ∈ E.

(iii) Let Y be the closure of {fe : e ∈ E}. By using Theorem 11.3.7, or

˜

otherwise show that Y with the metric d inherited from C(E) is complete.

Show that (E, d) has a completion by considering the map θ : E ’ Y given

by θ(e) = fe .

14

I take it from from a book [15] crammed with neat proofs.

597

Please send corrections however trivial to twk@dpmms.cam.ac.uk

Exercise K.301. [14.1, P] Results like Lemma 14.1.9 rely on a strong link

between the algebraic operation and the metric. From one point of view this

question consists of simple results dressed up in jargon but I think they shed

some light on the matter.

(i) (This just sets up a bit of notation.) Suppose that (X, d) is a metric

space. Show that d2 : E 2 ’ R given by

d2 ((x, y), (x , y )) = d(x, x ) + d(y, y )

de¬nes a metric on X 2 . Show that, if (X, d) is complete, so is (X 2 , d2 ).

(ii) Consider X = [0, ∞) with the usual Euclidean metric d and E =

(0, ∞). Show that (X, d) is complete, E is a dense subset of X and that,

if we write ME (x, y) = xy (that is if M is ordinary multiplication), then

(E, M ) is a group and ME : (E 2 , d2 ) ’ (E, d) is continuous. Show, however,

that there does not exist a continuous map M : (X 2 , d2 ) ’ (X, d) with

M (x, y) = ME (x, y) for all x, y ∈ E such that (X, M ) is a group.

(iii) Consider X = [0, ∞) with the usual Euclidean metric d and E =

(0, ∞) © Q. Show that (X, d) is complete, E is a dense subset of X and that,

if we write ME (x, y) = xy, then (E, M ) is a group and ME : (E 2 , d2 ) ’ (E, d)

is continuous. Show, however, that there does not exist a continuous map

M : (X 2 , d2 ) ’ (X, d) with M (x, y) = ME (x, y) for all x, y ∈ E such that

(X, M ) is a group.

(iv) Consider X = (0, ∞). Show that if we write

d(x, y) = | log x ’ log y|

then (X, d) is a metric space. Let E = (0, ∞) © Q. Show that (X, d) is

complete, E is a dense subset of X and that, if we write ME (x, y) = xy, then

(E, M ) is a group and ME : (E 2 , d2 ) ’ (E, d) is continuous. Show that there

exists a continuous map M : (X 2 , d2 ) ’ (X, d) with M (x, y) = ME (x, y) for

all x, y ∈ E such that (X, M ) is a group.

Exercise K.302. [14.1, P] (i) Observe that R is a group under addition. If

E is a subgroup of R which is also a closed set with respect to the Euclidean

metric, show that either E = R or

E = {n± : n ∈ Z}

for some ± ∈ R.

(ii) Observe that

S 1 = {» ∈ C : |»| = 1}

598 A COMPANION TO ANALYSIS

is a group under multiplication. What can you say about subgroups E of S 1

which are closed with respect to the usual metric?

(iii) Observe that Rm is a group under vector addition. What can you

say about subgroups E of Rm which are closed with respect to the Euclidean

metric?

Exercise K.303. [14.1, T, ‘‘ ] Exercise K.56 is not important in itself, but

the method of its proof is. Extend the result and proof to f : E ’ R where

E is a dense subspace of a metric space (X, d).

Can the result be extended to f : E ’ Y where E is a dense subspace

of a metric space (X, d) and (Y, ρ) is a metric space? (Give a proof or

counterexample.) If not, what natural extra condition can we place on (Y, ρ)

so that the result can be extended?

Exercise K.304. [14.1, P, S, ‘ ] Suppose that f : R ’ R satis¬es

|f (x) ’ f (y)| ¤ (x ’ y)2 for all x, y ∈ R.

Show that f is constant.

Show that the result remains true if we replace R by Q. Explain why this

is consistent with examples of the type given in Example 1.1.3

Exercise K.305. [14.1, P] Let (X, d) be a metric space with the Bolzano-

Weierstrass property. Show that, given any > 0, we can ¬nd a ¬nite set of

points x1 , x2 , . . . xn such that the open balls B(xj , ) centre xj and radius

cover X (that is to say, n B(xj , ) = X). (This result occurs elsewhere

j=1

both in the main text and exercises but it will do no harm to reprove it.)

Deduce that (X, d) has a countable dense subset.

Give an example of a complete metric space which does not have the

Bolzano-Weierstrass property but does have a countable dense subset. Give

an example of a metric space which is not complete but does have a countable

dense subset. Give an example of a complete metric space which does not

have a countable dense subset.

Exercise K.306. [14.1, P] (i) By observing that every open interval con-

tains a rational number, or otherwise, show that every open subset of R

(with the standard Euclidean metric) can be written as the countable union

of open intervals.

(ii) Let (X, d) be a metric space with a countable dense subspace. Show

that every open subset of X can be written as the countable union of open

balls.

599

Please send corrections however trivial to twk@dpmms.cam.ac.uk

(iii) Consider R2 . De¬ne ρ : R2 — R2 ’ R by

1 if y1 = y2

ρ((x1 , y1 ), (x2 , y2 )) =

min(1, |x1 ’ x2 |) if y1 = y2 .

Show that ρ is a metric and that ρ is complete. Show that, if we work in

(R2 , ρ),

V = {(x, y) : |x| < 1, y ∈ R}

is an open set that can not be written as the countable union of open balls.

(iv) Let (X, d) be a metric space with a countable dense subspace. Show

that every open ball can be written as the countable union of closed balls.

Show that every open set can be written as the countable union of closed

balls. Show that, if U is an open set, we can ¬nd bounded closed sets Kj with

Kj+1 ⊆ Kj [1 ¤ j] and ∞ Kj = U . [This result is useful for spaces like Rn

j=1

with the usual metric where we know, in addition, that bounded closed sets

have the Bolzano-Weierstrass property.]

Exercise K.307. [14.1, P] The previous question K.306 dealt with general

metric spaces. Apart from the last part this question deals with the particular

space R with the usual metric. We need the notion of an equivalence relation.

(i) Let U be an open subset of R. If x, y ∈ U , write x ∼ y if there is an

open interval (a, b) ‚ U with x, y ∈ (a, b). Show that ∼ is an equivalence

relation on U .

(ii) Write [x] = {y ∈ U : y ∼ x} for the equivalence class of some x ∈ U .

If [x] is bounded, show, by considering the in¬mum and supremum of [x], or

otherwise, that [x] is an open interval. What can we say about [x] if it is

bounded below but not above? Prove your answer carefully. What can we

say about [x] if it is bounded above but not below? What can we say if [x]

is neither bounded above nor below?

(iii) Show that U is the disjoint union of a collection C of sets of the form

(a, b), (c, ∞), (’∞, c) and R.

(iv) Suppose that U is the disjoint union of a collection C of sets of the

form (a, b), (c, ∞), (’∞, c) and R. If J ∈ C explain why there exists an

I ∈ C with I ⊇ J. Explain why, if a is an end point of J which is not an end

point of I, there must exist a J ∈ C with a ∈ J . Hence, or otherwise, show

that J = I. Conclude that C = C.

(v) Show that C is countable.

(vi) We saw in part (iv) that C is uniquely de¬ned and this raises the

possibility of de¬ning the ˜length™ of U to be the sum of the lengths of the

600 A COMPANION TO ANALYSIS

intervals making up C. However, this approach fails in higher dimensions.

The rest of this question concerns R2 with the usual metric.

Show that the open square (’a, a) — (’a, a) is not the union of disjoint

open discs. [It may be helpful to look at points on the boundary of a disc

forming part of such a putative union.]

Show that the open disc {(x, y) : x2 + y 2 < 1} is not the union of disjoint

open squares.

Exercise K.308. [14.1, T] We say that metric spaces (X, d) and (Y, ρ) are

homeomorphic if there exists a bijective map f : X ’ Y such that f and

f ’1 are continuous. We say that f is a homeomorphism between X and Y .

(i) Show that homeomorphism is an equivalence relation on metric spaces.

(ii) If f : X ’ Y is a homeomorphism, show that U is open in (X, d) if

and only if f (U ) is open in (Y, ρ).

(iii) By constructing an explicit homeomorphism, show that R with the

usual metric is homeomorphic to the open interval (’1, 1) with the usual

metric. Deduce that the property of completeness is not preserved under

homeomorphism.

(iv) By constructing an explicit homeomorphism show that I = (’1, 1)

with the usual metric is homeomorphic to

J = {z ∈ C : |z| = 1, z = 1}

with the usual metric. Show that [’1, 1] with the usual metric is a completion

of I. Find a completion of J.

Explain brie¬‚y why the completion of I adds two points but the comple-

tion of J adds only one.

Exercise K.309. [14.1, T] (i) Suppose (X, d) is a metric space with the

Bolzano-Weierstrass property and (Y, ρ) is a any metric space. If f : X ’ Y

is a bijective continuous function show that (Y, ρ) has the Bolzano“Weierstrass

property and that f ’1 : Y ’ X is uniformly continuous. (Note that we have

shown that, in the language of Exercise K.308, (X, d) and (Y, ρ) are homeo-

morphic.)

(ii) Look brie¬‚y at Exercise 5.6.8. Which results (if any) of that exercise

can can be obtained using (i)?

(iii) Consider R with the usual metric. Give an example of a uniformly

continuous bijective map f : R ’ R with f ’1 not uniformly continuous.

(iv) Let d be the usual metric on R and ρ the discrete metric on R. Let

f : (R, ρ) ’ (R, d) be given by f (x) = x. Show that f is is a bijective

continuous function but f ’1 is not continuous.

601

Please send corrections however trivial to twk@dpmms.cam.ac.uk

Exercise K.310. [14.1, T, ‘ ] Suppose that (X, d) is a metric space with

the Bolzano-Weierstrass property. Explain (by referring to Exercise K.305,

if necessary) why we can ¬nd a countable dense subset {x1 , x2 , x3 , . . . },

say, for X. Consider l 2 with its usual norm (see Exercise K.188). Show that

the function f : X ’ l2 given by

f (x) = (d(x, x1 ), 2’1 d(x, x2 ), 2’2 d(x, x2 ), . . . )

is well de¬ned, continuous and injective. Deduce that f (X) is homeomorphic

to X. [Thus l2 contains a subsets homeomorphic to any given metric space

with the Bolzano-Weierstrass property.]

Exercise K.311. [14.1, P] Suppose that (X, d) and (Y, ρ) are metric spaces

and f : X ’ Y is a continuous surjective map.

(i) Suppose that ρ(f (x), f (x )) ¤ Kd(x, x ) for all x, x ∈ X and some

K > 0. If (X, d) is complete, does it follow that (Y, ρ) is complete? If

(Y, ρ) is complete, does it follow that (X, d) is complete? Give proofs or

counterexamples as appropriate.

(ii) Suppose that ρ(f (x), f (x )) ≥ Kd(x, x ) for all x, x ∈ X and some

K > 0. If (X, d) is complete, does it follow that (Y, ρ) is complete? If

(Y, ρ) is complete, does it follow that (X, d) is complete? Give proofs or

counterexamples as appropriate.

Exercise K.312. [14.1, P] Let X be the space of open intervals ± = (a, b)

with a < b in R. If ±, β ∈ X, then the symmetric di¬erence ± β consists

of the empty set, one open interval or two disjoint open intervals. We de¬ne

d(±, β) to be the total length of the intervals making up ± β. Show that d

is a metric on X.

Show that the completion of (X, d) contains precisely one further point.

Exercise K.313. [14.1, P] Consider the set N of non-negative integers. If

n ∈ N and n = 0, then there exist unique r, s ∈ N with s an odd integer and

n = s2r . Write »(n) = r. If n, m ∈ N and n = m we write

d(n, m) = 2’»(|n’m|) .

We take d(n, n) = 0.

(i) Show that d is a metric on N.

(ii) Show that the open ball centre 1 and radius 1 is closed.

(iii) Show that no one point set {n} is open.

(iv) Show that the function f : N ’ N given by f (x) = x2 is continuous.

(v) Show that the function g : N ’ N given by f (x) = 2x is not continuous

at any point of N.

602 A COMPANION TO ANALYSIS

(vi) Show that d is not complete. (Be careful. You must show that your

Cauchy sequence does not converge to any point of N.)

Re¬‚ect on what the completion might look like. (You are not called to

come to any conclusion.)

Exercise K.314. [Appendix C, P] Show that there exists a continuous

function F : R2 ’ R which is in¬nitely di¬erentiable at at every point except

0, which has directional derivative zero in all directions at 0 but which is not

di¬erentiable at 0.

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Index

abuse of language, 422 Big Oh and little oh, 506

algebraists, dislike metrics, 127, 589 bijective function, 475

alternating series test, 78 binomial expansion

antiderivative, existence and uniqueness, for general exponent, 297

186 positive integral exponent, 562

area, general problems, 169“172, 214“217, binomial theorem, 297, 562

229“231 bisection, bisection search, see lion hunt-

authors, other ing

Beardon, 56, 395 Bishop™s constructive analysis, 415“419

Berlinski, 20 Bolzano-Weierstrass

Billingsly, 230 and compactness, 421

Boas, 60, 152, 211 and total boundedness, 274

Bourbaki, 376, 422 equivalent to fundamental axiom, 40

for R, 38“39