greatly amuses mathematicians and greatly annoys educational theorists.

45

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

is the Euclidean norm on Rm , then the following re-

Lemma 4.1.4. If

sults hold.

(i) x ≥ 0 for all x ∈ Rm .

(ii) If x = 0, then x = 0.

(iii) If » ∈ R and x ∈ Rm , then »x = |»| x .

(iv) (The triangle inequality.) If x, y ∈ Rm then x + y ¤ x + y .

Proof. The triangle inequality can be deduced from the Cauchy-Schwarz in-

equality as follows.

2

=(x + y) · (x + y) = x 2 + 2x · y + y 2

x+y

¤ x 2 + 2 x y + y 2 = ( x + y )2 .

The remaining veri¬cations are left to the reader.

Exercise 4.1.5. (i) By carefully examining the proof of Lemma 4.1.2, or

otherwise, show that we have equality in the Cauchy-Schwarz inequality (that

is, we have |x · y| = x y ) if and only if x and y are linearly dependent

(that is, we can ¬nd real » and µ, not both zero, such that »x = µy).

(ii) Show that we have equality in the triangle inequality (that is x +

y = x + y ) if and only if we can ¬nd positive » and µ, not both zero,

such that »x = µy.

We observe that

1/2

m

(xj ’ yj )2

x’y =

i=1

which (at least if m = 1, m = 2 or m = 3) is recognisably the distance (more

properly, the Euclidean distance) between x and y. If we set x = a ’ b and

y = b ’ c, then the triangle inequality of Lemma 4.1.4 (iv) becomes

a’c ¤ a’b + b’c

which is usually read as saying that the length of one side of a triangle is less

than or equal to the sum of the lengths of the other two sides.

Exercise 4.1.6. If x = (x1 , x2 , . . . , xm ) ∈ Rm show that

m

max |xi | ¤ x ¤ |xi | ¤ m max |xi |.

1¤i¤m 1¤i¤m

i=1

Looking at each of the 3 inequalities in turn, ¬nd necessary and su¬cient

conditions for equality.

46 A COMPANION TO ANALYSIS

Exercise 4.1.7. We proved the triangle inequality by going via the Cauchy-

Schwarz inequality. Try and prove the inequality

1/2 1/2 1/2

3 3 3

x2 2

(xj + yj )2

≥

+ yj

j

i=1 i=1 i=1

directly.

Although we shall not take the matter to extremes, we shall have a strong

preference for coordinate free methods and statements. So far as I am aware,

no one has found a set of labelled axes (perhaps carved in stone or beau-

tifully cast in bronze) bearing an attestation from some higher power that

these are ˜nature™s coordinate axes™. Coordinate free statements and methods

encourage geometric intuition and generalise more readily.

Maxwell who played a crucial role in the development of vector methods

wrote in the ¬rst chapter of his great Treatise on Electricity and Magnetism

For many purposes of physical reasoning, as distinguished from

calculation, it is desirable to avoid explicitly introducing the Carte-

sian coordinates, and to ¬x the mind at once on a point of space

instead of its three coordinates, and on the magnitude and di-

rection of a force instead of its three components. This mode of

contemplating geometrical and physical quantities is more prim-

itive and more natural than the other. (Chapter 1, [37])

We now turn towards analysis.

De¬nition 4.1.8. We work in Rm with the Euclidean norm. We say that a

sequence a1 , a2 , . . . tends to a limit a as n tends to in¬nity, or more brie¬‚y

an ’ a as n ’ ∞,

if, given > 0, we can ¬nd an n0 ( ) such that

an ’ a < for all n ≥ n0 ( ).

Notice that this shows that De¬nition 1.2.1 was about the distance be-

tween two points and not the absolute value of the di¬erence of two numbers.

We can prove the following results on sequences in Rm in exactly the same

way as we proved the corresponding results for R (and more general ordered

¬elds F) in Lemma 1.2.2.

47

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

Lemma 4.1.9. We work in Rm with the Euclidean norm.

(i) The limit is unique. That is, if an ’ a and an ’ b as n ’ ∞, then

a = b.

(ii) If an ’ a as n ’ ∞ and n(1) < n(2) < n(3) . . . , then an(j) ’ a as

j ’ ∞.

(iii) If an = c for all n, then an ’ c as n ’ ∞.

(iv) If an ’ a and bn ’ b as n ’ ∞, then an + bn ’ a + b.

(v) Suppose an ∈ Rm , a ∈ Rm , »n ∈ R and » ∈ R. If an ’ a and

»n ’ », then »n an ’ »a.

Proof. Left to the reader.

Exercise 4.1.10. The parts of Lemma 1.2.2 do not all have corresponding

parts in Lemma 4.1.9. Explain brie¬‚y why these di¬erences occur.

Exercise 4.1.11. (i) If an ’ a and bn ’ b as n ’ ∞, show that an · bn ’

a.b. (You may ¬nd the Cauchy-Schwarz inequality useful.)

(ii) If x, y ∈ Rm show that

2

+ x ’ y 2 )/4.

x.y = ( x + y

Prove part (i) following the method of Exercise 1.2.6.

Lemma 4.1.9 is, of course, merely algebra and applies to Qm as much as

to Rm . In order to do analysis we need a more powerful tool and, in keeping

with the spirit of our general programme, we extend the Bolzano-Weierstrass

theorem to Rm .

Theorem 4.1.12. (Bolzano-Weierstrass.) If xn ∈ Rm and there exists a

K such that xn ¤ K for all n, then we can ¬nd n(1) < n(2) < . . . and

x ∈ Rm such that xn(j) ’ x as j ’ ∞.

Once again ˜any bounded sequence has a convergent subsequence™.

Proof. We prove the result for m = 2, leaving it to the reader to prove the

general result. Let us write xn = (xn , yn ). Observe that, since xn ¤ K,

it follows that |xn | ¤ K. By the Bolzano-Weierstrass theorem for the reals

(Theorem 3.2.1), it follows that there exist a real number x and a sequence

m(1) < m(2) < . . . such that xm(k) ’ x as k ’ ∞.

Since xm(k) ¤ K, it follows that |ym(k) | ¤ K so, again by the Bolzano-

Weierstrass theorem for the reals, if follows that there exist a real number y

and a sequence k(1) < k(2) < . . . such that ym(k(j)) ’ y as j ’ ∞.

Setting n(j) = m(k(j)) and x = (x, y) we have

xn(j) ’ x ¤ |xn(j) ’ x| + |yn(j) ’ y| = |xm(k(j)) ’ x| + |ym(k(j)) ’ y| ’ 0 + 0 = 0,

and so xn(j) ’ x as j ’ ∞.

48 A COMPANION TO ANALYSIS

Exercise 4.1.13. Prove Theorem 4.1.12 for general m.

Exercise 4.1.14. We can also prove Theorem 4.1.12 by ˜multidimensional

lion hunting™. In this exercise we again consider the case m = 2, leaving

the general case to the reader. She should compare this exercise with Exer-

cise 3.2.5.

We assume the hypotheses of Theorem 4.1.12. Consider the square

S0 = [a0 , b0 ] — [a0 , b0 ] = [’K, K] — [’K, K].

Explain why xn ∈ S0 for all n.

Set c0 = (a0 + b0 )/2, c0 = (a0 + b0 )/2 and consider the four squares

T0,1 = [a0 , c0 ] — [a0 , c0 ], T0,2 = [c0 , b0 ] — [a0 , c0 ],

T0,3 = [a0 , c0 ] — [c0 , b0 ], T0,4 = [c0 , b0 ] — [c0 , b0 ].

Explain why at least one of the squares T0,p , say, must be such that xr ∈ T0,p

for in¬nitely many values of r. We set [a1 , b1 ] — [a1 , b1 ] = T0,p .

Show that we can ¬nd a sequence of pairs of intervals [an , bn ] and [an , bn ]

such that

xr ∈ [an , bn ] — [an , bn ] for in¬nitely many values of r,

an’1 ¤ an ¤ bn ¤ bn’1 , an’1 ¤ an ¤ bn ¤ bn’1 ,

and bn ’ an = (bn’1 ’ an’1 )/2, bn ’ an = (bn’1 ’ an’1 )/2,

for all n ≥ 1.

Show that an ’ c as n ’ ∞ for some c ∈ [a0 , b0 ] and an ’ c as

n ’ ∞ for some c ∈ [a0 , b0 ]. Show further that we can ¬nd m(j) with

m(j + 1) > m(j) and xm(j) ∈ [aj , bj ] — [aj , bj ] for each j ≥ 1. Deduce the

Bolzano-Weierstrass theorem.

The proof of Theorem 4.1.12 involves extending a one dimensional result

to several dimensions. This is more or less inevitable because we stated

the fundamental axiom of analysis in a one dimensional form. However the

Bolzano-Weierstrass theorem itself contains no reference as to whether we are

working in R or Rm . It is thus a excellent tool for multidimensional analysis.

4.2 Open and closed sets

When we work in R the intervals are, in some sense, the ˜natural™ sets to

consider. One of the problems that we face when we try to do analysis in

many dimensions is that the types of sets with which we have to deal are

49

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

much more diverse. It turns out that the so called closed and open sets

are both su¬ciently diverse and su¬ciently well behaved to be useful. This

short section is devoted to deriving some of their simpler properties. Novices

frequently ¬nd the topic hard but eventually the reader will appreciate that

this section is a rather trivial interlude in a deeper discussion.

The de¬nition of a closed set is a natural one.

De¬nition 4.2.1. A set F ⊆ Rm is closed if whenever xn ∈ F for each n

and xn ’ x as n ’ ∞ then x ∈ F .

Thus a set is closed in the sense of analysis if it is ˜closed under the

operation of taking limits™. An indication of why this is good de¬nition is

given by the following version of the Bolzano-Weierstrass theorem.

Theorem 4.2.2. (i) If K is a closed bounded set in Rm then every sequence

in K has a subsequence converging to a point of K.

(ii) Conversely, if K is a subset of Rm such that every sequence in K has

a subsequence converging to a point of K, then K is a closed bounded set.

Proof. Both parts of the proof are easy.

(i) If xn is a sequence in K, then it is a bounded sequence and so, by

Theorem 4.1.12, has a convergent subsequence xn(j) ’ x, say. Since K is

closed, x ∈ K and we are done.

(ii) If K is not closed, we can ¬nd xn ∈ K and x ∈ K such that xn ’ x

/

as n ’ ∞. Since any subsequence of a convergent subsequence converges to

the same limit, no subsequence of the xn can converge to a point of K.

If K is not bounded, we can ¬nd xn ∈ K such that xn > n. If x is any

point of Rm , then the inequality

xn ’ x ≥ x n ’ x > n ’ x

shows that no subsequence of the xn can converge.

When working in Rm , the words ˜closed and bounded™ should always elicit

the response ˜Bolzano-Weierstrass™. We shall see important examples of this

slogan in action in the next section (Theorem 4.3.1 and Theorem 4.5.5).

The following remark is sometimes useful.

Exercise 4.2.3. (i) If A is a non-empty closed subset of R with supremum

±, then we can ¬nd an ∈ A with an ’ ± as n ’ ∞.

(ii) If A is a non-empty closed subset of R, then, if supa∈A a exists,

supa∈A a ∈ A.

We turn now to the de¬nition of an open set.

50 A COMPANION TO ANALYSIS

De¬nition 4.2.4. A set U ⊆ Rm is open if, whenever x ∈ U , there exists

an > 0 such that, whenever x ’ y < , we have y ∈ U .

Thus every point of an open set lies ˜well inside the set™. The ability

to ˜look in all directions™ plays an important role in many proofs. The ¬rst

example we shall see occurs in the proof of Rolle™s theorem (Theorem 4.4.4)

and the idea will play a key part in the study of complex analysis.

Exercise 4.2.5. Consider sets in R. Prove the following results. The inter-

val [a, b] = {x ∈ R : a ¤ x ¤ b} is closed, the interval (a, b) = {x ∈ R : a <

x < b} is open, the interval [a, b) = {x ∈ R : a ¤ x < b} is neither open nor

closed, R is both open and closed.

Lemma 4.2.6. Consider sets in Rm . Let x ∈ Rm and r > 0.

(i) The set B(x, r) = {y ∈ Rm : x ’ y < r} is open.

¯

(ii) The set B(x, r) = {y ∈ Rm : x ’ y ¤ r} is closed.

Proof. This is routine3 . There is no loss of generality in taking x = 0.

(i) If z ∈ B(0, r), then z < r so, if we set δ = r ’ z , it follows that

δ > 0. If z ’ y < δ, the triangle inequality gives

y ¤ z + z ’ y < z + δ = r,

so that y ∈ B(0, r). Thus B(0, r) is open.

¯

(ii) If yn ∈ B(0, r) and yn ’ y as n ’ ∞, then, by the triangle inequal-

ity,

y ¤ yn + y ’ yn ¤ r + y ’ yn ’ r + 0 = r

¯

as n ’ ∞. Thus y ¤ r and we have shown that B(0, r) is closed.

¯

We call B(x, r) the open ball of radius r and centre x. We call B(x, r)

the closed ball of radius r and centre x. Observe that the closed and open

balls of R are precisely the closed and open intervals.

The following restatement of the de¬nition helps us picture an open set.

Lemma 4.2.7. A subset U of Rm is open if and only if each point of U is

the centre of an open ball lying entirely within U .

Thus every point of an open set is surrounded by a ball consisting only

of points of the set.

The topics of this section are often treated using the idea of neighbour-

hoods. We shall not use neighbourhoods very much but they come in useful

from time to time.

3

Mathspeak for ˜It may be hard the ¬rst time you see it but when you look at it later

you will consider it to be routine.™

51

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

De¬nition 4.2.8. The set N is a neighbourhood of the point x if we can

¬nd an r > 0 (depending on both x and N ) such that B(x, r) ⊆ N .

Thus a set is open if and only if it is a neighbourhood of every point that

it contains.

Returning to the main theme we note the following remarkable fact.

Lemma 4.2.9. A subset E of Rm is open if and only if its complement

Rm \ E is closed.

Proof. Again, this is only a question of writing things down clearly. We split

the proof into two parts.

Necessity Suppose that E is open. If xn ∈ Rm \ E for all n and xn ’ x

as n ’ ∞, then we claim that x ∈ Rm \ E. For, if not, we must have

x ∈ E and so, since E is open, we can ¬nd an > 0 such that, whenever

a ’ x < , it follows that a ∈ E. Since xn ’ x, we can ¬nd an N such that

xN ’ x < and so xN ∈ E, contradicting the statement that xn ∈ Rm \ E.

Thus x ∈ Rm \ E and we have shown that Rm \ E is closed.

Su¬ciency Suppose that Rm \ E is closed. We show that E is open. For,

if not, there must exist an a ∈ E such that, given any > 0, there exists

a y ∈ E with y ’ a < . In particular, we can ¬nd xn ∈ Rm \ E such

/

that xn ’ a < 1/n. By the axiom of Archimedes, this means that xn ’ a

as n ’ ∞ and so, since Rm \ E is closed, a ∈ Rm \ E, contradicting our

assumption that a ∈ E. Thus E is open.

We observe the following basic results on open and closed sets.

Lemma 4.2.10. Consider the collection „ of open sets in Rm .

(i) … ∈ „ , Rm ∈ „ .

(ii) If U± ∈ „ for all ± ∈ A, then ±∈A U± ∈ „ .

(iii) If U1 , U2 , . . . , Un ∈ „ , then n Uj ∈ „ .

j=1

Proof. This is routine.

(i) Since the empty set contains no points, every point in the empty set

has any property we desire (in this case, that of being the centre of an open

ball lying within the empty set). Thus the empty set is open. If x ∈ Rm

then B(x, 1) ⊆ Rm . Thus Rm is open.

(ii) If x ∈ ±∈A U± , then we can ¬nd a particular ±(0) ∈ A such that

x ∈ U±(0) . Since U±(0) is open, we can ¬nd a δ > 0 such that B(x, δ) ⊆ U±(0) .

Automatically, B(x, δ) ⊆ ±∈A U± . We have shown that ±∈A U± is open.

(iii) If x ∈ n Uj , then x ∈ Uj for each 1 ¤ j ¤ n. Since each Uj is

j=1

open we can ¬nd a δj > 0, such that B(x, δj ) ⊆ Uj for each 1 ¤ j ¤ n.

Setting δ = min1¤j¤n δj , we have δ > 0 (note that this part of the argument

52 A COMPANION TO ANALYSIS

requires that we are only dealing with a ¬nite number of open sets Uj ) and

B(x, δ) ⊆ Uj for each 1 ¤ j ¤ n. Thus B(x, δ) ⊆ n Uj and we have

j=1

shown that n Uj is open.

j=1

Lemma 4.2.11. Consider the collection F of closed sets in Rm .

(i) … ∈ F, Rm ∈ F.

(ii) If F± ∈ F for all ± ∈ A, then ±∈A F± ∈ F.

(iii) If F1 , F2 , . . . , Fn ∈ F, then n Fj ∈ F.

j=1

Proof. This follows from Lemma 4.2.10 by repeated use of Lemma 4.2.9.

(i) Observe that … = Rm \Rm and Rm = Rm \…. Now use Lemma 4.2.10 (i).

(ii) Observe that

F± = R m \ (Rm \ F± )

±∈A ±∈A

and use Lemma 4.2.10 (ii).

(iii) Observe that

n n

Fj = R m \ (Rm \ Fj )

j=1 j=1

and use Lemma 4.2.10 (iii).

Exercise 4.2.12. We proved Lemma 4.2.10 directly and obtained Lemma 4.2.11

by complementation. Prove Lemma 4.2.11 and obtain Lemma 4.2.10 by com-

plementation.

Exercise 4.2.13. (i) We work in R and use the usual notation for intervals

(see Exercise 4.2.5 if necessary). Show that

∞

(’1 ’ j ’1 , 1) = [’1, 1)

j=1

and conclude that the intersection of open sets need not be open. Why does

this not contradict Lemma 4.2.10?

(ii) Let U1 , U2 , . . . be open sets in R such that U1 ⊇ U2 ⊇ U3 ⊇ . . . .

Show, by means of examples, that ∞ Uj may be (a) open but not closed,

j=1

(b) closed but not open, (c) open and closed or (d) neither open nor closed.

(iii) What result do we get from (iii) by complementation?

(iv) Let Fj = [aj , bj ] and F1 ⊆ F2 ⊆ F3 ⊆ . . . Show, by means of exam-

ples, that ∞ Fj may be (a) open but not closed, (b) closed but not open,

j=1

(c) open and closed or (d) neither open nor closed.

53

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

(v) Let a < b and c < d. Show that, if we work in R2 , [a, b] — [c, d] is

closed, (a, b) — (c, d) is open and (a, b) — [c, d] is neither open nor closed.

(vi) Do part (ii) with R replaced by R2 .

(vii) If A is open in Rm and B is open in Rn , show that A — B is open

in Rm+n . State and prove the corresponding result for closed sets.

Of course, analysis deals with (reasonably well behaved) functions as

well as sets. The notion of continuity gives us a natural class of reasonably

well behaved functions. The de¬nition carries over unchanged from the one

dimensional case.

De¬nition 4.2.14. Let E ⊆ Rm . We say that a function f : E ’ Rp is

continuous at some point x ∈ E if, given > 0, we can ¬nd a δ( , x) > 0

such that, whenever y ∈ E and x ’ y < δ( , x), we have

f (x) ’ f (y) < .

If f is continuous at every point x ∈ E, we say that f is a continuous function

on E.

This may be the place to make a comment on vector notation. It is

conventional in elementary analysis to distinguish elements of Rm from those

in R by writing points of Rm in boldface when printing and underlining

them when handwriting. Eventually this convention becomes tedious and,

in practice, mathematicians only use boldface when they wish to emphasise

that vectors are involved.

Exercise 4.2.15. After looking at Lemma 1.3.2 and parts (iii) to (v) of

Lemma 4.1.9, state the corresponding results for continuous functions. (Thus

part (v) of Lemma 4.1.9 corresponds to the statement that, if » : E ’ R and

f : E ’ Rp are continuous at x ∈ E, then so is »f .) Prove your statements

directly from De¬nition 4.2.14.

Suppose that E ⊆ Rm and f : E ’ R is continuous at x. Show that, if

f (t) = 0 for all t ∈ E, then 1/f is continuous at x.

Once again we have the following useful observation.

Lemma 4.2.16. Let E be a subset of Rm and f : E ’ Rp a function.

Suppose that x ∈ E and that f is continuous at x. If xn ∈ E for all n and

xn ’ x as n ’ ∞, then f (xn ) ’ f (x) as n ’ ∞.

Proof. Left to the reader.

Another way of looking at continuity, which will become progressively

more important as we proceed, is given by the following lemma.

54 A COMPANION TO ANALYSIS

Lemma 4.2.17. The function f : Rm ’ Rp is continuous if and only if

f ’1 (U ) is open whenever U is an open set in Rp .

The reader may need to be reminded of the de¬nition

f ’1 (U ) = {x ∈ Rm : f (x) ∈ U }.

Proof. As with most of the proofs in this section, this is just a matter of