basic features of euclidean space

FUNCTIONS OF SEVERAL REAL VARIABLES World Scientific Publishing Co. Pte. Ltd.http://www.worldscibooks.com/mathematics/7672.html

Chapter 1

Basic Features of EuclideanSpace, Rn

This chapter is foundational. The reader who is familiar with this ma-terial can go on to Chapter 2. Those who are not can either go overthe chapter before going on, or can begin with the latter chapters andreturn to the parts of Chapter 1 as needed. Here we review basic fea-tures of Euclidean space Rn and prove results concerning orthogonality,convergence, completeness and compactness.

1.1 Real numbers

In this section we briefly give a review of the basic properties of realnumbers. There are a number of ways to deal with the real numbers,all more or less equivalent. The way we shall do it (and perhaps themost efficient way) is by means of the least upper bound axiom . Theintegers are denoted by Z = {...,2,1, 0, 1, 2, ...} and the positiveintegres by Z+ = {1, 2, 3, ...}. The set Z+ is also called the setof natural numbers and is sometimes denoted by N. The principleof mathematical induction states that: If a subset S N satisfies(i) 1 S (ii) n S (n + 1) S, then S = N. The set of ratio-nal numbers is denote by Q =

{mn : m,n Z, n 6= 0

}. It was known

since the times of Pythagoras that2 / Q. Such numbers can not

be written as fractions and are called irrational numbers. As we shall

1


2 Chapter 1 Basic Features of Euclidean Space, Rn

see shortly, there are infinitely many irrational numbers. The set ofrationals together with set of irrationals form the set of real numbersR. That is, R = Q Qc. We shall assume the standard arithmeticoperations and their properties with respect to the real numbers. Weshall also assume the ordering of real numbers ( and ).Definition 1.1.1. Let S be a nonempty subset of R.

1. A real number M is called an upper bound of S if x M for allx S. In this case S is said to be bounded above.

2. A real number m is called a lower bound of S if m x for allx S. In this case S is said to be bounded below.

3. The set S is said to be bounded if it is both bounded above andbounded below. Thus S is bounded if there exist real numbers mand M such that S [m,M ].

4. A set which is not bounded is called unbounded.

5. if M S we call M the maximal element of S. If m S we callm the minimal element of S.

Definition 1.1.2. Let S be a nonempty bounded subset of R.

1. If S has a least upper bound, then we call it the supremum of S,denoted by sup(S) (or sometimes lub(S)).

2. If S has a greatest lower bound, then we call it the infimum of S,denoted by inf(S) (or sometimes glb(S)).1

The Completeness Axiom. Every nonempty subset S of R whichis bounded above has a least upper bound. That is, there exists a realnumber, sup(S).

If S is a set of real numbers which is bouned below, then by consid-ering the set S = {x : x S} we may state the completeness axiomin the alternative form as:2 Every nonempty set of real numbers whichis bounded below has an infimum in R.

1lub, means least upper bound and glb, means greatest lower bound.2Changing sign reverses the order in R.


1.1 Real numbers 3

Observe that if S is bounded above, then M = sup(S) is the uniquenumber with the following two properties:(i) x M for all x S, and (ii) for every > 0, there exists x S withx > M .In other words, (i) tells us that M is an upper bound, whereas (ii) tellsus that there is no smaller upper bound. Similarly, if S is boundedbellow, then m = inf(S) is the unique number such that(i) m x for all x S and (ii) for every > 0, ther exist x S withx < m+ .

Note that sup(S) = inf(S) and inf(S) = sup(S). If S has noupper bound we shall define sup(S) = +, and if S has no lower bound,we shall define inf(S) = .

As the following example shows the sup(S) may or may not belongto S itself. However, we shall see in Section 1.5 (Corollary 1.5.8), italways belongs to the closure of S. Similarly for inf(S).Example 1.1.3. 1. Let S =

{1n : n = 1, 2, ...

}. Then the sup(S) = 1

and is the maximal element of S. On the other hand the inf(S) = 0and 0 / S.

2. Let a, b R with a < b. Thensup[a, b] = sup[a, b) = sup(a, b] = sup(a, b) = b

andinf[a, b] = inf[a, b) = inf(a, b] = inf(a, b) = a.

Theorem 1.1.4. (Archimedean property). Let a, b R with a > 0 andb > 0. Then there exists a positive integer n such that na > b.

Proof. Suppose the statement is false. That is, there exist a > 0 andb > 0 such that na b for all n Z+. Then, the set S = {na : n Z+}is a nonempty subset of R bounded above by b. By the completenessaxiom S has a supremum. Let M = sup(S). Hence, na M for alln Z+. Therefore (n+ 1)a M , or na M a for all n Z+. Thus,M a is also an upper bound of S. But M a < M . This contradictsthat M = sup(S). This contradiction proves the theorem.

Corollary 1.1.5. For any positive real number b there exists n Z+such that n > b. In particular, for every > 0 there exists n Z+ suchthat 1n < .



Proof. Taking a = 1 in Theorem 1.1.4, we get n = n1 > b. Settingb = 1 yields

1n < .

Corollary 1.1.6. For any positive real number x there exists a (unique)m Z+ such that m 1 x < m.Proof. By Corollary 1.1.5, there exists a positive integer n such thatx < n. Let m be the least such positive integer (and so unique). Thenm 1 x < m.Theorem 1.1.7. (Denseness of Q). Let a, b R with a < b. Thenthere exists a rational number r Q such that a < r < b.Proof. It is enough to prove the theorem for 0 < a < b. Since ba > 0,it follows from the Archimedean property that there exist n Z+ suchthat n(ba) > 1. That is, nb > na+1. By Corollary 1.1.6, there existsm Z+ such thatm1 na < m. Now we have na < m na+1 < nb.Hence, na < m < nb. Therefore, a < mn < b.

The following proposition tells us that there are infinitely many irra-tionals. In fact, there are so many that, as we prove in the next theorem,they are densely distributed on the real line.

Proposition 1.1.8. Let n Z+ with n not a perfect square. Thenn / Q.

Proof. Suppose thatn were rational. Then it can be written as

n =

ab with a, b Z+ and a + b be smallest possible. Let k be the uniquepositive integer such that k 0 and 2 > 0, by the Archimedean property, thereexist n Z+ such that n(b a) >

2. Therefore,

b > a+

2

n> a+

2

2n> a.

Since(a+

2n

)(a+

2

2n

)=

2n / Q, it follows that at least one of

the numbers 1 = a+2n or 2 = a+

2

2n is irrational.

Definition 1.1.10. The absolute value of a real number x is defined by

|x| ={x when x 0,x when x < 0.

Note that |x|2 = x2 and |x| =x2.

For x, y R the absolute value has the following properties.

1. |x| 0 and |x| = 0 if and only if x = 0.

2. |xy| = |x| |y|

3. |x+ y| |x|+ |y| (triangle inequality).

Moreover,

| |x| |y| | |x y|.We remark that |x+ y| = |x| + |y| if and only if xy 0, and |x + y| 0 we have

((a r, a+ r) {a}) S 6= .

It is worthwhile remarking



1. A finite set S = {s1, ..., sm} has no limit points. To see this taker = min {|si sj | : i, j = 1, ..,m}, then ((si r, si + r) {si}) S = .

2. If a is a limit point of S, then every open interval (a r, a + r)around a contains infinitely many points of S; for if (a r, a +r) contained a finite number of points s1, ..., sk of S distinctfrom a, then the open interval around a with radius r0 =min {|a sj| : j = 1, ..., k} contains no point of S distinct froma, which is impossible.

The following important result characterizes the sets in R whichhave limit points. In the next section we shall extend it to Rn.

Theorem 1.1.12. (Bolzano-Weierstrass)3 Every bounded infinite setin R has at least one limit point.

Proof. Let S be a bounded infinite set in R. Since S is bounded thereis a closed interval [a, b] containing S. Hence, both m = inf(S) andM = sup(S) exist. Consider the set

A = {x R : (, x) S is at most finite} .

Since m A it follows that A 6= . At the same time A is boundedabove by M . In fact, if x0 R with M < x0, then x < M < x0for all x S. Since S is infinite, this implies that x0 / A. Hence,x M for all x A. It follows that A has a least upper bound, callit y = sup(A). We show that y is a limit point of S. Let > 0 beany positive number. Then (y + ) / A and so (, y + ) S isinfinite. Since y = sup(A) there exists x1 A such that y < x1 y.Therefore (, y ) S (, x1) S and so (, y ) S is atmost finite. Hence, (y , y + ) contains an infinite number of pointsof S. Thus y is a limit point of S.

3B. Bolzano (1781-1848). Mathematician, logician and philosopher.K. Weierstrass (1815-1897). Mathematician who made many contribution in mathe-matical analysis and devoted a great deal of attention to the rigorous foundation ofanalysis.


1.1 Real numbers 7

EXERCISES

1. Show that if r is a nonzero rational number and x is irrational,then x r, r x, rx, rx and xr are all irrational.

2. Show the least upper bound of a bounded set is unique.

3. If 6= A B R and B is bounded, show thatinf(B) inf(A) sup(A) sup(B).

4. Let A and B be two nonempty bounded sets of real numbers andc R. If A + B = {a+ b : a A, b B} and cA = {ca : a A}.Show that

(a) sup(A + B) = sup(A) + sup(B) and inf(A + B) = inf(A) +inf(B).

(b) When c 0, then sup(cA) = c sup(A) and inf(cA) =c inf(A).When c < 0, then sup(cA) = c inf(A) and inf(cA) =c sup(A).

5. Let S ={x+ 1x : x > 0

}. Show that sup(S) = and inf(S) = 2.

6. Let S ={

mm+n : m,n N

}. Show that sup(S) = 1 and inf(S) =

0.Hint. Fix m = 1 (resp. n = 1).

7. Let S ={x Q+ : x2 < 2

}. Show that sup(S) / Q. Thus, Q is

not (order) complete.

8. Let x R and S = {r Q : r < x}. Show that sup(S) = x9. Let a, b R. Show that |ab| 12 (a2 + b2). Deduce that if a 0

and b 0, then ab 12 (a+ b).10. Show that if 0 a b c and b > 0, then 2ac ab + bc 1+ ac .11. (Bernoullis inequality). For any x R and k Z+, show that

(1 + x)k 1 + kx.



12. Let x, y R. Show that(a) |xy| = |x| |y|.(b) |x y| |x|+ |y|.(c) ||x| |y|| |x y|.

13. Let x, y R. Show that

max {x, y} = x+y+|yx|2 , min {x, y} = x+y|yx|2 .

14. Let x, y R with x 6= y. Show that |x||x+y| + |y||xy| 1.15. Let x, y R. Show that

|x+ y|1 + |x+ y|

|x|1 + |x| +

|y|1 + |y| .

1.1.1 Convergence of sequences of real numbers

If for every k Z+ there is an associated real number ak, the orderedlist {a1, a2, ...}, briefly written as {ak}k=1, is called a sequence. Given areal sequence {ak}k=1, the major question that one might ask about thesequence is how its terms behave for k arbitrarily large, i.e, as k .Definition 1.1.13. Let {ak}k=1 be a sequences of real numbers. Thesequence is said to converge to a R, if for every > 0 there existsa positive integer N(depending on ) such that |ak a| < wheneverk N . We write limk ak = a or ak a as k .

Equivalently, limk ak = a if for every > 0 there exists a positiveinteger N such that ak (a , a+ ) for all k N . If a sequence doesnot converge (has no limit), we say that the sequence diverges.

Proposition 1.1.14. If a sequence converges, then its limit is unique.

Proof. Suppose a convergent sequence {ak} had two limits, say ak aand ak b with a 6= b. Then for = |ab|2 > 0 there exist positiveintegers N1 and N2 such that |aka| < 2 for all k N1 and |akb| < 2for all k N2. But, then for all k N3 = max {N1, N2} we have|a b| = |a ak + ak b| |ak a|+ |ak b| < 2 + 2 = = |ab|2 , thatis, 1 < 12 which is impossible.


1.1 Real numbers 9

Example 1.1.15. 1. Let ak =1kp with p > 0. Then limk

1kp = 0.

2. Let x R with |x| < 1. Then limk xk = 0.Solution.

1. Let > 0 be any positive number. Then | 1kp 0| = 1kp < if andonly if kp > 1 or k > (

1 )

1p . Hence, taking N > (1 )

1p , we have

| 1kp 0| < for all k N .

2. If x = 0 this is trivial. Assume x 6= 0 and let > 0. Then

|xk| = |x|k < if and only if k log |x| < log() or k > log()log |x| .

If 0 < < 1, take N > log()log |x| . Then |xk| < for all k N .

Proposition 1.1.16. Let {ak}k=1 be a sequence. If ak a, then |ak| |a|. In particular, ak 0 if and only if |ak| 0.

Proof. The first assertion follows from the inequality ||ak||a|||aka|.In the particular case where a = 0, the converse is also true, for if|ak| |0| = 0, then for any > 0, there is N such that | |ak| 0| < for all k N , that is, |ak| < for all k N , or ak 0.

Definition 1.1.17. Let {ak}k=1 be a sequence of real numbers.

1. The sequence {ak}k=1 is called increasing (resp. decreasing) ifak ak+1 (resp. ak ak+1) for all k = 1, 2, .... When theseinequalities are strict, we say the sequence {ak}k=1 is strictlyincreasing (resp. strictly decreasing). A sequence that is ei-ther (strictly) increasing or (strictly) decreasing is called (strictly)monotone.

2. The sequence {ak}k=1 is called bounded if there existsM > 0 with|ak| M for all k = 1, 2, .... A sequence which is not bounded iscalled unbounded.

Proposition 1.1.18. Every convergent sequence is bounded.



Proof. Suppose ak a. Then for = 1 there exists positive integer Nsuch that |ak x| < 1 for all k N . So, for all k N we have

|ak| = |(ak a) + a| |ak a|+ |a| < 1 + |a|.Taking M = max {|a1|, ..., |aN |, 1 + |a|}, we have |ak| M for all k =1, 2, ...; That is, {ak}k=1 is bounded.Corollary 1.1.19. An unbounded sequence diverges.

The following result is fundamental.

Theorem 1.1.20. Every bounded monotone sequence in R is conver-gent. More precisely, the limit of an increasing (resp. decreasing) se-quence is the supremum (resp., infimum) of its set of values.

Proof. Suppose {ak}k=1 is increasing and bounded. Then the set{ak : k = 1, 2, ...} (the range) of the sequence is bounded above. It fol-lows from the completeness axiom that this set has a supremum, call itL. Since L is an upper bound, we have ak L for all k. Moreover, sinceL is the least upper bound, for any > 0 there is some N for whichaN > L . Since {ak}k=1 is increasing, for all k N we also haveak aN > L . Therefore, L < ak L < L+ for all k N andthis shows that ak L.

Similarly, a decreasing sequence bounded from below converges tothe infimum of the set of its values.

Example 1.1.21. Let ak =xk

k! for x R, where k! = 1 2 (k 1) k.Then ak 0 as k (that is, k! grows faster than exponentially ask ).Solution.

To see this we consider three cases; If |x| 1, then

0 |x|k

k! 1k! 1k 0.

For x > 1, we will argue using Theorem 1.1.20. Note thatak+1ak

=x

k+1 0 as k . Therefore, for = 1, we can find N such thatak+1 < ak for all k N . Hence, {ak} is (eventually) decreasing. More-over, since 0 < ak < max {a1, ..., aN}, the sequence is also bounded.


1.1 Real numbers 11

It follows, from Theorem 1.1.20, that {ak} converges to a number, saya 0. Now, a = limk ak+1 = limk(ak xk+1) = a 0 = 0. Thus,limk x

k

k! = 0 for all x 0. Finally, if x < 0, then by the aboveargument |ak| = |x|

k

k! 0 as k . But then also ak 0 as k .The following consequence of Theorem 1.1.20 is also an important

property of R.

Theorem 1.1.22. (Nested intervals property). Let In = [an, bn] be asequence of closed bounded intervals in R such that In+1 In for alln = 1, 2, ... and the length bn an of In tends to 0 as n . Thenthere is exactly one point contained in all of the intervals In. That is,there exists R such that n=1 In = {}.Proof. The condition I1 I2 I3 ... means that a1 a2 a3 ... and b1 b2 b3 .... So the sequence {an} is increasing andthe sequence {bn} is decreasing. Since a1 < an bn < b1, both arebounded. Therefore, by Theorem 1.1.20, both converge, say an and bn . Moreover, since bn an 0, it follows that = .Obviously is the supremum of {an : n = 1, 2, ...} and the infimum of{bn : n = 1, 2, ...} and so an bn for all n. Thus, belongs to allthe intervals. Suppose, that 1 6= were another point common to allthe intervals In. Suppose, for example, an 1 < bn for all n.Then bn an 1 6= 0. Letting n , this gives 1 whichcontradicts the fact 1 < . Similarly, if an < 1 bn we also get acontradiction.

Definition 1.1.23. A number R is called a limit point (or accu-mulation point) of a sequence {ak}k=1, if every open interval around contains an infinite number of terms of the sequence.

We remark that a limit point of the range {ak : k = 1, 2, ...} ofthe sequence is also a limit point of the sequence. The converse is nottrue; for instance, the points 1 and 1 are limit points of the sequenceak = (1)k. However, the range of the sequence is {1, 1} and has nolimit point.

Theorem 1.1.24. (Bolzano-Weierstrass for sequences) Every boundedsequence in R has a limit point.



Proof. Suppose {ak}k=1 is bounded. Then S = {ak : k = 1, 2, ...} is abounded set of real numbers. There are two possiblities. Either S isfinite or S is infinite. If S is finite, then there is some R for whichak = for an infinite number of values of k. Hence is a limit pointof {ak}k=1. On the other hand, if S is infinite, the Bolzano-Weierstrasstheorem, implies that S has a limit point, say . Therefore, for every > 0 we have ak ( , + ) for infinitely many k and thus is alimit point of {ak}k=1.Definition 1.1.25. Let {ak}k=1 be a sequence and {mk}k=1 be anincreasing sequence of positive integers, that is, m1 < m2 < ... < mk 0 there exists N suchthat |ak a| < for all k N . Since m1 < m2 < ... < mk < ..., itfollows by induction on k, that mk k. Hence, for all mk k N , wehave |amk a| < . Thus, amk a.

The converse is not true; For the sequence ak = (1)k, the subse-quences a2k = 1 and a2k1 = 1 converge to 1 and 1 respectively,but the sequence itself does not converge. However we ask the reader toprove in Exercise 2 below, that if a2k a and a2k1 a, then ak a.

The Bolzano-Weiertsrass Theorem has the following importantcorollary.

Corollary 1.1.28. Every bounded sequence in R has a convergent sub-sequence.

Proof. Suppose {ak}k=1 is bounded. By the Bolzano-Weierstrass theo-rem, {ak}k=1 has a limit point, say . Then, given any > 0 the openinterval ( , + ) contains infinitely many terms of the sequence. Letam1 , am2 , ..., amk , ... be the terms of the seuqence in ( , + ) thatcorrespond to the indices m1 < m2 < ... < mk < .... Then {amk}k=1 isa subsequence of {ak}k=1 and amk .


1.1 Real numbers 13

Definition 1.1.29. A sequence {ak}k=1 is called a Cauchy4 sequence if|ak aj | 0 as k, j , that is, for every > 0 there exists positiveinteger N such that |ak aj | < for all k, j N .

Note that every Cauchy sequence is bounded. In fact, for = 1 thereexists N such that for k, j N we have |akaj | < 1. Hence, for k > Nwe have |ak| = |akaN+1+aN+1| |akaN+1|+ |aN+1| < 1+ |aN+1|.Taking as beforeM = max {|a1|, ..., |aN |, 1 + |aN+1|}, we have |ak| Mfor all k = 1, 2, ....

Proposition 1.1.30. 1. Every convergent sequence is a Cauchysequence.

2. A Cauchy sequence converges, if it has a convergent subsequence.

Proof. 1. Suppose ak a. Then |ak aj | |ak a|+ |aj a| 0when k, j . Thus, {ak}k=1 is Cauchy.

2. Let > 0 be given. Since {ak} is Cauchy there exists N such that|akaj| < 2 for all k, j N . On the other hand suppose that theconvergent subsequence {amk} converges to a. Then we can find apositive integer k large enough so that mk > N and |amka| < 2 .So, for k > N we have |aka| |akamk |+ |amka| < 2+ 2 = .Thus, ak a.

The following theorem is also fundamental. This property of R isalso refered to as the completeness of R.

Theorem 1.1.31. Every Cauchy sequence in R converges.

Proof. Suppose {ak}k=1 is Cauchy. Then it is bounded. By Corollary1.1.28 {ak}k=1 has a convergent subsequence. Therefore from Proposi-tion 1.1.30 (2) {ak}k=1 converges.

This concludes our preliminary work on R. In the next sections wewill extend these results to Rn.

4A. Cauchy (1789-1857). Mathematician who made many important contributionsin real and complex analysis.



EXERCISES

1. Let {ak} be a sequence. Show that if ak a, then ak+m a forany m Z+.

2. Let {ak} be a sequence. Show that

if a2k a and a2k1 a, then ak a.

3. Let ck ak bk for all k > N . Show that, if ck a and bk a,then ak a. (This is known as the squeezing theorem).

4. Let {ak} be a sequence. Show that

if limkak+1ak

= l with 1 < l < 1, then limk ak = 0.

5. Let ak =kk. Show that limk ak = 1.

6. Let ak =k!kk. Show that limk ak = 0.

7. Let 0 < < 1 and ak = (+1k )

k. Show that limk ak = 0.

8. Let ak = (1+1k )

k. Show that limk ak = e (this is the definitionof the number e 2.718 called Eulers number). Hint. ApplyTheorem 1.1.20 to the sequence bk = (1 +

1k )

k+1 and note thatlimk bk = limk ak.

9. Show that limk(1 1k )k = 1e .10. Show that limk(1 1k2 )k = 1.11. Show that limk(1 + 12k )

k =e.

12. Show that limk(1 + 2k )k = e2.

13. Let ak = 1 +k

k+1 cos(k2 ). Show that {ak} is bounded. However,

the sequence diverges. Hint. Use subsequences.

14. Let x R with |x| < 1. Show that

limk

(1 + x+ x2 + ...+ xk) =1

x 1 .

Hint. See Example 1.1.15 (2).


1.2 Rn as a vector space 15

15. Show that

limk

[1

k2 + 1+

1k2 + 2

+ ...+1

k2 + k

]= 1.

Hint. Use the squeezing theorem.

1.2 Rn as a vector space

For n > 1, Rn will denote the set of all n-tuples (pairs, triples, quadru-ples,...) of real numbers, that is,

Rn = {x = (x1, x2, ..., xn) : xi R, i = 1, 2, ..., n} .

Each element x Rn is called a vector and the numbers x1, x2, ..., xnare called the first, second,..., nth component of x respectively. Any realnumber will be called a scalar. In the important case (from a geometricalpoint of view) when n = 2 or n = 3, we shall use the familiar notation

R2 = {(x, y) : x, y R} for the xy-plane andR3 = {(x, y, z) : x, y, z R} for the xyz-space.

Just as the real numbers are represented by points on the number line,similarly any x = (x1, ..., xn) Rn represents a point (or vector) inRn. For each i = 1, ..., n the real numbers xi are called the Cartesian

5

coordinates of x. The point (0, ..., 0) corresponds to origin in Rn. If weagree that all vectors have their initial point at the origin, then a pointin Rn determines a vector in Rn and vice versa. Thus we can identify(and we do so) a point in Rn with the vector having initial point theorigin and terminal point the given point in Rn.

The set Rn has both an algebraic (vector space) structure and ametric space structure. To give Rn its algebraic structure, we defineon Rn two operations, addition and scalar multiplication that make ita vector space over the field of real numbers (or a real vector space).Formally, let x, y Rn with x = (x1, ..., xn) and y = (y1, ..., yn). Thesum of x and y is defined by

5R. Descartes (1596-1650). Philosopher, mathematician and physicist who madefundamental contributions to scientific thought and knowledge.



u

u+vv

Figure 1.1: Sum of vectors

x+ y = (x1 + y1, ..., xn + yn),

that is, by adding corresponding components. Since addition in R isassociative and commutative with identity element the number 0, it isevident that addition in Rn is also associative and commutative and thezero vector 0 = (0, ..., 0) is the identity element for vector addition.Figure 1.1 in the plane shows that this addition is exactly the parallelo-gram law for adding vectors in Physics. Note that even in Rn the spacedetermined by x and y is a plane, so this picture is valid in general.

The second operation is multiplication of a vector x Rn by a scalarc R called scalar multiplication and is defined as follows;

cx = (cx1, ...cxn).

This gives a vector pointing in the same direction as x, but of a differentmagnitude. Here as is easily checked, for x, y Rn and c, d R we havec(x + y) = cx + cy, (c + d)x = cx + dx,c(dx) = d(cx) = (cd)x and1x = x. When all these properties are satisfied we say Rn is a realvector space. In general, if Rn above is replaced by any set V on whichan addition and a scalar multiplication can be defined so that all theabove properties hold V is called a real vector space.

Definition 1.2.1. Let V be a real vector space. Given v1, v2, ..., vk in Vand scalars c1, c2, ...ck R the sum

kj=1 cjvj = c1v1 + c2v2 + ...+ ckvk

is called a linear combination of the vectors v1, v2, ..., vk. Moreover, theset of vectors which are linear combinations of v1, v2, ..., vk is called thelinear span of {v1, v2, ..., vk} and we denote it by span {v1, v2, ..., vk}.


1.2 Rn as a vector space 17

Definition 1.2.2. A set of nonzero vectors {v1, v2, ..., vm} in V is said tobe linearly independent if the relation

mj=1 cjvj = 0 implies all cj = 0.

In general a set S of vectors in V is said to be linearly independent ifeach finite subset of S is linearly independent.

Clearly, any subset of a linearly independent set is linearly inde-pendent. A set of vectors {w1, w2, ..., wm} which is not linearly inde-pendent is called linearly dependent, that is, {w1, w2, ..., wm} is linearlydependent if there exists scalars c1, ..., cm with at least one cj 6= 0 withm

j=1 cjwj = 0.

Definition 1.2.3. A set of vectors B in a vector space V is called abasis for V if B is linearly independent and span(B) = V . The spaceV is said to be finite dimensional if it has a finite basis. The numberof vectors in the basis is called the dimension of V and is denoted bydim(V ).

We remark that any finite dimensional vector space has a finite basisand any two bases have the same number of elements. This number isthe dimension of the vector space. Since we wont need this level ofdetail we leave this fact as an exercise (Exercise 1.8.20).6

Example 1.2.4. In Rn the set of vectors

B = {e1 = (1, 0, 0, ..., 0), e2 = (0, 1, 0, ..., 0), ..., en = (0, 0, 0, ..., 0, 1)}

is a basis for Rn, called the standard basis of Rn. Indeed, supposec1e1 + c2e2 + ...+ cnen = 0. Then (c1, 0, 0, ..., 0) + (0, c2, 0, ..., 0) + ...+(0, 0, ..., 0, cn) = (0, 0, ..., 0), that is, (c1, c2, ..., cn) = (0, 0, ..., 0). Hencec1 = c2 = ... = cn = 0 and so B is linearly independent. At the sametime for any x = (x1, x2, ..., xn) Rn we have x = x1e1+x2e2+...+xnen.That is, span(B) = Rn. Thus, B is a basis and dim(Rn) = n.

Proposition 1.2.5. A set of vectors B = {v1, ..., vn} is a basis of V ifand only if every v V can be expressed as a unique linear combinationof the vj (j = 1, ...n).

6For a detailed account of vector spaces, see [14].



Proof. Suppose {v1, ..., vn} is a basis of V . Then any v V is a linearcombination of the vj . If this can be done in more than one way, sayn

j=1 cjvj = v =n

j=1 djvj . Thenn

j=1(cj dj)vj = 0. Since thevj are linearly independent, it follows that cj dj = 0 or cj = dj .Conversely, suppose that every v V can be expressed as a uniquelinear combination of the vj. This is in particular true for the zerovector. Hence the vj are linearly independent and so form a basis.

The unique scalars c1, c2, ..., cn R used to express v V as alinear combination of the basis elements B = {v1, ..., vn} are calledthe coordinates of v relative to the (ordered) basis B and the vector[v] = (c1, c2, ..., cn) Rn is called the coordinate vector of v V . Notethat when V = Rn, the coordinates of v V relative to the standardbasis are just the components of v, that is v = [v].

Definition 1.2.6. Let V be a vector space. A linear subspace of V isa subset W of V which is itself a vector space with the operations ofaddition and scalar multiplication on V .

Equivalently, a non-empty subset W of V is a linear subspace ofV if and only if for each u, v W and each scalar c R the vectoru+ cv W. That is, W is closed under the operations of addition andscalar multiplication.

In any vector space V the subset {0} is clearly a linear subspacecalled the trivial subspace of V . The largest linear subspace of V isV itself and the smallest is {0}. Note that linear subspaces are closedunder taking linear combinations. In a vector space V the span(S), ofany subset S V is a linear subspace called the subspace generated byS.

From Example 1.2.4, the dimension of Rn is n. The dimension ofthe trivial subspace is taken to be 0 and the dimension of any otherproper linear subspace of Rn is an integer strictly between 0 and n. Inparticular, in R2 and R3 the one-dimensional subspaces are lines passingfrom the origin. In R3 the two-dimensional subspaces are planes passingfrom the origin. In fact, any two linearly independent vectors u, v R3generate the plane {su+ tv : s, t R}.


1.3 Rn as an inner product space 19

1.3 Rn as an inner product space

1.3.1 The inner product and norm in Rn

Definition 1.3.1. Let V be a real vector space. An inner product on Vis an operation denoted by , which associates to each pair of vectorsx, y V a real number x, y and for all x, y, z V and c R satisfiesthe following properties;

1. x, x 0 and x, x = 0 if and only if x = 02. x, y = y, x3. x+ y, z = x, z+ y, z4. cx, y = cx, y

In this case, V is called an inner product space.

In Rn the usual inner product of two vectors x = (x1, ..., xn) andy = (y1, ..., yn) is defined by

x, y =ni=1

xiyi.

That this is an inner product can be readily verified by the reader. Itis often called the dot product of the vectors x, y and is denoted byx y = x, y.Definition 1.3.2. Let V be a vector space. A norm on V is a function||.|| : V R satisfying for all x, y V and c R the following properties

1. ||x|| 0 and ||x|| = 0 if and only if x = 02. ||cx|| = |c| ||x||3. ||x+ y|| ||x|| + ||y|| (triangle inequality).

The space V equipped with a norm is called a normed linear space.

Definition 1.3.3. The Euclidean norm on Rn is

||x|| =x, x =

(ni=1

x2i

) 12

.



The Euclidean norm on Rn clearly satisfies the conditions (1) and(2) of a norm. The triangle inequality will follow from the impor-tant Cauchy-Schwarz-Bunyakovsky7inequality which is valid in any innerproduct space and we prove it below. The reader should be aware thatthere are other norms in Rn which do not come from an inner prod-uct. In Theorem 1.3.7 we characterize the norms on a normed linearspace that come from an inner product. The importance of the Eu-clidean norm resides exactly in the fact that it comes from the usualinner product.

Theorem 1.3.4. (Cauchy-Schwarz inequality) Let (V, , ) be an innerproduct space. Then for all x, y V ,

|x, y| ||x|| ||y||.Moreover, equality occurs if and only if one of these vectors is a scalarmultiple of the other.

Proof. If x = 0, then both sides are zero and the inequality is true asequality. For x 6= 0, let t R. Then

0 ||tx+ y||2 = tx+ y, tx+ y = x, xt2 + 2x, yt+ y, y= ||x||2t2 + 2x, yt+ ||y||2.

This expression is a quadratic polynomial in t of the form (t) = at2 +bt + c, with a = ||x||2, b = 2x, y and c = ||y||2. Since (t) 0for all t R, its discriminant b2 4ac must be non positive, that is,x, y2 ||x||2 ||y||2. Therefore |x, y| ||x|| ||y||. Finally, note thatequality occurs if and only if b2 4ac = 0. But this means that thequadratic polynomial has a double (real) root t0 = b2a . That is, 0 =(t0) = ||x + t0y||2. It follows that x + t0y = 0 and so x = cy (withc = t0).

Now it is easy to prove the triangle inequality. In fact, we have

||x+ y||2 = ||x||2 + 2x, y + ||y||27H. Schwarz (1843-1921). A student of Gauss and Kummer. Professor at the Uni-

versity of Gottingen, he is known for his work in complex analysis. V. Bunyakovsky(1804-1889). Mathematician, member of the Petersburg Academy of Science. He iscredited of discovering the Cauchy-Schwarz inequality in 1859.



||x||2 + 2|x, y| + ||y||2 ||x||2 + 2||x|| ||y||+ ||y||2

= (||x|| + ||y||)2 using the Cauchy-Schwarz inequality.

Therefore,||x+ y|| ||x|| + ||y||.

Thus we see that Rn with the Euclidean norm is a normed linearspace.

Corollary 1.3.5. Let V be any normed linear space. Then for x, y V

|||x|| ||y||| ||x y||.

Proof. We have ||x|| = ||(x y) + y|| ||x y||+ ||y||. Therefore

||x|| ||y|| ||x y||.

Similarly||y|| ||x|| ||x y||

and thus the conclusion.

Proposition 1.3.6. (The parallelogram law). Let V be an inner productspace and x, y V . Then

||x+ y||2 + ||x y||2 = 2||x||2 + 2||y||2.

Proof. A direct computation gives

||x+y||2+||xy||2 = x+y, x+y+xy, xy= ||x||2+2x, y+||y||2

+||x||2 2x, y+ ||y||2 = 2||x||2 + 2||y||2.

Even for V = Rn everything takes place in a plane! In fact, Figure1.1 in R2 tells us that in a parallelogram the sum of the squares of thelengths of the diagonals is the sum of the squares of the sides.

The interest of the parallelogram law is that it characterizes thenorms that are derived from an inner product.



Theorem 1.3.7. Let V be a normed linear space with norm || .|| satis-fying the parallelogram law. Then there is a unique inner product , on V such that , 12 coincides with the given norm on V .Proof. For x, y V , define x, y = 14

[||x+ y||2 ||x y||2]. Thenx, x = 14 ||2x||2 = ||x||2. So x, x

12 = ||x||. To show that , is an

inner product on V , we need show that , satisfies the properties (1),(2), (3), (4) of the Definition 1.3.1 Properties (1) and (2) are obvious.To show (3) we have

x, z+ y, z = 14

[||x+ z||2 + ||y + z||2 (||x z||2 + ||y z||2)] .Setting x = u+w and y = u w, we get1

4

[||(u+z)+w||2+||(u+z)w||2(||(uz)+w||2+||(uz)w||2)] .By the parallelogram law (applied twice) this is equal to

[2||u+ z||2 + 2||w||2 2||w z||2 2||w||2] = 1

2

[||w + z||2 ||u z||2]= 2u, z = 2x+ y

2, z = x+ y, z.

Finally, for c R we have

cx, y = 14

[||cx+ y||2 ||cx y||2] = 14[4cx, y] = cx, y.

To see the uniqueness of this inner product, suppose that , 1 is anotherinner product such that x, x1 = ||x||2. Then x, x = x, x1 for allx V . In particular, for all x, y V we have x+y, x+y = x+y, x+y1and so x, y = x, y1

Besides the Euclidean norm on Rn, there are other norms on Rn ofwhich the following two are the most common.

Definition 1.3.8. For x = (x1, x2, ..., xn) Rn, we define||x|| = max {|x1|, |x2|, ..., |xn|} and ||x||1 =

ni=1 |xi|.



It is easy to see that both || || and || ||1 are norms in Rn. We askthe reader to show this in Exercise 4 below.

Example 1.3.9. For any x Rn, ||x|| ||x|| n||x||.

Solution. For each i = 1, ..., n we have

|xi| =x2i

x21 + ...+ x

2n = ||x||.

Hence, ||x|| ||x||. On the other hand, |xi| ||x||. Therefore,

||x||2 =ni=1

x2i n||x||2.

EXERCISES

1. Let x, y R3, where x = (1, 0, 2) and y = (3,1, 1).(a) Find x+ y, x y, ||x||, ||y||, ||x+ y||, ||x y||, x, y.(b) Verify, the Cauchy-Schwarz inequality, the triangle inequality

and the parallelogram law.

(c) Find ||x||, and verify ||x|| ||x|| 3||x||.

(d) Find ||x||1, and verify ||x||1 3||x|| 3||x||1.

(e) Is the paralellogram law valid using the norms || || and|| ||1?

2. For x, y Rn show that ||x+y|| = ||x||+ ||y|| if and only if x = cyfor some c > 0.

3. For x, y Rn show that

||x+ y|| ||x y|| ||x||2 + ||y||2

with equality if and only if x, y = 0.4. Show that

(a) || || and || ||1 are norms on Rn. Do they satisfy theparallelogram law?



(b) For x Rn, show that ||x||1 n||x|| n||x||1.

(c) Let v = (x, y) R2. Sketch the sets {v R2 : ||v|| = 1},{v R2 : ||v|| = 1

}and

{v R2 : ||v||1 = 1

}in R2 and ex-

tend to Rn.

5. Let V be a vector space. Show that any intersection of linear sub-spaces is again a linear subspace. Is the union of linear subspacesa subspace?

6. Let S V . Show that span(S) is the smallest linear subspacecontaining the set S.

1.3.2 Orthogonality

Let V be an inner product space. In view of the Cauchy-Schwarz in-equality since 1 x,y||x|| ||y|| 1, we can define the angle between anytwo nonzero vectors x, y V by

cos = x,y||x|| ||y|| or = cos1( x,y||x|| ||y||

), [0, ].

Hence the following definition is a natural consequence.

Definition 1.3.10. Let V be an inner product space and x, y V .The vectors x and y are called orthogonal (or perpendicular), denotedby xy, when x, y = 0.

The zero vector is orthogonal to every vector x V , since 0, x =00, x = 00, x = 0. Moreover, 0 is the only vector with this property.For if there were another vector v V with v, x = 0 for all x V ,then in particular ||v||2 = v, v = 0, and so v = 0.More generally, we say that two linear subspaces U and W of V areorthogonal if every vector in U is orthogonal to every vector in W . Wewrite UW . For a subset S of V , the orthogonal complement S of Sis defined to be

S = {y V : yx, for all x S} .

Clearly V = {0} and {0} = V . Note that S = xS {x}. Also ifxS, then x span(S). We denote by S = (S).



Proposition 1.3.11. Let S be a subset of V . Then

(a) S is a linear subspace of V .

(b) S S

(c) If S1 S2 V , then S2 S1(d) S = (span(S))

Proof.

(a) Let x, y V . If x and y are orthogonal to every vector of S, thenso is any linear combination of x and y. Therefore S is a linearsubspace of V .

(b) If x S, then for every y S we have xy. Hence x (S)

(c) S1 S2 and xS2, then also xS1.(d) Since S span(S), it follows from (3) that (span(S)) S.

The reverse inclusion is trivial.

Definition 1.3.12. We say that a set of nonzero vectors {v1, ..., vk) ofV is orthogonal if vi, vj = 0 for i 6= j. If in addition, ||vj || = 1 for allj = 1, ..., k, we say {v1, ..., vk) is an orthonormal set of vectors.Thus, orthonormality means vi, vj = ij =

{1 when i = j,0 when i 6= j

}, where

the symbol ij is the so-called Kroneckers delta.

Proposition 1.3.13. (The law of cosines). Let x, y be two nonzerovectors in Rn. Then

||x y||2 = ||x||2 + ||y||2 2||x|| ||y|| cos(),where is the angle between x, y.

Proof. We have

||x y||2 = x y, x y = ||x||2 + ||y||2 2x, y= ||x||2 + ||y||2 2||x|| ||y|| cos().



Corollary 1.3.14. (Pythagorean8 theorem and converse). Let x, y betwo nonzero vectors in Rn. Then x is orthogonal to y if and only if

||x+ y||2 = ||x||2 + ||y||2.

Proof. From the law of cosines we have

||x+ y||2 = ||x (y)||2 = ||x||2 + ||y||2 2||x|| ||y|| cos( )= ||x||2 + ||y||2 + 2||x|| ||y|| cos().

Hence, ||x+ y||2 = ||x||2+ ||y||2 iff cos() = 0 iff = 2 iff x, y = 0.

By induction we get,

Corollary 1.3.15. If {v1, ..., vk} is an orthogonal set of vectors of V ,then

||k

j=1

vj||2 =k

j=1

||vj ||2.

Geometry and algebra cooperate admirably in the followingproposition.

Proposition 1.3.16. An orthogonal set of nonzero vectors is linearlyindependent.

Proof. Supposek

j=1 cjvj = 0. Then

0 = k

j=1

cjvj, vi =k

j=1

cjvj , vi = ci||vi||2.

Since ||vi||2 > 0 for all i, it follows that ci = 0 for all i = 1, ...k.

Corollary 1.3.17. An orthonormal set of vectors B = {v1, ..., vn} of Vwhich spans V is a basis.

8Pythagoras, a famous mathematician and philoshoper of the 6th century B.C.He established the Pythagorean school first in the Greek island of Samos and thenin south Italy.



Observe that the standard basis B = {e1, ..., en} of Rn is an or-thonormal basis of Rn with the usual inner product.

We now introduce the Gram-Schmidt orthogonalization process9

which assures us that any subspace of V has an orthonormal basis,in fact many.

Theorem 1.3.18. (Gram-Schmidt) Let {v1, ..., vm} be a linearly inde-pendent set in V . Then there is an orthonormal set {w1, ..., wm} suchthat

span {v1, v2, ..., vj} = span {w1, w2, ..., wj}, for each j = 1, ...,m.Proof. The proof is by direct construction of the vectors wi and the pro-cedure is useful enough to have a name: The Gram-Schmidt process. Tobegin the construction, we observe that v1 6= 0 (since the vi are linearlyindependent) and we define w1 =

v1||v1|| . Clearly span {w1} = span {v1}.

The other vectors are then given inductively as follows: Supposew1, w2, ..., wk have been constructed so that they form an orthonormalset and for 1 j k, span {w1, ..., wj} = span {v1, ..., vj}. To constructthe next vector wk+1, let

uk+1 = vk+1 k

j=1

vk+1, wjwj .

Then uk+1 6=0, for otherwise vk+1 span {w1, ..., wk}=span {v1, ..., vk}which contradicts the linear independence of {v1, ..., vk+1}. At the sametime, for 1 i k we have

uk+1, wi=vk+1, wik

j=1

vk+1, wjwj , wi=vk+1, wjvk+1, wj=0.

Therefore, uk+1wi for all i = 1, ...., k. Moreover,span {v1, ..., vk+1} = span {w1, ..., wk , uk+1} .

9J. Gram (1850-1916). Actuary and mathematician. He is also know for the Grammatrix or Gramian. E. Schmidt (1876-1959). A student of Hilbert at the Universityof Gottingen. Together with Hilbert he made important contributions to functionalanalysis. The Gram-Schmidt process was first published in 1883.



To see this, set zk+1 =k

j=1vk+1, wjwj span {w1, ..., wk},so that vk+1 = uk+1 + zk+1 span {w1, ..., wk, uk+1}. Butalso zk+1 span {v1, ..., vk}. Hence uk+1 = vk+1 zk+1 span {v1, ..., vk+1} . Finally, choose wk+1 = uk+1||uk+1|| . Then ||wk+1|| = 1and span {w1, ..., wk , uk+1} = span {w1, ..., wk, wk+1}.This completes the induction and the proof.

Since every subspace of Rn has a basis we see that,

Corollary 1.3.19. Any subspace of Rn has an orthonormal basis.

Orthonormal sets have substantial computational advantages. Forif {v1, ..., vn} is a basis for V , then any vector v V is a unique linearcombination of the vectors v1, ..., vm. But actually finding this linearcombination can be laborius since it involves solving simultaneous linearequations i.e., a linear system. Working with an orthonormal basismakes the task much easier. Some of these advantages are illustratedin the next theorem.

Theorem 1.3.20. Let B = {v1, v2, ..., vn} be an orthonormal set in V .The following are equivalent.

1. Let x V . If xvj for all j = 1, ..., n, then x = 0.2. B = {v1, v2, ..., vn} is an orthonormal basis of V .3. Each x V can be written as x =nj=1 cjvj , where cj = x, vj.

The numbers cj = x, vj are called the Fourier10 coefficients of xwith respect to the orthonormal basis B

4. For all x, y V , x, y =nj=1x, vjy, vj.5. For x V , ||x||2 = nj=1 |x, vj|2 = nj=1 |cj |2 (Parsevals

identity)11

Proof. We will show that (1) (2) (3) (4) (5) (1).(1) (2): If B = {v1, ..., vn} were not an orthonormal basis, then

there is an orthonormal set A containing B. Let x A Bc. Then10J. Fourier (1768-1830). Mathematician and physicist best known for initiating

the study of Fourier series and their applications to problems of heat transfer.11M. Parseval (1755-1836), most famous for what is known as Parsevals equality.



x 6= 0 and xvj for all j = 1, ..., n, contradicting the assumption.

(2) (3): Since B is a basis of V , each x V can be uniquelywritten as x =

nj=1 cjvj. Then

x, vi = n

j=1

cjvj , vi =nj=1

cjvj , vi = ci.

(3) (4): Let y V . So that y =nj=1 djvj , where dj = y, vj.Then

x, y = x,nj=1

djvj =nj=1

djx, vj =nj=1

y, vjx, vj.

(4) (5): Taking x = y in (4) we get

||x||2 = x, x =nj=1

(x, vj)2 =nj=1

c2j .

(5) (1): Clearly, if xvj, that is, if x, vj = 0 for all j = 1, ..., n,then ||x||2 = 0 and so x = 0.

Remark 1.3.21. Note that Parsevals identity in the form

||k

j=1

cjvj ||2 =k

j=1

c2j

is a generalization of the Pythagorean theorem.

The following theorem is of considerable geometric interest.

Theorem 1.3.22. (Projection Theorem). Let V be an inner productspace and W a linear subspace of V . Then V = W W, that is, anyx V , can be written uniquely as x = y + z with y W and z W.Moreover, W =W .



Proof. Let dim(V ) = n. First observe W W = {0}. Suppose(W ) = k with 0 k n. Choose an orthonormal basis {w1, ..., wk} ofW and extend it to a basis for V . Use the Gram-Schmidt process toobtain an orthonormal basis {w1, ..., wk , wk+1, ..., wn} of V . Note thatwk+1, ..., wn W.

Given x V ,

x =nj=1

x,wjwj =k

j=1

x,wjwj +n

j=k+1

x,wjwj = y + z,

with y W and z W. If x W, then x,wj = 0 for all j = 1, ..., k,and so x =

nj=k+1x,wjwj . Hence {wk+1, ..., wn} is an orthonormal

basis of W and dim(W ) + dim(W) = dim(V ).To show that this decomposition is unique, suppose x = y1+z1, with

y1 W and z1 W. Then yy1 W and zz1 W. It follows thatz1z = xy1(xy) = yy1, that is, yy1 = zz1 WW = {0}.Therefore, y = y1 and z = z1.

Finally we show that W =W . We know that W W. To getthe reverse inclusion, let x W. Then x = y+z with y W Wand z W. So z = x y W W = {0}. Hence, z = 0 and sox = y W .Corollary 1.3.23. Let S be a subset of V . Then S = span(S).

Proof. From Proposition 1.3.11, S = (span(S)). Therefore S =(span(S)). Since span(S) is a subspace, by the Projection theorem(span(S)) = span(S). Hence, S = span(S).

Corollary 1.3.24. If W is a proper subspace of V , then there existsz0 6= 0 such that z0W .Proof. Let x0 V \W . Then x0 = y0 + z0 with y0 W and z0 Wand z0 6= 0, for if z0 = 0, then x0 = y0 W which is impossible.Definition 1.3.25. Let V =W W and v = w+ u with w W andu W as in the Projection theorem. The map PW : V V given byPW (v) = w is called the orthogonal projection onto W . Similarly, themap PW : V V given by PW(v) = u is the orthogonal projectiononto W. Observe that PW + PW = I.



Clearly,

R3 = R2 R.and the corresponding projections are P (x, y, z) = (x, y, 0), P (x, y, z) =(0, 0, z). More generally for k < n

Rn = Rk Rnk.Note that the proof of the Projection theorem tells us a way to

compute the projection P = PW . In fact, when {w1, w2, ..., wk} is anorthonormal basis for W , then P (v) = w =

kj=1v,wjwj . Note also

that for any v V , we have P 2(v) = P (P (v)) = P (w) = w = P (v).That is, P 2 = P .

Example 1.3.26. Let V = R3 and W = span{v1 = (1,1, 0),v2 = (0,1, 1)}. Find W, the projections PW onto W and the projec-tion PW onto W

.

Solution. Note that W is the plane generated by the vectors v1, v2.By the linearity of the inner productW =

{u R3 : uv1 and uv2

}.

Let u = (x, y, z). We have, uv1 if and only if u, v1 = 0, that is,x y = 0. Similarly, uv2, implies y + z = 0. Solving the system ofthese two equations we get u = t(1, 1, 1) where, t R. Thus,

W = span {u1 = (1, 1, 1)}

which is a line through the origin. To apply the Projection theorem andfind the projection PW , we need to construct an orthonormal basis forW . Note that {v1, v2} is a linear independent set, and so a basis forW . Let us apply Gram-Schmidt orthogonalization process to obtain anorthonormal basis {w1, w2} of W . Take w1 = vi||v1|| , where ||v1|| =

2.

That is, w1 = (12, 1

2, 0). To construct the next vector, let

u2 = v2 v2, w1w1 = v2 12w1 = (1

2,1

2, 0).

Now, ||u2|| =62 and normalizing u2 we get w2 = ( 16 ,

16, 2

6).

Since orthogonal vectors are linearly independent, taking w3 =u1||u1|| =



( 13, 1

3, 1

3), we obtain an orthonormal basis {w1, w2, w3} of V = R3.

Now, let v R3. By the projection theorem, we have v = w + u withw W and u W. Hence

PW (v) = w =2

j=1

v,wjwj = v,w1w1 + v,w2w2.

Therefore for v = (x, y, z), we have

PW (x, y, z) =x y

2(12, 1

2, 0) +

x y + 2z6

( 16, 1

6,26)

=1

3(2x y z,x+ 2y z,x y + 2z),

and PW(x, y, z) =13 (x+y+z, x+y+z, x+y+z), since PW +PW = I.

1.3.3 The cross product in R3

An important concept that distinguishes R3 from Euclidean spaces ofother dimensions is that of the cross product or vector product. InChapter 6, we will make extensive use of the cross product in the studyof surface integrals. Here, the usual notation for the elements of thestandard basis is i = e1 = (1, 0, 0), j = e2 = (0, 1, 0), k = e3 = (0, 0, 1).Thus, any vector a = (a1, a2, a3) R3 can be written as a = a1i+ a2j+a3k

Definition 1.3.27. Let a, b R3. The cross product of a = (a1, a2, a3)and b = (b1, b2, b3) is defined by

ab = det i j ka1 a2 a3b1 b2 b3

= (a2b3a3b2)i+(a3b1a1b3)j+(a1b2a2b1)k.Note that

i j = k, j k = i, k i = j and

j i = k, k j = i, i k = j.

The following proposition gives the basic properties of the cross product.



ba

ab

Figure 1.2: Cross product

Proposition 1.3.28. Let a, b, u R3 and , R. Then

1. a a = 0.

2. b a = (a b).

3. a b = (a b).

4. a (b+ u) = (a b)+ (a u) and (a+ b) u = (a u) + (b u).

5. (a b)a and (a b)b.

6. ||a b||2 = ||a||2||b||2 a, b2.

Proof. All these properties follow easily from the definition. We proveproperty (6).

||a||2||b||2 a, b2 = (a21+ a22+ a23)(b21+ b22+ b23) (a1b1+ a2b2+ a3b3)2.

Multiplying out and rearranging the terms we get

(a2b3 a3b2)2 + (a3b1 a1b3)2 + (a1b2 a2b1)2 = ||a b||2.

Property 5 is illustrated in Figure 1.2.

Example 1.3.29. Find two unit vectors perpendicular to both a =(2,1, 3) and b = (4, 3,5).



Solution. Since a b is perpendicular to both a and b, we compute

a b = det i j k2 1 34 3 5

= 4i 2j+ 2k.The lenght (norm) of the vector is ||a b|| = 16 + 4 + 4 = 26. Thedesired unit vector is

n =16[2i j+ k] .

If we had taken b a instead, we would have obtained the negative ofthis vector. Thus the two vectors are

n = 16[2i j+ k] .

Corollary 1.3.30. Let a, b R3 and [0, ] the angle between a andb. Then

||a b|| = ||a|| ||b|| sin .In particular, ||a b|| equals the area of the parallelogram with adjacentsides a, b.

Proof. We know that a, b = ||a|| ||b|| cos . Hence from (6) we get||a b||2 = ||a||2||b||2(1 cos2 ).

Thus, ||ab|| = ||a||||b|| sin . Furthermore, if a, b represent the adjacentsides of a parallelogram and we take b to be the base, then ||a|| sin isthe height. Hence the area of the parallelogram is equal to ||ab||.

Note that ab = 0 if and only if ||ab|| = 0 if and only if sin = 0 ifand only if = 0 or = , viz, a, b are colinear. Equivalently, a b 6= 0if and only if a and b are linearly independent.

Example 1.3.31. Find the area of the triangle whose vertices areA = (2,1, 3) and B = (1, 2, 4), and C = (3, 1, 1).

Solution. Two sides of the triangle are represented by the vectors

a = AB = (1 2)i+ (2 + 1)j+ (4 3)k = i+ 3j+ k,



b = AC = (3 2)i+ (1 + 1)j(1 3)k = i+ 2j 2k.The area of the triangle is clearly half of the erea of the parallelogramwith adjacent sides the vectors a and b. Hence the area of the triangleis 12 ||a b||. We compute

a b = det i j k1 3 1

1 2 2

= 8i j 5k.Since ||a b|| = 310, the area is 32

10.

Example 1.3.32. Find the equation of the plane passing from the pointa = (a1, a2, a3) and perpendicular to n = (n1, n2, n3). (The vector n iscalled the normal vector to the plane).

Solution. Let x = (x, y, z) be any other point on the plane. Thenthe vector x a lies on the plane and is perpendicular to n. Therefore

n,x a = 0is the vector equation of the plane under consideration. In terms of thecoordinates the equation is

n1(x a1) + n2(y a2) + n3(z a3) = 0.Example 1.3.33. Find the equation of the plane passing from thepoints A = (2,1, 3) and B = (1, 2, 4), and C = (3, 1, 1).

Solution. This is the plane containing the triangle of Example 1.3.31.The normal vector to this plane is n = (AB)(AC) = 8ij5k. Theproblem now is reduced to find the equation of the plane passing froma = (2,1, 3) and perpendicular to n = (8,1,5). So the equationis 8(x 2) (y + 1) 5(z 3) = 0 or 8x+ y + 5z = 30.Definition 1.3.34. Let a, b, c R3. The triple scalar product of a, band c is defined to be

a, (b c) = a1(b2c3 b3c2) + a2(b3c1 b1c3) + a3(b1c2 b2c1)

= det

a1 a2 a3b1 b2 b3c1 c2 c3

.



The following proposition gives a geometric interpretation of thetriple scalar product. We invite the reader here to draw a figure in R3.

Proposition 1.3.35. Let a, b, c R3. The volume V of the paral-lelepiped with adjacent sides a, b and c is |a, (b c)|. In particular, thevectors a, b and c are coplanar if and only if a, (b c) = 0.

Proof. We know, from Corollary 1.3.30, that ||b c|| is the area of theparallelogram with adjacent sides b and c. Moreover |a, (b c)| =||a|| ||b c|| cos, where is the acute angle that a makes with b c(the normal vector to the plane spanned by b and c). The parallelepipedwith adjacent sides a, b and c has height

h = ||a|| cos = |a,(bc)|||bc|| .From elementary geometry the volume of the parallelepiped is the prod-uct of the area of the base times the height. Hence, V = |a, (bc)|.

EXERCISES

1. Let a = 3i j+ 2k and b = i+ j+ 4k. Find a b, ||a b||, a, band verify properties (5) and (6).

2. Find the equation of the plane generated by the vectors a =(3,1, 1) and b = (1, 2,1).

3. Find the equation of the plane determined by the points a =(2, 0, 1), b = (1, 1, 3) and c = (4, 7,2).

4. Find a unit vector in the plane generated by the vectors a = i+2jand b = j+ 2k, perpendicular to the vector v = 2i+ j+ 2k

5. Find the equation the line passing through (1, 1, 1) and perpen-dicular to the plane 3x y + 2z 5 = 0.

6. Find the equation of the plane passing through (3, 2,1) and(1,1, 2) and parallel to the line l(t) = (1,1, 0) + t(3, 2,2).

7. Find the equation of the plane passing through (3, 4,1) parallelto the vectors a = i 3k and b = 2i+ j+ k.



8. Find the shortest distance from the point(3, 4, 5) to the linethrough the origin parallel to the vector 2i j+ 2k.

9. Find the volume of the tetrahedron determined by the points P1 =(2,1, 4), P2 = (1, 0, 3), P3 = (4, 3, 1) and P4 = (3,5, 0).Hint. The volume of the tetrahedron is one sixth the volume ofthe parallelepiped with adjacent sides P1P2, P1P3 and P1P4.

10. Let a, b, v R3 with v 6= 0. Show that

(a) If a, b = 0 and a b = 0, then either a = 0 or b = 0.(b) If a, v = b, v and a v = b v, then a = b.

11. Let a, b, c R3. Show that the area of the triangle with verticesa, b and c is given by 12 ||(a c) (b c)||. Find the area of thetriangle with vertices the points a, b and c of Exercise 3.

12. Let a, b, c R3. Show that a(bc) = (a, c)b(a, b)c. Deducethat

a (b c) + b (c a) + c (a b) = 0(the Jacobi identity).

13. Let a, b R3 be nonzero vectors. Show that the vector v =||a||b + ||b||a bisects the angle between a and b.

14. Let S = {x = (1, 0, 2), y = (3,1, 1)}

(a) Find the angle between the vectors x and y.

(b) Find an orthonormal basis for W = span(S).

(c) Find the orthogonal complement, W of W .

(d) Find the projections PW and PW .

Answers to selected Exercises

2. x 4y 7z = 0. 3. 17x y + 9z = 43. 4. 5

25 [5i+ 6j 8k].5. l(t) = (1, 1, 1)+t(3,1, 2). 6. yz+1 = 0. 7. 3x7y+z+20 = 0.8.34. 9. 1016 .



1.4 Rn as a metric space

Definition 1.4.1. A metric space is a pair (X, d) where X is a set andd is a function from XX into R, called a distance or metric satisfyingthe following conditions for x, y and z in X:

1. d(x, y) 0 and d(x, y) = 0 if and only if x = y2. d(x, y) = d(y, x) (symmetry)

3. d(x, z) d(x, y) + d(y, z) (triangle inequality)Since we have a norm in Rn it is natural to consider the distance

between points (vectors) x, y Rn to bed(x, y) = ||x y||.

The reader should draw a picture to see that this is really the distancebetween points of Rn. One easily verifies properties (1), (2) and (3) ofthe distance d : Rn Rn R. For instance, for (3) we haved(x, z) = ||xz|| = ||(xy)+(yz)|| ||xy||+||yz|| = d(x, y)+d(y, z).Thus, Rn equiped with the distance d is an example of a metric space.This would apply to any of the norms on Rn that we considered in Def-inition 1.3.8. However, we will be chiefly concerned with the Euclideannorm and the Euclidean distance

d(x, y) = ||x y|| =(

ni=1

(xi yi)2)1

2

.

Definition 1.4.2. Let a Rn and r R with r > 0. The open ballaround the point a is defined to be the set

Br(a) = {x Rn : d(a, x) < r} = {x Rn : ||a x|| < r}Br(a) is also called an r-neighborhood of a.

For instance, when n = 2 and a = (, ) R2 the open ball arounda with radius r > 0 is the open disk

Br(a) ={(x, y) R2 :

( x)2 + ( y)2 < r

},


1.4 Rn as a metric space 39

and when n = 1 and a R it is a (symmetric) open interval around a

Br(a) = {x R : |a x| < r} = (a r, a+ r).

We denote by

Br(a) = {x Rn : ||a x|| r} .The set Br(a) is called the closed ball around a. The reason for thename closed will become clear shortly.

Definition 1.4.3. Let Rn. The set is called an open set in Rn ifeach of its points has a sufficiently small open ball around it completelycontained in , that is, for each x there exists > 0 (which ingeneral may depend on x) such that B(x) .

The whole space Rn and the empty set are evidently open.Example 1.4.4. Any open ball Br(a) in R

n is an open set. To see this,let x Br(a). We have to find > 0 so that B(x) Br(a).Take = r||ax|| > 0. Then for any y B(x) the triangle inequalityimplies

||a y|| = ||a x+ x y|| ||a x||+ ||x y|| ||a x||+ = r,

that is, ||a y|| r. Hence, y Br(a).Definition 1.4.5. The boundary of a set Rn, denoted by , isthe set of points x Rn such that every open ball around x intersects and its complement c. That is, for every > 0

= {x Rn : B(x) 6= , B(x) c 6= } .

Clearly = (c).

Example 1.4.6. The boundary of Br(a) is the sphere Sr(a) centeredat a = (a1, ..., an) with radius r > 0,

Sr(a) = (Br(a)) = {x Rn : ||a x|| = r}

={(x1, .., xn) Rn : (a1 x1)2 + ...+ (an xn)2 = r2

}.



Figure 1.3: Unit sphere in R3: x2 + y2 + z2 = 1

Exercise 1.4.7. Show that an open set can not contain any of itsboundary points.

Definition 1.4.8. Let Rn. A point a is called an interiorpoint of if there exists > 0 such that B(a) . The set of allinterior points of is called the interior of and is denoted by .

Note that, and = if and only if is open.Definition 1.4.9. A set S in Rn is said to be closed if and only if itscomplement Sc in Rn is open.

The reader should be aware that this is not the same as not open!Note that Rn and are closed.Proposition 1.4.10. A set S is closed in Rn if and only if S S,that is, it contains its boundary.

Proof. Suppose S is closed. Then Sc is open. So, Sc S = . HenceS S. Conversely, let x Sc. Since S S, it follows that x is not aboundary point of S. Therefore, from the definition of boundary point,there exists an r > 0 such that Br(x) S = , that is, Br(x) Sc andso Sc is open.

Definition 1.4.11. Let S Rn. We define the closure of S, denotedby S, to be the set

S = S S.


1.4 Rn as a metric space 41

Thus, Proposition 1.4.10, tells us that a set S is closed if and only ifS = S. For example, Br(a) = Br(a) Sr(a), and thus we use the nameclosed ball.

Remark 1.4.12. A set need be neither open nor closed. For example,in R the set S = {x R : 0 x < 1} = [0, 1), whose boundary is S ={0, 1}, is not open in R as it contains the boundary point 0 nor isit closed since it does not contain its boundary. Its closure is S ={x R : 0 x 1} = [0, 1] and its interior S = {x R : 0 < x < 1} =(0, 1).

Similarly, in R2 the set ={(x, y) R2 : 0 x 1, 0 < y < 1} a

square, is neither open nor closed in R2. Here

={(x, y) R2 : 0 x 1, 0 y 1} = [0, 1] [0, 1]

and = (0, 1) (0, 1).

Example 1.4.13. 1. Consider the set Z R. We have Z = andZ = Z = Z. So Z is closed.

2. Consider the set Q R. Theorems 1.1.7, 1.1.9 tell us that Q = and Q = R = Q. Notice that Q is neither open nor closed andthus, Qc is also neither open nor closed (why?).

Exactly as in R, replacing | | by || ||, we have

Definition 1.4.14. Let Rn and a Rn (not necessarily in ).We call a an accumulation point or limit point of if every open ballaround a contains at least one point x distinct from a, that is, forevery r > 0 we have (Br(a) {a}) 6= .

A point a which is not an accumulation point of is called anisolated point of . Hence, a is an isolated point of if there is an openball Br0(a) such that Br0(a) = {a}, for some r0 > 0. We remarkthat as in R, any finite set = {v1, ..., vm} in Rn has no limit points.In fact, each point of is an isolated point. Note also that () = and so any finite set is closed.

The next proposition provides a useful characterization of closedness.



Proposition 1.4.15. A set S in Rn is closed if and only if S containsall its limit points.

Proof. Suppose S is closed and let a be a limit point of S. If a / S,then a Sc which is open. Therefore, there exists > 0 such thatB(a) Sc. It follows that B(a) S = . Since a is a limit point ofA this is impossible. Conversely, suppose S contains all its limit points.We show that Sc is open (and so S will be closed). Let x Sc. Byhypothesis x is not a limit point of S and consequently there exists r > 0such that Br(x) S = . Hence, Br(x) Sc and so Sc is open.

EXERCISES

1. Show that if 1 and 2 are open in Rn, then 1 2 and 1 2

are also open. Generalize to any finite number of open sets.

2. Show that if A1 and A2 are closed in Rn, then A1A2 and A1A2

are also closed. Generalize to any finite number of closed sets.

3. Give an example of a set S in Rn such that S = and S = Rn.4. Show that S = S \ S.5. Let x, y Rn with x 6= y. Show that there exist open sets U and

V such that x U , y V and U V = .12

6. Let S1 and S2 be sets in Rn. Show that

(S1 S2) = S1 S2 and S1 S2 (S1 S2)

7. Show that S1 S2 S1 S2 and S1 S2 = S1 S2.8. Show that (S1 S2) S1 S2 and (S1 S2) S1 S2.9. Let S Rn. Show that S = S Sc. Deduce that S is a closed

set.

10. Let S Rn. Show that (S)c = Sc. Deduce that S = S \ S.11. Let S Rn. Show that12A topological space with this property is called a Hausdorff space. In particular,Rn is a Hausdorff space.


1.5 Convergence of sequences in Rn 43

(a) (S) S.(b) (S) S.

12. Let S = Rn (such a set is called dense in Rn). Show that if isopen, then S = . Is this true if is not open?

13. Find the limit points of the sets: A ={(1 + 1m , sin

m2 ) : m N

}and B =

{([1 + 1m ] cos(m),

12m ) : m N

}14. Prove that the set of all limit points of any set is always closed.

1.5 Convergence of sequences in Rn

The significance of the metric d is that it enables us to talk aboutlimiting processes. This involves replacing | | by || ||. Here we studythe convergence of sequences in Rn.

We reserve the letter n for the dimension and use letters such asi, j, k for the index on a sequence, that is, we denote by {xk}k=1 asequence of vectors in Rn and the components of the vector xk will bexk = (xk1, xk2, ..., xkn). We now define the limit of a sequence in R

n

Definition 1.5.1. Let {xk}k=1 be a sequence of vectors in Rn andx Rn. We say that {xk}k=1 converges to a limit x if d(xk, x) =||xk x|| 0 in R as k . We write xk x.

Thus xk x if for every > 0 there exists a positive integer N suchthat ||xk x|| < whenever k N . If a sequence does not converge,we say that the sequence diverges. We shall denote a sequence {xk}k=1simply by {xk} and we shall omit writing k whenever this is clearfrom the context. We remark that xk x if and only if ||xkx||2 0.As in R, the limit of a convergent sequence is unique.

Example 1.5.2. The sequence {xk} ={(3 1k+1 , 1k )

}in R2 converges

to x = (3, 0). Indeed, letting k we have

||xkx||2 =(3 1

k + 1 3)2

+

(1

k 0)2

=1

(k + 1)2+

(1

2

)k 0.



Just as in R, a sequence {xk} in Rn is said to be bounded if thereexists M > 0 with ||xk|| M for all k = 1, 2, ..., and a sequence whichis not bounded is called unbounded. In addition, every convergent se-quence is bounded, and any unbounded sequence diverges.

The following theorem lists the basic properties of convergentsequences in Rn.

Theorem 1.5.3. Let {xk} and {yk} be two sequences in Rn with xk xand yk y, where x, y Rn. Then

1. For a, b R, axk + byk ax+ by2. ||xk|| ||x||3. xk, yk x, y4. xk x if and only if xki xi for each i = 1, ...n

Proof. 1. By hypothesis and the triangle inequality we have

0 ||(axk + byk) (ax+ by)|| = ||a(xk x) + b(yk y)|| |a| ||xk x||+ |b| ||yk y|| 0.

Therefore, axk + byk ax+ by2. From Corollary 1.3.5, | ||xk|| ||x|| | ||xk x|| 0.

Hence, ||xk|| ||x||3. First note that

|xk, yk x, y| = |xk, yk x, yk+ x, yk x, y|= |xk x, yk+ x, yk y| |xk x, yk|+ |x, yk y| .

Using the Cauchy-Schwarz inequality we obtain

|xk, yk x, y| ||xk x|| ||yk||+ ||x|| ||yk y||.Since {yk} converges, it is bounded. So there exists M > 0 suchthat ||yk|| M . Now, we have

|xk, yk x, y| ||xk x||M + ||x|| ||yk y|| 0.Thus, xk, yk x, y.



4. Suppose xk x. Then

|xki xi| =(xki xi)2

ni=1

(xki xi)2 = ||xk x|| 0.

Conversely, if xki xi for all i = 1, ..., n, then for every > 0 thereare positive integers Ni such that |xki xi| < n for all k Ni.Letting N = max {Ni : i = 1, 2, ..., n}, we have |xki xi| < n forall k N . Now,

||xk x|| = n

i=1

(xki xi)2 < n

i=1

2

n= .

Therefore, ||xk x|| < for all k N .By property (4) of the above proposition, we see that to study the

covergence of a sequence in Rn it suffices to study the convergence ofits component sequences in R.

Next, we would like to restate the definition of a Cauchy sequencefor a sequence in Rn, or more generally in a metric space.

Definition 1.5.4. Let (X, d) be a metric space.

1. A sequence {xk} of points in X is called a Cauchy sequence ifd(xk, xj) 0 as k, j , that is, for every > 0 there existspositive integer N such that d(xk, xj) < for all k, j N .

2. X is called a complete metric space if every Cauchy sequence inX converges to a point in X.

As in R, every Cauchy sequence is bounded. Moreover, the definitionof subsequence of a given sequence in R, as well as, Propositions 1.1.27,1.1.30 extend with identical proofs for sequences in Rn (all that is neededis to replace | | by || ||). Thus, for any {xk} in Rn we haveProposition 1.5.5. 1. If xk x, then every subsequence xkm x.

2. Every convergent sequence is a Cauchy sequence.



3. A Cauchy sequence converges, if it has a convergent subsequence.

Next we prove the completeness of Rn.

Theorem 1.5.6. Rn is complete.

Proof. Since {xk} is Cauchy, for each i = 1, ..., n we have

|xki xji| ||xk xj || 0

as k, j . Hence each component sequence {xki} is Cauchy in R.Since R is complete, there exist xi R such that xki xi for each i =1, ..., n. Taking x = (x1, ..., xn) Rn, Theorem 1.5.3 yields xk x.

The importance of completeness is that it enables us to prove thata sequence converges without a priori knowledge of the limit. It cantherefore be used to prove existence of the limit which would enableus to prove important existence theorems in the theory of differentialequations and elsewhere.

Sequential convergence can be used to characterize the closure of aset.

Proposition 1.5.7. Let S be a subset of Rn and x Rn. Then x Sif and only if there exists a sequence of points in S that converges to x.

Proof. Suppose x S. If x is in S iteslf, let xk = x for all k = 1.2.... Ifx / S, then x S. Therefore, for each k the open ball B 1

k(x) contains

points of S. So, for each k we can select a point xk S such that||xk x|| < 1k . Thus, in either case, xk S and xk x. Conversely,suppose {xk} is a sequence in S such that xk x. Then every openball around x contains elements of S. In fact, it contains all xk forsufficiently large k. Hence, x S or x S. That is, x S.

Of particular interest in R is the following

Corollary 1.5.8. Let 6= S R bounded from above. Suppose M =sup(S). Then M S. Hence M S if S is closed. Similarly, form = inf(S).



Proof. Since M = sup(S), for each k = 1, 2, ... there exists xk S suchthat M 1k < xk M . Hence, xk M . Thus, M S.Definition 1.5.9. Let S Rn. The diameter d(S) of S is

d(S) = sup {||x y|| : x, y S} .Definition 1.5.10. A set S in Rn is called bounded if there exists R > 0such that S BR(0). Equivalently, if its diameter is finite.

We conclude this section by generalizing the Bolzano-Weiestrass the-orem to Rn.

Theorem 1.5.11. (Bolzano-Weiertsrass theorem in Rn)

1. Every bounded sequence in Rn has a convergent subsequence.

2. Every bounded infinite set in Rn has a limit point.

Proof.

1. Suppose ||xk|| M for all k. Since |xki| ||xk|| for all i =1, ..., n, it follows that the sequences {xki}k=1 of the componentsare all bounded. Hence, from Corollary 1.1.28, we can extracta convergent subsequence from each component sequence. Thetrouble is that the indices on these subsequences might all bedifferent, so we can not put them together. (For example, wemight have chosen the even-numbered terms for i = 1 and the odd-numbered terms for i = 2). To construct the desired subsequence,we have to proceed step by step. First we choose a subsequence{xmk} such that the first components converge. This subsequenceis bounded. Hence it has a subsequence for which the secondcomponents converge. This new subsequence has the propertythat the first and the second components converge. Continuingthis way, after a finite number of steps we find a subsequencewhose components all converge.

2. Let S be an infinite bounded set in Rn. For k = 1, 2, ..., wecan select xk S with xk 6= xj for k 6= j. Then {xk}k=1 is abounded sequence in Rn of distinct elements of S. By part (1),{xk} has a subsequence which converges, say, to a. Hence, every



neighborhood of a contains infinitely many points of S and thusa is a limit point of S.

1.6 Compactness

The concept of compactness is of fundamental importance in analysisand in mathematics in general. Although there are several possible def-initions of compactness, these all coincide in Rn.

In this section I will denote any (infinite) index set.

Definition 1.6.1. By an open cover of a set S in Rn we mean a col-lection {Vi}iI of open set of Rn such that S

iI Vi. A subcover is

a subcollection which also covers S.

Lemma 1.6.2. (Lebesgue covering lemma)13 Let S be a closed andbounded subset of Rn and {Vi}iI an open cover of S. Then for eachx S, there exists some > 0 such that B(x) Vi for at least onei I. (Such a number > 0 is called a Lebesgue number of S for theopen cover {Vi}iI .)

Proof. Suppose that such a > 0 does not exist. Then for each k =1, 2, ..., there exists some xk S such that B 1

k(xk)V ci 6= for all i I.

Since S is bounded, {xk} is bounded and by the Bolzano-Weiretsrasstheorem it has a convergent subsequence, say, xmk x. Then x S =S, by the closedness of S. Since S iI Vi, it follows that x Vjfor some j I. Moreover, since Vj is open there exists r > 0 withBr(x) Vj . Now select some mk sufficiently large so that 1mk k. By the sequential compactness, the sequence {xk} hasa convergent subsequence, say {xmk}. However, since any convergentsequence is bounded, there is some M > 0 with xmk M for all mk.This contradicts xmk > mk.

14E. Heine (1821-1881). He is known for results on special functions (spherical har-monics, Legendre functions) and in real analysis. E. Borel (1871-1956). Mathemati-cian and politician in Paris. A student of Darboux, he made important contributionsin analysis.



Next we prove (3) (1). Let S be compact. First we show thatS is bounded. An open cover of S is given by the open balls Bk(0),k = 1, 2, .... Since S is compact, a finite number of these balls covers S,say, S mk=1Bk(0). Since B1(0) B2(0) ... Bm(0), it follows S Bm(0) and so S is bounded. Now we show that S is closed by showingthat Sc is open. Let y Sc. For each x S, let rx = 12 x y > 0.Then the open balls {Brx(x)}xS cover S. Since S is compact, there isa finite number of points x1, ..., xm S such that S

mj=1Brxj (xj).

Take r = min{rxj : j = 1, ...,m

}. Then Br(y)S = . Hence, Br(y)

Sc, so that Sc is open.

Conversely, suppose S is closed and bounded. Let S iI Vi be anopen cover of S. We claim that for each r > 0, there exist x1, ..., xm Ssuch that

S mj=1

Br(xj)

For suppose not; let r > 0 and fix some x1 S. Then choosex2 (SBr(x1)). Continuing this way, using induction, choose xk+1 (S\kj=1Br(xj)). Then, xkxm r for k 6= m. This implies that thesequence {xk} of points in S can not have any convergent subsequence.Since S is bounded, {xk} is also bounded and the Bolzano-Weierstrasstheorem implies that {xk} has a convergent subsequence. This contra-diction proves our claim. Now, by Lemma 1.6.2, let > 0 be a Lebesguenumber of S for the open cover {Vi}iI and choose x1, ..., xm S suchthat S mj=1B(xj). For each j = 1, ...,m pick some ij I suchthat B(xj) Vij . Then S

mj=1Br(xj)

mj=1 Vij . Hence, S is

compact.

Note that every finite subset of Rn is obviously bounded and beingclosed it is therefore compact. As the following exercise shows, boththe boundedness and the closedness of the subset S in Rn are neededto guarantee the compactness of S.

Exercise 1.6.6. 1. Let S = (0, 1] in R. Note that S is bounded butnot closed. Take Vk = (

1k , 2). Show that {Vk}kZ+ is an open

cover of S but contains no finite subcover of S.


1.6 Compactness 51

2. Let S = [0,) in R. Then S is closed but not bounded. TakeVk = (k 2, k). Show that {Vk}kZ+ is an open cover of S butcontains no finite subcover.

Remark 1.6.7. The proof of (3) (1) given above is valid in anymetric space. Thus, in any metric space, a compact set is closed andbounded. However, the converse, that is, the Heine-Borel theorem doesnot necessarily hold in any metric space. For example, let X be aninfinite set and d the discrete distance on X (that is, d(x, y) = 1 ifx 6= y and d(x, y) = 0 if x = y.) The set X is bounded since thedistance between any two of its points is at most 1. In addition, Xbeing the whole space is closed. Let r = 12 . For each x X, notethat Br(x) = {x} and X =

xX Br(x). But this open cover has no

finite subcover. Hence X is not compact. What goes wrong here isthe fact that no infinite subset of X (with the discrete distance) has alimit point (in other words, no sequence of distinct points can converge).This means that the Bolzano-Weierstrass theorem is not valid in thiscase. However, if the Bolzano-Weierstrass theorem were valid in somemetric space, then (as in the above proof of (1) (3)) the Heine-Boreltheorem would also be valid. Further nontrivial examples will followfrom Exercise 1.8.22.

Proposition 1.6.8. A closed subset A of a compact set S is itselfcompact.

Proof. Let {Vi}iI be an open cover of A. Then the collection{{Vi}iI , Ac} is an open cover of S. Since S is compact we can ex-tract a finite subcover of S. As A Ac = , it follows that the opencover {Vi}iI of A contains a finite subcover of A.

Corollary 1.6.9. If A is closed and S is compact, then AS is compact.

Next we consider the relationship of completeness to compactness.Notice that in a complete space we dont know anything about thelimit of the Cauchy sequence, just that there is one. So completeness isconsiderably weaker than compactness. For this reason when workingwith completeness we are not at all restricted to Rn (although from ourarguments it may seem we are).



Proposition 1.6.10. A closed subspace Y of a complete metric spaceX is complete.

Proof. Let {yn} be a Cauchy sequence in Y . Since the metric on Y isthe restriction of the one on X, {yn} is a Cauchy sequence in X. Bycompleteness it converges to a point x X. But since Y is closed,x Y .Definition 1.6.11. A locally compact metric space is a space in whicheach neighborhood of a point contains a compact neighborhood (of thatpoint).

Compact spaces are (obviously and trivially) locally compact. Everyinfinite set with the discrete metric is locally compact, but as we saw inRemark 1.6.7, it is not compact. Since closed balls are compact in Rn,Rn is locally compact.

We end this section with a proposition which is a special case of thecelebrated Tychonoff Theorem. This important result states that theproduct of an arbitrary collection of compact spaces is compact. Herewe require only the following,

Proposition 1.6.12. If X and Y are compact metric spaces, then sois X Y .Proof. A metric on X Y is given for instance as

d((x, y), (a, b)) = max(dX(x, a), dY (y, b)).

The reader should check that this is a metric and restricted to X or Ygives the original one. The rest is then a simple exercise in sequentialcompactness and is left to the reader.

1.7 Equivalent norms (*)

In Definition 1.3.8 we saw that || || and || ||1 are also norms on Rn.It can be shown that for 1 p


1.7 Equivalent norms (*) 53

is a norm on Rn called the p-norm (see Exercise 1.8.21). Of course, forp = 2 the 2-norm is just the Euclidean norm. In Example 1.3.9 we sawthat for each Rn, ||x|| ||x||

n||x||. When such an estimate

holds the norms || || and || || on Rn are called equivalent. Moregenerally we have

Definition 1.7.1. Given two norms || ||1 and || ||2 in Rn (or in anynormed linear space), the norms are called equivalent if there exist > 0and > 0 such that, for all x Rn,

||x||1 ||x||2 ||x||1.Note that if a sequence converges (or is Cauchy or is bounded) with

respect to one norm it also converges (is Cauchy or is bounded) withrespect to any other equivalent norm. However, it may happen that asequence converges with respect to one norm and diverges with respectto another. Fortunaltely, in Rn we do not have such worries. As wesee in the following theorem all norms in Rn are equivalent. In fact,we prove that all norms in a finite-dimensional normed linear space areequivalent.

Furthermore note that equivalent norms give equivalent metrics andso a set which is open with respect to one metric is also open withrespect to the other and vice versa.15

Theorem 1.7.2. Let V be a finite dimensional normed linear space.Then all norms on V are equivalent. In particular, this holds on Rn.

Proof. Let B = {v1, ..., vn} be a basis of V and x V . Then x can bewritten uniquely as x = a1v1 + ...+ anvn. Let

||x||1 =ni=1

|ai|.

This is easily seen to be a norm on V . Now let || || be any other normon V . We will show that || || is equivalent to || ||1. We have

||x||= ||a1v1+...anvn|

basic features of euclidean space

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