euclideananalysis_0607.pdf

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Course 214: Michaelmas Term 2006

Complex Analysis

Section 1: Complex Numbers and Euclidean

Spaces

David R. Wilkins

Copyright c David R. Wilkins 2006

Contents

1 Complex Numbers and Euclidean Spaces 2

1.1 The Least Upper Bound Principle . . . . . . . . . . . . . . . . 21.2 Monotonic Sequences . . . . . . . . . . . . . . . . . . . . . . . 2

1.3 The Complex Plane . . . . . . . . . . . . . . . . . . . . . . . . 31.4 Euclidean Spaces . . . . . . . . . . . . . . . . . . . . . . . . . 51.5 Convergence of Sequences in Euclidean Spaces . . . . . . . . . 71.6 The Bolzano-Weierstrass Theorem . . . . . . . . . . . . . . . . 71.7 Cauchy Sequences . . . . . . . . . . . . . . . . . . . . . . . . . 91.8 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.9 Convergent Sequences and Continuous Functions . . . . . . . 111.10 Components of Continuous Functions . . . . . . . . . . . . . . 131.11 Limits of Functions . . . . . . . . . . . . . . . . . . . . . . . . 141.12 The Intermediate Value Theorem . . . . . . . . . . . . . . . . 16

1.13 Uniform Convergence . . . . . . . . . . . . . . . . . . . . . . . 181.14 Open Sets in Euclidean Spaces . . . . . . . . . . . . . . . . . . 191.15 I nteriors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221.16 Closed Sets in Euclidean Spaces . . . . . . . . . . . . . . . . . 221.17 Closures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241.18 Continuous Functions and Open and Closed Sets . . . . . . . 251.19 Continuous Functions on Closed Bounded Sets . . . . . . . . . 261.20 Uniform Continuity . . . . . . . . . . . . . . . . . . . . . . . . 27

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1 Complex Numbers and Euclidean Spaces

1.1 The Least Upper Bound Principle

A widely-used basic principle of analysis, from which many important theo-rems ultimately derive, is the Least Upper Bound Principle.

Let D be a subset of the set Rof real numbers. A real number uis saidto be an upper boundof the set D ofxu for allxD. The set D is saidto be bounded aboveif such an upper bound exists.

Definition Let D be some set of real numbers which is bounded above.A real number s is said to be the least upper bound (or supremum) of D(denoted by sup D) if s is an upper bound of D and s u for all upperboundsu ofD .

Example The real number 2 is the least upper bound of the sets{x R:x 2} and{x R : x < 2}. Note that the first of these sets contains itsleast upper bound, whereas the second set does not.

The Least Upper Bound Principlemay be stated as follows:

ifD is any non-empty subset ofR which is bounded above then

there exists a least upper bound sup D for the set D.A lower bound of a set D of real numbers is a real number l with the

property that l x for all x D. A set D of real numbers is said to bebounded below if such a lower bound exists. If D is bounded below, thenthere exists a greatest lower bound (or infimum) infD of the set D. IndeedinfD = sup{x R :xD}.

1.2 Monotonic Sequences

An infinite sequence a1, a2, a3, . . . of real numbers is said to be strictly in-

creasing if an+1 > an for all n, strictly decreasing if an+1 < an for all n,non-decreasing ifan+1anfor alln, ornon-increasing ifan+1an for alln.A sequence satisfying any one of these conditions is said to be monotonic;thus a monotonic sequence is either non-decreasing or non-increasing.

Theorem 1.1 Any non-decreasing sequence of real numbers that is boundedabove is convergent. Similarly any non-increasing sequence of real numbersthat is bounded below is convergent.

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Proof Leta1, a2, a3, . . .be a non-decreasing sequence of real numbers that is

bounded above. It follows from the Least Upper Bound Principle that thereexists a least upper bound l for the set{an : n N}. We claim that thesequence converges to l.

Let > 0 be given. We must show that there exists some natural num-ber N such that|an l| < whenever n N. Now l is not an upperbound for the set{an : n N} (sincelis the least upper bound), and there-fore there must exist some natural numberNsuch thataN> l . But thenl < an l whenever n N, since the sequence is non-decreasing andbounded above by l. Thus|an l| < whenever nN. Therefore an las n+, as required.

If the sequence a1, a2, a3, . . . is non-increasing and bounded below thenthe sequencea1, a2, a3, . . . is non-decreasing and bounded above, andis therefore convergent. It follows that the sequence a1, a2, a3, . . . is alsoconvergent.

1.3 The Complex Plane

A complex number is a number of the form x+iy, where x and y are realnumbers, and i2 =1. The real numbersx and y are uniquely determinedby the complex number x + iy, and are referred to as therealandimaginary

parts of this complex number.The algebraic operations of addition, subtraction and multiplication aredefined on complex numbers according to the formulae

(x +yi) +(u+iv) = (x+u) +i(y +v), (x+yi)(u+iv) = (xu)+i(yv),(x+ yi) (u+ iv) = (xu yv) + i(xv+ yu),

where x,y, u and v are real numbers.We regard a real number x as coinciding with the complex number x+

i0. Note that the operations of addition, subtraction and multiplication ofcomplex numbers defined as above extend the corresponding operations on

the set of real numbers.The set Cof complex numbers, with the operations of addition and mul-

tiplication defined above, has the following properties:

(i) z1+ z2 = z2+ z1 for allz1, z2 C;(ii) z1+ (z2+ z3) = (z1+ z2) + z3 for all z1, z2, z3 C;

(iii) there exists a complex number 0 with the property thatz+0 = 0+z= zfor all complex numbers C;

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(iv) given any complex number z, there exists a complex number

z such

that z+ (z) = (z) + z= 0;(v) z1 z2=z2 z1 for allz1, z2 C;

(vi) z1 (z2 z3) =z1 (z2 z3) for all z1, z2, z3 C;(vii) there exists a complex number 1 with the property thatz1 = 1z= z

for all complex numbers C;

(viii) given any complex numberz satisfying z= 0, there exists a complexnumberz1 such that z z1 =z1 z= 1;

(ix) z1(z2+z3) = (z1z2)+(z1z3) and (z1+z2)z3 = (z1z3)+(z2z3)for all z1, z2, z3 C.

To verify property (viii), we note that ifz is a non-zero complex number,where z=x +iy for some real numbers x and y, and ifz1 is given by theformula

z1 = x

x2 + y2 i y

x2 + y2,

then z z1 =z1 z= 1.A field is a set X provided with binary operations + and, represent-

ing addition and multiplication respectively, which satisfy properties (i)(ix)above (where complex numbers are replaced in the statements of those prop-erties by elements of the set Xas appropriate). Thus properties (i)(ix) maybe summarized in the statement that the set of complex numbers, with theusual operations of addition and multiplication, is a field. The set R of realnumbers, with the usual algebraic operations, is also a field.

Given complex numberszandw, withw= 0, we define the quotientz/w(i.e.,zdivided by w) by the formula z/w = zw1.

Theconjugatezof a complex numberzis defined such thatx+ iy=xiyfor all real numbers x and y. The modulus|z| of a complex number z isdefined such that

|x+ iy

|= x2 + y2 for all real numbers x and y. Note

that|z|=|z|for all complex numbersz. Alsoz+ w= z+ wfor all complexnumbers z and w. The real part Re z of a complex number satisfies theformula 2 Re z= z+ z. Now| Re z| |z|. It follows that|z+ z| 2|z| forall complex numbers z.

Straightforward calculations show that zz =|z|2 for all complex num-bersz, from which it easily follows thatz1 =|z|2zfor all non-zero complexnumbersz.

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Let z and w be complex numbers, and let z = x+ iy and w = u+ iv,

where x,y, u and v are real numbers. Then

|zw|2 = (xu yv)2 + (xv+ yu)2= (x2u2 + y2v2 2xyuv) + (x2v2 + y2u2 + 2xyuv)= (x2 + y2)(u2 + v2) =|z|2 |w|2

It follows that|zw|=|z| |w| for all complex numbers zand w.Let zand w be complex numbers. Then

|z+ w|2 = (z+ w) (z+ w) =zz+ zw + wz+ ww=

|z|2 + 2 Re zw +

|w

|2

|z|2 + 2|zw| + |w|2 =|z|2 + 2|z| |w| + |w|2= (|z| + |w|)2.

It follows that|z+ w| |z| + |w|for all complex numbers zand w.We define thedistancefrom a complex numberzto a complex numberw

to be the quantity|w z|. Thus ifz=x+ iy and w= u+ iv then|w z|=

(x u)2 + (y v)2.

We picture the complex numbers as representing points of the Euclideanplane. A complex number x+ iy, where x and y are real numbers, repre-sents the point of the plane whose Cartesian coordinates (with respect to anappropriate origin) are (x, y). The fact that|w z| represents the distancebetween the points of the plane represented by the complex numbers zandw is an immediate consequence of Pythagoras Theorem.

Let z1,z2 and z3 be complex numbers. Then

|z3 z1|=|(z3 z2) + (z2 z1)| |z3 z2| + |z2 z1|.This important inequality is known as theTriangle Inequality. It correspondsto the geometric statement that the length of any side of a triangle in theEuclidean plane is less than or equal to the sum of the lengths of the othertwo sides.

1.4 Euclidean Spaces

We denote by Rn the set consisting of all n-tuples (x1, x2, . . . , xn) of realnumbers. The set Rn representsn-dimensionalEuclidean space(with respectto the standard Cartesian coordinate system). Let x and y be elements ofRn, where

x= (x1, x2, . . . , xn), y= (y1, y2, . . . , yn),

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and let be a real number. We define

x + y = (x1+ y1, x2+ y2, . . . , xn+ yn),

x y = (x1 y1, x2 y2, . . . , xn yn),x = (x1, x2, . . . , xn),

x y = x1y1+ x2y2+ + xnyn,|x| =

x21+ x

22+ + x2n.

The quantity x y is the scalar product (or inner product) of x and y, andthe quantity|x| is the Euclidean norm of x. Note that|x|2 = xx. TheEuclidean distance between two points x and y ofRn is defined to be theEuclidean norm|y x|of the vector y x.Lemma 1.2 (Schwarz Inequality) Let x and y be elements of Rn. Then|x y| |x||y|.ProofWe note that|x + y|2 0 for all real numbers and . But

|x + y|2 = (x + y).(x + y) =2|x|2 + 2x y+ 2|y|2.Therefore 2|x|2 + 2x y+ 2|y|2 0 for all real numbers and . Inparticular, suppose that =|y|2 and =x y. We conclude that

|y|4|x|2 2|y|2(x y)2 + (x y)2|y|2 0,so that (|x|2|y|2 (x y)2) |y|2 0. Thus ify=0 then|y|> 0, and hence

|x|2|y|2 (x y)2 0.But this inequality is trivially satisfied when y= 0. Thus|x y| |x||y|, asrequired.

It follows easily from Schwarz Inequality that|x + y| |x| + |y| for allx, y Rn. For

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

It follows that|z x| |z y| + |y x|

for all points x, y and z ofRn. This important inequality is known as theTriangle Inequality. It expresses the geometric fact the the length of anytriangle in a Euclidean space is less than or equal to the sum of the lengthsof the other two sides.

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1.5 Convergence of Sequences in Euclidean Spaces

Definition Let n be a positive integer, and let p1, p2, p3, . . . be an infinitesequence of points in n-dimensional Euclidean space Rn. This sequence ofpoints is said to convergeto some point r ofRn if, given any real number satisfying >0, there exists some positive integer Nsuch that|pj r|< wheneverjN.Lemma 1.3 Let p be a point of Rn, where p = (p1, p2, . . . , pn). Then asequence x1, x2, x3, . . . of points inR

n converges to p if and only if the ithcomponents of the elements of this sequence converge topi fori = 1, 2, . . . , n.

Proof Let xji and pidenote the ith components ofxjand p. Then |xjipi| |xj p| for all j. It follows directly from the definition of convergence thatifxj p as j+then xjipi as j+.

Conversely suppose that, for each i, xji pi as j +. Let >0 begiven. Then there exist natural numbersN1, N2, . . . , N nsuch that |xjipi| 1, the hypercube Hj definedby Hj = H(vj, M/2

j1) contains infinitely many members of the sequence(pj : j N), has sides of length M/2j1, and is contained in the hyper-cube Hj1. Let vj = (v

(1)j , v

(2)j , . . . , v

(n)j ) for each positive integer j. Then,

for each integeri between 1 andn, the sequencev(i)1 , v

(i)2 , v

(i)3 , . . .is a bounded

non-decreasing sequence of real numbers. Such a sequence is guaranteed toconverge to some real number ri (Theorem 1.1). Let r = (r1, r2, . . . , rn).Then the sequencev1, v2, v3, . . .converges to the point r. We claim that thesequence (pj :j

N) has a subsequence that converges to this point r.

We claim that, for every real number satisfying >0, there are infinitelymany members pj of the sequence (pj : j N) satisfying the inequality|pj r| < . Now, given > 0, there exists some integer j such thatM

n/2j1 < 1

2and|vj r|< 12. Now ifxis a point of the hypercube Hj

then

|x vj | M

n

2j1 mj and|pmj r| < 1/j for eachpositive integer j. This subsequence converges to the point r.

Each complex number determines a point of the Euclidean plane R2 whoseCartesian coordinates are the real and imaginary parts of the complex num-ber. Accordingly a sequence of complex numbers corresponds to a sequenceof points in R2, and converges if and only if the corresponding sequence

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of points in R2 converges. The following theorem is thus a special case of

Theorem 1.4.

Theorem 1.5 Every bounded sequence of complex numbers has a convergentsubsequence.

1.7 Cauchy Sequences

Definition A sequencex1, x2, x3, . . .of points in a Euclidean space is said tobe aCauchy sequenceif, given any >0, there exists some natural numberNsuch that|xj xk|< for all integers j and k satisfying jN and kN.

Lemma 1.6 Every convergent sequence in a Euclidean space is a Cauchysequence.

Proof Let x1, x2, x3, . . . be a sequence of points in a Euclidean space Rn

which converges to some point p ofRn. Given any >0, there exists somenatural number N such that|xj p| < /2 whenever j N. But then itfollows from the Triangle Inequality that

|xj xk| |xj p| + |p xk|< 2

+

2=

wheneverjN and kN.Theorem 1.7 Every Cauchy sequence inRn converges to some point ofRn.

Proof Let p1, p2, p3, . . .be a Cauchy sequence in Rn. Then, given any real

number satisfying > 0, there exists some natural number N such that|pj pk| < whenever j N and kN. In particular, there exists somenatural number L such that|pj pk| < 1 whenever j L and k L. LetR be the maximum of the numbers|p1|, |p2|, . . . , |pL1| and|pL| + 1. Then|pj| R whenever j < L. Moreover ifjL then

|pj| |pL| + |pj pL|


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integerMsuch that

|pkm

q

|< 1

2wheneverm

M. Also there exists some

positive integer N such that|pjpk| < 12 whenever j N and k N.Choosem large enough to ensure that mM and kmN. IfjN then

|pj q| |pj pkm| + |pkm q|< 12+ 12= .It follows that the Cauchy sequence (pj : j N) converges to the point q.Thus every Cauchy sequence is convergent, as required.

Theorem 1.8 (Cauchys Criterion for Convergence) A sequence of complexnumbers is convergent if and only if it is a Cauchy sequence.

ProofEvery convergent sequence of complex numbers is a Cauchy sequence(Lemma 1.6). Now every complex number corresponds to a point of theEuclidean plane R2. Moreover a sequence of complex numbers is a Cauchysequence if and only if the corresponding sequence of points ofR2 is a Cauchysequence. The result is therefore follows from Theorem 1.7.

1.8 Continuity

Definition LetXandYbe a subsets ofRm and Rn respectively. A functionf: X Y from X to Y is said to be continuousat a point p ofX if andonly if the following criterion is satisfied:

given any real number satisfying > 0 there exists some >0 such that|f(x)f(p)| < for all points x of X satisfying|x p|< .

The function f: X Y is said to be continuous on X if and only if it iscontinuous at every point pofX.

Lemma 1.9 The functionss:R2 R andp:R2 R defined bys(x, y) =x+ y andp(x, y) =xy are continuous.

ProofLet (u, v) R2. We first show that s:R2 Ris continuous at (u, v).Let >0 be given. Let= 1

2. If (x, y) is any point ofR2 whose distance

from (u, v) is less than then|x u|< and|y v|< , and hence|s(x, y) s(u, v)|=|x+ y u v| |x u| + |y v|< 2=.

This shows that s:R2 Ris continuous at (u, v).Next we show that p:R2 Ris continuous at (u, v). Nowp(x, y) p(u, v) =xy uv= (x u)(y v) + u(y v) + (x u)v

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for all points (x, y) ofR2. Thus if the distance from (x, y) to (u, v) is less

than then|xu| < and|yv | < , and hence|p(x, y)p(u, v)| 0 is given. If >0 is chosen to be the minimum of1 and /(1 + |u| + |v|) then 2 + (|u| + |v|) 0 be given. Then there exists some >0 such that|g(y) g(f(p))| < for all y Y satisfying|y f(p)| < . But then there existssome >0 such that|f(x) f(p)|< for all xX satisfying|x p|< .It follows that|g(f(x)) g(f(p))| < for all xX satisfying|x p| < ,and thus g fis continuous at p, as required.

Lemma 1.10 guarantees that a composition of continuous functions be-tween subsets of Euclidean spaces is continuous.

1.9 Convergent Sequences and Continuous Functions

Lemma 1.11 LetX andY be a subsets ofRm andRn respectively, and letf: X Y be a continuous function from X to Y. Let x1, x2, x3, . . . be asequence of points of X which converges to some point p of X. Then thesequencef(x1), f(x2), f(x3), . . . converges to f(p).

Proof Let > 0 be given. Then there exists some > 0 such that|f(x)f(p)| < for all x X satisfying|xp| < , since the func-tion f is continuous at p. Also there exists some natural number N such

that|xj p| < whenever j N, since the sequence x1, x2, x3, . . . con-verges to p. Thus if j N then|f(xj)f(p)| < . Thus the sequencef(x1), f(x2), f(x3), . . .converges to f(p), as required.

Proposition 1.12 Let a1, a2, a3, . . . and b1, b2, b3, . . . be convergent infinitesequences of complex numbers. Then the sum, difference and product of thesesequences are convergent, and

limj+

(aj+ bj) = limj+

aj+ limj+

bj,

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limj

+

(aj

bj) = lim

j

+

aj

limj+

bj,

limj+

(ajbj) =

limj+

aj

limj+

bj

.

If in addition bj= 0 for all n and limj+

bj= 0, then the quotient of thesequences(aj) and(bj) is convergent, and

limj+

ajbj

=lim

j+aj

limj+

bj.

ProofThroughout this proof let l= limj+ aj and m= limj+ bj .First of all, consider the case when aj and bj are real numbers for all

positive integers j. Then aj +bj = s(aj , bj) and ajbj = p(aj, bj), wheres:RR R andp:RR R are the functions given bys(x, y) =x +yandp(x, y) =xy for all real numbers x and y. Now the sequence ((aj, bj) :j N)is a sequence of points in R2 which converges to the point (l, m), since itscomponents are sequences of real numbers converging to the limits l and m(Lemma 1.3). Also the functions s and p are continuous (Lemma 1.9). Itnow follows from Lemma 1.11 that

limj+(aj+ bj) = limj+ s(aj, bj) =s

limj+(aj, bj)

=s(l, m) =l+ m,

and

limj+

(ajbj) = limj+

p(aj, bj) =p

limj+

(aj, bj)

=p(l, m) =lm.

Also the sequence (bj :j N) converges tom, and therefore limj+

(ajbj) = l m. Now the reciprocal function r:R \ {0} R is continuous onR \ {0}, where r(x) = 1/x for all non-zero real numbers x. It follows fromLemma 1.11 that ifbj= 0 for all positive integers j then 1/bj converges to1/mas j+. But then limj+(aj/bj) =l/m. This completes the proof ofProposition 1.12 in the case when (aj :j N) and (bj :j N) are convergentsequences of real numbers.

The result for convergent sequences of complex numbers now follows easilyon considering the real and imaginary parts of the sequences involved andusing the result that a sequence of complex numbers converges to a complexnumberu + iv, where uand v are real numbers, if and only if the real partsof those complex numbers converge to u and the imaginary parts convergeto v .

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1.10 Components of Continuous Functions

Letf: X Rn be a function mapping a mapping a set X inton-dimensionalEuclidean space Rn. Then

f(x) = (f1(x), f2(x), . . . , f n(x))

for all xX, where f1, f2, . . . , f n are functions from X to R, referred to asthe componentsof the function f.

Proposition 1.13 LetXbe a subset of some Euclidean space, and letp bea point ofX. A functionf: X Rn mappingX into the Euclidean spaceR

n

is continuous atp if and only if its components are continuous atp.

Proof Note that the ith component fi off is given by fi = pi f, wherepi:R

n Ris the continuous function which maps (y1, y2, . . . , yn) Rn ontoitsith coordinateyi. It therefore follows immediately from Lemma 1.10 thatiffis continuous the point p, then so are the components off.

Conversely suppose that the components offare continuous at pX.Let >0 be given. Then there exist positive real numbers1, 2, . . . , n suchthat|fi(x) fi(p)|< /

n for xX satisfying|x p| < i. Let be the

minimum of1, 2, . . . , n. IfxX satisfies|x p|< then

|f(x) f(p)|2 = ni=1

|fi(x) fi(p)|2 < 2,

and hence|f(x)f(p)| < . Thus the function f is continuous at p, asrequired.

Corollary 1.14 Let f: X C be a complex-valued function defined on asubsetXof some Euclidean space. Then the functionfis continuous if andonly if the real and imaginary parts offare continuous real-valued functionsonX.

Proposition 1.15 Let f: X R and g: X R be real-valued functionsdefined on some subsetX of a Euclidean space, and let p be a point ofX.Suppose that the functionsf andg are continuous at the point p. Then soare the functionsf+ g, f g andf g. If in additiong(x)= 0 for allxXthen the quotient functionf /g is continuous atp.

Proof Note that f + g = s h and f g = ph, where h: X R2,s:R2 R and p:R2 R are given by h(x) = (f(x), g(x)), s(u, v) = u+vand p(u, v) = uv for all x X and u, v R. If the functions f and g are

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continuous at p then so is the function h (Proposition 1.13). The functions

sand p are continuous on R2. It therefore follows from Lemma 1.10 that thecomposition functionss hand p hare continuous at p. Thus the functionsf+ g and f g are continuous at p. Nowf g= f+ (g), and both f andg are continuous. Therefore f g is continuous.

Now suppose that g(x)= 0 for all xX. Note that 1/g =r g, wherer:R \ {0} R is the reciprocal function, defined by r(t) = 1/t. Now thereciprocal function r is continuous. It now follows on applying Lemma 1.10that the function 1/gis continuous atp. The functionf /g, being the productof the functionsfand 1/g is therefore continuous at p.

Proposition 1.16 Letf: X

C andg: X

C be complex-valued functionsdefined on some subsetX of a Euclidean space, and let p be a point ofX.Suppose that the functionsf andg are continuous at the point p. Then soare the functionsf+ g, f g andf g. If in additiong(x)= 0 for allxXthen the quotient functionf /g is continuous atp.

Proof Let f(x) = u1(x) +iv1(x) and g(x) = u2 +iv2(x) for all x X,where u1, u2, v1 and v2 are real valued functions on X. It follows fromCorollary 1.14 that the functions u1, u2, v1 and v2 are all continuous at p.Now

f+ g= (u1+ u2) + i(v1+ v2), f

g= (u1

u2) + i(v1

v2),

f g= (u1v1 u2v2) + i(u1v2+ u2v1).Moreover

1/g= (u2 iv2)/(u22+ v22),provided that the functiong is non-zero at all points of its domain. It there-fore follows from repeated applications of Proposition 1.15 that the real andimaginary parts of the functions f+g, f g and f g are continuous at p.The same is true of the function 1/g, provided that the functiong is non-zerothroughout its domain. It then follows from Corollary 1.14 that the functionsf+ g,f

g and f g are themselves continuous at p. Also f /g is continuous

at p, provided that the function g is non-zero throughout its domain.

1.11 Limits of Functions

Let Xbe a subset of some Euclidean space Rn, and let p be a point ofRn.We say that the point p is a limit point ofX if, given any real number satisfying > 0, there exists a point x ofXsatisfying 0


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Definition Let X be a subset of a Euclidean space Rm, let f: X

Rn

mapping X into a Euclidean space Rn, let p be a limit point ofX, and letq be a point ofRn. We say that qis the limit off(x) as x tends to pin Xif, given any real number satisfying >0, there exists some real number satisfying > 0 such that|f(x) q| < for all points x of X satisfying00 such that the only point ofXwhose distance frompis less thanis thepoint p itself. It follows directly from the definition of continuity that anyfunction between subsets of Euclidean space is continuous at all the isolated

points of its domain.

Lemma 1.18 Let X, Y and Z be subsets of Euclidean spaces, let p be alimit point ofX, and letf: XY andg: YZbe functions. Suppose thatlimxp

f(x) = q. Suppose also that the functiong is defined and is continuous

atq. Then limxp

g(f(x)) =g(q).

ProofThe function g is continuous at q. Therefore there exists some >0such that|g(y) g(q)| < for all y Y satisfying|y q| < . But thenthere exists some > 0 such that|f(x)q| < for all x X satisfying0


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Then limxp

f(x) = q if and only if the function f is continuous at p. This

enables one to deduce basic results concerning limits of functions from thecorresponding results concerning continuity of functions.

The following result is thus a conseqence of Proposition 1.13.

Proposition 1.19 LetXbe a subset of some Euclidean space, letf: X Rnbe a function mapping X into n-dimensional Euclidean space Rn, let p bea limit point of the set X, and let q be a point in Rn. Let the real-valuedfunctionsf1, f2, . . . , f n be the components off, so that

f(x) = (f1(x), f2(x), . . . , f n(x))

for all xX, and let q= (q1, q2, . . . , q n). Then limxp

f(x) =q if and only if

limxp

fi(x) =qi fori= 1, 2, . . . , n.

The following result is a consequence of Proposition 1.16.

Proposition 1.20 Let X be a subset of some Euclidean space, let p be alimit point of X, and let f: X Rn and g: X Rn be functions on Xtaking values in some Euclidean spaceRn. Suppose that the limits lim

xpf(x)

and limxp

g(x) exist. Then

limxp

(f(x) + g(x)) = limxp

f(x) + limxp

g(x),

limxp

(f(x) g(x)) = limxp

f(x) limxp

g(x),

limxp

(f(x)g(x)) =

limxp

f(x)

limxp

g(x)

.

If moreover limxp

g(x)= 0 and the functiong is non-zero throughout its do-mainX then

limxp

f(x)g(x)

=limxp

f(x)

limxp

g(x).

1.12 The Intermediate Value Theorem

Proposition 1.21 Let f: [a, b] Z continuous integer-valued function de-fined on a closed interval [a, b]. Then the functionf is constant.

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Proof Let

S={x[a, b] :fis constant on the interval [a, x]},and let s= sup S. Nows[a, b], and therefore the function f is continuousat s. Therefore there exists some real number satisfying > 0 such that|f(x) f(s)|< 1

2 for all x[a, b] satisfying|x s|< . But the function f

is integer-valued. It follows that f(x) = f(s) for all x [a, b] satisfying|xs| < . Now s is not an upper bound for the set S. Thereforethere exists some element x0 of S satisfying s < x0 s. But thenf(s) =f(x0) =f(a), and therefore the functionfis constant on the interval[a, x] for allx

[a, b] satisfyings

x < s + . Thusx

[a, b]

[s, s + )

S.

In particularsS. NowScannot contain any elements x of [a, b] satisfyingx > s. Therefore [a, b] [s, s+ ) ={s}, and therefore s = b. This showsthat b S, and thus the function f is constant on the interval [a, b], asrequired.

Theorem 1.22 (The Intermediate Value Theorem)Leta andb be real num-bers satisfyinga < b, and letf: [a, b] R be a continuous function definedon the interval[a, b]. Letc be a real number which lies betweenf(a)andf(b)(so that eitherf(a)cf(b) or elsef(a)cf(b).) Then there existssomes

[a, b] for whichf(s) =c.

Proof Let c be a real number which lies between f(a) and f(b), and letgc:R\ {c} Z be the continuous integer-valued function on R\ {c} de-fined such that gc(x) = 0 whenever x < c and gc(x) = 1 ifx > c. Supposethat c were not in the range of the function f. Then the composition func-tion gc f: [a, b]Rwould be a continuous integer-valued function definedthroughout the interval [a, b]. This function would not be constant, sincegc(f(a))=gc(f(b)). But every continuous integer-valued function on the in-terval [a, b] is constant (Proposition 1.21). It follows that every real numberclying between f(a) and f(b) must belong to the range of the function f, as

required.

Corollary 1.23 Letf: [a, b][c, d]be a strictly increasing continuous func-tion mapping an interval[a, b] into an interval[c, d], wherea, b, c andd arereal numbers satisfyinga < bandc < d. Suppose thatf(a) =c andf(b) =d.Then the functionfhas a continuous inversef1: [c, d][a, b].

Proof Let x1 andx2 be distinct real numbers belonging to the interval [a, b]then either x1 < x2, in which case f(x1) < f(x2) or x1 > x2, in which casef(x1) > f(x2). Thusf(x1)= f(x2) whenever x1= x2. It follows that the

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function f is injective. The Intermediate Value Theorem (Theorem 1.22)

ensures that f is surjective. It follows that the functionfhas a well-definedinverse f1: [c, d][a, b]. It only remains to show that this inverse functionis continuous.

Let y be a real number satisfying c < y < d, and letx be the unique realnumber such thata < x < bandf(x) =y. Let >0 be given. We can thenchoose x1, x2[a, b] such that x < x1 < x < x2 < x+. Let y1 = f(x1)and y2 = f(x2). Then y1 < y < y2. Choose > 0 such that < yy1and < y2y. If v [c, d] satisfies|vy | < then y1 < v < y2 andtherefore x1 < f

1(v) < x2. But then|f1(v) f1(y)| < . We concludethat the function f1: [c, d][a, b] is continuous at all points in the interiorof the interval [a, b]. A similar argument shows that it is continuous at theendpoints of this interval. Thus the functionfhas a continuous inverse, asrequired.

1.13 Uniform Convergence

Definition LetXbe a subset of some Euclidean spaces, and let f1, f2, f3, . . .be a sequence of functions mapping X into some Euclidean space Rn. Thesequence (fj) is said to converge uniformlyto a function f: X Rn on Xas j + if, given any real number satisfying > 0, there exists some

positive integer N such that|fj(x

)f(x

)| < for all x

X and for allintegersj satisfying jN(where the value ofN is independent ofx).

Theorem 1.24 Letf1, f2, f3, . . .be a sequence of continuous functions map-ping some subsetXof a Euclidean space into Rn. Suppose that this sequenceconverges uniformly on X to some function f: X Rn. Then this limitfunctionf is continuous.

Proof Let p be an element of X, and let > 0 be given. If j is chosensufficiently large then|f(x)fj(x)| < 13 for all x X, since fj funiformly on X as j

+

. It then follows from the continuity offj that

there exists some >0 such that |fj(x)fj(p)|< 13for allxXsatisfying|x p|< . But then

|f(x) f(p)| |f(x) fj(x)| + |fj(x) fj(p)| + |fj(p) f(p)|< 1

3+ 1

3+ 1

3=

whenever |xp|< . Thus the functionfis continuous atp, as required.

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1.14 Open Sets in Euclidean Spaces

Let X be a subset of Rn. Given a point p of X and a non-negative realnumber r, the open ball BX(p, r) in X ofradius r about p is defined to bethe subset ofXgiven by

BX(p, r) {xX :|x p|< r}.

(ThusBX(p, r) is the set consisting of all points ofXthat lie within a sphereof radius r centred on the point p.)

Definition Let Xbe a subset ofRn. A subsetV ofX is said to be open

inXif and only if, given any point p ofV, there exists some >0 such thatBX(p, )V.

By convention, we regard the empty setas being an open subset ofX.(The criterion given above is satisfied vacuously in the case when V is theempty set.)

In particular, a subsetV ofRn is said to be anopen set(in Rn) if and onlyif, given any point p ofV, there exists some > 0 such that B(p, ) V,where B(p, r) ={x Rn :|x p|< r}.

Example Let H ={(x,y,z) R3 : z > c}, where c is some real number.Then His an open set in

R3

. Indeed let p be a point ofH. Then p= (u,v,w),where w > c. Let = wc. If the distance from a point (x,y,z) to thepoint (u,v,w) is less than then|zw| < , and hence z > c, so that(x,y,z)H. Thus B (p, )H, and therefore His an open set.

The previous example can be generalized. Given any integer i between 1and n, and given any real number ci, the sets

{(x1, x2, . . . , xn) Rn :xi > ci}, {(x1, x2, . . . , xn) Rn :xi< ci}

are open sets in Rn.

Example Let U be an open set in Rn. Then for any subset X ofRn, theintersectionU Xis open inX. (This follows directly from the definitions.)Thus for example, let S2 be the unit sphere in R3, given by

S2 ={(x,y,z) R3 :x2 + y2 + z2 = 1}

and let Nbe the subset ofS2 given by

N={(x,y,z)Rn :x2 + y2 + z2 = 1 and z >0}.

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Then N is open in S2, since N = H

S2, where H is the open set in R3

given byH={(x,y,z) R3 :z >0}.

Note thatNis not itself an open set in R3. Indeed the point (0, 0, 1) belongsto N, but, for any > 0, the open ball (in R3 of radius about (0, 0, 1)contains points (x,y,z) for which x2 +y2 +z2 = 1. Thus the open ball ofradius about the point (0, 0, 1) is not a subset ofN.

Lemma 1.25 LetXbe a subset ofRn, and letpbe a point ofX. Then, forany positive real numberr, the open ballBX(p, r) inX of radiusr about pis open inX.

Proof Let x be an element of BX(p, r). We must show that there existssome >0 such that BX(x, )BX(p, r). Let= r |x p|. Then >0,since|x p|< r. Moreover ifyBX(x, ) then

|y p| |y x| + |x p|< + |x p|= r,

by the Triangle Inequality, and hence y BX(p, r). Thus BX(x, ) BX(p, r). This shows that BX(p, r) is an open set, as required.

Lemma 1.26 LetXbe a subset ofRn, and letpbe a point ofX. Then, forany non-negative real numberr, the set{xX :|x p|> r} is an open setinX.

Proof Let x be a point ofX satisfying|x p|> r, and let y be any pointofX satisfying|y x|< , where =|x p| r. Then

|x p| |x y| + |y p|,

by the Triangle Inequality, and therefore

|y

p| |

x

p| |

y

x|>

|x

p|

= r.

Thus BX(x, ) is contained in the given set. The result follows.

Proposition 1.27 LetXbe a subset ofRn. The collection of open sets inXhas the following properties:

(i) the empty set and the whole setXare both open inX;(ii) the union of any collection of open sets inX is itself open inX;

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(iii) the intersection of any finite collection of open sets inX is itself open

inX.

ProofThe empty set is an open set by convention. Moreover the definitionof an open set is satisfied trivially by the whole set X. This proves (i).

LetAbe any collection of open sets in X, and let Udenote the union ofall the open sets belonging toA. We must show that Uis itself open in X.Let x U. Then x V for some set V belonging to the collectionA. Itfollows that there exists some > 0 such that BX(x, ) V. But V U,and thus BX(x, )U. This shows that U is open in X. This proves (ii).

Finally let V1, V2, V3, . . . , V k be a finite collection of subsets of X that

are open in X, and let Vdenote the intersection V1 V2 Vk of thesesets. Let x V. Now x Vj for j = 1, 2, . . . , k, and therefore thereexist strictly positive real numbers1, 2, . . . , k such thatBX(x, j)Vj forj = 1, 2, . . . , k. Let be the minimum of1, 2, . . . , k. Then >0. (This iswhere we need the fact that we are dealing with a finite collection of sets.)NowBX(x, )BX(x, j)Vj for j = 1, 2, . . . , k, and thus BX(x, )V.Thus the intersection V of the sets V1, V2, . . . , V k is itself open in X. Thisproves (iii).

Example The set{(x,y,z) R3 :x2 +y2 +z2 1} is an openset in R3, since it is the intersection of the open ball of radius 2 about the

origin with the open set{(x,y,z) R3 :z >1}.Example The set{(x,y,z) R3 :x2 + y2 + z2 1}is an open setin R3, since it is the union of the open ball of radius 2 about the origin withthe open set{(x,y,z)R3 :z >1}.Example The set

{(x,y,z) R3 : (x n)2 + y2 + z2 < 14

for some n Z}is an open set in R3, since it is the union of the open balls of radius 1

2about

the points (n, 0, 0) for all integers n.

Example For each natural number k , let

Vk ={(x,y,z)R3 :k2(x2 + y2 + z2)< 1}.Now each setVkis an open ball of radius 1/kabout the origin, and is thereforean open set in R3. However the intersection of the setsVk for all naturalnumbersk is the set{(0, 0, 0)}, and thus the intersection of the sets Vk for allnatural numbersk is not itself an open set in R3. This example demonstratesthat infinite intersections of open sets need not be open.

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Lemma 1.28 A sequencex1, x2, x3, . . .of points inRn converges to a pointp

if and only if, given any open set U which contains p, there exists somenatural numberN such thatxjU for allj satisfyingjN.ProofSuppose that the sequence x1, x2, x3, . . .has the property that, givenany open set U which contains p, there exists some natural number N suchthat xj U whenever jN. Let >0 be given. The open ball B(p, ) ofradius about p is an open set by Lemma 1.25. Therefore there exists somenatural numberNsuch thatxjB(p, ) wheneverjN. Thus |xjp|< wheneverjN. This shows that the sequence converges to p.

Conversely, suppose that the sequence x1, x2, x3, . . . converges to p. Let

Ube an open set which contains p. Then there exists some >0 such thatthe open ballB (p, ) of radius about p is a subset ofU. Thus there existssome >0 such that Ucontains all points x ofXthat satisfy|x p|< .But there exists some natural number Nwith the property that|xj p|< whenever j N, since the sequence converges to p. Therefore xj UwheneverjN, as required.

1.15 Interiors

Definition Let X be a subset of some Euclidean space, and let A be asubset ofX. The interior ofA in X is the subset ofA consisting of those

points p ofA for which there exists some positive real number such thatBX(p, )A. (HereBX(p, ) denotes the open ball in Xof radiuscentredon p.)

A straightforward application of Lemma 1.25 shows that ifX is a subsetof some Euclidean space, and ifA is a subset ofX then the interior ofA isopen in X.

1.16 Closed Sets in Euclidean Spaces

Let Xbe a subset ofRn

. A subset F ofX is said to be closed in X if andonly if its complement X\ F inXis open in X. (Recall that X\ F ={xX :xF}.)Example The sets{(x,y,z) R3 : z c},{(x,y,z) R3 : z c}, and{(x,y,z) R3 :z=c}are closed sets in R3 for each real numberc, since thecomplements of these sets are open in R3.

Example Let Xbe a subset ofRn, and let x0 be a point ofX. Then thesets{x X :|xx0| r} and{x X :|xx0| r} are closed for

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each non-negative real number r. In particular, the set

{x0

} consisting of

the single point x0 is a closed set in X. (These results follow immediatelyusing Lemma 1.25 and Lemma 1.26 and the definition of closed sets.)

LetAbe some collection of subsets of a set X. Then

X\SA

S=SA

(X\ S), X\SA

S=SA

(X\ S)

(i.e., the complement of the union of some collection of subsets ofX is theintersection of the complements of those sets, and the complement of theintersection of some collection of subsets ofX is the union of the comple-

ments of those sets). The following result therefore follows directly fromProposition 1.27.

Proposition 1.29 Let X be a subset of Rn. The collection of closed setsinXhas the following properties:

(i) the empty set and the whole setXare both closed inX;(ii) the intersection of any collection of closed sets inX is itself closed in

X;

(iii) the union of any finite collection of closed sets inX is itself closed inX.

Lemma 1.30 LetXbe a subset ofRn, and letF be a subset ofXwhich isclosed inX. Let x1, x2, x3, . . . be a sequence of points ofFwhich convergesto a pointp ofX. ThenpF.

Proof The complementX\ F ofF inXis open, sinceF is closed. Supposethat p were a point belonging toX\F. It would then follow from Lemma 1.28that xj X\F for all values of j greater than some positive integer N,contradicting the fact that xj

F for allj . This contradiction shows that p

must belong to F, as required.

Lemma 1.31 LetFbe a closed bounded set inRn, and letUbe an open setinRn. Suppose thatF U. Then there exists positive real number suchthat|x y| >0 for allxF andy Rn \ U.

Proof Suppose that such a positive real number did not exist. Thenthere would exist an infinite sequence (xj : j N) of points of F and acorrespondinding infinite sequence (yj : j N) of points of Rn \U such

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that

|xj

yj

|< 1/j for all positive integers j. The sequence (xj : j

N)

would be a bounded sequence of points ofRn, and would therefore have aconvergent subsequence (xmj : j N) (Theorem 1.4). Let p = lim

j+xmj .

Then p = limj+

ymj , because limj+

(xmj ymj) = 0. But then p F andp Rn \ U, because the sets F and Rn \ U are closed (Lemma 1.30). Butthis is impossible, as FU. It follows that there must exist some positivereal number with the required properties.

1.17 Closures

Definition Let Xbe a subset of some Euclidean spaceRn

, and let A be asubset ofX. TheclosureofA in Xis the subset ofXconsisting of all pointsx ofXwith the property that, given any real number satisfying > 0,there exists some point aofAsuch that|x a|< . We denote the closureofAin X byA.

Let X be a subset of some Euclidean space, and let A be a subset ofX. Note that a point x ofXbelongs to the closure ofA in X if and onlyif BX(x, )A is a non-empty set for all positive real numbers , whereBX(x, ) denotes the open ball in X of radius centred on x consisting ofall points ofXwhose distance from x is less than .

Lemma 1.32 LetXbe a subset of some Euclidean space, letA be a subsetofX, and letp be a point of the closureA ofA inX. Then there exists aninfinite sequence of points inA which converges to p.

ProofFor each positive integer j letxj be a point ofA satisfying|p xj|0 for which BX(p, ) F =. Butthen BX(p, )A = and therefore p A. Thus X\F X\A, andtherefore AF, as required.

1.18 Continuous Functions and Open and Closed Sets

Let X and Y be subsets of Rm and Rn, and let f: X Y be a functionfrom X to Y. We recall that the function f is continuous at a point pofXif, given any > 0, there exists some > 0 such that|f(u)f(p)| < for all points u ofX satisfying|u p| < . Thus the function f: X Yis continuous at p if and only if, given any > 0, there exists some > 0such that the function f maps BX(p, ) into BY(f(p), ) (where BX(p, )and BY(f(p), ) denote the open balls in X and Y of radius and aboutpand f(p) respectively).

Given any function f: X Y, we denote by f1(V) the preimage of asubset V ofY under the map f, defined by f1(V) ={xX :f(x)V}.Proposition 1.34 LetXandYbe subsets ofRm andRn, and letf: XYbe a function fromXtoY. The functionfis continuous if and only iff1(V)is open inX for every open subsetV ofY.

ProofSuppose that f: X Y is continuous. Let V be an open set in Y.We must show that f1(V) is open in X. Let p f1(V). Thenf(p)V. But V is open, hence there exists some > 0 with the property thatBY(f(p), ) V. But f is continuous at p. Therefore there exists some >0 such that f maps BX(p, ) into BY(f(p), ) (see the remarks above).Thus f(x)V for all xBX(p, ), showing that BX(p, )f1(V). Thisshows that f1(V) is open in Xfor every open set V inY.

Conversely suppose that f: X Y is a function with the property thatf1(V) is open in X for every open set V in Y. Let p X. We mustshow that f is continuous at p. Let > 0 be given. Then BY(f(p), ) is

an open set in Y, by Lemma 1.25, hence f1 (BY(f(p), )) is an open set

in Xwhich contains p. It follows that there exists some > 0 such thatBX(p, ) f1 (BY(f(p), )). Thus, given any > 0, there exists some > 0 such that f maps BX(p, ) into BY(f(p), ). We conclude thatf iscontinuous at p, as required.

Let Xbe a subset ofRn, let f: X Rbe continuous, and let c be somereal number. Then the sets{x X : f(x) > c} and{x X : f(x) < c}are open in X, and, given real numbers a and b satisfying a < b, the set{xX :a < f(x)< b} is open in X.

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1.19 Continuous Functions on Closed Bounded Sets

We shall prove that continuous functions between subsets of Euclidean spacesmap closed bounded sets to closed bounded sets.

Lemma 1.35 LetXbe a closed bounded subset of some Euclidean spaceRm,and letf: X Rn be a continuous function mappingX into some Euclideanspace Rn. Then there exists some non-negative real number M such that|f(x)| M for all xX.Proof Letf: X Rn be a continuous function defined on some closed set Xin Rm. Suppose that the function f is not bounded on X. We shall prove

that the domain X offmust then be unbounded.Now if the function f is not bounded on Xthen there must exist a se-

quence (xj :j N) of points ofXsuch that|f(xj)|> j for all positive inte-gersj. If (xmj :j N) were a subsequence of (xj :j N) converging to somepoint p ofRm, then p would belong to X, since X is closed (Lemma 1.30).Then lim

j+f(xmj) =f(p), and therefore|f(xmj)| |f(p)| + 1 for all suffi-

ciently large positive integers j. But this is impossible because|f(xj)| > jfor all positive integers j. Thus the sequence (xj : j N) cannot have anyconvergent subsequences. Now every bounded sequence of points in Rm hasa convergent subsequence (Theorem 1.4). It follows that (xj :j N) is nota bounded sequence of points in R

m

, and therefore X is an unbounded set.It follows from this that if the domain X of the continuous function

f: X Rn is both closed and bounded then the function fmust be boundedon X, as required.

Theorem 1.36 LetXbe a closed bounded set in some Euclidean space, andlet f: X Rn be a continuous function mapping X into some EuclideanspaceRn. Then the functionf mapsXonto a closed bounded setf(X) inRn.

Proof It follows from Lemma 1.35 that the set f(X) must be bounded.

Let q be a point belonging to the closure f(A) of f(A). Then thereexists a sequence (xj : j N) of points of X such that lim

j+f(xj) = q

(Lemma 1.32). Because the set Xis both closed and bounded, this sequenceis a bounded sequence in Euclidean space, and therefore has a convergentsubsequence (xmj :j N) (Theorem 1.4). Let p= lim

j+xmj . Then pX,

becauseXis closed (Lemma 1.30). But then q= limj+

f(xmj) =f(p), and

therefore qf(A). Thus every point of the closure off(A) belongs tof(A)itself, and therefore f(A) is closed, as required.

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1.20 Uniform Continuity

Definition LetXandYbe subsets of Euclidean spaces. A function f: XY from X to Yis said to be to be uniformly continuousif, given any >0,there exists some >0 (which does not depend on either x or x) such that|f(x) f(x)|< for all points x and x ofX satisfying|x x|< .

Theorem 1.37 Let X be a subset of Rm that is both closed and bounded.Then any continuous functionf: X Rn is uniformly continuous.

Proof Let >0 be given. Suppose that there did not exist any >0 suchthat

|f(x)

f(x)

|< for all points x, x

X satisfying

|x

x

|< . Then,

for each natural number j, there would exist points uj and vj in X suchthat|uj vj|< 1/j and|f(uj) f(vj)| . But the sequence u1, u2, u3, . . .would be bounded, sinceXis bounded, and thus would possess a subsequenceuj1, uj2, uj3, . . .converging to some point p (Theorem 1.4). Moreover pX,since X is closed. The sequence vj1, vj2, vj3, . . . would also converge to p,since lim

k+|vjk ujk |= 0. But then the sequences f(uj1), f(uj2), f(uj3), . . .

and f(vj1), f(vj2), f(vj3), . . . would converge to f(p), since f is continuous(Lemma 1.11), and thus lim

k+|f(ujk) f(vjk)|= 0. But this is impossible,

since uj and vj have been chosen so that|f(uj) f(vj)| for all j. Weconclude therefore that there must exist some >0 such that |f(x

)f(x)|

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