Analysis I MATH 521
Popular in Course
Popular in Mathematics (M)
This 67 page Class Notes was uploaded by Zechariah Hilpert on Thursday September 17, 2015. The Class Notes belongs to MATH 521 at University of Wisconsin - Madison taught by Staff in Fall. Since its upload, it has received 14 views. For similar materials see /class/205290/math-521-university-of-wisconsin-madison in Mathematics (M) at University of Wisconsin - Madison.
Reviews for Analysis I
Report this Material
What is Karma?
Karma is the currency of StudySoup.
You can buy or earn more Karma at anytime and redeem it for class notes, study guides, flashcards, and more!
Date Created: 09/17/15
Math 521 Lecture 8 Arun Rarn University of Wisconsin Madison 480 Lincoln Drive Madison7 WI 53706 ram mathwiscedu 1 The exponential function Let k E 220 de ne k factorial by 010 and klkk713217 ikaZZO Theorem 11 Let n7 k E 220 with k g n Let n7 k E 220 with k g n De ne a Let S be a set of cardinality n Then is the number of subsets ofS of cardinality k n 39 39 k n7k 39 n b k is the coe iCient ofz y in x y c 1 1 and if1 k n71 then n n i n 7 1 n 7 1 k 7 k 7 1 k 39 The exponential function is the element 51 of given by k em 9 kl39 keZZO Theorem 12 As elements of Qzyl 5152 5mi yh De ne k 1 1 71 Hi nlt zgt Zlt gt k kEZgtO Theorem 13 Let and g 2990 6 FM 190 0 G 2905 6 FM MO 1 a 1n1 em 71 511104 71 m b C is an abelian group under multiplication g is a commutative group under addition and GA g p gt gtep71 is an isomoijnhism of groups Math 521 7 Fall 2008 521 July 28 2009 The text references in these notes are to RC BuclgAdmmced Calculus7 3rd edition7 Waveland Press In these notes we shall often use the abbrevia tions gt for implies ltgt for if and only if 7 3 for there exists 7 V for for all These abbreviations help clarify the logical structure of the de nitions and theorerns The following standard notations are also used Z the set of positive integers natural numbers N the set of nonnegative integers Z the set of integers Q the set of rational numbers R the set of real numbers C the set of complex numbers The usual interval notation is used7 eg labzERa xltb7 foltgtbz R b7 etc We write to indicate that two objects are equal by de nition We also signal de nitions by writing z instead of if and only if Throughout R denotes the vector space of all ntuples of real nurnbers7 so R l R7 R2 R gtlt R7 R3 R gtlt R gtlt R7 etc Buck uses the term 71 space as a synonym for R 1 Sets Functions and Maps 1 The notation z E A means that the point z is an element of the set A The notation A Q B means that A is a subset of B ie for all x we have x E A gt z E B By de nition two sets are equal if each is a subset of the other AB ltgt AQBandBQA The notation x denotes the set of all z for which the property Pz is true The notation x E A denotes the set of all z E A for which the property Pz is true If A and B are sets then the sets AUBxx Aorx B A Bzx Aandz B are called respectively the union and intersection of A and B The empty set is denoted 0 z Q for all x Two sets are disjoint iff they have no elements in common ie iff A B 0 See Buck page 5 The set BAyEBy A is called the complement of A in B See page 30 The set AgtltBxyx Aandy B of all ordered pairs my with z E A and y E B is called the Cartesian product of A and B 2 An indexed family of sets is a function which assigns a set A to each element i of a set I The set I is called the index set of the family and the family is usually denoted Alha The union and intersection of the indexed family are de ned by zeUA ltgt Hie zeA ieI ze A ltgt Vie zeA id The notation 3i 6 I is an abbreviation for there exists i E I and the notation Vi E I is an abbreviation for for all i E I In these de nitions the set I can be in nite For nite sets I we recover the earlier de nitions eg for I 12 we have U AA1UA2 AA1 A2 id12 id12 This illustrates the logical principle that 3 is like an in nite or77 and V is like an in nite and From logic we know that not V ltgt 3 not and not 3 ltgt V not so for any set E and any indexed family Alha we have BUAi ltBAigti B AiUltBAigt id id id id In particular for I 12 we have BA1 UAZ BA1 BA2i BA1 Az BA1 U B A2 Also by logic we have set theoretic distributive laws B UAUB A Bu A BuA id id id id In particular for I 12 we have BUA1 A2 BUA1 BUAZ B A1UA2 B A1UB A2 3 A function is a rule which assigns a value f to every point z from a set called the domain of the function The set graphltfgt 96721 11 f96 of all pairs zy such that y f is called the graph of the function f 4 Let X and Y be sets We say that f is a map from X to Y and write f X a Y when f is a function which assigns a point y fx 6 Y to each point z E X Two rnaps f X a Y and f X a Y are said to be equal when X X Y Y and fx f for all z E X Thus if f and f equal rnaps then graphf graphf but not conversely because Y Y is part of the de nition of equality for maps See Buck page 23 3 5 WhenA X7BQY7andfXaYthesets fA y E Y Hm E A such that y f7 f 1B z e X me e B are called respectively the image of A by f and the pre image or inverse image of B by f See Buck page 76 Problem 6 Let f X a Y7 Alha be a family of subsets of X7 and BlLEI be a family of subsets of Y Say which of the following inclusions are always true Prove each true inclusion and give a counterexample for each false one r1 U 3 g U 16409 U f 1Bi g f 1 U 3 ieI ieI ieI ieI r1 H 3 g Bn f 1Bi g f 1 H 3 ieI ieI ieI ieI f U A g UfltAgt7 U m g f U Agt7 ieI id ieI id f H A g mm H M g f H Agt ieI id ieI id 7 If f X a Y and g Y a Z7 then the composition of f and g is the mapgofXaZde nedby g 0 M96 90 for z E X For any set X the identity map of X is the map idX X a X de ned by idX z for z E X Clearly idyoffandfoidXf for f X a Y 8 A mapg Ya X is said to be aleft inverse for the map f X aY iffgofidXie iffgf xfor allsEX AmapgYaXis 4 said to be a right inverse for the map f X a Y iff f o g idy7 ie iff fgy y for all y E Y A map 9 Y a X is said to be a two sided inverse to the map f X a Y iff it is both a left inverse and a right inverse to f If g is a left inverse to f and g is a right inverse to f then 9 9 Proof 9 g o idX g o f 09 idy og 9 In this case there is a unique inverse and it is denoted f l So if f X a Y has an inverse f l Y a X7 then 1 f96 ltgt 96 f 1y formEXandyEY 9 A map f X a Y is said to be oneone iff Vrlwg E X fx1 fx2 gt 1 2 and it is said to be onto iff Y fX7 ie VyEYHmEXyfx Thus 1 A map is one one if and only if it has a left inverse 2 A map is onto if and only if it has a right inverse 3 A map is one one and onto if and only if it has an inverse Remark 10 The 7only if7 part of item 2 is called the Axiom of Choice It was once controversial because one can imagine a situation where one can prove that a map f is onto but where one can not give an explicit formula for a right inverse Remark 11 The above assertions 1 3 are false if continuity de ned later is required There is a continuous one one and onto map whose inverse is not continuous and hence continuous one one map which does not have a continuous left inverse7 and a continuous onto map which does not have a continuous right inverse See Example 64 below Example 12 Consider the four maps f RaR f20700gtR7 f3iRgt0OO f40700gt0700 de ned by 2 Then fl is not one one and not onto7 f2 is one one but not onto7 f3 is onto but not one one7 and f4 is one one onto Any map 92R a 07 00 such that 92y uT for y 2 0 is a left inverse to f2 and any map 93 000 a R such that 93y i the i can depend on y is a right inverse to f3 The inverse map to f4 is f1y 2 Axioms for the Real Numbers We state here the axioms for the real number system R We shall accept these axioms without proof but it can proved from more general axioms that there is an essentially unique structure satisfying them 13 Algebraic Axioms The set R of real numbers is equipped with two operations RgtltR Rabgt gtab7 RXR Rabgt gtab such that the usual laws of grade school arithmetic hold Commutative Laws 1 b b a and a b b a Associative Laws 1 b c a b c and a b c a b c Distributive Law a b c a c b c ZeroOne There are necessarily unique distinct elements 01 E R such thata0aanda1afor allaER Inverses For every 1 E R there is a necessarily unique element 7a such that 1 7a 0 For every 1 E R0 there is a necessarily unique element 1 1 such that a a l 1 The standard notations from high school algebra are used in particular7 ab ab7 17ba7b7 and abab 1 14 Order Axioms The set R has an order relation denoted a lt b satis fying the following laws for all 17 b7 0 E R Trichotomy Exactly one of the alternatives 1 lt b7 1 b7 b lt 17 holds Addition altb gt acltbc 6 Multiplication 0 lt ab gt 0 lt ab The other order notations are de ned as usual7 ie a lt b ltgt b gt a and agb ltgt bZa ltgt eitheraltborab Remark 15 All the rules of algebra used in College Algebra Math 112 follow from the Algebraic Axioms 13 and Order Axioms 14 For example7 a b2 a2 2ab b27 a2 2 07 etc De nition 16 The set S of real numbers is said to be bounded above iff there is a number b E R such that x S b for all z E S the number b is then called an upper bound for S A number b E R is called a least upper bound for S iff it is an upper bound for S and b S b for every other upper bound b for S Similarly the set S is said to be bounded below iff there is an number a E R such that a S x for all z E S the element b is then called a lower bound for S An element a E R is called a greatest lower bound iff it is an lower bound for S and a S a for every other lower bound a for S The words in mum and greatest lower bound are synonymous as are the words supremum and least upper bound The least upperbound of the set S will be denoted supS and the greatest lower bound of the set S will be denoted infS 17 Completeness Axiom Every set S of real numbers which is bounded above has a least upper bound7 ie ifxgbfor alleS7thenp supS gbfor alleS Because multiplication by 71 reverses the order it is the same to say that every set which is bounded below has a greatest lower bound Thus ifagxfor alle Sthena infS gxfor alleS Theorem 18 Archimedean Property There is neither an in nite real number nor an in nitesimal real number More precisely 1 There is no real number which is larger than every integer 2 For every positive real number 8 gt 0 there is a positive integer n such that 1n lt 8 Problem 19 Prove Theorem 18 Hint If u gt n for every integer n what about u 7 1 Problem 20 Let R denote the set of real valued rational functions7 ie f E R iff f where p and qx are polynomials with real coef cients and qx is not the zero polynomial For g 6 R de ne an order relation by the condition that f gt 9 iff there exists an M such that f gt g for all z gt M Then the set R satis es the algebraic axioms and order axioms given above View R and hence Z as a subset of R by identifying the real number 0 with the constant function whose value is always 0 Exhibit in the lingo of Theorem 18 an in nite element f E R and an in nitesimal element 9 E R 3 Distance 21 The distance dpq between two points p 1727 7x and q y17y2 Wyn in R is de ned by dung 951 7 11 962 7 12 m i y The distance d17 0 from a vector 1 E R to the origin is called the norm of 1 denoted by W so dltP7Q3i10 7 Qi The norm satis es the following laws for 1110 6 R Zero Norm M 0 ltgt 1 07 Homogeneity avi a M if a gt 07 Symmetry i 7 vi w Triangle Inequality 1 w 3 W The Zero Norm Law holds because a sum of squares vanishes only if each summand vanishes and the Triangle Inequality is proved in Corollary on page 14 of Buck The laws for the norm imply that the distance function satis es the following Zero Distance dpq 0 ltgt p q7 Symmetry dp7 Q dq7p7 Triangle Inequality dp7 r lt dp7 q dq7 r 8 These are proved by reading 1 p 7 q and w q 7 r in the corresponding law for the norm Remark 22 In Chapters l V Buck writes lp 7 ql instead of dpq but uses the notation dpq starting in Chapter VI in a more general setting See Buck page 304 De nition 23 For p E R and 6 gt 0 the set 31 6 i q E R i 611 Q lt 5 is called the open ball centered at p with radius 6 Note that when n 1 the open ball is an open interval for a E R Bz6 x E R lx7al lt 6 x E R 176 lt z lt 16 a76a6 4 Topological Terminology The following de nitions are from section l5 of Buck In all these de nitions the term set means subset of R De nition 24 A set U Q R is open iff for every p E S there exists a 6 gt 0 such that B 75 Q U The interior of a set S Q R is the set of all p E S such that B 75 Q S for some 6 gt 0 Thus a set is open if and only if it equals its interior and the interior of a set is the largest open set contained in S A neighborhood of a point p is a set containing p in its interior The exterior of a set S is the interior of its complement R S The boundary of a set S is the set of points which are neither exterior or interior to S It is denoted by bdryS A set S Q R is closed iff its complement R S is open The closure of the set S is the set S S U bdryS It is the smallest closed set containing S De nition 25 A point p is a cluster point of a set S iff every neighbor hood of p contains in nitely many points of S The terms limit point and accumulation point are synonymous with cluster point but for Buck the term limit point has a slightly differnt meaning when applied to a sequence See De nition 34 and Remark 36 below A point p E S is an isolated point of S iff some neighborhood of p contains no other point of S De nition 26 A set S is disconnected iff there are disjoint open sets U and V such that S Q U U V and both S U and S V are nonempty A set is connected iff it is not disconnected Theorem 27 A subset S Q R of the real line is connected if and only z39fS is an interval 239e 17 Q S whenever ab E S Proof We prove only if Assume that there exist ab E S with 17 Q S Then there is a c E 17 with 0 S Let U 7000 and V 000 The point 0 lies in the open interval a7b as ab E S so a E U and b E V Hence both S U and S V are nonempty and clearly S Q UUV as 0 S Hence the open sets U and V separate S so S is disconnected as required We prove if Assume that S is disconnected7 ie that there exist open sets UVQRwithSQUUVS U7 07S V7 7andU V Choosea E S UandbE S V Thenay basU V Assume without loss of generality that a lt b The case b lt a is the same We must show that 17 Q S7 and for this it is enough as S Q U U V to show that cw Q U U V The set T x E 17 law Q U is nonempty as it contains a and bounded above as b is an upperbound so it has a supremum 0 Clearly 10 Q T Also a lt c as a 7 86 8 Q U for suf ciently small 8 gt 0 and c lt b as b 7 ab 8 Q V Q RU for suf ciently small 8 gt 0 If c 6 U7 then c 7 60 8 Q U for suf ciently small 8 gt 0 so 10 8 1 c U a 7 86 8 C U so 1 c 8 Q T contradicting the fact that c is an upperbound for T If c 6 V7 then 0787 06 Q V for suf ciently small 8 gt 0 so 10 7 8 V Q contradicting the fact that c is the least upperbound of T Hence 0 U U V as required D De nition 28 A set S is bounded iff it is contained in some large ball7 ie iff there exists M gt 0 such that lpl lt M for all p E S Thus a set or real numbers is bounded if and only if it is bounded above and bounded below See De nition 16 5 Limits 29 Let p0 be a cluster point of a set S and F be a function de ned on S but possibly not at p0 The notation lim Fp L 127120 10 means that for every 8 gt 0 there exists 6 gt 0 such that for all p E S Bp06 p0 we have lfp 7 Ll lt 8 ie iff Vegt036gt0VpeS0ltlp7p0llt6 gt lfp7Lllte When po 6 S and p0 is a cluster point of S we have that a function f de ned on S is continuous at p0 see De nition 48 below if and only if hm f10 f 190 127120 and the function is trivially continuous at a point po 6 S which is not a cluster point of S However7 the limit notation is usually used in situations where p0 is a cluster point of S but p0 S For example7 the derivative of a real valued function f I 7 R de ned on an open interval I Q R is de ned by WW hm we 7 mo 17mg z 7 0 The ratio in the limit is unde ned when x 0 but is de ned for nearby values of x 30 For a real valued function f de ned on a subset of R we can extend the de nition of the notation limmnaFQ L to include the cases where a ioo andor L ioo as follows Let R7ooURUoo consist of the set of real numbers together with two additional points which we think of as located at in nity The set R is sometimes called the set of extended real numbers For b E R7 a set U Q R is called neighborhood of b iff either b E R and U contains an open interval b 7 pb p for some p gt 07 or else b 00 and U contains an open interval M7 00 for some M gt 07 or else b 700 and U contains an open interval 7007 7M for some M gt 0 Because Bpp b 7 pb Ap this de nition agrees with the de nition in paragraph 24 A point a E R is called a cluster point of a subset S Q R iff every neighborhood of b intersect S in an in nite set If f S a R and a is a cluster point of S7 then the notation lim L 14in means that for every neighborhood V of L there is a neighborhood U of a such that fU S Q V 6 Sequences 31 A sequence is a function de ned on a subset of the integers Usually this subset is the set Z n E Z n gt 0 of positive integers It is customary to denote the value of a sequence at an integer n with a subscript rather than with parentheses and to denote a sequence with a notation like 10 or pnnez When n1 lt n2 lt 713 lt is an increasing sequence of positive integers7 the sequence pnkh is called a subsequence of the sequence De nition 32 The sequence pnh of points of R is said to converge to the point p E R iff lim pn p VLHOO ie iff for every neighborhood U of p there exists N gt 0 such that n 2 N gt pl 6 U This is sometimes abbreviated as pn a p as n a 00 We say a sequence converges or is convergent iff it converges to p for some p E R A sequence is said to diverge when it does not converge A sequence in R whose limit is in nite is also said to diverge Theorem 33 A set S is closed if and only if it is closed under limits of sequences ie whenever 10 is a sequence of points ofS and limiH00 pn p we have p E S Theorem 5 page 40 De nition 34 A point p is a limit point of a sequence pn iff some subsequence 10M converges to p Example 35 The limit points of the sequence pn 71 are 1 and 71 The limit points of the sequence qn 71 1n are also 1 and 71 Remark 36 Every sequence p7 determines a set pn n E Z Buck calls this set the trace of the sequence but that terminology is uncommon The trace can be nite for example the trace of the sequence pn 71 is the two element set 711 If the trace of a sequence is nite then there must be at least one constant subsequence and the common value is a limit point of the sequence By de nition only an in nite set has a cluster point Theorem 37 Bolzano Weierstrass Every bounded in nite subset of R has an cluster point Corollary 38 Every bounded sequence in R has a limit point Buck The orem 22 page 62 De nition 39 A sequence an of real numbers is said to be increasing iff a1 3 a2 lt is said to be decreasing iff a1 2 a2 2 and is called monotonic iff it is either increasing or decreasing Theorem 40 A bounded monotonic sequence is convergent Buck page 47 Proof Assume the sequence an is increasing and bounded above Let a supan Then a 2 an for all n as a is an upper bound but for 8 gt 0 aN gt a 7 8 for some N as a is the least upper bound Hence an gt a 7 8 for n 2 N as the sequence is increasing D 41 For any sequence akh of real numbers we have ak k 2 n Q ak k 2 m for m lt ii If the sequence akh is bounded above then the sequence 5 supak k 2 n is decreasing The limit of the latter sequence is denoted lim supan lim supak k 2 n7ioo 4 00 Similarly for a sequence which is bounded below lim inf an lim infak k 2 Hoe Hoe De nition 42 A sequence pn is called Cauchy iff lim lpn 7 pml 0 mn7ioo ie iff for every 8 gt 0 there exists N gt 0 such that n m 2 N gt lpn7pml lt 8 Theorem 43 Cauchy Convergence Criterion A sequence in R converges if and only if it is a Cauchy sequence Buck Theorem 23 and its corollary on pages 62 63 7 Compact Sets De nition 44 An open cover of a set S is a collection Uihd of open sets such that S Q Uid Ui The subset S is compact iff every compact cover Uihd of S has nite subcover7 ie there are indices i17i27 in E I such that S Q Uil Theorem 45 The closed interval a7b is compact Buck Theorem 24 page 65 Theorem 46 Heine Borel A subset of R is compact if and only if it is closed and bounded Buck Theorem 25 page 65 Remark 47 Call a set S sequentially compact iff for every sequence pl of points in S there is a point p E S and a subsequence pm which converges to p Combining Theorems 37 and 46 we see that a set is sequen tially compact if and only if it is compact if and only if it is closed and bounded It follows 8 Continuity Let f D a R where D Q R De nition 48 The function f is said to be continuous at a point p E D iff for every 8 gt 0 there exists a 6 gt 0 such that for all q E D lqipl lt8 5 lf 1f10llt 8 Theorem 49 The function f is continuous atp E D if and only iffor every sequence pl of points in D we have 32301 3330 gt up lt1 Proof We prove only if Assume f is continuous at p Choose sequence pl of points in D Assume 31530197 p 2 Choose 8 gt 0 Because f is assumed to be continuous at p there is a 6 gt 0 such that for all q E D lqipl lt5 5 lfQf10l lt6 3 14 By 2 there is an N such that lpn 710 lt 6 for n gt N Hence by 3 fp 7 fql lt 8 for n gt N This proves in MM 7 f p 4 as required We prove 239 Assume that f is not continuous at p E D Then there is an 8 gt 0 such that for every 6 gt 0 there is a q E D such that lq7pl lt5 but Ma 7 MM 2 6 ln particular7 for each n E Z there is a qn such that 1 lQn 7pl lt E but lfQn 7 MM 2 8 But then 2 holds but 4 fails This proves that 1 is false as required D De nition 50 The function f is said to be continuous iff it is continuous at every point of D iff VpEDV8gt036gt0Vq D lq7pl lt8 gt lfq7fpl lt8 The function f is said to be uniformly continuous iff V8gt036gt0VpEDVq D lq7pllt8 gt lfq7fpllt8 For continuity 6 5107 6 for uniform continuity 6 66 51 The function f is said to be Lipschitz iff there is a constant M such that WI 7 fQl S MP 7 Ql for all pq E D A Lipschitz function is uniformly continuous Proof 6 Problem 52 Let f z Show that f is Lipschitz on every compact interval 17 Q 000 For which values of p is f uniformly continuous on 07 oo Hint Use the Mean Value Theorem from calculus See Theorem 73 below Theorem 53 below may also help 15 Theorem 53 Assume that D is compact and f is continuous The f is uniformly continuous Proof Choose 8 gt 0 Then because f is continuous for every p E D there is a 6 6p gt 0 such that e q7pi lt 51 5 Ma 7 MM lt 5 Let Up Bp6p2 The sets Up cover D since p 6 Up ie D Q Up Up As D is compact nitely many of these sets cover D ie DQUmUUmUuUUM 5 De ne 6 imin6p16p2 6 Choose pq E D Assume q 7 p lt 6 By 5 we have that p E Um for some h Hence ipipki lt SCUM2 6 But 6 S 6pk2 by its de nition so iq 7m S iq 7p p 7m S 6 g g a 693 5pk 7 Hence by the de nition of and Equations 6 and 7 we have W 7 fqi mm 7 mm W 7 ram lt 3 3 e as required 1 Remark 54 A proof using the Bolzano Weierstrass theorem instead on the Heine Borel theorem is given in Buck page 85 The proof given here is like the proof sketched in Exercises 11 and 12 in Buck page 89 Theorem 55 Assume U Q R is open Then the map f U a R is continuous if and only if the preimage of every open subset of R is an open subset of R ie if and only if whenever V Q R is open the set f 1V is also open Buck Theorem 3 page 76 Theorem 56 Assume that D Q R E Q R and that f D 7 E and g E 7 R are continuous Then 9 o f D 7 R is continuous Buck Theorem 5 page 78 Theorem 57 The continuous image of a compact set is compact ff D a R is continuous and D is compact then fD is compact Buck Theorem 13 on page 93 Corollary 58 The continuous image of compact set is bounded Buck Theorem 10 on page 90 Theorem 59 ff D a R is continuous andD is compact then f assumes its maximum on D ie there epists p E D such that fq S fp for all q E D Similarly for the minimum Buck Theorem Z page 91 Theorem 60 The continuous image of a connected set is connected If f D a R is continuous and D is connected then fD is connected Buck Theorem 15 on page 94 Corollary 61 Intermediate Value Theorem Assume that S is connected and that f S a R is continuous Suppose that ab E fS and that a lt c lt h Then 0 E fS Buck Theorem 14 on page 93 Remark 62 The lntermediate Value Theorem from calculus is a special case It says that if f 043 a R is a real valued continuous function on the closed interval 046 Q R ab f04f and a S c S b then the equation u c has a solution x 6 046 Theorem 63 Let f D a E be continuous one one and onto and assume D and hence by Theorem 57 also E is compact Then f l E a D is continuous Proof Choose a convergent sequence qn in E and a let q lim 1 be its limit We will show that f 1Q giggle f 1qn the Theorem will then follow by Theorem 49 By Bolzano Weierstrass and Heine Borel there is a convergent subsequence f 1ank Let p i f 1an 17 be its limit Now f is assumed to be continuous so fp f giggle f 1qnigt giggle ff 1qni klgrgoqni q But up q p PM so lo 11mm PM If ltgt fails then there is a neighborhood U of f 1q such that for every N there ex ists n gt N with f 1qn U7 ie there is a subsequence f 1qmj with f 1qmj U As before choose a further subsequence again denoted f711mj which converges and let p hm f 1qmj 7400 denote the limit Then 19 U else we would have f71Qmj E U for suf ciently large j so 19 31 p But as before fp f 71330 f 1qmj 711 ff 1qmj 711330 1m 1 But now fp q fp which contradicts the fact that f is one one D Example 64 Let S E R2 2 y2 1 denote the unit circle in R2 and de ne f 027r a S by f0 cos 67sin0 Then f is one one onto and continuous but f 1 is not continuous The interval 07 27139 is not compact These example shows both that a continuous one one onto map whose inverse is not continuous See Remark 11 65 A real valued function f U a R de ned on a subset U of the real numbers R is called increasing iff 1 3 2 gt fz1 S fx2 for all 172 6 U7 decreasing iff 1 3 2 gt fz1 2 fx2 for all 172 6 U7 monotonic iff it is either increasing or decreasing The function f is called strictly increasing iff 1 lt 2 gt fz1 lt fx2 for all 172 6 U7 strictly decreasing iff 1 lt x2 gt fx1 gt fx2 for all 172 6 U7 strictly monotonic iff it is either strictly increasing or strictly decreasing Theorem 66 A continuous function f I a R de ned on an interval I Q R is one one if and only if it is strictly monotonic When these equivalent conditions hold the image J fI is again an interval and the inverse function is continuous Problem 67 Prove Theorem 66 This theorem is proved in Buck The orem 18 page 96 and Theorem 25 page 1147 but Buck assumes that the intervals are closed and bounded This assumption can be removed Theorem 68 Let S Q R be and f S a R be uniformly continuous Then the function f can be continuously emtended to the closure 5 of S ie there is a continuous function F S a R such that Fp fp forp E S Buck Theorem 25 on page 109 Example 69 The function f 071 a Rde ned by fx sin1x cannot be extended to a continuous function on the closure 01 of 071 Problem 70 Fix a positive number a E R The purpose of this problem is to de ne the exponential am for z E R De ne a0 1 and for n a positive integer de ne a aaa a7 1a Then 1 Prove that for every nonzero integer ii there is a unique solution b gt 0 to the equation b 1 De ne lln to be this unique solution7 ie Lil b ltgt b a 2 For a rational number q de ne 1 7 by 1 7 am1 where q Prove that this de nition is independent of the choice of the integers in and n such that q 3 Prove that there is a unique continuous function Ra 000 xi mzm such that am 1 7 when x is a rational number Q In your proof make clear which theorems from these notes you are appealing to Also make your proof self contained so that a person who doesnt have access to the statement of the problem can follow it You needn7t provide proofs for the theorems you use7 but do provide references to them In your proof of 3 you may use the inequality ap aqSM10Q 19 which is true when a gt 17 N is a positive integer7 1 S pq S N 17 and N 1 i 7 N M 7 k a k1 You need not prove this inequality but use calculus to show where it comes from Hint What is the de nition of ln x em7 and am used in calculus The natural logarithm function lnz is usually de ned as an integral How do you bound an integral by a sum How does the Mean Value Theorem Buck Theorem 3 page 118 give inequalities like this 9 Derivatives De nition 71 The function f I a R is said to be differentiable at the point 0 6 I iff the limit We hm M 4m maze z 7 0 exists we say that f is differentiable on a set iff it is differentiable at each point no in the set The function f is called the derivative of f Theorem 72 A di erentiable function is continuous Theorem 73 Mean Value Theorem Suppose that f is di erentiable of ab and continuous on lab The there eists c 6 ab such that f b i f a 7 f c 7 b 7 a Corollary 74 Assume that f is di erentiable on I Then the derivative f vanishes identically on I if and only iff is constant on I 10 The Integral 75 A partition of the closed interval a7b is an increasing nite sequence a ogkgn with x0 a and an b7 For any bounded function f de ned on ab and any partition P z oSkSn of ab de ne the upper sum by 30313 3 El n M71 It supfz zk1 g x 3 95k k1 20 and the lower sum by 0213 21mm 1k mum zm x 95k k1 Theorem 76 Assume that f db a R is continuous Then there is a unique number fjf called the de nite integral off on the interval db such that the inequality b 0213 s f s 5 0 P holds for every partition P of the interval db Theorem 77 The de nite integral satis es the following properties Constant b1 b 7 a Linearity abfgabfabg abcfcabfforc R Additivity abfbcff b 17 Order Preservation Iffx S 9a for allz 6 db then f S 9 if sablfl Theorem 78 Fundamental Theorem of Calculus Assume that I is an open interval a E I and that is de ned by f The F is di erentiable on I a7 00 and its derivative is f Hence Triangle Inequality b f Fltbgt 7 Fltagt for b 2 a The number fabf is called the integral off on the interval db 21 Remark 79 Henceforth we use the more traditional notation f dx for the integral The reader is reminded that the variable x in this expression is a dummy variable7 ie b b fa dab ft dt 11 Taylor s Formula Theorem 80 Taylor7s Formula 7 Lagrange Form Let I be an interval a E I and f I 7 R be of class 0 Then Wt Edam PMCW for any x E I where r k a w n 1 Pnaz 2 f 7 ak7 Rnaz 7 t dt x H k The polynomial Pnax is called the Taylor polynomial of degree n at a and Rna7 is called the nth Taylor remainder Proof By induction on n For n 0 this is the Fundamental Theorem of Calculus We assume the formula for n and integrate by parts 95 0M1 1 7 n t v 71W u f o 96 i t n Rama mudo 7odu LHW z 7 a 1 w 7 an f 2t dt do dt du f 2t dt UU n 1 n 1 n1 Adding Pna7 c to both sides gives f Pn1az Rn1a D 22 Corollary 81 If lf 1tl M fort e I then Mix 7 al 1 lt anazl 7 RH for z E I Proof Assume that z gt a The case x lt a is similar x mm imam fT zetrdtl Sm wait dt m 9570 xiaYLH Sa M ml 7 n1 39 D Remark 82 In Math 222 it is shown that that there is a number 0 between a and x such that f 10 n1 Rama 7 7 a This form of the remainder has the advantage that it is easy to remember the remainder is the next term in the series with f 1a replaced by f 1c However7 this version of the theorem only holds when f is real valued7 ie when m 1 For m gt 1 there will be a different value of c for each component of f 12 Series 83 A sequence determines a series and a series determines a sequence More precisely7 a sequence akh determines a series whose partial sums are Sn ak31112quot39an7 M w H H and the terms of the series may be recovered from the sequence of partial sums via the formula anSnisn717 1151 23 Convergence of the series is synonymous with convergence of the sequence of partial sums 00 V L E ak lim E ak Hoe k1 k1 Theorem 84 If the series 2k ak converges then the nth term converges to zero Buck Theorem 2 page 230 Example 85 The nth partial sum ka12n k0 of the geometric series is easy to compute 16 7 h1gt 17 zn1 17ka n hO h n o as the sum telesco es so dividin b 1 7 z ives p g y g 71 1 7 n1 2M 1 k0 T Hence if lt 1 we have the formula 1 gxk1iz for the in nite sum Theorem 86 Cauchy Convergence Criterion for Series A series 2k ak converges if and only if 71 mine 2 0160 hm1 Buck Theorem 3 page 265 Proof This follows immediately from the formula 2 akSniSm7 SnZak Z akam1an km1 h1 hm1 and Theorem 43 the Cauchy Convergence Criterion for Sequences D 24 87 The notation 00 Zak oo k1 means that for every M gt 0 there exists an integer N gt 0 such that ZZ1ak gt M for n gt N If ak 2 0 for all h then the sequence of par tial sums is monotonic increasing so by Theorem 40 either the limit 21 ak exists ie the sequence of partial sums is bounded or 21 oo ie the sequence of partial sums is unbounded De nition 88 The series 2k ak is said to converge absolutely 00 Z lakl lt oo k1 A series which converges but does not converge absolutely is said to converge conditionally Theorem 89 A series which converges absolutely converges Proof This is an immediate consequence of the inequality n E Z lakl hm1 and Theorem 86 D Theorem 90 Comparison Test If 0 S ak S by and the series Zkbk converges then the series 2k ak does Buck Theorem 5 page 23 Theorem 91 lntegral Test Assume that ak fk where f 1700 a 07 00 is monotonic decreasing Then the improper integral flee fp dx lt oo converges if and only if the sum 221 ak lt 00 does Buck Theorem 10 page Theorem 92 Root Test Let r lim sup lakllk haoo Then the series 2k ak converges absolutely ifr lt 1 and diverges does not converge ifr gt1 Ifak 1k the r 1 but Ekak oo Ifak 1k2 the r 1 and 2k ak lt 00 Buck Theorem 9 page 25 Remark 93 The proofs the convergence tests tell us how to estimate the error ie the difference between a partial sum and the in nite sum For example by the Integral Test the series Zk1k2 converges and f 1 4 h 7 n mzin39 Similarly if r lt 1 in the Root Test and r lt p lt 1 there is an N such that lakl lt pk for h gt N and hence iak iak 0 lale 0 pk pn Si k1 161 kn1 hn1 lip 1 P for n gt N Problem 94 Show that for p gt 1 the series 221 k p converges and that the estimate 00 V L szw k1 k1 holds for the difference error between the nth partial sum and the limit lt n11 p p71 Problem 95 Alternating harmonic series Show that the series 22171kk converges conditionally but not absolutely to iln 2 Hint Use Taylor7s Theorem See Buck section 35 page 147 Warning There is a mistake in the formula 3 37 for You will need to show that the absolute value of the integrand is S 1n De nition 96 A series 21 by is said to be a rearrangement of the series 221 ak iff there is a one one correspondence 0 Z1 a Z1 such that bk am Theorem 97 1 Any rear t of an L t buyout151170 series converges absolutely to the same limit 2 Assume that the series 21 ak converges conditionally and L E R see paragraph 30 Then there is a rearrangement 21 by of the series 22 ak such that Proof Part 1 is Theorem 13 page 239 of Buck Part 2 is proved on pages 238 9 of Buck in the special case where an 71 n and L 10 the general argument is much the same but uses Theorem 84 D 1 3 Uniform Convergence De nition 98 A sequence fn U a V of functions is said to converge pointwise to the function f U a R iff limiH00 fnp fp for all p 6 D7 ie iff Vp 6 Ne gt oann n gt N lfnp i fpl lt a A sequence fn U a V of functions is said to converge uniformly to the function f U a V iff limiH00 suppeU lfnp 7 fpl 07 ie iff v5 gt oanp e UVnn gt N Un t fpl lt 6 For a sequence uk D a Rmk of functions the series 2k uk of functions is said to converge pointwise or uniformly iff the sequence fn ZZ0uk of partial sum does Example 99 The sequence fn 01 a R de ned by fnz x converges pointwise but not uniformly to the function 0 for0 xlt1 f71 forz1 Theorem 100 If the sequence fn U a V of functions converges uni formly to the function f U a V and each fn is continuous then the limit f is also continuous Buck Theorem 3 page 266 Theorem 101 Weierstrass Comparison Test Assume that the functions uk U a R satisfy 3 My where 2k Mk lt 00 Then the series 2k uk converges uniformly Buck Theorem 2 page 266 Theorem 102 Assume that the sequence of continuous functions fn a7b a R converges uniformly to a function f Then the limit of the integrals is the integral of the limit ie dx abfz dz 27 Corollary 103 Let be an open interval and fn I a R be a sequence of dz erentlable functions Assume that the sequence fnn conuerqes uniformly to f I a R and that the sequence of derivatives also conuerqes uniformly Then the limit f is dl erentlable and the limit of the derivatives is the derivative of the limit 239e in M96 Hz force Problem 104 Assume that the sequence bnh is eventually bounded by the sequence 71 ie there is an N such that bnl lt nip for n gt N 1 Show that if p gt 1 the series u ansinnx 104 1 n1 converges uniformly 2 Show that ifp gt 1 then 2 7r bn 7 u s1nnx dz 104 2 77 0 3 Show that if p gt 2 then f is differentiable and that fa Z nbn cos nz n1 Hint See Problem 94 above You may use any of the theorems stated above but state which theorems you are using and verify that the hypotheses of the theorems are satis ed 28 Problem 105 Continue the notation of Problem 104 Show that the series ut Ze tbn sin 7w 105 1 n1 converges uniformly on 07 oo gtlt 077T ifp gt 1 and that the limit satis es the partial differential equation 3U 3211 a 1052 on the open set 07 oo gtlt 077139 Show also that u is continuous on the closed set 07 oo gtlt 077139 that it satis es the initial condition ultozgt we 1053 and the boundary condition ut7 0 ut7r 0 105 4 You may use any of the theorems stated above but state which theorerns you are using and verify that the hypotheses of the theorems are satis ed Hint 677 is very small if t gt 0 and 71 large It looks like this exercise proves that the solution of the partial differential equation 105 2 subject to the intial condition 105 3 and the boundary condition 105 4 is given by 105 1 where the coe cients are de ned by 104 2 ls there anything missing for a rigorous proof 14 Power Series 106 A series of form 220 ckx 7 111 is called a power series centered at a The radius of convergence of the power series is the number R de ned by 1 i lirnsup ck lk R kaoo lfthe lirn sup is in nite7 then R 0 and if the lirn sup is zero7 then R 00 1This PDE is called the Heat Equation 29 Problem 107 A formula for the radius of convergence Assume that the coef cients ck are nonzero Show that 1k hm i0k1i lim sup icki h7ioo knee lel if the limit on the right exists Theorem 108 Let 220 ckx 7 ak and R be its radius of convergence Then the series converges if ix 7 a lt R and diverges if ix 7 a gt B More precisely i If 0 lt r lt R the series converges uniformly the interval a 7 r7 a r ii I 7ai gt R then the nth term of the series is unbounded and hence does not converge to zero so the series does not converge by Theorem 84 Buck Theorem 14 page 240 Theorem 109 Let 220 ckx 7 ak be a power series and R be its radius of convergence Denote the sum by f chw 7 ak k0 for ix 7 a lt R Then f is di erentiable on the interval a 7 Ra R its derivative is given by term wise di erentiation ie fd Z hckw 7 akquotl7 k1 and the radius convergence of this last power series is also R Corollary 110 Taylor Series Continue the hypotheses of Theorem 109 Then f is in nitely di erentiable on the interval a 7 Ba R the jth derivative fm off is 77a7mra lek jl w H M8 fltigt96 ZMk 71 j1ckiakj k7 7 so that cj fjaj ie 0 7 rmZL 73970 x 7 a 30 Math 521 Lecture 30 Arun Rain University of Wisconsin Madison 480 Lincoln Drive Madison7 WI 53706 ram mathwiscedu 1 The space CX Let X be a topological space Let MX f XHC be the algebra of complex valued functions on X The operation on MX is the map 2 MX H MX given by Hz W for 6 X Let X be a topological space Let CX f X a R l f is continuous and bounded The supremum norm on CX is the function CX a R given by Hill 3 WW De ne d CX x CX HRZO by W79 Hf 7 9H Theorem 11 CX is a complete metric space 2 Sequences of functions Let X be a topological space Let MX be the algebra of complex valued functions on X Let f17 f2 be a sequence of functions in The sequence f17 f2 converges pointwise to f X a C if x f m for z E X The sequence f17 f2 converges uniformly to f X a C if it is such that if e E Rgt0 then there exists N E Zgt0 such that if n E 220 with n 2 N then lfn fl for allmeX Proposition 1 Let f17 f27 be a sequence of functions in CX Then f17 f27 conuerges in CX if and only if f17 f27 conuerges uniformly Theorem 21 Let X be a metric space and let E Q X Let m be a limit point of E Let f17 f27 be a sequence of functions in and suppose that lim exists for each n E Zgt0 tam Then lim lim fnt lim lim fnt ootaz taznaoo 3 The StoneWeierstrass theorem Theorem 31 If f a7 1 gt C is a continuous function then there exists a sequence of polyno mials 191192 such that 191192 conuerges uniformly to Let X be a metric space and let E Q X Let A be a subalgebra of The algebra A is self adjoint if it is such that if f E A then f E A The algebra A separates points if it is such that if 172 E E then there exists f E A such that 7g The algebra A vanishes at no point if it is such that if z E E then there exists f E A such that fz 7 0 Theorem 32 Let X be a metric space and let K be a compact subset of X Let A be a subalgebra of If A is self adjoint separates points and and uanishes at no point of K then A is dense in Math 521 Lecture 17 Arun Ram University of Wisconsin Madison 480 Lincoln Drive Madison7 WI 53706 ram mathwiscedu 1 Interiors and closures Let X be a topological space and let x E X A neighborhood of z is a subset N of X such that there exists an open subset U of X with z E U and U Q N Let X be a topological space and let E C X A neighborhood of E is a subset N of X such that there exists an open subset U of X with E Q U Q N Let X be a topological space and let E C X The interior of E is the subset E0 of E such that a E0 is open in X7 b If U is an open subset of E then U Q Equot Let X be a topological space and let E Q X The closure E of E is the subset E of X such that a E is closed7 b If V is a closed subset of X and V 2 E then V 2 E Let X be a topological space Let E Q X An interior point of E is a point z E X such that there exists a neighborhood Nm of z with Nm Q E Let X be a topological space Let E Q X A close point to E is a point z E X such that If Nm is a neighborhood of of x then Nm contains a point of E Theorem 11 Let X be a topological space Let E Q X a The interior ofE is the set of interior points of E b The closure ofE is the set of close points of E 2 Hausdorff spaces A Hausdorff space is a topological space X such that if m y E Y and z 7 y then there exist a neighborhood Nm of z and a neighborhood Ny of y such that Nm U N 0 Theorem 21 Let X be a topological space Show that the following are equivalent a Any two distinct points ofX haue disjoint neighborhoods b The intersection of the closed neighborhoods of any point of X consist of that point alone c The diagonal of the product space X X X is a closed set d For euery set I the diagonal of the product space Y X1 is closed in Y e No lter on X has more than one limit point If a lter 7 on X conuerges to m then m is the only cluster point of m 3 Limit points and cluster points Theorem 31 Let X be a topological space and let 17 2 be a sequence in X Then a y is a limit point of 1 2 if and only if if Ny is a neighborhood ofy then there exists no 6 Zgt0 such that M E Nm for all n E 220 n 2 no b y is a cluster point of 1 2 if and only if if Ny is a neighborhood ofy and n0 6 Zgt0 then there exists n E Zgt0 with n 2 no such that ml 6 Ny 4 Compact sets Let X be a set A lter f on X is convergent if it has a limit point Theorem 41 Let X be a topological space The following are equiualent a Euery lter on X has at least one cluster point b Euery ultra lter on X is conuergent c Euery family of closed subsets ofX whose intersection is empty contains a nite subfamily whose intersection is empty d Euery open couer ofX contains a nite subcouer Math 521 Lecture 4 Arun Ram University of Wisconsin Madison 480 Lincoln Drive Madlison7 WI 53706 ram mathwiscedu 1 Polynomials Let F be a eld If 0107 al a2 6 IF use the notation a0a1xa2x2 E Llxi 62120 The polynomial ring is the set Z l ai 6 IF and all but a nite number of the ai are equal to 0 62120 with operations given by Z Llmi Z bimi Z aibizi 62120 62120 62120 and Z aizi Z bjzj Z ckmi where ck Z aibj 62120 jeZZo 1662120 i7k Let a 6 IF The evaluation homomorphism is 1 2990 6 A S V where PaP0P1aP2a239 if PP0P1P2902 Let pz 6 FM pz po 1 plz 1 p22 1 The degree degpz of pz is the maximal nonnegative integer 1 such that pd 7 O The eld of rational functions in z is the set M l altzgt7bltzgt e 1m be y 0 1 with aw Cw i a m m m c m and with operations given by M C90 WWW MGM and WE C90 490M b90 190 WNW W 190 bd39 The ring of formal power series in z is with operations given by lt Z admit lt 2 ball lt Z ai bixigt 62120 62120 62120 lt2 lt2 62120 jEZZO Examples and lt Z Ckmi 7 Where Ck Z aibj k ijk 11im1z2m3n 132 133 131 251z2l3 72w ieZgtO 3 355 11359141 3 s1nzz7 gi12il 2i1 ZEZZO 2 4 im2i 400sz1ii im 0 2 3 11 lt5gt1nlt17zgtz7 m Z i6ZgtO The Laurent polynomials is the set M altmgt7bltzgt 6 Walk be y 0 with altzgt cltzgt and with operations given by cmamdmbmcm and altzgtcltxgt altzgtcltzgt be dltzgt bltzgtdltzgt be dltzgt bltzgtdltxgt Proposition 1 a is an integral domain b is an integral domain Proposition 2 a The invertible elements of are the invertible elements of lF b The invertible elements of are a0 alm ang 6 with a0 invertible in lF Corollary 1 l Mm E Fllzllmo 3 0 Let pz E The order 1pz of pz 19ng is the minimal integer t such that 6 Pl 739 0 The order function 1 a Z is a normalized discrete valuation see BouC Ch Vl 3 n0 6 Def 3 Math 521 Lecture Notes7 Fall 2004 Arun Ram University of Wisconsin Madison 480 Lincoln Drive Madison WI 53706 ram mathwiscedu 1 Numbers The positive integers is the set Zgt0 1 2 3 with the operation ZgtO X ZgtO T ZN m H z J The nonnegative integers is the set Z20 0 1 23 with the operation 220 2233 The advantage ofthe nonnegative integers Z20 over the positive integers Zgt0 is that Z20 contains an identity element for the operation and Zgt0 does not The integers is the set Z 7372710123 with the operations Z x Z a Z 2 7 3 H Hi The advantage of the integers Z over the nonnegative integers Z20 is that every element of Z has an inverse this is not true in Z20 There is another operation on Z Z x Z a Z m e 2 H but this operation is not very well behaved it is not associative and not commutative though it does have an identity There is another operation on Z Z x Z a Z Li H ii and this operation is associative commutative and has an identity but does not have inverses The rationals is the set a a C i QElabeZb7 0 where EE lfadibc with operations de ned by a 0 ad be a 0 ac 7 d 7 7 7 7 b d bd an b d bd The advantage of the rationals Q over the integers Z is that the multiplication has inverses well7 almost has inverses7the element 0 does not have an inverse By long division7 every rational number can be represented as a decimal expansion deT1 d1d0d1d2d3 where the idea is that 1414d1d0d1d2dg Z 1210 lemgr If a a a1a0a1a2 is a decimal expansion let as be the element of Q given by agn a a1a0a1a2 an1an The real numbers is the set R of decimal expansions R dTd1d0d1dg ldi E 0129 with a b for all n E Zgt0 agn 7 bgngn1 0 in Q7 and operations determined by a b c if7 for all n E Z agn bgngn1 an1 in Q7 and ab C for all 71 E Zgt07 agnbgngn1 an1 in An irrational number is a real number that is not a rational number Theorem 11 Q decimal expansions that repeat Theorem 12 Irrational numbers exist The complex numbers is the set Cabi l a7bER with operations given by OH bii a2 52 a1 a2 bi bz and a1 blia2 bgi alag 7 b1b2 a1b2 a2b1t Theorem 13 The fundamental theorem of algebra If 1901917 7pd E C with pd 7 0 then there are 17 7 Ad 6 C such that P0P1P2902 pd90d 1 2m7d The algebraic numbers is the set Q z E C l there exists pz E QM7 pz 7 07 with 192 0 A transcendental number is a complex number that is not an algebraic number References BouA Bourbaki7 N Algebra7 Springer7 New York7 7 BouC Bourbaki7 N Commutative Algebra7 Springer7 New York7 7 BouF Bourbaki7 N Functions of a Real Variable7 Springer7 New York7 777 Boul Bourbaki7 N Integration7 Springer7 New York7 7 BouS Bourbaki7 N Theory of Sets Springer7 New York7 7 BouT Bourbaki7 N General Topology7 Springer7 New York7 7 BouV Bourbaki7 N Topological Vector Spaces7 Springer7 New York7 7 Ru Rudin7 W Principles of Mathematical Analysis McGraw Hill7 New York7 7 DS Dunford and Schwartz7 92 Wiley7 New York7 7 Math 521 Lecture 9 Arun Ram University of Wisconsin Madison 480 Lincoln Drive Madison7 WI 53706 ram mathwiscedu 1 Derivations Let IF be a eld A vector space over IF is an abelian group V with a map lFxVgtV c711 gt gt 31 such that a lf01702 6 IF and v E V then 01 02v 011 021 b If c 6 IF and 121122 6 V then CUl 122 0121 0122 c If 01 62 6 IF and v E V then 0102v0102v d lfUEVthen1vv Let IF be a eld Let V7 W be vector spaces over IF An lFlinear map from V to W is a function 4p V a W such that a 4p is a group homomorphism7 b If c 6 IF and v E V then 0311 cltpv Let IF be a eld An algebra is a vector space A over IF with an operation A x A a A 017 a2 gt gt alag such that A is a ring and the scalar multiplication is the composition of the map lF A A E gt gt 5 1 and the multiplication in A Let IF be a eld Let A be an lF algebra A derivation of A is an lF linear map 1 A a A such that if a1 a2 6 A then da1a2 a1da2 da1a2 d d Theorem 11 a There is a unique deriuation UT of such that i 1 z z 17 Up 6 then if coe cient ofy in pm swam keZZO 0 Up 6 then at d There is a unique extension of CT to a deriuation of z d e There is a unique extension of CT to a derqution of z There is a unique extension of all to a deriuation of z 9 If E then if coe cient ofy in pz 2 59 W m If E then Math 521 Lecture 11 Arun Rarn University of Wisconsin Madison 480 Lincoln Drive Madison7 WI 53706 ram mathwiscedu 1 Ordered sets Let S be a set An partial order on S is a relation on S such that a lfzg2 Sandz yandygzthenzgz7 b lfzy Sandm yandy zthenzy Let S be a set An total order on S is a partial order on S such that c lfmy Sthenz y0ry z A partially ordered set or poset is a set S with a partial order on S Let S be a poset A lower order ideal of S is a subset E of S such that if y E E7 z E S and x g y then x E E Let S be a poset and let E be a subset of S An upper bound of E is an element 1 E S such that ify E E then b 2 y Let S be a poset and let E be a subset of S An lower bound of E is an element 6 E S such that ify E E then 6 g y Let S be a poset and let E be a subset of S The greatest lower bound of E is the element infE E S such that a infE is a lower bound of E7 b if 6 E S is a lower bound of E then 6 infE Let S be a poset and let E be a subset of S The least upper bound of E is an element supE E S such that a supE is a upper bound of E7 b if b E S is a upper bound of E then supE b A lattice is an poset S such that every pair of elements 71 6 S has a greatest lower bound and a least upper bound Let S be a poset The intervals in S are the sets 11zESlagmgb bm Sla zltb abm Slaltm b 1Im Slaltxltb7 la x65laz7 a7 m Slaltz 70071m6Slmgb 7007bz Slmltb F 00 00 for 071 E S The sets 11 011 6 S are closed intervals and the sets 11 011 6 S are open intervals HW Show that if S is a lattice then the intersection of two intervals is an interval Give an example to show that this is not necessarily true if S is not a lattice A poset S is left ltered if every subset E of S has an upper bound A poset S is right ltered if every subset E of S has an lower bound Let S be a poset and let E be a subset of S A minimal element of E is an element x E E such that if y E E then x g y A poset S is well ordered if every subset E of S has a minimal element HW Show that Every well ordered set is totally ordered HW Show that there exist totally ordered sets that are not well ordered Math 521 Lecture 16 Arun Rarn University of Wisconsin Madison 480 Lincoln Drive Madison7 WI 53706 ram mathwiscedu 1 Sequences Let X be a set A sequence 172737 of points in X is a function Zgt0 A X n gt gt xn Let X be a set and let 1 2 be a sequence in X A limit of the sequence 1 2 is a limit point of the sequence with respect to the F r chet lter on Zgt0 Write y lirn zn if y is a limit of the sequence 1727 Hoe The sequence 1727 converges if lirn mn exists and is unique Hoe The sequence 17 2 diverges if it does not converge Let X be a totally ordered set and let 1727 be a sequence in X The upper limit lim sup zn of the sequence 17 2 is the supremum of the set 17m2 7 lim sup zn supm1m27 Let X be a totally ordered set and let 1 2 be a sequence in X The lower limit lirn inf zn of the sequence 1 2 is the in mum of the set 1 2 7 lim infzn infz17 m2 A sequence 1727 is bounded if the set 17m2 is bounded A sequence 1 2 is monotonically increasing ifit is such that ifi E Zgt0 then xi n1 A sequence 1727 is monotonically decreasing if it is such that ifi E Zgt0 then ml 2 i1 2 Series 00 Let X be an abelian group and let a17a27 be a sequence in X The series Zan is n1 the sequence 317 32 33 7 where sk 51 52 sk Write 00 E ana if lim sna Hoe n1 00 The series E an converges if the sequence 517 32 converges n1 00 The series 2 an diverges if the sequence 317 32 diverges n1 00 00 The series 2 an converges absolutely if the series lanl converges 1 n1 n 00 Theorem 21 Suppose that Zan a and 2 an converges absolutely Then nil 7 n6ZgtO 00 a Every rearrangement of Zan converges to a n1 CO 0 b If 1 is a series and 1 b then is n Math 521 Lecture 15 Arun Ram University of Wisconsin Madison 480 Lincoln Drive Madison7 WI 53706 ram mathwiscedu 1 Neighborhoods Let X be a topological space and let x E X A neighborhood of z is a subset N of X such that there exists an open subset U of X With z E U and U Q N Let X be a topological space and let E C X A neighborhood of E is a subset N of X such that there exists an open subset U of X with E Q U Q N 2 Continuous functions Continuous functions are for comparing topological spaces Let X and Y be topological spaces A function f X a Y is continuous if it satis es the condition if V is an open subset of Y then f 1V is an open subset of X Let X and Y be topological spaces Let a E X A function f X a Y is continuous at a if it satis es the condition if V is a neighborhood of at in Y then f 1V is a neighborhood of a in X Theorem 21 Let X and Y be topological spaces and let a E X A function f X gt Y is continuous at a if and only if limw a at 3 Filters Let X be a set A lter on X is a collection f of subsets of X such that 1 a if E Q X such that there exists U E f with E Q U then E E f 2 b nite intersections of elements of F are in f 3 c0 f Let X be a set and let f1 and f2 be lters on X The lter fl is ner than f2 is f1 2 f2 Let X be a topological space and let x E X The neighborhood lter of z is the collection f neighborhoods of The Fr chet lter on Zgt0 is the collection f complements of nite subsets of Zgt0 Let f be a lter on a set X A lter base of f is a collection 3 of subsets of X such that f subsets of X that contain a set in B Let f be a lter on a set X A subbase of f is a collection 5 of subsets of X such that B nite intersections of elements of S is a base of the lter f 4 Limits points and cluster points Let X be a set and let f be a lter on X A limit point of f is a point z E X such that the neighborhood lter of z is ner than f Let X be a set and let 3 be a lter base of a lter f on X A cluster point of B is a point z E X such that z is in the closure of each set in B Let X be a set with a lter f and let Y be a topological space Let f X a Y be a function A limit point of f X a Y is a limit point of the lter base ff Write Write y lin x if y is a limit point of f A cluster point of f X a Y is a cluster point of the lter base ff Let X be a set A sequence 172737 of points in X is a function Zgt0 A X n gt gtmn Let X be a set and let 17 2 be a sequence in X A limit of the sequence 17 2 is a limit point of the sequence with respect to the Frechet lter on Zgt0 Write y nlggo 1 if y is a limit of the sequence 17 2 Let X be a set and let 17m27 be a sequence in X A cluster point of the sequence 17 2 is a cluster point of the sequence with respect to the Frechet lter on Zgt0 Let X and Y be topological spaces Let a E X A limit of fz as x approaches a is a limit point of f with respect to the neighborhood lter of a Write y lim 1 if y is a limit of fz as x approaches a Let X and Y be topological spaces and let a E X Let f X a Y be a function The function f is continuous at a if it sati es the condition7 if N is a neighborhood of fa in Y then f 1 is a neighborhood of a in X Theorem 41 Let X and Y be topological spaces and let a E X A function f X a Y is continuous at a if and only if limw a fa 5 Compact sets Let X be a set An ultra lter on X is a lter f such that there is no lter on X which is strictly ner that f Let X be a topological space The space X is quasicompact if every lter on X has a cluster point Theorem 51 Let X be a topological space The following are equivalent 1 a Every lter on X has at least one cluster point 2 b Every ultra lter on X is convergent 3 c Every family of closed subsets ofX whose intersection is empty contains a nite sub family whose intersection is empty 4 d Every open cover ofX contains a nite subcover A topological space is Hausdorff if any two distinct points of X have disjoint neighborhoods A topological space is compact if it is quasicompact and Hausdorff Math 521 Lecture 2 Arun Ram University of Wisconsin Madison 480 Lincoln Drive Madison7 WI 53706 ram mathwiscedu 1 Sets and functions The basic building blocks of mathematics are sets and functions Functions are for comparing sets Sets A set is a collection of elements Write s E S if s is an element of a set S The emptyset Q is the set with no elements A subset T of a set S is a set T such that ift E T then if E S Write T Q S if T is a subset of S Two sets S and T are equal if S Q T and T Q S Write T S if S and T are equal sets Let S and T be sets S is a proper subset of T if S Q T and S 7 T Write SiT if S is a proper subset of T Let S be a set and let A be a subset of S The complement of A in S is the set Acb Slb A Let S and T be sets The union of S and T is the set S U T of all u such that u E S or u E T SUTu l uESoruET Let S and T be sets The intersection of S and T is the set S O T of all u such that u E S and u E T S Tulu Sandu T Let S and T be sets The sets S and T are disjoint if S O T Q The product of two sets S and T is the set of all ordered pairs 372 where s E S and t 6 T7 SXTst l SESJET More generally7 given sets S17 Sn7 the product S1 is the set of all tuples 317 73 such that si 6 Si The elements of a set S are indexed by the elements of a set I if each element of S is labeled by a unique element of I lfi E I 3 denotes the corresponding element of S Example Let S T U and V be the sets S 1 2 U 12 T 123 and V 23 Then Functions Let S and T be sets A function or map f S a T is given by associating to each element 3 E S an element fs E T f S a T s H f8 Often in mathematics one will try to de ne a function without being exactly sure if what has been de ned really is a function In order to check that a function is well de ned one must check that a If s E S then fs E T 81 82 then flt81gt flt82 Let S and T be sets Two functions f S a T and g S a T are equal if fs 93 for all s E S Write f 9 if f and g are equal functions Let f S a T be a function Let K Q S The image of K is the set fK 1 l k 6K Let f S a T be a function Let L Q T The inverse image of L is the set f 1L 8 6 S l f8 6 L Let f S a T be a function The image of f is the set fS Let f S a T be a function and let t E T The ber of 1 over if is the set f 1t A function f S a T is injective if it satis es the condition If 8182 6 S and flt81gt flt82gt then 81 32 A map f S a T is surjective if it satis es the condition ift E T then there exists 3 E S such that fs t A function is bijective if it is both injective and surjective Examples It is useful to visualize a function f S a T as a graph with edges 5 fs connecting elements of s E S and fs E T With this idea in mind we have the following PICTURE In these pictures we are viewing the elements of the left column as elements of the set S and the elements of the right column as the elements of a set T In order to be a function the graph must have exactly one edge adjacent to each element of S A function is injective if there is at most one edge adjacent to each point of T A function is surjective if there is at least one edge adjacent to each point of T Let f S a T be a function and let R Q S The restriction of f to R is the function R given by R R a T r gt gt f Let S be a set7 let R be a subset of S and let 1 R a T be a function An extension of f to S is a function g S a T such that if r E R then gr fr Composition of functions Let f S a T and g T a U be functions The composition of f and g is the function g o 1 given by g o S gt U s gt gt Let S be a set The identity map on a set S is the map given by ids S H S S l gt 8 Let f S a T be a function An inverse function to f is a function f l T a S such that f o f 1 idT and f 1 o f ids where idT and id are the identity functions on T and S respectively If we visualize functions as graphs7 the identity function ids looks something like PICTURE In the pictures below7 if the left graph is a pictorial representation of a function f S a T then the inverse function to f f 1 T a S7 is represented by the graph on the right PICTURE Proposition 1 Let f S gt T be a function An inverse function to 1 exists if and only 239s bijective Pictorially7 the graph7 below left7 represents a function g S a T which is not bijective The inverse function to 9 does not exist in this case the graph of a possible candidate below right is not the graph of a function PICTURE Let f S gt T be a surjective function A section of f is a function 5 T a S such that f o s idT Let f S gt T be an injective function A retraction of f is a function f S a T such that r o f ids Math 521 Lecture 14 Arun Ram University of Wisconsin Madison 480 Lincoln Drive Madlison7 WI 53706 ram mathwiscedu 1 Topology A topological space is a set X with a speci cation of the open subsets of X where it is required that a Q is open and X is open7 b Unions of open sets are open7 c Finite intersections of open sets are open In other words7 a topology on X is a set T of subsets of X such that a QleTandXET7 b If Ui E T the Ui 6 T7 c If U17 U27 Un is a nite collection of elements of T then Ul E T A topological space is a set X with a topology T on X Let T be a topology on X An open set is a set in T A closed set is a subset E of X such that the complement E0 of E is open Let X be a topological space and let x E X A neighborhood of z is an open subset U of X such that z E U Let X be a topological space A subspace of X is a subset Y of X with the topology given by making the open sets be the sets L 1V such that V is an open subset of X7 where L Y a X is the inclusion 2 Examples Let X be a set The discrete topology on X is the topology such that every subset of X is open A metric space is a set X with a function 1 X x X a R20 such that a If x E X then dzz O7 1 b If my 6 X and dzy 0 then x y c If x 342 E X then o lx7 2 o lx7 y dyz Let X be a metric space Let x E X and let 6 E Rgt0 The ball of radius 6 at z is the set BM p 6 X l M729 S 6 Let X be a metric space The metric space topology on X is the topology generated by the sets B z7 for z E X and 6 E Rgt0 3 Continuous functions Continuous functions are for comparing topological spaces Let X and Y be topological spaces A function f X a Y is continuous if it satis es the condition if V is an open subset of Y then f 1V is an open subset of X Let X and Y be topological spaces An isomorphism or homeomorphism is a continuous function f X a Y such that the inverse function f 1 Y a X exists and is continuous 4 Interiors and Closures Let E C X The interior of E is the subset E0 of E such that a E0 is open in X7 b If U is an open subset of E then U Q Equot Let E Q X The closure E of E is the subset E of X such that a E is closed7 b If V is a closed subset of X and V 2 E then V 2 E LetECX ThesetEis denseinXifEX 5 Interior points and limit points A limit point of E is a point p E EE0 Let E Q X An interior point of E is an element p E E such that there exists a neighborhood UofpwithUQE 6 Compact sets Let K Q X A sequence of points of K is a subset k17 k2 k3 of elements K indexed by the elements of Zgt0 A compact subset of X is a subset K of X such that any open cover of K has a nite subcover In other words7 if U0 is a collection of open subsets of X such that K Q UDLUD then there is a nite subset U17 7 Un of U0 such that K Q U1 U U U Let K Q X A sequence of points of K is a subset k1 k2 k3 of elements K indexed by the elements of Zgt0 Theorem 61 Let X be a topological space and let K Q X Then K is compact if and only if every sequence k17 k2 7 of points of K has a limit point in K Math 521 Lecture 5 Arun Ram University of Wisconsin Madison 480 Lincoln Drive Madison7 WI 53706 ram mathwiscedu 1 Operations An operation on a set S is a map S x S a S Let 0 S X S gt S 317 32 gt gt 31 o 32 be an operation on S The operation 0 is associative if it satis es the condition If 31 32 33 E S then 31 o 32 o 33 31 o 32 o 33 The operation 0 is commutative if it satis es the condition If 31 32 E S thensl o 32 32 o 31 Examples The operation Z x Z a Z M H Hi is both commutative and associative The operation Z x Z a Z Li H iii is noncommutative and nonassociative 2 Monoids groups rings and elds A monoid without identity is a set C with an operation GXGHG 23 H y such that a 7 is associative if Lj k E G then i7jk fjk7 A monoid is a set C with an operation T suchthat w H 7 a 7 is associative if Lj k E G then i7jk fjk7 b C has an identity There exists an element X E G such that if y E G then X7y y7 y An commutative monoid is a set C with an operation 3 such that a G is a monoid7 b ifij E Gthenij ji A group is a set C with an operation G X G T such that M H i7j a 7 is associative if Lj k E G then i7jk fjk7 b C has an identity There exists an element X E G such that if y E G then X7y y7 y t A c G has inverses if y E G there is an element gt1 6 G such that My1 y 7y I where X is he identity in G An abelian group is a set C with an operation 5 such that a G is a group7 b ifij E Gthenij ji The identity element of an abelian group is denoted O A ring without identity is a set R with two operations R X R gt R R X R gt R and m H J m H U such that a R with the operation is an abelian group7 b is commutative If Lj E R then i j j i c multiplication is associative if i j k E R then ijk7 d distributive laws if Lj k E R then ij k ij g and jk m jk A ring is a ring without identity R such that there is an element 1 E R such that if y E R then 1y yl y A commutative ring is a ring such that if 71 6 R then my gm A eld is a commutative ring IF such that if y 6 IF and y 7 0 then there is an element y 1 6 IF With yyil yily 1 A division ring is a ring D such that if y E D and y 7 0 then there is an element y 1 E D with yy l y 1 The integers Z with the addition operation is an abelian group The integers Z with the addition and multiplication operations is a ring The rationals Q with the operations addition and multiplication is a eld Blath 521 Lecture 7 Arun Ram University of Wisconsin Madison 480 Lincoln Drive Madison7 WI 53706 ram mathwiscedu 1 Fields of fractions Let A be a commutative ring A zero divisor is an element a E A such that there exists I E A such that b 7 0 and ab O A integral domain is a commutative ring A with no zero divisors except 0 Let A be an integral domain A eld of fractions of A is the set rlmbeAb 0 with a G g a if ad bc7 and operations given by a c ad bc a c ac t t a bd and t a 3 Theorem 11 Let A be an integral domain Let lF be the eld of fractions of A a The operations on lF are well de ned and lF is a eld 17 The map L AA a gt gt Mo is an injectiue ring homomorphism 0 UK is a eld with an injectiue ring homomorphism C A gt K then there is a unique ring homomorphism 4p IF gt K such that C 4p 0 L hiath 521 Lecture 10 Arun Rarn University of Wisconsin Madison 480 Lincoln Drive Madlison7 WI 53706 ram mathwiscedu 1 Cardinality How big is a set Let S and T be sets S and T have the same cardinality7 CardS CardT7 if there is a bijective map from S to T Notation Let S be a set Then Q u5amp CardS n if CardS Card127 7b 00 otherwise Note Even if CardS 00 and CardT 007 one may have that CardS 7 CardT A set S is nite if CardS 7 00 A set S is in nite if S is not nite A set S is countable if either S is nite or if CardS CardZgt0 A set S is uncountable if S is not countable Let X be a topological space A perfect set is a subset E of X such that a E is closed7 b if z E E the z is a limit point of E The Cantor set is k Czequlm 4brkezgt0muiezgtowmialt3 Theorem 11 CardZgt0 CardZZO CardZ CardQ 7 CardR CardC HW Show that subsets of countable sets are countable HW Show that countable unions of countable sets are countable HW Show that if S is countable and n E Zgt0 then S is countable HW Show that 2Zgt0 is uncountable HW Show that 071 is uncountable HW Show that R is uncountable HW Show that the Cantor set is uncountable HW Show that every perfect subset of Rk is uncountable HW Show that the Cantor set is perfect Math 521 Lecture 18 Arun Ram University of Wisconsin Madison 480 Lincoln Drive Madison7 WI 53706 ram mathwiscedu 1 Interiors and closures Let X be a topological space and let x E X A neighborhood of z is a subset N of X such that there exists an open subset U of X with z E U and U Q N Let X be a topological space and let E C X A neighborhood of E is a subset N of X such that there exists an open subset U of X with E Q U Q N Let X be a topological space and let E C X The interior of E is the subset E0 of E such that a E0 is open in X7 b If U is an open subset of E then U Q Equot Let X be a topological space and let E Q X The closure E of E is the subset E of X such that a E is closed7 b If V is a closed subset of X and V 2 E then V 2 E Let X be a topological space Let E Q X An interior point of E is a point z E X such that there exists a neighborhood Nm of z with Nm Q E Let X be a topological space Let E Q X A close point to E is a point z E X such that If Nm is a neighborhood of of x then Nm contains a point of E Theorem 11 Let X be a topological space Let E Q X a The interior ofE is the set of interior points of E b The closure ofE is the set of close points of E 2 Hausdorff spaces A Hausdorff space is a topological space X such that if m y E Y and z 7 y then there exist a neighborhood Nm of z and a neighborhood Ny of y such that Nm U N 0 Theorem 21 Let X be a topological space Show that the following are equivalent a Any two distinct points ofX haue disjoint neighborhoods b The intersection of the closed neighborhoods of any point of X consist of that point alone c The diagonal of the product space X X X is a closed set d For euery set I the diagonal of the product space Y X1 is closed in Y e No lter on X has more than one limit point If a lter 7 on X conuerges to m then m is the only cluster point of m 3 Limit points and cluster points Theorem 31 Let X be a topological space and let 17 2 be a sequence in X Then a y is a limit point of 1 2 if and only if if Ny is a neighborhood ofy then there exists no 6 Zgt0 such that M E Nm for all n E 220 n 2 no b y is a cluster point of 1 2 if and only if if Ny is a neighborhood ofy and n0 6 Zgt0 then there exists n E Zgt0 with n 2 no such that an E Ny 4 Compact sets Let X be a set A lter f on X is convergent if it has a limit point Theorem 41 Let X be a topological space The following are equiualent a Euery lter on X has at least one cluster point b Euery ultra lter on X is conuergent c Euery family of closed subsets ofX whose intersection is empty contains a nite subfamily whose intersection is empty d Euery open couer ofX contains a nite subcouer Proof b a a Let f be a lter and let 7 be an ultra lter containing f Let x be a limit point of T Then z is a limit point of f a a b Let T be an ultra lter Let x be a cluster point of T Then z is a limit point of T So T is convergent a a c Let C be a closed family with empty intersection If every nite subfamily has empty intersection then C generates a lter f Let x be a cluster point of f Then z E C for every set C in C This is a contradiction So there exists a nite subfamily that does not have empty intersection c a a If there exists a lter f without a cluster point then C F l f E f is a family of closed sets contradicting c ltgt d by taking complements Theorem 42 Let X be a metric space and let E be a subset of X The set E is compact if and only if euery in nite subset ofE has a limit point in E Proof lt Let K be a compact set and let E be a in nite subset of K If there is no limit point of E in K then7 for each p E K there is a neighborhood Np which contains no other element of E Then the open cover N Np l P E KL of K has no nite subcover Let S be a in nite subset of E The metric space E has a countable base So every open cover of E has a countable subcover C 01 Cg lfC does not have a nite subcover then7 for each n7 03mg 7 0 but n Ci 0 Let S be a set which contains a point from each Ci Then S has a limit point But this is a contradiction 1 Theorem 43 Let X be a Hausdor topological space and let K be a compact subset of X Then K is closed Proof Let x E K The neighborhood lter 3a of z induces a lter BK on K which has a cluster point y E K Since 3a is coarser than BK considered as a lter base on X the point y is a cluster point of So y z since X is Hausdorff The proof in baby Rudin We will show that K0 is open Let p 6 KC Let N be the open cover of K given by N Nq l q 6K7 where Nq Bdpqq l q 6K Let Nq17 WNW be a nite subcover of K Then M Mq1 qu where Mg 13d17qp7 is an open set such that p E M Q Kc So p is an interior point of K0 So K is open 1 Theorem 44 Let X be a metric space and let E be a compact subset of X Then E is closed and bounded Proof Since a metric space is Hausdorff7 E is closed If E is not bounded then there is an in nite sequence in E that does not have a limit point D Theorem 45 a A k cell is compact b Let E be a subset of Rk IfE is closed and bounded then E is compact Proof If E is closed and bounded then E is a closed subset of a k cell Since closed subsets of compact sets are compact E is compact El
Are you sure you want to buy this material for
You're already Subscribed!
Looks like you've already subscribed to StudySoup, you won't need to purchase another subscription to get this material. To access this material simply click 'View Full Document'