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# Topics in Analysis MATH 300

Yale

GPA 3.81

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Chapter 4 Structure of Point Sets 41 Open and Closed Sets For each p E R and E gt 0 Denote by N5p x E R lx ipl lt 6 C R the e neighborhood of p in R De nition 11 Let A C R A point p E A is called an interior point of A in R if there exists an E gt 0 such that N5p C A Denote by IntA the set of interior points of A Example 6 H l 1 Let A 01 C R Then Int01 01 Becauce for each p E 01 if E mmp 1 7 10 then N5p C 01 On the other hand for all 6 gt 0 N51 17 61 6 C 01 Hence 1 Int 0 2 Let A 01 Q Then IntA 0 Because for each p E A and E gt 0 there exists q E Q such that lp 7 ql lt 6 le q E N5p but q A Therefore N5p C A De nition 12 1 A subset A C R is open in R if every point in A is an interior point ofA le A mtA 2 A subset B C R is closed in R if R BC is open in R Example 7 1 R is open in itself Pick any p E R and E gt 0 N5p C R Since R is open the empty set 0 is closed in R However 0 is also open since Int Q which then shows that R is closed Thus 0 and R are subsets of R that are both open and closed in R 2 N5p is open 3 01 is closed in R 34 4 If S is any nte set in R then S is closed 5 S Q is neither closed nor open in R Proposition 16 Let A C R tntA is the largest open set contained in A Proof To show tntA is open Pick any p E tntA Let E gt 0 such that N5p C A Now for any q E N4107 there exists 6 gt 0 such that Now C NAP C A This is possible since N5p is open Therefore7 N5p C z ntA7 which shows tntA is open To show tntA is the largest Now if 0 C A7 then tntO C tntA It further that O is open7 then 0 tntO C tntA D Theorem 5 1 Let 0030164 be any t39 t L or t L of open subset sets of R Then UDLEAOD is open in R 2 Let Oi1 be any nite collection of open sets Then 10i is open in R Proof 1 For x E UWEAOW7 there exists 04 E A such that z 6 00 0a being open implies there exists 6 gt 0 such that New c 00 c U 00 046A Hence UDLEAOD is open E0 z E LIOi irnplies z 6 O for all 239 17 n For each 239 let 6 gt 0 such that N52 C Oi This is possible since Oi is open Now let 6 rnin61en Then NEW C Nellx C 01 W Hence New C 1th which shows that LIOi is open D Remark 11 Note that the intersection of an in nite collection of open sets may not be open For exarnple7 consider 0 71 7 123971 12 C R7 which is open for all 239 but Oi 7171l7 D8 1 which is closed in R 35 Theorem 6 1 Let Faa6A be any quot t 739 or t L of closed subset sets of R Then aeAFa is closed in R 2 Let FlEL1 be any nite collection of closed sets Then ULIF is closed in R Proof Note that FD U F5 and U FD F5 046A 046A 046A 046A Hence these results follow the results from the previous theorem D De nition 13 Let A C R A point p E R is a limit point of A if for every 6 gt 0 there exist q E A q 31 p such that q E N5p In other words N410 W A 7Q Q A point p E A is called an isolated point of A if it is not a limit point of A Example 8 Let A 01 U Then the limit points of A is the set 01 Note that 2 is an isolated point of A since for E 12 N42 W A Q Limit points can be characterized in terms of sequences Theorem 7 Let A C R z E R is a limit point ofA if and only if there erists a sequence zn C A xn 31 z for all n such that M a x Proof Excercise D Theorem 8 A C R is closed if and only ifA contains all its limit points Proof Suppose A is closed Pick any x 6 Ac which is open There exists 6 gt 0 such that New C Ac This implies mm A 0 a Mac m m A 0 which shows that z is not a limit point of A Thus A must contain all its limit points Suppose A contains all its limit points Then by the previous arguments any x E A0 is not a limit point Thus there exists 6 gt 0 such that 1Ve9696 A But z 6 A0 S0 NEW A Q which shows NEW C Ac Therefore Ac is open D De nition 14 Suppose A C R Denote by A the set of limit points of A Then the closure of A denoted by clA is de ned as clA A U A Theorem 9 Let A C R clA is the smallest closed set containing A Proof Homework D 36 42 Relatively Open and Closed Sets De nition 15 Let X be a subset of R 1 A subset O C X is said to be open in X or relatively open in X if for each x E 0 there exists 6 6x gt 0 such that N5z X C O 2 A subset F C X is said to be closed in X or relatively closed in X if X F is open in X Example 9 1 121 is open in 01 2 01 Q is open in Q 3 01 Q is open in RQ Theorem 10 Let X C R Z O C X is open in X if and only lfO X A for some A that is open in R 2 F C X is closed in X if and only lfF X B for some B that is closed in R Proof Excercise D De nition 16 1 A subset set nonempty C C R is separated if C can be written as a disjoint union of two nonempty relatively open sets in C In other words there exist 0 31 0 C C and Q 31 U C C such that O and U are relatively open in C and O U andOOUU 2 A nonempty C C R is connected if it is not separated Example 10 H Let C 0 1 U Then G is saparated for 01 and 2 are disjoint and open in C D Let C 01 U 12 U 23 Then G is saparated by 01 and 12 U 23 which are disjont and both open in C 3 01 is connected Theorem 11 Every interval I in R is connected 37 Proof Suppose I is separated Let 0 and U be disjoint nonempty subsets of I that are relatively open in I and I O U U LethOandyEU WLOG7WMAxlty Let Bz Ozlty Clearly z E B and B is bounded We have b supB E I since I is an interval ie z E I forallzltzltyand b yl I being a disjoint union of O and U irnplies either I E O or b E U If I supB 6 07 then b lt y This implies for all 0 lt E lt y 7 b7 b 62 0 and hence Neb 1 0 which contradicts the fact that O is relatively open in I One obtains the same contradiction if b supB E U Hence I is connected D Remark 12 It is an excercise to show that every nonernpty connected sets in R is an interval 43 Compact Sets De nition 17 Let K C R A collection of open sets in R7 OJWEA7 is an open cover for K if K c U 0a 046A Example 11 Suppose K C R The for any 6 gt 07 the collection N5z z E K is an open cover for K De nition 18 A subset K C R is compact if every open cover for K has a nite open subcover In other words7 if OaheA is an open cover for K7 then there exist 0417 7 an in A such that n K c U 0 i1 To show a subset K is not cornpact7 we only need to nd an open cover with no nite open subcover Example 12 1 Let K 0717 which is open and bounded Clearly7 On 0717171 n 273 is an open cover for 071 Moreover7 if n S m then On C Om There exists no nite open subcover and hence 071 is not cornpact Suppose there exists a nite open subcover7 say m K c U 0M k1 38 Let N rnaxnL7 7nm7 then OM C ON for all k 17m This shows that KC UOM ON0171N k1 which is absurd 2 Suppose now K 000 which is closed and unbounded Clearly7 On fl7171 n 237 is an open cover for K and there is no nite open subcover 3 Any nite set is compact Theorem 12 IfK C R is compact then K is closed and bounded in R Proof Fix 6 gt 0 Clearly7 N5p z E K is an open cover for K Since K is cornpact7 there exists 1 may 6 K such that i1 Each New is bounded and any nite union of bounded sets is bounded7 and hence K is bounded To show K is closed le Kc R K is open Fix z 6 KC We need to show there exists 6 gt 0 st New C Kc Now for each y 6 K7 let 6y lx iylQ Observation 1 Newt MM 0 ii K C UyEKNEyy Since K is cornpact7 there exist yl my 6 K such that K c U Newly U i1 Let 0 LINEW Then by observation U 0 0 which implies O C Kc sinse K C U Moreover7 O is open and contains x Hence there exists 6 gt 0 such that New C O C Kc 1 Remark 13 As a consequence of the previous theorern7 the set A n E N is not compact for it is not closed A does not contain its limit point 0 Theorem 13 Any closed subset of a compact set is compact 39 Proof Suppose K C R is compact and A C K is closed Let 003054 be any open cover for A Then Ac U 003054 is clearly an open cover for K K being compact implies there exists 041 an E A such that KCACUOM UmLJODW Thus AA KA AcUOmUUOUA Om CUOD i1 i1 which shows that A is compact 1 Theorem 14 Heme Bord IfK is closed and bounded in R then K is compact Proof By Theorem 13 it suf ces to show that PM M is compact By contradiction suppose K is not compact Then there exists an open cover 003054 that has no nite subcover Let a0 7M and b0 M Consider the following recursion Suppose ambn C a0b0 for some n 2 0 such that ambnl cannot be covered by any nite subcover le lf 041oz is any nite subset of A n i1 For short notation we say ambn has property Clearly a0b0 has this property Let Tn 1 b2 Next we divide ambnl into amml U rmbn Either amml or rmbn must have property P and so we let that interval be an1bn1l And continue In the end one obtains a sequence of intervals ambnl satisfying property Note that an1bn1 C ambnl C a0b0 bn1 7 an 7 b0 7 102n The sequences on C a0b0 and on C a0b0 are monotone and Cauchy and by the previous equation lim an z lim b Hoe Hoe for some z E R Since a0b0 is closed we have x E a0b0 Since 003054 is an open cover for a0b0 there exists oz 6 A such that z E 00 Since 00 is open there exists 6 gt 0 such that New C 00 Now pick 71 large enough such that be 7 a02 lt 62 We have ambnl C 00 This contradicts the fact that ambnl cannot be covered by any nite collection of Oa7s 1 40 In summary7 compact subsets of R can be Characterized as follows Theorem 15 Let K C R The followings are equivalent 1 K is compact 2 K is closed and bounded 3 Every in nite subset of K has a limit point in K Proof Excercise 1 Chapter 3 Real Sequences 31 Boundedness and Convergence A sequence is an in nitely collection of objects 117 127 137 indexed by 17 27 37 We denote it by anff1 or simply an De nition 5 A real sequence on is said to be bounded if there exists a positive K E R such that lanl S K for all n A real sequence is said to be bounded above if there exists a real number M such that an S M for all n It is said to be bounded below if there exists a real number L such that L S on for all n Example 1 The sequence on 71 717 17 71 17 is alternating between 1 and 71 The sequence an 171 is getting closer to 0 as n getting bigger and bigger Both of these sequences are bounded by 1 De nition 6 The real sequence on is said to be a convergent sequence if there is a E R such that for each 6 gt 07 there exists an N 6 N7 which depends on 67 such that lan 7 al lt E for all n 2 N a is said to be the limit of an De nition 7 A real sequence on is said to be increasing nondecreasing if 11 3 a2 3 S on S 7 for all n A real sequence on is said to be decreasing nonincreasing if 7 7 for all n A real sequence on is said to be monotone if it is either increasing or decreasing Proposition 7 Every bounded and monotone sequence in R is convergent Proof Suppose an is decreasing Since an is bounded7 the greatest lower bound 1 infnan gt foo For each 6 gt 07 we have a E gt 1 Thus a E is not a lower bound for 19 20 an Therefore7 there exists N Ne such that aN lt a 6 Since an is decreasing7 we havea6 gtaN 2 an 2 agta7e7 for alln 2 N This implies lan7al lt efor alln 2 N Thus an 7 a as n 7 00 If an is increasing7 then the proof is the same by using least upper bound D Recall a few usefull properties of R For any 171 E R we have 1 la bl S lal lbl By the de nition of 1 17 a S lal for all a E R Now if ab 2 07 then labl 05 S 101 151 On the otherhand if a b lt 0 then lablababSlial ibl al bl 2 la 719 2 Hal 7 M We have lallbablSlbllaibl laiblilalilbl By interchange a and b7 we have laibl ll 01 2 151 lal lal 151 Thus7 la 512 max allblr al 151Halilbll Properties of Convergent Sequences Proposition 8 Every convergent sequence is bounded Proof Suppose an C R and an 7 a as n 7 00 Pick any 6 17 then there exists N E N such that loan 7 al lt 1 for all n 2 N This implies Hanl 7 laH 3 loan 7al lt17Vn 2 N Hence7 lanl S lal 1 for all n 2 N Now let A rnax all7 laN1l lal 1 Then lanl S A for all n 2 1 This shows that an is bounded D Proposition 9 Suppose an and bn are sequences in R such that lirnyH00 an a and lirnyH00 bn h Then the followings hold 239 lirnyH00 can ca7 Vc E R it lirnnnoomn bn a b 21 ill limlH00 anbn ab iv Iffur39lher that bn 31 0 for all n and b 31 0 then 1 an 1 lm i woo bn 1 Proof 0 WMA c 31 0 Fix 6 gt 0 an 7 1 implies there exists N E N such that lan 7 al lt elcl for all n 2 N This implies loan 7 col lcllan 7 al lt lclelcl 67 V71 2 N Hence7 can 7 ca 0 Fix 6 gt 0 Let N1 and N2 be positive integers such that lan 7 al lt 62 for all n 2 N1 and lbn 7 bl lt 62 for all n 2 N2 Let N maxN17N27 then lanbn 7 abl S lan7al lbn7bl lt6262 67 for alan N Hence an bn 7 a b 0 Sketch lanbn 7 abl lanbn 7 anb anb 7 abl S lanl lbn 7 bl lbllan 7 bl S Albn 7 bl lbllan 7 al7 where A supn21an 0 Sketch l1bn 7 1bl lb 7 bnbnbl Since b 31 07 pick 71 large enough so that lbn 7bl S lbl27 which implies lbnl 2 lbl2 Therefore7 lb7bnbnbl S lb7bnl2lbl2 for large n 1 De nition 8 We say the sequence can C R diverges to 00 and write limlH00 an 00 if for each M E R7 there exists an N Ne such that an 2 M for all n 2 N We say the sequence can diverges to 700 and write limlH00 an 700 if for each M E R7 there exists an N such that an lt M for all n 2 N Example 2 Let r gt 1 The sequence Tn diverges to 00 and 77 diverges to 700 On the other hand7 7r diverges neither to 00 nor 700 Proposition 10 If an gt bn for all n and limlH00 bn 00 then limlH00 an 00 Proof Excercise D 22 Example 3 Let an xnn 1 Find a and Ne for each 6 gt 0 such that lanial lt 67 V71 2 NE Let a 1 Fix 6 gt 0 We have i nn171 1 1 xnn171 7 W n1lt lte if n gt 16 Thus7 one can pick Ne l1617 where denotes the smallest integer 2 x Example 4 Let an 1 Prove directly that an a 00 Denote by the greatest integer n gyg f 1EDMHZ2AL n n 1 i E 4W2 2 8 724472 whenever n 2 rnax8716M 23 32 TwoState Markov Chains Let us consider a problem of checking the phone status free or busy at each interval period of time Denote by pn the probability that the phone is free on the 71 check and by 1 the probability that the phone is busy on the 71 check 0 Let p0 and go 1 7 p0 be the probabilities that the phone is free and busy initially respectively 0 Suppose we are at the 71 step for some n 2 0 and the phone is free We assume that with the probability 0 S p S 1 the phone will remain free at the n 1 step Since the phone is either busy or free 1 7 p is the probability that the phone will be busy 0 Suppose next that we are at the 71 step for some n 2 0 and the phone is busy We assume that with the probability 0 S q S 1 the phone will remain busy which shows that 1 7 q is the probability that the phone will be free These assumptions provide a recursive relation between the status at the n 1 and 71 steps In other words pm 7 ppm 17 m and Qn1 17 pm qqn 31 Note that 1 10 1 for all n 2 0 In fact we have p0 qo 1 and suppose that pk qk 1 for some k 2 0 then pm gm ppk 17 QM 17ppk 9 7 pk 9k 1 Hence by induction 1 10 1 for all n 2 0 Thus pm ppn17Q17pn 17Qp71Qpn Let a 1 7 q and b p q 7 1 Apply recursively we have 10 L a bpn a ba bpn1 a ab bzpn1 a Zbk bn po k0 1 7 bn1 7 W1 a 17 b J 100 This shows that the sequence pn converges if and only if lbl lt 1 Similarly Qn1 17p17 1n qqn 17Pq 7 1 18 Let c17p and b pq71 Then Qn1 C MC bqyki C Eb bZQn71 C bk DWIQO k0 17b 1 C7 17 b l bn qm 24 which shows that qn converges if and only if lbl lt 1 Suppose we make a futher assumption that lbl lp q 7 ll lt 17 then lim lima1ibnb iiiii weep woo 14 po 14 17p1iq p7 and 17b n i c i 17p 77 JLHSO A W JEEoGliiblbqo 1ib 1ip1iqiq Note that the limits of pn and qn do not depend on the inital state p0 and go Write 31 in matrix form7 we have pm p 1 i q n Qn1 1 i p q an Let A 1 5p 1 21 and 1 for n 2 0 Recursively7 we have 11 A1 32 17 lim 11 lim Ayn A lim 1 AT7 V LHOO V LHOO V LHOO which shows that 17 is invariant under A or a xed point if A In otherwords7 if p0 16 and go 1 then pl 16 and qn j for all n The probabilistic system 32 is called a two state Markov Chain becauce time is discrete and at each stage7 the system has only two possible states free and busy The domain for this Markov Chain is where the system converges7 which is precisely Dp7Q6R20p7q17lpq71llt1 Mathematics 300b Midterm 1 22107 Name Please answer FOUR of the FIVE questions each question is worth TEN points Books notes calculators computers cellphones may NOT be used 1 2 3 4 5 Total 1 Prove that if an 7 a as n 7 00 and an lt b for all n E N then a S b ls it necessarily true that a lt b Give a proof or counterexample Suppose a gt b Since an 7 b as n 7 00 for any number 8 gt 0 there is a number N E N such that lan 7al 3 8 for all n 2 N Set 8 a7b gt 0 Then a 7 IN 3 law 7 al 3 a 7 b so IN 2 b contradicting the hypothesis that an lt b for all n E N Therefore a S b It isn7t necessarily true that a lt b For example let an 7 then an lt 0 for all nENandan70asn7oo 2 Let an be a sequence of real numbers Prove that if there is a number 7 lt 1 such A 00 that 6anL 7 anl S r for all n 6 N7 then an is a Cauchy sequence First note that r 2 0 since 0 S lag 7a1l S 7 Suppose mm 6 N with m gt n Then am 7 an am 7 am1 an1 7 an7 so by the generalized triangle inequality lamianl S lamiam71lquot lan1ianl S Tm 1r 1r Tn7rm17r Since 0 S r lt1rm 7 r 1 7 r S rn17 r so lam 7 anl S rn17 r for all mm ENwithm 271 Since 7quot 7 0 as n 7 007 for each 8 gt 0 there is a number N E N such that r S 8177 for all n 2 N lfm 2 n 2 Nthen lam7anl S rn17r S 8 Similarly lam 7 anl S 8 if n 2 m 2 N It follows that an is a Cauchy sequence Let an be a bounded7 rnonotone increasing sequence of real numbers Prove that an7supS as 717007 where Sann N Since S is nonernpty and bounded above7 it has a least upper bound b sup S Since b is an upper bound for S an S b for all n E N For any number 8 gt 07 b 7 8 is not an upper bound of S so there exists N E N such that b 7 8 lt 1N Since an is monotone increasing7 b 7 8 lt IN 3 can for all n 2 N Thereforeb76 ltan gbfor alln 2 N7 so lan7bl S efor alln 2 N It follows that an 7 b supS as n 7 oo 4 Suppose that f 1b 7 R is not bounded on 1b Prove that there is a point c in A U V 1 b such that for each real number 5 gt 0 f is unbounded on x E 1 b 17d 3 5 Hint use the Bolzano Weierstrass theorern Since f is not bounded on 1 b for each number n E N there is a point 1 E 1 b such that 2 71 By the Bolzano Weierstrass theorem the sequence of points in 1b has a subsequence that converges to a point c E 1b For any number 5 gt 0 there is a number K E N such that 1 7 c S 5 for all k 2 K Since znk E x E 1b x 7c S 5 for all k 2 K and 2 71k 2 k f is not bounded on x E 1b x 7 c S 5 Note this is the rst part of the proof that every continuous function on 1b is bounded It follows from that theorem that the function f in the question can7t be continuous everywhere on 1b Suppose that f 0 oo 7 R is uniformly continuous on 0 oo Prove that if 1 gt 0 and 1 7 0 as n 7 00 then is a Cauchy sequence Since f is uniformly continuous on 0 00 for each number 8 gt 0 there is a number 5 gt 0 such that 7 S 8 for all Ly 6 000 such that x 7y S 5 Since 1 7 0 as n 7 00 there is a number N E N such that 1 7 0 S 52 for all n 2 N If 77171 2 N then S 52 and S 52 so by the triangle inequality 1m 7 1n 3 S 5 It follows that 7 S 8 proving that the sequence is Cauchy Chapter 5 Continuity and the Intermediate Value Theorem Motivation So far we have studied basic properties and different types of subsets in R that will enable us to study continuous functions on R Given D C R nonempty and a function f D a R Suppose fa 31 fb for some a7b 6 D7 1 lt b We ask 1 ls it possible to nd a lt c lt b such that fc is between fa and fb7 2 What are the necessary and suf cient requirements on f for this condition to hold 3 If f is bounded on D7 What is the constraint on D so that the maximum and minimum of f are attained In this chapter7 we will study the necessary and suf cient conditions on f and D for these properties to hold 51 Continuity and Uniform Continuity De nition 19 Continuity Let D be a subset of R7 and f D a R is a real valued function de ned on D Sometimes7 we refer to D7 denoted by D0mf7 as the domain of f We say f is continuous at z E D if for each 6 gt 0 there exists 6 6Le gt 0 such that lfz 7 lt 6 whenever y E D and 1x iyl lt 6 In other words7 f is continuous at z E D if for each 6 gt 0 there exists 6 6Le gt 0 such that y 6 N5fx7 whenever y E D N5z We say f is continuous if it is continuous at all z E D 42 Example 13 Let D R and f zz Fix an x E R and E gt 0 lffyl1952 921lyyllyllyl lt12 ll iyllt67 whenever7 lx 7 pl lt 6 rnine7 62z The following result provides another characterization of continuity which can be easier to check Theorem 16 Continuity and Sequences Suppose f D 7 R The followings are equiva lent 1 f is continuous at z E D 2 Ifn C D and xn 7 z E D then fan 7 Proof Suppose f is continuous at x Fix 6 gt 0 Let 6 6W 6 gt 0 such that lfz 7 lt 6 whenever y E D 7yl lt 6 xn 7 z implies there exists N N6 gt 0 such that 1 7 xl lt 6 whenever n 2 N This irnplies7 7 lt 67 for all n 2 N Therefore7 x 7 Now suppose that every sequence C D that converges to z irnplies fzn converges to x and suppose that f is not continuous at x This implies there exists 6 gt 0 such that for all on gt 07 there exists xn such that 1 7 xl lt 6 and lff9 l 26 Let 6 7 0 Then xn 7 x but fzn 74gt x which is a contradiction Hence f is continuous at x Algebra of Continuous functions Theorem 17 Suppose f and g are continuous functions de ned on D C R and a E R Then 1 of is continuous on D 2 f g is continuous on D 3 fg is continuous on D 4 g is continuous at all z E D such that g 31 0 Proof Use Theorem 16 and the properties of convergent sequences from Proposition 9 D 43 Proposition 17 Suppose D C R Let g D 7 R and f gD 7 R where gD C R is the image of g Ifg and f are continuous then the composition function de ned by f 09 D 7 R by f 09W 7 mm 96 6 D7 is also continuous Proof Fix z E D and suppose C D such that xn 7 x Let yn 9xn and y 9z7 then yn 7 y and hence y 7 Fix 6 gt 07 f 0 mm 7 f o 9 7 WWW 7 f996 7 Wyn 7 MIN lt 67 for large n Hence7 f o 7 f o gp7 which shows that f o g is continuous at x Since z is arbitrary7 f o g is continuous on D 1 Topological Properties of Continuous Functions Suppose f D 7 R For each B C R7 de ne f71B x e D fx 6 B c D Proposition 18 Continuity and Open Sets f D 7 R is continuous if and only if f 1B is open in D for every B open in R Proof Suppose f is continuous abd B C R is open in R Pick any p E f 1B Let q fp E B Since B is open in R7 there exists an E gt 0 such that N5q C B Since f is continuous at p7 there exists a 617 gt 0 such that fz 7 q lt 6 whenever z E D and x 7p lt 67 In other wordls7 if z E N5pp D then u E N5q C B This implies N5pp D C f 1B Since ng p D is relatively open in D7 the union N5p10 D Pef WB is also open in D7 which shows that f 1B is open because PUB U mum 176F103 Suppose f 1B is open in D for every B open in R Pick any p 6 D7 we need to show that f is continuous at p Fix 6 gt 0 and let B N5fp which is open in R By assurnption f 1B is open in D and contains p This implies there exists 6 gt 0 such that N5p D C f 1B Moreover7 for each q E N5p D C f 1B7 we have fq E B N5fp In other wordls7 fq 7 fp lt 6 whenever q E D and p 7 q lt 6 So f is continuous at p D 44 Proposition 19 Continuity and Closed Sets f D a R is continuous if and only if f 1B is closed in D for every B closed in R Proof Note that if f D a R is any function and B C R then f 1Bc f 1B0 lndeed pefRFie meBWa m Bem fd rep UABD9 Now the proof follows from the previous Proposition D Suppose f D a R is continuous Then for any S C D the restriction map flg S a R by fl5x fp for all z E S is de ned and continuous on S De nition 20 A real value function f D a R is said to be bounded on S C D if S M Vs E S In other word the image fl5S fS x E S is a bounded set in R Theorem 18 Suppose D C R and f D a R is continuous Then fS is compact closed and bounded in R whenever S C D is compact In other words for continuous functions the image of a compact set is compact Proof Let 003064 be an open cover for fS Since f is continuous f 10a is open for each 04 and clearly f 10aa6A is an open cover for S Since S is compact there exists 041 oz 6 A such that SCUf ao This implies fS C f f10agt Uff 10a C U004 Hence fS is compact D Remark 14 Suppose S is compact and f is continuous on S If y E fS is any a limit point of fS then y E fS since fS is closed Thus there exists z E S such that y lf fS is bounded above below then sup fS inffS is a limit point of fS lf further that fS is closed then there exists b E S a E S such that b supS a inf S If S is compact then fS is compact closed and bounded However the converse is not true For each f D a R that is bounded De ne spfsupfp xED and igfffp xED 45 Proposition 20 Suppose f D 7 00 is continuous and D is compact Then the supremun and the in mum off are attained In other words there exist ab 6 D such that fa supf and fb inff D D De nition 21 Uniform Continuity Let f D 7 R for some nonempty D C R We say f is uniformly continuous on D if for each 6 gt 0 there exists 6 66 gt 0 such that ifx 7 lt 67 whenever xy 6 D7 and ix 7yi lt 6 Example 14 The function fx x is uniformly continuous on R because for each 6 gt 07 ifx 7 ix 7 yi lt 67 whenever ix 7yi lt 6 6 Thus our choice of 6 here does not depend on x It only depends on 6 Next consider the function f 071 7 R by fx 1x We would like to show that f is not uniformly continuous By contradiction7 suppose that it is le Fix an E gt 0 there exists 6 66 gt 0 such that ifx 7 lt 6 whenever ix 7yi lt 6 Let 0 lt x a lt 6 and y a27 then clearly ix 7 yi lt 6 However7 1 a for small enough a When D is compact closed and bounded in R7 being continuous on D implies being uniformly continuous Theorem 19 Suppose D C R is compact and f D 7 R f is continuous on D if and only iff is uniformly continuous on D Proof The reverse direction is clearly true Suppose f is continuous on D7 which is compact in R Fix 6 gt 0 For each x 6 D7 there exists 61 gt 0 such that fy E NE2fx whenever y E N5wx D 51 le ify 7 lt 62 whenever y E D and iy 7 xi lt 6w Cle arly7 D c U Nam2a m D xeD 46 Since D is cornpact7 there exist 1 may 6 D such that i1 Let 6 rnin6w1277622 Now for any x 6 D7 we have x E Nth2a D for some i7 and N5z C N5 Hence whenever y E N5z D C N5 D7 we have by 51 lfW fyl S lfW f39l lf39 fyl lt 6 Thus f is uniformly continuous D Theorem 20 Intermediate Value Theorem Suppose D ab and f 17 7 R is continuous such that fa 31 fb Then for any y between fa and fb either fa lt y lt fb or fb lt y lt fa there epists c E a7b such that fc y Proof WLOG7 WMA fa lty lt fb Let A x E a7b f lty C a7b Clearly A is bounded and nonempty for a E A Thus c supA E R and since a7 b is closed7 c E a7b Moreover7 we have a lt c lt b lndeed7 fb 7 y gt 0 so for 6 fb 7 y27 there exists 6 gt 0 such that lfb 7 lt 6 whenever lx 7 bl lt 6 In other words7 f gt y 6 for all lx 7 bl lt 6 Hence7 c lt b A similar argument shows that a lt c To show fc y It is clear that fc S y On the other hand7 f is continuous at c Fix 6 gt 07 there exists 6 gt 0 such that lfc 7 lt 6 whenever lx 7 cl lt 6 Note we can pick 6 gt 0 small enough such that N5c 6 a7 b We have c 62 is not a least upper bound for A This implies fc S y S fc 62 Thus la 7 f0l S lfc 62 f0l lt 6 which shows that fc y D The above Theorern is the special case of the following result Theorem 21 Suppose D is connected in Rie D is an interval in R and f D 7 R is continuous Then fD is also connected in R Proof Suppose fD is separated le there exist nonernpty disjoint sets 0 and U that are open in fD and fD O U U Then clearly f 1O and f 1U are nonernpty disjoint sets that are open in D and D f 1O Uf 1U This shows that D is disconnected7 which is a contradiction 1

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