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## Intro to Real Analysis 2

by: Webster Green Sr.

35

0

7

# Intro to Real Analysis 2 MATH 5152U

Webster Green Sr.

GPA 3.86

Staff

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COURSE
PROF.
Staff
TYPE
Class Notes
PAGES
7
WORDS
KARMA
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## Popular in Mathematics (M)

This 7 page Class Notes was uploaded by Webster Green Sr. on Sunday October 11, 2015. The Class Notes belongs to MATH 5152U at Columbus State University taught by Staff in Fall. Since its upload, it has received 35 views. For similar materials see /class/221196/math-5152u-columbus-state-university in Mathematics (M) at Columbus State University.

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Date Created: 10/11/15
Section 44 1 Prove that f a b 7 R is right continuous at p 6 ab if and only if fp exists and equals fp Similarly f is left continuous at p if and only fp exists and equals fp Proof Suppose that fp exists and equals fp then hm ffp 742 So for any 6 gt 0 there exists a 6 gt 0 such that for all z E a b with p lt z lt p 6 we have W96 7 fPl lt 8 Certainly if z p 7 fpl 0 lt 8 Then by de nition of right continuity f is right continuous at p Conversely if f is right continuous at p by de nition of right continuity limJr x fp The left 171p continuous case can be proved in a similar way D 2 For each of the following functions 1 de ned on R0 nd limJr x and lim x 1710 1710 provided the limits exist c M 7 17 m2 Sincewhen 71ltzlt0 0lt 17z2lt 1 h 17352 hm 00 170 1710 On the other hand when 0 lt z lt 1 0 lt17 2 lt1 hm 17352 hm 0 0 170 710 d M 7 m2 71 Sincewhen71ltzlt0 71ltz271lt0 hm 271 hm 7171 170 170 On the other hand when 0 lt z lt1 71 lt z2 71lt O hm 271 hm 7171 710 710 1 1 1 1 Since lim 7 7oo lim 7 700 Similarly lim 7 00 because lim 7 00 710 m 710 m 170 m 710 m Notethatifngiltn17forn N7then 1 lt lt m n1 and 1 1 1 7 lt7 1 n1ltngtltzinn Asma07naooSincelim n 17wehave n1 1 limz7 1 140 35 Now if 771 1 g i lt in for n e N then 1 1 77ltlt7 n n1 and 1 1 1 71 1777 1lt 7 777 17 n1ltltngtgtzLjlt ltltngtgt n As z HO n a foo Since n1 lim 1 we have 3 For each of functions f in Exercise 27 determine whether 1 has a removable discontinuity a jump discontinuity or a discontinuity of second kind7 at z 0 If f has a removable discontinuity at O7 specify how f0 should be de ned so that f is continuous at O o m i Since h 17352 lim 00 140 1410 lim 17352 lim 0 0 haw haw and 0 17 021 f has a removable discontinuity at O Rede ne f0 0 Then 1 is continuous at O Since hm 271 hm 7171 x70 x70 hm 271 hm 7171 x70 x70 and ND 1 f is continuous at O e m7Bl i i 1 i 1 i 1 i i i 1 Since lim 7 700 lim 7 700 and lim 7 007 f has a discontinuity of second kind x70 x x70 x x701 x Since linbx 1 and O is undefined7 f has a removable discontinuity at O Rede ne O 1 1 x7 m becomes continuous at O 5 Let x x 7 Discuss continuity of 1 Sketch the graph of 1 Since gx is discontinuous at n E Z7 1 is discontinuous at all integers 10 Let f be a real valued function on 0171 a if f is continuous on 0171 and limJr x exists Prove that is uniformly continuous on 0171 x7 Proof Since lim x exists7 let limJr x L for some L E R For any 6 gt 07 there exists a 61 such 704 1 I that Whenever 0 lt x 7 a lt 61 we have 6 mm7mlt Since 1 is continuous on 0171 f is continuous on a c b for all c with 0 lt c lt b 7 a Let c min6127 b 7 a2 Notethatcltb7a Since 1 is uniformly continuous ona 317 there exists a 62 such that 7 lt 6 provided lx7yl lt 62 and xy 6 1717071 Now let 6 minc62 For any xy E 0171 with lx 7y lt 6 if xy 6 01 317 then 7 lt 6 because 6 g 62 If xy E 170 c7 then lfx7fyl lfz7LL7fyl lfx7Lllfy7Ll lt Section 32 1 LetAn12 a Show that the set A is not compact by constructing an open cover of A that does not have a nite subcover of A Proof Note that the sequence is monotone decreasing For each n 6 N7 let i 1 1 1 1 1 em1n77777 7 n n 7 7 nn 1 1 n n n1 Then Ngn 7m7m l Let OnNg Note that On is open 1 n n and On A So On il is an open cover of A Since On contains only one point of A7 any union of a nite number of On contains only nite number of points of A Therefore7 the open cover On il of A does not have a nite subcover of A D b Prove directly using the de nition that K A U 0 is compact Proof Let OaaEA be any open cover of K Since K C UaEAOa7 the point 0 6 On for some a E A Since 01 is open7 there exists an 8 such that Ng0 766 C 01 Note that lim771L 0 So there exists a no 6 N such that for all n 2 no 77 lt 8 which implies that i 6 788 C On for all n 2 no Using the fact that K C UaEAOa again7 for k 12 7710 717 we have i 6 Oak for some ak E A Then Kc OaUUZO10akOaUOa1U0a2UUO Since OaheA is an arbitrary open cover of K7 the set K is compact D an07139 2 Suppose pn is a convergent sequence in R with limpn p Prove7 using the de nition7 that the set A p U pn n E N is a compact subset of R Proof Let OaheA be any open cover of A Since A C UaEAOa7 the point p 6 On for some a E N Since 01 is open7 there exists an 8 such that Ngp C 01 Note that limpn p So there exists a no 6 N such that for all n 2 no lpn 7 pl lt 8 which implies that pn E Ngp C On for all n 2 no Using the fact that A C UaEAOa again7 for k 127710 7 17 we have pk 6 Oak for some ak E A Then ACOaU 211amOaUOmUOazLLuUO an07139 Since OaheA is an arbitrary open cover of A7 the set A is compact D 3 Show that 01 is not compact by constructing an open cover of 01 that does not have a nite subcover Proof Let On i 2 for n E N with n 2 2 Note that 01 C UzilOn 02 So On io is an open cover of 071 But for a nite number open sets of the family7 say On17 On27 OM7 let m maxn17n27 771k Since the union ULlOnk 7 27 z ULIOM if 0 lt z lt i Therefore7 071 that does not have a nite subcover from the open cover Onf7 0 of 071 TheTnterval 071 is not compact D 4 Suppose that A and B are compact subsets of R a Prove using only the de nition that A U B is compact Proof Let OnneM any open cover of AUB Then Onn6M is open cover of both A and B be cause A and B are subsets of AUB Since A and B are compact subsets of R7 there exists a nite col lection 07717 07727 7 Om whose union covers A and there exists a nite collection 07717 07727 7 Om whose union covers B Certainly7 the union of 07717 07727 7 Om and Om17 07727 7 Om covers AUB7 ie7 A u B c 11371077 u Meow Therefore7 A U B is compact D b Prove that A B is compact Proof We can prove this directly by the de nition as we did for part a But we prove this by Theorem 325 Since A and B are compact7 both A and B are closed by Theorem 325 a Hence A B is closed Since A B C A7 A B is compact by Theorem 325 D 5 Let K be a nonempty compact subset of R Prove that supK and ian exist and are in K Proof Since K is compact7 K is closed and bounded by Theorem 325 a Since K is nonempty and bounded7 supK and ian exist by the least upper bound property Let Oz sup K Since K is nonempty7 for any 8 gt 07 there is z E K such that 04 7 8 lt z lt 047 ie7 z E N404 K with z 31 a Thus 04 is a limit point of K Since K is closed7 K contains all its limit points by Theorem 319 Therefore 04 E K Similarly we can show that ian E K D 6 Let K be a compact subset of R Use Bolzano Weierstrass theorem to prove that every in nite subset of K has a limit point in K Proof By Bolzano Weierstrass theorem7 every bounded and in nite subset of R has a limit point Since K is compact7 K is bounded by Theorem 325 a Thus every subset ofK is bounded Therefore7 every in nite subset of K has a limit point p The point p is in K because K is closed and it contains all its limit points D 7 Prove Theorem 3210 Let K be a nonempty compact subset of R Then every sequence in K has a convergent subsequence that converges to point in K Proof Since K is compact7 K is closed and bounded by Theorem 325 a Thus every sequence in K is bounded By Corollary 2411 Bolzano Weierstrass7 every bounded sequence in R has a convergent subsequence that converges to p Since K is closed7 the point p E K D 8 Prove that a compact set has only a nite number of isolated points Section 43 Remark 1 To show that a function is not uniformly continuous on a set E we can choose a sequence mn of E so that lzn 7 13n1l can be arbitrarily small but 7 flt13n1 2 6 for some xed number 6 gt 0 or we can choose two sequences and of E so that W 7 M can be arbitrarily small but lfmn 7 2 6 for some xed number 6 gt 0 Remark 2 To show that a function is uniformly continuous on a set E we can show that f is a Lipschitz function on E ie work with 7 with zy E E and nd an M gt 0 such that lfm 7 lt Mlz 7 Note that sometimes it may not work see problem 5 2 Show that the following functions are not uniformly continuous on the given domain b gz Dom g 000 Proof Let 6 1 Choose zn i for n E N Note that lz 7 m 1 l 7 lt l n 1 7 n 71 1 71 and i can be abortively small But for all n E N lgn iglt n1gtl ln2 7 77 12l 2n123gt18 Therefore 9 is not uniformly continuous D c hz sini dom h 000 Proof Let 6 12 Choose mn and yn 112 for n E N Note that 2 l l 1 m y 27m 1 27m 1 474 1 27m can be arbitrarily small But for all n E N 7r 1 lbw 7 lsin 27m 7 s1n 2mrl1gt 5 8 Therefore h is not uniformly continuous D 3 Prove that each of the following functions is uniformly continuous in the indicated set b 9m 2 m E N Proof Since for 6 12 if zy E N with lz 7 yl lt 6 then x y Hence 7 0 lt 6 for any 6 gt 0 Therefore 9 is uniformly continuous on N D d cosm7 z E R Proof For m y E R7 follow the example 422f to estimate cosmicosy72sinmlysin 2 2 e 5m 17m 0oo Proof for my 0oo7 2 7 yz 7 2y27y27y2 M1 y17 z1y1 wyzyy m1y1 yy ltz1gtlty1gt y Sincezygt07 zymy zyy lt1 m1y1 zyzy1 Wehave m2 yZ l5 5yl m m zyzy y m1y1 lt lmiyl The function 5z is an Lipschitz function and hence is uniformly continuous D 4 Show that each of the following is a Lipschitz function 10 990 1100111 9 0700 Proof For m y E 07 07 zyil m y k i 4 m y 902 1 242 1 902 1 242 1 S M yl where we used lmyl lt 1 1 21y21quotm21y21 So 9 is a Lipschitz function E 1

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