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# Intro to Real Analysis 2 MATH 5152G

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This 4 page Class Notes was uploaded by Webster Green Sr. on Sunday October 11, 2015. The Class Notes belongs to MATH 5152G at Columbus State University taught by Staff in Fall. Since its upload, it has received 37 views. For similar materials see /class/221194/math-5152g-columbus-state-university in Mathematics (M) at Columbus State University.

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Date Created: 10/11/15

Section 42 1 For each of the following7 determine whether the given function is continuous at the indicated points x0 2x275x73 a x 3 m7 g atx03 13 Note that f is continuous at p if and only if lim x p Since 1717 i 2x275x73 i x732x1 i m i i 11313971 77 N 7 6 x is discontinuous at x 3 2 lt dkxm atxo2 1 1 24 new and 1 1 47 2 mEEfltzgt new 90gt So lirn2 x does not exist x is discontinuous at x 3 x7 8x7 x E Q 2LetfR7Rbede nedby x 2m28 Q a Prove7 using 6 and 6 that f is continuous at 2 Proof Note that 2 16 If x E Q then x7168x 716 8x 7 2 Ifx 27 then x7162x287162x2782x2Hx72 Soifx Qand0lt x72lt17 then xlt3and2x22x724lt2x728lt10 Nowforany 6gt01et6min1610 For0lt x72lt6ifx C27 x7168x72lt86 8ltgt lt5 mm mains 2x2Hx72lt 10x72lt 106 10 8 Therefore lim x 16 2 x72 f is continuous at 2 D b ls 1 continuous at 1 Justify your answer NO Note that f1 8 and that there is an irrational sequence pn with limpn 1 So lim fpn hm2pi 8 10 7g f1 6 Let E C R7 and suppose that f is continuous at p E E a Prove that lfl is continuous at p ls converse true Proof Since 1 is continuous at 197 for any 6 gt 07 there exists a 6 gt 0 such that when x E E with 196 7191 lt 67 11 fpl lt 6 Then HfMl 7 WW S ffpl lt6 By de nition of continuity7 lfl is continuous at p D The converse is NOT true For example7 71 0 f17 0 is discontinuous at z 0 But 1 is continuous at z O 14 Use the intermediate value theorem to prove that every polynomial of odd degree has at least one real zero Proof Let pz anz an1z 1 1 am 1 a0 be a polynomial of odd degree Pm is continuous on R Note that when x approaches 00 or 700 the leading term anz determine the sign of If an gt 07 then limmace anz 00 and limmLLOQ anz 700 If an lt 07 then limmace anz 700 and limmLLOO anz 00 In either case7 there exist a number M gt 0 such that PM and P7M have the opposite signs pz is continuous on 7M7 By the MVT7 there is a c E 7M7 M such that 190 O D 15 Prove that there exists z E Om2 such that cosz m Proof Let x cosz 7 x f is continuous on Om2 and f0 cosO 7 0 1 gt 07 f7r2 cos7r2 7 7r2 77r2 lt 0 By the MVT7 there exists 0 E Om2 such that fc 0 Hence cosc c D 16 Suppose that f 717 1 7 R is continuous and satis es f0 f1 Prove that there exists 39y E 07 such that y f y Proof De ne the function gz x 7 fz Since the domain of f is 7117 the domain of fmis 761 lt71x1gt So the function g is de ned and continuous on 07 Note that Section 31 1 Prove that every open interval in R is an open subset of R Proof Let a7b be a nite open interval of R For any point z E 17 choose 8 so that 0 lt 8 lt minx 7 ab 7 Then Ngz C 17 Therefore cab is open Now for the open interval 100 and any point z E 1007 choose 8 so that 0 lt 8 lt xia Then NEW C 100 Therefore cab is open Similarly we can show that the open interval 7007b is open Finally7 since any x E R 70000 and 8 gt 07 Ngz C 700007 the interval 70000 is open 2 Show that every nite subset of R is closed Proof Let A be a nite subset of R having 71 points 117027 7an We can assume that a1lta2ltltan Then Ac 700411 U 11702 U a27a3 U U awhan U amoo Thus Ac is a union of open intervals Therefore7 A is closed D 3 Show that the intervals 70041 and 100 are closed subsets of R Proof Since R7ooa 100 and Ra7 oo 7007a are open7 the intervals 70041 and 1 00 are closed subsets of R D 4 For the following subsets E of R7 ll in the chart nE 5 a Let F be a closed subset of R and let pn be a sequence of F which converges to p E R Prove that p E F Proof We prove this by contradiction Assume that the point p F Then the point p 6 F0 Since F is closedl7 the set F0 is open Thus there exists 8 gt 0 such that Ngp C F0 So Ngp H F 0 This contradicts the fact that point p is a limit point of F Therefore7 p E F D Proof using Theorem 3111 Since limpn 107 the point p is a limit point of F Thus p E F Since F is closedl7 by Theorem 31117 F F FU F Therefore7 p E F D 6 b Give an example of countable collection Fn il of closed subsets of R such that Uf Fn is not closed Let Fn The set Fn is a closed subset of R for n E N But Uf Fn n E N is not closed because it does not contain its limit point 0 Another example is Q of the set of rational numbers Since Q is countable7 we let Q Tn n E N Let Fn The set Fn is a closed subset of R for n E N But Uf1Fn Q is not closed 8 Let E be subset of R a Prove that lntE is open Proof Let x E lntE We show that there exists 8 gt 0 such that Ngx C lntE and hence lntE is open Since z E lntE7 by de nition of interior points of E7 there exists 8 gt 0 such that Ngz C E Next we show that every point y E Ngz is an interior point of E Since New x 7 87 8 is open7 if y E NEW7 then there exists 6 gt 0 such that N5y C Ngx C E Thus y is an interior point of E This means that Ngx C lntE We have proved that every point of lntE is an interior point of lntE Therefore7 lntE is open D b Prove that E is open if and only if E lntE Proof If E is open7 then by de nition of a open set7 E C lntE Certainly every interior point of E is a point of E lntE C E Thus E lntE On the other handl7 if E lntE7 by part a we know that lntE is open Therefore E is open D c If G C E and G is open7 prove that G C lntE Proof Let x E G Since G is open7 here exists 8 gt 0 such that Ngx C G Since G C E7 Ngx C E Therefore z E lntE We have proved that G C lntE D 9 Let A and B be subsets of R b Show that A B C K E Proof Since A B C A CE and A B C B CE7 A B CZ E Since Z Eis closedl7 by Theorem 3111 c7 A B CZ E D c Give an example for which the containment in part b is proper Example 1 Let AQ and BRQ ThenA B 0 A B ButZRE Hence Z PR a

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