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# Intro to Real Analysis I MATH 4423

Marketplace > Idaho State University > Mathematics (M) > MATH 4423 > Intro to Real Analysis I
Oswald Boyle
ISU
GPA 3.94

Dennis Stowe

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COURSE
PROF.
Dennis Stowe
TYPE
Class Notes
PAGES
2
WORDS
KARMA
25 ?

## Popular in Mathematics (M)

This 2 page Class Notes was uploaded by Oswald Boyle on Monday October 12, 2015. The Class Notes belongs to MATH 4423 at Idaho State University taught by Dennis Stowe in Fall. Since its upload, it has received 18 views. For similar materials see /class/222162/math-4423-idaho-state-university in Mathematics (M) at Idaho State University.

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Date Created: 10/12/15
Double Negation 7NP P Comm forConjun PQ QP Commfordisj PVQQVP DeMorgan39s Laws NP v Q MP A NQ NP A Q NP v NQ Phrasologyforthe conditional P a Q If P thenQ P impliesQ Qfollows from P NotP unlessQ QifP QonlyifP Whenever PQ QwheneverP P is sufficient for Q Q is necessary for P P is a sufficient condition forQ Q is a necessaw condition for P Converse Q a P Contrapositive NQ 7 MP DEFINITIONS 1 Dedekind Cuts A Dedekind cut in Q is a splitting on intotwo nonempty setsA and B A n B 0A U B QA 03 0 such that Ile EAandb E BthenL1 ltb b A has no largest element 2 Sequence of an E IR converges to b E IR if Vs gt 03N such that LINL1N1 are all within s of b 3 A sequence an E IR is Cauchy if Vs gt 03N such that LINL1N1 are within s of each other 4 Inner Products a 35 y y 35 symmetric operation b 35y z 35y 352 and 35y c35y exist because has bilinear properties c 3535 2 0 and 3535 0 iff35 0 Positive Definite 5 CauchySchwartz inequality a lelyl S W S MM 6 Cardinality deals with determining if given sets have the same number of elements a Does one set have more than the other b quotSame Numbel pair up elements in different sets There isa bijection fA a B 7 Bijections a njectiveifVL1L1O E A Eifafao E B such thatLl L1O gt fL1 fao b Surjective if for each I E B there is at least one L1 E A such that fL1 b That is the range offis B 8 Set terminology a Finite ifS is the empty set orSN123 n for some n E N b Infinite otherwise c Denumerable ifSNN S has the same cardinality as N d Countable ifdenumerable orfinite e Uncountable otherwise 9 Continuity Afunction fS 7 IRwhereS Q IR is continuous atthe pointa E S if Vs gt 035 gt 0Vx E SI35 7 LII lt 5 7fal lt s a quotContinuous at aquot ornotonly discussed if L1 E S in domain b f is continuousquot if continuous at all points L1 E S THEOREMS 1 IfS Q IR is nonempty and has an upper bound in IR then it has a least upper bound in IR Evew convergent sequence is Cauchy 2 3 In IR evew Cauchy sequence is convergent 4 N 39 is an equivalence relation meaning a ANA forall setsA b ANB 7 RNA c ANB and BNC impliesANC 5 IfAAo are denumerable sets not necessarily disjoint then the set UL1 A A1 U AZ U U A71 is denumerable There is no surjection f N a N 7 SCHROEDERBERNSTEIN THEOREM IfAB are sets and A B and B lt Athen A N B In other words IffA a B and gB a A are injections then there existsa bijection hA a B 8 PROPERTIES OF CONTINUOUS FUNCTIONS lff L1b 7 IR is continuous then a fis bounded EIM such that S M for all35 E L1b fattains maximum and minimum values 3350351 E L1b such that f350 S S f351 for all35 E L1b f ins evenvalue between fL1 and fb inclusive Iffa S y S orfb S y S fa then yforsome35 E L1b fis uniformly continuous Vs gt 035 gt 0Vst E L1bs 7 t lt 5 7 lt s 9 9957 EXAMPLES f17 is a value ofa continuous function f L1b 7 IR use the lub property to prove thatthere is a smallest 35 E L1b such thatf35 17 Proof LetS 35 E L1 bf35 17 We are given thatS is nonempty Since Sis bounded below by a it has a greatest lower bound 350 We must show that 350 E L1b and that f350 17 Because a is a lower bound forS and 350 is the greatest lower bound L1 S 350 On the other hand we are given thatS has at least one element 35 and since 350 is a lower bound forS it follows that 350 S 35 S b Therefore 350 E L1b t remainsto show that f350 17 Suppose otherwise Then f350 7 17 gt 0 By continuity offat 350 it followsthat there exists 5 gt 0 such that X E arbIrIX 7360i lt57 fx 7fXo lt fXo 717i Because x0 5 gt x0 it is not a lower bound for 5 Therefore there exists 3 E 112 with x lt x0 5 and since x0 is a lower bound forS one also has 3 2 30 The numberx therefore meets the hypothesis in the display above but since Vthe conclusion 7fxo lt fxo 7 17 is clearly false The contradiction shows thatf0co 17 Provethatthe interval 112 Llb39 where L1b x E 1R L1 3 S band L112 x E 1R 35t E 112 withs t 1 and 5L1 tb E 1R1 Proof Suppose thatx E 112 Then x 7 23 312 The cofactors ofa and b here sum to b 7 L1b 7 L1 1 and because 0 S b 7x S b 7 L1 and 0 S x 7 L1 S b 7 L1 each is in 01 Thereforex E L1 b39 The reverse containment suppose thatx E Llb39 thusx so tb forsome St E 01 with s t 1 Writing asa17sasa17sbsatbx17tatb17tbtbb ShowsthatLl S x S b which isto saythatx E L1b If an E HR is a convergent sequence show that there exists M such that LL1 S M for all n Proof Since the sequence converges lets 1 Then ILLl 7 1 lt 1 ifngtN fwe fix N then by the triangle inequality 1na 7aa S Ian 7 1 Ial lt 1 a For all ngtN Let M max l1IIleIIL13I ILINI ILlN1 tfoows that lt M for all n ie the sequence is bounded as required Proof of the intermediate value theorem based on first principles Proof To be definite suppose that f01 S y S LetX t E L1b S y This set includes aand is bounded above by b so it has a least upper bound 6 E L1b Suppose thatfc 11 Becausefis continuous at c there exists 5 gt 0 such that t E ab n s 765 6 ft 7fcI lt fc 71 An application of the triangle inequality shows that lt y for all such t iffc lt y and gt y for all such t if gt 11 Again some elementx E X satisfiesx gt c 7 5 and because c is an upper bound forX one also has 3 S c Thereforex E L1b n c 7 5 5 and since S 11 one concludes that the former of the two possibilities occurs lt y thatminbc 52 S c sothat bc and one concludes that lt y contrawto the hypothesis that 2 y The contradiction shows that y Proofthat every convergent sequence in HR is Cauchy Proof Suppose xk 7 x Lets gt 0 be given Define71 52 Since xk a x 3101 such that Vj 2 101 we have Ix 7x lt 71 It follows by using the triangle inequality that for alljl 2 101 we have xk 7x1 ka 73 x 7x1 3 lxk Xl lx le lxk xllx1 xl lt77 77 5 Thus setting 1 101 the convergent sequence xk is Cauchy by definition of Cauchy Sequences

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