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Intro To Adv Math

by: Ms. Norwood Gerhold

Intro To Adv Math MAA 3200

Ms. Norwood Gerhold
GPA 3.83


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This 3 page Class Notes was uploaded by Ms. Norwood Gerhold on Monday October 12, 2015. The Class Notes belongs to MAA 3200 at Florida International University taught by Staff in Fall. Since its upload, it has received 27 views. For similar materials see /class/221801/maa-3200-florida-international-university in Mathematics (M) at Florida International University.


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Date Created: 10/12/15
MAA 32007 Construction of the Real Numbers7 Part Two see also Ch 93 of llorash7 on reserve Part One was an outline of the main results about R that are not already in Chs 1 2 of Wade7s book The recent lectures have included proofs of many of those results7 though we don7t have time to do every single step in great detail Here in Part Two7 l7ll summarize what we have done in class7 and provide notes on some proofs7 of trichotomy7 etc Thm The product of two Cauchy sequences is also Cauchy Proof in class 11 1903 It is based on the fact that Cauchy sequences are bounded7 and uses the triangle inequality Thm R is a eld On 1112037 we notedproved the closure7 commuta tive and inverse properties in Postulate 17 but didn7t cover the distributive or associative properties Def an is positive means Ele gt 0 and Eln E N such that Vm 2 717 am 2 6 Def 3 lt 3 means 3 a is positive This should have been included in Part One Thm R is an ordered eld We proved the additive property on 11 127 and the transitive property is HW Trichotomy is harder7 but we started it on 11 1903 We will skip the multiplicative property Proof of trichotomy Using the additive property7 we can replace a lt y by y a gt 07 and so on So7 ETS exactly ONE of these is true a 0 or a lt 0 or 0 lt a Also7 a lt 0 is equivalent to a gt 0 which means we can use the de nition of positive on it Part I ETS at least one is true We assume that a gt 0 and a gt 0 are false ETS a 0 Since a E R7 we know a an where an is a Cauchy sequence in Q The goal is to show an gt 0 Let 6 gt 0 Since an is Cauchy7 there is an N so that m7 n 2 N implies an am lt 62 Assume n 2 N ETS on lt e which is the same as 6 lt on lt 6 Since a gt 0 is false7 we take the negation of its de nition and we get 1 Ve gt 07VN7 Elm 2 N7 am lt 6 We apply this with 6 62 and with the same N as above So using the Cauchy assumption7 an lt am 62 lt 6 e 2 6 This proves the second part of our ETS above Likewise7 since a gt 0 is false7 this reasoning shows that an lt 67 which implies 6 lt an7 which is the rst part of the ETS Part II ETS Two of the phrases cannot both true This will involve three steps Part ILA ETS a 0 and a gt 0 cannot both true Part ILB ETS a 0 and a gt 0 cannot both true Proof If a 0 then a 07 so apply ILA to a Part llC ETS a gt 0 and a gt 0 cannot both true Proof very sim ilar to the proof of HA Omitted This nishes the proof of the trichotomy property Thm R is a complete ordered eld Outline of proof of completeness Let E be a nonempty bounded set of real numbers ETS EIL E R7 L sup E Of course7 L must be of the form an7 so our rst job is to de ne a sequence of rational numbers qn that seems likely to work Let n E N Let Sn mn m E Z7 mn is an upper bound of Claim 1 Sn is nonempty7 because E has an upper bound M E R7 and there is some mn gt M The existence of mn follows easily from the AP in R7 but that would be circular reasoning The correct proof is pretty involved and well omit it unless you request it But assuming Claim 17 Sn has a least element gm 6 Q by WOP Claim 2 The sequence qn is Cauchy7 and sup E DONE We won7t prove this in detail either unless you request it But consider an example Let E 07 C R and sup E 2 m 1414 Then ql 2 and q2 32 and q3 53 etc Note7 for example7 that 0 lt 32 2 lt 12 if not7 qg 32 is wrong In general7 0 lt 171 sup E lt ln7 so that qn gt 2 sup E I hope this makes Claim 2 seem believable Thm If F is a complete ordered eld then F is isomorphic to R This is a little beyond the scope of the course so you can treat it as optional Discussion This means there is a 1 1 correspondence f F gt R which preserves structure For example imagine that F is the same as R but that 4 is written the Roman way IV It is debatable whether we can say 774 IV but can certainly say that IV corresponds to 4 That is we can set f I V 4 Since we require f to preserve structure we require that a lt b iff fa lt fb and that fa b fa fb etc Constructing f takes a while so I will only show you the idea of it Step 1 Since F is a eld it has a zero element which we will denote 0F and likewise it has a 1F We set f0p 0 and f1F 1 By closure F contains 1F 1F which we denote 2 F and we are forced by the rule fa b fa to set f2p 2 Continuing this way we see that F has to contain a copy of Z and we can de ne f ZF gt Z in a unique way just by following the rule above Step 2 Since F is a eld it contains fractions and contains a copy of Q We de ne f Q F gt Q according to isomorphism rules such f a b fafb We know that 0F lt 1F etc and can show that f preserves the lt relation for all integers and fractions Step 3 Since F is complete its Cauchy sequences converge Also we have a density theorem for F So every 3 E F corresponds to a Cauchy sequence in Q F We use f to pair it with a Cauchy sequence in Q which converges to some number in R which we de ne to be f This completes the construction of f F gt R though there are a lot of details to be checked Remark The idea of isomorphism is often useful in algebra For exam ple we claimed without much proof that Q C R It would be more precise to say that R contains a set of equivalence classes which is isomorphic to Q But it is generally easier and harmless to treat 77isomorphic77 as 77equals


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