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# 468 Class Note for MATH 403 at PSU

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This 5 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at Pennsylvania State University taught by a professor in Fall. Since its upload, it has received 30 views.

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Date Created: 02/06/15
Math 403 Lecture Notes for Week 1 1 Ordered Fields De nition 1 Field Axioms A eld is a set X together with two binary functions and de ned on X which satisfy the following properties 1 I y y I 2 1y2 z WM 3 There is an element7 denoted 07 such that z 0 z for all z 4 For all I there is a w7 denoted 71 such that z w 0 5 z y y z 6 1y2Iyz 7 There is an element7 denoted 17 such that z 1 z for all x 8 For all I f 0 there is a w7 denoted 1 1 such that z w 1 5 zyzzyzz The real numbers R with the usual addition and multiplication is a eld So are the rationals Q and the complex numbers C with the usual operations Example 2 We can de ne a somewhat more exotic eld R as follows ng 01 has two elements7 with and de ned as follows 0 1 0 1 0 0 1 0 0 0 1 1 0 1 0 1 De nition 3 A strictly totally ordered set consists of a set X and a binary relation lt which satis es 111 2 For all z and y exactly one of the three possibilities holds 1 lt y z y y lt z 3 lfzltyandyltzthenzlt2n De nition 4 Order Axioms An ordered eld consists of a eld X7 and a set P Q X of positive elements satisfying the following 1 lszPandy6P7thenzy P 21f16Pandy6P7thenzyEP 3i lszPthenizg Pi 4 For each I exactly one of the three possibilities holds 1 0 z E P 71 6 Pi Example 5 1 The reals with the set P z z z gt 0 is an ordered eld 2 The reals with the set P z z z lt 0 is not an ordered eld since 71 E P but 71 71 Pl 3 The rationals Q with P z E Q z gt 0 is an ordered eld 4 The eld ng cannot be ordered ie there is no set P which makes it an ordered eld To see this note that 71 1 so if 1 E P then we need 71 P iiei l P whereas if 1 P we need 716 P ie 16 Pl De nition 6 We can de ne a strict totally ordering on an ordered eld by setting zltyifandonlyifyizEP Wewritez Syfor zyorz ltyi Note 7 We can restate the ordering axioms in terms of the order lt as follows 1 lfzltyandzltwthenzzltywi 2i lf0ltzltyand0ltzltwthenzzltywi 3i Exactly one of z lt y z y y lt I 2 Bounded Sets De nition 8 Let A be a subset of an ordered eld 1 We say that b is an upper bound of A if I S b for all z 6 Al 2 We say that c is a lower bound of A if c S I for all z 6 Al 3 We say that b is a least upper bound of A if b is an upper bound of A and for any other upper bound 5 of A we have b S cl 4 We say that c is a greatest lower bound of A if c is a lower bound of A and for any other lower bound d of A we have d S cl 5 We say that b is the greatest element of A if b is an upper bound of A and b e A 6 We say that c is the least element of A if c is a lower bound of A and c 6 Al Example 9 1 In R the set I z 0 S I lt 1 has 1 as an upper bound and 0 as a lower bound These are respectively a least upper bound and a greatest lower bound The set has a least element 0 but no greatest element 2 In R the set I z z lt 0 does not have a lower bound 3 ln Q the set I E Q lt z lt has an upper bound for instance 2 It does not have a least upper bound since any upper bound 1 must be rational so V3 lt b and hence there is another rational c with V3 lt c lt b This 5 is an upper bound which is less than 12 De nition 10 Completeness Axiom We say an ordered eld X is com plete or satis es the least upper bound property if whenever S is a nonempty subset of X which has an upper bound then S has a least upper bound We denote this least upper bound sup 5 Example 11 The reals are complete The rationals Q is not complete To see this consider the set I E Q lt This has an upper bound but no least upper bound Note 12 The least upper bound property implies the corresponding property for lower bounds Any nonempty set which has a lower bound has a greatest lower bound Note 13 The reals satisfy the least upper bound property In fact the reals are the unique complete ordered eld We will derive many properties of the reals just from the assumption that they form a complete ordered eld Proposition 14 Let L and U be nonempty subsets ofR with R L U U and such that for all l E L and u E U we have l lt u Then either L has a greatest element or U has a least element Proof Let u be any element of U Then u is an upper bound of L Hence by completeness L has a least upper bound b If I E L then L has a greatest element and we are done So suppose not then b E L We claim that b is the least element of U If not there is some u E U with u lt I But then u is an upper bound for L which is less than I contradicting that b was the least upper bound So I is the least element of U in this case B De nition 15 In an ordered eld an integer is a number that is either 0 or of the form 1 1 or of the form 71 1 A rational is an element of the form p q 1 where p and q are integers with q f 0 Proposition 16 Axiom of Archimedes In a complete ordered eld for any I there is an integer n with z lt n Proof If I lt 0 we can take n to be 0 So suppose I 2 0 Let S be the set S h z k is an integer with k S I Then S is nonempty since it includes 0 and has an upper bound namely 1 Hence 5 has a least upper bound y The element y 7 is then not an upper bound for 5 so there is some integer k E S with y 7 lt kl But then y lt k lt k 1 so k 1 is not in 5 Since k 1 is an integer this means that z lt k 1 so setting n k 1 works D Corollary 17 The rationals are dense in a complete ordered eld ie ifz lt y then there is a rational r with z lt r lt y Proof First suppose I 2 0 By the Axiom of Archimedes there is an integer q with q gt y 7 z 1 so y 7 z gt 5 Consider the set S with n S n z n is an integer with y S 7 q S is nonempty and bounded below by y 4 so it has a greatest lower bound pi Since 5 consists only of integers p is in fact an element of 5 so p is the least element of 5 This means that 71 Lltygg q q Wealso havezy7y7zlt 7 Lgli Solettingrwehave z lt r lt y If I lt 0 we can nd an integer n with n gt 7r so n z gt 0 As above we then nd a rational r with n z lt r lt n y The number r 7 n is then a rational with z lt r 7 n lt y D 3 In nite Decimals For de niteness we will de ne the reals to be in nite decimals By an in nite decimal we mean a number of the form nidodldg or the negation of such a number where n and each di is an integer and 0 S di S 9 for each i 2 1 This sequence represents the real number di 10139 These are added and multiplied with the expected carrying of digits The one point to be careful about is that we must identify eventually 9 sequences and eventually 0 sequences such as 0999 1 000 a With this understanding the collection of in nite decimals forms a complete ordered eldi 4 Limits De nition 18 A sequence is a list of real numbers indexed by the positive integers We use the notation ltfngtn21 to denote the sequence 111213 De nition 19 Ne de nition of limit Let Ingtn21be a sequence and y a real number We say that the sequence has limit y and write limnH00 In y if For any 6 gt 0 there is a natural number N such that lIn 7 yl lt e for all n 2 N De nition 20 Let Ingtn21be a sequence We say that the sequence has limit 007 and write limnH00 In 00 if For any h there is a natural number N such that In gt h for all n 2 N We de ne limit 700 similarly Note 21 We shall try to be careful about what we mean when we say a sequence has a limit We shall generally say a sequence has a limit as a real number to indicate that there is a real number y which is the limit of the sequence We will say a sequence has a limit as an eItended real number if we also allow the limit to be ioo Example 22 Consider the sequence with I n Then limnH00 0 To see this7 let 6 gt 0 be given We need to nd an N such that lIn 7 0 lt e for all n 2 N7 is i lt s This is equivalent to ensuring n gt so choosing N gt suf ces De nition 23 We say that a sequence ltIngtn21is bounded above if the set In n 2 l is bounded above We de ne bounded below similarly De nition 24 We say a sequence is monotone increasing if In S In1 for all n 2 1 We de ne monotone decreasing similarly Proposition 25 Let ltfngtn21 be a sequence which is bounded above and mono tone increasing Then ltfngtn21hll8 a limit as a real number Proof Consider the set In n 2 1 It is nonempty and bounded above7 so it has a least upper bound y We claim that limnH00 In y Let 6 gt 0 be given We want to nd an N as required in the de nition of limit Suppose there is none Then for any N there is an n 2 N with lIn 7 yl 2 s This means In S y7 s Since the sequence is monotone increasing7 we have IN S In S y7e as well Since this holds for any N7 we have that y 7 e is an upper bound for the sequence7 contradicting that y was a least upper bound D Note 26 The same is true for a sequence which is bounded below and monotone decreasing Note 27 Both conditions in the proposition are necessary For instance7 the sequence 17 7117 717 is bounded7 but has no limit The sequence with In n is monotone but not bounded above7 and has no limit as a real number Note7 though7 that a monotone increasing sequence always has a limit as an extended real number7 since if it is not bounded above it has limit 00

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