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# Foundations of Analysis MATH 4200

Utah State University

GPA 3.77

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This 12 page Class Notes was uploaded by Darren Schulist on Wednesday October 28, 2015. The Class Notes belongs to MATH 4200 at Utah State University taught by E. Heal in Fall. Since its upload, it has received 11 views. For similar materials see /class/230412/math-4200-utah-state-university in Mathematics (M) at Utah State University.

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Date Created: 10/28/15

MATHEMATICAL INDUCTION Principle of Mathematical Induction Let S1 82 83 be a sequence of statements satisfying i S1 is true and ii For each positive integer k Sk gt Sk1 Then Sn is true for every positive integer n Principle of Mathematical Induction Generalized Let SQ Sj1 Sj2 be a sequence of statements satisfying i SQ is true and ii For each positive integer k 2 j Sk gt Sk1 Then Sn is true for every positive integer n 2j Principle of Strong Induction Let SQ Sj1 Sj2 be a sequence of statements satisfying i SQ is true and ii For each positive integer k 2 j If SQ Sj1 Sk are all true then Sk1 is true Then Sn is true for every positive integer n 2 j Well Ordering Principle for Natural Numbers Every non empty collection of positive integers contains a least element Math 4200 Linear Approximation Theorem Suppose f is de ned in an open interval containing 10 Then f is differentiable at 10 if and only if there exists a linear affine function Ax ax 7 10 fxo that approximates f near 10 in the sense that hm WWW 0 z7z0 z zo Proof 1 Suppose f is differentiable at x 10 Let Ax f xox 7 10 fxo hm f A lim i zlifl fzo 0 S0 z7z0 E IO z7z0 hm f A 0 z7z0 11 Suppose there exists Ax ax 7 10 fxo such that fzAz 139 0Thl39 M0 d willow 1 320 as a 7m lim z7z0 fz7fzo7az7zo 0 z7z0 lim z7z0 W 7 a 0 This implies that lim My a 0 z7z0 z to Riemann Integral Let f be a bounded function on 61 b Let P 61 x0 lt cl lt lt xquot b be a partition of 61b The norm ofthe partition denoted IP is equal to maxq x171 1 12 n For 139 l 2 n let c1 6 xH xn fm q x171 is called aRz39emann Sum forfover 61 b 11 The function f is said to be Riemann integrable on the interval 61 b provided g fmm xlL That is for each 8 gt 0 there exists 5 gt 0 such that given any partition P 61 x0 lt cl lt lt xquot b with Pl lt 5 and given any choice of c e xH xn i1 2 n it follows that lt8 i mm aa L b Notation The limit is usually denoted by J fxdx a Fundamental Theorem of Arithmetic Every positive integer greater than 1 is either a prime number or can be written as a product of prime numbers Furthermore this factorization is unique except for the order Fore example we can write 69362233472 or 1200224 352 and there are no other possible factorizations of 6936 or 1200 into prime numbers if we ignore the ordering of the factors That is for each positive integer n there exists unique prime numbers 171 p2 pk and positive integers i1 i2 ik such that lt lt lt lt and n 1 5 i3 ik I71 I72 I73 Pk I71 I72 I73 Pk The proof of this theorem is obtained by using mathematical induction Math 4200 Theorem If A is a finite set containing exactly n elements then the power set of A PA contains exactly 2quot elements Proof We will prove this by mathematical induction For each n let Sn be the statement quotIf a set contains exactly n elements then its power set contains exactly 2quot elementsquot Since the only subsets of singleton set are the empty set and the set itself the statement Sl is true We must show that for each n Sn implies Snl Suppose Sn is true Let A a1a2 an an be a set containing n1 elements Any subset of A either contains an or it doesn t The number of subsets of A which do not contain a is exactly the same as the number of subsets of 11 a2 an If Sn is true this number is 2quot A subset of A containing am is obtained by forming the union of a subset of 11 a2 an with the set arm Thus the number ofsubsets of A containing a is again 2quot The total number of subsets of A is then equal to 2quot 2quot 2 2quot 2 1 Hence Snl is true whenever Sn is true By the principle of mathematical induction Sn is true for all n Theorem Product Rule Suppose L and 91 M Then LM Proof Let 6 gt 0 There exists 61 gt O suchthat if 0 lt 1 7 1 lt 61 then 7L lt 666 Then L 7 666 lt lt L666 whenever O lt 1 7 1 lt 61 Let B max L 666 L 7 666 We now have lt B whenever O lt 1 7 1 lt 61 There exists 62 gt 0 such that if 0 lt 1 7 1 lt 62 then M56 M lt There exists 63 gt 0 such that if 0 lt 1 7 1 lt 63 then fz 7m lt Let 6 min616263 If 0 lt 1 7a lt 6 then fxgx7LMUta917fIMfxM7LMS f9 H99 M WWW M S B91MMW1L lt B Mm mlt e Fundamental Theorem of Arithmetic Every integer greater than one is either prime or is a product of primes Proof For each positive interger n gt 1 let S01 be the statement n is a prime or is a product of primes l S2 is true since 2 is a prime number 2 Let kbe any integer k 22 We will show that ifS2 S3 Sk are true then Sk1 is true If k1 is a prime number then Sk1 is obviously true If k1 is not a prime number then k1 a b where a and b are integers 2 S a S k 2 S b S k Now ifS2 S3 Sk are true then Sa and Sb are true So both a and b are either prime or the product of prime numbers Hence k1 a b can be written as a product of primes By the principle of strong induction S01 is true for all n 2 2 Completeness Axiom for the Real Line The axioms for the real line R include the eld axioms the order axioms and the following completeness axiom Axiom Every nonempty subset S of R that is bounded above has a least upper bound sup S Note An equivalent statement is that every nonempty subset S of R that is bounded below has a greatest lower bound inf S Back Away Principle for the Supremum Suppose S is a nonempty subset of R which is bounded above Let M sup S the least upper bound of S For each 8 gt 0 there exists x in S suchthat M gltxSM Back Away Principle for In mum Suppose S is a nonempty subset of R which is bounded below Let m inf S the greatest lower bound of S For each 8 gt 0 there exists x in S suchthat m S xltM8 Archimedean Principle For every real number x there exists a natural number n such that n gt x Dedekind Principle Suppose R H UK H K and VerVye Kxlt y Then either H has a last point or K has a rst point Note The Dedekind Principle is equivalent to the completeness axiom Math 4200 What is meant when we write lim f L 9H a Intuitive De nition Limit of x as x approaches a equals L means that x gets closer and closer to L as x gets nearer and nearer to a What do you mean by quotcloser and closerquot and quotnearer and nearerquot b Suppose f is a function de ned in a neighborhood ofthe point x a This means that f is defined in some open interval conaining x a7 except perhaps at the point x a Then If hasa limit as x approaches a provided there eXists a number L such that for each 6 gt 07 there eXists 6 gt 0 such that if Oltlxiallt6 then lfx7Ll lt6 We write lim L c lim L provided for each 6 gt 07 there eXists 6 gt 0 such that if x is not equal to a and the distance from x to a is less than 6 then the distance from x to L is less than 6 d lim L provided for each 6 gt 07 there eXists 6 gt 0 such that ifxy a andai 6 ltxlta6thenLieltfxltL e e How can lim fail to exist f What is meant when we write lim f L What is meant when we write lim f L zaa Math 4200 Theorem The least upper bound principle is equivalent to the Dedekind cut principle Proof 1 We will rst show that if we assume the least upper bound principle then we can obtain the Dedekind cut principle Suppose R HUK H K and VerVyeKxlty The setHis bounded above since every element of K is an upper bound of H Let w be the least upper bound ofH Since RH UK weH or weK If wthhen clearly w is the last point of H Suppose w e K Let y e K If y lt w theny is an upper bound ofH but less than the least upper bound of H a contradiction So w S y and w is therefore the rst element ofK 11 Now suppose that the Dedekind cut principle is true Let S be a subset of R that is nonempty and bounded above We must show that S has a least upper bound Let K y y is an upper bound ofS let H R K Now bothH andK are nonempty and their union is R Let x e H y e K Then x is not an upper bound ofS and so there exists 2 in H such that x lt 2 Since Since y is an upper bound for H we have x lt z S y hence x lt y By the Dedekind property either H has a last element orK has a rst element If H had a last element say w then w would also be an upper bound of H a contradiction since w is not in K So K must have a rst element call it v and v is the least upper bound of S Theorem 5 is an irrational number Proof Suppose that 5 is arational number Then 5 where a and b are 2 integers 1 3E 0 Then 5 g and 5b2 a2 Consider both sides of the last equation The prime factorization of a2 is obtained by multiplying the prime factorization of a by itself So all of the exponents of the primes after simplifying in the factorization of a2 are even The prime factorization of 5 b2 is obtained by multiplying 5 by the prime factorization of b2 Since the exponents of the primes in the factorization of b2 after simplifying are all even the exponent of 5 in the factorization of 5 2 must be odd This is a logical contradiction since the two prime factorizations must agree The supposition that 5 is an irrational number led to this contradiction Therefore 5 is an irrational number

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