mathAnalysis3-07-draft - ARE211, Fall 2007 LECTURE #3: THU,...

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Preliminary draft only: please check for Fnal version ARE211, Fall 2007 LECTURE #3: THU, SEP 6, 2007 PRINT DATE: AUGUST 21, 2007 (ANALYSIS3) Contents 1. Analysis (cont) 1 1.6. Two Preliminary Results 1 1.7. Cauchy Sequences 2 1. Analysis (cont) 1.6. Two Preliminary Results Theorem: (Theorem A for future ref): Given two subsets, P and Q , of the real line, if P Q then sup( P ) sup( Q ) and inf( P ) inf( Q ). Obvious just by looking at a picture: if P is smaller than Q then then every upper bound to Q must also be an upper bound to P and there may be upper bounds to P that aren’t upper bounds to Q . Doing the proof is just a matter of saying the obvious in symbols. Proof: We’ll just prove the Frst half about sups. Pick P,Q R such that P Q . There are two cases to consider: Case A : sup( Q ) = ; in this case, there’s nothing to prove because any possible sup for P has to be less than or equal to inFnity. Case B : sup( Q ) = ¯ b . We’ll consider b > ¯ b and prove that b cannot be a least upper bound for P . This will establish that sup( P ) sup( Q ). Well, ¯ b q , for every q Q . Since P Q , ¯ b p , for 1
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2 LECTURE #3: THU, SEP 6, 2007 PRINT DATE: AUGUST 21, 2007 (ANALYSIS3) every p P . Hence ¯ b is an upper bound for P . Since b > ¯ b , b cannot be a least upper bound for P . s Notice that if the statement of Theorem A had omitted any mention of the real line, then the theorem would have been false. Why? Theorem: (Theorem B for future ref): Given S R , b R is the least upper bound (supremum) of S iF b is an upper bound for S and if for every e > 0, there exists s S such that b - s < e . Proof: : Need to prove this in both directions, i.e., (a) if b satis±es the requirements of the theorem then it is a supremum; (b) if it doesn’t satisfy the requirements of the theorem, then it isn’t a supremum. Proof of (a): Assume that for every e > 0, there exists s S such that b - s < e . We need to show that b b p for every upper bound b p for S . We’ll do this by showing that if b p < b then b p cannot be an upper bound for S . Pick b p = b - e , for some e > 0. By assumption, s S s.t. b - s < e = b - b p , i.e., - s < - b p or s > b p proving that b p isn’t an upper bound for S . Proof of (b): Now assume that one of the requirements of the theorem isn’t satis±ed. The ±rst requirement is that b is an upper bound for S . If this requirement is violated, then trivally b can’t be a supremum; Now suppose that the e condition fails, i.e., e > 0 such that s S , b - s e (see ²ig. 1). We’ll show that b cannot be a least upper bound for S , i.e., we’ll show that there exists b p < b such that b p is an upper bound for S . Pick b p = b - e . Since b - s e , for all s S , or, equivalently, b - e s , for all s S , then b p = b - e s , for all s S . Hence b p is an upper bound for S and is also smaller than b . So we’ve proved that b isn’t a least upper bound.
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This note was uploaded on 08/01/2008 for the course ARE 211 taught by Professor Simon during the Fall '07 term at University of California, Berkeley.

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mathAnalysis3-07-draft - ARE211, Fall 2007 LECTURE #3: THU,...

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