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mathAnalysis3-07-draft

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

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Preliminary draft only: please check for final 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 first 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 infinity. 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 . square 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 iff b is an upper bound for S and if for every epsilon1 > 0, there exists s S such that b - s < epsilon1 . Proof: : Need to prove this in both directions, i.e., (a) if b satisfies 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 epsilon1 > 0, there exists s S such that b - s < epsilon1 . We need to show that b b prime for every upper bound b prime for S . We’ll do this by showing that if b prime < b then b prime cannot be an upper bound for S . Pick b prime = b - epsilon1 , for some epsilon1 > 0. By assumption, s S s.t. b - s < epsilon1 = b - b prime , i.e., - s < - b prime or s > b prime proving that b prime isn’t an upper bound for S . Proof of (b): Now assume that one of the requirements of the theorem isn’t satisfied. The first 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 epsilon1 condition fails, i.e., epsilon1 > 0 such that s S , b - s epsilon1 (see Fig. 1). We’ll show that b cannot be a least upper bound for S , i.e., we’ll show that there exists b prime < b such that b prime is an upper bound for S . Pick b prime = b - epsilon1 . Since b - s epsilon1 , for all s S , or, equivalently, b - epsilon1 s , for all s S , then b prime = b - epsilon1 s , for all s S . Hence b prime 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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mathAnalysis3-07-draft - ARE211 Fall 2007 LECTURE#3 THU SEP...

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