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mathAnalysis2-07-draft

# mathAnalysis2-07-draft - ARE211 Fall 2007 LECTURE#2 THU...

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Preliminary draft only: please check for final version ARE211, Fall 2007 LECTURE #2: THU, AUG 30, 2007 PRINT DATE: AUGUST 21, 2007 (ANALYSIS2) Contents 1. Analysis (cont) 1 1.4. Convergence of Sequences 1 1.4.1. Boundedness of Sequences 2 1.4.2. Boundedness of Sequences (cont) 3 1.5. Least Upper bounds and Greatest Lower Bounds 4 1. Analysis (cont) 1.4. Convergence of Sequences Definition: We say that a sequence { x n } in X converges to an element x X in the metric d if epsilon1 > 0 N N such that for all n > N , d ( x n , x ) < epsilon1 . A sequence that converges to a point is called a convergent sequence. The point that it converges to is called a limit of the sequence. (It’s also the limit of the sequence, i.e., there’s only one, but that’s a result not a definition.) Example: { x n } defined by x n = 1 /n converges to zero in the Pythagorian metric, but not in the discrete metric. 1

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2 LECTURE #2: THU, AUG 30, 2007 PRINT DATE: AUGUST 21, 2007 (ANALYSIS2) Proof: We need to show that for all epsilon1 , there exists N such that for all n > N , d ( x n , 0) < epsilon1 . Now in the Pythagorian metric, d ( x n , 0) = radicalbig ( x n - 0) 2 = | x n | , i.e., we just need to show that for n sufficiently large, | x n | < epsilon1 . Pick N > 1 /epsilon1 . Then 1 /N < epsilon1 and for n > N , 1 /n < 1 /N < epsilon1 . Done On the other hand, consider the discrete metric. To show that x n doesn’t converge to zero, we need to show that there exists epsilon1 > 0 such that for all N , there exists n > N such that d ( x n , 0) > epsilon1 . Pick epsilon1 = 1 / 2. Let N be arbitrary and observe that d ( x N +1 , 0) = 1 > epsilon1 . Done Note that what we did here was an example of a general proof technique: to prove that an assertion involving ’s and ’s is false, go through and change the ’s to and the ’s to . Theorem: A sequence { x n } converges to α R in the discrete metric if and only if there exists N such that for all n > N , x n = α . 1.4.1. Boundedness of Sequences. Definition: A sequence is bounded above by b R if x n b , for all n N . A sequence is bounded above if it is bounded above by some b R . Defn of bounded below is parallel. Example: The sequence { x n } defined by x n = ( - 1) n is bounded above by 1 and below by -1. (It’s bounded above by lots of other things too, e.g, 2, π , etc, and similarly bounded below by lots of other things. The point is that in order to be bounded above, a sequence just has to be bounded above by something ).
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mathAnalysis2-07-draft - ARE211 Fall 2007 LECTURE#2 THU...

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