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Unformatted text preview: P r e l i m i n a r y d r a f t o n l y : p l e a s e c h e c k f o r fi n a l v e r s i o n 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 > 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. (Its also the limit of the sequence, i.e., theres only one, but thats 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 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 doesnt 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 R in the discrete metric if and only if there exists N such that for all n > N , x n = . 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. (Its bounded above by lots of other things too, e.g, 2, , 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...
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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.
 Fall '07
 Simon

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