117-021909-1

117-021909-1 - MATH 117 LECTURE NOTES FEBRUARY 19, 2009 DR....

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Unformatted text preview: MATH 117 LECTURE NOTES FEBRUARY 19, 2009 DR. JULIE ROWLETT 1. Subsequences and Cauchy Sequences Definition 1.1. A subsequence of a sequence { s n } n N of real numbers is an ordered subset of { s n } n N indexed by n 1 < n 2 < n 3 < ... < n k < n k +1 ... N , such that n k . 1.1. Examples of sequences and subsequences. (1) Lets take our favorite sequence, the harmonic one with s n = 1 n . Then, a subsequence is s k = 1 2 k . However, u 1 = 1 5 , u 2 = 1 4 , u 3 = 1 6 , u 3+ k = s k for k 1 , is not a subsequence since the order of terms is not preserved. (2) The main idea in the preceding example is that, although a subsequence may omit (or skip) terms, it may not rearrange them. (3) Exercise: Construct a sequence which is not monotone but which has a monotone subsequence. 1.2. Subsequence Theorems. Theorem 1.2. If a sequence { s n } n =1 converges to a real number s, then every subsequence also converges to s. Proof: Let { s n k } be a subsequence of the original sequence { s n } . Let > . Then, since s n s (this is another way to write s n converges to s ), there exists N N such that if n N, | s n- s | < . By the definition of subsequence, n k . Hence, for all n k N, | s n k- s | < . This is the definition of convergence for the subsequence, so s n k s also. Theorem 1.3. If every subsequence of a sequence of real numbers { s n } n N con- verges to a real number s, then s n s. Proof: The proof is by contradiction. Assume that s n 9 s. Then, there exists > , such that for any N N , there exists n > N such that | s n- s | . So, there exists n > 1 such that (1.4) | s n- s | . 1 2 DR. JULIE ROWLETT Lets be clever and call this particular n, n 1 . Now, there also exists n 2 > n 1 such that | s n 2- s | . By induction, assume we have found n 1 < n 2 < ... < n k such that | s n j- s | for j = 1 ,...,k. Then, again, there exists n k +1 > n k such that | s n k +1- s | > , because otherwise, wed have | s n- s | < for all n n k , which contradicts (1.4). By induction, there is a subsequence s n k constructed this way so that n k < n k +1 and | s n k- s | > . This is a contradiction to the hypothesis that every subsequence converges to s....
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This note was uploaded on 03/05/2009 for the course MATH 117 taught by Professor Akhmedov,a during the Spring '08 term at UCSB.

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117-021909-1 - MATH 117 LECTURE NOTES FEBRUARY 19, 2009 DR....

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