117-021909-1 - MATH 117 LECTURE NOTES DR JULIE ROWLETT 1 Subsequences and Cauchy Sequences Denition 1.1 A subsequence of a sequence{sn}nN of real

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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) Let’s 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 > 0 . 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 s. Then, there exists > 0 , 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 Let’s 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, we’d 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 | ≥ > 0 . This is a contradiction to the hypothesis that every subsequence converges to s. 1.3. Cauchy Sequences. Recall the definition of a Cauchy Sequence. Definition 1.5. A sequence { x n } n =1 in a metric space ( X, d ) is Cauchy if, for any > 0 , there exists N N such that d ( x n , x m ) < for any n, m N.  #### You've reached the end of your free preview.

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• Fall '08
• Akhmedov,A
• Real Numbers, Limit of a sequence, Cauchy sequence, subsequence, DR. JULIE ROWLETT
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