117-021909-1 - MATH 117 LECTURE NOTES DR JULIE ROWLETT 1...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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.
Image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern