Facebook Page
Twitter
RSS
+ Reply to Thread
Results 1 to 2 of 2
  1. MHB Craftsman
    Alexmahone's Avatar
    Status
    Offline
    Join Date
    Jan 2012
    Posts
    262
    Thanks
    125 times
    Thanked
    70 times
    #1
    Prove that if a subsequence of a Cauchy sequence converges then so does the original Cauchy sequence.

    I'm assuming that we're not allowed to use the fact that every Cauchy sequence converges. Here's my attempt:

    Let $\displaystyle\{s_n\}$ be the original Cauchy sequence. Let $\displaystyle \{s_{n_k}\}$ be the convergent subsequence.

    Given $\epsilon>0$,

    $\exists N_1\in\mathbb{N}$ such that $\displaystyle|s_n-s_m|<\frac{\epsilon}{2}$ whenever $n\ge N_1$ and $m\ge N_1$.

    $\{s_{n_k}\}$ converges, say, to $L$.

    So $\exists N_2\in\mathbb{N}$ such that $\displaystyle|s_{n_k}-L|<\frac{\epsilon}{2}$ whenever $k\ge N_2$.

    $\displaystyle|s_n-L|=|s_n-s_{n_k}+s_{n_k}-L|\le |s_n-s_{n_k}|+|s_{n_k}-L|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$ whenever $n\ge N_1$, $n_k\ge N_1$ and $k\ge N_2$.

    How do I wrap up this proof by finding the $N$ such that $|s_n-L|<\epsilon$ holds whenever $n\ge N$?

  2. MHB Master
    MHB Global Moderator
    MHB Math Scholar
    MHB POTW Director
    Euge's Avatar
    Status
    Offline
    Join Date
    Jun 2014
    Posts
    1,387
    Thanks
    1,367 time
    Thanked
    3,484 times
    Thank/Post
    2.512
    Awards
    MHB Advanced Algebra Award (2016)  

MHB Topology and Advanced Geometry Award (2015)  

MHB Analysis Award (2015)  

Graduate POTW Award (2014)  

MHB Advanced Algebra Award (2014)
    #2
    Hi Alexmahone,

    It'll be useful to use the fact that for all $k\in \Bbb N$, $n_k \ge k$. Let $N = \max\{N_1,N_2\}$. If $n \ge N$, then $n\ge N_1$ and $n_N \ge N \ge N_1$, which implies $\lvert s_n - s_{n_N}\rvert < \epsilon/2$. Also, $n \ge N_2$ and $n_N \ge N \ge N_2$, so that $\lvert s_{n_N} - L\rvert < \epsilon/2$. Thus $\lvert s_n - L\rvert \le \lvert s_n - s_{n_N}\rvert + \lvert s_{n_N} - L\rvert < \epsilon/2 + \epsilon/2 = \epsilon$.

Similar Threads

  1. The proof of local convergence of the following function...
    By majidyusefi in forum Advanced Applied Mathematics
    Replies: 2
    Last Post: August 10th, 2014, 13:31
  2. Pointwise convergence implies uniform convergence
    By Siron in forum Topology and Advanced Geometry
    Replies: 1
    Last Post: January 29th, 2014, 16:22
  3. Absolute convergence proof
    By Fermat in forum Analysis
    Replies: 0
    Last Post: January 13th, 2014, 07:39
  4. Proof of convergence theory in optimization
    By ianchenmu in forum Advanced Applied Mathematics
    Replies: 0
    Last Post: April 16th, 2013, 18:10
  5. Proof the convergence of a gamma sum
    By ZaidAlyafey in forum Analysis
    Replies: 5
    Last Post: February 5th, 2013, 08:01

Tags for this Thread

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •  
Math Help Boards