Results 1 to 5 of 5
Like Tree1Thanks
  • 1 Post By HallsofIvy

Thread: How do I complete this convergence proof?

  1. #1
    MHF Contributor alexmahone's Avatar
    Joined
    Oct 2008
    Posts
    1,111
    Thanks
    7

    How do I complete this convergence proof?

    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 $n\ge N$?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Apr 2005
    Posts
    18,587
    Thanks
    2596

    Re: How do I complete this convergence proof?

    Take N equal to the larger of N_1 and N_2.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor alexmahone's Avatar
    Joined
    Oct 2008
    Posts
    1,111
    Thanks
    7

    Re: How do I complete this convergence proof?

    Quote Originally Posted by HallsofIvy View Post
    Take N equal to the larger of N_1 and N_2.
    Why would that imply that $n_k\ge N_1$ and $k\ge N_2$?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    20,491
    Thanks
    2330
    Awards
    1

    Re: How do I complete this convergence proof?

    Quote Originally Posted by alexmahone View Post
    Why would that imply that $n_k\ge N_1$ and $k\ge N_2$?
    Here is another way. $N_1+N_2$ is an integer greater than either $N_1\text{ or }N_2$.
    So let $N=N_1+N_2$.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Apr 2005
    Posts
    18,587
    Thanks
    2596

    Re: How do I complete this convergence proof?

    Quote Originally Posted by alexmahone View Post
    Why would that imply that $n_k\ge N_1$ and $k\ge N_2$?
    Given that N is the larger of N_1 and N_2, N\ge N_1 and N\ge N_2. So if n_k\ge N we have n_k\ge N\ge N_1. And if k\ge N we have k\ge N\ge N_2.
    Thanks from alexmahone
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. NP-complete Proof
    Posted in the Discrete Math Forum
    Replies: 0
    Last Post: May 27th 2015, 10:02 PM
  2. Infinite Primes Proof is complete ?
    Posted in the Number Theory Forum
    Replies: 3
    Last Post: Nov 2nd 2009, 05:49 AM
  3. Complete Induction proof
    Posted in the Discrete Math Forum
    Replies: 2
    Last Post: Oct 22nd 2009, 07:41 AM
  4. Complete Interval of Convergence Question(s)
    Posted in the Calculus Forum
    Replies: 10
    Last Post: Apr 16th 2008, 09:03 PM
  5. proof by complete induction
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Sep 25th 2007, 02:09 PM

/mathhelpforum @mathhelpforum