Results 1 to 10 of 10

Math Help - Subsequence of a Cauchy Sequence is Cauchy

  1. #1
    Junior Member
    Joined
    Oct 2009
    Posts
    72

    Subsequence of a Cauchy Sequence is Cauchy

    Show that a subsequence of a Cauchy sequence is a Cauchy sequence. The catch is that you can't use the fact that every Cauchy subsequence is convergent.

    I really don't know where to begin without assuming what the problem says I can't assume.

    Any direction would be appreciated.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by magus View Post
    Show that a subsequence of a Cauchy sequence is a Cauchy sequence. The catch is that you can't use the fact that every Cauchy subsequence is convergent.

    I really don't know where to begin without assuming what the problem says I can't assume.

    Any direction would be appreciated.

    Let \{x_n\}\,,\,\,\{x_{n_k}\} be the Cauchy seq. and one of its subsequences, and let \epsilon>0

    1) There exists M_\epsilon\in\mathbb{N}\,\,s.t.\,\,n,m>M_\epsilon\  Longrightarrow |x_n-x_m|<\epsilon

    2) By the very definition of "infinite subsequence", there exist K\in\mathbb{N}\,\,s.t.\,\,\forall k>K\,,\,n_k>M_\epsilon

    Now end the proof.

    Tonio
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Oct 2009
    Posts
    72
    Thanks. Doesn't that nearly end the proof? Not that I'm mad at you or anything for taking me that far :P

    I can't think of anything after that but then going right to |x_{n_k}-x_m|\leq \epsilon which completes the proof.

    Is this a valid step or do I have more work to do?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1574
    Awards
    1
    The ‘trick’ comes from the nature of subsequences and the definition of Cauchy Sequences.

    If \left(x_{n_k}\right) is a subsequence of  \left(x_n\right) then  \left(n_k\right)\ge k .

    So if k~\&~j~\ge N then n_k~\&~n_j~\ge N.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Oct 2009
    Posts
    72
    Two questions though. Doesn't that statement near complete the proof and also can k~\&~j~\ge N be the same N as in n_k~\&~n_j~\ge N?

    It would seem to me that the subsub scripts would need to be greater then some other number. Maybe I'm being too nitpicky but just to be sure I'm getting what you two are saying I'm going to write it all out.

    We start with the known that {s_n} is convergent and therefore a Cauchy sequence. Then we know from that that there exists some m,n \ge N such that |s_n-s_m|\le \epsilon.


    We then choose some j,k\le Msuch that the subsequences of {s_n} s_{n_j} and s_{m_k} such that n_j, n_k\ge N

    We're then want to prove that s_{n_j}-s_{m_k} \le \epsilon

    since {s_n} is convergent it is bounded and therefore so is s_{n_j} and s_{m_k}.

    If we choose a large enough N the sequences will be bounded into smaller and smaller intervals because they are bound by the sequences bounds. The process of tightening this window is by choosing larger values of N and therefore forcing larger values of M. This is exactly the Cauchy criterion therefore the subsequence of a Cauchy sequence is a Cauchy sequence.


    Does that fly?
    Last edited by magus; September 27th 2010 at 02:14 PM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1574
    Awards
    1
    Quote Originally Posted by magus View Post
    Two questions though. Doesn't that statement near complete the proof and also can k~\&~j~\ge N be the same N as in n_k~\&~n_j~\ge N? It would seem to me that the subsub scripts would need to be greater then some other number.
    The point is: that for Cauchy Sequences it is true that for each  \varepsilon  > 0 there is one N that works for k\ge N~\&~j\ge N then  \left| {x_k  - x_j } \right| < \varepsilon .

    I am puzzled by this question.

    If the sequence \left( {x_n } \right) converges then each subsequence \left( {x_{n_k} } \right) converges to the same limit.

    Moreover, any convergent sequence is a Cauchy Sequence.
    So any subsequence of a Cauchy Sequence must be a Cauchy Sequence.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Oct 2009
    Posts
    72
    Quote Originally Posted by Plato View Post
    Moreover, any convergent sequence is a Cauchy Sequence.
    So any subsequence of a Cauchy Sequence must be a Cauchy Sequence.
    Right but for some reason they don't want me to use that information even though it seems the simplist way to prove it.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1574
    Awards
    1
    Nevertheless, do you understand why for each  \varepsilon  > 0 one N works?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Oct 2009
    Posts
    72
    Yes I do now. Thank you all for all your help.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    Feb 2010
    Posts
    15
    Quote Originally Posted by magus View Post
    Right but for some reason they don't want me to use that information even though it seems the simplist way to prove it.
    Maybe because a Cauchy sequence is not necessarily convergent unless it is defined in a complete metric space.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. cauchy subsequence question
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: October 21st 2011, 04:25 PM
  2. cauchy sequence
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: November 1st 2010, 05:40 AM
  3. Cauchy sequence w/ unbounded subsequence
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: November 13th 2009, 03:58 PM
  4. Cauchy subsequence and a limit
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 8th 2007, 02:44 PM
  5. Cauchy sequence
    Posted in the Calculus Forum
    Replies: 5
    Last Post: October 24th 2006, 01:03 PM

/mathhelpforum @mathhelpforum