Results 1 to 4 of 4

Math Help - Cauchy Sequence in a Metric Space

  1. #1
    Newbie
    Joined
    Nov 2009
    Posts
    4

    Exclamation Cauchy Sequence in a Metric Space

    Problem:

    Let { x_{i}} be a Cauchy sequence in a metric space (M,D). Let A={ x_{1}, x_{2}, x_{3}, ...}.
    Suppose that { x_{i}} doesn't converge in M. Prove that A is a closed subset of (M,D).

    What I have done:

    So we know { x_{i}} is Cauchy, so let \epsilon>0 be given. Then there exists a positive integer N such that the D( x_{i}, x_{j})< \epsilon for all i,j \geq N. Also, and this may be where I'm stuck.. Since { x_{i}} doesn't converge in M, then we know that for any y \inM, \epsilon>0, and N, we can find an n>N such that D( x_{n}, y)> \epsilon.

    That last statement could be wrong/unnecessary, so any help you can give me on where to go would be very helpful.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by tidus89 View Post
    Problem:

    Let { x_{i}} be a Cauchy sequence in a metric space (M,D). Let A={ x_{1}, x_{2}, x_{3}, ...}.
    Suppose that { x_{i}} doesn't converge in M. Prove that A is a closed subset of (M,D).

    What I have done:

    So we know { x_{i}} is Cauchy, so let \epsilon>0 be given. Then there exists a positive integer N such that the D( x_{i}, x_{j})< \epsilon for all i,j \geq N. Also, and this may be where I'm stuck.. Since { x_{i}} doesn't converge in M, then we know that for any y \inM, \epsilon>0, and N, we can find an n>N such that D( x_{n}, y)> \epsilon.

    That last statement could be wrong/unnecessary, so any help you can give me on where to go would be very helpful.
    Maybe the best way to do this is an argument by contradiction. Suppose that y is in the closure of A but not in A. Then there must be a sequence in A (in other words, a subsequence of \{x_i\}) converging to y. But if a subsequence of a Cauchy sequence converges, then the whole sequence converges. There's your contradiction.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Nov 2009
    Posts
    4
    So are we supposing that A is open, and then the contradiction comes from the fact that the sequence would then have to converge in M?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by tidus89 View Post
    So are we supposing that A is open, and then the contradiction comes from the fact that the sequence would then have to converge in M?
    No, we are not supposing that A is open, we are supposing that A is not closed. That means that there is a point in the closure of A that is not in A.

    Yes, the contradiction comes from the fact that we then prove that the sequence converges, contrary to the statement of the question.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Cauchy Sequence in Metric Space
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: October 8th 2011, 10:29 AM
  2. All Cauchy Sequences are bounded in a metric space
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: February 1st 2011, 04:18 PM
  3. Cauchy sequence convergence in a general metric space
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: November 29th 2010, 10:31 PM
  4. metric space: convergent sequence
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 11th 2009, 12:40 PM
  5. Cauchy Sequence in metric space problem
    Posted in the Calculus Forum
    Replies: 5
    Last Post: September 17th 2007, 11:40 AM

Search Tags


/mathhelpforum @mathhelpforum