Results 1 to 2 of 2

Math Help - under standing metric spaces and cauchy sequences... (analysis/topology)

  1. #1
    Member
    Joined
    Apr 2008
    From
    Seoul, South Korea
    Posts
    128

    under standing metric spaces and cauchy sequences... (analysis/topology)

    hmm so i have this metric defined by x,y in R^k such that d_1(x,y)=max{abs (x_j-y_j):j=1,2,...,k} and d_2(x,y)= sigma (from j=1 to k) abs (x_j-y_j). i did the first part of the question which was to verify that these are indeed metrics for R^k. the second part of the question asks to prove that d_1 and d_2 are complete. from the book definition, the metric space (S,d) is said to be complete if every cauchy sequence in S converges to some element in S. the hint given is to consider d_1(x,y)<=d(x,y)<=sqrt(k)d_1(x,y) where d is the standard metric defined by d(x,y)=sqrt (sigma from j=1 to k) (x_j-y_j)^2. can i get some help? thanks !
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by squarerootof2 View Post
    hmm so i have this metric defined by x,y in R^k such that d_1(x,y)=max{abs (x_j-y_j):j=1,2,...,k} and d_2(x,y)= sigma (from j=1 to k) abs (x_j-y_j). i did the first part of the question which was to verify that these are indeed metrics for R^k. the second part of the question asks to prove that d_1 and d_2 are complete. from the book definition, the metric space (S,d) is said to be complete if every cauchy sequence in S converges to some element in S. the hint given is to consider d_1(x,y)<=d(x,y)<=sqrt(k)d_1(x,y) where d is the standard metric defined by d(x,y)=sqrt (sigma from j=1 to k) (x_j-y_j)^2. can i get some help? thanks !
    1) d_1 is complete. Let \{ \bold{x}_n \} be a Cauchy sequence, so it means, d(\bold{x}_n , \bold{x}_m) < \epsilon whenever n,m>N. Thus, \max \{ |x^{(i)}_n - x^{(i)}_m | \} < \epsilon. This means that, | \bold{x}_n - \bold{x}_m| < \sqrt{k}|\max\{ |x^{(i)}_n - x^{(i)}_m| \} < \epsilon \sqrt{k}. Thus, we see that if \{ \bold{x}_n\} is a Cauchy with respect to d_1 then it is Cauchy with respect to the ordinary Euclidean metric. But Euclidean metric spaces are complete, thus d_1 is also complete.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. All Cauchy Sequences are bounded in a metric space
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: February 1st 2011, 03:18 PM
  2. [SOLVED] Non-convergent sequences in metric spaces
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: September 22nd 2010, 11:34 AM
  3. Metric Spaces and Topology
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: October 13th 2009, 05:59 PM
  4. topology/metric spaces questions
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 26th 2008, 03:38 AM
  5. Metric Spaces/Topology help
    Posted in the Advanced Math Topics Forum
    Replies: 2
    Last Post: November 25th 2008, 11:49 PM

Search Tags


/mathhelpforum @mathhelpforum