1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Complete metric space proof

  1. Dec 27, 2016 #1
    1. The problem statement, all variables and given/known data
    Let ##E## be a metric subspace to ##M##. Show that ##E## is closed in ##M## if ##E## is complete. Show the converse if ##M## is complete.

    2. Relevant equations
    A set ##E## is closed if every limit point is part of ##E##.
    We denote the set of all limit points ##E'##.

    A point in ##p\in M## is a limit point to ##E\subseteq M## if ##\forall \epsilon > 0## ##\exists q \in E \cap B(p,\epsilon)##

    3. The attempt at a solution
    We want to show that ##E' \subseteq E##. Take ##p \in E'## then clearly we can choose
    ##p_n \in B(p,1/n)## so that ##p_n \in E##.
    But then for all ##\epsilon > 0## ##d(p_n,p_m)\le d(p_n,p)+ d(p_m,p)<\epsilon## for ##n,m \ge N_\epsilon = 2/\epsilon## i.e. ##(p_n)## is a cauchy sequence.

    But then it must converge to some ##\lambda \in E##. However ##\lambda = p## since ##d(p,\lambda) = d(p_n,p)+d(p_n,\lambda)< \epsilon##

    Is this a correct proof? Should I use a similar approach to the second part?
     
  2. jcsd
  3. Dec 27, 2016 #2

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    Can you elaborate?
     
  4. Dec 27, 2016 #3
    I tried to expand the argument a bit, hopefully this works better:

    We have ##d(p_n,p)<1/n## where ##p_n \in E##. By the triangle inequality we have
    ##d(p_n,p_m) \le d(p_n,p)+d(p_m,p) < 1/n+1/m \le 2/N_\epsilon = \epsilon## for ##n,m\ge N_\epsilon##.
    So ##\forall \epsilon > 0## we have ##d(p_n,p_m) < \epsilon## for ##n,m \ge N_\epsilon = 2/\epsilon##.

    Btw. I'm unclear about if using ##1/n## is neccesary at all. It seemed like a good idea at the time but perhaps starting from ##d(p_n,p)< \epsilon/2## we get ##n,m## can take any values in ##Z^+## instead,.
     
  5. Dec 27, 2016 #4

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    Seems ok. The only minor concern is that ##2/\varepsilon## is not necessarily an integer, so you might not be able to put it equal to ##N_\varepsilon##.

    I think it is definitely necessary, I don't see how you can do without.
     
  6. Dec 28, 2016 #5
    So choosing ##N_\epsilon = floor(1+2/\epsilon)## should do it.

    Thanks, thats good to know!


    Moving on to the next part I take it I'm meant to show that
    ##M## complete, ##E## closed ##\Longrightarrow ## ##E## complete.

    Let's consider an arbitrary CS ##(p_n)## in ##E##. Since ##M## is complete it converges to a point ##p\in M##.
    This means that ##\forall \epsilon > 0## ##\exists N## so that ##d(p_n,p)< \epsilon## ##\forall n \ge N##.
    If ##p_n = p## for some ##n## we have ##p\in E##. If ##p_n \ne p## it means that ##p## is a limit point of ##E## since ##p_n \in E## and ##\epsilon## is arbitrary small. And since ##E' \subseteq E## since ##E## is closed we have that ##E## is complete.
     
  7. Dec 28, 2016 #6

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    Seems ok!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Complete metric space proof
  1. COMPLETE metric space (Replies: 28)

Loading...