1. Limited time only! Sign up for a free 30min personal 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!

Cauchy sequence problem

  1. Jun 5, 2013 #1
    1. The problem statement, all variables and given/known data
    Let (M,d) be a complete metric space and define a sequence of non empty sets F1[itex]\supseteq[/itex]F2[itex]\supseteq[/itex]F3[itex]\supseteq[/itex] such that diam(Fn)->0, where diam(Fn)=sup(d(x,y),x,y[itex]\in[/itex]Fn). Show that there [itex]\bigcap[/itex]n=1Fn is nonempty (contains one element).


    2. Relevant equations



    3. The attempt at a solution
    We wonna use the completeness of M somehow. Let (xn) be a sequence of elements such that xn[itex]\in[/itex]Fn. Then as diam(Fn)->0 we must have for a specific N that lxn - xml < ε for all m,n>N. Thus the sequence of (xn) is a Cauchy sequence and must be convergent in M due to the assumed completeness. Denote the limit by x. We must show that x[itex]\in[/itex]Fn for all n. But I am unsure how to this. And is this even the right approach? I don't have a lot of experience with proofs.
     
  2. jcsd
  3. Jun 5, 2013 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You are almost there. Now use that the sets Fn are closed. What does closed mean in terms of sequences?
     
  4. Jun 5, 2013 #3
    I dont know what you are referring to sorry.
    The only thing I can come up with is something like: Assume x is not a member of all Fn. Then we can pick diam(Fn)<ε and lx-xnl ≥ ε. But that is a contradiction. But I don't know if that is the right way to do it.
    What did you mean by closed in terms of sequences?
     
  5. Jun 5, 2013 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    If a set F is closed and {xn} is a sequence in F that converges to x, then x is also in F. That's what I mean.
     
  6. Jun 5, 2013 #5
    how do you prove that?
     
  7. Jun 5, 2013 #6

    LCKurtz

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    @aaa202 Your theorem is false as stated. Consider ##F_n = (0,\frac 1 n)##. Given the discussion so far, haven't you noticed that you haven't assumed ##F_n## is closed?
     
  8. Jun 5, 2013 #7
    oops I forgot to say they were haha. But you are right I didn't say it and assumed it all along.
     
    Last edited: Jun 5, 2013
  9. Jun 5, 2013 #8

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    How did you define closed and open?
     
  10. Jun 5, 2013 #9
    Something like: An element is said to be on the boarder (I dont know the appropriate term in english) in a set if each sphere around it contains at least one element of the set. A set is closed if it contains all its boarder elements. Im sorry if this is not well translated - i hope you can understand.
     
  11. Jun 5, 2013 #10

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Approximately right word, but it's spelled 'border'. 'boarder' is something else. In the sequence argument, isn't x on the border of F? And the F's being closed is so important, I guess I didn't even notice the problem statement didn't say that.
     
  12. Jun 5, 2013 #11
    Maybe? what makes you say that. I don't know
     
  13. Jun 5, 2013 #12

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Every neighborhood of x contains points in F. Use that a sequence of points in F converges to x.
     
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: Cauchy sequence problem
  1. Cauchy Sequence (Replies: 10)

  2. Cauchy Sequence (Replies: 9)

  3. Cauchy sequence (Replies: 2)

Loading...