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!

Bounded Subsets of a Metric Space

  1. Feb 5, 2013 #1
    1. The problem statement, all variables and given/known data

    Let X be a metric space and let E be a subset of X. Show that E is bounded if and only if there exists M>0 s.t. for all p,q in E, we have d(p,q)<M.

    2. Relevant equations

    Use the definition of bounded which states that a subset E of a metric space X is bounded if there exists a q in X and an M>0 s.t. d(p,q)<M for all p in E.

    3. The attempt at a solution

    Proof (so far):
    Let E be bounded. Then there exists M/2>0 and x in X s.t. d(p,x)<M/2 for all p in E.
    Now, take arbitrary p,q in E and observe that:
    d(p,q) ≤ d(p,x)+d(x,q) < M/2+M/2= M
    Thus, d(p,q)<M for all p,q in E.

    Now, I'm getting hung up on the second part of the proof, but I feel as if it shouldn't be hard. I think I've just been staring at this for too long at this point. Any advice as to how I ought to start going in the opposite direction would be greatly appreciated. I considered trying to show that E is compact (and would therefore be closed and, more importantly, bounded), but I'm not sure that's the best route or if it's awfully easy to do.

    Last edited: Feb 5, 2013
  2. jcsd
  3. Feb 5, 2013 #2
    A bounded set is not compact in general. So you won't be able to show that E is compact.

    Try to pick an arbitrary element p in E. What is the distance from p to the other points in E?
  4. Feb 5, 2013 #3
    You're right, not sure what I was thinking there.

    Yes, as I looked at it again I just reached that conclusion before I saw your response. I suppose I overlooked that because it's so obvious!

    Is it as simple as picking any q in E? Then q is in X and d(p,q)<M for all p in E.

  5. Feb 5, 2013 #4
    Doesn't work if E is empty though, so you want to treat that in a special case.
  6. Feb 5, 2013 #5
    Ah, of course.

    Thanks for your help!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Bounded Subsets of a Metric Space