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Prove that if A is contained in some closed ball, then A is bounded.

  1. Sep 3, 2012 #1
    1. The problem statement, all variables and given/known data
    Let M be a metric space and A[itex]\subseteq[/itex]M be any subset:
    Prove that if A is contained in some closed ball, then A is bounded.


    2. Relevant equations
    Def of closed-ball: [itex]\bar{B}[/itex]R(x) = {y[itex]\in[/itex]M:d(x,y)≤R} for some R>0
    Def of bounded: A is bounded if [itex]\exists[/itex]R>0 s.t. d(x,y)≤R [itex]\forall[/itex]x,y[itex]\in[/itex]A
    Empty set is defined to be bounded in for these problems

    3. The attempt at a solution
    Spse that A is contained in some closed ball.
    Let that ball be [itex]\bar{B}[/itex]R(y0) = {y[itex]\in[/itex]M:d(y,y0)≤R}, for some arbitrary fixed y0
    1) If A=[itex]\phi[/itex], vacuously true.
    2)Let A[itex]\subseteq[/itex][itex]\bar{B}[/itex]R(y0)
    Let x1,x2[itex]\in[/itex]A.
    The d(x1,x2)≤d(x1,y0) + d(x2,y0)≤2R.
    Thus, diam(A)≤2R and we see that A is bounded.
    Q.E.D.
     
    Last edited: Sep 3, 2012
  2. jcsd
  3. Sep 3, 2012 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    So this is homework but you don't know any relevant formulas or facts? And you haven't made any attempt to do the problem yourself? Are you sure you are taking this class!?

    What is the definition of "bounded set"? What is the definition of "closed ball"? If you know those, then this should be relatively easy. If you don't, nothing I say will help you until you have looked up those definitions (different texts may have slightly different definitions- you need to know definitions you are to use.
     
  4. Sep 3, 2012 #3
    Sorry, I accidentally pressed "Enter" after the title lol. Please, see my attempt. I'm pretty sure it is correct; but, I want to make sure that I don't have to say anymore or elaborate on the empty set bit.

    Thank you!
     
  5. Sep 3, 2012 #4
    Yes, your proof is correct. I don't think you need to elaborate on the empty set case.
     
  6. Sep 3, 2012 #5
    Thank you very much! I appreciate your help!
     
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