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Prove the open ball B(x,r) is a subset of B(x',r') iff d(x,x')<r'-r

  1. May 21, 2013 #1
    As stated, the problem is to prove B(x,r) is a subset of B(x',r') if and only if d(x,x')<r'-r, where d is a measure function. The proof going one direction seems pretty simple, but I can not figure out the other direction

    Attempt at answer

    <=

    Assume d(x,x') < r'-r.
    Assume a is an element of B(x,r).
    Then d(x,x') < r'-r implies d(x,x') + r < r' implies d(x,x') + d(x,a) < r' as d(x,a) < r.
    This implies that d(x',a) < r' by use of the triangle inequality, and thus a is an element of B(x',r').

    =>

    Let a be an element of B(x,r).
    Then a is an element of B(x',r').
    So d(x,a)<r and d(x',a)<r'.
    This implies that d(x',a)-d(x,a) < r'-r

    This would almost suggest a use of the reverse triangle inequality, but the inequality is pointing the wrong direction to use here. I am not sure how to proceed.
     
  2. jcsd
  3. May 21, 2013 #2

    Fredrik

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    Since this is a textbook-style problem, I'm moving the thread to the calculus & beyond homework forum. You should post questions about textbook-style questions there, and questions about concepts, definitions and theorems in the topology & analysis forum.

    Your proof of the first implication is fine, but can be made a bit prettier:

    Suppose that ##d(x,x')<r'-r##. Let ##a\in B(x,r)## be arbitrary. Then
    $$d(a,x')\leq d(a,x)+d(x,x')<r+(r'-r)=r'.$$ To prove the other implication, use that the point at the center of an open ball is an element of the open ball.
     
  4. May 21, 2013 #3
    Sorry about putting the thread in the wrong place. I will be sure to put it in the right place next time.

    Ok, so since B(x,r) is a subset of B(x',r'), we know that x is an element of B(x',r').
    This implies that d(x,x') < r'.

    But we need d(x,x') < r' - r

    I'm not exactly sure how to proceed from that point. Intuitively, it seems like we want to use the point in B(x,r) that is farthest from x'. However, I have no idea how to find that point.
     
  5. May 21, 2013 #4

    Fredrik

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    You're right, that was a mistake on my part. I forgot what we were trying to prove, exactly as your post suggests. Unfortunately I don't have time to think this through right now. I'll have another look tomorrow if this isn't solved by then.
     
  6. May 21, 2013 #5
    No sweat. Thank you for the help so far!
     
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