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Proof of equality of diameter of a set and its closure

  1. Mar 24, 2014 #1
    In showing diam(cl(A)) ≤ diam(A), (cl(A)=closure of A) one method of proof* involves letting x,y be points in cl(A) and saying that for any radius r>0, balls B(x,r) and B(y,r) exist such that the balls intersect with A.

    But if x,y is in cl(A), isn't there the possibility that x,y are isolated points? Shouldn't we let x,y be in the set of limit points of A?

    But doing that, I get (a is in the intersection of B(x,r) and A, b is in the intersection of B(y,r) and A).

    d(x,y)≤d(x,a)+d(a,b)+d(b,y)<2r+d(a,b)
    Then taking the supremum of both sides
    ∴ sup{d(x,y): x,y in the set of limit points of A} ≤ 2r+sup{d(a,b): a,b in A}
    ∴ sup{d(x,y): x,y in the set of limit points of A} ≤ 2r+diam(A)

    The problem being the left hand side doesn't become the definition of diam(cl(A))


    *https://www.physicsforums.com/showthread.php?t=416201 (post #6)
     
  2. jcsd
  3. Mar 24, 2014 #2
    Sure, ##x## and ##y## can be isolated points. But then you know that ##x,y\in A## for sure. So ##B(x,r)## and ##B(y,r)## intersect ##A##. Hence the proof continues like usual.
     
  4. Mar 29, 2014 #3
    Ah, thanks micromass. If x and y were assumed to be isolated points it would just follow that diam(A)=diam(cl(A)) cause they're in A right?
     
  5. Mar 29, 2014 #4
    Yeah, clearly the only problem is with points which are in ##\mathrm{cl}(A)## but not in ##A##.
     
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