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Diameter of a union of metric spaces

  1. Jan 27, 2012 #1
    1. The problem statement, all variables and given/known data
    suppose that a metric space A is a union A = B U C of two subsets of finite diameter. Prove A has finite diameter.


    2. Relevant equations

    The Diameter of a metric space M is sup D(a,b) for all a,b in M.

    3. The attempt at a solution
    Really, no idea where to begin. I just began this subject, and I'm not too well versed in thinking of what these objects really are. If I'm understanding this right, I can imagine having say two disjoint rectangles that are far apart, and these are my B and C, with A being the union of them. Then each B and C may have a diameter, which represents the maximum distance between any two points in them, and part of me wants to take the farthest point of one and connect it to the other, but that seems like that wouldn't always work.

    The only thing I can think of is that since both sets have finite diameter, then neither set is infinite. Therefore there must exist a set of points that are the farthest apart...

    But like I said, I'm new to this subject (metric spaces) and have no idea where to begin. Any help would be greatly appreciated. (In case it matters, I haven't studied topology, and balls begin in the next section (although I have learned a little about neighborhoods, not sure if that's the same thing)
     
  2. jcsd
  3. Jan 27, 2012 #2
  4. Feb 1, 2012 #3
    No, I'm sorry. This stuff is completely foreign to me. I have no idea how I can show this.
     
  5. Feb 2, 2012 #4

    jbunniii

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    I suggest that you start by picking two reference points, say b and c, where [itex]b \in B[/itex] and [itex]c \in C[/itex]. Being points in a metric space, there is by definition some finite distance, say d(b,c), between them.

    Now take arbitrary points x and y in [itex]B \cup C[/itex]. See if you can find an upper bound for d(x,y) that doesn't depend on x and y.
     
  6. Feb 2, 2012 #5
    You've got the solution in the thread I sent you! It's shwon in problem 3.
     
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