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Metric space question

  1. Dec 4, 2011 #1
    1. The problem statement, all variables and given/known data

    Consider a metric space (X,d) with subsets A and B of X, where A and B have non-zero intersection. Show that diam(A[itex]\bigcup[/itex]B) [itex]\leq[/itex] diam(A) + diam(B)

    2. Relevant equations



    3. The attempt at a solution

    A hint would be very much appreciated. :smile:


    Let x[itex]\in[/itex]A, y[itex]\in[/itex]B, z[itex]\in[/itex]A[itex]\bigcup[/itex]B

    diam(A[itex]\bigcup[/itex]B) = sup[d(x,y)] for every x,y.
    So d(x,y) [itex]\leq[/itex] diam(A[itex]\bigcup[/itex]B)

    Either z[itex]\in[/itex]A, or z[itex]\in[/itex]B, or z in both.

    Consider firstly z[itex]\in[/itex]A.
    d(x,z) [itex]\leq[/itex] diam(A)

    If z[itex]\in[/itex]B,
    d(z,y) [itex]\leq[/itex] diam(B)

    Therefore, d(x,z) + d(z,y) [itex]\leq[/itex] diam(A) + diam(B)

    But d(x,y) [itex]\leq[/itex] d(x,z) + d(z,y), hence
    d(x,y) [itex]\leq[/itex] diam(A) + diam(B)


    As far as I've gotten...
     
  2. jcsd
  3. Dec 4, 2011 #2

    hunt_mat

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    Examine a point in [itex]z\in A\cap B[/itex] and then examine the distance from [itex]x\in A[/itex] and [itex]b\in B[/itex] and examine the distance from x to y.and apply the definition of diameter.
     
  4. Dec 4, 2011 #3
    Wait, I have these two results

    d(x,y) ≤ diam(A⋃B)

    d(x,y) ≤ diam(A) + diam(B)

    The latter holds for all x,y, and the maximum value of d(x,y) is diam(A⋃B), which still has to be less than or equal to diam(A) + diam(B).

    So, diam(A⋃B) ≤ diam(A) + diam(B)


    Is my reasoning correct?
     
  5. Dec 4, 2011 #4

    hunt_mat

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    I think so.
     
  6. Dec 4, 2011 #5

    Dick

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    That doesn't sound right to me. Where did you get d(x,y)<=diam(A)+diam(B)? Isn't that what you are trying to prove? And where did you use that the intersection of A and B is nonempty? The result is false if that isn't true.
     
  7. Dec 4, 2011 #6
    I showed it in my first post, using the triangle inequality. And no, I'm trying to prove that
    diam(A[itex]\bigcup[/itex]B) [itex]\leq[/itex] diam(A) + diam(B)

    As for using the fact that the intersection is non-empty, if it was empty then the inequality
    d(x,y) [itex]\leq[/itex] diam(A) + diam(B) is not necessarily true, but it is non-empty, so it is correct.
     
  8. Dec 4, 2011 #7

    Dick

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    Ah, indeed you did. Sorry, I didn't read the initial post thoroughly enough.
     
  9. Dec 4, 2011 #8
    No problem, so is my proof correct?
     
  10. Dec 4, 2011 #9

    hunt_mat

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    Suppose [itex]z\in A\cap B[/itex], [itex]x\in A[/itex] and [itex]y\in B[/itex], the from the triangle inequality:
    [tex]
    d(x,y)\leqslant d(x,z)+d(z,y)
    [/tex]
    we also know that [itex]d(x,z)\leqslant\textrm{diam}(A)[/itex] and [itex]d(z,y)\leqslant\textrm{diam}(B)[/itex], so we just put these together.
     
  11. Dec 4, 2011 #10

    Dick

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    I think you need to be a little more precise. diam(AUB) is the LEAST UPPER BOUND for d(x,y). diam(A)+diam(B) is another upper bound. Just saying they are both upper bounds won't do it. I think that's what you were trying to say with "the maximum value of d(x,y) is diam(A⋃B)" but that's not quite right. There may be no x and y with d(x,y)=diam(AUB).
     
  12. Dec 4, 2011 #11

    I like Serena

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    I'm afraid you did not apply the triangle property correctly in your opening post.

    If you assume that z is in A, then
    d(x,z) ≤ diam(A),
    but you can not assume that d(z,y) ≤ diam(B).

    So you cannot conclude that d(x,z)+d(z,y) ≤ diam(A)+diam(B)


    To get your triangle inequality straight you need to use what Dick suggested.
     
  13. Dec 4, 2011 #12

    Dick

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    Ach, I'm starting mix up who wrote what. But hunt_mat gave the correct argument. I think maybe Maybe_Memory meant to as well but muddled up the cases a bit and didn't mention the triangle inequality.
     
    Last edited: Dec 4, 2011
  14. Dec 4, 2011 #13
    I'm assuming z is in A, and getting d(x,z) ≤ diam(A)
    Then considering the case where z is in B, and getting d(z,y) ≤ diam(B).

    Then adding the results.
     
  15. Dec 4, 2011 #14

    hunt_mat

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    You're considering two different possibilities and you need both possibilities to be true at the same time, you can only do that if z is in the intersection of A and B.
     
  16. Dec 4, 2011 #15

    I like Serena

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    You cannot combine 2 choices for z in 1 equation.
     
  17. Dec 4, 2011 #16
    Aaah! Right, I see the problem! So as hunt_mat said

    I understand this now.

    As for the issue with combining my results to get the desired expression;

    d(x,y) [itex]\leq[/itex] diam(A[itex]\bigcup[/itex]B)

    d(x,y) [itex]\leq[/itex] diam(A) + diam(B)

    diam(A[itex]\bigcup[/itex]B) is the least upper bound for d(x,y) and is contained in d(x,y) so d(x,y) can take the value diam(A[itex]\bigcup[/itex]B)
    so diam(A[itex]\bigcup[/itex]B) [itex]\leq[/itex] diam(A) + diam(B)
     
  18. Dec 4, 2011 #17

    I like Serena

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    You have the right idea, but I'm afraid it's not a proper proof.

    First off, you need to split the possibilities for x and y into 4 cases, which can be reduced to 2 cases: x and y in the same set, or x and y in different sets.

    Furthermore, there is no guarantee that d(x,y) can take the value diam(A[itex]\cup[/itex]B).
    What you can do, is start with your triangle inequality for x,y, and z.
    And then take the sup on both sides, yielding the diameters.
     
  19. Dec 5, 2011 #18
    Okay, if I take d(x,y) [itex]\leq[/itex] d(x,z) + d(z,y), take the sup on both sides,
    sup [d(x,y)] [itex]\leq[/itex] sup [d(x,z) + d(z,y)]

    Is it correct to say sup [d(x,z) + d(z,y)] [itex]\leq[/itex] sup [d(x,z)] + sup[d(z,y)]
    hence sup [d(x,y)] [itex]\leq[/itex] sup [d(x,z)] + sup[d(z,y)] which is what i need?
     
  20. Dec 5, 2011 #19

    I like Serena

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    It's still slightly more subtle.
    What you say is correct btw, but it is not sufficient.

    First we have to assume that sup[d(x,y)]=diam(AuB) for x in A and y in B.
    (You have to prove the other cases separately.)
    Then you can say what you said with an arbitrary z in AnB.
    Then you can conclude that sup[d(x,z)] ≤ diam(A) and sup[d(z,y)] ≤ diam(B).
    And from this follows your inequality (for this case).
     
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