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Diameter of Union of two sets

  1. Jun 26, 2011 #1
    Hi, I am stuck with the following proofs. In metric space
    here, A,B,C are subset of metric space (X,d) and C is bounded

    Problem 1.) d(A,B) <=d(A,C)+d(B,C)+diam(C)
    Problem 2.)|d(b,A)-d(c,A)| <= d(b,c) where 'b' belongs to 'B' and 'c' belongs to 'C'.

    Problem 3)- diam(A U B)<= diam A+ diam B+ d(A,B) ,Here d(A,B)= inf {d(a,b)| a belong to A and b to B}

    I think if i get some clue even about one i can handle other one.

    Thanks in advance..:)
  2. jcsd
  3. Jun 26, 2011 #2
    Hi nirajkadiyan6! :smile:

    For the first one. Take [itex]a\in A,b\in B, c\in C,c^\prime \in C[/itex] arbitrary. It suffices to show

    [tex]d(A,B)\leq (a,c)+d(b,c^\prime)+d(c,c^\prime)[/tex]

    Can you show that?
  4. Jun 26, 2011 #3
    This was the thing i proved in very first attempt but it is not helping me
    This can be proved easily but, for arbitrary a and c, d(a,c) can't be simply written is d(A,C) as d(A,C) is smallest distance between some a',c'
    so d(a,c)< d(A,C) for arbitrary a,c
    Same holds for d(A,B)
  5. Jun 26, 2011 #4
    Do you mean me to take c and c' to be those elements of C which makes d(A,C) and d(B,C) shortest.

    If it is the case then i am done with my problem.
    Can you suggest something about others.
  6. Jun 26, 2011 #5
    No, such a c and c' won't exist in general. But if you show the inequality in my post for all c and c', then you've shown it for the infimum too.
  7. Jun 26, 2011 #6
    That's what the problem is. I am not able to show it for all c and c'.
  8. Jun 26, 2011 #7
    just apply the triangle inequality for a certain choice of c and c'. This works for every choice of c and c'...
  9. Jun 26, 2011 #8
    It works for all c and c' but then how can we prove the original problem with all c and c'
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