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Math Help - show that A is bounded iff diam(A) is finite

  1. #1
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    show that A is bounded iff diam(A) is finite

    We define the diameter of a nonempty subset A of M by diam(A)=sup{ d(a,b): a,b \in A}. Show that A is bounded if and only if diam(A) is finite.

    this is seems obvious, if diam(A) is not finite, then the distance from a to b can be infinite, which says that A is not bounded.

    However I cannot think of an official proof of this, can anyone help?
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  2. #2
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    Re: show that A is bounded iff diam(A) is finite

    Quote Originally Posted by wopashui View Post
    We define the diameter of a nonempty subset A of M by diam(A)=sup{ d(a,b): a,b \in A}. Show that A is bounded if and only if diam(A) is finite.
    I agree with you here. It seems that you are being asked to prove a definition.

    So I guess there is some other definition of a set being bounded.
    What is the definition of a bounded set in your text?
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  3. #3
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    Re: show that A is bounded iff diam(A) is finite

    A is bounded if there is some x\inM and some constant C<infinity s.t d(a,x)<=C for all a\inA

    So if We know A is bounded, we have d(a,x)<=C and d(b,x)<=C which implies d(a,b)<= d(a,x)+d(x,b)<=C+C=2C< infinity
    so this shows that diam(A) is less than infinity, hence its finite. Is this correct?

    How about prove from the other direction?
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    Re: show that A is bounded iff diam(A) is finite

    Quote Originally Posted by wopashui View Post
    A is bounded if there is some x\inM and some constant C<infinity s.t d(a,x)<=C for all a\inA So if We know A is bounded, we have d(a,x)<=C and d(b,x)<=C which implies d(a,b)<= d(a,x)+d(x,b)<=C+C=2C< infinity
    so this shows that diam(A) is less than infinity, hence its finite. Is this correct?
    How about prove from the other direction?
    Well OK. Suppose that \delta(A)<\infty
    Let x_0\in A then \left( {\forall x \in A} \right)\left[ {d(x_0 ,x) \le \delta (A)} \right].
    What can you with that?
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    Re: show that A is bounded iff diam(A) is finite

    Quote Originally Posted by Plato View Post
    Well OK. Suppose that \delta(A)<\infty
    Let x_0\in A then \left( {\forall x \in A} \right)\left[ {d(x_0 ,x) \le \delta (A)} \right].
    What can you with that?
    what is \delta(A) here? Is it the diamater of A? Should we let x_0 \in M?
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    Re: show that A is bounded iff diam(A) is finite

    Quote Originally Posted by wopashui View Post
    what is \delta(A) here? Is it the diamater of A? Should we let x_0 \in M?
    Yes \delta(A) stands for the diameter of A.
    Suppose A is not bounded but \delta(A)<\infty.
    If c>0 then \left( {\exists x_1  \in A} \right)\left( {\exists x_2  \in A} \right)\left[ {d(x_1 ,x_2 ) > \delta (A) + c} \right]. Why is that?
    What does the contradict?
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  7. #7
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    Re: show that A is bounded iff diam(A) is finite

    Quote Originally Posted by Plato View Post
    Yes \delta(A) stands for the diameter of A.
    Suppose A is not bounded but \delta(A)<\infty.
    If c>0 then \left( {\exists x_1 \in A} \right)\left( {\exists x_2 \in A} \right)\left[ {d(x_1 ,x_2 ) > \delta (A) + c} \right]. Why is that?
    What does the contradict?
    this is contridiction since \delta (A) = max d(x_1,x_2), so d(x_1,x_2) should be less than \delta (A) right?

    but how did you get \delta (A) less than d(a,b), if we wanna prove by contridcition

    we have if A is not bounded, there exists a,b \inA, d(a,x) > C for all  x \in M and d(b,x) > C  for all x \in M

    thus d(a,b) <= d(a,x)+d(b,x) >2C  , but we can't say that d(a,b) > 2C  from this inequality
    Last edited by wopashui; October 17th 2011 at 02:30 PM.
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    Re: show that A is bounded iff diam(A) is finite

    Quote Originally Posted by wopashui View Post
    this is contridiction since \delta (A) = max d(x_1,x_2), so d(x_1,x_2) should be less than \delta (A) right?
    That is correct.
    Here we would have \delta(A)+c< d(x_1,x_2)\le \delta(A)
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    Re: show that A is bounded iff diam(A) is finite

    Quote Originally Posted by Plato View Post
    That is correct.
    Here we would have \delta(A)+c< d(x_1,x_2)\le \delta(A)

    but how did you get \delta (A) less than d(a,b), if we wanna prove by contridcition



    we have if A is not bounded, there exists a,b \inA, d(a,x) > C for all x \in M and d(b,x) > C for all x \in M

    thus d(a,b) <= d(a,x)+d(b,x) >2C , but we can't say that d(a,b) > 2C from this inequality
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    Re: show that A is bounded iff diam(A) is finite

    Quote Originally Posted by wopashui View Post
    but how did you get \delta (A)+c less than d(a,b), if we wanna prove by contridcition
    Don't you see that C=\delta (A)+c>0 so some pair x_1~\&~x_2 that have the property \delta (A)+c\le d(x_1,x_x_2)?
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    Re: show that A is bounded iff diam(A) is finite

    yes, I see that, but i just not sure how you got the last inequality, since by definition x1 x2 cannot both be an element of A, one of them must be in M, thats wy i have d(a,b) <= d(a,x)+d(b,x) >2C ,
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    Re: show that A is bounded iff diam(A) is finite

    Quote Originally Posted by wopashui View Post
    yes, I see that, but i just not sure how you got the last inequality, since by definition x1 x2 cannot both be an element of A, one of them must be in M, thats wy i have d(a,b) <= d(a,x)+d(b,x) >2C ,
    You have a problem with negations.

    Here is the definition of bounded set that you gave.
    The set A is bounded if and only if
    \left( {\exists C > 0} \right)\left( {\forall x \in A} \right)\left( {\forall y \in A} \right)\left[ {d(x,y) \leqslant C} \right].

    So to say A is not bounded if and only if
    \left( {\forall C > 0} \right)\left( {\exists x \in A} \right)\left( {\exists y \in A} \right)\left[ {d(x,y) > C} \right]

    So you see that x_1\in A~\&~x_2\in A.
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  13. #13
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    Re: show that A is bounded iff diam(A) is finite

    ok , isee now, I forgot to use the face we have proved that if A is bounded, then d(a,b)<= C for some C

    thank you
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