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Thread: 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 $\displaystyle diam(A)=sup${$\displaystyle d(a,b): a,b \in A$}. 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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    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 $\displaystyle diam(A)=sup${$\displaystyle d(a,b): a,b \in A$}. 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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    Re: show that A is bounded iff diam(A) is finite

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

    So if We know A is bounded, we have $\displaystyle d(a,x)<=C $and $\displaystyle d(b,x)<=C$ which implies $\displaystyle d(a,b)<= d(a,x)+d(x,b)<=C+C=2C< infinity$
    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 $\displaystyle x\inM $and some constant C<infinity s.t $\displaystyle d(a,x)<=C $for all $\displaystyle a\inA$ So if We know A is bounded, we have $\displaystyle d(a,x)<=C $and $\displaystyle d(b,x)<=C$ which implies $\displaystyle d(a,b)<= d(a,x)+d(x,b)<=C+C=2C< infinity$
    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 $\displaystyle \delta(A)<\infty$
    Let $\displaystyle x_0\in A$ then $\displaystyle \left( {\forall x \in A} \right)\left[ {d(x_0 ,x) \le \delta (A)} \right]$.
    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 $\displaystyle \delta(A)<\infty$
    Let $\displaystyle x_0\in A$ then $\displaystyle \left( {\forall x \in A} \right)\left[ {d(x_0 ,x) \le \delta (A)} \right]$.
    What can you with that?
    what is $\displaystyle \delta(A)$ here? Is it the diamater of A? Should we let $\displaystyle x_0 \in$$\displaystyle 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 $\displaystyle \delta(A)$ here? Is it the diamater of A? Should we let $\displaystyle x_0 \in$$\displaystyle M$?
    Yes $\displaystyle \delta(A)$ stands for the diameter of $\displaystyle A.$
    Suppose $\displaystyle A$ is not bounded but $\displaystyle \delta(A)<\infty$.
    If $\displaystyle c>0$ then $\displaystyle \left( {\exists x_1 \in A} \right)\left( {\exists x_2 \in A} \right)\left[ {d(x_1 ,x_2 ) > \delta (A) + c} \right]$. Why is that?
    What does the contradict?
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    Re: show that A is bounded iff diam(A) is finite

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

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

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

    thus $\displaystyle d(a,b) <= d(a,x)+d(b,x) >2C $, but we can't say that $\displaystyle d(a,b) > 2C $from this inequality
    Last edited by wopashui; Oct 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 $\displaystyle \delta (A)$ = $\displaystyle max d(x_1,x_2), so d(x_1,x_2) $should be less than $\displaystyle \delta (A)$ right?
    That is correct.
    Here we would have $\displaystyle \delta(A)+c< d(x_1,x_2)\le \delta(A)$
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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 $\displaystyle \delta(A)+c< d(x_1,x_2)\le \delta(A)$

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



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

    thus $\displaystyle d(a,b) <= d(a,x)+d(b,x) >2C $, but we can't say that $\displaystyle 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 $\displaystyle \delta (A)+c$ less than d(a,b), if we wanna prove by contridcition
    Don't you see that $\displaystyle C=\delta (A)+c>0$ so some pair $\displaystyle x_1~\&~x_2$ that have the property $\displaystyle \delta (A)+c\le d(x_1,x_x_2)$?
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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 $\displaystyle 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 $\displaystyle 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 $\displaystyle A$ is bounded if and only if
    $\displaystyle \left( {\exists C > 0} \right)\left( {\forall x \in A} \right)\left( {\forall y \in A} \right)\left[ {d(x,y) \leqslant C} \right]$.

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

    So you see that $\displaystyle x_1\in A~\&~x_2\in A$.
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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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