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Math Help - need help for proving

  1. #1
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    need help for proving

    Let A be a nonempty subset of R that is bounded above. Show that there is a sequence (x_n) of elements of A that converges to sup A.

    Can we say that since A is bounded above, so is ( x_n), then ( x_n) is nondecreasing since it's bounded above, and so x= sup{ x_n; n \in N} exist.
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  2. #2
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    Re: need help for proving

    Quote Originally Posted by wopashui View Post
    Let A be a nonempty subset of R that is bounded above. Show that there is a sequence (x_n) of elements of A that converges to sup A.
    Can we say that since A is bounded above, so is ( x_n), then ( x_n) is nondecreasing since it's bounded above, and so x= sup{ x_n; n \in N} exist.
    Suppose that \alpha=\sup(A).
    If \alpha\in A use a constant sequence consisting of \alpha.

    If \alpha\notin A then \alpha-1<\alpha so
    \left( {\exists a_1  \in A} \right)\left[ {\alpha  - 1 < a_1  < \alpha } \right]
    \left( {\exists a_2  \in A} \right)\left[ {\max\{a_1,\alpha-2^{-1}\}  < a_2  < \alpha } \right]
    For n>2 we have \left( {\exists a_n  \in A} \right)\left[ {\max\{a_{n-1},\alpha-n^{-1}\}  < a_n  < \alpha } \right]
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  3. #3
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    Re: need help for proving

    Quote Originally Posted by wopashui View Post
    Let A be a nonempty subset of R that is bounded above. Show that there is a sequence (x_n) of elements of A that converges to sup A.

    Can we say that since A is bounded above, so is ( x_n), then ( x_n) is nondecreasing since it's bounded above, and so x= sup{ x_n; n \in N} exist.
    No, you can't say anything at all about \{x_n\} until after you prove that it exists at which point you are done.

    What you can do is assert that, with a= sup A, for any positive integer n, there exist a point of the set A in the interval (a, a+1/n). (Why?) Define x_n to be that number.
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    Re: need help for proving

    Quote Originally Posted by HallsofIvy View Post
    What you can do is assert that, with a= sup A, for any positive integer n, there exist a point of the set A in the interval (a, a+1/n). (Why?)
    I hope you meant "there exist a point of the set A in the interval (a-1/n,a)"?
    Because a=\sup(A) we have A\cap (a,a+1/n)=\emptyset.

    If a\in A and is an isolated point there is some k such that
    A\cap (a-1/k,a)=\emptyset.
    Last edited by Plato; September 17th 2011 at 01:57 PM.
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  5. #5
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    Re: need help for proving

    Oh, yes, we are talking about sup, not inf. Thanks, Plato.
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  6. #6
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    Re: need help for proving

    So what Plato did is to prove the sequence a_n = x_n exist, right, now we just need to show that it converges to supA?
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  7. #7
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    Re: need help for proving

    Quote Originally Posted by wopashui View Post
    So what Plato did is to prove the sequence a_n = x_n exist, right, now we just need to show that it converges to supA?
    That is simple to do.
    \alpha  - n^{ - 1}  < x_n  < \alpha . BUT \left( {\alpha  - n^{ - 1} } \right) \to \alpha
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  8. #8
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    Re: need help for proving

    Quote Originally Posted by Plato View Post
    That is simple to do.
    \alpha  - n^{ - 1}  < x_n  < \alpha . BUT \left( {\alpha  - n^{ - 1} } \right) \to \alpha
    so by the squzze theorem, this is ture right?
    Last edited by wopashui; September 19th 2011 at 01:35 PM.
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