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Thread: 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 $\displaystyle (x_n) $ of elements of A that converges to sup A.

    Can we say that since A is bounded above, so is ($\displaystyle x_n$), then ($\displaystyle x_n$) is nondecreasing since it's bounded above, and so $\displaystyle x=$ sup{$\displaystyle x_n; n \in N$} 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 $\displaystyle (x_n) $ of elements of A that converges to sup A.
    Can we say that since A is bounded above, so is ($\displaystyle x_n$), then ($\displaystyle x_n$) is nondecreasing since it's bounded above, and so $\displaystyle x=$ sup{$\displaystyle x_n; n \in N$} exist.
    Suppose that $\displaystyle \alpha=\sup(A)$.
    If $\displaystyle \alpha\in A$ use a constant sequence consisting of $\displaystyle \alpha$.

    If $\displaystyle \alpha\notin A$ then $\displaystyle \alpha-1<\alpha$ so
    $\displaystyle \left( {\exists a_1 \in A} \right)\left[ {\alpha - 1 < a_1 < \alpha } \right]$
    $\displaystyle \left( {\exists a_2 \in A} \right)\left[ {\max\{a_1,\alpha-2^{-1}\} < a_2 < \alpha } \right]$
    For $\displaystyle n>2$ we have $\displaystyle \left( {\exists a_n \in A} \right)\left[ {\max\{a_{n-1},\alpha-n^{-1}\} < a_n < \alpha } \right]$
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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 $\displaystyle (x_n) $ of elements of A that converges to sup A.

    Can we say that since A is bounded above, so is ($\displaystyle x_n$), then ($\displaystyle x_n$) is nondecreasing since it's bounded above, and so $\displaystyle x=$ sup{$\displaystyle x_n; n \in N$} exist.
    No, you can't say anything at all about $\displaystyle \{x_n\}$ 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 $\displaystyle x_n$ 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 $\displaystyle a=\sup(A)$ we have $\displaystyle A\cap (a,a+1/n)=\emptyset.$

    If $\displaystyle a\in A$ and is an isolated point there is some k such that
    $\displaystyle A\cap (a-1/k,a)=\emptyset.$
    Last edited by Plato; Sep 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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    Re: need help for proving

    So what Plato did is to prove the sequence $\displaystyle a_n = x_n$ exist, right, now we just need to show that it converges to supA?
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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 $\displaystyle a_n = x_n$ exist, right, now we just need to show that it converges to supA?
    That is simple to do.
    $\displaystyle \alpha - n^{ - 1} < x_n < \alpha $. BUT $\displaystyle \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.
    $\displaystyle \alpha - n^{ - 1} < x_n < \alpha $. BUT $\displaystyle \left( {\alpha - n^{ - 1} } \right) \to \alpha $
    so by the squzze theorem, this is ture right?
    Last edited by wopashui; Sep 19th 2011 at 01:35 PM.
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