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Math Help - open inertvals (proof)

  1. #1
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    open inertvals (proof)

    Hello all,

    Problem:

    Let A be a bounded subset of \mathbb{R}, which contains infinitely many different points. Prove that there exists at least one point x_{0} such that all open intervals around x_{0} contains infinitely many different points from A.
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  2. #2
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    This doesn't seem true to me. If A=\{ \frac{1}{ 2^n}|n\in \mathbb{N}\} then the open interval (\frac{1}{2^n}-\frac{1}{2^{n+1}},\frac{1}{2^n}+\frac{1}{2^{n+1}}) contains only \frac{1}{2^n}.

    Is there a flaw in this counterexample I'm not seeing?
    Last edited by DrSteve; January 12th 2011 at 05:49 AM. Reason: Fixed counterexample
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  3. #3
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    Quote Originally Posted by DrSteve View Post
    This doesn't seem true to me. If A=\{ 2^n|n\in \mathbb{N}\} Is there a flaw in this counterexample I'm not seeing?
    In the given, A is a bounded set.
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  4. #4
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    Oops - I meant \frac{1}{2^n}. I'm going to edit my post. Now tell me if there's something wrong with it.
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  5. #5
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    Now that works nicely as a counter-example.
    I suspect the OP meant to add that \math{A} is closed.
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  6. #6
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    Quote Originally Posted by DrSteve View Post
    This doesn't seem true to me. If A=\{ \frac{1}{ 2^n}|n\in \mathbb{N}\} then the open interval (\frac{1}{2^n}-\frac{1}{2^{n+1}},\frac{1}{2^n}+\frac{1}{2^{n+1}}) contains only \frac{1}{2^n}.

    Is there a flaw in this counterexample I'm not seeing?
    It doesn't say that the point x_0 has to be in A. In this example, take x_0=0.
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  7. #7
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    Quote Originally Posted by Opalg View Post
    It doesn't say that the point x_0 has to be in A. In this example, take x_0=0.
    Oops I missed that too.
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  8. #8
    Senior Member roninpro's Avatar
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    In this case, I would try to construct a limit point.

    First, A can be contained in a closed interval I=[a,b]. Split I in half. One of the parts must contain infinitely many points from A. Take that one and call it A_1. Then, split A_1 in half. One of those parts must contain infinitely many points from A, so take that one and call it A_2. Continue this procedure indefinitely to receive a sequence of closed intervals A_1, A_2, A_3,\ldots such that A_1\subset A_2\subset A_3,\ldots whose sizes are \frac{b-a}{2},\frac{b-a}{2^2},\frac{b-a}{2^3},\ldots, respectively.

    Now, by the Nested Interval theorem, \bigcap A_n contains a single point, and that point is the limit point we are looking for.
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  9. #9
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by surjective View Post
    Hello all,

    Problem:

    Let A be a bounded subset of \mathbb{R}, which contains infinitely many different points. Prove that there exists at least one point x_{0} such that all open intervals around x_{0} contains infinitely many different points from A.
    Merely note that since A is bounded we have that \overline{A} is bounded and thus compact. Thus, we have that there is an infinite subset of \overilne{A} and by compactness (or the Bolzano-Weierstrass property to be exact) this sequence must have a limit piont. In any T_1 space (if this doesn't mean anything ignore it and replace it with \mathbb{R}) a point is a limit point if and only if every neighborhood contains infintely many points of the set.
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