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Thread: open inertvals (proof)

  1. #1
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    open inertvals (proof)

    Hello all,

    Problem:

    Let A be a bounded subset of $\displaystyle \mathbb{R}$, which contains infinitely many different points. Prove that there exists at least one point $\displaystyle x_{0}$ such that all open intervals around $\displaystyle x_{0}$ 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 $\displaystyle A=\{ \frac{1}{ 2^n}|n\in \mathbb{N}\}$ then the open interval $\displaystyle (\frac{1}{2^n}-\frac{1}{2^{n+1}},\frac{1}{2^n}+\frac{1}{2^{n+1}})$ contains only $\displaystyle \frac{1}{2^n}$.

    Is there a flaw in this counterexample I'm not seeing?
    Last edited by DrSteve; Jan 12th 2011 at 04:49 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 $\displaystyle A=\{ 2^n|n\in \mathbb{N}\}$ Is there a flaw in this counterexample I'm not seeing?
    In the given, $\displaystyle A$ is a bounded set.
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    Oops - I meant $\displaystyle \frac{1}{2^n}$. 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 $\displaystyle \math{A}$ 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 $\displaystyle A=\{ \frac{1}{ 2^n}|n\in \mathbb{N}\}$ then the open interval $\displaystyle (\frac{1}{2^n}-\frac{1}{2^{n+1}},\frac{1}{2^n}+\frac{1}{2^{n+1}})$ contains only $\displaystyle \frac{1}{2^n}$.

    Is there a flaw in this counterexample I'm not seeing?
    It doesn't say that the point $\displaystyle x_0$ has to be in A. In this example, take $\displaystyle x_0=0$.
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  7. #7
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    Quote Originally Posted by Opalg View Post
    It doesn't say that the point $\displaystyle x_0$ has to be in A. In this example, take $\displaystyle x_0=0$.
    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, $\displaystyle A$ can be contained in a closed interval $\displaystyle I=[a,b]$. Split $\displaystyle I$ in half. One of the parts must contain infinitely many points from $\displaystyle A$. Take that one and call it $\displaystyle A_1$. Then, split $\displaystyle A_1$ in half. One of those parts must contain infinitely many points from $\displaystyle A$, so take that one and call it $\displaystyle A_2$. Continue this procedure indefinitely to receive a sequence of closed intervals $\displaystyle A_1, A_2, A_3,\ldots$ such that $\displaystyle A_1\subset A_2\subset A_3,\ldots$ whose sizes are $\displaystyle \frac{b-a}{2},\frac{b-a}{2^2},\frac{b-a}{2^3},\ldots$, respectively.

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