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Math Help - Dense-in-itself set

  1. #1
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    Dense-in-itself set

    Let E\subset \mathbb{R}^1 be a non-empty countable set. Suppose E has no isolated points, prove that \bar E\backslash E is dense in \bar E.

    It is difficult, isn't it?
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  2. #2
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    Re: Dense-in-itself set

    I think the best approach might be to suppose that \bar E\backslash E is not dense in \bar E. Start unraveling the definitions and see what you can prove about the set E.

    - Hollywood
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  3. #3
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    Re: Dense-in-itself set

    I could not see any property of E, if I assume \bar E'\backslash E is not dense in E.

    I argue as follows. We need only prove the closure of \bar E'\backslash E contains \bar E.
    Suppose then x not in the closure of \bar E'\backslash E, then there exists a \tilde \delta>0 , such that for any \delta\in (0,\tilde \delta), the ball B(x,\delta)\subset (E'\backslash E)^c=E^{'c}\cup E. Since E is countable, we see B(x,\delta)\cap E^{'c}\neq \emptyset. This, by definition, means x\in E^{'c-} (the closure of E^{'c}).

    Here, I could only prove x\in E^{'c-}, but not x\in E^{'c}.

    Would you help me out? Thank you.
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    Re: Dense-in-itself set

    Quote Originally Posted by xinglongdada View Post
    Let E\subset \mathbb{R}^1 be a non-empty countable set. Suppose E has no isolated points, prove that \bar E\backslash E is dense in \bar E.
    Another way to define being dense-in-itself to to say that the set contains no isolated points.

    Thus is E dense-in-itself?

    Can \overline{E}\setminus E be dense-in-itself?
    Last edited by Plato; October 26th 2012 at 02:35 PM.
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    Re: Dense-in-itself set

    Yes, indeed they are equivalent. E\subset E' is equivalent to saying E has no isolated points.

    I do not see how to prove the statement.
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    Re: Dense-in-itself set

    Quote Originally Posted by xinglongdada View Post
    Prove that \overline{E}\backslash E is dense in \overline{E}.
    Well frankly I was confused by the title of the thread and the question above.

    If you want to prove the above question, can you show that

    \overline{\overline{E}\setminus E}=\overline{E}~?
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    Re: Dense-in-itself set

    E has no isolated points so every point x in cl(E) is limit of a sequence xn of different elements of E.E is countable so in the neigborhood of each xn there is yn in cl(E)-E.yn is converging to x,so we have our claim.
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  8. #8
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    Re: Dense-in-itself set

    Yes, I want to prove this.
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  9. #9
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    Re: Dense-in-itself set

    Quote Originally Posted by hedi View Post
    E has no isolated points so every point x in cl(E) is limit of a sequence xn of different elements of E.E is countable so in the neigborhood of each xn there is yn in cl(E)-E.yn is converging to x,so we have our claim.

    Why " in the neigborhood of each xn there is yn in cl(E)-E"? I know only
    B(x_n,1/n,x_n+1/n)\subset E^{-c}\cup (\bar E\backslash E)\cup E. Thus, by the countability of E, I could only see B(x_n-1/n,x_n+1/n) contains an element y_n lies in \bar E^c or \bar E\backslash E?
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  10. #10
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    Re: Dense-in-itself set

    Quote Originally Posted by xinglongdada View Post
    Why " in the neigborhood of each xn there is yn in cl(E)-E"? I know only
    B(x_n,1/n,x_n+1/n)\subset E^{-c}\cup (\bar E\backslash E)\cup E. Thus, by the countability of E, I could only see B(x_n-1/n,x_n+1/n) contains an element y_n lies in \bar E^c or \bar E\backslash E?
    I HATE THESE TRICKY PROOFS.
    Notation: \mathcal{B}(x;\delta)=\left( {x - \delta ,x + \delta } \right) where \delta>0.
    Note that if t\in E then \mathcal{B}(t;1) contains a point x_1\in E\setminus\{t\} because E has no isolated points.
    But \mathcal{B}(t;1) is uncountable so \left( {\exists y_1 \notin E} \right)\left[ {y_1 \in \mathcal{B}(t;1)} \right]

    So if \delta_1=1 then \exists\delta_2 such that \mathcal{B}(t;\delta_2)}\subseteq\mathcal{B}(t; \delta_1)} so that x_1\notin \mathcal{B}(t;\delta_2)} AND y_1\notin \mathcal{B}(t;\delta_2)}.

    Can you see why I HATE this?

    There is a sequence of points \left( {x_n } \right) in E such that \left( {x_n } \right) \to t.

    There is a sequence of points \left( {y_n } \right) in E^c such that \left( {y_n } \right) \to t.
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    Re: Dense-in-itself set

    Then you prove that E\subset \partial E. How to prove the statement then?
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    Re: Dense-in-itself set

    Quote Originally Posted by xinglongdada View Post
    Then you prove that E\subset \partial E. How to prove the statement then?
    Read reply #7.
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  13. #13
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    Re: Dense-in-itself set

    since E has no isolated points, the open sets in cl(E) are uncountable so they must intersect cl(E)-E,because E is countable.(notice that cl(E) is uncountable).
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  14. #14
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    Re: Dense-in-itself set

    A friend of mine came up with this proof having to do with perfect sets. By definiton, a set E is perfect if E=E', the set of limit points of E. A set is perfect if it is closed and has no isolated points. And if a set is perfect, it is uncountable. This Wikipedia article has some background:
    Derived set (mathematics) - Wikipedia, the free encyclopedia
    You can also find a proof that a perfect set is uncountable on the web.

    Ok. Suppose that \bar E\backslash E is not dense in \bar E. Then there exists x\in\bar{E} and \epsilon>0 such that B(x,\epsilon)\cap(\bar E\backslash E) is empty. So \bar{B}(x,\frac{\epsilon}{2})\cap(\bar E\backslash E) is empty. This means that in \bar{B}(x,\frac{\epsilon}{2}) every point of \bar{E} is also a point of E. So \bar{B}(x,\frac{\epsilon}{2})\cap\bar{E} is a subset of E. But \bar{B}(x,\frac{\epsilon}{2})\cap\bar{E} is closed, and since it is a subset of E, contains no isolated points. So \bar{B}(x,\frac{\epsilon}{2})\cap\bar{E} is perfect, and therefore uncountable. This is a contradiction, so \bar E\backslash E is dense in \bar E.

    - Hollywood
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