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Math Help - Closed sets

  1. #1
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    Closed sets

    Hey, I was just trying out some questions from Rudin, when I had this doubt as to whether my answer is right, can you please tell me ifI am right?

    Let E' be the set of all limit points of a set E. Prove that E' is closed.

    Let x be a limit point of E'. So every nbd of x contains infinitely many points of E'. But in every nbd of every point of E' are infinitely many points of E, since every point of E' is a limit point of E.
    Thereby every nbd of x contains infinitely many points of E. So x is a limit point of E (ie) x is in E'.
    Hence E' is closed. Am i right?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by poorna View Post
    Hey, I was just trying out some questions from Rudin, when I had this doubt as to whether my answer is right, can you please tell me ifI am right?

    Let E' be the set of all limit points of a set E. Prove that E' is closed.

    Let x be a limit point of E'. So every nbd of x contains infinitely many points of E'. But in every nbd of every point of E' are infinitely many points of E, since every point of E' is a limit point of E.
    Thereby every nbd of x contains infinitely many points of E. So x is a limit point of E (ie) x is in E'.
    Hence E' is closed. Am i right?
    This is true in any T_1 space, but since it's Rudin it must be that you are speaking of a metric space.

    What you said is pretty much it, but I think you need to say it slightly better. Every neighborhood U contains a point y of E' different from itself but since U is also a neighborhood of y it follows (since this is a metric space) that U contains infinitely many points of E, in particular it must contain a point of E distinct from x and thus x is a limit point of E so x\in E'
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  3. #3
    Junior Member nimon's Avatar
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    I think it is best just to show that the complement of E^{\prime} is open. So let the surrounding topological space (or metric space) be denoted by X and assume that E^{\prime}\subset X is a proper subset. Then let x\in X\backslash E^{\prime}, so that x is not a limit point of E. Then by definition there must be a neighbourhood U of x such that U\cap E^{\prime}= \emptyset, i.e. x\in U \subseteq X\backslash E^{\prime}. Since this holds for any such x\in X\backslash E^{\prime}, this means that X\backslash E^{\prime} is open, so that E^{\prime} is closed.
    Last edited by nimon; May 14th 2010 at 02:37 AM. Reason: Extra assumption
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by nimon View Post
    Then by definition there must be a neighbourhood U of x such that U\cap E^{\prime}= \emptyset, i.e. x\in U \subseteq X\backslash E^{\prime}.
    That isn't true at all. It is true that there is a neighborhood such that U\cap E\subseteq\{x\}
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