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Thread: 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 $\displaystyle T_1$ 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 $\displaystyle U$ contains a point $\displaystyle y$ of $\displaystyle E'$ different from itself but since $\displaystyle U$ is also a neighborhood of $\displaystyle y$ it follows (since this is a metric space) that $\displaystyle U$ contains infinitely many points of $\displaystyle E$, in particular it must contain a point of $\displaystyle E$ distinct from $\displaystyle x$ and thus $\displaystyle x$ is a limit point of $\displaystyle E$ so $\displaystyle x\in E'$
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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 $\displaystyle E^{\prime}$ is open. So let the surrounding topological space (or metric space) be denoted by $\displaystyle X$ and assume that $\displaystyle E^{\prime}\subset X$ is a proper subset. Then let $\displaystyle x\in X\backslash E^{\prime}$, so that $\displaystyle x$ is not a limit point of $\displaystyle E$. Then by definition there must be a neighbourhood $\displaystyle U$ of $\displaystyle x$ such that $\displaystyle U\cap E^{\prime}= \emptyset$, i.e. $\displaystyle x\in U \subseteq X\backslash E^{\prime}$. Since this holds for any such $\displaystyle x\in X\backslash E^{\prime}$, this means that $\displaystyle X\backslash E^{\prime}$ is open, so that $\displaystyle E^{\prime}$ is closed.
    Last edited by nimon; May 14th 2010 at 01:37 AM. Reason: Extra assumption
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by nimon View Post
    Then by definition there must be a neighbourhood $\displaystyle U$ of $\displaystyle x$ such that $\displaystyle U\cap E^{\prime}= \emptyset$, i.e. $\displaystyle x\in U \subseteq X\backslash E^{\prime}$.
    That isn't true at all. It is true that there is a neighborhood such that $\displaystyle U\cap E\subseteq\{x\}$
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