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Thread: Closed set

  1. #1
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    Closed set

    I am having trouble solving this problem from Principles of Mathematical Analysis by Rudin.

    *Let $\displaystyle E'$ be the set of all limit points of the set $\displaystyle E$. Prove that $\displaystyle E'$ is closed.

    What I am trying to do is prove that the complement of the set is open thus making the set itself closed. So far I have

    $\displaystyle A=(E')^c$. if $\displaystyle x\in A$ then $\displaystyle x$ is not a limit point of $\displaystyle E$ so there exist a neighborhood $\displaystyle N_r(x)$ such that $\displaystyle N_r(x)\subset E^c$ for some $\displaystyle r>0$. Therefore, $\displaystyle \forall x\in A$ we have that $\displaystyle x$ is a limit point of $\displaystyle E^c$.

    Any help or pointers on what I should do next would be greatly appreciated.
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  2. #2
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    There is a similar question here.
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  3. #3
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    Quote Originally Posted by putnam120 View Post
    I am having trouble solving this problem from Principles of Mathematical Analysis by Rudin.

    *Let $\displaystyle E'$ be the set of all limit points of the set $\displaystyle E$. Prove that $\displaystyle E'$ is closed.

    What I am trying to do is prove that the complement of the set is open thus making the set itself closed. So far I have

    $\displaystyle A=(E')^c$. if $\displaystyle x\in A$ then $\displaystyle x$ is not a limit point of $\displaystyle E$ so there exist a neighborhood $\displaystyle N_r(x)$ such that $\displaystyle N_r(x)\subset E^c$ for some $\displaystyle r>0$. Therefore, $\displaystyle \forall x\in A$ we have that $\displaystyle x$ is a limit point of $\displaystyle E^c$.

    Any help or pointers on what I should do next would be greatly appreciated.
    You have gone slightly to far. What you have shown is that for all $\displaystyle x \in A$ there exists a neighbourhood of $\displaystyle x$ which contains no points of $\displaystyle E'$, so is a subset of $\displaystyle A$, which proves that $\displaystyle A$ is open.

    RonL
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