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Let $X$ be any topological space and let $C \subseteq X$ be any subset. If $C$ is a closed subset of $X$ does it follow that the set of all limit points of $C$ is closed as well?

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    @Berci: In an arbitrary topological space, a point $x \in X$ is a **limit point** if every open set $U \ni x$ contains a point of $X \backslash \{x\}$.2012-10-28

2 Answers 2

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You can prove the statement for any set in a $T_1$ space; it doesn't have to be closed.

Let $p$ be a limit point of the set of limit points of $C$. Let $U$ be an open set containing $p$. Since $p$ is a limit point of the set of limit points of $C$, $U$ must contain a limit point of $C$, $q$. Hence $U - \{p\}$ is a neighborhood of $q$ and must contain a point from $C$. Thus $p$ is a limit point of $C$ and is contained in the set of limit points of $C$. We conclude that the set of limit points of $C$ is closed.

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The limit points of $C$ equals $C$ since $C$ is closed.

edit: this is not a good answer. Read the comments.

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    @M.B: sorry didn't read it until now, thanks.2012-10-28