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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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    I think yes, even if $C$ is not closed.2012-10-28
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    @Berci: I know this is true in metric spaces but not sure if it holds in any topological space.2012-10-28
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    Ok. What do you mean by limit point then?2012-10-28
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    @Berci: accumulation point.2012-10-28
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    Prove that any limit point of the set of limit points of $C$ is also a limit point of $C$.2012-10-28
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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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    That depends on the definition of *limit point*. In particular, if *limit point* means *accumulation point* or *cluster point*, a closed discrete set has no limits points and is therefore not the set of its limit points.2012-10-28
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    This is not true if $C$ contains an isolated point.2012-10-28
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    @user10: Yes, if $C$ is closed, $C'$ is also closed. The answer to your original question is *yes*; Ayman and I are just pointing out that this argument is incorrect, because it uses the wrong definition of *limit point*.2012-10-28
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    Why not accept Ayman's answer? He *shows* that the derived set is closed.2012-10-28
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    @M.B: sorry didn't read it until now, thanks.2012-10-28