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Let $L$ be the set of limit points of set $A$. How do you show that the set $L$ is closed? Or, is this statement not necessarily always true?

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  • $\begingroup$ What does $A$ lie inside? A metric space, or any topological space? $\endgroup$ – Hayden Aug 8 '14 at 11:30
  • $\begingroup$ You show that if $x$ is a limit point pf $L$ then it is also a limit point of $A$. $\endgroup$ – MJD Aug 8 '14 at 11:41
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The following proof works if singletons are closed.

Let $x\notin L$.

Then an open set $U$ exists such that $\left(U\backslash\left\{ x\right\} \right)\cap A=\emptyset$.

Then $U\backslash\left\{ x\right\} $ is open and from $\left(U\backslash\left\{ x\right\} \right)\cap A=\emptyset$ it follows that $\left(U\backslash\left\{ x\right\} \right)\cap L=\emptyset$.

(This because for any $y\in U\backslash\left\{ x\right\} $ we have open set $U\backslash\left\{ x\right\} $ with $\left(U\backslash\left\{ x\right\} \right)\cap A=\emptyset$, showing that $y\notin L$)

Combined with $x\notin L$ we now have $U\cap L=\emptyset$, so for $x\notin L$ we have found open set $U$ with $x\in U$ and $U\cap L=\emptyset$.

This can be done for any $x\notin L$ so allows the conclusion that $L$ is closed.

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