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Let $X\subset \mathbb{R}$.I have to prove,for all $X$ that $X´$,i.e, the set of accumulation points, is a closed set.

Well,I know , by definition, that every accumulation point is a point of closure.

How a set is said to be closed if $X = \overline X$ ,and how the accumulations points are

points of closure,then:

$X´=\overline X$.

So $X´ $ is closed.

Is that right?

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    Let $X=\{0\}$. Then $X'=\varnothing$, and $\overline X=X$, so in this case $X'\ne\overline X$.2012-07-19
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    Oh!My bad!Thanks!!!2012-07-19

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HINT: Suppose that $x\notin X'$. Then $x$ has an open nbhd $N$ such that $N\cap X\subseteq\{x\}$. (Why?) What can you say about $N\cap X'$?

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    You mean $N\cap X\subseteq\{x\}$.2012-07-19
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    @Asaf: Yep; thanks.2012-07-19
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    Is $N\cap X´ = \emptyset$?2012-07-19
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    @MeAndMath What you want to prove is that **no** $x$ outside of $X'$ can be a limit point of $X'$, which will let you conclude that any limit point must be contained in $X'$, from where it follows $X'$ must be closed.2012-07-19
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    @PeterTamaroff THANK YOU SO MUCH!!!2012-07-19
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    @MeAndMath Could you prove it?2012-07-19
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    @PeterTamaroff Think I can...2012-07-19