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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 $X'\ne\overline X$. – Brian M. Scott Jul 19 '12 at 20:50
Oh!My bad!Thanks!!! – HipsterMathematician Jul 19 '12 at 20:51
up vote 5 down vote accepted

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 $N\cap X\subseteq\{x\}$. – Asaf Karagila Jul 19 '12 at 20:55
@Asaf: Yep; thanks. – Brian M. Scott Jul 19 '12 at 20:59
Is $N\cap X´ = \emptyset$? – HipsterMathematician Jul 19 '12 at 21:48
@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. – Pedro Tamaroff Jul 19 '12 at 21:56
@PeterTamaroff THANK YOU SO MUCH!!! – HipsterMathematician Jul 19 '12 at 21:58

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