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We know that $p\in X$ is an accumulation point of subset $E$ of metric space $(X,d)$ if every neighborhood of $p$, meet $E$ on uncountably many points. Now in $\mathbb R^k$ with standard metric it is true that "If $E$ be uncountable and $F$ be the set of all accumulation points of $E$ then $F$ is perfect.(i.e $F'=F$)"

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  • $\begingroup$ If $F'$ is the Cantor-Bendixon derivative (i.e. removing the isolated points) then completeness is not equivalent to $F'=F$. Such set $F$ for which $F=F'$ is called perfect. $\endgroup$ – Asaf Karagila Mar 12 '13 at 11:02
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    $\begingroup$ "Assumption point"? Is that standard terminology? $\endgroup$ – Gerry Myerson Mar 12 '13 at 11:48
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    $\begingroup$ It's normally called a condensation point, in my terminology. $\endgroup$ – Henno Brandsma Mar 12 '13 at 12:08
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    $\begingroup$ "Accumulation" should be the word the OP is looking for. $\endgroup$ – Pedro Tamaroff Mar 12 '13 at 15:03
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HINT: Let $\mathscr{B}$ be a countable base for $\Bbb R^k$; for instance, you could take $\mathscr{B}$ to be the set of open balls with rational radius whose centres are in $\Bbb Q^k$. Let $\mathscr{C}=\{B\in\mathscr{B}:B\cap E\text{ is countable}\}$, and let $C=\bigcup\mathscr{C}$. Show that every point of $E\setminus C$ is the set of condensation points of $E$, and that $(E\setminus C)'=E\setminus C$.

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