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Let $E'$ be the set of limit points of $E$, and $\overline E \triangleq E'\cup E$ be its closure, in some metric space. Is it true that $E'=\overline E'$? That $\overline E' \subset E'$ is shown in Limit Points of closure of A is subset of limit points of A. And I think the converse ($E' \subset \overline E'$) is clearly also true. So it appears that we should have $E'=\overline E'$. Did I mess up somewhere?

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    $\begingroup$ Yes, your conclusion is correct. The opposite inclusion is trivial: if every nbhd of $x$ hits $E$, then certainly every nbhd of $x$ hits $\operatorname{cl}E$, since $E\subseteq\operatorname{cl}E$. $\endgroup$ – Brian M. Scott Nov 19 '16 at 4:04
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This is correct since $\overline{E}=E\cup E'$, there is $$ \overline{E}'=(E\cup E')'=E'\cup E^{'^{'}}=E' $$ last step is because $E^{'^{'}}\subset E'$.

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