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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?

Community
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syeh_106
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    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 $E\subseteq\operatorname{cl}E$. – Brian M. Scott Nov 19 '16 at 04:04

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This is correct since $\overline{E}=E\cup E'$, there is $$ \overline{E}'=(E\cup E')'=E'\cup (E')'=E' $$ The last step is because $(E')'\subset E'$.

Eugene Zhang
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Yes, you are correct.

$\overline E' \subset E'$ holds because if $x \in \overline E',$ then any neighbourhood of $x,$ say $N_\varepsilon(x)$, has an element of either $E$ (which directly implies $x \in E'$ ) or it has an element of $E'$, say $y$.
Now since $y$ also has a smaller neighbourhood (which contains an element of $E$) in $N_\varepsilon(x)$ concludes that $x$ is in $E'$. Hence, $\overline E' \subset E'$.

The other side is trivial since the non-empty intersection of any neighbourhood of $x (\in E')$ with $E$ implies non-empty intersection of that neighbourhood with $\overline E$. Thus, $E' \subset \overline E'$.
Hence, $E' = \overline E'$.

Singh_Gunjeet
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