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Let $A'$ be the set of the limit points of $A$, let $A=\{\frac{1}{n}:{n\in\mathbb{N}}\}$ find $A'$ and $A''$

because all the elements of of the sequence except for the limit point itself need to be contained in the set the only limit point of $A$ is $0$ and therefore $A'$ is $0$, does that mean the $A''$ is $\emptyset$?

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    $\begingroup$ Yes.${}{}{}{}{}{}$ $\endgroup$ – Omnomnomnom Nov 28 '16 at 19:37
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    $\begingroup$ Well, $A'$ is $\{0\}$, not $0$. $\endgroup$ – Daniel Fischer Nov 28 '16 at 19:40
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We know $A'' \subset \overline {A'} = \{0\}$. But $0\notin A''$ because every deleted neighborhood of $0$ contains no elements of $A'$. So yes, $A'' = \varnothing$.

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