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We have the theorem:

Let $E′$ be the set of all limit points of a set $E$. Prove that $E′$ is closed.

I understand the proof by Roberto Cardona given in the link, what I still don't grasp is essentially the closure of $E'$. In other words, why I have to suppose that $x_0\in E'$ is a limit point of $E'$? If I assume this, then by definition (*) $E'$ is closed and the only thing that I prove is that $x_0$ is also a limit of $E$ (i.e. that I don't create somehow limit points that are not limit points of $E$).

(*) A set $E'\subseteq \mathbb{R}$ is closed if it contains all its limit points.

user2820579
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    I don't understand your confusion. The proof does not assume that $x_0\in E'$, and you seem to acknowledge this when you say you still need to prove that $x_0$ is a limit point of $E$. – Eric Wofsey May 29 '18 at 23:22
  • I see now where I misunderstood the argument. Thanks! – user2820579 May 29 '18 at 23:30
  • The guy says "Thus $x_0$ is a limit point of $E$ and by definition is contained in $E'$, so I understood that $x_0\in E'$ from the beginning. I think is better to write "Thus $x_0$ is a limit point of $E$ and therefore $x_0\in E'$". I still get lost in the semantics... – user2820579 May 29 '18 at 23:36

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The proof starts by : let $x_0$ be a limit point of $E'$. Then (insert argument), $x_0 \in E'$. So we have shown that any limit point of $E'$ is already in $E'$ (or $E'' \subseteq E'$) so the criterion $(\ast)$ allows one to conclude that $E'$ is closed. The criterion $(\ast)$ is better formulated as

A set $A \subseteq \mathbb{R}$ is closed if it contains all its limit points.

Or

A set $A \subseteq \mathbb{R}$ is closed if $A' \subseteq A$

which we apply to $A =E'$.

Henno Brandsma
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