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According to Rudin's principles of mathematical analysis 2.24 (a):

If a set $A$ and $B$ are open sets then $A /cup B$ is open. Therefore we have 4 possible scenarios letting $A=E$ and $B=E'$ where $E'$ is the set of limit points of $E$:

1.) $E$ and $E'$ are closed. Therefore $\overline{E}=E$ by theorem 2.27 (a)

2.) $E$ is closed and $E'$ is open.

3.) $E$ is open and $E'$ is closed (okay since $E'$ is closed.

4.) $E$ and $E'$ are open. Which is a contradiction of theorem 2.24 (a).

In the case of 2.) by definition, if $E$ is closed, then every limit point of $E$ is in $E$. That is, $E' \subset E$. Since every point of $E'$ is in $E$ (a closed set) then $E'$ must be closed. (Unsure about whether this is sufficient)

This is my attempt so far to prove that $E'$ is closed using axioms. If there is something I am missing or anything that needs to be modified I would appreciate it.

EDIT: using theorem 2.27 (b) if $E$ is closed then $E = \overline{E}$. Thus for 2.) if we take the set of limit points of the equality, we obtain $E' = (\overline{E})'$ then I must show that $E'= (E \cup E')' = \overline{E'}$.

Eric L
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  • Thank you for bringing this to my attention. I voted to close as well – Eric L Jul 17 '17 at 17:22
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    "Since every point of E′ is in E (a closed set) then E′ must be closed. (Unsure about whether this is sufficient)". Every point if $(0,1)$ is in $[0,1]$ and $[0,1]$ is closed so $(0,1)$ is closed? No. It's not sufficient or even true. – fleablood Jul 17 '17 at 17:30
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    Hint: Prove that if $e$ is a limit point of $E'$ it is also a limit point of $E$. Thus $E'' \subset E'$ and $E'$ is closed. – fleablood Jul 17 '17 at 17:32
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    ... every neighborhood of $e$ has a point $f \in E'$ so $f$ is a limit point of $E$ so every neighborhood of $f$ has a point of $E$ including then neighborhoods of $f$ that are completely within the neighborhood of $e$. – fleablood Jul 17 '17 at 17:36
  • I realize that is not sufficient looking back. Neighborhoods and the closure of a neighborhood is a clear contradiction. But since $E'$ is a subset of $E$ could we not take the union of $E$ and $E'$ such that $\overline{E} \subset E$ ( since $E \cup E = E$) then prove that $(\overline{E})' \subset E'$? – Eric L Jul 17 '17 at 18:29

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