My friend and I are learning reading Baby Rudin currently and we both have reached the problem set for chapter 2, and my friend asks me to solve this problem and here is my solution:
Original Problem: Let $E'$ be the set of all limit points of a set $E$. Prove that $E'$ is closed. Prove that $ E$ and $\overline{E}$ have the same limit points. (Recall that $\overline{E} = E \cup E'$.) Do $ E $ and $E'$ always have the same limit points?
Let $ x \in E'.$ If we have any neighborhood N(x), it contains a $x' \in E'$ with $ x \ne x' $.
By Theorem 2.22, we have:$(\bigcup_{\alpha} E_{\alpha})^{c} = \bigcap_{\alpha} (E_{\alpha}^{c}) \Longleftrightarrow N \setminus \{x\} = N \cap \{x\}^c$
This is open, which implies the containment of a neighborhood $N'(x')$.
$N' $ intersects $ E $ since $ x' \in E' \implies N $ does too. Then that implies that ( x \in E' ) because this is the same for every limit point. We get the conclusion that ( E' ) is closed.
Since $E \subseteq \overline{E} $, a limit point belonging to one will belong to the other. If $ x $ is a limit point of $\overline{E} = E \cup E'$, $ N_{\frac{1}{n}} (x) $ must intersect either $ E $ for an infinite number of values of $ n \in \mathbb{N} $, or $ E' $ for an infinite number of values of $ n $ (anything otherwise leads to a contradiction).
We then see, since$N_{\epsilon} (x) \subseteq N_{\epsilon'} (x) \text{, when } \epsilon < \epsilon',$All neighborhoods of ( x ) intersect either $ E $ or $ E' $. So in both cases $ x \in E' $$\blacksquare$
For the third part of this problem the answer is no
My friend said my proof is incorrect, although I believe it is correct, so I would like to know if this solution is incorrect or correct, and how to fix it if it is the former; thank you if you decide to answer my question