Let $E'$ be the set of all limit points of a set $E$. Prove that $E'$ is closed.
I'm studying baby Rudin, and this is a partial problem of prob 6 of chap 2. I found this problem is hard, so I want to check if my argument is valid.
My argument:
I will show that $E'^c$ is open.
Take $x\in E'^c$. Then it implies that $x$ is not a limit point of $E$.
(a point $p$ is a limit point of a set $A$ if every nbhd of $p$ contains a point $q\in A$ such that $p\not= q$.)
So there exists an open nbhd $U$ such that $x\in U$ and $(U\setminus \{x\})\cap E =\varnothing$.
Since $U\setminus \{x\}$ is also open, and it doesn't intersect $E$, none of elements of $U\setminus \{x\}$ is a limit point of $E$. i.e., $(U\setminus \{x\})\cap E'=\varnothing.$
Since $x\in E'^c$, we have $U\cap E'=\varnothing$, thus $U\subset E'^c$. Hence every point $x\in E'^c$ is an interior point of $E'^c$, it follows that $E'^c$ is open.
PS: is it possible to show it directly? i.e., showing every limit point of $E'$ is a point of $E'$? If yes, How?
Best Answer
Your proof is correct. If you want to show it directly, then let $\ell$ be a limit point of $E'$ and let $U$ be any neighborhood of $\ell$. To show that $\ell$ is also a limit point of $E$, you need to find an element of $E$ inside $U \smallsetminus \{\ell\}$. Well, first, by definition of $\ell$ being a limit point of $E'$, there is an element $\ell^\ast$ of $E'$ inside $U \smallsetminus \{\ell\}$.
But then as $U \smallsetminus \{\ell\}$ is also open (Rudin only deals with metric spaces where singletons are closed), it qualifies as a neighborhood of $\ell^\ast$. So since $\ell^\ast$ is a limit point of $E$, there is an element $e$ of $E$ inside $U \smallsetminus \{\ell, \ell^\ast\}$. In particular, this element $e$ of $E$ is inside $U \smallsetminus \{\ell\}$ as desired.
Thus, all limit points of $E'$ are also limit points of $E$ and so they must belong to $E'$ too.