This is part of Exercise 2.6 in Baby Rudin.
Relevant definition: If $X$ is a metric space, if $E \subset X$, and if $E'$ denotes the set of all limit points of $E$ in $X$, then the closure of $E$ is the set $\bar E = E \cup E'$.
So, what I'm trying to show here, essentially, is that $\left(\bar E\right)' = E'$. I have that $\left(\bar E\right)' \subset E'$ (which is, arguably, the more "difficult" direction), but I'm having trouble showing that $\left(\bar E\right)' \supset E'$. I've seen a number of arguments using the definition above and the fact that $\bar E$ is closed and thus contains all of its limit points, i.e., since $\bar E = E \cup E'$, if $x$ is in $E'$, then it is in $\bar E$, and since $\bar E$ is closed it contains all of its limit points, therefore $x$ is in $\left(\bar E\right)'$. But, from just this information, that does not follow. We do have that, if $x$ is in $\left(\bar E\right)'$, then $x$ is in $\bar E$, but what is needed for this proof is the reverse.
Is it the case that $x$ is a limit point of $\bar E$ if and only if $x$ is in $\bar E$? Put another way, is it possible that $\bar E$ contains isolated points? Or am I simply overlooking something more obvious…
Best Answer
It is not the case that $x$ is a limit point of $\bar{E}$ iff $x$ is in $\bar{E}$. Consider $E:=\{0\}$. Then $\bar{E}=\{0\}$ and $(\bar{E})'=\emptyset$. Thus $0$ is an element of $\bar{E}$ that is not a limit point of $\bar{E}$.
To solve your question, just use the definition of limit point. Suppose $x\in E'$. We want to show that $x\in(\bar{E})'$. Fix a neighborhood $U$ of $x$. Since $x$ is a limit point of $E$, we know that $(U\cap E)\setminus\{x\}$ is nonempty. Thus $(U\cap\bar{E})\setminus\{x\}$ is nonempty. We have just shown that $x$ is a limit point of $\bar{E}$. Therefore $E'\subseteq (\bar{E})'$.