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I'm working through Rudin's Principles of Mathematical Analysis, and came across the following problem.

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

My proof: Consider $F = E \setminus E'$ and $y \in F$. Then since $y$ is not a limit point of $E$, $\exists \ \delta > 0$ such that $B_{\delta}(y) \cap E' = \emptyset$. But then $F$ is open, so $E'$ is closed. $\square$

Naturally I wanted to convince myself this was correct, and upon doing some looking around online (see question asked here on MSE, or UCSD homework solutions paper) it seems to me like the "normal" way to prove this is to show that every limit point of $E'$ is in fact a point of $E'$.

This has made me question whether there is a flaw in my argument, or whether it looks correct and is reasonable enough to just show that the complement is open?

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Eric Hansen
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    It's correct, and better than the standard argument. – Sarvesh Ravichandran Iyer Oct 20 '16 at 01:40
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    @Eric Hansen: I don't understand your proof. The set $E'$ is not necessary a subset of $E$. Do you mean $F = E \setminus (E \cap E')$ or $F = X \setminus E'$ where $X$ is the space in which we work? – levap Oct 20 '16 at 02:30
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    @levap Ahh no you're not misunderstanding you're completely right, for some reason I just assumed $E$ already contained all of its limit points. This is just incorrect then I suppose. – Eric Hansen Oct 20 '16 at 02:34

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As levap pointed out in the comments, you’ve implicitly assumed that $E'\subseteq E$, which need not be true. However, it is still half of a correct argument. Since $\operatorname{cl}E=E\cup E'$, we know that

$$\begin{align*} (E\setminus E')\cup(X\setminus\operatorname{cl}E)&=(E\setminus E')\cup\big(X\setminus(E\cup E')\big)\\ &=(E\setminus E')\cup\big((X\setminus E)\cap(X\setminus E')\big)\\ &=(E\setminus E')\cup\big((X\setminus E)\setminus E'\big)\\ &=X\setminus E'\;. \end{align*}$$

You’ve shown that if $x\in E\setminus E'$, then $x$ has an open nbhd disjoint from $E'$. And $X\setminus\operatorname{cl}E$ is open and disjoint from $E'$, so every point of $X\setminus\operatorname{cl}E$ certainly has an open nbhd disjoint from $E'$. Thus, every point of $X\setminus E'$ has an open nbhd disjoint from $E'$, and $E'$ is therefore closed.

Brian M. Scott
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