Principles of Mathematical Analysis, Rudin, 3th ed, Chapter 2, Problem 6

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?

Answer

In Rudin's book, the definition of "limit point" is

A point $p$ is a limit point of the set $E$ if every neighborbood of $p$ contains a point $q \neq p$ such that $q \in E$.

(1) $E'$ is closed.

Let $x$ be a limit point of $E'$. For $\epsilon >0$, there exists $y \in E'$ such that $y \in N_\epsilon(x)$ and $x \neq y$. Since $y \in E'$, there exists $z \in E$ such that $z \in N_{\epsilon - |x-y|}(y)$ and $z \neq y$. Note that $z \neq x$ and $z \in N_\epsilon(x)$. This implies $x \in E'$.

(2) $E$ and $\overline{E}$ have the same limit points.

Obviously, $E' \subset \overline{E}'$. Conversely, $\overline{E}' = E' \cup (E')' \subset E'$ by (1) and problem 7.

(3) For $E=\{1/n \;|\; n \in \mathbb{N}\}$, $E'=\{0\}$ and $E'=\emptyset$. Therefore, $E' \neq (E')'$.

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