[Math] Prove that the set of limit points (derived set) is closed

general-topologyreal-analysis

Let $E$ be a non-empty subset of $\Bbb{R}$. Let $E'$ be its derived set, i.e. the set of all limit points of $E$. Prove that $E'$ is a closed set.

I think that we should prove that its complement is open and so $E'$ is closed. How can I do this?

Best Answer

If $p$ is not a limit point of $E$ (so it's in the complement of $E'$) this can mean one of two things:

  1. There is an open ball $N_r(p)$ such that $N_r(p) \cap E = \{p\}$.

  2. There is an open ball $N_r(p)$ such that $N_r(p) \cap E= \emptyset$.

In both cases it's easy to show that any point $q \in N_r(p)$ (because we can find an $r'>0$ such that $N_{r'}(q) \subseteq N_r(p)$) also is not in $E'$ so that $N_r(p) \subseteq \mathbb{R}\setminus E'$ and the latter set is open.

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