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

Question

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?

Answer 1

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.

Answer 2

This is a direct proof to see that E' is closed.

Suppose x is in the closure of E', then consider an open set U that contains x. Using the definition of a cluster point, U must contain infinite point of E' (as we can form a sequence converging to x of elements of E'), let y one of those points different from x:

• If y is in E, we have finished as the intersection of U and E has an element different from x.

• If y isn't in E, as U contains y (and y is in the derivation set), there must be infinite points of E in U different from y. So the intersection of U and E also have an element diferent from x; and x is in E'

Thus, as E' contains all of its cluster points, it's closed.