# Reviewing for Analysis Exam- can't figure out pf. about Set of Lim. Pts. being closed

## Homework Statement

Theorem: Given a metric space $\left(X,d\right)$, the set of all limit points of a subset $E\subset X$, denoted $E'$ is a closed set.

I have an Analysis Exam tomorrow and have been studying for quite awhile and last week, my professor gave us a list of Theorems to know the proofs of. This is one of them. I perused my notes and could not find the proof. I also searched through three or four Analysis texts and could not find a good proof. It doesn't seem like it would be hard to prove, but I can't think of a really rigorous one.

I have the following ideas:

- Show that $(E')^c$ is open.
- I let there be a $p \in (E')^c$ and a $q\in E'$
- From there I thought that possibly considering the two cases of $q \in E$ and $q \notin E$.
- From there I am stuck. I think I might need to use something about an open neighborhood and use distances between points, but I am not sure.

I really appreciate any helpful clues or hints. I am just looking to be pushed in a good direction, not for the solution to be given out. I would ask my professor, but I am not on campus today and he doesn't have office hours today.

Thanks Much

Well, if $p \in (E')^c$, then it is not a limit point. Thus, there exists a neighborhood around $p$ that does NOT contain any points of $E$. Your job is to make the radius of this neighborhood sufficiently small so that this occurs. How does this imply that the set is open?

Another option is to explicitly prove that $E'$ is closed. This may be easier....

Thank you for your response. I will go from there and see what I can come up with. I'll try both routes suggested.

I had the Exam today, and a variation of this problem appeared. The professor wanted us to do the proof for sub sequential limits instead of general limit points. And I figured it out last night after lineintegral's advice (I took the second option of direct proof- turned out to be painfully obvious). So thanks much!