Why is the Set of Limit Points Closed in a Metric Space?

Click For Summary

Homework Help Overview

The discussion revolves around a theorem in metric spaces, specifically addressing why the set of limit points of a subset is closed. The original poster is preparing for an analysis exam and seeks a rigorous proof for this theorem, having difficulty finding one in their notes or textbooks.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster considers showing that the complement of the set of limit points is open and contemplates different cases regarding points in the set. They express uncertainty about using open neighborhoods and distances.
  • One participant suggests that if a point is not a limit point, there exists a neighborhood around it that does not contain points of the subset, prompting a discussion on how this leads to the conclusion that the set is open.
  • Another participant proposes that explicitly proving the set of limit points is closed might be an easier approach.

Discussion Status

The discussion has provided some helpful guidance, with participants suggesting different routes to explore for the proof. The original poster expresses intent to pursue both suggested approaches, indicating a productive direction in the conversation.

Contextual Notes

The original poster mentions the urgency of preparing for an exam and the constraints of not being able to consult their professor directly due to being off-campus. This context adds to the pressure of finding a solution quickly.

TheEigenvalue
Messages
45
Reaction score
0

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
 
Physics news on Phys.org


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!
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
3K
Replies
1
Views
2K
Replies
8
Views
2K
  • · Replies 9 ·
Replies
9
Views
6K
  • · Replies 18 ·
Replies
18
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 1 ·
Replies
1
Views
7K