# Closed set

1. Sep 25, 2013

### Lee33

1. The problem statement, all variables and given/known data

Prove that if lim n→∞ (p_n ) = p in a metric space then the set of points {p,p_1,p_2, ...,} are closed.

2. Relevant information

The definition of close in my book is "a set is closed if and only if its complementary is open." So I want to prove this by contradiction. I can't prove it by using accumulation points or compactness.

A theorem in my book states that a set S in a metric space is closed if and only if whenever {q_1,q_2 ,... } is a sequence of points in S that is convergent then the lim n→∞ q_n ∈S .

3. The attempt at a solution

Suppose the set S is not closed which will imply that S^c is not open then there exists a point in S^c; lets denote this point p ∈ S^c such that there is an open ball B(p,r)∈ S^c ...

But I cant continue any further because I don't know what to do next.

Last edited: Sep 25, 2013
2. Sep 25, 2013

### brmath

I have a feeling this problem is not stated quite right. They haven't given you any way to determine whether that limit is in the space S or not.

Here's an example where it doesn't work: Let Q be the metric space of rational numbers where the metric is usual distance |x - y|. Let $p_n$ be a sequence of rational numbers whose limit is $\sqrt 2$. Clearly the limit is not in Q and the sequence is not a closed set.

If you were told the metric space is complete, you would have a chance.

3. Sep 25, 2013

### Lee33

If that is the case then I am sure that the metric space must be complete then.

4. Sep 25, 2013

### gopher_p

In the metric space $\mathbb{Q}$, a sequence $p_n$ of positive elements such that $p_n^2\rightarrow 2$ does not have a limit. It's not that there is a limit, just not in $\mathbb{Q}$; there isn't a limit. Such a sequence is closed in $\mathbb{Q}$; every element of $\mathbb{Q}$ not in the sequence has an open neighborhood which does not contain any elements of the sequence.

5. Sep 25, 2013

### gopher_p

You want to use that theorem in your book. What kinds of sequences in your set can converge? What are the limit points of your set?

6. Sep 25, 2013

### Lee33

Sequence that can converge are bounded. The limit point will just be the point p because limit points are unique.

7. Sep 25, 2013

### gopher_p

What about the sequence $p_1,p_1,p_1,...$?

8. Sep 25, 2013

### Lee33

They are in our set S.

9. Sep 25, 2013

### gopher_p

Does the sequence $p_1,p_1,p_1,...$ converge? To what?

10. Sep 25, 2013

### Lee33

Yes the sequence $p_1,p_1,p_1,...$ does converge to the point p.

11. Sep 25, 2013

### gopher_p

$p_1,p_1,p_1,...$ is a constant sequence. It converges to $p_1$.

12. Sep 25, 2013

### Lee33

I didn't carefully look at your sequence. I thought it was p1, p2, ... pn. But yes it converges to p1, but then how can I finish my proof?

13. Sep 25, 2013

### gopher_p

Well you gotta first decide if you want to continue trying to prove that the compliment of your set is open or if you want to take my advice an use the theorem from your book. If you want to use the theorem from your book, then you need to think about the types of sequences that you can get from the set $\{ p,p_1,p_2,...\}$ and the potential limits of those sequences. One further hint that I'll give is to consider convergent sequences which contain only finitely many distinct elements from your set and then those that contain infinitely many distinct elements.

Also, and I can only speculate, but you might find some inspiration in the proof of the theorem from your book. You might be able to borrow some ideas from that proof and use them in yours.

14. Sep 25, 2013

### Office_Shredder

Staff Emeritus
If you want to use the complement of a closed set is open definition, then I would do proof by contradiction. Suppose the complement is not open, and you have a point q which is not p or p1, p2,..., and for every $\epsilon$, there exists some n such that $d(q,p_n) \leq \epsilon$.

Can you derive a contradiction from this?

15. Sep 26, 2013

### brmath

You certainly are right -- thank you for pointing it out to me.