# Real Analysis Question: Sequences and Closed Sets

1. Nov 29, 2011

### Szichedelic

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

Let {xn} be a sequence of real numbers. Let E denote the set of all numbers z that have the property that there exists a subsequence {xnk} convergent to z. Show that E is closed.

2. Relevant equations
A closed set must contain all of its accumulation points.
Sets with no accumulation points are closed.
The Bolzano-Weierstrass theorem: every bounded sequence has a convergent subsequence.

3. The attempt at a solution
What I've done for this problem is I've split it up into several cases.

Case 1

{xn} is convergent to some number z. Then all subsequences converge to that number z. Thus, E = {z}. Since E is a finite set (consisting of one isolated point), it is closed.

Case 2
{xn} is unbounded and divergent. Then, by the Bolzano-Weierstrass theorem, there are no convergent subsequences. Thus, E=∅, which is both open and closed.

Case 3
{xn} is bounded and divergent. Then, by the Bolzano-Weierstrass theorem, there exists at least one convergent subsequence. If there are finitely many convergent subsequences, then E is a finite set and thus, is closed.

The problem I am having is what if there are infinitely many convergent subsequences? The idea in my head is that the set E would contain elements which were isolated from one another. However, I am having a pretty hard time clarifying this idea in my head.

Any hints would be much appreciated! Thank you.

2. Nov 29, 2011

### Dick

Try showing that the complement of E is open. You don't really need all those cases.

3. Nov 29, 2011

### Szichedelic

So the complement of E would be all such numbers that don't have a subsequence which converges to it?

4. Nov 29, 2011

### Dick

Sure. If you can show that set is open, then E is closed.