# Real Analysis 1 Question

1. Mar 21, 2017

### PsychonautQQ

1. The problem statement, all variables and given/known data
Let (a_n) be a bounded sequence. Prove that the set of subsequential limit points of (a_n) is a subsequentially compact set

2. Relevant equations
To be a subsequentutially compact set, every sequence in the set of limit points of (a_n) must have a convergent subsequence.

3. The attempt at a solution
I need a hint to help get me started >.< haha. My attempt at the solution is just thoughts, hard to get the pencil to the paper if you know what I mean.

So first of all i'm trying to think of what sequences in the set of subsequential limit points will look like. Yeah, any insight what-so-ever is appreciated, analysis is hard >.<

2. Mar 21, 2017

### Staff: Mentor

Without having the solution in mind ... it's always a good idea, to start with what one has. This means to write down the definitions. I counted three: limit point, subsequential limit point and subsequential compact. I suppose this involves a lot of subsequences, so it might be possible to concentrate on only them. I guess, you will need to construct a (convergent) sequence out of the ones in the set of subsequential limit points and you will be able to use the ordinary compactness to find one.

3. Mar 21, 2017

### nuuskur

You have a family of convergent subsequences that are all bounded. Let $D$ be the set that contains the limit points of the subsequences. What can be said about $D$?

4. Mar 22, 2017

### PsychonautQQ

So I have a family of convergent subsequences because I have a set of subsequential limit points. If we let D be the set that contains the limit points of the subsequences, we can say that D is bounded, because (a_n) is bounded. Therefore we can use the bolzano-weirstrass theorem which states that every bounded sequences has a convergent subsequence, and thus the set of subsequential limit points of (a_n) is a subsequentially compact set, because any sequence in this set will have a convergent subsequence.
Boom shacka lacka?

5. Mar 22, 2017

### nuuskur

Your end result is correct, but you have convergent subsequences because the initial sequence is bounded. The Bolzano-Weierstrass theorem is a very powerful tool, as you can see :)

So I have a family of convergent subsequences because I have a set of subsequential limit points

This is a bit odd to consider. If I have a point, I can immediately construct a sequence that converges to that point. B-W guarantees the existence of at least one convergent subsequence of a bounded sequence. There are probably more. Without loss of generality, a family of such subsequences.

Last edited: Mar 22, 2017