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