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Real Analysis 1 Question

  1. Mar 21, 2017 #1
    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. jcsd
  3. Mar 21, 2017 #2

    fresh_42

    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.
     
  4. Mar 21, 2017 #3
    You have a family of convergent subsequences that are all bounded. Let [itex]D[/itex] be the set that contains the limit points of the subsequences. What can be said about [itex]D[/itex]?
     
  5. Mar 22, 2017 #4
    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?
     
  6. Mar 22, 2017 #5
    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
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