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Real Analysis Question: Sequences and Closed Sets

  1. Nov 29, 2011 #1
    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. jcsd
  3. Nov 29, 2011 #2

    Dick

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    Try showing that the complement of E is open. You don't really need all those cases.
     
  4. Nov 29, 2011 #3
    So the complement of E would be all such numbers that don't have a subsequence which converges to it?
     
  5. Nov 29, 2011 #4

    Dick

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