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Homework Help: Convergence of Sequences and closed sets

  1. Jan 18, 2010 #1
    1. The problem statement, all variables and given/known data
    This is the Theorem as stated in the book:
    Let S be a subset of a metric space E. Then S is closed if and only if, whenever p1, p2, p3,... is a sequence of points of S that is convergent in E, we have:
    lim(n->inf)pn is in S.


    2. Relevant equations
    From "introduction to Analysis" Rosenlicht, page 47.


    3. The attempt at a solution
    I understand the "only if" portion of this theorem, in that a closed subset implies the limit will lie in the subset. However, I'm missing something in the "if" portion, in that if a subset contains a convergent sequence and the limit is contained in the subset, then the subset must be closed. Maybe I am reading this wrong, but could it not be the case that a convergent sequence (and its limit) lie completely in an open subset. For example, the sequence 1/n^2 is contained completely in (-1,1), an open subset of the metric space R.

    BTW... I've used the info in this forum for a long time... glad to finally be a part of it :)
     
  2. jcsd
  3. Jan 18, 2010 #2
    The statement asserts that whenever you have a convergent sequence of terms in A, the limit point for the sequence is also in A. One such sequence is not sufficient.
     
  4. Jan 18, 2010 #3
    Does the statement not also imply that a subset containing a convergent sequence AND its limit must be a CLOSED subset in the metric space? This is the part I have a question about.

    Why could a subset contain a convergent sequence and its limit, but be an OPEN subset?

    Thanks in advanced for your help....
     
  5. Jan 18, 2010 #4

    LCKurtz

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    I assume your definition of a set S being closed is that it contains all of its limit points. So you are looking at the theorem that a set is closed if and only if the limit of any convergent sequence of points of S is itself in S. The "only if" part of this says:

    If every converent sequence in S has its limit point in S, then S is closed. I think you are missing the "every". The problem with your example is there is a sequence in S that doesn't have the property: {1 - 1/n}, so the interval isn't closed.
     
  6. Jan 18, 2010 #5
    I see! Quite an oversight on my part... thanks for the help!
     
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