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Clarification needed

  1. Aug 30, 2012 #1
    1. The problem statement, all variables and given/known data
    Prove that a non-empty set of complex numbers F is closed iff every convergent sequence of elements of F converges to an element of F.


    2. Relevant equations



    3. The attempt at a solution

    I don't understand the second part of the iff statement. If someone could clarify what it means about convergent sequences, I could go from there.
     
  2. jcsd
  3. Aug 30, 2012 #2

    LCKurtz

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    Not sure what you call the "second part". You have to show two statements:

    1. If every convergent sequence of elements of F converges to an element of F, then F is closed.

    2. If F is closed, then every convergent sequence of elements in F converges to an element of F.
     
  4. Aug 30, 2012 #3
    I don't understand what "every convergent sequence of elements of F converges to an element of F" means.
     
  5. Aug 30, 2012 #4

    Bacle2

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    Well, {1/n} n=1,.... is a sequence of elements in the interval (0,1), and the

    sequence converges to ___ which is in ____.......
     
  6. Aug 30, 2012 #5
    it converges to 0, which is in the interval, or the border of the interval. But 0 isn't an element of that sequence though.
     
  7. Aug 30, 2012 #6

    Bacle2

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    Actually, 0 is _not_ in the interval. Remember:

    (0,1):={ x: 0<x<1 } , which does not include 0 . Tho this is just for the

    real line.
     
  8. Aug 30, 2012 #7
    So how does that example work in this situation then. We needed a sequence who's limit is an element of the sequence.
     
  9. Aug 30, 2012 #8

    Bacle2

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    I was trying to illustrate a familiar case in which a sequence defined on S does not

    converge to a point in S. What is your book's definition of closed set ?
     
  10. Aug 30, 2012 #9
    We don't have a book, but basically a closed set is one where the boundary of the set is contained in the set.
     
  11. Aug 30, 2012 #10

    Bacle2

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    Maybe this is the nicest , more general way of doing things:

    If F is closed in ℂ , then U:= ℂ\F is open . Since U is ( U are? ) open, if x is in U ,

    then there is a 'hood ( neighborhood) of x contained in U , i.e., ( in this context) a

    ball B(x,r) contained in U . What would happen if x was a limit point of some sequence

    {fn} in F , could you find such a ball ?
     
  12. Aug 30, 2012 #11

    Bacle2

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    Sorry, LCKurtz, I did not mean to strong-arm myself into the post and take-it over.
     
  13. Aug 30, 2012 #12

    LCKurtz

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    No problem. I have been gone all afternoon anyway. And judging from the responses, it may take both of us to explain the concept.
     
  14. Aug 30, 2012 #13

    LCKurtz

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    F is a subset. Every convergent sequence in F has a limit. This limit may or may not be in F. Look at Bacle's example on the real line. If F = (0,1) and ##x_n=\frac 1 {n+1}## then ##x_n\rightarrow 0##, which is not in F. So F isn't closed. Such limits must all be in F if F is to be closed. If F = [0,1], then it is closed, as an example.
     
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