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Show that lim S is a closed set

  1. Sep 14, 2017 #1
    1. The problem statement, all variables and given/known data
    Suppose that pn → p and each pn lies in lim S. We claim that p ∈ lim S. Since
    pn is a limit of S there is a sequence (pn,k)k∈N in S that converges to pn as k →∞.
    Thus there exists qn = pn,k(n)∈ S such that
    d(pn, qn) <1/n.
    Then, as n→∞ we have
    d(p, qn) ≤ d(p, pn) + d(pn, qn) → 0
    which implies that qn → p, so p ∈ lim S, which completes the proof that lim S is a
    closed set.

    2. Relevant equations


    3. The attempt at a solution
    The only part of the proof I don't understand is the following:

    Since
    pn is a limit of S there is a sequence (pn,k)k∈N in S that converges to pn as k →∞.
    Thus there exists qn = pn,k(n)∈ S such that
    d(pn, qn) <1/n.

    Where does the 1/n come from? Why can we be sure that such a qn exists?
     
  2. jcsd
  3. Sep 14, 2017 #2

    Dick

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    You are given that ##p## is a limit of points of ##lim S##. You want to show that ##p## itself is an element of ##lim S##. To do this you have show that there is a sequence of point in ##S## whose limit is ##p##. The argument shows that for every ##n## there is a point of ##S## whose distance from ##p## is less than ##1/n##. Which goes to zero as ##n \rightarrow \infty##. Hence ##p## is in ##lim S##. Clearer?
     
  4. Sep 14, 2017 #3

    FactChecker

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    Anything that goes to 0 will do the trick. You can use 1/n and n→∞ or you can use ε>0 and ε→0. They will both make the distance go to 0.
    Because pn is a limit of points in S, there is a qn at least that close to pn.
     
  5. Sep 16, 2017 #4
    I guess my main confusion lies in the following: Thus there exists qn = pn,k(n)∈ S such that d(pn, qn) <1/n.

    Does this somehow come from the definition of convergence? Also, I am confused by the notation. The proof says that there is a sequence that converges to pn. So are we saying that n is some fixed number in this instance? Because obviously a sequence can't converge to another sequence.
     
  6. Sep 17, 2017 #5

    FactChecker

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    Yes. It comes from the sentence before: pn is a limit of S there is a sequence (pn,k)k∈N in S that converges to pn as k →∞.
    For that statement, n is fixed. But the logic will work for any n, which is important since n →∞.
    I agree. and it doesn't even have to converge to the same limit as the other sequence. It is converging to the limit of the points, pn that the sequences (pn,k)k∈N converge to.
     
  7. Sep 17, 2017 #6

    FactChecker

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    Start with the blue sequence of points, pn, converging to the red point, A. Each blue point is in S, so there is a sequence of green points converging to each one. Select the sequence of circled points converging to A. That circled sequence proves that A is also in S.
    limitOfLimits.png
     
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