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Thread: Closed Sets and Convergent Sequences

  1. #1
    Member RedBarchetta's Avatar
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    Closed Sets and Convergent Sequences

    I was reading this theorem in my analysis text and came across something I could not quite follow:

    Theorem: Let S be a subset of the metric space E. Then S is closed if and only if, whenever $\displaystyle p_1, p_2, ...$ is a sequence of points of S that is convergent in E, we have $\displaystyle \lim_{n\to\infty} p_n \in S $.

    I understand the "only if" part. Now for the "if" part the text states:

    "Suppose S is not closed. Then the complement is not open, and there exists a point p in the complement of S such that any open ball of center p contains points of S. *Hence for each positive integer n we can choose $\displaystyle p_n \in S$ such that $\displaystyle d(p, p_n)<\frac{1}{n}$.Then $\displaystyle \lim_{n\to\infty} p_n = p $, with each $\displaystyle p_n \in S$ and $\displaystyle p \notin S$.*This shows that if the hypothesis on convergent sequences holds, then S must be closed, completing the proof. "

    ^^ I've bolded the text that I don't understand. I've thought about this and I can't figure out where the 1/n comes from as well as the conclusion $\displaystyle \lim_{n\to\infty} p_n = p $. Thanks for the help!
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  2. #2
    A Plied Mathematician
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    The $\displaystyle 1/n$ is like an epsilon. If a set is not open, then there is a point in that set such that every $\displaystyle \varepsilon$-neighborhood around the point contains points not in the set, correct? In this case, we're considering $\displaystyle S^{c}$, which by hypothesis is not open. The point $\displaystyle p\in S^{c}.$ But now the author wants to create a sequence of points in a sequence of index-able neighborhoods, so instead of using an $\displaystyle \varepsilon,$ he uses the equivalent sequence $\displaystyle 1/n$ in order to be able to index them. The sequence $\displaystyle p_{n}\in S,$ and hence is not in $\displaystyle S^{c},$ by construction.

    Does that make sense?
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  3. #3
    Member RedBarchetta's Avatar
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    I think I am getting confused on the use of n. Specifically, are we choosing a positive integer n and then considering all p_n such that $\displaystyle d(p, p_n)<\frac{1}{n}$ ?
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    Quote Originally Posted by RedBarchetta View Post
    I think I am getting confused on the use of n. Specifically, are we choosing a positive integer n and then considering all p_n such that $\displaystyle d(p, p_n)<\frac{1}{n}$ ?
    Actually you are constructing a sequence of points in $\displaystyle S$ such that the sequence converges to $\displaystyle p\notin S$ which contradicts the given.

    Now why? If $\displaystyle S^c$ is not open then for some $\displaystyle p\in S^c$ there is no ball at $\displaystyle p$ which is a subset of $\displaystyle S^c$.
    For each $\displaystyle n$ the ball $\displaystyle B\left( {p;\frac{1}{n}} \right)$ is not a subset of $\displaystyle S^c$.
    Thus $\displaystyle \left( {\exists p_n \in \left( {B\left( {p;\frac{1}
    {n}} \right) \cap S} \right)} \right)$

    Does that help?
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  5. #5
    Member RedBarchetta's Avatar
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    I guess my issue is taking these sets and using them to prove that this point p is the limit of the sequence of $\displaystyle p_n$.

    So by definition: a point $\displaystyle p \in E$ is called a limit of the sequence $\displaystyle p_1, p_2, ...$ if, given any real number $\displaystyle \epsilon > 0 $, there is a positive integer $\displaystyle N$ such that $\displaystyle d(p, p_n)< \epsilon$ whenever $\displaystyle n>N$.

    How do we take this collection of open balls and prove using the definition that the limit is indeed p? So why can we use just 1/n rather than just some arbitrary real number? how do we know we can find big N? Can S be a finite set? or is an infinite set? I guess I have too many questions!

    So are you saying when you consider a set B(p, 1/n) for some n, then there exists some p_n in S that is in that open ball? Say we let n=1, then are you saying that p_1 is in B(p, 1)? or just some arbitrary p_n? could it be just one p_n say p_3? or possibly multiple p_n's say p_3, p_4?

    Sorry for not getting this, I feel like i'm missing something.
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  6. #6
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    Quote Originally Posted by RedBarchetta View Post
    So are you saying when you consider a set B(p, 1/n) for some n, then there exists some p_n in S that is in that open ball? Say we let n=1, then are you saying that p_1 is in B(p, 1)? or just some arbitrary p_n? could it be just one p_n say p_3? or possibly multiple p_n's say p_3, p_4?
    Sorry for not getting this, I feel like i'm missing something.
    You are missing a great deal. Or maybe you donít understand sequential limits.

    First, did you notice we said for each n?
    That means for each n we get a $\displaystyle p_n$.

    Moreover, for each $\displaystyle \varepsilon >0$, $\displaystyle \left( {\exists n} \right)\left[ {\frac{1}{n} < \varepsilon } \right]$.
    Thus $\displaystyle d\left( {p,p_n } \right) < \frac{1}{n} < \varepsilon $.
    That means $\displaystyle \left( {p_n } \right) \to p \notin S$.
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