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Thread: [SOLVED] Complex limit points

  1. #1
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    [SOLVED] Complex limit points

    This is the first of a probably numerous series (pun intended) of questions about complex series. My series work has never been good, and I don't write proofs well.

    So here we go with the first:
    A point $\displaystyle \alpha$ is said to be a limit point of a set E of points in the complex plane (as opposed to a sequence $\displaystyle z_n$) if every neighborhood of $\displaystyle \alpha$ contains infinitely many (distinct) points of E. Show that a sequence $\displaystyle z_n$ can have a limit point $\displaystyle \alpha$ in the sense that, given any $\displaystyle \epsilon$ the inequality $\displaystyle |z_n - \alpha | < \epsilon$ holds for infinitely many points of $\displaystyle z_n$, without being a limit point of the set of points corresponding to the terms of $\displaystyle z_n$.
    I can't even think of an example that fits this situation, much less writing a proof for it. Any pointers (not full solutions please) would be appreciated.

    -Dan
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    Dan, are you quite sure that you have quoted this problem correctly?
    It clear that $\displaystyle \alpha$ can be a limit point of the set $\displaystyle \left\{ {z_n } \right\}$ without being the limit of the sequence $\displaystyle \left( {z_n } \right)$. Unless you textbook is making some unusual point, I don’t think that statement is true. (Or maybe I have misread the statement!)

    Post Script
    I have just thought of a reading of the phrase “given any $\displaystyle \varepsilon >0$ the inequality $\displaystyle \left| {z_n - \alpha } \right| < \varepsilon $ holds for infinitely many points of $\displaystyle {z_n }$”. If we read that as meaning infinitely many terms of the sequence then any sequence that is constant from some index on will fit that reading. Here is a trivial example: $\displaystyle \left\{ {z_n :\forall n\left[ {z_n = i} \right]} \right\}$. The number i is limit of the sequence but not a limit point.
    Last edited by Plato; Feb 16th 2008 at 09:29 AM.
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Plato View Post
    Dan, are you quite sure that you have quoted this problem correctly?
    It clear that $\displaystyle \alpha$ can be a limit point of the set $\displaystyle \left\{ {z_n } \right\}$ without being the limit of the sequence $\displaystyle \left( {z_n } \right)$. Unless you textbook is making some unusual point, I don’t think that statement is true. (Or maybe I have misread the statement!)

    Post Script
    I have just thought of a reading of the phrase “given any $\displaystyle \varepsilon >0$ the inequality $\displaystyle \left| {z_n - \alpha } \right| < \varepsilon $ holds for infinitely many points of $\displaystyle {z_n }$”. If we read that as meaning infinitely many terms of the sequence then any sequence that is constant from some index on will fit that reading. Here is a trivial example: $\displaystyle \left\{ {z_n :\forall n\left[ {z_n = i} \right]} \right\}$. The number i is limit of the sequence but not a limit point.
    I'm going to have to "chew" on this for a bit. I'll get back to you. Thank you!

    -Dan
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    This is not true. Consider $\displaystyle z_n = 0$ then there is no limit point for the set $\displaystyle \{z_1,z_2,...\}$ because it is a finite set. And no neighborhood has will have infinitely many point of this set because the set is finite. So you might say what happens if $\displaystyle z_n $ has infinitely many point. But that is still not good enough because consider $\displaystyle z_n = n$ then $\displaystyle \{1,2,3....\}$ and the points are never close enough together so that within a certain small neighrhood they accumulate. However, if $\displaystyle z_n$ is a sequence of distinct point that converges then $\displaystyle \{z_1,z_2,z_3,...\}$ has an accumulation point. In fact, the accumation point (limit point) is the limit of the sequence. Let $\displaystyle z_n \to z$. Then for any $\displaystyle \epsilon > 0$ we can find $\displaystyle N$ so that if $\displaystyle n\geq N\implies |z_n - z|<\epsilon$ thus $\displaystyle z_N,z_{N+1},z_{N+2},...$ are all within the $\displaystyle \epsilon$-neighborhood of $\displaystyle z$ and the crucial step here is that all these point are distinct hence there are infinitely many of them.
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    It is true if one reads the statement as “each neighborhood contains infinitely many terms of the sequence”.
    I suspect that was the intention of the author of the problem.
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    Forum Admin topsquark's Avatar
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    I see it now. Thank you both.

    -Dan
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