Results 1 to 6 of 6

Math Help - [SOLVED] Complex limit points

  1. #1
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,843
    Thanks
    320
    Awards
    1

    [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 \alpha is said to be a limit point of a set E of points in the complex plane (as opposed to a sequence z_n) if every neighborhood of \alpha contains infinitely many (distinct) points of E. Show that a sequence z_n can have a limit point \alpha in the sense that, given any \epsilon the inequality |z_n - \alpha | < \epsilon holds for infinitely many points of z_n, without being a limit point of the set of points corresponding to the terms of 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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1574
    Awards
    1
    Dan, are you quite sure that you have quoted this problem correctly?
    It clear that \alpha can be a limit point of the set \left\{ {z_n } \right\} without being the limit of the sequence \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 \varepsilon >0 the inequality \left| {z_n  - \alpha } \right| < \varepsilon holds for infinitely many points of {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: \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; February 16th 2008 at 09:29 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,843
    Thanks
    320
    Awards
    1
    Quote Originally Posted by Plato View Post
    Dan, are you quite sure that you have quoted this problem correctly?
    It clear that \alpha can be a limit point of the set \left\{ {z_n } \right\} without being the limit of the sequence \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 \varepsilon >0 the inequality \left| {z_n  - \alpha } \right| < \varepsilon holds for infinitely many points of {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: \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
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    This is not true. Consider z_n = 0 then there is no limit point for the set \{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 z_n has infinitely many point. But that is still not good enough because consider z_n = n then \{1,2,3....\} and the points are never close enough together so that within a certain small neighrhood they accumulate. However, if z_n is a sequence of distinct point that converges then \{z_1,z_2,z_3,...\} has an accumulation point. In fact, the accumation point (limit point) is the limit of the sequence. Let z_n \to z. Then for any \epsilon > 0 we can find N so that if n\geq N\implies |z_n - z|<\epsilon thus z_N,z_{N+1},z_{N+2},... are all within the \epsilon-neighborhood of z and the crucial step here is that all these point are distinct hence there are infinitely many of them.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,605
    Thanks
    1574
    Awards
    1
    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.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,843
    Thanks
    320
    Awards
    1
    I see it now. Thank you both.

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Limit points
    Posted in the Calculus Forum
    Replies: 0
    Last Post: November 8th 2009, 08:53 AM
  2. Replies: 3
    Last Post: January 11th 2009, 11:49 AM
  3. Replies: 1
    Last Post: March 3rd 2008, 07:17 AM
  4. [SOLVED] Complex limit points (Take II)
    Posted in the Calculus Forum
    Replies: 8
    Last Post: February 19th 2008, 08:16 AM
  5. [SOLVED] Three colinear complex points
    Posted in the Advanced Math Topics Forum
    Replies: 4
    Last Post: January 27th 2008, 04:53 PM

Search Tags


/mathhelpforum @mathhelpforum