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Topology in the Complex Plane

  1. Jan 17, 2008 #1
    I'm having some trouble understanding the distinction between closed sets, open sets, and those which are neither when the set itself involves there not being a finite boundary. For example, the set { |z - 4| >= |z| : z is complex}. This turns out to be the inequality 2>= Re(z). On the right side it does have a boundary at Re(z) = 2. To the left, up and down it just continues for every possible value of Im(z). Would this even be considered a boundary? Is the set open, closed or neither?

    Edit: Nevermind, instead I just want to make sure I'm going down the right road. A set which would extend infinitely would not have a boundary point, as there is no point exterior to the set in that direction, and therefore there isn't a point which is both an exterior and an interior point in the set to be a boundary point. This makes talk about boundaries in this case meaningless and the example above would come out to be a closed set because its only boundary is contained within the set. Correct?
    Last edited: Jan 18, 2008
  2. jcsd
  3. Jan 18, 2008 #2
    dear sculptured
    there are many ways in topology to prove that the set A={ |z - 4| >= |z| : z is complex} is closed. I will show you 2 ways that don't require you to think of boundarys, interior points or exterior points.

    1. you can prove that complement of A is an open set in C. Lets call this set B. we will have to show that B is the union of open sets. what is the typical open set in C? well the open disk of course! Can you think of a way to show that B is the union of open discs?

    2. you can prove that B is open by the use of a continuous map f! I think you have a good understanding of what are open sets in R, so lets find a continuous map f: C->R where B is the preimage of an open set.
    First note that A={|z-4| - |z| >= 0 : z is complex}
    Lets take f(z)=|z-4| -|z| which is continuous and
    [tex] A = f^-1( [0,\infty)) [/tex]
    [tex] B = f^-1( (-\infty,0) ) [/tex]
  4. Jan 31, 2009 #3
    is it true that,"the entire C plane is closed"???
    pls explain..
  5. Feb 1, 2009 #4
    In any topological space X, the empty set is (by the definition of a topology) open; therefore, its complement (which is X) is closed.
  6. Feb 1, 2009 #5
    thnks for ur reply...
  7. Feb 7, 2009 #6
    is zero a boundary point??explain,,
    C\{x+iy | x>0,y=0} is neither open nor closed.?explain...
  8. Feb 7, 2009 #7


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    A boundary point of what set? "Boundary Point" is only defined with respect to a given set. A point is a boundary point of set A if and only if it is neither an interior point of A nor an interior point of the complement of A ('exterior' point of A). That is the same as saying every neighborhood of the point contains some points of A and some points not in A.

    Neither. A set is closed if it contains all of its boundary points. A set is open if it contains none of its boundary points. This is the complex numbers with the positive real axis removed. Every neightborhood of a point on that axis contains some points not in the set (nearby points on the axis) and some points in the set (points with small imaginary point) so the set does not contain all of its boundary points and is not closed. But (0,0) is also a boundary point so the set does not contain "none" of its boundary points and is not open.

    By the way, you left out a possibility: there exist sets that are both open and closed.
  9. Feb 7, 2009 #8
    Mr. HallsofIvy, thanks for your reply. but the point, it is not closed is not that clear for me, could you gimme an example to make it clear.
  10. Feb 7, 2009 #9
    Every closed set contains its closure (the set in union with its boundary or frontier points: points for which every neighborhood contains some point of the set in question). The closure of your set is the entire complex plane, since your set consists of the complex plane with the positive real axis removed. Every point on that axis is a boundary point, as you can verify. Since your set does not contain its closure, it is not closed.
  11. Feb 7, 2009 #10


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    I assume you are referring to the set C- {(x,y)| x> 0, y= 0}, the complex plane with the positive real axis removed. As I said before, every point on {(x,y)| x> 0, y= 0} is a boundary point because a neighborhood of any such point, a tiny disk with (x,0) at its center, contains some points of the set as well as some points ((x,0) itself, for example) that are not in the set. Since a set is closed if and only if it contains all of its boundary points (which is the same as saying it contains its closure as slider142 says) this set is not closed.

    One can also argue that it is not closed by showing that its complement, here the set {(x,y)|x> 0 and y=0}, is not open. That is true because, again, every neighborhood of every point on that line contains points outside the set: Since this set contains some of its boundary points, it is not open.
  12. Feb 7, 2009 #11
    A consists of the set of points that have both coordinates rational,
    then every point in the plane is a limit point of A.

    A set is said to be closed if it contains all of its limit points...

    but,S={z : z=x+iy with x,y are rational} is not closed..!:confused: why is it so..?explain..
  13. Feb 7, 2009 #12
    The set A doesn't contain all of its limit points.
  14. Feb 7, 2009 #13
    How can a null set and C plane be closed and open,,,??
  15. Feb 7, 2009 #14
    As you said Mr.Adriank, The set A doesn't contain all of its limit points.
    which points are missing and how .?can you explain.?
  16. Feb 7, 2009 #15
    In addition, the entire space X is (by the definition of a topology) open, so its complement (the empty set) is closed. This is using the abstract definition of a topology.

    You might given the following definition of an open set in C: If A is a subset of C, then A is open if for every z in A, there is some r > 0 such that the set B(z, r) = {w in C with |z - w| < r} is contained in A. We also say that A is closed if the set C - A is open.

    Using this definition we can show that the empty set and C are open. If A is the empty set Ø, then there is no z in A, so the condition for an open set is vacuously true. If A is the entire space C, then clearly for every z in C, B(z, r) is a subset of C.

    Since Ø is open, C - Ø = C is closed; similarly, since C is open, the set C - C = Ø is closed.

    Let S be the set of points in C with both coordinates rational. As you say, every point in C is a limit point of S. Thus, any point in C but not S (i.e. any point with at least one irrational coordinate) is a limit point of S that is not in S.
    Last edited: Feb 8, 2009
  17. Feb 8, 2009 #16


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    ??? Didn't you just answer that question? As you say in your first sentence, The limit points of A are all points in the plane so A does not contain all its limit points and so is not closed.

    By the definition of "open" and "closed". In any topology both the entire set and the empty set are open. Since each is the complement of the other and complements of open sets are closed they are also both closed.

    A space is "connected" if and only if those are the only sets that are both open and closed.
  18. Feb 8, 2009 #17
    If z0 is a limit point of A, then every neighborhood of z0 contains
    infinitely many points of A.
    thats a theorem, for which the proof is not clear..
    Could somebody gimme the proof..with explanation..!
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