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Topological Properties of Closed Sets in the Complex Plane

  1. Mar 2, 2008 #1
    [SOLVED] Topological Properties of Closed Sets in the Complex Plane

    The problem statement, all variables and given/known data
    1. Show that the boundary of any set D is itself a closed set.
    2. Show that if D is a set and E is a closed set containing D, then E must contain the boundary of D.
    3. Let C be a bounded closed convex set and let D be the complement of C. Show that D is a domain.

    The attempt at a solution
    I would like some closure concerning the correctness of my answers.

    1. I will denote the complement of X by ~X. Let B be the boundary of D. Consider ~B. If D is closed, then ~B is the union of D - B and ~D. Both D - B and ~D are open and thus ~B is open so B is closed. If D is open, then ~B is the union of D and ~D - B. ~D is closed so ~D - B is open. ~B is thus open and B is closed.

    2. Suppose p is a boundary point of D not in E. By definition, any open disc d centered at p contains points in D and not in D. The point p in d is not in E. Thus, d contains points in E (namely thos in D) and not in E (namely p), so p is a boundary point of E. Since E is closed, p is contained in E. Contradiction. Thus, p is in E. Since p is an arbitrary boundary point of D, E contains D's boundary.

    3. D is a domain if it is open and connected. It is open because C is closed. To show that D is connected, I'm thinking of the following flaky argument: Let p and q be two points in D. Join them with a line segment. If that line segment is entirely within D, then all is well. If part of the line segment goes through C, stretch the segment up or down, left or right, until it hugs the exterior of C (which I think is possible because C is bounded). The "connection" between p and q is now entirely within C.
     
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  3. Mar 2, 2008 #2

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    For 1, you've only considered the cases where D is either open or closed, but D can in general be neither open nor closed. Try finding a single argument that works for all D. 2 looks ok. For 3, you're not using the fact that C is convex, and since the statement is false when C isn't convex, your proof can't be correct. You're on the right track though, just try to be a little more careful.
     
  4. Mar 2, 2008 #3
    Ah. I was under the impression that all sets are either open or closed. I'm imagining a set that looks like a disc with half of its circumference in the set and half not. This set is neither opened nor closed right because the boundary is the entire circumference. I understand how my argument in 1 fails now.

    So, B is closed if it contains all of its boundary points. Let B' be the boundary of B and let p be an arbitrary point in B'. If I prove that p is in B, I'm done. I may prove that p is in B by proving that p is a boundary point of D. Let d be any open disc centered at p. By definition, d contains points in B and not in B. I must prove that d contains points in D and not in D. By way of contradiction, suppose there is a d for which all its points are in D. Then p is an interior point of D so p is not in B. Wait a moment! This is not what I wanted to prove. I don't see any flaw so where did I go wrong.

    I did not use the fact that C is convex because I couldn't think of an appropriate example to prove its necessity. After some thought, I determined a set that is bound, closed but not convex for which my argument fails: a ring. The convexity of C prevents it from enclosing any points in D. Thus, the "line-segment stretching" argument is guaranteed to work.
     
  5. Mar 3, 2008 #4

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    Nothing went wrong, you just haven't reached your contradiction yet. Actually, a proof by contradiction probably isn't the best way. Remember that d must also contain some point b in B. Then apply the condition that b is a boundary point of D.

    Well, sort of, but that's nothing like a proof. What do you mean when you say C can't "enclose a point"? The only way I can interpret this is something like "prevent you from finding a path from the point to the area far away from C", which is basically what you're trying to prove, so it doesn't make sense to just state it as a fact.
     
    Last edited: Mar 3, 2008
  6. Mar 3, 2008 #5
    Keeping in mind that I need to prove that d contains points in D and points not in D, how does knowing that b is a boundary point of D help?

    It's difficult for me to describe the situation formally. Here's what I'm imagining: Since C is bounded and convex, then D is comprised of the complex-plane minus a big hole somewhere (which is C). Thus, if two points are on opposite sides of the hole, then one may always trace a path from one point to the other around the hole.
     
  7. Mar 3, 2008 #6
    I think I've figured out 1 with your suggestion. Since b is a boundary point of D inside d, then any disc centered at b contains points in D and points not in D. Thus, d must contain points in D and points not in D and so p is a boundary point of B. Is this correct?
     
  8. Mar 3, 2008 #7

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    Yea, you've got 1 now. For 3, one way would be to note that since C is bounded, it is contained in a ball (ie, a filled in circle) of radius R. You should be able to argue every point can be connected to the circle bounding this ball (roughly, if the point is outside (or on) the circle its easy, and if its inside, draw a straight line through the point and note C can't block both directions along this line by convexity).
     
  9. Mar 3, 2008 #8
    I think I understand your what you wrote. Since C is bounded, I may encircle it. Call the circle c. I may connect any two points in D by connecting them to the circle which is also in D. There are no problems if the points are outside c. If one of the points, say p, is inside c, then the line connecting p to c may pass through C. In this case, one may connect p to c by using the line opposite to the one that passes through C. By convexity, this line will not pass through C. Is this correct?
     
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