[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.