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[SOLVED] Seperation of a Point and Convex Set
Let C be a closed convex set and let r be a point not in C. It is a fact that there is a point p in C with |r - p| l<= |r - q| for all q in C.
Let L be the perpendicular bisector of the line segment from r to p. Show that no point of C lies on L or in the half-plane, determined by L, which contains r.
2. The attempt at a solution
The point p happens to be the closest point in C to r. I can imagine how the point s in the middle of the line segment from r to p and some of the points near s on L could not be in C because then there would be a closer point to r in C than p. But what about points far away from s? I'm guessing the fact that C is closed and convex comes into play at this point, but how?
Homework Statement
Let C be a closed convex set and let r be a point not in C. It is a fact that there is a point p in C with |r - p| l<= |r - q| for all q in C.
Let L be the perpendicular bisector of the line segment from r to p. Show that no point of C lies on L or in the half-plane, determined by L, which contains r.
2. The attempt at a solution
The point p happens to be the closest point in C to r. I can imagine how the point s in the middle of the line segment from r to p and some of the points near s on L could not be in C because then there would be a closer point to r in C than p. But what about points far away from s? I'm guessing the fact that C is closed and convex comes into play at this point, but how?