- #1
Poopsilon
- 288
- 1
Let A be a non-empty subset of the complex plane and let b ∈ ℂ be an arbitrary point not in A. Now define d(A,b) := inf{|z-b| : z ∈ A}. Show that if A is closed, then there is an a ∈ A such that d(A,b) = |a-b|.
Ok so basically what I did was begin by choosing some arbitrary element of A and labeling it z_1 and then set |z_1 - b|=x_1. Then I defined a rule such that we find some other element in A, z_2, such that |z_2 - b|=x_2 < x_1. So in this way we have defined a sequence of real numbers which is always positive, decreasing, and bounded below by zero, hence it converges to some real number call it x. Now there is a plurality of z ∈ ℂ such that |z-b| = x, the difficult part for me is showing that one of these z is in A.
I mean I can see that the sequence (z_n) need not converge, since it may simply continue bouncing around a circle of radius x in the complex plane. What I originally wanted to use to prove that there is a z ∈ A such that |z-b|=x was that convergent sequences in closed sets converge to limits within that set. But since (z_n) need not necessarily converge for (x_n) to converge, that puts a damper on things. I was thinking of maybe finding a subsequence which converges, or assuming by contradiction that if none of the z such that |z-b|=x are in A than that would cause (x_n) not to converge, but I can't find a way to set that up properly either.
Ok so basically what I did was begin by choosing some arbitrary element of A and labeling it z_1 and then set |z_1 - b|=x_1. Then I defined a rule such that we find some other element in A, z_2, such that |z_2 - b|=x_2 < x_1. So in this way we have defined a sequence of real numbers which is always positive, decreasing, and bounded below by zero, hence it converges to some real number call it x. Now there is a plurality of z ∈ ℂ such that |z-b| = x, the difficult part for me is showing that one of these z is in A.
I mean I can see that the sequence (z_n) need not converge, since it may simply continue bouncing around a circle of radius x in the complex plane. What I originally wanted to use to prove that there is a z ∈ A such that |z-b|=x was that convergent sequences in closed sets converge to limits within that set. But since (z_n) need not necessarily converge for (x_n) to converge, that puts a damper on things. I was thinking of maybe finding a subsequence which converges, or assuming by contradiction that if none of the z such that |z-b|=x are in A than that would cause (x_n) not to converge, but I can't find a way to set that up properly either.
