Sequences in Complex Plane which Converge Absolutely

1. Nov 7, 2011

Poopsilon

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.

2. Nov 7, 2011

LCKurtz

Say you call K = {z: |z - b| ≤ 1}. Then $K\cap A$ is compact and you can get your sequence in there...

3. Nov 7, 2011

Poopsilon

I'm sorry but this is simply too terse and cryptic for me to comprehend. The distance from the closest point in A to b is unknown so your intersection of K and A may be empty. Are you describing a way to find a convergent subsequence? I just really can't tell, please elaborate.

4. Nov 7, 2011

Dick

Saying K = {z: |z - b| ≤ 1} is probably not what LCKurtz meant to write. How about K = {z: |z - b| ≤ 2*d(A,b)}? You know how having a compact set would solve your 'points bouncing around' problem, yes?

5. Nov 7, 2011

LCKurtz

Yes, careless slip there. Thanks Dick.

6. Nov 8, 2011

Poopsilon

Ah, yes, excellent, thanks.

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