Does Minimum of Complex Set Subset Exist?

Bashyboy
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Homework Statement


The following doesn't come from a textbook, and I am very uncertain whether it is true or false. Suppose that ##B \subseteq \mathbb{C}## is a convex set, and consider the set ##L_B := \{|b|: b \in B \}##.

Homework Equations

The Attempt at a Solution


My question is, will ##min~L_B## exist? My thought was that ##B## being convex implied that ##L_B## is convex; but I am unsure whether convexity of ##L_B## is sufficient to conclude that ##min~L_B##. Please refrain from giving me an entire answer, but I would appreciate a few hints.
 
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Think about an open set in ##\mathbb{R}##.
 
Ah, a counterexample! For instance, if we have ##B = (0,1)##, then ##L_B = B##, yet ##B## does not have a minimum. What if we stipulate that ##B## must also be compact?
 
Does a compact set always have a minimum?
 
A compact set is both bounded and closed.
 
Since a compact set is bounded and closed, the infimum is the minimum, and the minimum exists.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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