Problem 8 - Complex Analysis

Question

For any subsets \(A, B \subset C\) $$ d(A, B):=\inf \\{|z-w| ; \quad z \in A, w \in B\\} $$ is called the distance between \(A\) and \(B\). If \(B=\\{w\\}\), then one simply writes \(d(A, w)\) instead of \(d(A,\\{w\\})\) Show: (a) If \(A \subset \mathbb{C}\) is a closed subset and \(b \in \mathbb{C}\) is arbitrary, then there is an \(a \in A\) with $$ d(A, b)=|a-b| $$ (b) If \(A \subset \mathbb{C}\) is a closed subset and \(B \subset \mathbb{C}\) is compact, then there are elements \(a \in A\) and \(b \in B\) such that $$ d(A, B)=|a-b| $$

Step-by-Step Solution

Answer

For (a), an element \( a \in A \) exists such that \( d(A, b) = |a-b| \). For (b), elements \( a \in A \), \( b \in B \) exist such that \( d(A, B) = |a-b| \).

Step 1: Understand and Define the Problem

The exercise involves proving statements about the distance between a subset and a point or another set in complex numbers. For the first part, we need to prove that for any closed set \( A \) and a point \( b \), there exists an element \( a \) such that the distance \( d(A, b) = |a-b| \). For the second part, we have to prove the existence of elements \( a \in A \) and \( b \in B \) such that the distance \( d(A, B) = |a-b| \), where \( A \) is closed and \( B \) is compact.

Step 2: Use Properties of Closed Sets for Part (a)

Since \( A \) is closed and \( b \) is any complex number, consider the function \( f(z) = |z-b| \) where \( z \in A \). The function \( f(z) \) reaches its infimum because \( A \) is closed and the function \( |z-b| \) is continuous. Therefore, there exists an \( a \in A \) such that \( f(a) = d(A, b) = |a-b| \). This yields the required result for (a).

Step 3: Apply Compactness for Part (b)

Since \( A \) is closed and \( B \) is compact, consider the function \( g(z,w) = |z-w| \) where \( z \in A \) and \( w \in B \). Due to the compactness of \( B \), the infimum of \( g(z,w) \) over \( A \times B \) is achieved by the Extreme Value Theorem, which states that a continuous function on a compact set reaches its minimum. Thus, there are points \( a \in A \) and \( b \in B \) such that \( g(a,b) = d(A,B) = |a-b| \). This completes the proof for (b).

Key Concepts

Closed Subset

In the realm of complex analysis, a closed subset is a set that contains all its boundary points. This means, if you take any point that 'almost' belongs to the set, it actually does belong to the set. Practically, this means the set is 'closed off' like a sealed container.

• Closed sets are essential in analysis because they ensure limits of sequences within the set remain inside the set.
• For example, if you are approaching the boundary of a closed set from within, you'll still remain in the set.

This property is key in the exercise because it ensures the function \( f(z) = |z-b| \) achieves its lowest value purely due to this closure property. This is possible since the set does not exclude potential minimum points near the boundary. In complex analysis, closed sets frequently help in demonstrating continuous functions achieve their extreme values on them, as shown in the first part of the solution.

Compact Set

A compact set in mathematical analysis is like a 'finite' set, although it might 'seem' infinite. This means you can contain the entire set within a bounded region, and every sequence in the set has a limit that is also within the set.

• Compactness implies both closedness and boundedness, crucial for ensuring certain functions behave well on such sets.
• Think of compactness as a combination of being completely sealed and fitting into a finite box.

In the second part of the original exercise, this concept is used when discussing the set \( B \). The fact that \( B \) is compact is crucial in applying the Extreme Value Theorem, ensuring that the function \( g(z,w) = |z-w| \) reaches its minimum on \( A \times B \). This is possible because compact sets do not allow values to 'escape' the defined region, guaranteeing the existence of this minimum.

Extreme Value Theorem

The Extreme Value Theorem is a fundamental concept in analysis, asserting that if a function is continuous over a compact set, it must achieve a maximum and a minimum value on that set.

• This theorem is crucial for proving the existence of extreme points (highest or lowest values) within a certain space.
• Without compactness, it might be possible for a function to keep getting smaller (or larger) indefinitely without reaching a definite minimum (or maximum).

In the context of the exercise, the theorem helps ensure that for the function \( g(z,w) = |z-w| \) defined on the product of sets \( A \) and \( B \), there exist points \( a \in A \) and \( b \in B \) where the minimum distance \( |a-b| \) is achieved. This is because the spaces considered (due to compactness and closedness) limit how the function's values can behave, ensuring it stays within specific bounds and reaches these extremes.