Problem 22 - An Introduction to Abstract Analysis

Question

Let \(A\) be a subset of the normed linear space \(X\) satisfying \(\lambda a \in A\) whenever \(a \in A\) and \(\lambda \geq 0\). Show that \(A\) is closed if and only if \(A \cap\\{x:\|x\| \leq 1\\}\) is closed.

Step-by-Step Solution

Answer

In a normed linear space, prove that a subset \(A\) with the property \(\lambda a \in A, \forall \lambda \geq 0, a \in A\) is closed if and only if the intersection \(A \cap \{x: \|x\| \leq 1\}\) is closed.

Step 1: Part 1: If \(A\) is closed, then \(A \cap \{x: \|x\| \leq 1\}\) is closed

Suppose \(A\) is a closed set. Recall that the intersection of two closed sets is also closed. To show that \(A \cap \{x: \|x\| \leq 1\}\) is closed, we need to prove that the set \(\{x: \|x\| \leq 1\}\) is closed. Denote the set \(\{x: \|x\| \leq 1\}\) by \(B\). Let \((x_n)_{n \in \mathbb{N}}\) be a convergent sequence in \(B\) with limit \(x \in X\). Since the norm function \(\|\cdot\|\) is continuous, we have $$\|x\| = \lim_{n \to \infty} \|x_n\| \leq 1.$$ Thus, \(x \in B\) and we have shown that \(B\) is closed. So, \(A \cap \{x: \|x\| \leq 1\}\) is the intersection of two closed sets, making it closed as well.

Step 2: Part 2: If \(A \cap \{x: \|x\| \leq 1\}\) is closed, then \(A\) is closed

Now suppose \(A \cap \{x: \|x\| \leq 1\}\) is closed. Let \((a_n)_{n \in \mathbb{N}}\) be a convergent sequence in \(A\) with limit \(a \in X\). We need to show that \(a \in A\). We know that for each \(a_n\), there exists a scalar \(\lambda_n \geq 0\) such that \(\lambda_n a_n \in A \cap \{x: \|x\|\leq 1\}\). We can write \(a_n = \frac{1}{\lambda_n} (\lambda_n a_n)\), where \(\frac{1}{\lambda_n} \geq 0\). Notice that the sequence \((\lambda_n a_n)_{n \in \mathbb{N}}\) is a sequence in the closed set \(A \cap \{x: \|x\|\leq 1\}\), which converges to \(\frac{1}{\lambda} a\), where \(\lambda = \lim_{n \to \infty} \lambda_n \geq 0\). Since \(A \cap \{x: \|x\|\leq 1\}\) is closed, this limit point \(\frac{1}{\lambda} a\) also belongs to \(A \cap \{x: \|x\|\leq 1\}\). Moreover, since \(\lambda \geq 0\), we have that \(a = \lambda (\frac{1}{\lambda} a) \in A\). This shows that \(A\) is closed. Thus, we have proven that \(A\) is closed if and only if \(A \cap \{x: \|x\|\leq 1\}\) is closed.

Key Concepts

Closed Sets

In the context of normed linear spaces, a set is described as *closed* if it contains all its limit points. This means that whenever a sequence within the set converges to a limit, that limit is also part of the set.

Closed sets have important properties in analysis such as:

• They include all their boundary points and limits of sequences within them.
• The intersection of closed sets is also closed.

In the given problem, we discuss how a set \(A\) in a normed space can be deemed closed if and only if the intersection of \(A\) with another closed set is also closed. This is proved by showing that if \(A\) is closed, then by the property of intersections, \(A \cap \{x: \|x\| \leq 1\}\) is closed. And vice versa, if the intersection is closed, so must be \(A\). Understanding this back-and-forth relationship helps in grasping the structural properties of closed sets.

Convergent Sequences

A sequence \((x_n)\) in a normed space is said to be *convergent* if there exists a point \(x\) such that as \(n\) approaches infinity, \(x_n\) tends towards \(x\). In simple terms, as you take more terms in the sequence, they get closer and closer to a single, fixed point known as the limit.

• The sequence has a limit \(x\) if for every \(\epsilon > 0\), there exists a number \(N\) such that for all \(n > N\), \(\|x_n - x\| < \epsilon\).
• Convergent sequences are always bounded, meaning they don't "escape" to infinity.

This concept is key in the problem because the closure of a set \(A\) is defined in terms of limit points arising from convergent sequences contained in \(A\). If every sequence within \(A\) that has a limit in space \(X\) has that limit also in \(A\), then \(A\) is closed.

Intersection of Sets

The *intersection* of two sets \(A\) and \(B\), denoted \(A \cap B\), is the set containing all elements that are present in both \(A\) and \(B\). When dealing with normed linear spaces, the intersection of sets aids in investigating properties such as closedness.

• If both sets in the intersection are closed, their intersection is also closed.
• Intersecting with a norm-bounded set like \(\{x: \|x\| \leq 1\}\) helps to localize analysis to a unit ball, which is itself closed.

In the problem you're working on, the condition involving the intersection with \(\{x: \|x\| \leq 1\}\) enables one to verify the closed nature of \(A\). The closed properties of the intersection essentially "test" for the enclosed properties of the larger set, \(A\), allowing for deeper insights into the structure of both.