Problem 23 - An Introduction to Abstract Analysis

Question

For any set \(A\) in a normed linear space \(X\), show that \(\operatorname{dist}(x, A)=\) \(\operatorname{dist}(x, \bar{A})\) for all \(x\) in \(X\).

Step-by-Step Solution

Answer

Question: Show that the distance between any point \(x\) in a normed linear space \(X\) and a set \(A\) in \(X\) is the same as the distance between \(x\) and the closure of the set \(A\), denoted by \(\bar{A}\). Answer: To prove that the distance between a point \(x\) and a set \(A\) is the same as the distance between \(x\) and the closure of the set \(A\), we need to show that \(\operatorname{dist}(x, A) = \operatorname{dist}(x, \bar{A})\). This is achieved by showing that \(\operatorname{dist}(x, A) \leq \operatorname{dist}(x, \bar{A})\) and \(\operatorname{dist}(x, \bar{A}) \leq \operatorname{dist}(x, A)\). The distance between a point and a set is defined as the infimum of the norm of the difference between the point and elements in the set. By comparing the infimum of the distances, we can conclude that the distance between \(x\) and \(A\) is equal to the distance between \(x\) and \(\bar{A}\).

Step 1: Define distance of a point and a set in a normed linear space

The distance between a point \(x\) in a normed linear space \(X\) and a set \(A \subseteq X\) is defined as: $$\operatorname{dist}(x, A) = \inf_{y \in A} \|x - y\|$$ where the infimum (inf) is taken over all \(y\) in \(A\).

Step 2: Show that \(\operatorname{dist}(x, A) \leq \operatorname{dist}(x, \bar{A})\)

For an arbitrary point \(x \in X\), since \(A \subseteq \bar{A}\), the infimum for \(\operatorname{dist}(x, A)\) will be taken over a possibly smaller set compared to \(\operatorname{dist}(x, \bar{A})\). As a result, we obtain: $$ \operatorname{dist}(x, A) \leq \operatorname{dist}(x, \bar{A}) $$

Step 3: Show that \(\operatorname{dist}(x, \bar{A}) \leq \operatorname{dist}(x, A)\)

Let's take an arbitrary element \(z \in \bar{A}\). Then, there exists a sequence \(\{y_n\}\) in \(A\) such that \(y_n \to z\) as \(n \to \infty\). Using the triangle inequality, we get: $$ \|x-z\| \leq \|x-y_n\|+\|y_n-z\| $$ Now, let's take the infimum over all \(y_n \in A\), and we'll have: $$ \|x-z\| \leq \operatorname{dist}(x, A) + \|y_n-z\| $$ Since \(y_n \to z\), as \(n \to \infty\), \(\|y_n-z\| \to 0\). Therefore, $$ \operatorname{dist}(x, \bar{A}) = \inf_{z \in \bar{A}} \|x-z\| \leq \operatorname{dist}(x, A) $$

Step 4: Prove the equality \(\operatorname{dist}(x, A) = \operatorname{dist}(x, \bar{A})\)

Combining Steps 2 and 3, we have shown that \(\operatorname{dist}(x, A) \leq \operatorname{dist}(x, \bar{A})\) and \(\operatorname{dist}(x, \bar{A}) \leq \operatorname{dist}(x, A)\). Therefore, we can conclude that: $$ \operatorname{dist}(x, A) = \operatorname{dist}(x, \bar{A}) $$

Key Concepts

Distance Between Point and Set

Understanding the concept of distance in a normed linear space is crucial for grasping many advanced topics in mathematical analysis. Simply put, the distance between a point and a set provides a measure of how close or far the point is from the set.

When we talk about the distance between a point, say \( x \), and a set \( A \) in a space \( X \), we are seeking the smallest 'gap' between \( x \) and any point in \( A \). Mathematically, this is expressed as the infimum of the norms of the differences between \( x \) and every point in \( A \), written as:\[ \operatorname{dist}(x, A) = \inf_{y \in A} \|x - y\| \].

Here, the infimum is like the 'minimum' but is more general. It comes into play especially when the set \( A \) doesn't actually contain a point that achieves this minimum distance—perhaps because \( A \) is not closed, or because the closest points to \( x \) merely get arbitrarily close to \( x \) without being elements of \( A \). This concept becomes significant when dealing with sets that might be unbounded or that don't have a well-defined 'edge'.

Infimum in Analysis

The infimum, often abbreviated as 'inf', is a fundamental concept in real analysis that describes the greatest lower bound of a set. To fully appreciate its role, consider a non-empty set \( S \) of real numbers. The infimum is the largest number that is less than or equal to all numbers in \( S \). If such a number is in the set \( S \), it is also the minimum.

In the context of our distance function, the infimum allows us to handle cases where the minimum distance is not achieved by any particular point in the set. It's a more comprehensive approach compared to simply finding the minimum because it takes into account the whole neighborhood around points in \( A \), rather than relying on discrete values. Calculating the infimum can be abstract, but it's essential for pinpointing precise distances in mathematical spaces where we can't necessarily rely on our intuitive sense of 'nearest'.

Triangle Inequality

The triangle inequality is a cornerstone of normed linear spaces and a rule that governs the norms of vectors. It states that for any vectors \(a\), \(b\) in a normed space, the norm of their sum is less than or equal to the sum of their norms.

This can be written as:\[ \|a + b\| \leq \|a\| + \|b\| \].

In practical terms, the triangle inequality tells us that if we are measuring distances via norms, we cannot take a 'shortcut'. The direct route (the norm of the sum) is always shorter than or equal to the detour (the sum of the norms). This principle is helpful in the exercise at hand to show the relationship between the distance to set \(A\) and its closure \(\bar{A}\), providing the underpinning to assert that a point's distance to a set and its closure are the same. It also helps to comprehend how norms operate in ways that parallel our everyday understanding of distances while incorporating the more abstract dimensions of analysis.