Question

Prove: (a) If \(S_{1}\) and \(S_{2}\) have zero content, then \(S_{1} \cup S_{2}\) has zero content. (b) If \(S_{1}\) has zero content and \(S_{2} \subset S_{1}\), then \(S_{2}\) has zero content. (c) If \(S\) has zero content, then \(\bar{S}\) has zero content.

Short answer

Question: Prove the following statements about sets with zero content: (a) the union of two sets with zero content also has zero content, (b) if a set has zero content and another set is its subset, the subset also has zero content, and (c) if a set has zero content, its complement also has zero content. Solution: (a) If \(S_1\) and \(S_2\) are sets with zero content, for any positive number \(\varepsilon > 0\), there exist finite collections of intervals covering \(S_1\) and \(S_2\) with total lengths less than \(\frac{\varepsilon}{2}\). The union of these collections covers \(S_1 \cup S_2\) and has total length less than \(\varepsilon\), so \(S_1 \cup S_2\) has zero content. (b) If \(S_1\) has zero content and \(S_2\) is a subset of \(S_1\), for any positive number \(\varepsilon > 0\), there exists a finite collection of intervals covering \(S_1\) with total length less than \(\varepsilon\). Since \(S_2\) is a subset of \(S_1\), this collection also covers \(S_2\) and has total length less than \(\varepsilon\), so \(S_2\) has zero content. (c) If \(S\) has zero content, for any positive number \(\varepsilon > 0\), there exists a finite collection of intervals covering \(S\) with total length less than \(\varepsilon\). The complements of these intervals cover the entire space except for the original intervals, and the total length of any finite cover of \(\bar{S}\) must also be less than \(\varepsilon\). Thus, \(\bar{S}\) has zero content.

Step-by-step solution

Proof for (a)

Let \(S_1\) and \(S_2\) be sets with zero content. By definition, for any positive number \(\varepsilon > 0\), there exist finite collections of intervals \(\{I_{1i}\}\) and \(\{I_{2j}\}\) covering \(S_1\) and \(S_2\) respectively, such that \(\sum_{i} |I_{1i}| < \frac{\varepsilon}{2}\) and \(\sum_{j} |I_{2j}| < \frac{\varepsilon}{2}\). Consider the union of these two collections: \(\{I_{1i}\} \cup \{I_{2j}\}\). This collection covers \(S_1 \cup S_2\), and its total length is given by \(\sum_{i} |I_{1i}| + \sum_{j} |I_{2j}| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon\). Since we have found a finite collection of intervals covering \(S_1 \cup S_2\) with total length less than \(\varepsilon\), we can conclude that \(S_1 \cup S_2\) has zero content.

Proof for (b)

Let \(S_1\) be a set with zero content, and let \(S_2\) be a subset of \(S_1\). By definition, for any positive number \(\varepsilon > 0\), there exists a finite collection of intervals \(\{I_i\}\) covering \(S_1\), such that \(\sum_{i} |I_i| < \varepsilon\). Since \(S_2\) is a subset of \(S_1\), the collection of intervals \(\{I_i\}\) also covers \(S_2\). Moreover, the total length of these intervals for \(S_2\) remains the same, so \(\sum_{i} |I_i| < \varepsilon\). Since we have found a finite collection of intervals covering \(S_2\) with total length less than \(\varepsilon\), we can conclude that \(S_2\) has zero content.

Proof for (c)

Let \(S\) be a set with zero content, and let \(\bar{S}\) denote its complement. We want to show that \(\bar{S}\) also has zero content. For any positive number \(\varepsilon > 0\), there exists a finite collection of intervals \(\{I_i\}\) covering \(S\), such that \(\sum_{i} |I_i| < \varepsilon\). Now, let us consider the complement of each interval in the collection, denoted by \(\{\bar{I_i}\}\). The union of these complements covers the entire space, except for the original intervals \(\{I_i\}\). That is, \(\bigcup_{i} \bar{I_i} = \bar{S} \cup \bigcup_{i} (I_i \cap \bar{S})\). Since \(\bigcup_{i} (I_i \cap \bar{S})\) is a finite collection of intervals with total length at most equal to the total length of \(\{I_i\}\), its total length is less than \(\varepsilon\). Additionally, since the union of all complements covers the entire space and all complements are closed intervals, any finite cover of \(\bar{S}\) must also have a total length less than \(\varepsilon\). Thus, we have demonstrated that for any positive \(\varepsilon > 0\), there exists a finite collection of intervals with total length less than \(\varepsilon\) covering \(\bar{S}\). Therefore, we can conclude that \(\bar{S}\) has zero content.

Key concepts

Real Analysis

When delving into the enticing world of Real Analysis, we explore the rigorous study of real numbers and real-valued functions. At its heart lies the concept of limits, which is critical for understanding concepts like continuity, derivatives, and integrals.

Within Real Analysis, the notion of a set having zero content is particularly interesting. It refers to a set that can be covered by intervals whose total length can be made arbitrarily small. In basic terms, no matter how fine a magnifying glass you use, the set's 'size' in terms of length can be made as close to zero as you wish without ever actually being zero. This is invaluable when considering the properties of functions and their graphs, especially when discussing the finer points of continuity and differentiability.

Applying this to a real-world example, imagine trying to measure the coastline of a very irregularly shaped island. If we were to consider every tiny inlet and outlet for our measurement—the 'set' in this analogy—using an increasingly finer level of detail, we might say imprecisely that the 'island's boundary has zero content', emphasizing how complicated and intricate the edge is despite the actual island being a finite size.

Measure Theory

Ambitiously diving into the realm of Measure Theory, we encounter a powerful tool for abstracting and generalizing the concept of 'size'. At its core, Measure Theory allows us to assign a 'measure' to a set, giving us a way to precisely define and analyze ideas such as length, area, volume, and beyond, even in the context of more abstract spaces.

In the context of the exercise, having zero content relates closely to having 'zero measure'. A set with zero measure can be thought of as taking up no 'space' within the real number line, despite possibly containing infinitely many points just like a set with zero content. This assists us in understanding concepts such as almost everywhere convergence and the negligibility of certain sets in integration theory. For instance, in the vast universe of real numbers, the sum of lengths of all intervals covering the rational numbers—a set dense in real numbers and infinite in count—has zero measure, exemplifying the mind-bending nature of Measure Theory.

Set Theory

The fundamental framework of Set Theory provides the building blocks for nearly all of mathematical discourse and formulation. It deals with the nature and relations of sets, which are essentially collections of objects. From the perspective of set theory, operations such as union and subset formation become essential in understanding and solving more complex mathematical problems.

The concept of a set having zero content is about considering the 'size' or 'extent' of a set under these operations. When proving that the union of two sets with zero content also has zero content, or that a subset of a set with zero content inherits this property, we're leveraging set theoretic principles. Set Theory guides us through the process of combining or refining sets while analyzing their fundamental characteristics, like size and containment relationships. For example, consider a set of all points within a circle (excluding the boundary). In set theory, it's not just about listing these points, but exploring how they relate to other sets, such as a smaller concentric circle or the entire plane, which enriches our understanding of spatial relationships and size considerations.