Question

Use the Bolzano-Weierstrass theorem to show that if \(S_{1}, S_{2}, \ldots, S_{m}, \ldots\) is an infinite sequence of nonempty compact sets and \(S_{1} \supset S_{2} \supset \cdots \supset S_{m} \supset \cdots,\) then \(\bigcap_{m=1}^{\infty} S_{m}\) is nonempty. Show that the conclusion does not follow if the sets are assumed to be closed rather than compact.

Short answer

In conclusion, if we have an infinite sequence of nonempty compact sets with nested inclusions, then their intersection is nonempty. We provided a proof based on the Bolzano-Weierstrass theorem and demonstrated that this conclusion doesn't hold for closed sets in general by providing a counterexample with a sequence of closed sets where the intersection is not empty.

Step-by-step solution

Prove the intersection of compact sets is nonempty

We are given an infinite sequence of nonempty compact sets \(S_1, S_2, \ldots, S_m, \ldots\) with \(S_1 \supset S_2 \supset \cdots \supset S_m \supset \cdots\). Since each \(S_i\) is compact, by Bolzano-Weierstrass theorem every sequence in \(S_i\) will have a convergent subsequence. Let \(x_1 \in S_1\). Then since \(S_1 \supset S_2\), there exists \(x_2 \in S_2\) such that \(x_2\) is a limit point of the sequence \(\{x_1\}\). In general, for any \(n \in \mathbb{N}\), we can pick \(x_{n+1} \in S_{n+1} \) that is a limit point of the sequence \(\{x_1, \dots, x_n\}\). Now, consider the sequence \(\{x_n\}\) formed by picking such \(x_n\) for every \(n \in \mathbb{N}\). Since \(S_1\) is compact, there exists \(x\) and a subsequence \(\{x_{n_k}\}\) of \(\{x_n\}\) such that \(x_{n_k} \rightarrow x\) as \(k \rightarrow +\infty\). We will now show that \(x\) belongs to the intersection of the sets. Let \(j \in \mathbb{N}\). Then there is some \(k\) such that \(j(k) \geq j\). Since \(x_{n_j} \in S_j\) for every \(j \in \mathbb{N}\) and \(S_{j(k)} \subset S_j\), we have \(x_{n_k} \in S_j\). Since \(x_{n_k} \rightarrow x\) and \(S_j\) is closed, \(x \in S_j\). Therefore, \(x\) belongs to the intersection \(\bigcap_{m=1}^{\infty} S_{m}\). Hence, \(\bigcap_{m=1}^{\infty} S_{m}\) is nonempty.

Provide a counterexample for closed sets

Now, let us consider the following example. Define a sequence of closed sets in \(\mathbb{R}\) as follows: \(S_n = \left[0, \frac{1}{n}\right]\), for every \(n \in \mathbb{N}\). Clearly, each \(S_n\) is nonempty, closed, and bounded in \(\mathbb{R}\). We also have \(S_1 \supset S_2 \supset \cdots \supset S_n \supset \cdots\). Let's inspect the intersection: $$\bigcap_{n=1}^{\infty} S_n = \left\{x \in \mathbb{R} : x \in \left[0, \frac{1}{n}\right], \forall n \in \mathbb{N}\right\}.$$ Observe that if \(x > 0\), then there exists a natural number \(N\) such that \(\frac{1}{N} < x\). As a result, \(x \notin S_N\). Thus, \(x\) doesn't belong to the intersection. Hence, \(\bigcap_{n=1}^{\infty} S_n = \{0\}\). This counterexample shows the intersection might not be empty if the sets are closed, rather than compact.

Key concepts

Compact Sets

In mathematics, a compact set is one of the fundamental concepts in topology and analysis. A set is called compact if it is both closed and bounded. This means that the set does not 'stretch' to infinity and all of its limit points are contained within the set itself.

A helpful way to understand compact sets is to think of them as mathematical 'containers' that hold all of their limit points; no point can escape the boundary of the set. The significance of compactness lies in various properties that make analysis work smoothly, such as every sequence within a compact set having a convergent subsequence, as guaranteed by the Bolzano-Weierstrass theorem.

This property comes into play when proving the Bolzano-Weierstrass theorem itself, which states that every bounded sequence in \(\mathbb{R}^n\) has a convergent subsequence. Therefore, if you have an infinite sequence of nonempty compact sets that are nested inside each other, there has to be at least one point that all sets have in common - this is due to the property that each set contains all of its limit points.

Intersection of Sets

The intersection of sets is a basic concept in set theory and represents the set containing all the elements that are common to every set being intersected. The symbol for intersection is \(\cap\). When intersecting a series of sets, if there's even a single element common to all, the intersection is nonempty.

Regarding the nested sequence of compact sets \(S_1, S_2, \ldots, S_m, \ldots\), the intersection \(\bigcap_{m=1}^{\infty} S_m\) refers to all the points shared by these infinite sets. When these sets are compact, the nonemptiness of their intersection can be guaranteed by the presence of a convergent subsequence from the Bolzano-Weierstrass theorem. The nested nature of these sets further implies that if a point is a limit point in one, it remains within all subsequent sets due to their decreasing (or at least, non-increasing) size.

Convergent Subsequence

A convergent subsequence is a sequence derived from another sequence in which its terms tend to a specific value, known as the limit, as the sequence progresses indefinitely. In the context of the Bolzano-Weierstrass theorem, the ability to find a convergent subsequence within any bounded sequence is central.

In terms of understanding how this applies to compact sets, imagine you're taking snapshots of points within a compact set over time. As time goes on, you notice some of these points get repeatedly closer to a specific location. This pattern reflects a convergent subsequence, suggesting the presence of a limit point within the set. Given the nested compact sets in our exercise, by applying the Bolzano-Weierstrass theorem, we can illustrate that there's always a sequence that will have a subsequence converging to a point that is common to all sets, hence ensuring the intersection is nonempty.

Closed Sets

A closed set in topology is a set that includes all its limit points; in other words, it's 'closed off' from the outside. Essentially, any convergent sequence that starts in a closed set has its limit also contained within the set.

However, being closed doesn't necessarily mean a set is compact. Compactness requires the set to be bounded as well as closed. The step-by-step solution presents a sequence of closed sets \(S_1, S_2, \ldots, S_n, \ldots\) with each set a closed interval shrinking to zero. The intersections \(\bigcap_{n=1}^{\infty} S_n\) show that for closed sets, the intersection can be a single point. It illustrates that while a closed set will contain the limits of sequences starting within it, without the boundedness factor, we can't use the Bolzano-Weierstrass theorem to guarantee the presence of a convergent subsequence, potentially leading to an empty intersection. This highlights the unique property of compact sets concerning intersections, which closed sets do not necessarily share.