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Chapter 2: Problem 13

Construct a compact set of real numbers whose limit points form a countable set.

Short Answer

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Question: Construct a compact set of real numbers whose limit points form a countable set. Answer: We can construct the compact set of real numbers C as the union of intervals around rational numbers (q_n) in the following way: $$C = \bigcup_{n=1}^{\infty} \left( q_n - \frac{1}{2^n}, q_n + \frac{1}{2^n} \right)$$ Here, C is compact because it is closed (contains all its limit points, which are rational numbers) and bounded (has an upper and lower bound). The set of limit points for C is the countable set of rational numbers, \(\mathbb{Q}\).

Step by step solution

01

Choose the countable set of limit points

Let's use the set of rational numbers \(\mathbb{Q}\) as our countable set of limit points. This is because rational numbers can be listed in a sequence, making them countable.
02

Construct intervals around the limit points

For each rational number \(q_n\) in our sequence, we can create an interval around it to include more real numbers. To do this, let's consider the following intervals: $$I_n = \left( q_n - \frac{1}{2^n}, q_n + \frac{1}{2^n} \right)$$ We are choosing these intervals because they get smaller as n increases. This ensures that the total length of the intervals remains finite and the resulting set will be bounded.
03

Construct the compact set

Now, we will construct a compact set of real numbers by taking the union of all intervals \(I_n\) centered around the rational numbers \(q_n\): $$C = \bigcup_{n=1}^{\infty} I_n$$
04

Prove that C is compact

We need to show that C is both closed and bounded in order to prove that it is compact. First, let's show that C is bounded. Since each interval \(I_n\) has a length of \(\frac{1}{2^{n-1}}\), we can find an upper and lower bound for C by summing all of these lengths: $$\sum_{n=1}^{\infty} \frac{1}{2^{n-1}} = \sum_{n=0}^{\infty} \frac{1}{2^n}$$ This geometric series converges to 2. We can find an interval of length 2 that contains the original set of rational numbers. Since our constructed set C only includes numbers within these intervals, it must be bounded as well. Additionally, C is closed because every limit point of C is a rational number, and we have included all of them. So, C contains all its limit points. Since C is both closed and bounded, it is a compact set of real numbers. Its limit points form the countable set of rational numbers, \(\mathbb{Q}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Limit Points
Limit points, also known as accumulation points, are an essential concept in topology and analysis. A point is considered a limit point of a set if every neighborhood of this point contains at least one other point from the set. It's important to understand that a limit point does not need to be part of the set.

For instance, in the rational numbers \(\mathbb{Q}\), every real number is a limit point. This is because between any two real numbers, no matter how close, there are infinitely many rationals. Therefore, even if we take a small interval around a real number not in \(\mathbb{Q}\), we will always find a rational number within this interval.
Countable Set
A countable set is one whose elements can be put in one-to-one correspondence with the natural numbers. This means the set is either finite or countably infinite; that is, we can list the elements as \(e_1, e_2, e_3, ...\) indefinitely. For example, the rational numbers form a countable set because we can list them in a sequence, even though there are infinitely many.

This may seem counterintuitive since the rational numbers are dense in the real numbers; however, the clever arrangement of fractions in a sequence demonstrates their countability. On the other hand, the real numbers are uncountable, as proven by Cantor's diagonal argument.
Rational Numbers
Rational numbers are numbers that can be expressed as the quotient of two integers, where the denominator is not zero. They encompass all fractions and can also be represented as terminating or repeating decimals. Not all real numbers are rational; those that cannot be expressed as a simple fraction are called irrational, such as \(\sqrt{2}\) or \(\pi\).

The fact that rational numbers are dense in the real line means that they are present in any interval, no matter how small, which makes them an interesting choice when we want to construct sets with specific properties, like the countable set of limit points in the exercise.
Bounded and Closed Sets
A set is bounded if it fits within a certain range and does not stretch out to infinity. Formally, a set S is bounded if there exists a real number M such that \( |s| < M \) for all \( s \in S \). For example, the interval \[0, 10\] is bounded because all numbers in this set are less than some finite distance (in this case, 10) from 0.

A set is closed if it contains all its limit points. If you have a closed interval like \[a, b\], it's closed because it includes the endpoints \(a\) and \(b\), which are limit points. In contrast, an open interval \(a, b\) does not include its limit points and thus is not closed. The combination of bounded and closed sets is important in analysis because these sets, under certain conditions, guarantee the existence of limit points within the same set, which is pivotal when discussing compactness.
Geometric Series Convergence
A geometric series is a series in the form \(a + ar + ar^2 + ar^3 + \ldots\) where \(a\) is the initial term and \(r\) is the common ratio between terms. The convergence of such a series depends on the value of \(r\). If \(|r| < 1\), the series converges to \(\frac{a}{1 - r}\).

For the exercise, we consider the geometric series \(\sum_{n=0}^{\infty} \frac{1}{2^n}\), having \(a = 1\) and a common ratio \(r = \frac{1}{2}\). Since \(r\) is less than 1, the series converges to 2. This outcome is leveraged to demonstrate that certain constructions of sets, as in the compact set exercise, result in a bounded set which is an integral component in proving the set's compactness.

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Most popular questions from this chapter

Prove that every open set in \(R^{1}\) is the union of an at most countable collection of disjoint segments. Hint: Use Exercise \(22 .\)

(a) If \(A\) and \(B\) are disjoint closed sets in some metric space \(X\), prove that they are separated. (b) Prove the same for disjoint open sets. (c) Fix \(p \in X, 8>0\), define \(A\) to be the set of all \(q \in X\) for which \(d(p, q)<\delta\), define \(B\) similarly, with \(>\) in place of \(<\). Prove that \(A\) and \(B\) are separated. (d) Prove that every connected metric space with at least two points is uncountable. Hint: Use (c).

A metric space is called separable if it contains a countable dense subset. Show that \(R^{k}\) is separable. Hint: Consider the set of points which have only rational coordinates.

Let \(E^{\prime}\) be the set of all limit points of a set \(E\). Prove that \(E^{\prime}\) is closed. Prove that \(E\) and \(E\) have the same limit points. (Recall that \(\left.E=E \cup E^{\prime} .\right)\) Do \(E\) and \(E^{\prime}\) always have the same limit points?

Let \(E^{\circ}\) denote the set of all interior points of a set \(E\). [See Definition \(2.18(e)\); \(E^{\circ}\) is called the interior of \(E\) ] (a) Prove that \(E^{\circ}\) is always open. (b) Prove that \(E\) is open if and only if \(E^{\circ}=E\). (c) If \(G \subset E\) and \(G\) is open, prove that \(G \subset E^{\circ}\). (d) Prove that the complement of \(E^{\circ}\) is the closure of the complement of \(E\). (e) Do \(E\) and \(E\) always have the same interiors? (f) Do \(E\) and \(E^{\circ}\) always have the same closures?
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