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Chapter 3: Problem 8

Prove that \(\mathbb{Z}\) (the integers) is a closed subset of \(\mathbb{R}\).

Short Answer

Expert verified
Question: Prove that the set of integers is a closed subset of the set of real numbers. Answer: To prove that the set of integers is a closed subset of the set of real numbers, we showed that the complement of integers in real numbers is an open set. We considered an element in the complement, found open intervals around it, and demonstrated that these intervals are contained entirely in the complement of integers. This demonstrates that the complement of integers is open, and therefore, the set of integers is a closed subset of real numbers.

Step by step solution

01

Define and identify the complement of \(\mathbb{Z}\)

We first want to find the complement of \(\mathbb{Z}\) in \(\mathbb{R}\). By definition, the complement of \(\mathbb{Z}\) in \(\mathbb{R}\), denoted as \(\mathbb{R} \setminus \mathbb{Z}\), is the set of all real numbers that are not integers. Symbolically, we can express it as follows: $$\mathbb{R} \setminus \mathbb{Z} = \{x \in \mathbb{R} : x \notin \mathbb{Z}\}$$
02

Consider an element in the complement of \(\mathbb{Z}\)

Let \(x \in \mathbb{R} \setminus \mathbb{Z}\). Since \(x\) is not an integer, it must be between two consecutive integers, say \(m\) and \(m+1\), where \(m \in \mathbb{Z}\). Mathematically, this can be expressed as: $$m < x < m+1$$
03

Construct open intervals around \(x\)

We want to find an open interval around \(x\) that is contained entirely within the complement of \(\mathbb{Z}\). To do this, we look at the distance between \(x\) and its neighboring integers \(m\) and \(m+1\). Specifically, we consider the minimum of the distances \((x - m)\) and \((m + 1 - x)\). Let \(\epsilon = \min\{(x - m), (m + 1 - x)\}\). Now we will construct an open interval centered at \(x\) with a radius of \(\epsilon\). This interval is defined as \((x-\epsilon,x+\epsilon)\). We can guarantee that this open interval is contained within the complement of \(\mathbb{Z}\) because it is constructed to be smaller than the distance between \(x\) and its nearest integer.
04

Show that the complement of \(\mathbb{Z}\) is open

By showing that for every element \(x \in \mathbb{R} \setminus \mathbb{Z}\) there exists an open interval \((x-\epsilon,x+\epsilon)\) contained entirely in the complement of \(\mathbb{Z}\), we have demonstrated that the complement of \(\mathbb{Z}\) is an open set. Therefore, according to the definition of closed sets, we can conclude that \(\mathbb{Z}\) is a closed subset of \(\mathbb{R}\).

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Key Concepts

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

Integers
Integers are the numbers we often count with or use for simple calculations. They include:
  • Positive numbers like 1, 2, 3, and so on.
  • Zero, which acts as a neutral number between the positive and negative.
  • Negative numbers like -1, -2, -3, etc.
Integers are symbolically represented by the letter \( \mathbb{Z} \), originating from the German word "Zahlen," meaning numbers.
This set of numbers is crucial in arithmetic and everyday mathematics.
One key characteristic of integers is that they do not include fractions or decimals. For example, numbers like 3.5 or \(-\frac{2}{3}\) are not integers. Understanding the concept of integers helps in differentiating between whole numbers and other number types.
Open Set
An open set in real analysis is a vital concept. It is a set that contains no boundary points. In simple words, for every point in an open set, you can find nearby points which are also included in the set.
This means each point is surrounded by an interval that is fully contained within the set.One essential property of open sets is that they are made up of open intervals. An open interval, like \((a, b)\), includes all the numbers between \(a\) and \(b\) but not the endpoints \(a\) and \(b\) themselves.
Think of standing at a spot where you can move a bit in any direction and still remain within the area. Open sets are crucial when discussing the topology of the real numbers and have many applications in understanding continuity and limits in calculus.
Closed Set
Closed sets play an essential role in real analysis, often as complements to open sets. A closed set includes its boundary points. If you are at a boundary point of a closed set, you can't step out without leaving the set. In this way, closed sets can be defined by including all points where sequences of points inside the set converge.For example, in the context of integers \( \mathbb{Z} \), every integer can be seen as a point that is included within the boundary. And since the complement of integers in real numbers \( \mathbb{R} \) is open, this implies \( \mathbb{Z} \) itself is closed.
This is a key aspect when considering limits and understanding how functions behave near the edges of intervals.Another way to describe closed sets is through their relationship with open sets; specifically, a set is closed if its complement in some larger set (like the real numbers) is open.
Real Numbers
Real numbers encompass both rational and irrational numbers. They include integers, fractions, and numbers that cannot be expressed as simple fractions, like the square root of 2 or \(\pi\).
  • Rational numbers: These can be expressed as a fraction \(\frac{a}{b}\) where \(a\) and \(b\) are integers, and \(b eq 0\).
  • Irrational numbers: These cannot be neatly expressed as a fraction and have non-repeating, non-terminating decimal expansions.
Real numbers are represented by the symbol \(\mathbb{R}\). They form a continuum, meaning there are no gaps between numbers on the real number line. This makes them crucial for calculus and analysis, providing a foundation for limits, integrals, and derivatives.Understanding real numbers is a stepping stone for grasping more advanced mathematical concepts and their applications in the real world.

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Most popular questions from this chapter

The set \(\\{x:\|x-a\| \leq r\\}\) is closed in a normed linear space \(X\).

The set \(\\{x\\}\), which consists of a single point in the normed linear space \(X\), is closed.

Let \(f\) and \(g\) be continuous mappings from a normed linear space \(X\) into a second normed linear space \(Y\). Prove that \(\\{x: f(x)=\) \(g(x)\\}\) is closed in \(X\).

Let \(A\) and \(B\) be disjoint closed sets in a normed linear space \(X\). Construct a continuous function \(f: X \rightarrow[0,1]\) such that \(f(a)=0\) for all \(a\) in \(A\), and \(f(b)=1\) for all \(b\) in \(B\).

The following constructs the Cantor set in \(\mathbb{R}\) : begin with the unit interval \([0,1] .\) The set \(F_{1}\) is obtained from this set by removing the middle third, so that $$ F_{1}=\left[0, \frac{1}{3}\right] \cup\left[\frac{2}{3}, 1\right] $$ Now \(F_{2}\) is obtained from \(F_{1}\) by removing the middle third from each of the constituent intervals of \(F_{1} .\) Hence $$ F_{2}=\left[0, \frac{1}{9}\right] \cup\left[\frac{2}{9}, \frac{1}{3}\right] \cup\left[\frac{2}{3}, \frac{7}{9}\right] \cup\left[\frac{8}{9}, 1\right] $$ In general, \(F_{n}\) is the union of \(2^{n}\) intervals, each of which has the form $$ \left[\frac{k}{3^{n}}, \frac{k+1}{3^{n}}\right] $$ where \(k\) lies between 0 and \(3^{n} .\) The Cantor set \(F\) is what remains after this process has been carried out for all \(n\) in \(\mathbb{N}\). Show that (i) \(F\) is closed in \(\mathbb{R}\) (ii) int \(F\) is empty (iii) \(F\) contains no non-empty open set. (iv) the complement of \(F\) can be expressed as a countable union of open intervals. (If you are unfamiliar with the definition of a countable set, then Appendix A gives a brief introduction to this concept.)
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