/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 8 Consider the subspace \(\left(Q,... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 8

Consider the subspace \(\left(Q, d_{Q}\right)\) (the rational numbers) of \((R, d) .\) Let \(a_{1}, a_{2}\), \(\ldots\) be a sequence of rational numbers such that \(\lim _{n} a_{n}=\sqrt{2}\). Prove that, given \(\varepsilon>0\), there is a positive integer \(N\) such that for \(n, m>N,\left|a_{n}-a_{m}\right|<\) \(\varepsilon\). Does the sequence \(a_{1}, a_{2}, \ldots\) converge when considered to be a sequence of points of \(\left(Q, d_{Q}\right)\) ?

Short Answer

Expert verified
The sequence \(a_n\) is Cauchy in \(\mathbb{Q}\) but does not converge in \(\mathbb{Q}\) as it approaches the irrational \(\sqrt{2}\).

Step by step solution

01

Understand the Problem Context

We are given a sequence \(a_n\) of rational numbers such that \(\lim_{n \to \infty} a_n = \sqrt{2}\). Even though \(\sqrt{2}\) is irrational, we need to show the sequence is Cauchy in \(\mathbb{Q}\). This will also address its convergence within \(\mathbb{Q}\).
02

Define Limit Properties of the Sequence

Given \(\lim_{n \to \infty} a_n = \sqrt{2}\), for any \(\varepsilon>0\), there exists a positive integer \(N_1\) such that for all \(n>N_1\), \(|a_n - \sqrt{2}| < \frac{\varepsilon}{2}\). This follows from the definition of convergence.
03

Construct the Cauchy Condition

Since \(a_n\) converges to \(\sqrt{2}\), for \(n, m > N_1\), we can write: \(|a_n - a_m| = |a_n - \sqrt{2} + \sqrt{2} - a_m| \leq |a_n - \sqrt{2}| + |a_m - \sqrt{2}| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon\).
04

Establish Cauchy Sequence

Since \(|a_n - a_m| < \varepsilon\) for all \(n, m > N_1\), the sequence \(a_n\) is a Cauchy sequence in \(\mathbb{Q}\). Because all Cauchy sequences converge in the real numbers and \(a_n\) is converging to an irrational limit (\(\sqrt{2}\)), it does not converge to an element within \(\mathbb{Q}\).
05

Conclusion on Convergence in \(\mathbb{Q}\)

The sequence \(a_n\) is Cauchy in \(\mathbb{Q}\), meaning for any positive \(\varepsilon\), there is \(N\) where \(|a_n - a_m| < \varepsilon\) for all \(n,m > N\). However, \(a_n\) does not converge to a rational number because it approaches the irrational number \(\sqrt{2}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Rational Numbers
Rational numbers are numbers that can be expressed as a fraction of two integers. In mathematical notation, a rational number is of the form \( \frac{p}{q} \), where \( p \) and \( q \) are integers and \( q eq 0 \). These are numbers like \( \frac{1}{2}, -3, \text{or even } 4 \).Rational numbers are part of the set \( \mathbb{Q} \), which includes both positive and negative numbers, as well as zero. Importantly, the rational numbers are dense in the real numbers, meaning between any two rational numbers, you can find another rational number.However, not all real numbers are rational. For example, numbers like \( \sqrt{2} \) or \( \pi \) cannot be expressed as fractions with integer numerators and denominators, making them irrational. That's why they are outside the scope of \( \mathbb{Q} \). Despite this, rational numbers are fundamental in mathematics because they are easier to work with computationally compared to irrational numbers.
Convergence
Convergence in sequences refers to the property where the terms of a sequence get arbitrarily close to a specific value as you progress further along the sequence. This specific value is known as the limit. For a sequence \( a_n \), if \( a_n \to L \), we say that the sequence converges to \( L \). To define this formally, a sequence \( a_n \) converges to a limit \( L \) if for every \( \varepsilon > 0 \), there exists a positive integer \( N \) such that for all \( n > N \), \( |a_n - L| < \varepsilon \). This means the terms become arbitrarily close to \( L \) as \( n \) becomes larger.Convergence is a fundamental concept when discussing sequences, especially within the realm of rational numbers. This is because
  • It tells us about the stability of a sequence.
  • It helps in mathematical analysis as it lets us know when a process approaches a predictable outcome.
In many cases, identifying whether a sequence converges can simplify the understanding of more complex mathematical relationships and behaviors.
Subspace
A subspace is an inherently important idea in mathematics, especially in the context of linear algebra and topology. When we speak of the subspace \( (\mathbb{Q}, d_{\mathbb{Q}}) \), we are referring to the rational numbers equipped with a specific distance function, forming a subset of the real numbers.A subspace must satisfy certain conditions:
  • It must be non-empty.
  • It must be closed under addition 鈥� if two elements are in the subspace, their sum must also be within the subspace.
  • It must be closed under scalar multiplication 鈥� multiplying an element by a real number should still yield an element within the subspace.
In our context, the subspace \( \mathbb{Q} \) consists of all rational numbers and forms a mathematical structure that lets us focus on certain properties of numbers that are fractions of integers.Considering sequences like \( a_n \) within the framework of subspaces like \( \mathbb{Q} \) helps highlight how certain numbers behave distinctively, especially when dealing with limits and convergence.
Irrational Limit
The concept of an irrational limit arises when a convergent sequence approaches a value that is not among rational numbers. In mathematical terms, an irrational number cannot be expressed as a simple fraction of two integers.To understand \( \sqrt{2} \) as an irrational number, consider this: there is no fraction \( \frac{p}{q} \) (where \( p \) and \( q \) are integers) that exactly equals \( \sqrt{2} \). This is because \( \sqrt{2} \) has a non-repeating, non-terminating decimal expansion.When we have a sequence of rational numbers, like \( a_n \), converging to an irrational limit such as \( \sqrt{2} \), it shows that while the elements of the sequence are rational, the closer they converge, the more the limit moves towards an irrational number.This distinction is critical because:
  • It highlights the dense nature of rational numbers in real numbers.
  • It demonstrates that a sequence in \( \mathbb{Q} \) can be Cauchy (and thus very tightly clustered) yet still not converge within \( \mathbb{Q} \).
This concept underlies much of mathematical analysis and is key to understanding the larger structure of real numbers versus rationals.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(A\) be a non-empty subset of a metric space \((X, d)\). Define the function \(f: X \rightarrow R\) by \(f(x)=d(x, A)\). Prove that \(f\) is continuous.

Let \(a\) be a point on the real line \(R\). Prove that each of the following collections of subsets of \(R\) constitute a basis for the system of neighborhoods at \(a\) : i) All closed intervals of the form \([a-\varepsilon, a+\varepsilon], \varepsilon>0\); ii) All open balls \(B(a ; \varepsilon), \varepsilon\) a positive rational number; iii) All open balls \(B\left(a ; \frac{1}{n}\right), n\) a positive integer; iv) All open balls \(B\left(a ; \frac{1}{n}\right), n\) a positive integer larger than some fixed integer \(k\). Show that no finite collection of subsets of \(R\) can be a basis for the system of neighborhoods at \(a\).

Let \(X\) be a set. For \(x, y \in X\) define the function \(d\) by $$ d(x, x)=0, $$ and $$ d(x, y)=1 $$ if \(x \neq y .\) Prove that \((X, d)\) is a metric space.

Let \(a\) be a point in a metric space \(X\). Let \(N\) be the set of positive integers. Prove that there is a collection \(\left\\{B_{n}\right\\}_{n \in N}\) of neighborhoods of \(a\) which constitutes a basis for the system of neighborhoods at \(a\).

For each pair of points \(a, b \in R^{n}\), prove that there is a topological equivalence between \(\left(R^{n}, d\right)\) and itself defined by inverse functions \(f: R^{n} \rightarrow\) \(R^{n}\) and \(g: R^{n} \rightarrow R^{n}\) such that \(f(a)=b\). [Hint: If \(a=\left(a_{1}, a_{2}, \ldots, a_{n}\right), b=\left(b_{1}\right.\), \(\left.b_{2}, \ldots, b_{n}\right)\), define \(f\) by setting \(f\left(x_{1}, x_{2}, \ldots, x_{n}\right)=\left(x_{1}+b_{1}-a_{1}, x_{2}+b_{2}-\right.\) \(\left.a_{2}, \ldots, x_{n}+b_{n}-a_{n}\right)\).]
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Applied Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Decision Maths

Read Explanation

Logic and Functions

Read Explanation

Discrete Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.